\section{Introduction}
\label{sec:intr}
It is well known that the second homology of a Lie (super)algebra $\mathfrak{g}$ is identified with the kernel of its universal central extension, and thus classifies all central extