\section{Introduction}
Given a field $k$, consider the polynomial ring $\operatorname{\mathrm{Pol}}_n≔k[x_1,\dotsc,x_n]$,
on which the symmetric group $S_n$ acts by interchanging indeterminates.
For $\alpha_i=x_{i+1}-x_i$, $\operatorname{\mathrm{Pol}}_n$ has a direct sum decomposition $\operatorname{\mathrm{Pol}}_n=\operatorname{\mathrm{Pol}}_n^{s_i}\oplus \alpha_i\operatorname{\mathrm{Pol}}_n^{s_i}$
into summands on which the simple transposition $s_i$ acts by $\id$ and $-\id$ respectively.

The \term{NilHecke algebra} $\operatorname{\mathsf{NH}}_n$ of type $\mathrm{A}_n$ is the $k$-subalgebra of $\End_k(\operatorname{\mathrm{Pol}}_n)$ 
generated by multiplication with $x_i$ and the \term{Demazure operators}
$\partial_i\colon \operatorname{\mathrm{Pol}}_n\to\operatorname{\mathrm{Pol}}_n^{s_i}, f\mapsto \frac{f-s_i(f)}{x_{i+1}-x_i}$,
which assign to a polynomial its $s_i$-invariant part
and have $\ker\partial_i=\alpha_i\operatorname{\mathrm{Pol}}_n^{s_i}$
\autocites[§2]{Demazure:Invariants}.
They are $s_i$-derivations
(\ie, $\partial_i(fg)=\partial_i(f)g+s_i(f)\partial(g)$)
and satisfy $\partial_i^2=0$.
Thus, $\operatorname{\mathsf{NH}}_n$ is the $k$-algebra with presentation
\begin{equation}
	\operatorname{\mathsf{NH}}_n=\Biggl\langle x_i,\partial_i\Bigg|
	\begin{aligned}
	\partial_i x_i - x_{i+1}\partial_i &= -1, &
	x_ix_j &= x_jx_i &&\forall i,j\\[-1ex]
	\partial_i x_{i+1} - x_i\partial_i &= 1,& 
	\partial_ix_j &= x_j\partial_i && \forall j\neq i, i+1\\[-1ex]
	\partial_i\partial_{i+1}\partial_i &= \partial_{i+1}\partial_i\partial_{i+1},&
	\partial_i\partial_j &= \partial_j\partial_i && \neq i, i+1
	\end{aligned}
	\Biggr\rangle,
\end{equation}
which acts faithfully on the polynomial ring $\operatorname{\mathrm{Pol}}_n$.
The subalgebra $\operatorname{\mathsf{NC}}_n$ of $\operatorname{\mathsf{NH}}_n$ generated by the $\partial_i$'s is called \term{NilCoxeter algebra}.

The invariant algebra $\mathop{\Lambda\operator@font Pol}\nolimits_n≔\operatorname{\mathrm{Pol}}_n^{S_n}$,
whose elements are called \term{symmetric polynomials},
is the largest subalgebra of $\operatorname{\mathrm{Pol}}_n$ annihilated by $\operatorname{\mathsf{NC}}_n$.
As a $k$-algebra, $\mathop{\Lambda\operator@font Pol}\nolimits_n$ is isomorphic to the polynomial rings
$k[\varepsilon_1^{(n)},\dotsc,\varepsilon_n^{(n)}]$ and
$k[h_1^{(n)},\dotsc,h_n^{(n)}]$, where
\begin{equation}
	\varepsilon^{(n)}_k = \sum_{\mathclap{1\leq i_1<\cdots<i_k\leq n}} x_{i_1}\dotsm x_{i_k}\quad\text{and}\quad
	h^{(n)}_k           = \sum_{\mathclap{1\leq i_1\leq \cdots\leq i_k\leq n}} x_{i_1}\dotsm x_{i_k}
\end{equation}
in the \term{elementary} and \term{complete symmetric polynomials} of degree $0\leq 1\leq n$, respectively.
As a $\mathop{\Lambda\operator@font Pol}\nolimits_n$-algebra, $\operatorname{\mathrm{Pol}}_n$ is a free of rank $n!$.

\Citeauthor{EKL:odd-NilHecke} have developed a theory of
\emph{odd} symmetric polynomials and have constructed an odd analogue $\operatorname{\mathsf{ONH}}_n$ of the NilHecke algebra \autocite{EKL:odd-NilHecke}.
By odd, it is meant that $x_i x_j = -x_jx_i$ whenever $i\neq j$.
Our goal is to mimick the theory of symmetric and odd symmetric polynomials
for the \term{super polymial algebra} $\operatorname{\widetilde{\mathrm{Pol}}}_I$
associated to a graded index set $I=I_0\sqcup I_1$.
Let the degree function on $I$ be denoted by $|-|$.
Then $\operatorname{\widetilde{\mathrm{Pol}}}_I$ is the $\mathbf{Z}/2\mathbf{Z}$-graded algebra generated by $x_i$ for $i\in I$ and
subject to the relations $x_ix_j=(-1)^{|i||j|}x_jx_i$.

In \cref{sec:superalgebra}, we start with a brief review on super algebras
and the construction of a quiver Hecke Clifford superalgebra $\operatorname{\mathsf{NH}\mathfrak C}(C)$
from \autocite{KKT:Quiver-Hecke-Superalgebras}
in order to construct a faithful polynomial for $\operatorname{\mathsf{NH}\mathfrak C}(C)$
along the lines of \autocite{KL:quantum-group-I} in \cref{sec:super-klr}.

In \cref{sec:clifford-symmetric-polynomials},
we develop our theory of Clifford polynomials $\operatorname{\mathrm{Pol}\mathfrak C}(I)$ and Clifford symmetric polynomials $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits(I)$.
We show that the latter are generated by a generalisation of elementary symmetric polynomials
and that $\operatorname{\mathrm{Pol}\mathfrak C}(I)$ is a free $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits(I)$-module.
Our main results, which are stated in \cref{prop:Clifford-Polynomials-free},
lead us to a super-analogue for the construction of cyclotomic quotients of the NilHecke algebra
and for the cohomology rings of Grassmannians and flag varieties.
 

\section{Superalgebras, supermodules and supercategories}
\label{sec:superalgebra}

A \term{super vector space} $V$ is a $\mathbf{Z}/2\mathbf{Z}$-graded vector space $V=V_0⊕V_1$.
The $\mathbf{Z}/2\mathbf{Z}$-degree of an homogeneous element is called its \term{parity}, which is denoted by $\pty{-}$.
A homogeneous element is called \term{even} and \term{odd} according to its parity.
The dimension of $V$ is written $\dim_k V=\dim_k V_0|\dim_k V_1$,
and we write $k^{m|n}$ for the canonical $m|n$-dimensional $k$-super vector space.
Morphisms of super vector spaces are degree-preserving morphisms of vector spaces.

The thus defined category $k\mhyphen\mathrm{sVect}$ of super vector spaces
has an internal Hom, denoted by $\hom_k(-,-)$.
It is the super vector space 
whose even and odd parts consist of parity-preserving and -exchanging morphisms, respectively.

Together with $\hom_k(-,-)$,
the tensor product $(V⊗_k W)_p ≔ ⨁_{q\in\mathbf Z/2\mathbf Z}V_q⊗W_{p-q}$
and the braiding $V⊗W\to W⊗V, v⊗w ↦ (-1)^{\pty{v}\pty{w}}w⊗v$
turn $k\mhyphen\mathrm{sVect}$ into a closed symmetric monoidal category;
\ie, the tensor product of morphisms satisfies
\begin{equation}
	\label{eq:super-interchange}
	\begin{aligned}
	(f⊗g)(v⊗w)&=(-1)^{\pty{g}\pty{v}}f(v)⊗g(w)\\ 
	(f⊗g)\circ(f'⊗g') &= (-1)^{\pty{f'}\pty{g}}f\!f'⊗gg'
	\end{aligned}
\end{equation}
for all elements $v\in V, w\in W$ 
and morphisms $V''\xto{f'} V'\xto{f} V$ and $W''\xto{g'}W'\xto{g}W$.


\begin{remark}
	Given finite dimensional super vector spaces $V=k^{n|m}$ and $W=k^{p|q}$ with homogeneous bases,
	a homomorphism in $\hom_k(V, W)$ can be written as an $(m|n)\times (p|q)$-block matrix
	\[
		\bordermatrix{
			& m	& n\cr
			p & T_{0,0} & T_{0,1}\cr
			q & T_{1,0} & T_{1,1}
		}
	\]
	where $\hom_k(V,W)_0$ and $\hom_k(V,W)_1$ comprise the diagonal and the antidiagonal blocks respectively.
\end{remark}

\subsection{Superalgebras}
\begin{definition}
	A \term{$k$-superalgebra} $A$ is a super vector space $A$ with a graded $k$-algebra structure.
	It is called \term{supercommutative} or simply \term{commutative}
	if the multiplication commutes with the braiding;
	\ie, $ab = (-1)^{\pty{a}\pty{b}}ba$ for all $a,b∈A$.
	In particular, this means that in a commutative superalgebra,
	every odd element squares to zero.
	The $k$-tensor product of two superalgebras $A$ and $B$
	carries a superalgebra structure by $(a⊗b)⋅(a'⊗b')=(-1)^{\pty{a'}\pty{b}}aa'⊗bb'$.
\end{definition}
\begin{example}
	For every super vector space $V$,
	the space $\hom_k(V,V)$ is a superalgebra.
	It is generally non-commutative, unless $V=k^{1\mid 0}$ or $V=k^{0\mid 1}$.
	For super vector spaces $V$ and $W$, the space $\hom_k(V,V)\otimes\hom_k(W,W)\subseteq\hom_k(V\otimes W, V\otimes W)$
	is a super subalgebra, according to \eqref{eq:super-interchange}.
\end{example}
\begin{example}
	The exterior algebra on $n$ generators
	is the $k$-algebra 
	\[\Lambda[\omega_1, \dotsc, \omega_n] ≔ k⟨\omega_1, \dotsc, \omega_n⟩/(\omega_i\omega_j + \omega_j\omega_i)_{1≤i,j≤n}.\]
	It is a $1|n$-dimensional commutative superalgebra.
	The \term{polynomial superalgebra} on $m|n$ indeterminates is the commutative superalgebra
	$k[y_1,\dotsc, y_m|\omega_1,\dotsc, \omega_n] ≔ k[y_1,\dotsc, y_m] ⊗_k \Lambda[\omega_1,\dotsc, \omega_n]$.
\end{example}
\begin{example}
	\label{def:super-algebra:clifford-algebra}
	The \term{Clifford algebra} on $n$ generators is the $k$-superalgebra 
	\begin{equation}
	\mathfrak{C}_n≔⟨\mathfrak{c}_1,\ldots, \mathfrak{c}_n\mid \mathfrak{c}_i^2 = 1\ \text{and}\ \mathfrak{c}_i \mathfrak{c}_j = -\mathfrak{c}_j \mathfrak{c}_i\ \text{if $i\neq j$}⟩
	\end{equation}
	all of whose generators $\mathfrak{c}_i$ are odd.
	It is not supercommutative.
\end{example}
\begin{definition}
	Given two $k$-superalgebras $A$ and $B$, an \term{$A$-$B$-super bimodule} $M$
	is a super vector space endowed with \emph{even} superalgebra morphisms
	$A\to\hom_k(M,M)$ and $B\to\Hom_k(M,M)$.
	According to the super interchange law for vector space homomorphisms,
	a morphism $f$ of bimodules satisfies $f(amb) = (-1)^{\pty{f}\pty{a}} af(m)b$ for $a \in A$, $b \in B$.
	We denote the category of $A$-$B$-super bimodules by $\Mod[s]{A}[B]$.
	
	One-sided modules are defined in the obvious way.
	Over a supercommutative superalgebra,
	a left module obtains right module structure by
	$a\mathbin{.} m = (-1)^{\pty{a}\pty{m}} m\mathbin{.} a$.
	Super bimodules admit the obvious tensor product over their superalgebras.
\end{definition}
 
\subsection{Supercategories}

\begin{definition}
	A $k$-linear \term{supercategory}, \term{superfunctor} and \term{supernatural transformation}
	respectively are a category, a functor and a natural transformation enriched in the monoidal category $k\mhyphen\mathrm{sVect}$.
	This means that in a supercategory $\mathcal C$, all Hom-spaces are $k$-super vector spaces
	and composition is an even map
	\[
		\circ_{V,X,W}\in\Hom_{k\mhyphen\mathrm{sVect}}\bigl(\Hom_\mathcal C(X,W)\otimes_k\Hom_\mathcal C(V,X), \Hom_\mathcal C(V,W)\bigr).
	\]
	A superfunctor induces even maps of super vector spaces,
	and a supernatural transformation $\eta\colon F\Rightarrow G$ consists of morphisms
	$\eta_X\in\Hom_\mathcal C(FX, GX)$,
	whose homogeneous parts $\eta_{X,p}$ of parity $p$
	are natural themselves and satisfy
	$\eta_{Y,p}\circ F(f)=(-1)^{p\pty{f}}\circ G(f)\circ \eta_{X,p}$
	for homogeneous morphisms $f\colon X\to Y$.
	
	A \term{monoidal supercategory} $\mathcal C$
	is a monoidal structure on the underlying category of $\mathcal C$,
	whose associators and unitors are even supernatural transformations
	that satisfy the super interchange law \eqref{eq:super-interchange};
	see \autocite{Bru:Supercategories} for a comprehensive treatment.
\end{definition}
\begin{example}
	The super vector spaces $\hom_k(-,-)$ turn $k\mhyphen\mathrm{sVect}$ 
	into a closed symmetric monoidal supercategory.
\end{example}
\begin{example}
	The category $\Mod[s]{A}$ of modules over a super $k$-algebra $A$
	is a closed symmetric monoidal supercategory,
	whose the Hom-spaces
	\[
		\hom_A(M, N) ≔ \operatorname{eq}\Bigl(
			\hom_k(M, N)\xrightrightarrows[a_*]{a^*}\prod_{a\in A}\hom_k(M,N)
		\Bigr)
	\]
	are constructed as usual.
	The forgetful functor $\Mod[s]{A}\tok\mhyphen\mathrm{sVect}$ is a superfunctor.
\end{example}

\subsection{Super-Diagrams}
\label{sec:super-diagrams}
We draw vertical lines for morphisms; these are
placed next to each other to represent their tensor product.
The super interchange law \eqref{eq:super-interchange} is thus depicted as
\begingroup
\tikzset{every diag/.append style={smaller}}
\begin{equation*}
	\begin{array}{c@{{}={}}c@{{}={}}c}
	f⊗g 
	& (f⊗1)\circ(1⊗g)
	& (-1)^{\pty{f}\pty{g}} (1⊗g)\circ(f⊗1)\\
	\tikz[diag]\draw	(0,-1) node[below]{$V$}  --node[fill=white, draw, inner sep=1pt]{$f$} (0,1) node[above]{$W$}
						(2,-1) node[below]{$V'$} --node[fill=white, draw, inner sep=1pt]{$\mathstrut g$} (2,1) node[above]{$W'$};
	&
	\tikz[diag]\draw	
		(0,-1) node[below]{$V$} 
			--node[near end, fill=white, draw, inner sep=1pt]{$f$}
		
		
			(0,1) node[above]{$W$}
		(2,-1) 
			node[below]{$V'$} 
			--node[near start, fill=white, draw, inner sep=1pt]{$\mathstrut g$}
			 (2,1) node[above]{$W'$};
	&
	(-1)^{\pty{f}\pty{g}}
	\tikz[diag]\draw	
		(0,-1) node[below]{$V$}  
			--node[near start, fill=white, draw, inner sep=1pt]{$f$}
			 (0,1) node[above]{$W$}
		(2,-1) node[below]{$V'.$} 
			--node[near end, fill=white, draw, inner sep=1pt]{$\mathstrut g$}
			 (2,1) node[above]{$W'$};
	\\
	\end{array}
\end{equation*}
\endgroup
Such diagrams are to be understood as \emph{first vertical}
and \emph{then horizontal composition} of morphisms \autocite[cf.][]{Bru:Supercategories}.

\section{Super-KLR algebras}
\label{sec:super-klr}
We now recall the notion of KLR-superalgebras
following \autocite{KKT:Quiver-Hecke-Superalgebras}.
This is a generalisation of the ordinary KLR-algebra 
from \autocites{KL:quantum-group-I}{Rou:2-Kac-Moody}.
We start with an index set $I$ with a partition $I=I_\text{even}\sqcup I_\text{odd}$
that turns it into a superset.
Our goal is to extend the classical theory
to the super-setting.


\subsection{NilHecke Superalgebra}
Recall the Clifford algebra $\mathfrak{C}_n$ from \cref{def:super-algebra:clifford-algebra}. 	
The following definitions are motivated by the definition of the quiver Hecke Clifford superalgebra
from \autocite[§3.3]{KKT:Quiver-Hecke-Superalgebras}; see \cref{quiver Hecke superalgebra:graphically}.
\begin{definition}
	\label{def:super-klr-algebra:nil-hecke-clifford-algebra}
	The \term{NilHecke Clifford superalgebra} $\operatorname{\mathsf{NH}\mathfrak C}_n$  of type $\mathrm{A}_n$ is the $k$-superalgebra
	with even generators $y_i$ and $\mathfrak d_i$, and odd generators $\mathfrak{c}_i$,
	which are subject to the relations
	\settowidth\algnRef{$\mathfrak d_i \mathfrak d_{i+1} \mathfrak d_i$}
	\begin{equation}
		\label{eq:Definition-NilHecke-Clifford-algebra}
		\begin{aligned}
			\mathmakebox[\algnRef][r]{y_iy_j} &=
				y_jy_i
				&&\forall i,j,\\
			\mathmakebox[\algnRef][r]{\mathfrak{c}_i\mathfrak{c}_j} &=
				-\mathfrak{c}_j\mathfrak{c}_i
				&&\forall i\neq j,\\
			\mathmakebox[\algnRef][r]{\mathfrak{c}_i^2} &=
				1&&\forall i,\\
			\mathmakebox[\algnRef][r]{y_i \mathfrak{c}_j} &= 
				(-1)^{\delta_{i,j}} \mathfrak{c}_j y_i
				&&\forall i,j,
		\end{aligned}
		\hfil
		\settowidth\algnRef{$-1-c_ic_{i+1}$}
		\begin{aligned}
			\mathfrak d_i y_j - y_{s_i(j)}\mathfrak d_i &= 
			\begin{cases}
				-1-\mathfrak{c}_i \mathfrak{c}_{i+1}					& \text{if $j=i$}\\
				\phantom{-}1-\mathfrak{c}_i \mathfrak{c}_{i+1}		& \text{if $j=i+1$}\\
				0									& \text{otherwise}
			\end{cases}\\
			\mathfrak d_i\mathfrak{c}_j &= \casesGap\mathmakebox[\algnRef][l]{\mathfrak{c}_{s_i(j)}\mathfrak d_i}
			\quad\forall i,j,
		\end{aligned}
		\hfil
	\end{equation}
	and the NilCoxeter-relations
	\settowidth{\algnRef}{$(-1)^{\delta_{i,j}} \mathfrak{c}_j y_i$}
	\begin{equation}
		\label{eq:NilCoxeter--Clifford-relations}
		\begin{aligned}
			\mathfrak d_i^2 &= 0\\
			\mathfrak d_i \mathfrak d_j &= \mathmakebox[\algnRef][l]{\mathfrak d_j \mathfrak d_i} &&\text{for $|i-j|>1$}\\
			\mathfrak d_i \mathfrak d_{i+1} \mathfrak d_i &= \mathfrak d_{i+1} \mathfrak d_i \mathfrak d_{i+1}.
		\end{aligned}
	\end{equation}	
	The \term{NilCoxeter Clifford algebra} $\operatorname{\mathsf{NC}\mathfrak C}_n\subseteq \operatorname{\mathsf{NH}\mathfrak C}_n$ is the subalgebra
	 generated by the symbols $\mathfrak d_i$ and $\mathfrak{c}_j$.
\end{definition}
\begin{definition}
	\label{def:noncommutative-algebra}
	An \term{algebra} $B$ over a (not necessarily commutative) ring $A$
	is an $A$-bimodule equipped with a multiplication map
	$B⊗_A B\to B$
	that is an even $A$-$A$-bimodule homomorphism.
\end{definition}

We can consider $\operatorname{\mathsf{NH}\mathfrak C}_n$ as a $\mathfrak{C}_n$ algebra
generated by the symbols $y_i$ and the $\mathfrak d_i$.
We introduce the following analogues to the polynomial representation 
of the classical NilHecke algebra:

\begin{definition}
	\label{def:Polynomial-Clifford-Algebra}
	The \term{polynomial Clifford superalgebra} is the $\mathfrak{C}_n$-algebra
	with generators and relations
	\begin{equation*}
		\operatorname{\mathrm{Pol}\mathfrak C}_n ≔ \bigl\langle y_1,\dotsc, y_n \bigm|
		y_i y_j = y_j y_i,\quad
	
		y_i \mathfrak{c}_j = (-1)^{\delta_{ij}} \mathfrak{c}_j y_i\quad∀i,j\bigr\rangle_{\mathfrak{C}_n}.
	\end{equation*}
	As a Clifford-analogue for the Demazure operator $\partial_i$ on $\operatorname{\mathrm{Pol}}_n$,
	we define an even endomorphism $\mathfrak d_i$ of $\operatorname{\mathrm{Pol}\mathfrak C}_n$ by
		\begin{alignat*}{3}
			\mathfrak d_i(y_i) 		&≔ -1-\mathfrak{c}_i\mathfrak{c}_{i+1}, 				\qquad	&	\mathfrak d_i(y_j) &≔ 0 & \qquad & ∀j\neq i,i+1,\\
			\mathfrak d_i(y_{i+1}) 	&≔ \phantom{-}1-\mathfrak{c}_i\mathfrak{c}_{i+1},				&	\mathfrak d_i(\mathfrak{c}_j) &≔ 0 && ∀j,\\
		\intertext{such that $\mathfrak d_i$ is a $s_i$-derivation, \ie,}
			\mathfrak d_i(fg)			&≔ \mathrlap{\mathfrak d_i(f)g+s_i(f)\mathfrak d_i(g)} &&&& ∀j\neq i,i+1,
		\end{alignat*}
	where the symmetric group $S_n$ acts on $\operatorname{\mathrm{Pol}\mathfrak C}_n$
	by permuting the $y_i$'s and $\mathfrak{c}_j$'s independently.
\end{definition}
\begin{lemma}
	\label{lem:PolCn-is-free}
	$\operatorname{\mathrm{Pol}\mathfrak C}_n$ is a free as left and right $\mathfrak{C}_n$-module
	with basis $\{y_1^{\alpha_1}\dotsm y_n^{\alpha_n} \mid \alpha_i ∈ \mathbf{N}\}$.
	In particular, $\operatorname{\mathrm{Pol}\mathfrak C}_n$ has the same graded rank
	as left and as right $\mathfrak{C}_n$-module.
\end{lemma}
\begin{proof}
	Take any monomial from $\operatorname{\mathrm{Pol}\mathfrak C}_n$.
	According to the defining relation $y_i\mathfrak{c}_j=(-1)^{\delta_{ij}}\mathfrak{c}_jy_i$ of $\operatorname{\mathrm{Pol}\mathfrak C}_n$,
	we may move all $\mathfrak{c}_i$'s to the left (or right),
	possibly at the cost of introducing a sign when sliding them past $y_i$'s with the same index.
	Since the $y_i$'s commute,
	we may sort powers of $y_i$'s by their index.
	As a left (or right) $\mathfrak{C}_n$-module,
	$\operatorname{\mathrm{Pol}\mathfrak C}_n$ is thus isomorphic to $\mathfrak{C}_n⊗_k \operatorname{\mathrm{Pol}}_n$ (or $\operatorname{\mathrm{Pol}}_n⊗_k\mathfrak{C}_n$).
\end{proof}
\begin{lemma}
	The Clifford Demazure operator $\mathfrak d_i$ is a well-defined right $\mathfrak{C}_n$-module homomorphism.
\end{lemma}
\begin{proof}
	To show the statement inductively take an $f∈\operatorname{\mathrm{Pol}\mathfrak C}_n$ and assume w.\,l.\,o.\,g\@ifnextchar.{}{.\@}\ that $i=1$.
	We show that $\mathfrak d_1$ is compatible with the relations of $\operatorname{\mathrm{Pol}\mathfrak C}_n$.
	For the commutativity relation of the $y$'s, we compute
	\begin{align*}
		\mathfrak d_1(y_1 y_2 f) &= 
			 y_1y_2 \mathfrak d_1 (f) +
			\underbrace{\mathfrak d_1 (y_1 y_2)}_{
				\mathclap{
					\subalign{
						&= (-1-\mathfrak{c}_1\mathfrak{c}_2)y_2 + y_2 (1-\mathfrak{c}_1\mathfrak{c}_2)\\
						&= 0
					}
				}
			} f,&
		\mathfrak d_1(y_2 y_1 f) &= 
			y_2y_1 \mathfrak d_1 (f)+
			\underbrace{\mathfrak d_1 (y_2 y_1)}_{
				\mathclap{
					\subalign{
						&=  (1-\mathfrak{c}_1\mathfrak{c}_2)y_1 + y_1 (-1-\mathfrak{c}_1\mathfrak{c}_2)\\
						&= 0
					}
				}
			} f;\\
	\intertext{%
		hence $\mathfrak d_1 (y_1y_2 f)=\mathfrak d_1(y_2y_1 f)$.
		For the relations involving $\mathfrak{c}$'s and $y$'s, one calculates for instance
	}
		\mathfrak d_1(\mathfrak{c}_1 y_1 f) &= 
			\vphantom{{}-{}}
			\underbrace{\mathfrak d_1(\mathfrak{c}_1 y_1)}_{
				\mathclap{
					\subalign{
						&= \mathfrak{c}_2(-1-\mathfrak{c}_1 \mathfrak{c}_2)\\
						&= \mathfrak{c}_1-\mathfrak{c}_2\\
						&= -(-1-\mathfrak{c}_1\mathfrak{c}_2) \mathfrak{c}_1
					}
				}
			} f + \mathfrak{c}_2 y_2 \mathfrak d_1 (f)
		&
		%
		\mathfrak d_1(\mathfrak{c}_2 y_1 f) &= 
			\vphantom{{}-{}}
			\underbrace{\mathfrak d_1(\mathfrak{c}_2 y_1)}_{
				\mathclap{
					\subalign{
						&= \mathfrak{c}_1(-1-\mathfrak{c}_1 \mathfrak{c}_2)\\
						&= -\mathfrak{c}_1-\mathfrak{c}_2\\
						&= (-1-\mathfrak{c}_1\mathfrak{c}_2) \mathfrak{c}_2
					}
				}
			} f + \mathfrak{c}_2 y_2 \mathfrak d_1 (f)\\[-\jot]
		-\mathfrak d_1( y_1 \mathfrak{c}_1 f) &= 
					\overbrace{-\mathfrak d_1(y_1 \mathfrak{c}_1)}_{} f -  y_2 \mathfrak{c}_2 \mathfrak d_1 (f);
		&
		\mathfrak d_1( y_1 \mathfrak{c}_2 f) &= 
			\overbrace{\mathfrak d_1( y_1\mathfrak{c}_2)}_{} f -  y_2 \mathfrak{c}_2 \mathfrak d_1 (f);\\
	\shortintertext{The remaining relations}
		\mathfrak d_1(y_2\mathfrak{c}_1 f)&=\mathfrak d_1(\mathfrak{c}_2y_2 f)&
		\mathfrak d_1(\mathfrak{c}_2y_2 f)&=-\mathfrak d_1(y_2\mathfrak{c}_2 f)
	\end{align*} are shown similarly.
	The operator $\mathfrak d_i$ is a right $\mathfrak{C}_n$-module homomorphism
	because $\mathfrak d_i(f\mathfrak{c}_j) = \mathfrak d_i(f)\mathfrak{c}_j + s_i(f)\mathfrak d_i(\mathfrak{c}_j) = \mathfrak d_i(f)\mathfrak{c}_j$
	for all $f∈\operatorname{\mathrm{Pol}\mathfrak C}_n$ and $1\leq j\leq n$.
\end{proof}
\begin{remark}
	\label{lem:Kernel-of-Clifford-Demazure-is-Algebra}
	The kernel $\ker\mathfrak d_i$ is a $\mathfrak{C}_n$-subalgebra of $\operatorname{\mathrm{Pol}\mathfrak C}_n$
	containing $\mathfrak{C}_n$
	because $\mathfrak d_i(fg) = \mathfrak d(f)g + s_i(f)\mathfrak d_i(g) = 0$ for $f,g\in\ker\mathfrak d_i$	
\end{remark}
\begin{lemma}
	\label{lem:super-klr-algebra:clifford-demazure-relations}
	\begin{thmlist}
		\item The Clifford Demazure operators $\mathfrak d_i$ and multiplications by $y_i$ and $\mathfrak{c}_i$
			render $\operatorname{\mathrm{Pol}\mathfrak C}_n$ a representation of the NilHecke Clifford superalgebra.
		\item $\ker\mathfrak d_i=\im\mathfrak d_i$.	 
	\end{thmlist}
\end{lemma}
\begin{proof}
	\begin{thmlist}
	\item 
	We have to make sure that the Clifford Demazure operators
	satisfy the relations from \cref{def:super-klr-algebra:nil-hecke-clifford-algebra}.
	From the definition of $\mathfrak d_i$ one can see
	that the relations \cref{eq:Definition-NilHecke-Clifford-algebra} are satisfied.
	We show that the $\mathfrak d_i$'s satisfy \cref{eq:NilCoxeter--Clifford-relations}:
	\begin{description}
		\item[$\mathfrak d_i^2 (f)=0$]
		Assume that the statement holds for $f,g∈\operatorname{\mathrm{Pol}\mathfrak C}_n$;
		it follows from
		\settowidth{\algnRef}{$\mathfrak d_1 \mathfrak d_2 \mathfrak d_1 (y_3 f)$}
		\begin{align*}
			\mathmakebox[\algnRef][r]{\mathfrak d_1^2(fg)} &= \mathfrak d_i \big( \mathfrak d_i(f)g + s_i(f) \mathfrak d_i (g) \big)\\
			&= s_i(\mathfrak d_i(f))\, \mathfrak d_i(g) + \mathfrak d_i(s_i f)\, \mathfrak d_i (g) 
		\end{align*}
		that it suffices to show that $s_i\mathfrak d_i=-\mathfrak d_i s_i$.
		To show this inductively, assume that $s_i\mathfrak d_i (f)=-\mathfrak d_i(s_i f)$.
		We need to show that $s_i\mathfrak d_i (y_j f)=-\mathfrak d_i \bigl(s_i(y_j f)\bigr)$,
		which is obvious for $j\neq i$.
		For $j=i$, one inductively calculates
		\begin{align*}
			\mathmakebox[\algnRef][r]{s_i\mathfrak d_i(y_i f)} &= s_i\bigl[(-1-\mathfrak{c}_i\mathfrak{c}_{i+1}) f 
				+ y_{i+1}\,\mathfrak d_i f\bigr]\\
			&\mathbin{\smash{\stackrel{*}{=}}} (-1 +\mathfrak{c}_i\mathfrak{c}_{i+1}) \, s_i f 
				- y_{i+1}\, \mathfrak d_i s_i f\\
			&= -\mathfrak d_i(y_{i+1}\, s_i f),
		\end{align*}
		applying induction hypothesis in $*$.
		The calculation for $j=i+1$ is similar.
		%
		\item[$\mathfrak d_i\mathfrak d_j (f)=\mathfrak d_j \mathfrak d_i (f)$ for $|i-j|>1$]
		Clear from the definition of $\mathfrak d_i$.  
		\item[$\mathfrak d_i\mathfrak d_{i+1}\mathfrak d_i(f)= \mathfrak d_{i+1}\mathfrak d_{i}\mathfrak d_{i+1}(f)$]
		Assume the statement holds for $f∈\operatorname{\mathrm{Pol}\mathfrak C}_n$.
		It is clear that the statement then also holds true for $\mathfrak{c}_j f$ and $y_jf$ 
		for all $j\neq i,i+1,i+2$.
		For these cases, we assume w.\,l.\,o.\,g\@ifnextchar.{}{.\@}\ that $i=1$. 
		Applying the induction hypothesis in $*$, we compute:
		\begin{align*}
			\mathfrak d_1 \mathfrak d_2 \mathfrak d_1 (y_1 f)
			&= \mathfrak d_1 \mathfrak d_2 (-1-\mathfrak{c}_1\mathfrak{c}_2)f + y_2\, \mathfrak d_2f)\\*
			&= \mathfrak d_1 \big((-1-\mathfrak{c}_1\mathfrak{c}_3)\mathfrak d_2 f + (-1-\mathfrak{c}_2\mathfrak{c}_3)\mathfrak d_1 f + y_3\, \mathfrak d_2 \mathfrak d_1 f\big)\\
			&= (-1-\mathfrak{c}_2\mathfrak{c}_3)\mathfrak d_1 \mathfrak d_2 f + y_3 \,\mathfrak d_1 \mathfrak d_2 \mathfrak d_1 f\\
			%
			&\mathbin{\smash{\stackrel{*}{=}}}
			   \mathfrak d_2 \big((-1-\mathfrak{c}_1\mathfrak{c}_2)\mathfrak d_2 f + y_2\, \mathfrak d_1 \mathfrak d_2 f\big)\\
			&= \mathfrak d_2 \mathfrak d_1 (y_1\, \mathfrak d_2 f)\\*
			&= \mathfrak d_2 \mathfrak d_1 \mathfrak d_2 (y_1 f);\\
			%
			\mathfrak d_1 \mathfrak d_2 \mathfrak d_1 (y_2 f)
			&= \mathfrak d_1 \mathfrak d_2 \big((-1-\mathfrak{c}_1\mathfrak{c}_2)f + y_1\,\mathfrak d_1 f\big)\\*
			&= \mathfrak d_1 \big((1-\mathfrak{c}_1\mathfrak{c}_3) \mathfrak d_2 f + y_1 \,\mathfrak d_2 \mathfrak d_1 f\big)\\
			&= (-1-\mathfrak{c}_2\mathfrak{c}_3)\mathfrak d_1 \mathfrak d_2 f + (-1-\mathfrak{c}_1\mathfrak{c}_2)\mathfrak d_2 \mathfrak d_1 f + y_2\, \mathfrak d_1 \mathfrak d_2 \mathfrak d_1 f\\
			%
			&\mathbin{\smash{\stackrel{*}{=}}}
			   \mathfrak d_2 \big((-1-\mathfrak{c}_1\mathfrak{c}_3)\mathfrak d_1 + y_3\,\mathfrak d_1 \mathfrak d_2 f\big)\\			
			&= \mathfrak d_2 \mathfrak d_1 \big((-1-\mathfrak{c}_2\mathfrak{c}_3)f + y_3\, \mathfrak d_2 f)\\*
			&= \mathfrak d_2 \mathfrak d_1 \mathfrak d_2 (y_2 f);\\
			%
			\mathfrak d_1 \mathfrak d_2 \mathfrak d_1(y_3 f)
			&= \mathfrak d_1 \mathfrak d_2 (y_3 \,\mathfrak d_1 f)\\*
			&= \mathfrak d_1 \big( (1-\mathfrak{c}_2\mathfrak{c}_3) \mathfrak d_1 f + y_2 \,\mathfrak d_2 \mathfrak d_1 f)\\
			&= (1-\mathfrak{c}_1\mathfrak{c}_2) \mathfrak d_2 \mathfrak d_1 f + y_1\, \mathfrak d_1 \mathfrak d_2 \mathfrak d_1 f\\
			%
			&\stackrel{*}{=}
			   \mathfrak d_2 \big((1-\mathfrak{c}_1\mathfrak{c}_3)f + (1-\mathfrak{c}_1\mathfrak{c}_2)\mathfrak d_2 f + y_1\, \mathfrak d_1 \mathfrak d_2 f\big)\\
			&= \mathfrak d_2 \mathfrak d_1 \big((1-\mathfrak{c}_2\mathfrak{c}_3)f + y_2 \,\mathfrak d_2 f\big)\\*
			&=\mathfrak d_2 \mathfrak d_1 \mathfrak d_2 (y_3\, f).
		\end{align*}
	\end{description}
	Therefore, $\operatorname{\mathrm{Pol}\mathfrak C}_n$ is a representation of $\operatorname{\mathsf{NH}\mathfrak C}_n$.
	We show in \cref{lem:Clifford-NilHecke-acts-Faithfully} that it is faithful.
	
	\item 
	Because $\mathfrak d_i^2=0$, we may regard $(\operatorname{\mathrm{Pol}\mathfrak C}_n, \mathfrak d_i)$ as a chain complex of right $\mathfrak{C}_n$-modules,
	using its polynomial grading in the $y_i$'s.
	We claim that this chain complex $\operatorname{\mathrm{Pol}\mathfrak C}_n$ is contractible by the homotopy
		\[
			h_k\colon  (\operatorname{\mathrm{Pol}\mathfrak C}_n)_k \to (\operatorname{\mathrm{Pol}\mathfrak C}_n)_{k+1}, 
			f\mapsto
			\begin{cases}
				\frac{1}{2}(-1-\mathfrak{c}_i\mathfrak{c}_{i+1})y_i f & \text{if $2\mid k$,}\\
				\frac{1}{2}(+1-\mathfrak{c}_i\mathfrak{c}_{i+1})y_{i+1}f & \text{if $2\nmid k$.}
			\end{cases}
		\]
	Assume $2\mid \deg f$. We obtain
	\begin{align*}
		&\mathrel{\phantom{=}} h_{k-1}(\mathfrak d_i f) + \mathfrak d_i h_k (f)\\
		&= \tfrac 12(1-\mathfrak{c}_i\mathfrak{c}_{i+1})y_{i+1} (\mathfrak d_i f)
			+\tfrac{1}{2}\mathfrak d_i\bigl( (-1-\mathfrak{c}_i\mathfrak{c}_{i+1}) y_if\bigr)\\
		&= \tfrac 12(1-\mathfrak{c}_i\mathfrak{c}_{i+1})y_{i+1} (\mathfrak d_i f)
			+\tfrac{1}{2}(-1+\mathfrak{c}_i\mathfrak{c}_{i+1}) \bigl[(-1-\mathfrak{c}_i\mathfrak{c}_{i+1}) f + y_{i+1}\mathfrak d_i(f)\bigr]\\
		&= \tfrac{1}{2}\bigl[ \underbrace{(1-\mathfrak{c}_i\mathfrak{c}_{i+1})+(-1+\mathfrak{c}_1\mathfrak{c}_{i+1})}_0\bigr] y_{i+1} \mathfrak d_i (f)
			+ \underbrace{\tfrac{1}{2}(-1+\mathfrak{c}_1\mathfrak{c}_{i+1})(-1-\mathfrak{c}_i\mathfrak{c}_{i+1})}_{-1} f. 
 	\end{align*}
 	The computation is similar if $2\nmid\deg f$.
	Therefore, $h$ indeed is a homotopy from $\id$ to $0$,
 	and the chain complex $(\operatorname{\mathrm{Pol}\mathfrak C}_n, \mathfrak d_i)$ is acyclic.
 	This proves the statement.\qedhere
	\end{thmlist}
\end{proof}


\subsection{Quiver Hecke (Clifford) superalgebra}
\label{sec:quiver-hecke-superalgebra}
Let $I=I_0\sqcup I_1$ be a finite set as above.

\begin{definition}[{\autocite[§3]{KKT:Quiver-Hecke-Superalgebras}}]
	A \term{generalised Cartan matrix} on $I$ is a matrix $C=(d_{ij})∈𝐙^{I×I}$ such that
	\begin{thmlist}
		\item $d_{ii}=2$ for all $i∈I$;
		\item $d_{ij}≤0$ for distinct $i,j∈I$;
		\item $d_{ij}=0$ if and only if $d_{ji}=0$; and
		\item $2\mid d_{ij}$ if $i∈I_\text{odd}$.
	\end{thmlist}
	We define a $𝐙$-valued bilinear form on $𝐍^I$ by $⟨i, j⟩≔-d_{ij}$.
	Equivalently, we may describe the same datum by a graph $\Gamma_C$
	with vertex set $I$ and $d_{ij}$ directed edges from $i$ to $j$ for $i\neq j$.
	For distinct indices $i,j$, we thus write $i\mathrel{-} j$ if $d_{ij}\neq 0$
	and $i\not\conn j$ if $-d_{ij}=0$.
\end{definition}
For the remainder of this section, let $C∈\mathbf{Z}^{I\times I}$ be a symmetrisable generalised Cartan matrix
and fix an orientation for each edge $i\mathrel{-} j$ of $\Gamma_C$.
We write $i\to j$ if the edge is oriented from $i$ to $j$.

The following diagrammatic calculus for the quiver Hecke super algebra from \autocite[§3]{KKT:Quiver-Hecke-Superalgebras}
mimics the constructions from \autocites{KL:quantum-group-I}{KL:quantum-group-II}.

\begin{definition}
	A \term{string diagram} with $n$ strings consists of $n$ continuous paths
	$𝜙_k:[0,1]↪𝐑×[0,1]$ in a strip, $1≤k≤n$, called strings,
	such that
	\begin{thmlist}
		\item each string starts in $𝐍×\{0\}$ and ends in $𝐍×\{1\}$;
		\item the projection of any string to the second coordinate is strictly monotonically increasing;
		\item each endpoint has precisely one string attached to it; and
		\item at most two strings may intersect in a single point.
	\end{thmlist}
	A string diagram is defined up to isotopy.
	Strings in a diagram may be endowed with certain point-like decorations,
	whose positions are also specified up to isotopy.
	In the super-setting, crossings and decorations are endowed with parities
	such that the remarks in \cref{sec:super-diagrams} concerning vertical positioning apply.
	Strings are numbered from the left at their bottom endpoints.
\end{definition}

\begin{definition}[($\operatorname{\tilde{\mathsf{H}}}_n(C)$, diagrammatically)]
	\label{quiver Hecke superalgebra:graphically}
	The \term{quiver Hecke superalgebra} $\operatorname{\tilde{\mathsf{H}}}_n(C)$ is the $k$-linear supercategory
	consisting of the following data.
	\begin{itemize}
		\item Its objects are sequences $\nu∈I^n$.
		\item F $\nu'$ is not permutation of $\nu$, $\Hom(\nu, \nu')$ is the zero space.
			Otherwise, $\Hom(\nu, \nu')$ is generated by string diagrams on $n$ strings
			which connect identical entries of $\nu$ and $\nu'$.
			We say that the $k$-th string is labelled by $\nu_k$
			and usually write its label below the string.
			Strings may be decorated by an arbitrary non-negative number of dots,
			distant from the crossings.
		\item 
			The dots 
			$x_{k,\nu} ≔
				\tikz[diag, inline] \draw (0,-1) node[right]{$\nu_k$} -- node[rot, deco]{$\bullet$} (0,1);$
			and crossings
			$\tau_{k,\nu} ≔ 
				\tikz[diag, inline] \draw
						(-1,-1) node[left]{$\nu_k$} -- (1,1)
						(1,-1) node[right]{$\nu_{k+1}$} --(-1,1);
			$
			have parities
			$|\tikz[diag, inline] \draw (0,-1) node[right]{$\nu_k$} -- node[rot, deco]{$\bullet$} (0,1);| ≔ |\nu_k|$
			and 
			$| 
				\tikz[diag, inline] \draw
						(-1,-1) node[left]{$\nu_k$} -- (1,1)
						(1,-1) node[right]{$\nu_{k+1}$} --(-1,1);
			 |≔|\nu_k||\nu_{k+1}|$.
		\item Composition is given by vertically stacking diagrams
		and subject to the following local relations:
	\settowidth{\algnRef}{$%
		\displaystyle 
	
		\tikz[diag] \draw (-1,-1) node[below]{$i$} to node[label=10:$d_{ij}$]{$\bullet$} (-1,1) (1,-1)node[below]{$i$} -- (1,1);
	
		+\tikz[diag] \draw (-1,-1) node[below]{$i$} to (-1,1) (1,-1)node[below]{$i$} to node[label=10:$d_{ji}$]{$\bullet$} (1,1);
	$}
	\begin{align}
	
		\tikz[diag] \draw 
			(-1,-1) node[below]{$i$} -- node[deco] {$\bullet$} (-1,1) 
			(1,-1) node[below]{$j$} -- node[deco] {$\bullet$} (1,1);
		=
		\tikz[diag] \draw 
			(-1,-1) node[below]{$i$} -- node[deco, near end] {$\bullet$} (-1,1) 
			(1,-1) node[below]{$j$} -- node[deco, near start] {$\bullet$} (1,1);
		&=
		\casesGap
		\mathmakebox[\algnRef]{
			\mathllap{(-1)^{\pty{i}\pty{j}}}
			\tikz[diag] \draw 
				(-1,-1) node[below]{$i$} -- node[deco, near start] {$\bullet$} (-1,1) 
				(1,-1) node[below]{$j$} -- node[deco, near end] {$\bullet$} (1,1);
		}
		\quad
		\forall i,j\\
		%
	
	
		  \tikz[diag] \draw (-1,-1) node[below]{$i$} -- (1,1) (1,-1) node[below]{$j$} to[near start] node[deco]{$\bullet$} (-1,1);
		- (-1)^{|i||j|}
		  \tikz[diag] \draw (-1,-1) node[below]{$i$} -- (1,1) (1,-1) node[below]{$j$} to[near end] node[deco]{$\bullet$} (-1,1);
		&=\begin{cases}
			\mathmakebox[\algnRef]{0}
			& \text{if $i\neq j$}\\
			\mathmakebox[\algnRef]{%
				\tikz[diag] \draw (-1,-1) node[below]{$i$} -- (-1,1) (1,-1)node[below]{$i$} -- (1,1);
			}
			& \text{if $i=j$}
		\end{cases}
		\\
		  \tikz[diag] \draw (-1,-1) node[below]{$i$} to[near start] node[deco]{$\bullet$} (1,1) (1,-1) node[below]{$j$} -- (-1,1);
		- (-1)^{|i||j|}
		  \tikz[diag] \draw (-1,-1) node[below]{$i$} to[near end] node[deco]{$\bullet$} (1,1) (1,-1) node[below]{$j$} -- (-1,1);
		&=\begin{cases}
			\mathmakebox[\algnRef]{0}
			& \text{if $i\neq j$}\\			
			\mathmakebox[\algnRef]{%
				\mathllap{-{}}\tikz[diag] \draw (-1,-1) node[below]{$i$} -- (-1,1) (1,-1)node[below]{$i$} -- (1,1);
			}
			& \text{if $i=j$}
		\end{cases}
		\\
		%
	
	
		%
		\tikz[diag, curved] \draw (-1,-2) node[below]{$i$} to[out=+45, in=-45] (-1,2)
			(1,-2) node[below]{$j$} to[out=135,in=-135] (1,2);
		&= \begin{cases}
			\mathmakebox[\algnRef]{%
				0
			}
			& \text{if $i=j$}\\
			\mathmakebox[\algnRef]{%
				\tikz[diag] \draw (-1,-1) node[below]{$i$} -- (-1,1) (1,-1)node[below]{$j$} -- (1,1);
			}
			& \text{if $i\not\conn j$};\\
			\mathmakebox[\algnRef]{%
			
				\tikz[diag] \draw (-1,-1) node[below]{$i$} to node[deco, label=10:$d_{ij}$]{$\bullet$} (-1,1) (1,-1)node[below]{$j$} -- (1,1);
			
				+\tikz[diag] \draw (-1,-1) node[below]{$i$} to (-1,1) (1,-1)node[below]{$j$} to node[deco, label=10:$d_{ji}$]{$\bullet$} (1,1);
			}
			& \text{if $i\mathrel{-} j$}
		\end{cases}
		\\
		%
	
	
		%
		\tikz[diag, scale=.6] \draw (-2,-2) node[below]{$i$} -- (2,2)
			(2,-2) node[below]{$k$} -- (-2,2)
		(0,-2) node[below]{$j$} to[out=135,in=-135, curved] (0,2);
		-(-1)^{|i||j|}
		\tikz[diag, scale=.6] \draw (-2,-2) node[below]{$i$} -- (2,2)
			(2,-2) node[below]{$k$} -- (-2,2)
			(0,-2) node[below]{$j$} to[out=45,in=-45, curved] (0,2);
		&=
		\begin{cases}
			\mathmakebox[\algnRef]{%
			
				\tikz[diag] \draw (-2,-1) node[below]{$i$} --(-2,1) (0,-1) node[below]{$j$} --(0,1) (2,-1) node[below]{$i=k$} --(2,1);
			}
			& \text{if $i=k$ and $i\mathrel{-} j$}\\
			\mathmakebox[\algnRef]{0}
			& \text{otherwise.}
		\end{cases}
	\end{align}
	\end{itemize}
\end{definition}
\begin{remark}
	There is an embedding of $\operatorname{\tilde{\mathsf{H}}}_n(C)$
	into Brundan's diagrammatic super Kac-Moody 2-category
	\autocite{Bru:Super-Kac-Moody}
	by adding an arrow tip pointing upwards to every string.
\end{remark}


For calculations it is advantageous
to \emph{adjoin} odd generators for all odd indices $i∈I_\text{odd}$
instead of endowing the $x_i$'s with a super grading themselves.
This is captured by the algebra defined in the following diagramatically
along the lines of \autocite{KKT:Quiver-Hecke-Superalgebras}.
\begin{definition}[($\operatorname{\mathsf{H}\mathfrak C}_n(C)$, diagrammatically)]
	\label{def:super-klr-algebra:HC(C):diagramatically}
	With the same data as above, the \term{quiver Hecke Clifford superalgebra} $\operatorname{\mathsf{H}\mathfrak C}_n(C)$
	is the supercategory with the same objects $\nu∈I^n$ as $\operatorname{\tilde{\mathsf{H}}}_n(C)$,
	the even generators
	$y_{ik,\nu}=\tikz[diag, inline] \draw (0,-1) node[right]{$\nu_k$} -- node[rot, deco]{$\blacklozenge$} (0,1);$,
	$\sigma_{k,\nu}=%
		\tikz[diag, inline] \draw
			(-1,-1) node[left]{$\nu_k$} -- (1,1)
			(1,-1) node[right]{$\nu_{k+1}$} -- node[deco]{$\whitebullet$} (-1,1);%
	$	
	the odd generator $\mathfrak{c}_{k,\nu}=\tikz[diag, inline] \draw (0,-1) node[right]{$\nu_k$} -- node[rot, deco]{$\whitelozenge$} (0,1);$,
	subject to the relations
	$\mathfrak{c}_{k,\nu}=0$ if $|\nu_k|=0$,
	$\tikz[diag, inline] \draw (0,-1) node[right]{$i$} -- 
		node[very near start, rot, deco]{$\whitelozenge$}
		node[very near end, rot, deco]{$\whitelozenge$} 
		(0,1);
	=\tikz[diag, inline] \draw (0,-1) node[right]{$i$} -- (0,1); $,
	the commutativity relations%
	$\tikz[diag, inline] \draw (0,-1) node[left]{$i$} -- 
		node[very near start, rot, deco]{$\whitelozenge$}
		node[very near end, rot, deco]{$\blacklozenge$} 
		(0,1);
	=-
	\tikz[diag, inline] \draw (0,-1) node[right]{$i$} -- 
		node[very near start, rot, deco]{$\blacklozenge$}
		node[very near end, rot, deco]{$\whitelozenge$} 
		(0,1);
	$
	, 
	$\tikz[diag, inline] 
		\draw 
		(-2,-1) node[left]{$i$} -- node[very near start, rot, deco]{$\whitelozenge$}(-2,1)
		(2,-1) node[right]{$j$} -- node[very near end, rot, deco]{$\whitelozenge$} (2,1)
		(0,0) node{$\cdots$};
	=-
	\tikz[diag, inline] 
		\draw 
		(-2,-1) node[left]{$i$} -- node[very near end, rot, deco]{$\whitelozenge$}(-2,1)
		(2,-1) node[right]{$j$} -- node[very near start, rot, deco]{$\whitelozenge$} (2,1)
		(0,0) node{$\cdots$};$
	,
	$\tikz[diag, inline] 
		\draw 
		(-2,-1) node[left]{$i$} -- node[very near start, rot, deco]{$\whitelozenge$}(-2,1)
		(2,-1) node[right]{$j$} -- node[very near end, rot, deco]{$\blacklozenge$} (2,1)
		(0,0) node{$\cdots$};
	=
	\tikz[diag, inline] 
		\draw 
		(-2,-1) node[left]{$i$} -- node[very near end, rot, deco]{$\whitelozenge$}(-2,1)
		(2,-1) node[right]{$j$} -- node[very near start, rot, deco]{$\blacklozenge$} (2,1)
		(0,0) node{$\cdots$};$,
	and the following relations for the interaction with the crossing:
	\settowidth{\algnRef}{$\displaystyle%
		\quad
	 	(-1)^{d_{ij}/2}
	
		\tikz[diag] \draw
			(-1,-1) node[below]{$i$} -- 
				node[rot,label=10:$d_{ij}$, deco]{$\blacklozenge$}
				(-1,1)
			(1,-1) node[below]{$j$} -- (1,1);
		+
		(-1)^{d_{ji}/2}
		t_{ji}\tikz[diag] \draw 
			(-1,-1) node[below]{$i$} -- (-1,1)
			(1,-1) node[below]{$j$} --
				node[label=10:$d_{ji}$,deco,rot]{$\blacklozenge$}
				(1,1);
	$}
	\begin{align}
	
	
		%
	
	
		\label{eq:qHC:sliding-black-diamonds}
		\tikz[diag] \draw 
		  	(-1,-1) node[below]{$i$} -- (1,1)
		  		(1,-1) node[below]{$j$} --
		  			node[very near start, rot, deco]{$\blacklozenge$}
		  			node[deco]{$\whitebullet$}
		  			(-1,1);
		- 
		\tikz[diag] \draw 
		  	(-1,-1) node[below]{$i$} -- (1,1)
		  	(1,-1) node[below]{$j$} --
		  		node[very near end, rot, deco]{$\blacklozenge$}
		  		node[deco]{$\whitebullet$}
		  		(-1,1);
		&=\begin{cases}
			\mathmakebox[\algnRef]{0}
			& \text{if $i\neq j$}\\
			\mathmakebox[\algnRef]{%
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (-1,1)
						(1,-1) node[below]{$i$} -- (1,1);
				- 
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- 
						node[rot, deco]{$\whitelozenge$} 
						(-1,1) 
					(1,-1) node[below]{$i$} --
						node[rot, deco]{$\whitelozenge$} 
						(1,1);
			}
			& \text{if $i=j$}
		\end{cases}
		\\
		\tikz[diag] \draw 
		  	(-1,-1) node[below]{$i$} --
		  		node[very near start, rot, deco]{$\blacklozenge$}
		  		(1,1)
		  	(1,-1) node[below]{$j$} --	
		  		node[deco]{$\whitebullet$}
		  		(-1,1);
		-
		\tikz[diag] \draw 
		  	(-1,-1) node[below]{$i$} --
		  		node[very near end, rot, deco]{$\blacklozenge$}
		  		(1,1)
		  	(1,-1) node[below]{$j$} --	
		  			node[deco]{$\whitebullet$}
		  			(-1,1);
		&=\begin{cases}
			\mathmakebox[\algnRef]{0}
			& \text{if $i\neq j$}\\			
			\mathmakebox[\algnRef]{%
				\mathllap{-{}}
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (-1,1) 
					(1,-1)node[below]{$i$} -- (1,1);
				- 
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- 
						node[rot, deco]{$\whitelozenge$} 
						(-1,1) 
					(1,-1) node[below]{$i$} -- 
						node[rot, deco]{$\whitelozenge$} 
						(1,1);
			}
			& \text{if $i=j$}
		\end{cases}
		\\
		\label{eq:qHC:sliding-white-diamonds}
		\tikz[diag] \draw 
		  	(-1,-1) node[below]{$i$} -- (1,1)
		  		(1,-1) node[below]{$j$} --
		  			node[very near start, rot, deco]{$\whitelozenge$}
		  			node[deco]{$\whitebullet$}
		  			(-1,1);
		- 
		\tikz[diag] \draw 
		  	(-1,-1) node[below]{$i$} -- (1,1)
		  		(1,-1) node[below]{$j$} --
		  			node[very near end, rot, deco]{$\whitelozenge$}
		  			node[deco]{$\whitebullet$}
		  			(-1,1);
		&= \casesGap\mathmakebox[\algnRef]{		
		\tikz[diag] \draw 
		  	(-1,-1) node[below]{$i$} --
		  		node[very near start, rot, deco]{$\whitelozenge$}
		  		(1,1)
		  	(1,-1) node[below]{$j$} --	
		  			node[deco]{$\whitebullet$}
		  			(-1,1);
		-
		\tikz[diag] \draw 
		  	(-1,-1) node[below]{$i$} --
		  		node[very near end, rot, deco]{$\whitelozenge$}
		  		(1,1)
		  	(1,-1) node[below]{$j$} --	
		  			node[deco]{$\whitebullet$}
		  			(-1,1);
			= 0
		}
		\quad ∀i,j
		\\
		%
	
	
		%
		\label{eq:qHC:double-crossing}
		\begin{tikzpicture}[diag]
			\draw[curved, name path=stringA] 
				(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] (-1,2);
			\draw[curved, name path=stringB]
				(1,-2) node[below]{$j$} 	to[out=135,in=-135] (1,2);
			\draw[name intersections={of=stringA and stringB}]
				(intersection-1) node[deco]{$\whitebullet$}
				(intersection-2) node[deco]{$\whitebullet$};
		\end{tikzpicture}
		&= \begin{cases}
			\mathmakebox[\algnRef]{%
				0
			}
			& \text{if $i=j$}\\
			\mathmakebox[\algnRef]{%
				\tikz[diag] \draw (-1,-1) node[below]{$i$} -- (-1,1) (1,-1)node[below]{$j$} -- (1,1);
			}
			& \text{if $i\not\conn j$};\\
			\mathmakebox[\algnRef]{%
				(-1)^{d_{ij}/2}
			
				\tikz[diag] \draw
					(-1,-1) node[below]{$i$} -- 
						node[rot,label=10:$d_{ij}$, deco]{$\blacklozenge$}
						(-1,1)
					(1,-1) node[below]{$j$} -- (1,1);
				+
				(-1)^{d_{ji}/2}
			
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (-1,1)
					(1,-1) node[below]{$j$} --
						node[label=10:$d_{ji}$,deco,rot]{$\blacklozenge$}
						(1,1);
			}
			& \text{if $i\mathrel{-} j$}
		\end{cases}
		\\
		%
	
	
		%
		\begin{tikzpicture}[diag, scale=.6]
			\draw[name path=stringA]
				(-2,-2) 	node[below]{$i$} -- (2,2)
				(2,-2) 		node[below]{$k$} -- (-2,2)
				(0,0)		node[deco]{$\whitebullet$};
			\draw[name path=stringB] 
				(0,-2) 		node[below]{$j$} to[out=135, in=-135, curved] (0,2);
			\draw[name intersections={of=stringA and stringB}]
				(intersection-1) node[deco]{$\whitebullet$}
				(intersection-2) node[deco]{$\whitebullet$};
		\end{tikzpicture}
		-
		\begin{tikzpicture}[diag, scale=.6]
			\draw[name path=stringA]
				(-2,-2) 	node[below]{$i$} -- (2,2)
				(2,-2) 		node[below]{$k$} -- (-2,2)
				(0,0)		node[deco]{$\whitebullet$};
			\draw[name path=stringB] 
				(0,-2) 		node[below]{$j$} to[out=45, in=-45, curved] (0,2);
			\draw[name intersections={of=stringA and stringB}]
				(intersection-1) node[deco]{$\whitebullet$}
				(intersection-2) node[deco]{$\whitebullet$};
		\end{tikzpicture}
		&=
		\begin{cases}
			\mathmakebox[\algnRef]{%
			
				\tikz[diag] \draw
					(-2,-1) node[below]{$i$} --(-2,1)
					(0,-1) node[below]{$j$} --(0,1)
					(2,-1) node[below]{$i=k$} --(2,1);
			}
			& \parbox{3cm}{if $i=k$\\ and $i\mathrel{-} j$}\\
			\mathmakebox[\algnRef]{0}
			& \text{otherwise.}
		\end{cases}

	\end{align}
	For every odd index, $\operatorname{\mathsf{H}\mathfrak C}_n(C)$ contains a copy of $\operatorname{\mathsf{NH}\mathfrak C}_n$.
\end{definition}
\begin{lemma}%
	\label{lemma:quiver-Hecke-super-embedding}%
	There is an embedding
	\begin{align*}
			\iota:\operatorname{\tilde{\mathsf{H}}}_n(C) &\into \operatorname{\mathsf{H}\mathfrak C}_n(C),\\*
			x_{k,\nu}	
				& \mapsto (\mathfrak{c}_{k,\nu})^{\pty{\nu_k}} y_{k,\nu}\\*
			\tau_{k,\nu} 
				&\mapsto \gamma_{\nu_k, \nu_{k+1}} (\mathfrak{c}_{k,\nu}-\mathfrak{c}_{k+1,\nu})^{\pty{\nu_k}\pty{\nu_{k+1}}} \sigma_{k,\nu}
	\end{align*}
	with factors $\gamma_{ij}$ such that $\gamma_{ij}=1$ if at least one index is even,
	$\gamma_{ii}=\frac{1}{2}$ if $i$ is odd and $\gamma_{ij}\gamma_{ji} = - \frac{1}{2}$ otherwise.
\end{lemma}
\begin{proof}
	We verify the well-definedness of this map
	and refer to \autocite[thm.\ 3.3]{KKT:Quiver-Hecke-Superalgebras} for injectivity.
	We have to check compatibility with the following relations:
	\begin{description}
	\item[{$
		\tikz[diag,inline] \draw 
			(-1,-1) node[left]{$i$} -- (1,1) 
			(1,-1) node[right]{$j$} --
				node[very near start, rot, deco]{$\bullet$} 					
				(-1,1);
		\pm	
		\tikz[diag,inline] \draw 
			(-1,-1) node[left]{$i$} -- (1,1) 
			(1,-1) node[right]{$j$} --
				node[very near end, rot, deco]{$\bullet$} 					
				(-1,1);
	$}]
		If one of the indices $i$ and $j$ is even,
		domain and codomain locally reduce to the ordinary KLR-algebra,
		so nothing remains to prove.
		Let us thus assume that both $i$ and $j$ are odd indices.
		\begingroup
		\tikzset{every diag/.append style={smaller}}
		\delimiterfactor=600
		\delimitershortfall=20ex
		\begin{align*}
				&\mathrel{\phantom{=}} \tfrac 1{\gamma_{ij}}\iota(\tau_{i,\nu}x_{k+1,\nu}  + \tau_{i,\nu}x_{k+1,\nu})\\
				&= \tfrac 1{\gamma_{ij}} \iota\left(
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (1,1) 
					(1,-1) node[below]{$j$} --
						node[very near start, rot, deco]{$\bullet$} 					
						(-1,1);
				+	
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (1,1) 
					(1,-1) node[below]{$j$} --
						node[very near end, rot, deco]{$\bullet$} 					
						(-1,1);
				\right)\\
				&=
				\left(
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (1,1) 
					(1,-1) node[below]{$j$} --
						node[at end, rot, deco]{$\whitelozenge$}
						node[very near start, rot, deco]{$\whitelozenge$} 
						node[at start, rot, deco]{$\blacklozenge$}
						node[deco]{$\whitebullet$}					
						(-1,1);
				-
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- 
						node[at end, rot, deco]{$\whitelozenge$} (1,1) 
					(1,-1) node[below]{$j$} --
						node[very near start, rot, deco]{$\whitelozenge$} 
						node[at start, rot, deco]{$\blacklozenge$}
						node[deco]{$\whitebullet$}					
						(-1,1);
				\right)
				+
				\left(
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (1,1) 
					(1,-1) node[below]{$j$} --
						node[at end, rot, deco]{$\whitelozenge$}
						node[near end, rot, deco]{$\whitelozenge$} 
						node[very near end, rot, deco]{$\blacklozenge$}
						node[deco]{$\whitebullet$}
						(-1,1);
				-
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- node[at end, rot, deco]{$\whitelozenge$} (1,1) 
					(1,-1) node[below]{$j$} --
						node[near end, rot, deco]{$\whitelozenge$} 
						node[very near end, rot, deco]{$\blacklozenge$}
						node[deco]{$\whitebullet$}
						(-1,1);
				\right)
				\\
				&=
				\left(
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- (1,1) 
						(1,-1) node[below]{$j$} --
							node[very near start, rot, deco]{$\blacklozenge$}
							node[at end, rot, deco]{$\whitelozenge$}
						 	node[very near end, rot, deco]{$\whitelozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
					-
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} --
							node[at end, rot, deco]{$\whitelozenge$}
							(1,1) 
						(1,-1) node[below]{$j$} --
							node[very near start, rot, deco]{$\blacklozenge$}
						 	node[very near end, rot, deco]{$\whitelozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
				\right)		
				+
				\left(
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- (1,1) 
						(1,-1) node[below]{$j$} to
							node[at end, rot, deco]{$\whitelozenge$}
							node[very near end, rot, deco]{$\blacklozenge$}
						 	node[near end, rot, deco]{$\whitelozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
					-
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} --
							node[near end, rot, deco]{$\whitelozenge$}
							(1,1) 
						(1,-1) node[below]{$j$} to
							node[at end, rot, deco]{$\whitelozenge$}
							node[very near end, rot, deco]{$\blacklozenge$}
							node[deco]{$\whitebullet$}
						 	(-1,1);
				\right)
				\\
			\intertext{%
				by \cref{eq:qHC:sliding-white-diamonds}.
				If $i\neq j$, then \cref{eq:qHC:sliding-black-diamonds} asserts
				that we can slide all diamonds across the crossing.
				According to the commutativity relations for 
				$\tikz[diag, inline] \draw (0,-1) -- node[rot, deco]{$\blacklozenge$} (0,1);$'s
				and $\tikz[diag, inline] \draw (0,-1) -- node[rot, deco]{$\whitelozenge$} (0,1);$'s,
				we obtain for $i\neq j$:
			}
				&=
					\left(
					-
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- (1,1) 
						(1,-1) node[below]{$j$} to
							node[at end, rot, deco]{$\whitelozenge$}
							node[near end, rot, deco]{$\whitelozenge$}
						 	node[very near end, rot, deco]{$\blacklozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
					+
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} --
							node[near end, rot, deco]{$\whitelozenge$}
							(1,1) 
						(1,-1) node[below]{$j$} --
							node[very near end, rot, deco]{$\blacklozenge$}
						 	node[at end, rot, deco]{$\whitelozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
					\right)
					+
					\left(
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- (1,1) 
						(1,-1) node[below]{$j$} to
							node[at end, rot, deco]{$\whitelozenge$}
							node[near end, rot, deco]{$\whitelozenge$}
						 	node[very near end, rot, deco]{$\blacklozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
					-
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} --
							node[near end, rot, deco]{$\whitelozenge$}
							(1,1) 
						(1,-1) node[below]{$j$} --
							node[very near end, rot, deco]{$\blacklozenge$}
						 	node[at end, rot, deco]{$\whitelozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
					\right)
				\\
				&= 0.
				\\
			\intertext{%
				If $i=j$, we obtain from \cref{eq:qHC:sliding-black-diamonds}:
			}
				&=
				\left(
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- (1,1) 
						(1,-1) node[below]{$i$} to
							node[at end, rot, deco]{$\whitelozenge$}
							node[very near end, rot, deco]{$\whitelozenge$}
						 	node[near end, rot, deco]{$\blacklozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
					+
				
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- 
							node[pos=.625,rot, deco]{$\whitelozenge$}		
							node[pos=.325,rot, deco]{$\whitelozenge$}
							(-1,1)
						 (1,-1) node[below]{$i$} --			 	
						 	(1,1);
					-
				
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- 
							node[near start,rot, deco]{$\whitelozenge$}		
							node[midway,rot, deco]{$\whitelozenge$}
							node[near end,rot, deco]{$\whitelozenge$}
							(-1,1)
						 (1,-1) node[below]{$i$} --
							node[near start,rot, deco]{$\whitelozenge$}				 	
						 	(1,1);
					-
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- 
							node[at end, rot, deco]{$\whitelozenge$}
							(1,1) 
						(1,-1) node[below]{$i$} to
							node[very near end, rot, deco]{$\blacklozenge$}
						 	node[near end, rot, deco]{$\whitelozenge$}
						 	node[deco]{$\whitebullet$}
						 	(-1,1);
					-
				
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- 
							node[pos=.325,rot, deco]{$\whitelozenge$}		
							(-1,1)
						 (1,-1) node[below]{$i$} --
							node[pos=.625,rot, deco]{$\whitelozenge$}		 	
						 	(1,1);
					+
				
					\tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- 
							node[pos=.325,rot, deco]{$\whitelozenge$}
							node[pos=.625,rot, deco]{$\whitelozenge$}		
							(-1,1)
						 (1,-1) node[below]{$i$} --
						 	node[very near start,rot, deco]{$\whitelozenge$}
							node[very near end,rot, deco]{$\whitelozenge$}		 	
						 	(1,1);
				\right)+
				\left(
					-
					\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (1,1) 
					(1,-1) node[below]{$i$} to
						node[at end, rot, deco]{$\whitelozenge$}
						node[very near end, rot, deco]{$\whitelozenge$}
					 	node[near end, rot, deco]{$\blacklozenge$}
					 	node[deco]{$\whitebullet$}
					 	(-1,1);
					 +
					 \tikz[diag] \draw 
						(-1,-1) node[below]{$i$} -- 
							node[at end, rot, deco]{$\whitelozenge$}
							(1,1) 
						(1,-1) node[below]{$i$} to
							node[very near end, rot, deco]{$\blacklozenge$}
					 		node[near end, rot, deco]{$\whitelozenge$}
					 		node[deco]{$\whitebullet$}
					 		(-1,1);
				\right)
				\\
				&=2
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- 		
						(-1,1)
					 (1,-1) node[below]{$i$} --	 	
					 	(1,1);.
		\end{align*}
		\endgroup%
		The computation for 
		
			\tikz[diag, inline] \draw 
				(-1,-1) node[left]{$i$} --
					node[very near end, rot, deco]{$\bullet$} 
					(1,1) 
				(1,-1) node[right]{$j$} -- 					
					(-1,1);
			+	
			\tikz[diag, inline] \draw 
				(-1,-1) node[left]{$i$} --
					node[very near start, rot, deco]{$\bullet$} 
					(1,1) 
				(1,-1) node[right]{$j$} -- 					
					(-1,1);
		$
		is similar.
	\item[{$
		\smash{
		\begin{tikzpicture}[diag, inline]
			\draw[curved, name path=stringA] 
				(-1,-2) node[left]{$i$} 	to[out=+45, in=-45] (-1,2);
			\draw[curved, name path=stringB]
				(1,-2) node[right]{$j$} 	to[out=135,in=-135] (1,2);
		\end{tikzpicture}
		}
	$}]
		If $i=j$, this relation is obviously satisfied.
		For $i\neq j$, nothing remains to be proven if one index is even;
		we thus assume $|i||j|=1$.
		\begingroup
		\tikzset{every diag/.append style={smaller}}
		\delimiterfactor=700
		\delimitershortfall=20ex
		\begin{align*}
			\underbrace{\frac{1}{\gamma_{ij}\gamma_{ji}}}_{-2}\iota\left(
			\begin{tikzpicture}[diag,smaller,smaller,smaller]
				\draw[curved, name path=stringA] 
					(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] (-1,2);
				\draw[curved, name path=stringB]
					(1,-2) node[below]{$j$} 	to[out=135,in=-135] (1,2);
			\end{tikzpicture}
			\right)
		
			&= 
			\frac{1}{\gamma_{ji}}
			\left(
				\quad
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] (-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135] node[deco,rot]{$\whitelozenge$}(1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node{\clap{$\mathllap{\iota}\Big(\qquad\Big)$}};
				\end{tikzpicture}
				-
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] node[deco,rot]{$\whitelozenge$}(-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135] (1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node{\clap{$\mathllap{\iota}\Big(\qquad\Big)$}};
				\end{tikzpicture}
			\right)
		
			=
			\left(
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] node[deco,rot,at end]{$\whitelozenge$}(-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135] node[deco,rot]{$\whitelozenge$}(1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node[deco]{$\whitebullet$};
				\end{tikzpicture}
				-
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] (-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135] 
							node[deco,rot]{$\whitelozenge$} 
							node[deco,rot,at end]{$\whitelozenge$} 
							(1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node[deco]{$\whitebullet$};
				\end{tikzpicture}
			\right)+\left(
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45]
							node[deco,rot]{$\whitelozenge$}
							node[deco,rot,at end]{$\whitelozenge$}
							(-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135] (1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node[deco]{$\whitebullet$};
				\end{tikzpicture}
				+
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45]
							node[deco,rot]{$\whitelozenge$}(-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135]
							node[deco,rot,at end]{$\whitelozenge$} (1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node[deco]{$\whitebullet$};
				\end{tikzpicture}
			\right)
			\\&
			=
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] node[deco,rot,at end]{$\whitelozenge$}(-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135] node[deco,rot,very near end]{$\whitelozenge$}(1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node[deco]{$\whitebullet$};
				\end{tikzpicture}
				-2
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] (-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135] 
							(1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node[deco]{$\whitebullet$};
				\end{tikzpicture}
				-
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45]
							node[deco,rot, at end]{$\whitelozenge$}
							(-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135]
							node[deco,rot,very near end]{$\whitelozenge$}
							 (1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node[deco]{$\whitebullet$};
				\end{tikzpicture}
			\\
		\intertext{%
			because we may drag white diamonds across crossings by \cref{eq:qHC:sliding-white-diamonds}.
			If $i\not\conn j$, this equals
			$-2\tikz[diag, scale=1.25, inline] \draw (-1,-1) node[left]{$i$} -- (-1,1) (1,-1)node[right]{$j$} -- (1,1);$
			because we may resolve the double crossings by \cref{eq:qHC:double-crossing}.
			If $i\mathrel{-} j$, we calculate
		}
				\begin{tikzpicture}[diag]
					\draw[curved, name path=stringA] 
						(-1,-2) node[below]{$i$} 	to[out=+45, in=-45] (-1,2);
					\draw[curved, name path=stringB]
						(1,-2) node[below]{$j$} 	to[out=135,in=-135] 
							(1,2);
					\draw[name intersections={of=stringA and stringB}]
						(intersection-1) node[deco]{$\whitebullet$}
						(intersection-2) node[deco]{$\whitebullet$};
				\end{tikzpicture}
			&=
				(-1)^{\frac{d_{ij}}{2}}
			
				\cdot
				\tikz[diag] \draw
					(-1,-1) node[below]{$i$} -- 
						node[rot,deco,label=-80:$d_{ij}$]{$\blacklozenge$}
						(-1,1)
					(1,-1) node[below]{$j$} -- (1,1);
				+
				(-1)^{\frac{d_{ji}}{2}}
				t_{ji}
			
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (-1,1)
					(1,-1) node[below]{$j$} --
						node[deco,rot,label=-80:$d_{ji}$]{$\blacklozenge$}
						(1,1);
			\\&
			=
			
				\tikz[diag] \draw
					(-1,-1) node[below]{$i$} -- 
						node[rot, deco,pos=.625]{$\whitelozenge$}
						node[rot, deco,pos=.375]{$\blacklozenge$}
						node{$\biggl(\quad\biggr)^{\mathrlap{d_{ij}}}$}
						(-1,1)
					(1,-1) node[below]{$j$} -- (1,1);
				+
			
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (-1,1)
					(1,-1) node[below]{$j$} --
						node[rot, deco,pos=.625]{$\whitelozenge$}
						node[rot, deco,pos=.375]{$\blacklozenge$}
						node{$\biggl(\quad\biggr)^{\mathrlap{d_{ji}}}$}
						(1,1);
			\\
			&=
				\iota\left(
			
				\tikz[diag] \draw
					(-1,-1) node[below]{$i$} -- 
						node[deco]{$\bullet$}
						node[deco]{$\smash{\phantom{{}^{d_{ij}}\bullet}^{d_{ij}}}\vphantom{\bullet}$}
						(-1,1)
					(1,-1) node[below]{$j$} -- (1,1);
				+
			
				\tikz[diag] \draw 
					(-1,-1) node[below]{$i$} -- (-1,1)
					(1,-1) node[below]{$j$} --
						node[deco]{$\bullet$}
						node[deco]{$\smash{\phantom{{}^{d_{ij}}\bullet}^{d_{ij}}}\vphantom{\bullet}$}
						(1,1);
				\right).
		\end{align*}
		\endgroup
	\item[{$\smash{
		\begin{tikzpicture}[diag, inline, scale=.6]
			\draw[name path=stringA]
				(-2,-2) 	-- (2,2)
				(2,-2) 		-- (-2,2);
			\draw[name path=stringB] 
				(0,-2) 		to[out=135, in=-135, curved] (0,2);
		\end{tikzpicture}
		-
		\begin{tikzpicture}[diag, inline,scale=.6]
			\draw[name path=stringA]
				(-2,-2) 	-- (2,2)
				(2,-2) 		-- (-2,2);
			\draw[name path=stringB] 
				(0,-2) 		to[out=45, in=-45, curved] (0,2);
		\end{tikzpicture}
		}
	$}]
		Assume w.\,l.\,o.\,g\@ifnextchar.{}{.\@}\ that $k=1$.
		The braid relation then follows from the computation
		\begin{align*}	
			&\mathrel{\phantom{=}}\tau_1 \tau_2 \tau_{1,\nu}\\
			&=	(\mathfrak{c}_{1}-\mathfrak{c}_{2})^{\pty{s_2s_1(\nu)_1}\pty{s_2s_1(\nu)_2}} \sigma_1\,
				(\mathfrak{c}_{2}-\mathfrak{c}_{3})^{\pty{s_1(\nu)_2}\pty{s_1(\nu)_3}} \sigma_2\,
				(\mathfrak{c}_{1}-\mathfrak{c}_{2})^{\pty{\nu_1}\pty{\nu_2}} \sigma_{1,\nu}\\
			&=  (\mathfrak{c}_{1}-\mathfrak{c}_{2})^{\pty{\nu_1}\pty{\nu_2}}
				(\mathfrak{c}_{1}-\mathfrak{c}_{3})^{\pty{\nu_1}\pty{\nu_3}}
				(\mathfrak{c}_{2}-\mathfrak{c}_{3})^{\pty{\nu_2}\pty{\nu_3}}
				\sigma_1 \sigma_2 \sigma_{1,\nu}\\
			&\stackrel{*}{=}
				(\mathfrak{c}_{2}-\mathfrak{c}_{3})^{\pty{\nu_2}\pty{\nu_3}}
				(\mathfrak{c}_{1}-\mathfrak{c}_{3})^{\pty{\nu_1}\pty{\nu_3}}
				(\mathfrak{c}_{1}-\mathfrak{c}_{2})^{\pty{\nu_1}\pty{\nu_2}}
				\sigma_2 \sigma_1 \sigma_{2,\nu}\\
			&=	(\mathfrak{c}_{2}-\mathfrak{c}_{3})^{\pty{s_1s_2(\nu)_2}\pty{s_1s_2(\nu)_3}} \sigma_2\,
				(\mathfrak{c}_{1}-\mathfrak{c}_{2})^{\pty{s_2(\nu)_1}\pty{s_2(\nu)_2}} \sigma_1\,
				(\mathfrak{c}_{2}-\mathfrak{c}_{3})^{\pty{\nu_2}\pty{\nu_3}} \sigma_{2,\nu}\\
			&=	\tau_2\tau_1\tau_{2,\nu}.
		\end{align*}
		The equation $*$ is clear if at least one index is even
		and is verified by multiplying out the $\mathfrak{c}_k$'s if every index is odd.
	\end{description}
	For injectivity see \autocite[thm.\ 3.3]{KKT:Quiver-Hecke-Superalgebras}.
\end{proof}



\subsection{Faithful polynomial representation}
\label{sec:faithful-super-polynomial-rep}
As a generalisation of the polynomial representation $\operatorname{\mathrm{Pol}\mathfrak C}_n$ of the NilHecke Clifford algebra $\operatorname{\mathsf{NH}\mathfrak C}_n$
defined in \cref{lemma:quiver-Hecke-super-embedding},
we define a representation of the quiver Hecke Clifford algebra $\operatorname{\mathsf{NH}\mathfrak C}_n(C)$
as an analogue to the construction in \autocite[§2.3]{KL:quantum-group-I}.
\begin{definition}
	Let $\operatorname{\mathrm{Pol}\mathfrak C}_n(C)$ be the $k$-linear supercategory defined as follows.
	\begin{itemize}
		\item Its objects are the free $k$-super vector spaces $\operatorname{\mathrm{Pol}\mathfrak C}_\nu ≔ k[y_{1,\nu},\ldots, y_{n,\nu}, \mathfrak{c}_{1,\nu},\dotsc, \mathfrak{c}_{n,\nu}]$,
		indexed by sequences $\nu\in I^n$.
		The $y_{k,\nu}$'s and $\mathfrak{c}_{k,\nu}$ have the same parities
		and satisfy the same relations 
		as in \cref{def:super-klr-algebra:HC(C):diagramatically}. 
		\item Its morphisms super vector space homomorphisms.
	\end{itemize}
\end{definition}
Given a sequence $\nu\in I^n$,
$\operatorname{\mathrm{Pol}\mathfrak C}_n(C)$ and $\operatorname{\mathsf{H}\mathfrak C}_n(C)$ both have full subcategories $\operatorname{\mathrm{Pol}\mathfrak C}(\nu)$ and $\operatorname{\mathsf{NH}}_n(\nu)$
with objects $\{\operatorname{\mathrm{Pol}\mathfrak C}_{\mu}\mid \mu∈S_n\mathbin{.}\nu\}$ for  $\operatorname{\mathrm{Pol}\mathfrak C}(\nu)$
and string diagrams between these for $\operatorname{\mathsf{NH}}_n(\nu)$.

\begin{proposition}
	\label{prop:super-polynomials-are-faithful-rep}
	$\operatorname{\mathsf{H}\mathfrak C}(\nu)$ acts faithfully on $\operatorname{\mathrm{Pol}\mathfrak C}(\nu)$,
		where $y_{k,\nu}$ and $\mathfrak{c}_{k,\nu}$ act by multiplication and $\sigma_{k,\nu}$ acts by
			\begin{equation}
			\label{eq:Definition-of-Faithful-Action}
				\sigma_{k,\nu} =
				\begin{cases}
					s_k 			& \text{if $\nu_k\not\conn \nu_{k+1}$ or $\nu_k\leftarrow\nu_{k+1}$,}\\
					\mathfrak d_k 		& \text{if $\nu_k = \nu_{k+1}$,}\\
					\begin{multlined}
						\Big(
						
							(\mathfrak{c}_{k,\nu}y_{k,\nu})^{d_{\nu_k,\nu_{k+1}}} + \\[-3ex]
						
							+(\mathfrak{c}_{k+1,\nu}y_{k+1,\nu})^{d_{\nu_{k+1},\nu_k}}
						\Big)
						s_k
					\end{multlined}
									& \text{if $\nu_k\rightarrow\nu_{k+1}$.}
			\end{cases}
		\end{equation}
\end{proposition}
\begin{proof}
	For the non-super case,
	faithfulness of the polynomial representation of the quiver Hecke algebra
	has been proven in \autocite[§2.3]{KL:quantum-group-I};
	Our super-algebraic setup necessitates only minor alterations,
	which we make explicit in the following.
	
	\proofsection{Spanning set}
	As a vector space, $\operatorname{\mathsf{H}\mathfrak C}_n(\nu)$ 
	is isomorphic to a direct sum $⨁_{\mu, \mu' ∈ S_n\mathbin{.}\nu}({}_{\mu'}\!\operatorname{\mathsf{H}\mathfrak C}_{\mu})$
	of subspaces ${}_{\mu'}\!\operatorname{\mathsf{H}\mathfrak C}_{\mu}$,
	each of which contains diagrams connecting the bottom sequence $\mu$
	with the top sequence $\mu'$.
	Let ${}_{\mu'}S_\mu$ be the subgroup ${}_{\mu'}S_\mu = \{w ∈ S_n \mid \mu_{w(k)} = \mu'_k\}$ of$S_n$.
	
	Given a diagram in ${}_{\mu'}\!\operatorname{\mathsf{H}\mathfrak C}_{\mu}$,
	assume there are two strings intersecting more than once.
	According to the relation \cref{eq:qHC:double-crossing},
	it can be replaced by a linear combination of diagrams
	with fewer crossings, possibly at the cost of introducing more diamonds.
	Diamonds can be slid across crossings
	by \crefrange{eq:qHC:sliding-black-diamonds}{eq:qHC:sliding-white-diamonds},
	possibly at the cost of introducing	summands with less crossings,
	and past other diamonds, possibly introducing signs.
	
	The vector space ${}_{\mu'}\operatorname{\mathsf{H}\mathfrak C}_{\mu}$ therefore is spanned by diagrams
	with any two strands intersecting at most once,
	all black diamonds below all crossings,
	and all white diamonds below the black ones
	such that there is at most one white diamond per string.
	The white ones may be arranged descending to the right.
	A typical diagram in this spanning set looks as follows:
	\begin{equation*}
		\tikz[diag]\draw
			(0,-2) node [below]{$\mu_1$} -- 
				node [rot, at end, deco]{$\blacklozenge$} 
				node [rot, midway, deco]{$\whitelozenge$} 
				(0,-1) 
				to[out=90,in=-90] (1,2)
			(1,-2) node [below]{$\mu_2$}  -- (1,-1) to[out=90,in=-90]  (2,2)
			(2,-2) node [below]{$\mu_3$}  -- 
				node [rot, at end, deco]{$\blacklozenge$}
				node [rot, near end, deco]{$\blacklozenge$}
				node [rot, near start, deco]{$\whitelozenge$}
				(2,-1) 
				to[out=90,in=-90](4,2)
			(3,-2) node [below]{$\mu_4$} -- 
				node [rot, at end, deco]{$\blacklozenge$}
				node [rot, at start, deco]{$\whitelozenge$}
				(3,-1)
				 to[out=90,in=-90] (0,2)
			(4,-2)  node [below]{$\mu_5$} -- (4,-1)
				to[out=90, in=-90] (3,2);
	\end{equation*}
	This spanning set can be written as
	\newcommand{\operatorname{\mathsf{B}\mathfrak{C}}}{\operatorname{\mathsf{B}\mathfrak{C}}}
	\begin{equation}
		\begin{split}	
			{}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu &= \{
				\sigma_{\pi,\mu}^{}
				y_{\mu}^{\bm{\alpha}}
				\mathfrak{c}_{\mu}^{\bm{\beta}}
			\}
			\\
			&=\{
				\sigma_{\pi,\mu'}^{}
			
				y_{1,\mu}^{\alpha_1}\dotsm y_{n,\mu}^{\alpha_n}\,
				\mathfrak{c}_{1,\mu}^{\beta_1}\dotsm \mathfrak{c}_{n,\mu}^{\beta_n}
			\},
		\end{split}
	\end{equation}

	where $\mu$ and $\mu'$ are permutations of $\nu$,
	$\pi$ is a reduced expression for an element of ${}_{\mu'}S_\mu$,
	$\bm{\alpha}∈\mathbf{N}^n$, $\bm{\beta}∈\{0,1\}^n$ are multiindices,
	and $\mathfrak{c}_\mu^{\bm{\beta}}$ is an ordered monomial.
	Choose a complete order $\leq$ on $I$ such that $i<j$ whenever there is an edge $i\to j$.
	This order induces a lexicographic order on $I^n$.
	We show by induction on $\mu$ \wrt\ this order on $I^n$
	that ${}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu$ is a $k$-basis on which $\operatorname{\mathsf{H}\mathfrak C}(\nu)$ acts faithfully.
	
	\proofsection{Base of induction}
	Let $n_i$ be the number of entries $i$ in $\nu$.
	Let
	\[
		\mu=(\underbrace{i_1, \dotsc, i_1}_{n_{1}}, \underbrace{i_2,\dotsc}_{n_2}, \dotsc)
	\]
	such that $i_1<i_2<\dotsb$.
	The tuple $\mu$ is the lowest element in the orbit $S_n\nu$ \wrt\ the order $\leq$.
	We may write a permutation $w ∈ {}_{\nu'}S_\nu$ as $w=uv$
	such that $u ∈ S_{n_1} \times S_{n_2} \times \dotsb$ permutes the $i_1$, $i_2$ etc.\ independently,
	and $v$ is of minimal length,\;
	\ie, it does not interchange identically labelled strings and interchanges distinct labels at most once. 
	The spanning set can thus be written as
	\[
		{}_{\nu'}\operatorname{\mathsf{B}\mathfrak{C}}_\nu = \{ \sigma_{u,\mu}^{} \sigma_{v,\mu}^{} y_\mu^{\bm{\alpha}}  \mathfrak{c}_\mu^{\bm{\beta}}
		\}.
	\]
	We make sure that these elements act linearly independently on $\operatorname{\mathrm{Pol}\mathfrak C}(\nu)$.
	\begin{itemize}
		\item The terms $y_\mu^{\bm{\alpha}}\mathfrak{c}_\mu^{\bm{\beta}}$ take $y_\mu^{\bm{\alpha}'}\mathfrak{c}_\mu^{\bm{\beta}'}$
		to $\pm y_\mu^{\bm{\alpha}+\bm{\alpha}'} \mathfrak{c}_\mu^{\bm{\beta}+\bm{\beta}'}$.
		\item Since $v$ permutes strings with distinct labels,
			$\sigma_{v,\mu}$ acts by the permutation $v$.
		\item Since the permutation $u = u_1\times\cdots\times u_m ∈ S_{n_1} \times\dotsb\times S_{n_m}$
			permutes identically only labelled strings,
			$\sigma_{u,\nu}$ acts by 
			$\mathfrak d_{u_1}\times\cdots\times\mathfrak d_{u_n}\in\operatorname{\mathsf{NC}\mathfrak C}_{n_1}\times\operatorname{\mathsf{NC}\mathfrak C}_{n_2}\times\dotsb\times\operatorname{\mathsf{NC}\mathfrak C}_{n_m}$.
			We shall prove in \cref{lem:Clifford-NilHecke-acts-Faithfully} 
			that the action of $\operatorname{\mathsf{NC}\mathfrak C}_n$ on polynomials is faithful.
	\end{itemize}
	
	\proofsection{Induction step}
	It suffices to show that if ${}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu$ is linearly independent.
	We have that ${}_{s_k(\mu')}\operatorname{\mathsf{B}\mathfrak{C}}_\mu$ is linearly independent
	if $\mu'_k$, $\mu'_{k+1}$ are distinct and connected by an edge;
	otherwise, ${}_{s_k(\mu')}\operatorname{\mathsf{B}\mathfrak{C}}_\mu$ maps bijectively to ${}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu$ by $\sigma_{k,\mu'}$.
	The multiplication map $(\sigma_k\cdot): {}_{s_k(\mu')}\operatorname{\mathsf{B}\mathfrak{C}}_\nu \into {}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu$
	is seen to be injective by the same argument as in \autocite{KL:quantum-group-I}:
	
	Let ${}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu$ be endowed with a partial order $\preceq$
	such that if we assign to an element $\sigma_{w,\mu}  y_\mu^{\bm{\alpha}} \mathfrak{c}_\mu^{\bm{\beta}}$
	of the spanning set the tuple $(\ell(w), \bm{\alpha}, \bm{\beta})$,
	then the order $\preceq$ coincides with the lexicographic ordering $\leq$ on these tuples.
	Define a map
	\begin{equation}
		\begin{split}
			\varsigma\colon  {}_{s_k(\mu')}\operatorname{\mathsf{B}\mathfrak{C}}_\mu &\to {}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu\\
			
			& \mapsto
			\begin{cases}
				s_k D &
				\parbox{4cm}{if the $k$-th and $(k+1)$-st strand do not intersect}\\[1em]
				%
				t_{\mu_k,\mu_{k+1}} D^* (\mathfrak{c}_{k,\mu}y_{k,\mu})^{d_{\mu_k,\mu_{k+1}}}  &
				\text{otherwise,}
			\end{cases}
		\end{split}
	\end{equation}
	where $D^*$ is obtained from $D$ by removing the crossing.
	The map $\varsigma$ is injective,
	and multiplication by $\sigma_{k,\mu'}$ satisfies
	\[
		\sigma_{k,\mu'} \cdot D ∈ \{±\varsigma(D)\} + \sum_{\mathclap{D'\prec\varsigma(D)}} \mathbf ZD'
		\subseteq {}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu.
	\]
	By the induction hypothesis on ${}_{\mu'}\operatorname{\mathsf{B}\mathfrak{C}}_\mu$, 
	we obtain that the multiplication map $(\sigma_{k,\mu'}⋅)$ must be injective.
\end{proof}
\begin{definition}
	We endow $\operatorname{\mathrm{Pol}\mathfrak C}(\nu)$ with a polynomial grading 
	by setting $\deg(y_{k,\mu})=1$ and $\deg(\mathfrak{c}_{k,\mu})=0$.
	Additionally setting $\deg(\sigma_{k,\mu})=-1$,
	we also endow $\operatorname{\mathsf{H}\mathfrak C}(\nu)$ with a grading.
\end{definition}
\begin{corollary}
	With this grading, $\operatorname{\mathrm{Pol}\mathfrak C}(\nu)$ is a faithful graded $\operatorname{\mathsf{H}\mathfrak C}(\nu)$-module.
\end{corollary}



\section{Clifford symmetric Polynomials}
\label{sec:clifford-symmetric-polynomials}
\newcommand{\hat{\mathfrak{C}}}{\hat{\mathfrak{C}}}
Let the index set $I=\{1,\dotsc, n\} = I_0\sqcup I_1$ be endowed with parities as above.
From now on, 
we consider the algebra $\mathfrak{C}_{I}≔\mathfrak{C}_n/(\mathfrak{c}_i)_{i\in I_0}$.
$\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ and $\mathfrak d_i$ are defined analogously to \cref{def:Polynomial-Clifford-Algebra}.
\begin{definition}
	The intersection 
	$\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}≔⋂_{k=1}^{n-1}\ker \mathfrak d_k \overset{\text{\cref{lem:super-klr-algebra:clifford-demazure-relations}}}{=} ⋂_{k=1}^{n-1}\im \mathfrak d_i\subseteq\operatorname{\mathrm{Pol}\mathfrak C}_n(I)$
	is a $\mathfrak{C}_{I}$-subalgebra and is called the algebra of \term{$\mathfrak d$-symmetric polynomials}%
	\footnote{%
		One may be tempted to call these supersymmetric polynomials;
		this term has already been coined for another notion though
		\autocite{Stembridge:supersymmetric-polynomials}.
	}.
\end{definition}

In this section,
we want to investigate some of their properties.
To this end, we shall introduce the notion of elementary $\mathfrak d$-symmetric polynomials
and show in \cref{prop:Clifford-Polynomials-free} that they generate $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$ as  $\mathfrak{C}_{I}$-algebra.



\subsection{Interlude: counting graded ranks}
\label{sec:graded-ranks}
\begin{definition}
	Given a graded ring $R$ and a free $\mathbf{Z}$-graded $R$-module $M=⨁_{k∈\mathbf{Z}} M_k$,
	its \term{graded rank} or \term{Poincaré series}
	is the formal power series $\rk_{q,R}(M) ≔ \sum_{k∈\mathbf{Z}} \rk_R (M_k) q^k$,
	seen as an element of the localisation $\mathbf{Z}[[q]]_{(1-q)}$.
	We write $\rk_{q,\mathbf{Z}}$ for the graded rank
	of graded free abelian groups.
	With respect to a field $k$, we denote the graded dimension by $\dim_{q,k}$.
\end{definition}

\begin{definition}
	Let $(W,S)$ be a Coxeter system with a fixed generating set $S$.
	which determines a length function $\ell$ on $W$.
	We define $\ord_q(W) ≔ \sum_{w∈W} q^{\ell(w)}$.
\end{definition}

We compute some examples that we shall use later on.
Note that we are working with classical, non-super polynomial rings in this section.
Since $1-q$ is a unit in $\mathbf{Z}[[q]]_{(1-q)}$,
the \term{q-integers}$(n)_q ≔ 1 + \dotsb + q^{n-1}$ can be expressed as $(n)_q = \frac{1-q^n}{1-q}$;
see \autocite{LeStum:q-Integers}
for a conversion between the notations $(n)_q$ and $[n]_q$
and a general introduction to $q$-integers.

\begin{lemma}[{\autocite[cf.][§1]{Winkel:Poincare}}]
	\label{lem:poincare-polynomials-coxeter-groups}
	The Coxeter group of type $\mathrm{A}_n$,  $\mathrm{BC}_n$ and  $\mathrm{D}_n$
	have the following $q$-orders:
	\begin{thmlist}
		\item $\ord_q(\mathrm{A}_n) = (n)_q!$ with the \term{$q$-factorial} $(n)_q! ≔ (n)_q (n-1)_q\dotsm (1)_q$
		\item $\ord_q(\mathrm{BC}_n) = (2n)_q!!$ with the \term{$q$-double factorial} $(2n)_q!! ≔ (2n)_q (n-2)_q \dotsm (2)_q$
		\item $\ord_q(\mathrm{D}_n) = (2n-2)_q!!(n)_q$
	\end{thmlist}
	We have included $\ord_q(\mathrm{BC}_n)$	and $\ord_q(\mathrm{D}_n)$ for the sake of completeness
	but will not need them in the following.
\end{lemma}
\begin{proof}
	\newcommand{\operatorname{inv}}{\operatorname{inv}}
	\begin{prooflist}
	\item
	\label{proof:orders-of-Coxeter-groups:Sn}
	The length $\ell(w)$ of a permutation $w∈S_n$
	equals the number of \term{inversions} \autocite[prop.\ 1.5.2]{Brenti:Coxeter-Groups}, 
	\ie, the cardinality of $\operatorname{inv} w ≔ \{(i,j) \mid 1\leq i<j\leq n, w(i)>w(j)\}$.
	Assume that $\sum_{w∈S_n} p(w) = (n)_q!$.
	Consider the permutations
	\[
		\begin{split}
			\pi_0\colon  (1,\dotsc, n+1) &\mapsto (1,\dotsc, n , n+1),\\
			\pi_1\colon  (1,\dotsc, n+1) &\mapsto (1,\dotsc , n+1,  n),\\
			\shortvdotswithin{\mapsto}
			\pi_{n}\colon  (1,\dotsc, n+1) &\mapsto (n+1, 1,\dotsc, n)
		\end{split}
	\]
	contained in $S_{n+1}$, where $\pi_0=e$ is the trivial permutation.
	Moving the rightmost entry in $\pi_0$ to the left will subsequently create new inversions
	such that $\pi_k$ has the $k$ inversions
	\[
		\operatorname{inv}\pi_k = \bigl\{(n-k+1,n-k+2), \dotsc, (n-k+1, n+1)\bigr\}.
	\]
	and $\sum_{k} q^{\ell(\pi_k)} = (1+q+\dotsb q^{n}) = (n+1)_q$.
	Since any $w∈S_n$ interchanges only the first $n$ slots,
	the inversions of $w$ and $\pi_k$ do not interfere,
	which gives $\operatorname{inv} (\pi_kw) = w^{-1}(\operatorname{inv} \pi_k) \sqcup \operatorname{inv}(w)$.
	Therefore, $\sum_{k} q^{\ell(\pi_k w)}=(n+1)_q q^{\ell(w)}$.
	By the induction hypothesis,
	letting $w$ traverse all of $S_n$ proves the claim.
	
	\item
	The Coxeter group $\mathrm{BC}_n$,
	called the \term{hyperoctahedral group} or \term{signed permutation group},
	is the group of permutations $\pi$ of the set $\{\pm 1,\dotsc, \pm n\}$
	such that $\pi(-k)=-\pi(k)$.
	It has as generators the simple transpositions $s_k\colon  \pm k\mapstofrom \pm k+1$ for $1\leq k\leq n-1$
	and the additional generator $s_0\colon  1\mapstofrom -1$.
	A signed permutation $w$ has set of inversions
	\[
		\operatorname{inv}(w) ≔ 
		\bigl\{ (i,j) \bigm| 1\leq i<j \leq n, w(i)>w(j)\bigr\}
		\cup
		\bigl\{ (-i,j) \bigm| 1\leq i\leq j \leq n, w(-i)>w(j)\bigr\}.
	\]
	We count the number of inversions as in \cref{proof:orders-of-Coxeter-groups:Sn}.
	Assume that $\ord_q \mathrm{BC}_n = (n)_q!!$,
	and consider the permutations
	\begin{align*}
			\pi_0\colon  (1,\dotsc, n+1) &\mapsto (1,\dotsc, n , n+1),\\*
			\shortvdotswithin{\mapsto}
			\pi_n\colon  (1,\dotsc, n+1) &\mapsto (n+1, 1,\dotsc, n),\\
			\pi_{n+1}\colon  (1,\dotsc, n+1) &\mapsto (-n-1, 1,\dotsc, n),\\*
			\shortvdotswithin{\mapsto}
			\pi_{2n+1}\colon  (1,\dotsc, n+1) &\mapsto (1,\dotsc, w(n), -n-1).
	\end{align*}
	They have inversions
	\[
		\operatorname{inv} \pi_k =
		\begin{cases}
		\bigl\{(n-k+1, n-k+2),\dotsc, (n-k+1, n+1)\bigr\} & \text{if $0\leq k\leq n$,}\\[1em]
		\begin{multlined}[c][\widthof{$\bigl\{(n-k+1, n-k+2),\dotsc, (n-k+1, n+1)\bigr\}$}]
		\bigl\{
				\underbrace{(1,l),\dotsc, (l-1,l)}_{\text{$l-1$ many}},
				\underbrace{(-1,l),\dotsc, (l+1,l)}_{\text{$l-1$ many}}\\[-1.4em]
				\underbrace{(-l,l),\dotsc, (l, n+1)}_{\text{$n-l+2$ many}}
		\bigr\}
		\end{multlined}
		& \text{if $1<l\leq n+1$ for $l≔k-n$.}
		\end{cases}
	\]
	Hence, 	$|\operatorname{inv} \pi_k| = k$
	and $\sum_k q^{\ell(\pi_k)} = (2n+2)_q$.
	The rest of the argument is as in \cref{proof:orders-of-Coxeter-groups:Sn}.

	
	\item 
	The Coxeter group $\mathrm{D}_n$, called \term{demihypercube group},
	is the subgroup of $\mathrm{BC}_n$ generated by $s_1,\dotsc, s_n$
	and the additional generator $\tilde{s}_0 ≔ s_0s_1s_0\colon (1,2,3,\dotsc)\mapstofrom (-2,-1,3,\dotsc)$.
	It is the subgroup of signed permutations
	that flip an even number of signs.
	A permutation $w∈ \mathrm{D}_n$ has inversions
	\[
		\operatorname{inv}(w) ≔
		\bigl\{ (i,j) \bigm| 1\leq i<j \leq n, w(i)>w(j) \bigr\}
		\cup
		\bigl\{ (-i,j) \mid 1\leq i \lneq j \leq n, w(-i)>w(j)\bigr\}.
	\]
	Assume by induction that $\ord_q \mathrm{D}_n = (2n-2)_q!!(n)_q$.
	Consider the permutations 
	\begin{align*}
			\pi_0\colon  (1,\dotsc, n+1) &\mapsto (w(1),\dotsc, w(n) , n+1),\\*
			\shortvdotswithin{\mapsto}
			\pi_{n-1}\colon  (1,\dotsc, n+1) &\mapsto (1, n+1,\dotsc, w(n)),\\
			\pi_{n}\colon  (1,\dotsc, n+1) &\mapsto (n+1, 1,2,\dotsc, w(n)),\\
			\pi'_{n}\colon  (1,\dotsc, n+1) &\mapsto (-n-1, -1,2,\dotsc, w(n)),\\*
			\shortvdotswithin{\mapsto}
			\pi_{2n}\colon  (1,\dotsc, n+1) &\mapsto (1,\dotsc, n, -n-1).
	\end{align*}
	Their set of inversions
	\[
		\operatorname{inv}\pi^{(\prime)}_k =
		\begin{cases}
			\bigl\{(n-k+1, n-k+2),\dotsc, (n-k+1, n+1)\bigr\} & \text{if $0\leq k\leq n$,}\\[1em]
			\begin{multlined}[c][\widthof{$\Bigl\{(n-k+1, n-k+2),\dotsc, (n-k+1, n+1)\Bigr\}$}]
				\bigl\{
						\underbrace{(1,l),\dotsc, (l-1,l)}_{\text{$l-1$ many}},
						\underbrace{(-1,l),\dotsc, (l+1,l)}_{\text{$l-1$ many}}\\[-1.4em]
						\underbrace{(-l,l+1),\dotsc, (l, n+1)}_{\text{$n-l+1$ many}}
				\bigr\}
			\end{multlined}
			& \text{if $1<l\leq n+1$ for $l≔k-n$}
		\end{cases}
	\]
	has cardinality $|\operatorname{inv}\pi^{(\prime)}_k|=k$
	and therefore
	\begin{align*}
		\sum_k q^{\ell(\pi^{(\prime)}_k)}
		&= 1+\dotsb+q^n+q^n+\dotsb+q^{2n}= (1+q^n)(n+1)_q.\\
	\shortintertext{%
		As before, we see that
	}
		\ord_q \mathrm{D}_{n+1
		&= (2n-2)_q!! \underbrace{(1+q^n)(n)_q}_{=(2n)_q} (n+1)_q
		= (2n)_q!! (n+1)_q.\qedhere
	\end{align*}
	\end{prooflist}
\end{proof}
\begin{lemma}
	\label{rmk:interesting-graded-ranks}
	The following abelian groups have the respective graded ranks
	\begin{alignat*}{4}
		\rk_{q,\mathbf{Z}} \operatorname{\mathrm{Pol}}_n &= \mathbf Z[y_1,\dotsc, y_n] &&= \frac{1}{(1-q)^n},\qquad &
		\rk_{q,\mathbf{Z}} \mathop{\Lambda\operator@font Pol}\nolimits_n &= \operatorname{\mathrm{Pol}}_n^{S_n} &&= \frac{1}{(n)_q! (1-q)^n},\\
		\rk_{q,\mathbf{Z}} \operatorname{\mathsf{NC}}_n &= \mathbf Z[\partial_1,\dotsc,\partial_n] &&= \frac{1}{(n)_q!}, &
		\rk_{q,\mathbf{Z}} \operatorname{\mathsf{NH}}_n &= \operatorname{\mathrm{Pol}}_n \otimes_\mathbf Z\operatorname{\mathsf{NC}}_n &&= \frac{1}{(n)_q!(1-q)^n},
	\end{alignat*}
	where we set $\deg\partial_i=-1$.
\end{lemma}
\begin{proof}
\begin{description}
	\item[$\operatorname{\mathrm{Pol}}_n$] Every indeterminate $y_i$ independently generates a free abelian group
	$⟨1,y_i,y_i^2,\dotsc⟩_{\mathbf{Z}}$ of graded rank $1+q+q^2+\dots = \frac{1}{1-q}$.
	Since $\operatorname{\mathrm{Pol}}_n \cong \mathbf{Z}[y]^{⊗n}$ as abelian groups,
	$\operatorname{\mathrm{Pol}}_n$ has graded rank $\rk_{q,\mathbf{Z}}\operatorname{\mathrm{Pol}}_n = (1-q)^{-n}$.
	%
	\item[$\mathop{\Lambda\operator@font Pol}\nolimits_n$]
	Recall that $\mathop{\Lambda\operator@font Pol}\nolimits_n\cong \mathbf{Z}\bigl[\varepsilon^{(n)}_1,\dotsc, \varepsilon^{(n)}_n\bigr]$,
	where $\varepsilon^{(n)}_m$ denotes
	the elementary symmetric polynomials in $n$ indeterminates of degree $m$.
	Each polynomial $\varepsilon^{(n)}_m$ generates a subgroup of graded rank $\frac{1}{1-q^m}$;
	$\mathop{\Lambda\operator@font Pol}\nolimits_n$ thus has graded rank 
	$\rk_{q,\mathbf{Z}} \mathop{\Lambda\operator@font Pol}\nolimits_n = \prod_{m=1}^{n} \frac{1}{1-q^m}$.
	Since $(1-q^m) = (m)_q (1-q)$ (telescoping sum),
	induction shows that
	$\rk_{q,\mathbf{Z}} \mathop{\Lambda\operator@font Pol}\nolimits_n = \frac{1}{(n)_q! (1-q)^n}$.
	%
	\item[$\operatorname{\mathsf{NC}}_n$] 
	By the relations of the NilCoxeter-algebra $\operatorname{\mathsf{NC}}_n$,
	there is a bijection $\mathbf Z S_n\to\operatorname{\mathsf{NC}}_n, s_n\mapsto\partial_n$
	such that we can write $\partial_w≔\partial_{i_1}\dotsm\partial_{i_n}$
	for any reduced expression $w=s_{i_1}\dotsm s_{i_n}$ in $S_n$.
	In particular, $\deg\partial_w = -\ell(w)$.
	By \cref{lem:poincare-polynomials-coxeter-groups} ,
	this shows that $\rk_{q,\mathbf{Z}} \operatorname{\mathsf{NC}}_n = (n)_q!$\,.
	%
	\item[$\operatorname{\mathsf{NH}}_n$] This follows from the fact that $\operatorname{\mathsf{NH}}_n \cong \operatorname{\mathrm{Pol}}_n ⊗_\mathbf{Z} \operatorname{\mathsf{NC}}_n$ as abelian groups.
	\qedhere
\end{description}
\end{proof}
\begin{remark}
	Similar statements hold for the invariants under Coxeter groups of type $\mathrm{BC}_n$ and $\mathrm{D}_n$:
	\begin{thmlist}
		\item The hyperoctahedral group $\mathrm{BC}_n$
		acts on the polynomial ring $\operatorname{\mathrm{Pol}}_n$ by permuting the indeterminates
		and flipping their signs.
		It thus has invariants
		$\operatorname{\mathrm{Pol}}_n^{\mathrm{BC}_n} = \mathbf{Z}[y_1^2, \dotsc, y_n^2]^{S_n}$,
		which yields
		\begin{equation*}
			\rk_{q, \mathbf{Z}}\operatorname{\mathrm{Pol}}_n^{\mathrm{B}_n} = \frac{1}{(1-q^2)(1-q^4)\dotsm(1-q^{2n})}
			=\frac{1}{(2n)_q!!(1-q)^n}.
		\end{equation*}
		\item The demihypercube group $\mathrm{D}_n$
		acts on $\operatorname{\mathrm{Pol}}_n$ by permuting the indeterminates
		and flipping two signs at once.
		The invariants thus are
		\begin{align*}
			\operatorname{\mathrm{Pol}}_n^{\mathrm{D}_n} &= \mathbf{Z}\bigl[\varepsilon_1(y_1^2,\dotsc,y_n^2), \dotsc,\varepsilon_{n-1}(y_1^2,\dotsc,y_n^2), \varepsilon_n(y_1, \dotsc, y_n)\bigr]\\
		\shortintertext{with graded rank}
			\rk_{q, \mathbf{Z}}(\operatorname{\mathrm{Pol}}_n) &= 
			\frac{1}{(1-q^2)\dotsm(1-q^{2n-2})(1-q^n)}
			= \frac{1}{(2n-2)_q!! (n)_q (1-q)^n}.
		\end{align*}
	\end{thmlist}
	We notice that $\rk_{q,\mathbf{Z}} \operatorname{\mathrm{Pol}}_n/\rk_{q,\mathbf{Z}} \operatorname{\mathrm{Pol}}_n^W = \ord_q W$
	for $W∈\{S_n, \mathrm{BC}_n, \mathrm{D}_n\}$.
	In fact, graded ranks satisfy $\rk_{q,\mathbf{Z}} \bigl(\operatorname{\mathrm{Pol}}_n/ \operatorname{\mathrm{Pol}}_n^W\bigr) = \ord_q W$ 
	for general finite reflection groups \autocite{Chevalley:Invariants-of-finite-groups}.
	Even stronger, $\operatorname{\mathrm{Pol}}_n$ is a free $\mathop{\Lambda\operator@font Pol}\nolimits_n$-module
	of graded rank $\ord_q W$ for other finite reflection groups $W$
	\autocite[thm.\ 6.2]{Demazure:Invariants}.
	We shall show a respective statement in the Clifford setup for $W=S_n$ in \cref{prop:Clifford-Polynomials-free}.
\end{remark}
\begin{corollary}
	\label{cor:graded-rank-Clifford-polynomials}
	We endow $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ with a (polynomial) grading
	such that the odd generators $\mathfrak{c}_i$ are of degree zero.
	\Cref{rmk:interesting-graded-ranks} then shows that
	\begin{equation*}
		\rk_{q,\mathfrak{C}_{I}} \operatorname{\mathrm{Pol}\mathfrak C}_{I} = \frac{1}{(1-q)^n},\quad
		\rk_{q,\mathfrak{C}_{I}} \operatorname{\mathsf{NH}\mathfrak C}_{I} = \frac{(n)_q!}{(1-q)^n}
	\end{equation*}
	both as left and right $\mathfrak{C}_{I}$-modules.
\end{corollary}


\subsection{\texorpdfstring{Elementary $\mathfrak d$-symmetric polynomials}{Elementary 𝔡-symmetric polynomials}}
\label{sec:elementary-d-symmetric-polynomials}
Our aim is to construct analogues for the elementary symmetric polynomials for $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$
that coincide with the ordinary elementary symmetric polynomials
if all indices are even.
\begin{lemma}
	\label{lem:Elementary-Clifford-Polynomials:two-indeterminates}
	The operator $\mathfrak d_i$ on Clifford-polynomials has as kernel the
	$\mathfrak{C}_{I}$-subalgebra
	\begin{equation}
		\label{eqn:Clifford-Symmetric:Demazure-Kernel}
		\ker\mathfrak d_i = ⟨\gamma_{i,i+1} y_i + \gamma_{i+1, i} y_{i+1},\ y_i y_{i+1},\ y_j \mid j\neq i,i+1 ⟩
	\end{equation}
	of $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$, where we define
	\begin{equation}
		\gamma_{i,i\pm 1} ≔ 
		\begin{cases}
			\pm\mathfrak{c}_i & \text{if $i, i\pm 1$ both are of odd parity,}\\
			1 & \text{otherwise}.
		\end{cases}	
	\end{equation}
	Using the convention $0^0≔1$,
	we can write $\gamma_{i,i\pm 1}≔(\pm\mathfrak{c}_i)^{\pty{i}\pty{i\pm 1}}$.
	Note that $\gamma_{i,i\pm 1}^2=1$ for all $i$.
\end{lemma}
	\begin{proof}
	In the purely even case, the kernel of the Demazure operator $\partial_i$
	is the subalgebra
	\begin{equation*}
		\ker\partial_i=⟨y_i + y_{i+1}, y_iy_{i+1}, y_j \mid j\neq i,i+1⟩⊆\operatorname{\mathrm{Pol}}_n.
	\end{equation*}
	In the super-setting, 
	we have already seen that $\ker\mathfrak d_i⊆\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ is a $\mathfrak{C}_{I}$-subalgebra
	in \cref{lem:super-klr-algebra:clifford-demazure-relations}.
	Assume w.\,l.\,o.\,g\@ifnextchar.{}{.\@}\ that $i=1$ and $I=\{1, 2\}$.
	The only indeterminates are thus $y_1,y_2$ and $\mathfrak{c}_1$,$\mathfrak{c}_2$.
	Let $\Lambda≔⟨y_1y_{2},\: \gamma_{1,2}y_1 + \gamma_{2,1}y_{2}⟩\subseteq\operatorname{\mathrm{Pol}\mathfrak C}_{\{1, 2\}}$ as $\mathfrak{C}_{\{1, 2\}}$-algebra.
	We have to show that $\Lambda = \ker\mathfrak d_{\{1, 2\}}$ as subalgebras of $\operatorname{\mathrm{Pol}\mathfrak C}_{\{1, 2\}}$.
	\begin{itemize}
	\item[($⊆$)]
		In general, $y_1y_{2}∈\ker\mathfrak d_1$ because
		\[
			\mathfrak d_1(y_1 y_{2}) 
			= (-1-\mathfrak{c}_1 \mathfrak{c}_{2}) y_{2} + y_{2}(1-\mathfrak{c}_1 \mathfrak{c}_{2})
			= 0.
		\]
		Furthermore, we have $\mathfrak d_1(y_1)y_1 - \mathfrak d_1(y_{2})y_{2} ∈ \ker\mathfrak d_1$ because
		\begin{equation*}
			\mathfrak d_1\big(\mathfrak d_1(y_1)y_1 - \mathfrak d_1(y_{2})y_{2}\big)
			= -\mathfrak d_1(y_{2})\mathfrak d_1(y_1)+\mathfrak d_1(y_1)\mathfrak d_1(y_{2})
			= 0.
		\end{equation*}
		Therefore, $\ker \mathfrak d_1$ also contains the polynomial%
		\footnote{%
			It seems more natural to put all Clifford-generators on the right,
			because $\mathfrak d_1$ is $\mathfrak{C}_{I}$-right linear.
			We shall stick to putting them on the left though
			in order to preserve compatibility with the notation used in
			\cite{KKT:Quiver-Hecke-Superalgebras}.%
		}
		\begin{align*}
			&\mathrel{\phantom{=}} \big(-\tfrac 12 (\mathfrak{c}_1 - \mathfrak{c}_{2})\big)^{\pty{1}\pty{2}}
				\left( \mathfrak d_1(y_1) y_1 - \mathfrak d_1(y_{2}) y_{2} \right)\\
			&= \big(-\tfrac 12 (\mathfrak{c}_1 - \mathfrak{c}_{2})\big)^{\pty{1}\pty{2}}
				\big( (-1-\mathfrak{c}_1 \mathfrak{c}_{2})y_1 - (1-\mathfrak{c}_1 \mathfrak{c}_{2})y_1 \big)\\
			&= \begin{cases}
				\mathfrak{c}_1 y_1 - \mathfrak{c}_{2} y_{2} & \text{if both $1$, $2$ are odd,}\\
				y_1 + y_{2} & \text{otherwise}
			\end{cases}\\
			&= \gamma_{1,2}y_1 + \gamma_{2,1}y_{2}.
		\end{align*}
	\item[($⊇$)] 
		The algebra $\operatorname{\mathrm{Pol}\mathfrak C}_{\{1, 2\}}$ contains an element $\alpha_1 ≔ y_1-y_{2}$,
		which satisfies $\mathfrak d_1(\alpha_1)=-2$.
		Recall that $\operatorname{\mathrm{Pol}\mathfrak C}_{\{1, 2\}}$ is a free left and right $\mathfrak{C}_{\{1, 2\}}$-module
		of graded rank $\rk_{q,\mathfrak{C}_{\{1, 2\}}} \operatorname{\mathrm{Pol}\mathfrak C}_{\{1, 2\}} = \frac{1}{(1-q)^2}$.
		Its subalgebra $\Lambda$
		is also $\mathfrak{C}_{\{1, 2\}}$-free of graded rank $\rk_{q,\mathfrak{C}_{\{1, 2\}}}\Lambda=\frac{1}{1-q}\cdot\frac{1}{1-q^2}$.
		The $\mathfrak{C}_{\{1, 2\}}$-submodule $\Lambda\alpha_1 ⊆ \operatorname{\mathrm{Pol}\mathfrak C}_{\{1, 2\}}$ does not intersect $\ker\mathfrak d_1$
		because for any $\lambda∈\Lambda$ non-zero, we have $\mathfrak d_1(\lambda\alpha_1)=-2s_1(\lambda) \neq 0$.
		It has graded rank $\rk_{q,\mathfrak{C}_{\{1, 2\}}} \Lambda\alpha_1 = q\rk_{q,\mathfrak{C}_{\{1, 2\}}}\Lambda$.
		Since
		\[
			\rk_{q,\mathfrak{C}_{\{1, 2\}}} \Lambda + \rk_{q,\mathfrak{C}_{\{1, 2\}}} \Lambda\alpha_1
			= \frac{1+q}{(1+q)(1-q^2)} = \frac{1}{(1-q)^2} = \rk_{q,\hat{\mathfrak{C}}_2} \operatorname{\mathrm{Pol}\mathfrak C}_{\{1, 2\}},
		\]
		we obtain that $\operatorname{\mathrm{Pol}\mathfrak C}_{\{1, 2\}}=\Lambda⊕\Lambda\alpha_s$ as $\mathfrak{C}_{\{1, 2\}}$-module
		and in particular that the inclusion $\Lambda⊆\ker\mathfrak d_1$ in fact is an equality.
		\qedhere
	\end{itemize}
\end{proof}
Let $\gamma_{n,n+1} ≔ 1$ if $n+1$ exceeds the number of indeterminates.
Since
\begin{align}
	 y_i y_{i+1}
	 &= \gamma_{i+1,i+2}\gamma_{i,i+1}(\gamma_{i+1}\gamma_{i+1,i+2}\gamma_{i+1,i+2}\, y_i y_{i+1})\\
	 &= (-1)^{\pty{i}\pty{i+1}+\pty{i+1}\pty{i+2}}
		 (y_i y_{i+1} \, \gamma_{i+1,i+2}\gamma_{i+1,i+2}) \gamma_{i+1,i+2}\gamma_{i,i+1}
\end{align}
we may replace the second generator by $\gamma_{i,i+1} y_1\,\gamma_{i+1,i+2} y_{i+1}$
without changing the $\mathfrak{C}_{I}$-span.
\begin{definition}
	\label{def:Elementary-Clifford-Polynomials:two-indeterminates}
	We denote the generators of $\ker\mathfrak d_i$  by
	\begin{equation*}
		\mathfrak e^{(i-1,i+1)}_1≔\gamma_{i,i+1} y_i + \gamma_{i+1, i} y_{i+1};\qquad
		\mathfrak e^{(i-1,i+1)}_2≔\gamma_{i,i+1} y_i\,\gamma_{i+1,i+2} y_{i+1}.
	\end{equation*}
	If $i=1$, we simply write $\mathfrak e^{(2)}_m ≔ \mathfrak e^{(i-1,i+1)}_m$.
	Wee can thus write $\ker\mathfrak d_1 = ⟨\mathfrak e^{(2)}_1, \mathfrak e^{(2)}_2⟩⊆\operatorname{\mathrm{Pol}\mathfrak C}_2$ as $\mathfrak{C}_{\{1, 2\}}$-algebra.
\end{definition}
Our goal is to show that there are polynomials $\mathfrak e^{(n)}_m$, $1\leq m\leq n$
of polynomial degree $m$ 
that generate $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$ of $\mathfrak{C}_{I}$-algebras.
These are meant to serve as a Clifford-replacement
for the ordinary elementary symmetric polynomials.

Our first task is to find a recursive formula for polynomials lying in $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$.
Proving that they are indeed generators requires some more work.
We shall prove this in \cref{prop:Clifford-Polynomials-free}.

Before coming to the $n$-fold intersection $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}=\bigcap_{k=1}^{n-1}\ker\mathfrak d_k$,
we consider the intersection $\ker\mathfrak d_1 \cap \ker\mathfrak d_2 ⊆ \operatorname{\mathrm{Pol}\mathfrak C}_{\{1,2,3\}}$
of just the first two kernels.
We can multiply the first generator of $\ker\mathfrak d_2$
with the unit $\gamma_{2,1}\gamma_{2,3}$ from the left
without changing its span.
Thus
\begin{equation}
	\ker\mathfrak d_1 \cap \ker\mathfrak d_{2}
	= \begin{multlined}[t]\Bigl\langle
		\overunderbraces{&\br{2}{\mathfrak e^{(0,2)}_1}}%
			{&\gamma_{1,2} y_1 + &\gamma_{2,1} y_{2}& + \gamma_{2,1}\gamma_{2,3}\,\gamma_{3,2}\ y_3}%
			{&& \br{2}{\gamma_{2,1}\gamma_{2,3}\, \mathfrak e^{(1,3)}_1}},
	\\[.5em]	

	%
		\overunderbraces{&\br{2}{\gamma_{1,2}\,y_1\, \gamma_{2,1}\gamma_{2,3}\,\mathfrak e^{(1,3)}_1}}%
			{
				& \gamma_{1,2}\,y_1 \ \gamma_{2,1}\, y_2 
				+ & \gamma_{1,2}\,y_1 \ \gamma_{2,1}\gamma_{2,3}\,\gamma_{3,2}\, y_3 &
				+ \gamma_{2,1}\,y_2 \ \gamma_{2,1}\gamma_{2,3}\,\gamma_{3,2}\, y_3
			}
			{&&\br{2}{\mathfrak e^{(0,2)}_1 \gamma_{2,1}\gamma_{2,3}\,\gamma_{3,2}\, y_3}},
	\\[.5em]


	\overunderbraces{&\br{1}{\mathfrak e^{(0,2)}_2}}%
		{&\gamma_{1,2}\,y_1\ \gamma_{2,3}\, y_2\ & \gamma_{3,4}\, y_3}%
		{}
	\Bigr\rangle
	\end{multlined}
\end{equation}
as a $\mathfrak{C}_{I}$-algebra.
This depiction serves as a template for the following lemma:
\begin{lemma}
	\label{lem:Elementary-Clifford-Polynomials:Recursively}
	The recursively defined polynomials
	\begin{equation}
		\label{eqn:Elementary-Clifford-Polynomials:Recursively}
		\begin{split}
		\mathfrak e^{(1)}_{\mathstrut 1} &=  \gamma_{1,2} y_1\\
			& \shortvdotswithin{=}
			\mathfrak e^{(n)}_1 &= \mathfrak e^{(n-1)}_1 
				+ (\gamma_{2,1}\gamma_{2,3}\ \gamma_{3,2}\gamma_{3,4}\dotsm \gamma_{n-1,n-2}\gamma_{n-1}{n})
				\gamma_{n,n-1}\ y_n,\\
			\mathfrak e^{(n)}_{\mathstrut 1}2 &=
				 \mathfrak e^{(n-1)}_1
				 + \mathfrak e^{(n)}_1 (\gamma_{3,2}\gamma_{3,4}\dotsm \gamma_{n-1,n-2}\gamma_{n-1}{n})
				\gamma_{n,n-1}\ y_n,\\
			& \shortvdotswithin{=}
			\mathfrak e^{(n)}_m &=
				 \mathfrak e^{(n-1)}_{m} 
				 + \mathfrak e^{(n-1)}_{m-1} (\gamma_{m+1,m}\gamma_{m+1,m+2}\dotsm \gamma_{n-1,n-2}\gamma_{n-1}{n})
				\gamma_{n,n-1}\ y_n,\\
			\mathfrak e^{(n)}_{n} &= \mathfrak e^{(n-1)}_n\,\gamma_{n+1,n+2}\ y_n.
		\end{split}
	\end{equation}
	are contained in the common kernel $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I} ⊆ \operatorname{\mathrm{Pol}\mathfrak C}_{I}$ 
	of the Clifford Demazure operators.
	We shall refer to these polynomials as \term{elementary $\mathfrak d$-symmetric polynomials}
	of degree $m$ in $n$ indeterminates.
	In other words, the lemma states that $\Lambda_I≔⟨\mathfrak e^{(n)}_1, \ldots, \mathfrak e^{(n)}_n⟩\subseteq\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$
	is a subalgebra.
\end{lemma}
\newcommand{\mathit{o\varepsilon}}{\mathit{o\varepsilon}}
\begin{table}
	\caption{%
		Elementary $\mathfrak d$-symmetric polynomials $\mathfrak e^{(n)}_m$
		as defined in \cref{lem:Elementary-Clifford-Polynomials:Recursively},
		spelt out explicitly for $m\leq n\leq 4$.
	}
	\label{tab:elementary-d-symmetric-polynomials}
		\footnotesize
		\centering
		$\begin{array}{@{}l@{}llll@{}}
			\toprule				
			m\backslash n & \hfil 1 & \hfil 2 & \hfil 3 & \hfil 4\\
			\midrule
			1
			& \gamma_{1,2} y_1
			& \begin{array}{@{}>{{}}l@{}}
				\phantom{{}+{}}\gamma_{1,2}\, y_1 \cr
				+ \gamma_{2,1}\, y_2
			\end{array}
			& \begin{array}{@{}>{{}}l@{}}
				\phantom{{}+{}} \gamma_{1,2}\, y_1\cr
				+ \gamma_{2,1}\, y_2\cr
				+ \gamma_{2,1}\gamma_{2,3}\gamma_{3,2}\ y_3
			\end{array}
			& \begin{array}{@{}>{{}}l@{}}
				\phantom{{}+{}}\gamma_{1,2}\, y_1\cr
				+ \gamma_{2,1}\, y_2\cr
				+ \gamma_{2,1}\gamma_{2,3}\gamma_{3,2}\ y_3\cr
				+ \gamma_{2,1}\gamma_{2,3}\, \gamma_{3,2}\gamma_{3,4}\gamma_{4,3}\ y_4
			\end{array}
			\\\addlinespace
			2
			&& \phantom{{}+{}} \gamma_{1,2},y_1 \ \gamma_{2,3}\, y_2
			& \begin{array}{@{}>{{}}l@{}}
				\phantom{{}+{}}\gamma_{1,2}\, y_1\ \gamma_{2,3}\, y_2\cr
				+ \gamma_{1,2}\,y_1\ \gamma_{3,2}\, y_3\cr
				+ \gamma_{2,1}\,y_2\ \gamma_{3,2}\, y_3
			\end{array}
			& \begin{array}{@{}>{{}}l@{}}
				\phantom{{}+{}}\gamma_{1,2}\, y_1\ \gamma_{2,3}\, y_2\cr
				+ \gamma_{1,2}\,y_1\ \gamma_{3,2}\, y_3\cr
				+ \gamma_{2,1}\,y_2\ \gamma_{3,2}\, y_3\cr
				+ \gamma_{1,2}\,y_1\ \gamma_{3,2}\gamma_{3,4}\gamma_{4,3}\, y_4\cr
				+ \gamma_{2,1}\,y_2 \ \gamma_{3,2}\gamma_{3,4}\gamma_{4,3}\ y_4\cr
				+ \gamma_{2,1}\gamma_{2,3}\gamma_{3,2}\,y_3 \ \gamma_{3,2}\gamma_{3,4}\,\gamma_{4,3}\ y_4
			\end{array}
			\\\addlinespace
			3 &&& \phantom{{}+{}}\, \gamma_{1,2}\,y_1\,\gamma_{2,3}\,y_3\,\gamma_{3,4}\,y_3
			& \begin{array}{@{}>{{}}l@{}}
				\phantom{{}+{}}
				  \gamma_{1,2}\,y_1\ \gamma_{2,3}\, y_2\ \gamma_{3,4}\,y_3\cr
				+ \gamma_{1,2}\,y_1\ \gamma_{2,3}\, y_2\ \gamma_{4,3}\,y_4\cr
				+ \gamma_{1,2}\,y_1\ \gamma_{3,2}\, y_3\ \gamma_{4,3}\,y_4\cr
				+ \gamma_{2,1}\,y_2\ \gamma_{3,2}\, y_3\ \gamma_{4,3}\,y_4
			\end{array}
			\\\addlinespace
			4 &&&& \phantom{{}+{}}\gamma_{1,2}\,y_1\ \gamma_{2,3}\,y_2\ \gamma_{3,4}\,y_3\ \gamma_{4,5}\,y_4
			\\	
			\bottomrule
		\end{array}$
\end{table}
\begin{table}
	\caption{%
		Ordinary elementary symmetric polynomials $\varepsilon^{(n)}_m$ for $m\leq n\leq 4$.
		They are a specialisation of the ones listed in \cref{tab:elementary-d-symmetric-polynomials}
		at $\gamma_{i,i\pm 1}=1$ for all $i$, obtained when all $i$ have \emph{even} parity.
	}
	\label{tab:elementary-symmetric-polynomials}
	\centering
	\footnotesize
	$\begin{array}{@{}l@{}llll@{}}
			\toprule
			m\backslash n 
			& \mathmakebox[\widthof{$\gamma_{1,2} y_1$}]{1} 
			& \mathmakebox[\widthof{$\phantom{{}+{}} \gamma_{1,2},y_1 \ \gamma_{2,3}\, y_2$}]{2} 
			& \mathmakebox[\widthof{$\phantom{{}+{}}\, \gamma_{1,2}\,y_1\,\gamma_{2,3}\,y_3\,\gamma_{3,4}\,y_3$}]{3}  
			& \mathmakebox[\widthof{$\gamma_{2,1}\gamma_{2,3}\gamma_{3,2}\,y_3 \ \gamma_{3,2}\gamma_{3,4}\,\gamma_{4,3}\ y_4$}]{4}
			\\
			\midrule
			1
			& y_1
			& y_1 + y_2
			& y_1 + y_2 + y_3
			& \phantom{{}+{}} y_1 + y_2 + y_3 + y_4
			\\\addlinespace
			2
			&& y_1y_2
			& y_1y_2 + y_1y_3 + y_2y_3
			& \begin{array}{@{}>{{}}r<{{}}@{}l}
				& y_1y_2 + y_1y_3\\  + & y_1y_4 + y_2y_3\\ + & y_2y_4 + y_3y_4 
			\end{array}
			\\\addlinespace
			3 
			&&& y_1 y_2y_3
			& \begin{array}{@{}>{{}}r<{{}}@{}l}
				& y_1y_2y_3 + y_1y_2y_4\\ + & y_1y_3y_4 + y_2y_3y_4
			\end{array}
			\\\addlinespace
			4 &&&& \phantom{{}+{}} y_1y_2y_3y_4
			\\
			\bottomrule
	\end{array}$
\end{table}
\begin{table}
	\caption{%
		Odd elementary symmetric polynomials $\mathit{o\varepsilon}^{(n)}_m$ for $m\leq n\leq 4$.
		They are a specialisation of the ones listed in \cref{tab:elementary-d-symmetric-polynomials}
		at $\gamma_{i,i\pm 1}=\pm 1$ for all $i$, obtained when all $i$ have \emph{odd} parity.
	}
	\label{tab:odd-elementary-symmetric-polynomials}
	\centering
	\footnotesize
	$\begin{array}{@{}l@{}llll@{}}
			\toprule
			m\backslash n 
			& \mathmakebox[\widthof{$\gamma_{1,2} y_1$}]{1} 
			& \mathmakebox[\widthof{$\phantom{{}+{}} \gamma_{1,2},y_1 \ \gamma_{2,3}\, y_2$}]{2} 
			& \mathmakebox[\widthof{$\phantom{{}+{}}\, \gamma_{1,2}\,y_1\,\gamma_{2,3}\,y_3\,\gamma_{3,4}\,y_3$}]{3}  
			& \mathmakebox[\widthof{$\gamma_{2,1}\gamma_{2,3}\gamma_{3,2}\,y_3 \ \gamma_{3,2}\gamma_{3,4}\,\gamma_{4,3}\ y_4$}]{4}
			\\
			\midrule
			1
			& \phantom{{}+{}} x_1
			& \phantom{{}+{}} x_1 - x_2
			& \phantom{{}+{}} x_1-x_2+x_3
			& \phantom{{}+{}} x_1-x_2+x_3-x_4
			\\\addlinespace
			2
			&& \phantom{{}+{}} x_1 x_2
			& \begin{array}{@{}>{{}}l@{}}
				\phantom{{}+{}}x_1 x_2
				- x_1 x_3\cr
				+ x_2 x_3
			\end{array}
			& \begin{array}{@{}>{{}}c@{}}
				\phantom{{}+{}}x_1 x_2
				- x_1 x_3
				+ x_2 x_3\cr
				+ x_1 x_4
				- x_2 x_4
				- x_3 x_4
			\end{array}
			\\\addlinespace
			3 &&& \phantom{{}+{}} x_1 x_3 x_3
			& \begin{array}{@{}>{{}}l@{}}
				\phantom{{}+{}} x_1 x_3 x_3
				- x_1 x_2 x_3\cr
				+ x_1 x_3 x_4
				-  x_2  x_3 x_3
			\end{array}
			\\\addlinespace
			4 &&&& \phantom{{}+{}}x_1 x_3 x_3 x_4\\
			\bottomrule
	\end{array}$
\end{table}

\begin{example}
	\label{example:Elementary-Clifford-Polynomials}
	Before proving the lemma let us make the definition of the $\mathfrak e$'s explicit
	for small numbers of indeterminates.
	\begin{thmlist}%
		\item For $n=1,\dotsc,4$ variables,
		the elementary $\mathfrak d$-symmetric polynomials are listed explicitly
		in \cref{tab:elementary-d-symmetric-polynomials}.
		
	\item 
		If all indices are even, then all $\gamma_{i,i\pm1}=1$.
		In this case, the $\mathfrak d$-elementary symmetric polynomials 
		from \cref{tab:elementary-d-symmetric-polynomials}
		specialise to those from \cref{tab:elementary-symmetric-polynomials} at $\gamma_{i,i\pm1}=1$,
		which we recognise at the ordinary elementary symmetric polynomials $\varepsilon^{(n)}_m$.
		The induction formula from \cref{lem:Elementary-Clifford-Polynomials:Recursively}
		indeed gives
		\begin{equation}
		\begin{split}
		\varepsilon^{(1)}_1 &=  x_1\\		
		\varepsilon^{(n)}_m &=
		\varepsilon^{(n-1)}_{m} 
		+ \varepsilon^{(n-1)}_{m-1} x_n
		\end{split}
		\end{equation}
		by setting each of the $\gamma$'s in \cref{eqn:Elementary-Clifford-Polynomials:Recursively} to $1$.
		This is a well-known recursive formula for the ordinary elementary symmetric polynomials
		\begin{equation}
		\varepsilon^{(n)}_m =\sum_{\mathclap{1\leq k_{1}<\dotsb <k_{m}\leq n}} x_{k_{1}}\dotsm x_{k_m}.
		\end{equation}

	\item\label{example:Elementary-Clifford-Polynomials:odd}
		If all indices are odd, then $\gamma_{i,i\pm 1}=\pm \mathfrak{c}_i$.
		We set $x_i≔\gamma_{i,i+1}y_i = \mathfrak{c}_i y_i$
		(cf.\ \cref{lemma:quiver-Hecke-super-embedding}).
		The polynomials from \cref{tab:elementary-d-symmetric-polynomials} 
		then specialise at $\gamma_{i,i\pm 1}=\pm \mathfrak{c}_i$
		to the ones listed in \cref{tab:elementary-symmetric-polynomials}.
		We refer to these as \term{odd elementary symmetric} polynomials
		and denote them by $\mathit{o\varepsilon}^{(n)}_m$.
		The recursion formula reduces to
		\begin{equation}
			\begin{split}
			\mathit{o\varepsilon}^{(1)}_1 &=  x_1\\		
			\mathit{o\varepsilon}^{(n)}_m &=
			\mathit{o\varepsilon}^{(n-1)}_{m} 
			+ (-1)^{n-m}\mathit{o\varepsilon}^{(n-1)}_{m-1}\, x_n\\
			\mathit{o\varepsilon}^{(n)}_n &=
			\mathit{o\varepsilon}^{(n-1)}_{m-1}\, x_n
			\end{split}
		\end{equation}
		Consider also the odd elementary symmetric polynomials
		\begin{equation}		
			\prescript{}{\mathrm{EKL}}{o\varepsilon}^{(n)}_m ≔ \sum_{\mathclap{1\leq k_{1}<\dotsb <k_{m}\leq n}} (-1)^{k_m-1} x_{k_{1}}\dotsm (-1)^{k_1-1} x_{k_m}
		\end{equation}		
		as defined in \autocite[(2.21–2.23)]{EKL:odd-NilHecke},
		which satisfy the recursion formula
		\begin{equation*}
			\prescript{}{\mathrm{EKL}}{o\varepsilon}^{(n)}_m =
				\prescript{}{\mathrm{EKL}}{o\varepsilon}^{(n-1)}_m
				+ (-1)^{n-1}\prescript{}{\mathrm{EKL}}{o\varepsilon}^{(n-1)}_m x_n.
		\end{equation*}
		Our odd elementary symmetric polynomials
		indeed coincide with those from \autocite{EKL:odd-NilHecke}
		up to an overall sign $(-1)^{\frac{m(m-1)}{2}}$:
		\begin{align}
		\label{eqn:Odd-Elementary-Polynomial:EKL-Sign}
		\mathit{o\varepsilon}^{(n)}_m &=
		\mathit{o\varepsilon}^{(n-1)}_{m} 
		+ (-1)^{n-m} \mathit{o\varepsilon}^{(n-1)}_{m-1}\, x_n
		\notag\\
		&= 	(-1)^{\frac{m(m-1)}{2}} \cdot \prescript{}{\mathrm{EKL}}{o\varepsilon}^{(n-1)}_m
		+ (-1)^{\big(\frac{(m-1)(m-2)}{2}\big) + (n-m)}
		\cdot \prescript{}{\mathrm{EKL}}{o\varepsilon}^{(n-1)}_m\, x_n
		\notag\\
		&= (-1)^{\frac{m(m-1)}{2}}  \cdot
		\prescript{}{\mathrm{EKL}}{o\varepsilon}^{(n-1)}_m
		+ (-1)^{n-1}x_n \cdot \prescript{}{\mathrm{EKL}}{o\varepsilon}{(n-1)}_m
		\notag\\
		&=  (-1)^{\frac{m(m-1)}{2}} 
		\cdot \prescript{}{\mathrm{EKL}}{o\varepsilon}^{(n)}_m.
		\end{align}
		According to our sign convention,
		the first monomial $x_1\dotsm x_m$ of the odd polynomials 
		always has $+1$ as coefficient.
		Odd symmetric functions have already been treated extensively in
		\autocites{EKL:odd-NilHecke}{EK:odd-NilHecke-Hopf}{EQ:odd-NilHecke-dg}.
	\end{thmlist}
\end{example}
\begin{proof}[Proof of \cref{lem:Elementary-Clifford-Polynomials:Recursively}]
	The statement is proven by induction on $n$ and $m$.
	It is convenient to set $\mathfrak e^{(n)}_0 = 1$ for all $n\geq 0$.
	The claim then trivially holds for $n=1$ and $m=0,1$.
	
	\proofsection{Induction step on $n$ for $m=1$}
	Assume that $\mathfrak e^{(k)}_1 ∈ \mathfrak d_{1}\cap\dotsb\cap\mathfrak d_{n-1}$ for all $k\leq n$.
	We want to show that $\ker\mathfrak d_{1}\cap\dotsb\cap\mathfrak d_{n}$ contains the polynomial
	\begin{align*}
		\mathfrak e^{(n+1)}_1
		&= \mathfrak e^{(n)}_1 + (\gamma_{2,1}\gamma_{2,3} \dotsm \gamma_{n,n-1}\gamma_{n-1,n}) \gamma_{n+1,n}\, y_{n+1}.\\
	\intertext{%
		The first summand $\mathfrak e^{(n)}_1$ is contained in $\ker\mathfrak d_1 \cap \dotsb \cap\ker\mathfrak d_{n-1}$ 
		by the induction hypothesis.
		The second summand contains no $y_{k}$ for $k\leq n$
		and thus also lies in $\ker\mathfrak d_1 \cap \dotsb \cap\ker\mathfrak d_{n-1}$
		(cf.\ the definition of $\mathfrak d$ in \cref{eq:Definition-NilHecke-Clifford-algebra}).
		In order to show that $\mathfrak e^{(n+1)}_1 ∈ \ker \mathfrak d_{n}$,
		we expand the first summand $\mathfrak e^{(n)}_1$ and obtain
	}
		\mathfrak e^{(n+1)}_1 &=
	
			\mathfrak e^{(n-1)}_1
			+ (\gamma_{2,1} \dotsm \gamma_{n-1,n-2}\gamma_{n-1,n}) \gamma_{n,n-1}\, y_{n}+{}
			(\gamma_{2,1} \dotsm \gamma_{n,n-1}\gamma_{n,n+1}) \gamma_{n+1,n}\, y_{n+1}.
		\\%\end{multlined}\\
	\shortintertext{Regrouping the terms gives}
		&= \mathfrak e^{(n-1)}_1  + 
		\underbrace{
			\gamma_{2,1} \dotsm \gamma_{n,n-1}\gamma_{n-1,n}\ \gamma_{n,n-1}
		}_{∈\mathfrak{C}_{I}^*}
		\underbrace{
			( y_{n} + \gamma_{n,n+1}\gamma_{n+1,n} \ y_{n+1})
		}_{
			= \gamma_{n,n+1}\mathfrak e^{(n-1,n+1)}_{1}
		}.
	\end{align*}
	The first summand $\mathfrak e^{(n-1)}_1 $ is clearly contained in $\ker \mathfrak d_{n}$
	because it contains no $y_{n}$, $y_{n+1}$.
	The second summand lies in the $\mathfrak{C}_{I}$-algebra $\bigl\langle\mathfrak e^{(n-1,n+1)}_{1}\bigr\rangle ⊆ \ker \mathfrak d_{n}$
	(see \cref{lem:Elementary-Clifford-Polynomials:two-indeterminates,def:Elementary-Clifford-Polynomials:two-indeterminates}).
	Therefore, ${\mathfrak e^{(n+1)}_1 ∈ \ker\mathfrak d_{1}\cap\dotsb\cap\mathfrak d_{n}}$.
	
	\proofsection{Induction step for $m>1$}
	Assume $\mathfrak e^{(k)}_{m} ∈ \ker\mathfrak d_1 \cap\dotsb\cap \ker\mathfrak d_{n-1}$ for all $k\leq n$ and $m\leq k$.
	We want to show that $\ker\mathfrak d_1 \cap\dotsb\cap \ker\mathfrak d_n$ contains
	\begin{equation}
		\label{eq:proof:Elementary-Clifford-Polynomials:resolve-e-n+1}	
		\mathfrak e^{(n+1)}_m = \mathfrak e^{(n)}_m + \mathfrak e^{(n)}_{m-1}
		(\gamma_{m+1,m}\gamma_{m+1,m+2} \dotsm \gamma_{n,n-1}\gamma_{n-1,n})\gamma_{n+1,n}\,y_{n+1}.
	\end{equation}
	\settowidth{\algnRef}{$\mathfrak e^{(n+1)}_m\mathrel{\phantom{=}}$}
	\begin{description
	\item[$\mathfrak e^{(n+1)}_m ∈ \ker\mathfrak d_{k}$ for $k\leq n-1$]
		By the induction hypothesis, $\mathfrak e^{(n)}_m ∈ \ker\mathfrak d_{k}$.
		The second summand satisfies
		\begin{equation*}
			\begin{split}
			\mathmakebox[\algnRef][l]{
			\mathfrak d_k \bigl( 
				\underbrace{\mathfrak e^{(n)}_{m-1}}_{∈\ker\mathfrak d_k}
				\underbrace{\gamma_{m+1,m}\gamma_{m+1,m+2} \dotsm \gamma_{n,n-1}\gamma_{n-1,n}}_{∈\mathfrak{C}_{I}^*}
				\gamma_{n+1,n}y_{n+1}\bigr)
			} & \\
			&= s_i \bigl( 
					\mathfrak e^{(n)}_{m-1}\cdot
					\gamma_{m+1,m}\gamma_{m+1,m+2} \dotsm \gamma_{n,n-1}\gamma_{n-1,n}\gamma_{n+1,n}
			\bigr)\,
			\mathfrak d_k (y_{n+1})\\
			&= 0
			\end{split}
		\end{equation*}
		by the induction hypothesis on $\mathfrak e^{(n)}_{m-1}$.
	
	\item[$\mathfrak e^{(n+1)}_m ∈ \ker\mathfrak d_{n}$]
		For $\mathfrak d_{n}$, we expand $\mathfrak e^{(n)}_m$ and $\mathfrak e^{(n)}_{m-1}$ 
		in \cref{eq:proof:Elementary-Clifford-Polynomials:resolve-e-n+1}	
		by the recursion formula and obtain
		\setlength{\algnRef}{\linewidth-1em-\mathindent-\algnRef-\widthof{${}={}$}}
		\begin{align}
			\mathfrak e^{(n+1)}_m
			&=
			\setlength{\multlinegap}{0pt}
			\begin{multlined}[t][\algnRef]
				\Bigl[
					\overbrace{
						\mathfrak e^{(n-1)}_m 
						+ \mathfrak e^{(n-1)}_{m-1} \bigl(\gamma_{m+1,m}\dotsm \gamma_{n-1,n-2}\gamma_{n-1,n} \bigr) \gamma_{n,n-1}\,y_n
					}^{
						\mathfrak e^{(n)}_m
					}
				\Bigr] + {}\\[1ex]
				{}+
				\Big[
				\underbrace{
					\mathfrak e^{(n-1)}_{m-1}
					+ \mathfrak e^{(n-1)}_{m-2}(\gamma_{m,m-1} \dotsm \gamma_{n-1,n})\gamma_{n,n-1}\, y_{n})
				}_{\mathfrak e^{(n)}_{m-1}}
				\Big]\cdot{}\\[-2ex]
				\cdot(\gamma_{m+1,m} \dotsm \gamma_{n,n-1}\gamma_{n,n+1})\gamma_{n+1,n}\, y_{n+1}.
			\end{multlined
		\intertext{We regroup both terms containing $\mathfrak e^{(n-1)}_{m-1}$ and get}
			\label{eq:recursion of e-m-n+1 final}
			&=
			\begin{multlined}[t][\algnRef]
				\mathfrak e^{(n-1)}_m +
				\mathfrak e^{(n-1)}_{m-2} \,
				\overbrace{
					(\gamma_{m,m-1} \dotsm \gamma_{n-1,n})\gamma_{n,n-1}\, y_n \,
					(\gamma_{m+1,m} \dotsm \gamma_{n,n+1})\gamma_{n+1,n-1}\, y_{n+1} \,
				}^{
					=\pm (\gamma_{m,m-1} \dotsm \gamma_{n,n-1})(\gamma_{m+1,m}\dotsm \gamma_{n+1,n}) 
					\smash{\overbrace{\scriptstyle y_{n} y_{n+1}}^{\scriptstyle\mathclap{=\gamma_{n+1,n+2}\gamma_{n,n+1}\,\mathfrak e^{(n-1,n+1)}_2}}}
				}
				 + {}\\[2ex]
				{} + 
				\mathfrak e^{(n-1)}_{m-1}
				\underbrace{\bigl(\dotsm \gamma_{n-1,n-2}\gamma_{n-1,n}\, \gamma_{n,n-1} \bigr)}%
				_{∈\mathfrak{C}_{I}^*}
				\bigl(
					\underbrace{\gamma_{n,n-1} y_n +\gamma_{n+1,n}y_{n+1}}_{=\mathfrak e^{(n-1,n+1)}_1}
				\bigr).
			\end{multlined}
		\end{align}
		The operator $\mathfrak d_n$ maps the terms $\mathfrak e^{(n-1)}_m$, $\mathfrak e^{(n-1)}_{m-2}$ and $\mathfrak e^{(n-1)}_{m-1}$ to zero
		because they do not contain $y_n$ and $y_{n+1}$.
		The second and the last summand of \eqref{eq:recursion of e-m-n+1 final} are mapped to zero by $\mathfrak d_n$
		because they are contained in the $\mathfrak{C}_{I}$-bimodule
		$\mathfrak e^{(n-1)}_{m-1}\cdot⟨\mathfrak e^{(n-1,n+1)}_2⟩ ⊆ \ker\mathfrak d_n$.
	\end{description}
	Therefore, $\mathfrak e^{(n+1)}_m ∈ \ker\mathfrak d_n$.
	We have already seen that $\mathfrak e^{(n+1)}_m ∈ \ker\mathfrak d_1\cap\dotsb\cap\ker\mathfrak d_{n-1}$.
	This proves the assertion.
\end{proof}
\begin{remark}
	\label{rmk:graded-rank-d-symmetric-polynomials}
	The $\mathfrak{C}_{I}$-algebra $\Lambda_I$
	has a basis $\bigl\{\bigl(\mathfrak e^{(n)}_1\bigr)^{\alpha_i}, \ldots, \bigl(\mathfrak e^{(n)}_n\bigr)^{\alpha_n}\bigr\}$,
	$\alpha_i ∈ \mathbf{N}$,
	as left and as right $\mathfrak{C}_{I}$-module
	(cf.\ the proof of \cref{lem:PolCn-is-free}).
	With the same argumentation as in \cref{rmk:interesting-graded-ranks},
	we see that its graded $\mathfrak{C}_{I}$-rank is $\frac{1}{(n)_q! (1-q)^n}$
	both as left and right $\mathfrak{C}_{I}$-module.
\end{remark}



\subsection{\texorpdfstring{Schubert Polynomials; freeness of $\operatorname{\mathrm{Pol}\mathfrak C}_n$}{Schubert Polynomials; freeness of Polℭn}}
\newcommand{\mathfrak{s}}{\mathfrak{s}}
\label{sec:Clifford-complete-Schubert-polynomials}
\begin{definition}
	The \term{$\mathfrak d$-Schubert polynomial} $\mathfrak{s}_w ∈ \operatorname{\mathrm{Pol}\mathfrak C}_{I}$
	associated to a reduced word $w∈S_n$ is
	\begin{equation}
		\mathfrak{s}_w ≔ \mathfrak d_{w^{-1}w_0} (y_1^{n-1}\dotsm y_{n-1}^1).
	\end{equation}
	This makes sense, since according to \cref{lem:super-klr-algebra:clifford-demazure-relations},
	the $\mathfrak d_i$ satisfy the relations of the symmetric group.
	This is the definition of the ordinary Schubert polynomials
	\autocite[§2.3]{Manivel}
	with $\mathfrak d$ instead of $∂$.
	Hence, in the purely even case, the $\mathfrak d$-Schubert polynomials coincide with the ordinary ones.
\end{definition}
\begin{lemma}
	The $\mathfrak d$-Schubert polynomials have the following properties:
	\begin{thmlist}
	\item\label{lem:divided-difference-on-schubert-polynomials} 
		If $wv^{-1}$ is a reduced expression, then $\mathfrak d_v\mathfrak{s}_w = \mathfrak{s}_{wv^{-1}}$.
	\item\label{lem:relations-schubert-polynomials} 
		For any tuple $\bm{\alpha}\in\mathbf{N}^n$,
		\begin{equation}
		\label{eq:relations-schubert-polynomials}
			y^{\bm{\alpha}}\mathfrak d_v\mathfrak{s}_w
			\begin{cases}
				∈ \mathfrak{C}_{I}^* y^{\bm{\alpha}} & \text{if $v=w^{-1}$,}\\
				= 0 & \text{if $\ell(u)=\ell(w)$ but $v\neq w^{-1}$,}\\
				= 0 & \text{if $\ell(v)>\ell(w)$.}
			\end{cases}
		\end{equation}
		In particular, $\mathfrak{s}_e$ is non-zero.
	\end{thmlist}
\end{lemma}
\begin{proof}
	The first statement is clear from the definition of $\mathfrak{s}_w$.
	For the second one, we argue as follows:
	\begin{description}
	\item[$\ell(v)>\ell(w)$]
		Every $\mathfrak d_i$ reduces the polynomial degree by one.
		Recall that the longest element $w_0$ of the symmetric group has 
		$\ell(w_0)=\frac{n(n-1)}{2}=\deg(y^{n-1}_1\dotsm y^1_{n-1})$.
		Thus, $\mathfrak{s}_w$ has polynomial degree $\ell(w)$.
		Since $\mathfrak d_i$ acts on constants by zero, the assertion follows.
	\item[$\ell(v)=\ell(w), v\neq w^{-1}$]
		In this case, $v(w^{-1}w_0)$ is not a reduced expression, 
		which implies that $\mathfrak d_v\mathfrak d_{w^{-1}w_0}=0$.
	\item[$v=w^{-1}$]
		Take the reduced expression $w_0 = s_{n-1} \dotsm s_2 s_1 s_{n-2}\dotsm s_3 s_2 s_1 s_2 s_1$.
		We have
		\begin{align*}
			&\mathrel{\phantom{=}} \mathfrak d_{w_0} (y^{n-1}_1\dotsm y^1_{n-1})\\
			&= \mathfrak d_{w_0s_1} \mathfrak d_1( y_1 \, y_2 y_1 \, y_3 y_2 y_1 \, y_4 y_3 y_2 y_1\dotsm y_{n-1} y_{n-2}\dotsm y_1)\\
			&=
					\mathfrak d_{w_0s_1} \bigl[\mathfrak d_1(y_1) \, (y_1 \, y_2 y_1 y_3 y_2 y_1 \dotsm) +
					 y_2 \mathfrak d_{y_1 \, y_2 y_1 \, y_3 y_2 y_1 \dotsm}
					\bigr].
		\intertext{%
		The second summand vanishes because in the argument of $\mathfrak d_1$,
		the indeterminates $y_1$, $y_2$ only occur in products $y_1y_2$,
		which lie in $\ker\mathfrak d_1$ (see \cref{lem:Elementary-Clifford-Polynomials:two-indeterminates}).
		We continue with the next operators.
		It turns out that in every step, we can apply $\mathfrak d_i$ to precisely one factor $y_i$,
		since the remaining $y_i$, $y_{i+1}$'s occur pairwise. We obtain
		}
			&= \mathfrak d_{w_0s_1s_2} s_2(\mathfrak d_1 y_1) \bigl[ (\mathfrak d_2y_2) \,(y_1 \, y_3 y_2 y_1 \dotsm) + \overbrace{\mathfrak d_2(y_1 \, y_3 y_2 y_1 \dotsm)}^0\bigr]\\
			&= \mathfrak d_{w_0s_1s_2s_1} s_1s_2(\mathfrak d_1 y_1) \, s_1(\mathfrak d_2y_2) \, \bigl[\mathfrak d_1(y_1)\, y_3 y_2 y_1 \dotsm\bigr]\\
			\shortvdotswithin{=}
			&= (w_0s_1)(\underbrace{\mathfrak d_1 y_1}_{∈\mathfrak{C}_{I}^*}) (w_0s_1s_2)(\underbrace{\mathfrak d_2 y_2}_{∈\mathfrak{C}_{I}^*}) (w_0s_1s_2s_1)(\underbrace{\mathfrak d_1 y_1}_{\cdots})\dotsm s_1 (\mathfrak d_2 y_2) \cdot \mathfrak d_1(y_1).
		\end{align*}
		Since the permutation action of $S_n$ preserves $\mathfrak{C}_{I}^*$,
		we obtain that $\mathfrak{s}_e ∈\mathfrak{C}_{I}^*$ is non-zero.
		\qedhere
	\end{description}
\end{proof}
\Needspace{3\baselineskip} 
\begin{lemma}
	\label{lem:Clifford-NilHecke-acts-Faithfully}
	The action of $\operatorname{\mathsf{NH}\mathfrak C}_{I}$ on $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ is faithful.
\end{lemma}
\begin{proof}
	We recall briefly the proof of \autocite[prop.\ 2.11]{EKL:odd-NilHecke}
	which can be applied nearly literally.	
	By the defining relations,
	it is clear that $\{y^{\bm{\alpha}}\mathfrak d_v\}_{\bm{\alpha}, v∈S_n}$
	is a generating set of  $\operatorname{\mathsf{NH}\mathfrak C}_{I}$ as $\mathfrak{C}_{I}$-algebra.
	We show by induction on $\ell(v)$ that
	$y^{\bm{\alpha}}\mathfrak d_v \mathfrak{s}_w$ are linearly independent elements of the $\mathfrak{C}_{I}$-bimodule $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$.
	This proves that the generating set $\{y^{\bm{\alpha}}\mathfrak d_v\}$ acts linearly independently in $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$.
	
	\proofsection{Induction base}
	The only element of length $1$ is $e$.
	According to \cref{eq:relations-schubert-polynomials}, we obtain that
	$y^{\bm{\alpha}}\mathfrak d_v \mathfrak{s}_e \neq 0$ only if $v=e$.
	In this case, $y^{\bm{\alpha}}\mathfrak d_v ∈ \mathfrak{C}_{I} y^{\bm{\alpha}}$.
	Thus, $y^{\bm{\alpha}}\mathfrak d_e$ cannot be a linear combination of $\{y^{\bm{\alpha'}}\mathfrak d_v\}_{v>e}$.
	
	\proofsection{Induction step}
	Assume that $\{y^{\bm{\alpha'}}\mathfrak d_{v'}\}_{v'<v}$ is linearly independent 
	for a fixed $v∈S_n$.
	Assume $y^{\bm{\alpha}}\mathfrak d_{v}$ were a linear combination of $\{y^{\bm{\alpha'}}\mathfrak d_{v'}\}_{\ell(v')<\ell(v)}$.
	But $y^{\bm{\alpha}}\mathfrak d_{v} \mathfrak{s}_{v^{-1}} ∈ \mathfrak{C}_{I}^* y^{\bm{\alpha}}$
	by the first case in \cref{eq:relations-schubert-polynomials},
	whereas a linear combination of $\{y^{\bm{\alpha'}}\mathfrak d_{v'}\}$ maps $\mathfrak{s}_{v^{-1}}$ either to zero
	or to a polynomial of degree strictly larger than $0$.
	This is a contradiction;
	there can thus be no linear relations between $y^{\bm{\alpha}}\mathfrak d_{v}$ 
	and elements $y^{\bm{\alpha'}}\mathfrak d_{v'}$ for $\ell(v')<\ell(v)$.
	Now, assume $y^{\bm{\alpha}}\mathfrak d_{v}$ were a non-trivial linear combination
	of $\{y^{\bm{\alpha'}}\mathfrak d_{v'}\}_{\ell(v')\geq\ell(v)}$.
	According to the second two cases in \cref{eq:relations-schubert-polynomials}, this is also a contradiction.
\end{proof}

Using Schubert polynomials, we may prove the following analogue
of the respective theorem for ordinary symmetric polynomials.
Recall that the respective Clifford-analogues $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ and $\Lambda_{I}$ of $\operatorname{\mathrm{Pol}}_{I}$ and $\mathop{\Lambda\operator@font Pol}\nolimits_{I}$
have graded ranks $\rk_{q,\mathfrak{C}_{I}} \Lambda_{I} = (n)_q! (1-q)^{-n}$
and $\rk_{q,\mathfrak{C}_{I}} \operatorname{\mathrm{Pol}\mathfrak C}_{I} = (1-q)^{-n}$;
see \cref{rmk:graded-rank-d-symmetric-polynomials,cor:graded-rank-Clifford-polynomials}.

\begin{theorem}
	\label{prop:Clifford-Polynomials-free}
	The following facts, which are well-known for the purely even and the purely odd case,
	remain true for the Clifford-setting:
	\begin{thmlist}
		\item\label{lem:Clifford-Polynomials-free-over-symmetric}
			$\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ is a free $\Lambda_{I}$-module of graded rank $(n)_q!$.
		\item\label{lem:Clifford-Symmetric-polynomials-are-kernel-of-Demazure}
			 The inclusion $\Lambda_{I} ⊆ \mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$
			from \cref{lem:Elementary-Clifford-Polynomials:Recursively} is in fact an equality. 
		\item The faithful action of $\operatorname{\mathsf{NH}\mathfrak C}_{I}$ on $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$
		gives an isomorphism $\operatorname{\mathsf{NH}\mathfrak C}_{I}\xrightarrow{\cong}\End_{\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}}(\operatorname{\mathrm{Pol}\mathfrak C}_{I})$.
	\end{thmlist}
\end{theorem}
\begin{proof}
	Since Schubert polynomials behave completely analogously to the classical case
	(apart from $\mathfrak{C}_{I}$),
	the classical proof applies.
	Nevertheless, we briefly explain the argument.
	For details, consider \autocite[§3.2]{Lauda:Categorification-Quantum-sl2} for the even
	and \autocite[prop.\ 2.15]{EKL:odd-NilHecke} for the odd case.
	\begin{prooflist}
		\item\label{proof:Clifford-Polynomials-free-over-symmetric}
		Define the $\mathfrak{C}_{I}$-bimodule
		\(\mathfrak{H}_{I} ≔ ⟨y_1^{a_1} \dotsm y_{n-1}^{a_{n-1}} \mid a_i \leq n-i⟩_{\mathfrak{C}_{I}}\).
		This module has a basis given by the Schubert polynomials $\mathfrak{s}_w$;
		see the proof of \autocite[prop.\ 2.12]{EKL:odd-NilHecke}.
		For every index $i$, one may chose an exponent $0\leq a_i \leq n-i$,
		which yields a generator $y_i^{a_i}$ of (polynomial) degree $a_i$.
	
		The submodule generated by $y_i^\bullet$ thus has graded rank $(n-i)_q$;
		therefore, $\rk_{q,\mathfrak{C}_{I}}(\mathfrak{H}_n) = (n)_q!$.
		The same proof as in the even (or odd) case shows that the multiplication
		\[
		\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I} ⊗_{\mathfrak{C}_{I}} \mathfrak{H}_{I} \to \operatorname{\mathrm{Pol}\mathfrak C}_{I}
		\]
		is injective.
		As both sides have the same graded $\mathfrak{C}_{I}$-rank $(1-q)^{-n}$
		(see \cref{rmk:interesting-graded-ranks,rmk:graded-rank-d-symmetric-polynomials}),
		it is an isomorphism of $\mathfrak{C}_{I}$-modules.
		This exhibits $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ as a free $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$-module of graded rank $(n)_q!$.
		
	\item In \ref{proof:Clifford-Polynomials-free-over-symmetric}, we have shown
		that $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ has a $\Lambda_{I}$-basis by Schubert polynomials $\mathfrak{s}_w$.
		Given a polynomial $p∈\operatorname{\mathrm{Pol}\mathfrak C}_{I}$,
		we take a linear combination
		$\sum_{w∈S_n} q_w \mathfrak{s}_w$ for $q_w ∈ \Lambda_{I}$
		with $q_w$ non-zero for some $w>e$.
		Assume the image
		\[
			\mathfrak d_i(p)=\mathfrak d_i\Bigl(\sum_w \mathfrak d_i(q_w)\mathfrak{s}_w\Bigr) 
			= \sum_w s_i(q_w) \mathfrak{s}_{s_iw}
		\]
		were zero.
		By assumption, there was some non-zero $q_w$ for $w>e$, so
		there is still a non-zero term $s_i(q_w) \mathfrak{s}_{s_iw}$ in $\mathfrak d_i(p)$.
		However, since $\mathfrak d_i \mathfrak{s}_v = \mathfrak{s}_{s_iv}$
		if $s_iv$ is a reduced expression by \cref{lem:divided-difference-on-schubert-polynomials},
		$\mathfrak d_i$ maps no other Schubert polynomial to $\mathfrak{s}_{s_iv}$.
		Therefore, $\mathfrak d_i(p)$ cannot be zero.
		There are thus no $\Lambda_{I}$-linear combinations of Schubert polynomials in $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$
		not already contained in $\ker \mathfrak d_i$.
		This proves the statement.
		
		\item We have already seen in \cref{lem:Clifford-NilHecke-acts-Faithfully}
		that $\operatorname{\mathsf{NH}\mathfrak C}_n$ acts faithfully on the free $\mathfrak{C}_{I}$-module $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$.
		This action is compatible with the $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$-module structure
		from \ref{proof:Clifford-Polynomials-free-over-symmetric}:
		as $\operatorname{\mathsf{NH}\mathfrak C}_{I} = \operatorname{\mathsf{NC}\mathfrak C}_{I} ⊗_{\mathfrak{C}_{I}} \operatorname{\mathrm{Pol}\mathfrak C}_{I}$,
		we see that $\operatorname{\mathrm{Pol}\mathfrak C}_{I}$ acts on itself,
		$\operatorname{\mathsf{NC}\mathfrak C}_{I}$ acts trivially on $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$ by the very definition of the latter
		and that $\operatorname{\mathsf{NC}\mathfrak C}_{I}$ indeed acts on $\mathfrak{H}_n$ because it acts by decreasing exponents.
		A graded rank computation
		\begin{equation*}
			\rk_{q,\mathfrak{C}_{I}} \End_{\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}}(\operatorname{\mathrm{Pol}\mathfrak C}_{I})
			= \rk_{q,\mathfrak{C}_{I}}(\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}) \cdot \rk_{q,\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}}(\operatorname{\mathrm{Pol}\mathfrak C}_{I})^2 = \frac{(n)_q!^2}{(n)_q! (1-q)^n}
			= \rk_{q,\mathfrak{C}_{I}} \operatorname{\mathsf{NH}\mathfrak C}_{I}
		\end{equation*}
		shows the asserted isomorphism.\qedhere
	\end{prooflist}
\end{proof}


\subsection{Complete symmetric polynomials}
\label{sec:Clifford-complete-symmetric-polynomials}
\newcommand{h}{h}
We give another collection of generators for $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$,
namely a Clifford-analogue for the complete homogeneous symmetric polynomials.
We beg the reader's pardon for the confusing amount of new notation.
\begin{definition}
	\label{def:clifford-complete-symmetric-polynomial}
	Let $M_{(n)}$ be the $n\times n$-matrix
	\begin{align}
		M_{(n)}&≔\begin{psmallmatrix}
			\mathfrak e^{(n)}_1 & \tilde{\kappa}_{1,2}\\
			- \mathfrak e^{(n)}_2 && \tilde{\kappa}_{1,3}\\
			\vdots &&& \ddots\\
			(-1)^{n-2} \mathfrak e^{(n)}_{n-1} &&&& \tilde{\kappa}_{1,n}\\
			(-1)^{n-1} \mathfrak e^{(n)}_{n} &&&& 0
		\end{psmallmatrix}\\
	\shortintertext{where for $k<l-1$ we set}
		\notag
		\kappa_{k,l}
		&≔\gamma_{k+1,k}\gamma_{k+1,k+2}\,\gamma_{k+2,k+1}\gamma_{k+2,k+3}
			\dotsm\gamma_{l-2,l-3}\gamma_{l-2,l-1}\,\gamma_{l-1,l-2}\gamma_{l-1,l}\,\gamma_{l,l-1}\\
	\shortintertext{and}
		\kappa_{l-1,l}&≔\gamma_{l,l-1}, \kappa_{l,l}≔\gamma_{l,l+1}.\\
	\intertext{Furthermore, let}
			\notag
		\tilde{\kappa}_{k,l}
		&≔\kappa_{k,l}\kappa_{l,l} = \gamma_{k+1,k}\gamma_{k+1,k+2}\dotsm \gamma_{l-1,l-2}\gamma_{l-1,l}\,\gamma_{l,l-1}\gamma_{l,l+1}.
		\intertext{Note that we can “split” the $\kappa$'s by}
		\notag
		\kappa_{k,l} &\within{≔}{=} 
		\underbrace{\gamma_{k+1,k}\gamma_{k+1,k+2}\dotsm\gamma_{m-1,m-2}\gamma_{m-1,m}\,\gamma_{m,m-1}}_{\kappa_{k,m}}
		\gamma_{m,m+1}
		\underbrace{\gamma_{m+1,m}\gamma_{m+1,m+2}\dotsm\gamma_{l-1,l-2}\gamma_{l-1,l}\,\gamma_{l,l-1}}_{\kappa_{m,l}}\\
		\label{eq:splitting-kappa-factors}
		&\within{≔}{=}  \tilde{\kappa}_{k,m}\kappa_{m,l}.
	\end{align}
	The \term{complete symmetric polynomial} $h^{(n)}_m$
	of (polynomial) degree $m$ on $n$ indeterminates
	is the top left entry of the matrix power $M_{(n)}^m$.
	In particular, $h^{(n)}_1=\mathfrak e^{(n)}_1$ and $h^{(n)}_{0}=\mathfrak e^{(n)}_0=1$.
\end{definition}
\begin{remark}
	The top row of $M_{(n)}^m$ has entries
	\[
		\bigl(M_{(n)}^m\bigr)_{1,\bullet}
		=
		\bigl(h^{(n)}_m,\: h^{(n)}_{m-1}\tilde{\kappa}_{1,2},\: h^{(n)}_{m-2}\tilde{\kappa}_{1,2}\tilde{\kappa}_{1,3},\dotsc,
		\tilde{\kappa}_{1,2}\dotsm\tilde{\kappa}_{1,m+1}, 0,\dotsc, 0\bigr).
	\]
\end{remark}
\begin{example}
	\begin{thmlist}
		\item If all indices are even and thus $\mathfrak e^{(n)}_m = \varepsilon^{(n)}_m$ and $\kappa_{k,l}=1$ for all $k,l$,
			these polynomials coincide with the ordinary complete symmetric polynomials
			\[
				h^{(n)}_m = \sum_{\mathclap{1\leq k_{1}\leq\dotsb \leq k_{m}\leq n}} x_{k_{1}}\dotsm x_{k_m}.
			\]
		\item If all indices are odd,
		the resulting complete homogeneous symmetric polynomials coincide
		with those of \autocite{EKL:odd-NilHecke}
		up to a renormalisation involving the $\kappa$'s.
	\end{thmlist}
\end{example}
\begin{proposition}
	\label{prop:Complete-Clifford-Polynomials:Vanishing-Relation}
	Using the Kronecker-$\delta$, there are identities
	\begin{equation*}
	\sum_{l=0}^{m} (-1)^l\,h^{(n)}_{m-l}\,\tilde{\kappa}_{1,2}\dotsm\tilde{\kappa}_{1,l}\mathfrak e^{(n)}_l
	=
	\sum_{l=0}^{m} (-1)^l\,\tilde{\kappa}_{1,2}\dotsm\tilde{\kappa}_{1,l}\mathfrak e^{(n)}_l \,h^{(n)}_{n-l}
	= \delta_{m, 0}.
	\end{equation*}
\end{proposition}
\begin{proof}
	The complete homogeneous symmetric polynomial $h^{(n)}_{m}$
	is the top left entry of the matrix power $M_{(n)}^m=M_{(n)}^{m-1}\cdot M_{(n)}$
	and therefore equals
	\begin{equation*}
		h^{(n)}_{m}=
		\left(
			h^{(n)}_{m-1},\: h^{(n)}_{m-2}\tilde{\kappa}_{1,2},\dotsc,
			\tilde{\kappa}_{1,2}\dotsm\tilde{\kappa}_{1,m+1}, 0,\dotsc, 0
		\right)
		\cdot
		\left(
			\mathfrak e^{(n)}_1, 
			-\mathfrak e^{(n)}_2,
			\dotsc,
			(-1)^{n}\mathfrak e^{(n)}_n
		\right)^T.
	\end{equation*}
	We obtain a recursive description
	$
		h^{(n)}_{m} = \sum_{l=1}^{m} (-1)^{l-1} h^{(n)}_{m-l} \tilde{\kappa}_{1,2}\dotsm\tilde{\kappa}_{1,l} \mathfrak e^{(n)}_l
	$
	of the complete homogeneous symmetric polynomials for $m\geq 1$.
	Moving $h^{\smash{(n)}}_{m}$ to the other side proves the statement.
	%
	For the second equality, we denote the remaining entries of $M_{(n)}^m$ by
	\begin{equation*}
		\begin{psmallmatrix}
			h^{(n)}_m & h^{(n)}_{m-1}\tilde{\kappa}_{1,2} & h^{(n)}_{m-2}\tilde{\kappa}_{1,2}\tilde{\kappa}_{1,3} & \cdots\\
			h^{(n)}_{(1),m+1} & h^{(n)}_{(1),m}\tilde{\kappa}_{1,2} & \ddots\\
			h^{(n)}_{(2),m+2} & \ddots
		\end{psmallmatrix}
		≔M_{(n)}^m,
	\end{equation*}
	such that every $h^{(n)}_{(k),l}$ is a polynomial of degree $l$.
	The polynomial $h^{(n)}_{m}$ is the top left entry of
	${M_{(n)}^m=M_{(n)}\cdot M_{(n)}^{m-1}}$
	and is thus given by
	\begin{align*}
	h^{(n)}_{m} &= \mathfrak e_1 h^{(n)}_{m-1} + \tilde{\kappa}_{1,2}h^{(n)}_{(1),m}\\
	h^{(n)}_{(1),m} &= -\mathfrak e_2 h^{(n)}_{m-2} + \tilde{\kappa}_{1,3}h^{(n)}_{(2),m}\\
	\shortvdotswithin{=}
	h^{(n)}_{(k-1),m} &= (-1)^k\mathfrak e_k h^{(n)}_{m-k} + \tilde{\kappa}_{1,k+1}h^{(n)}_{(k),m}.
	\end{align*}
	We obtain a recursion formula
	\begin{align*}
		h^{(n)}_{m} &= \mathfrak e^{(n)}_1 h^{(n)}_{m-1} + \tilde{\kappa}_{1,2}
		\left(
			-\mathfrak e^{(n)}_2 h^{(n)}_{m-2} + \tilde{\kappa}_{1,3}\bigl(\cdots(
				\pm\mathfrak e^{(n)}_{m-1} h^{(n)}_{1} \mp \tilde{\kappa}_{1,m}\mathfrak e^{(n)}_{m}
			)\cdots\bigr)
		\right)\\
		&= \sum_{l=1}^m (-1)^{l-1} \tilde{\kappa}_{1,2}\dotsm \tilde{\kappa}_{1,l} \mathfrak e^{(n)}_{l} \, h^{(n)}_{m-l}.
	\end{align*}
	Again, moving $h^{(n)}_{m}$ to the other side proves the claim.
\end{proof}


\subsection{Interlude: cohomology of Grassmannians and partial flag varieties}
We recall some classical (non-super) theory on the cohomology rings of Grassmannians
and partial flag varieties.
Our aim is to build super-analogues for these rings.
\begin{definition}
	Fix natural numbers $k\leq n$.
	The set of all $k$-dimensional subspaces
	$F⊆k^n$ of the $n$-dimensional $k$-vector space
	forms an algebraic variety, called \term{Grassmann variety} and denoted by $\operatorname{Gr}(k,n)$.
	This is a special case of the following construction:
\end{definition}
\begin{definition}
	Fix natural number $\ell\leq n$.
	A \term{partial flag} $F_\bullet$ of length $\ell$ is a chain $F_\bullet$ of $k$-vector subspaces
	$0 ⊆ F_0 ⊆ F_1 ⊆ \dotsb ⊆ F_\ell = k^n$.
	The set of all partial flags of length $\ell$ is the
	\term{(partial) flag variety} $\operatorname{Fl}(\ell; n)$.
	If $\ell=n$, then  $\operatorname{Fl}(n)≔\operatorname{Fl}(n; n)$ is called the \term{full flag variety}.
	To a flag $F_\bullet$, one associates its \term{dimension vector} 
	$\bm{k}=(\dim_k F_0, \dotsc, \dim_k F_\ell)$.
	The connected components of $\operatorname{Fl}(\ell)$
	are the varieties $\operatorname{Fl}(\bm{k})$ of partial flags with dimension vector $\bm{k}$.
	In particular, $\operatorname{Gr}(k,n)=\operatorname{Fl}(0, k, n)$.
\end{definition}
\begin{proposition}[{\autocites[prop.\ 3.6.15, rmk.\ 3.6.16]{Manivel}[or][prop.\ 3]{Fulton:Young-Tableaux}}]
	\label{prop:Flag-variety-cohomology}
	The full flag variety $\operatorname{Fl}(n)$
	has cohomology ring $H^*\operatorname{Fl}(n)\cong\operatorname{\mathrm{Pol}}_n/(\mathop{\Lambda\operator@font Pol}\nolimits_n)_+$.
	For the dimension vector $\bm{k}=(k_1,\dotsc, k_\ell=0)$,
	the partial flag variety $\operatorname{Fl}(\bm{k})$ has cohomology ring
	\begin{equation}
		H_{\bm{k}}≔H^*\operatorname{Fl}(\bm{k})\cong \frac{\mathop{\Lambda\operator@font Pol}\nolimits_{k_1-k_0}⊗_k\dotsb⊗_k\mathop{\Lambda\operator@font Pol}\nolimits_{k_\ell-k_{\ell-1}}}{(\mathop{\Lambda\operator@font Pol}\nolimits_n)_+}.
	\end{equation}
	We denote the nominator by $\mathop{\Lambda\operator@font Pol}\nolimits_{\bm{k}}$.
\end{proposition}
\begin{corollary}
	\label{cor:Grassmannian-cohomology}
	The Grassmannian $\operatorname{Gr}(k,n)$ has cohomology ring
	\begin{equation}
		H_{(k,n)}≔H^*(\operatorname{Gr}(k,n))\cong \frac{k\bigl[\varepsilon^{(n)}_1, \dotsc, \varepsilon^{(n)}_m\bigr]}%
		{\bigl(h^{(n)}_{n-k+1}, \dotsc, h^{(n)}_{n}\bigr)}.
	\end{equation}
	The $\varepsilon$'s and $h$'s are the elementary and the complete symmetric polynomials, respectively.
\end{corollary}

Recall the notion of graded ranks from \cref{sec:graded-ranks}.
\begin{lemma}
	\label{lem:graded-dimension-Grassmann-cohomology}
	The cohomology ring $H(k,n)≔H^*(\operatorname{Gr}(k,n))$ of the Grassmannian 
	has graded dimension
	$
		\dim_{q,k} H(k,n) = \binom{n}{k}_q
	$
	with the \term{$q$-binomial coefficient}
	 $\binom{n}{k}_q ≔ \frac{(n)_q!}{(k)_q! (n-k)_q!}$.
\end{lemma}
\begin{proof}
	Without quotienting out the ideal, the polynomial ring 
	$\mathop{\Lambda\operator@font Pol}\nolimits_k≔k[\varepsilon_1, \dotsc, \varepsilon_k]$  in the symmetric polynomials
	has graded dimension
	$\dim_{q,k} \mathop{\Lambda\operator@font Pol}\nolimits_k = (1-q)^{-1}\dotsm(1-q^k)^{-1}$.
	To count the graded dimension of the ideal $\mathfrak{a}=(h_{n-k+1}, \dotsc, h_n)$ to be quotiented out,
	we note that each generator $h_m$ contributes a principal ideal with graded dimension
	$\dim_{q,k} (h_m)=q^m \dim_{q,k}\mathop{\Lambda\operator@font Pol}\nolimits_k$.
	We then proceed by the inclusion\-/exclusion principle to obtain
	\begin{align*}
		&\mathrel{\phantom{=}}\dim_{q,k} (h_{n-k+1}, \dotsc, h_n)\\
		&=
		\Bigl(
			\bigl(q^{n-k+1} + \dotsb q^n\bigr) - \bigl(q^{2(n-k)+3} + \dotsb + q^{2n-1}\bigr) + \dotsb \pm q^{(n-k+1)+\dotsb + n}
		\Bigr) \dim_{q,k}\mathop{\Lambda\operator@font Pol}\nolimits_k \\
		&= \bigl(1- (1-q^{n-k+1})\dotsm(1-q^n)\bigr)\dim_{q,k}\mathop{\Lambda\operator@font Pol}\nolimits_k.
	\end{align*}
	The quotient thus has graded dimension
	\begin{align*}
		\dim_{q,k} (\mathop{\Lambda\operator@font Pol}\nolimits_k/\mathfrak a) &= \frac{(1-q^{n-k+1})\dotsm(1-q^n)}{ (1-q)\dotsm(1-q^k)}
		= \frac{(1-q)^{n-(n-k)} (n)_q! / (n-k)_q!}{(1-q)^k (k)_q!}
		= \tbinom{n}{k}_q.\qedhere
	\end{align*}
\end{proof}
For the cohomology ring of partial flag varieties, this yields:
\begin{corollary}
	\label{cor:graded-dimension-flag-cohomology}
	Let $\bm{k}≔(0=k_0\leq \dotsc\leq k_r=n)$ be a dimension vector.
	The cohomology ring $H(\bm{k})$ of the partial flag variety $\operatorname{Fl}(\bm{k})$
	has as graded dimension the \term{$q$-multinomial coefficient}
	\begin{equation*}
		\dim_{q,k} H(\bm{k}) = \dbinom{n}{k_1, k_2-k_1, \dotsc, k_r-k_{r-1}}_q
		≔ \frac{(n)_q!}{(k_1)_q! (k_2-k_1)_q! \dotsm (k_r-k_{r-1})_q!}.
	\end{equation*}
\end{corollary}
\begin{proof}
	The cohomology ring is the quotient
	\[
		\frac{
			\mathop{\Lambda\operator@font Pol}\nolimits_{k_1}⊗\dotsb ⊗\mathop{\Lambda\operator@font Pol}\nolimits_{k_l-k_{l-1}}
		}{
			\left(
				1-\bigl(1+\varepsilon_1^{(k_1)}+\dotsb+\varepsilon^{(k_1)_{k_1}}\bigr)
				\dotsm
				\bigl(1+\varepsilon_1^{(k_l-k_{l-1})}+\dotsb+\varepsilon_{(k_l-k_{l-1})}^{k_l-k_{l-1}}\bigr)
			\right)
		}
	\]
	of $\mathop{\Lambda\operator@font Pol}\nolimits_{\bm{k}} ≔ \mathop{\Lambda\operator@font Pol}\nolimits_{k_1}⊗\dotsb ⊗\mathop{\Lambda\operator@font Pol}\nolimits_{k_l-k_{l-1}}$.
	Denote the ideal quotiented out by $\mathfrak{a}$ and proceed
	as in the proof of \cref{lem:graded-dimension-Grassmann-cohomology}.
	We see that the nominator has graded dimension
	\begin{align*}
		\dim_{q,k} \Lambda
		 &= \prod_{a=1}^{l}\frac{1}{1-q}\dotsm\frac{1}{1-q^{k_a-k_{a-1}}}
		= \frac{1}{(1-q)^n}\prod_{a=1}^{l}\frac{1}{(k_a-k_{a-1})_q!},
	\end{align*}
	and the ideal $\mathfrak{a}$ quotiented out has graded  dimension
	\[
		\dim_{q,k}\mathfrak a = \bigl((1-q)^n(n)_q!-1\bigr)\dim_{q,k} \Lambda.
	\]
	This shows the assertion.
\end{proof}



\subsection{Cyclotomic quotients}
\label{sec:cyclotomic-quotients}
It is known that in the non-super case,
the cohomology ring $H_{(k,n)}$ of Grassmann varieties (see \cref{cor:Grassmannian-cohomology})
is Morita equivalent to the \term{cyclotomic quotient} of the NilHecke algebra
\cite[§5]{Lauda:Introduction-to-Diagrammatic-Algebra}.
For the odd setting,
the respective Morita equivalence has already been established in \cite[§5]{EKL:odd-NilHecke}.
We introduce analogues of both rings in the Clifford setting
and prove that they are Morita equivalent.
\begin{definition}
	The \term{$m$-th cyclotomic quotient} of the Hecke Clifford algebra $\operatorname{\mathsf{NH}\mathfrak C}_n$
	is the quotient $\operatorname{\mathsf{NH}\mathfrak C}_n^m ≔ \operatorname{\mathsf{NH}\mathfrak C}_n/\bigl((\kappa_{1,n}y_n)^m\bigr)$.
	\label{def:Clifford-Grassmann-Cohomology}%
	The quotient $\mathop{\mathit{H}\mathfrak C}\nolimits_{(m,n)} ≔ \mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}/\bigl(h^{(n)}_{k}\bigr)_{k>n-m}$
	is called the \term{Clifford cohomology ring}
	of the Grassmannian $\operatorname{Gr}(m,n)$.
\end{definition}
\begin{remark}
	In contrast to the standard definition of the cyclotomic quotient,
	we mod out some power of $y_n$ instead of $y_1$;
	the latter would necessitate taking the transpose of the matrix $M_{(n)}$
	in \cref{def:clifford-complete-symmetric-polynomial}.
\end{remark}
\begin{theorem}
	\label{prop:grassmann-cyclotomic-morita-eq}
	The rings $\operatorname{\mathsf{NH}\mathfrak C}_n^m$ and $\mathop{\mathit{H}\mathfrak C}\nolimits_{(m,n)}$ are Morita-equivalent.
\end{theorem}
\begin{proof}
	The proof in \autocite[§5]{Lauda:Introduction-to-Diagrammatic-Algebra}
	does not rely on commutativity of the rings involved and thus applies immediately in our setting.
	We review the argument briefly.
	The reader is strongly encouraged to verify the following calculation
	for $n=3$.
	Let us start with
	\begin{align}\notag
		\mathfrak e^{(n)}_n
		&= \gamma_{1,2}y_1 \, \gamma_{2,3}y_2 \dotsm \gamma_{n,n+1}y_n\\\notag
		&= \bigl(\gamma_{1,2}y_1 \, \gamma_{2,3}y_2 \dotsm \gamma_{n-1,n}y_{n-1}\bigr)\gamma_{n,n+1}y_n\\\notag
		&= \bigl(\mathfrak e^{(n)}_{n-1} - \tilde{\mathfrak e}^{(n)}_{n-1}\bigr)
			\smash{\underbrace{\gamma_{n,n+1}}_{\kappa_{n,n}}}y_n\\
	\shortintertext{where we set}\notag
		\tilde{\mathfrak e}^{(n)}_{n-1} 
		&= \gamma_{1,2}y_1\dotsm\gamma_{n-2,n-1}y_{n-2}\,\gamma_{n,n-1}y_n
		+ \dotsb
		+ \gamma_{2,1}y_2\,\gamma_{3,2}y_3\dotsm \gamma_{n,n-1}y_n\\\notag
		&= \bigl(\gamma_{1,2}y_1\dotsm \gamma_{n-2,n-1}y_{n-2}
			+\dotsb
			+\gamma_{2,1}y_2\dotsb \gamma_{n-1,n-2}y_{n-1}
			\bigr) \gamma_{n,n-1}y_n\\\notag
		&= \bigl(\mathfrak e^{(n)}_{n-2} - \tilde{\mathfrak e}^{(n)}_{n-2}\bigr)
		\smash{\underbrace{\gamma_{n,n-1}}_{\kappa_{n-1,n}}} y_n\\
	\shortintertext{where we set}\notag
		\tilde{\mathfrak e}^{(n)}_{n-2}
		&= \bigl(\gamma_{1,2}y_1\dotsm \gamma_{n-3,n-2}y_{n-3}
		+ \dotsb
		+ \gamma_{3,2}y_3\dotsm \gamma_{n-1,n-2}y_{n-1}\bigr) \gamma_{n-1,n-2}\gamma_{n-1,n}\gamma_{n,n-1} y_n\\\notag
		&= \bigl(\mathfrak e^{(n)}_{n-3}-\tilde{\mathfrak e}^{(n)}_{n-3}\bigr) 
			\smash{\underbrace{\gamma_{n-1,n-2}\gamma_{n-1,n}\gamma_{n,n-1}}_{\kappa_{n-2,n}}} y_n\\\notag
	&\vdotswithin{=}\\\notag
		\tilde{\mathfrak e}^{(n)}_{2}
		&= \bigl(\mathfrak e^{(n)}_{1} - \tilde{\mathfrak e}^{(n)}_{1}\bigr)
			\smash{\underbrace{\gamma_{3,2}\gamma_{3,4}\dotsm\gamma_{n-1,n-2}\gamma_{n-1,n}\gamma_{n,n-1}}_{\kappa_{2,n}}} y_n\\
	\shortintertext{where}\notag
		\tilde{\mathfrak e}^{(n)}_{1} &= 
			\underbrace{\gamma_{2,1}\gamma_{2,3}\dotsm\gamma_{n-1,n-2}\gamma_{n-1,n}\gamma_{n,n-1}}_{\kappa_{1,n}} y_n.\\
	\intertext{%
	
		To state the above calculation differently, we have that
	}\notag
		0 &=
		\mathfrak e^{(n)}_n - \Bigl(\mathfrak e^{(n)}_{n-1} - \bigl(\cdots\bigl(\mathfrak e^{(n)}_1 - \kappa_{1,n}y_n\bigr)\cdots\bigr)\kappa_{n-1,n}y_n\Bigr) \kappa_{n,n} y_n\\\notag
		&= \sum_{k=0}^{n} (-1)^k \mathfrak e^{(n)}_{n-k} \kappa_{n-k+1,n}y_n\dotsm \kappa_{n,n}y_n\\
		&= \label{eq:relation-of-bi-basis} 
			(-1)^n\sum_{k=0}^{n} (-1)^{k} \mathfrak e^{(n)}_k \smashoperator{\prod_{l=k+1}^{n}}\kappa_{l,n}y_n.
	\end{align}
	Set $b_l ≔ \prod_{l=k+1}^{n}\kappa_{l,n}y_n$ for $1 \leq k\leq n$;
	in particular $b_{n} ≔ 1$.
	Recall the $\mathfrak{C}_{I}$-bimodule $\mathfrak{H}_{I}$
	from the proof of \cref{lem:Clifford-Polynomials-free-over-symmetric}.
	We define a similar free $\mathfrak{C}_{I}$-bimodule of graded rank $(n)_q!$
	with direct sum decomposition
	\begin{equation}
		\label{eq:cyclotomic-quotients:H-tilde}
		\tilde{\mathfrak{H}}_{I} ≔ ⟨y_1^{\alpha_1} \dotsm y_{n-1}^{\alpha_{n-1}} \mid \alpha_i <i⟩_{\mathfrak{C}_{I}}
		=
		 ⨁_{\bm{\alpha}}
			\underbrace{⟨y^{\bm{\alpha}} b_1, \dotsc, y^{\bm{\alpha}} y_n^{n-1} \cdot b_n⟩_{\mathfrak{C}_{I}}}_{\eqqcolon B_{\bm{\alpha}}}
	\end{equation}
	for multiindices $\bm{\alpha}$.
	It is analogous to the proof of \cref{lem:Clifford-Polynomials-free-over-symmetric}
	that multiplication gives an isomorphism
	$\tilde{\mathfrak{H}}_{I} ⊗_{\mathfrak{C}_{I}} \mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I} \xto{\cong} \operatorname{\mathrm{Pol}\mathfrak C}_{I}$.	
	Recall from \cref{eq:splitting-kappa-factors} how to split the $\kappa$'s.
	Multiplication from the left by $\kappa_{1,n}y_n$ acts on this basis of $B_{\bm{\alpha}}$ by
	\begin{alignat*}{3}
		(\kappa_{1,n}y_n\cdot)\colon B_{\bm{\alpha}} &\to B_{\bm{\alpha}},
		\\*
		1 = b_{n} 
			&  \mapsto \kappa_{1,n}y_n 
			&& \overset{\mathclap{\raisebox{3pt}[0pt][0pt]{\scriptsize{\cref{eq:splitting-kappa-factors}}}}}{=} \kappa_{1,n}\kappa_{n,n}\kappa_{n,n} y_n 
			&&=\smash{\overbrace{\kappa_{1,n}\kappa_{n,n}}^{=\tilde{\kappa}_{1,n}}} b_{n-1}
		\\*
		b_{n-1}
			&  \mapsto \kappa_{1,n}y_n b_{n-1} 
			&&= \tilde\kappa_{1,n-1} \kappa_{n-1,n}y_n b_{n-1}	
			&&= \tilde\kappa_{1,n-1} b_{n-2}
		\\*
			& \vdotswithin{\mapsto}
		\\
		b_2 & \mapsto \kappa_{1,n}y_n b_2 &&= \tilde\kappa_{1,2}\kappa_{2,n} y_n b_2
		&&= 
			\tilde\kappa_{1,2}
			b_1 \\*
		b_1 &\mapsto \kappa_{1,n}y_nb_1 &&\overset{\mathclap{\raisebox{3pt}[0pt][0pt]{\scriptsize{\cref{eq:relation-of-bi-basis}}}}}{=}
			\textstyle\sum_{k=1}^{n} (-1)^k \mathfrak e^{(n)}_k 
			\underbrace{
				\textstyle \prod_{l=k+1}^{n}\kappa_{l,n}y_n
			}_{=b_k}.
			\span\span
	\end{alignat*}
	This shows that multiplication with $\kappa_{1,n}y_n$ from the left
	acts on the basis from \cref{eq:cyclotomic-quotients:H-tilde}
	by the matrix
	\begin{equation*}
		\begin{psmallmatrix}
			\mathfrak e^{(n)}_1 & \tilde{\kappa}_{1,2}\\
			- \mathfrak e^{(n)}_2 && \tilde{\kappa}_{1,3}\\
			\vdots &&& \ddots\\
			(-1)^{n-2} \mathfrak e^{(n)}_{n-1} &&&& \tilde{\kappa}_1,n\\
			(-1)^{n-1} \mathfrak e^{(n)}_{n} &&&& 0
		\end{psmallmatrix} = M_{(n)}.
	\end{equation*}
	Quotienting out the two-sided ideal $(\kappa_{1,n}y_n)^m \trianglelefteq \operatorname{\mathsf{NH}\mathfrak C}_n$
	is the same as requiring that $M_{(n)}^m=0$.
	\begin{claim}
		The ideal $(h_{n-m+1},\dotsc, h_{n}) \trianglelefteq \operatorname{\mathrm{Pol}\mathfrak C}_{I}$
		is also generated by the first column of $M_{(n)}^{m+1}$.
	\end{claim}
	By definition, $h_{k+1}$ is the top left entry of $M_{(n)}\cdot M_{(n)}^{k}$ for any $k$.
	Recall that we denoted the entries of the first column $\bigl(M_{(n)}^{k+1}\bigr)_{\bullet,1}$ of $M_{(n)}^{k+1}$
	by 
	\[
		\bigl(M_{(n)}^{k+1}\bigr)_{\bullet,1} \eqqcolon \bigl(h_{k+1}, h_{(1), k+2}, h_{(2), k+3}, \dotsc, h_{(n-k-1), n}\bigr)^T.
	\]
	We thus have
	\begin{alignat*}{2}
		h_{n-m+2} &= \mathfrak e^{(n)}_1 h_{n-m+1} + \tilde{\kappa}_{1,2} h_{(1),n-m+2}\\
		&\equiv \tilde{\kappa}_{1,2} h_{(1),n-m+2} &&\pmod{h_{n-m+1}}\\
		h_{n-m+3} &= 
		\mathfrak e^{(n)}_1 h_{n-m+2} + \tilde{\kappa}_{1,2} h_{(1),n-m+3}\\
		&= \mathrlap{\mathfrak e^{(n)}_2 \left(
			\mathfrak e^{(n)}_1 h_{n-m+1} + \tilde{\kappa}_{1,2} h_{(1),n-m+2}
		\right)+
		\tilde{\kappa}_{1,2}\left(
			-\mathfrak e^{(n)}_2 h_{n-m+1} + \tilde{\kappa}_{1,3} h_{(2),n-m+2}
		\right)}\\
		&\equiv \tilde{\kappa}_{1,2} \tilde{\kappa}_{1,3}  h_{(2), n-m+3} &&\pmod{h_{n-m+1}, h_{n-m+2}}\\
		\shortvdotswithin{=}
		h_{n} &\equiv \tilde{\kappa}_{1,2} \dotsm \tilde{\kappa}_{1,m}  h_{(m-1), n} &&\pmod{h_{n-m+1}, \dotsc, h_{n-1}}.
	\end{alignat*}
	This proves the claim, since all $\tilde{\kappa}$'s are units.
	Taking the product  $M_{(n)}^{m} = M_{(n)}^{m-n+1} \cdot  M_{(n)}^{n-1}$
	shows that the last column of $M_{(n)}^{m}$ has entries
	\[
		\bigl(M_{(n)}^{m}\bigr)_{\bullet,n} = 
		\bigl(h_{n-m+1}, h_{(1), n-m+2}, \dotsc, h_{(m-1,n)}\bigr)^T \tilde{\kappa}_{1,2}\dotsm\tilde{\kappa}_{1,n}.
	\]
	Thus, the entries of $\bigl(M_{(n)}^{m}\bigr)_{\bullet,n}$
	also generate $(h_{n-m+1},\dotsc, h_{n}) \trianglelefteq \operatorname{\mathrm{Pol}\mathfrak C}_{I}$.
	\begin{claim}
		All entries of $M_{(n)}^{m}$ are $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$-linear combinations
	of entries of the last column $\bigl(M_{(n)}^{m}\bigr)_{\bullet,n}$.
	\end{claim}
	Since $M_{(n)}^{k+1} = M_{(n)}^{1} M_{(n)}^{k}$ for any $k$,
	the entries of $\bigl(M_{(n)}^{k+1}\bigr)_{\bullet,1}$
	are $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I}$-linear combinations of entries of $\bigl(M_{(n)}^{k}\bigr)_{\bullet,1}$.
	Since $M_{(n)}^{k+1} = M_{(n)}^{k} M_{(n)}^{1}$,
	we have $\bigl(M_{(n)}^{k+1}\bigr)_{\bullet,2}=\tilde\kappa_{1,2} \bigl(M_{(n)}^{k}\bigr)_{\bullet,1}$.
	The entries of the first column $\bigl(M_{(n)}^{k+1}\bigr)_{\bullet,1}$ thus are linear combinations 
	of entries of the second column $\bigl(M_{(n)}^{k+1}\bigr)_{\bullet,2}$;
	the claim follows iteratively.
	
	Altogether, we have shown that requiring $M_{(n)}^{m}=0$ is the same
	as quotienting out $(h_{n-m+1},\dotsc, h_{n}) \trianglelefteq \operatorname{\mathrm{Pol}\mathfrak C}_{I}$.
	Applying this to each summand $B_{\bm{\alpha}}$ of \cref{eq:cyclotomic-quotients:H-tilde}
	yields
	\[
		\operatorname{\mathsf{NH}\mathfrak C}_n^m \cong \operatorname{Mat}_{n!}(\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{I})/\bigl(M_{(n)}^{m}\bigr) \cong \operatorname{Mat}_{n!}(\operatorname{\mathsf{H}\mathfrak C}_{(m,n)}).
	\]
	This establishes the asserted Morita equivalence.
\end{proof}


\subsection{Clifford algebras associated to partial flag varieties}
\label{sec:partial-flag-varieties}
Let $\mathfrak e^{(k,n)}_m$ be the polynomial of degree $m$
in the indeterminates $y_k,\dotsc,y_n$
that is obtained from $\mathfrak e^{(n-k)}_m$ by replacing every index $i$ by $i+k$.
Recall from \cref{lem:Elementary-Clifford-Polynomials:Recursively}
the recursion formula for the elementary $\mathfrak d$-symmetric polynomials.
We can easily derive the following corollary 
by regrouping the terms of \cref{eqn:Elementary-Clifford-Polynomials:Recursively}:
\begin{corollary}
	\label{cor:Elementary-Clifford-Polynomials:Recursively-Left}
	The elementary $\mathfrak d$-symmetric polynomials satisfy the recursion formula

	\begin{equation}
		\label{eqn:Elementary-Clifford-Polynomials:Recursively-Left}
		\mathfrak e^{(n)}_m = \gamma_{1,2}\,y_1\, \mathfrak e^{(1,n)}_{m-1} + \gamma_{2,1}\gamma_{2,3}\,\mathfrak e^{(1,n)}_m
	\end{equation}
	for the elementary $\mathfrak d$-symmetric polynomials,
	where $\gamma_{i,i\pm 1}$ is as defined in \cref{lem:Elementary-Clifford-Polynomials:two-indeterminates}.
\end{corollary}
\begin{table}
	\caption{%
		Elementary $\mathfrak d$-symmetric polynomials $\mathfrak e^{(k,n)}_m$ for $n=3$,
		as defined in \cref{def:Elementary-Clifford-Polynomials:two-indeterminates}.
		The leftmost column is the same as in \cref{tab:elementary-d-symmetric-polynomials}.
		The braces illustrates how one can construct the polynomials recursively
		as stated in \cref{cor:Elementary-Clifford-Polynomials:Recursively-Left},
		starting with the rightmost polynomial $\mathfrak e^{(2,3)}_3 = \gamma_{3,4}y_3$.
	}
	\label{tab:elementary-d-symmetric-polynomials:Recursively-Left}
	\centering
	\small
	\addtolength{\arraycolsep}{1em}
	$\begin{array}{@{}l@{}lll@{}}
		\toprule				
		m\backslash k+1 & \hfil 1 & \hfil 2 & \hfil 3\\
		\midrule
		1
		& 
		\gamma_{1,2}\, y_1
		+ \underbrace{
	  \gamma_{2,1}\, y_2
			+ \gamma_{2,1}\gamma_{2,3}\gamma_{3,2}\ y_3
		}_{\gamma_{2,1}\gamma_{2,3}\mathfrak e^{(1,3)}_1}
		& 
		\gamma_{2,3}\, y_2
		+ \underbrace{\gamma_{3,2}\,y_3}_{\mathclap{\gamma_{3,2}\gamma_{3,4}\,\mathfrak e^{(1,3)}_1}}
		& \gamma_{3,4}\,y_3
		\\
		2
		&
		\underbrace{
	  \gamma_{1,2}\,y_1\gamma_{2,3}\, y_2
			+ \gamma_{1,2}\,y_1\gamma_{3,2}\, y_3 
		}_{\gamma_{1,2}\,y_1\mathfrak e{(1,3)}_1}	
		+ \gamma_{2,1}\,y_2\ \gamma_{3,2}\, y_3
		& \gamma_{2,3}\, y_2\,\gamma_{3,2}\, y_3\\
		3
		& \gamma_{1,2}\,y_1\,\underbrace{\gamma_{2,3}\,y_3\,\gamma_{3,4}\,y_3}_{\mathfrak e^{(1,3)}_2}\\
		\bottomrule
	\end{array}$		
\end{table}	
\begin{example}
	The “regrouping” of terms is best made explicit
	by considering the first $\mathfrak d$-symmetric polynomials
	listed explicitly in \cref{tab:elementary-d-symmetric-polynomials}.
	We can start with the polynomial $\mathfrak e^{(2,3)}_1≔\gamma_{3,4} y_3$
	as defined in \cref{def:Elementary-Clifford-Polynomials:two-indeterminates}
	and construct the polynomials $\mathfrak e^{(k,3)}_m$
	for $0\leq k<3$ and $m\leq n-k$
	as described in the corollary.
	The resulting polynomials are listed in \cref{tab:elementary-d-symmetric-polynomials:Recursively-Left}
	with the grouping indicated by braces.
\end{example}
The two recursive expansions
\cref{eqn:Elementary-Clifford-Polynomials:Recursively,eqn:Elementary-Clifford-Polynomials:Recursively-Left}
from \cref{lem:Elementary-Clifford-Polynomials:Recursively,cor:Elementary-Clifford-Polynomials:Recursively-Left}
are subsumed in the following definition:
\begin{definition}
	\label{definition:Clifford-Symmetric-Polynomials:Lambda-Coefficients}
	For $1<k<n$, let $\lambda^{(0,k,n)}_{m,l}∈\mathfrak{C}_{I}$ be the coefficients defined by
	\begin{equation}
		\label{eqn:Clifford-Flag-Varieties:Expansion-Coefficients}
		\mathfrak e^{(n)}_m = \sum_{\mathclap{0\leq l\leq m}}
		\mathfrak e^{(k)}_l \lambda^{(0,k,n)}_{m,l}\, \mathfrak e^{(k,n)}_{m-l}.
	\end{equation}
\end{definition}
\begin{example}
	\label{example:Clifford-Symmetric-Polynomials:Lambda-Coefficients}
	By \cref{eqn:Elementary-Clifford-Polynomials:Recursively,eqn:Elementary-Clifford-Polynomials:Recursively-Left}
	we know the coefficients $\lambda^{(0,k,n)}_{m,l}$ for two special cases:
	\begin{thmlist}
	\item $\lambda^{(0,n-1,n)}_{m,l}$ is just coefficient of the expansion
		$\mathfrak e^{(n)}_m = \sum_l \mathfrak e^{(n-1)}_l\lambda_l \gamma_{n,n+1} y_n$. 
		The recursion formula from \cref{lem:Elementary-Clifford-Polynomials:Recursively} gives 
		\settowidth{\algnRef}{$\gamma_{m+1,m}\gamma_{m+1,m+2}\dotsm \gamma_{n-1,n-2}\gamma_{n-1}{n}$}
		\begin{align*}
			\lambda^{(0,n-1,n)}_{m,l} &= 
			\begin{cases}
				1 & \text{if $l=m$}\\
				\gamma_{m+1,m}\gamma_{m+1,m+2}\dotsm \gamma_{n-1,n-2}\gamma_{n-1}{n} & \text{if $l=m-1$}\\
				0 & \text{otherwise.}
			\end{cases}
		\end{align*}
	\item $\lambda^{(0,1,n)}_{m,l}$ is the coefficient of the expansion
		$\mathfrak e^{(n)}_m = \sum \gamma_{1,2} y_1 \lambda_l \mathfrak e^{(1, n-1)}_l$. 
		By \cref{cor:Elementary-Clifford-Polynomials:Recursively-Left} these are given by
		\begin{align*}
			\mathmakebox[\widthof{$\lambda^{(0,n-1,n)}_{m,l}$}][r]{\lambda^{(0,1,n)}_{m,l} }
			&= 
			\begin{cases}
				\mathmakebox[\algnRef][l]{\gamma_{2,1}\gamma_{2,3}} & \text{if $l=0$}\\
				\mathmakebox[\algnRef][l]{1} & \text{if $l=1$}\\
				\mathmakebox[\algnRef][l]{0} & \text{otherwise.}
			\end{cases}
		\end{align*}
	\end{thmlist}
	In general, however, it seems to be difficult to provide an explicit formula for $\lambda^{(0,k,n)}_{m,l}$
	for arbitrary $k$.
\end{example}
\begin{definition}
	Recall from \cref{lem:Elementary-Clifford-Polynomials:Recursively}
	that we defined $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{n}≔⟨\mathfrak e^{(n)}_1,\dotsc, \mathfrak e^{(n)}_n⟩$
	as the $\mathfrak{C}_{I}$-subalgebra of $ \operatorname{\mathrm{Pol}\mathfrak C}_{I}$
	generated by the elementary $\mathfrak d$-symmetric polynomials.
	We have shown $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{n} = ⋂_{i=1}^{n-1}\ker \mathfrak d_i$
	in \cref{lem:Clifford-Symmetric-polynomials-are-kernel-of-Demazure}.
	We additionally define the $\mathfrak{C}_{I}$-algebras
	$\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(k, n)} ≔ ⟨\mathfrak e^{(k,n)}_m\mid 1\leq m\leq m-k⟩ ⊆ \bigcap_{i=k+1}^{n-1}\ker \mathfrak d_i$
	and more generally
	\begin{align*}
		\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{( k_0, \dotsc, k_l)} 
		&≔
			\mathfrak{C}_{I}\bigl\langle\mathfrak e^{(k_0,k_1)}_{m_0},\mathfrak e^{(k_1,k_2)}_{m_1},
			\dotsc, \mathfrak e^{(k_{l-1},k_l)}_{m_{l-1}} \bigm| 1\leq m_j\leq k_{j+1}-k_j \bigr\rangle\\*
		&\within{≔}{\stackrel{*}{=}}
		\ker\mathfrak d_{k_0+1} 
		\cap\dotsb\cap \widehat{\ker\mathfrak d_{k_1}}
		\cap\dotsb\cap \widehat{\ker\mathfrak d_{k_j}}
		\cap\dotsb\cap \mathfrak d_{k_l-1}
	\end{align*}
	where all hatted factors $\widehat{\ker\mathfrak d_{k_j}}$ for $1\leq j\leq l$ are to be omitted.
	In particular, the $\mathfrak{C}_{I}$-algebra of $\mathfrak d$-symmetric polynomials 
	defined in \cref{lem:Elementary-Clifford-Polynomials:Recursively}
	is $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{n} = \mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,n)}$.
\end{definition}
\begin{remark}
	\label{rmk:General-Clifford-Symmetric-polynomials-are-kernel-of-Demazure}
	Note that \cref{lem:Clifford-Symmetric-polynomials-are-kernel-of-Demazure}
	applies to each $\bigl\langle\mathfrak e^{(k_j,k_{j+1})}_{m_j} \mid 1\leq m_j\leq k_{j+1}-k_j\bigr\rangle$.
	The obvious inclusion at $*$ is thus indeed an equality.
\end{remark}



\subsubsection{One step flag varieties}
\label{sec:Clifford-one-step-flag-varieties}%
For every $j$, there are the following inclusions of $\mathfrak{C}_{I}$-algebras:
consider
\begin{align}
	&\mathrel{\phantom{=}} \mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{( k_0, \dotsc, \widehat{k_{j}}, \dotsc, k_l)} \notag\\*
	&= \ker\mathfrak d_{k_0+1} 
		\cap\dotsb\cap \widehat{\ker\mathfrak d_{k_{j-1}}}
		\cap\dotsb\cap \ker\mathfrak d_{k_{j}}
		\cap\dotsb\cap \widehat{\ker\mathfrak d_{k_{j+1}}}
		\cap\dotsb\cap \mathfrak d_{k_l-1},\notag\\
\intertext{%
where the intersection with the $k_j$-th kernel is \emph{not} omitted.
Clearly there is an inclusion
}
	&\within{=}{⊆}
	\ker\mathfrak d_{k_0+1} 
		\cap\dotsb\cap \widehat{\ker\mathfrak d_{k_{j-1}}}
		\cap\dotsb\cap \widehat{\ker\mathfrak d_{k_{j}}}
		\cap\dotsb\cap \widehat{\ker\mathfrak d_{k_{j+1}}}
		\cap\dotsb\cap \mathfrak d_{k_l-1}\notag\\
	&= \mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{( k_1, \dotsc, k_l)}\label{eqn:Clifford-Symmetric-Rings:Inclusion}
\end{align}
of algebras, where the $k_j$-th kernel \emph{is} omitted.
\begin{lemma}
	Let $\bm{k}=(0,k, k+1,\dotsc, n)$ for some $k$,
	so that the there are inclusions
	\begin{equation}
		\label{eqn:Clifford-Symmetric-Rings:One-step-Inclusion}
		\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, \dotsc, k, \dotsc, n)}
		\subseteq\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,\dotsc, k,k+1,\dotsc, n)}
		\supseteq\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,\dotsc,  k+1,\dotsc,  n)}.
	\end{equation}
	by \cref{eqn:Clifford-Symmetric-Rings:Inclusion}.
	The two inclusions can be written explicitly in terms of $\mathfrak d$-symmetric polynomials,
	namely
	\begin{align}
		\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k, n)} &\longinto \mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,k,k+1,n)}\\
		\mathfrak e^{(0,k)}_{m}  &\longmapsto \mathfrak e^{(0,k)}_m \notag\\
		\mathfrak e^{(k,n)}_{m}  &\longmapsto \gamma_{k+1, k+2}\,y_{k+1} \, \mathfrak e^{(k+1,n)}_{m-1}
			+ \gamma_{k+1,k}\gamma_{k+1,k+2}\, \mathfrak e^{(k+1,n)}_m \notag
		\\[.5\baselineskip]
		\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k+1, n)} &\longinto \mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,k,k+1,n)}\\
		\mathfrak e^{(0,k+1)}_{m}  &\longmapsto
			\mathfrak e^{(0,k)}_m + \mathfrak e^{(0,k)}_{m-1}\cdot \gamma_{m+1,m}\dotsm\gamma_{k+1,k}\,y_{k+1} \notag\\
		\mathfrak e^{(k+1,n)}_{m}  &\longmapsto \mathfrak e^{(k+1,n)}_{m}. \notag
	\end{align}
\end{lemma}
\begin{proof}
	A polynomial from $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k, n)}$
	is contained in all $\ker\mathfrak d_i$ for $i∈\{1,\dotsc, k-1, k+1,k+2, \dotsc,n\}$
	by \cref{lem:Elementary-Clifford-Polynomials:Recursively}.
	It is a fortiori contained in all $\ker\mathfrak d_i$ for $i∈\{1,\dotsc, k-1, k+2\dotsc,n\}$;
	\ie, in $\ker\mathfrak d_1\cap\dotsb\ker\mathfrak d_{k-1}\cap\ker\mathfrak d_{k+2}\cap\dotsb\cap\ker\mathfrak d_n$.
	
	We know from
	\cref{lem:Clifford-Symmetric-polynomials-are-kernel-of-Demazure,rmk:General-Clifford-Symmetric-polynomials-are-kernel-of-Demazure}
	that this intersection equals $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k, k+1, n)}$,
	which has generators $\mathfrak e^{(0, k, k+1, n)}_m$.
	The coefficients for writing elementary $\mathfrak d$-symmetric polynomials from $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k, n)}$
	in terms of those polynomials from $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,k,k+1,n)}$
	are given in \cref{example:Clifford-Symmetric-Polynomials:Lambda-Coefficients}.
	This proves the statement.
\end{proof}
\begin{remark}
	\label{rmk:Clifford-Symmetric-Polynomials:Bimodule-structure}
	These inclusions turn $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,k,k+1,n)}$ into a $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k, n)}$-$\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k+1, n)}$-bimodule.
	In the even case, this bimodule structure corresponds to the one
	described in \autocite[(5.17)]{KL:quantum-group-III}.
	In contrast to the notation in \autocite{KL:quantum-group-III},
	we let $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k, n)}$ act from the left
	and $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0, k+1, n)}$ from the right.
\end{remark}



\subsubsection{Clifford Cohomology of partial flag varieties}
\label{sec:Clifford-Cohomology-Flag-Variety}
Recall the Clifford Grassmann cohomology ring $\mathop{\mathit{H}\mathfrak C}\nolimits_{(m,n)}$
from \cref{def:Clifford-Grassmann-Cohomology}
and the ordinary cohomology ring of partial flag varieties from \cref{prop:Flag-variety-cohomology}.
The following definition generalises both:
\begin{definition}
	\label{def:clifford-flag-cohomology}
	For any sequence $\bm{k}=(0=k_0\leq k_1\leq\dotsb\leq k_\ell=n)$,
	let the \term{Clifford cohomology ring} of the flag variety $\operatorname{Fl}(\bm{k})$
	be the quotient
	\begin{equation}
		\mathop{\mathit{H}\mathfrak C}\nolimits_{\bm{k}} ≔ \quot{\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(k_1, \ldots, k_l)}}{(\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,n)})_+}
	\end{equation}
	by the two-sided ideal generated by non-constant $\mathfrak d$-symmetric polynomials.
\end{definition}
\begin{remark}
	The graded dimension computations carried out in \cref{lem:graded-dimension-Grassmann-cohomology,cor:graded-dimension-flag-cohomology}
	remain valid for $\mathop{\mathit{H}\mathfrak C}\nolimits_{\bm{k}}$ if one replaces $\dim_{q,k}$ by $\rk_{q,\mathfrak{C}_{I}}$.
\end{remark}
\begin{proposition}
	\label{exm:clifford-flag-cohomology-generalises-grassmannians}
	The quotient $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,k,n)}/(\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,n)})_+$ is isomorphic to the
	Clifford cohomology ring $\mathop{\mathit{H}\mathfrak C}\nolimits_{(k,n)}$ of Grassmannians defined in \cref{def:Clifford-Grassmann-Cohomology}.
\end{proposition}
\begin{proof}
	We may expand the generators $\mathfrak e^{(0,n)}_k$ 
	as in \cref{eqn:Clifford-Flag-Varieties:Expansion-Coefficients}.
	In the quotient, we therefore obtain relations
	\[
		\sum_{l=0}^{m}
		\mathfrak e^{(0,k)}_l \ \lambda^{(0,k,n)}_{m,l}\, \mathfrak e^{(k,n)}_{m-l}
		= \delta_{m,0}.
	\]
	Comparing coefficients with those of the identity
	\begin{equation*}
		\sum_{l=0}^{m} (-1)^l\,h^{(k,n)}_{m-l}\,\tilde{\kappa}_{k,k+1}\dotsm\tilde{\kappa}_{k, k+l}\mathfrak e^{(k,n)}_l = \delta_{m,0}
	\end{equation*}
	from \cref{prop:Complete-Clifford-Polynomials:Vanishing-Relation}
	implies that
	\begin{equation}
		(-1)^l\,h^{(k,n)}_{m-l}\,\tilde{\kappa}_{k,k+1}\dotsm\tilde{\kappa}_{k, k+l} = \mathfrak e^{(0,k)}_{m-l}\,\lambda^{(0,k,n)}_{m,l};
	\end{equation}
	we thus indeed have an isomorphism $\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,k,n)}/(\mathop{\Lambda{\operator@font Pol}\mathfrak C}\nolimits_{(0,n)})_+ \cong \mathop{\mathit{H}\mathfrak C}\nolimits_{(k, n)}$.
\end{proof}
We have arrived at a super-generalisation
for the cohomology rings of Flag varieties.
In the ordinary set-up, it has been proven in \autocite{KL:quantum-group-III}
that these rings admit an action
by the Kac-Moody 2-category from \autocite{Bru:Kac-Moody}.
The latter has a super-analogue constructed in \autocite{Bru:Super-Kac-Moody}.
This motivates:
\begin{question}
	The Clifford cohomology rings $\mathop{\mathit{H}\mathfrak C}\nolimits_{\bm{k}}$ from \cref{def:clifford-flag-cohomology}
	are endowed with a bimodule structure by \cref{rmk:Clifford-Symmetric-Polynomials:Bimodule-structure}.
	Can one construct an action of the Kac-Moody 2-supercategory from \autocite{Bru:Super-Kac-Moody}
	on these Clifford cohomology rings?
\end{question}

\subsection*{Acknowledgements}
This article compiles the results of a Master's thesis
written with the advice of Catharina Stroppel,
whom the author wants to thank for her continuous support
as well as the opportunity provided to dive into this topic.

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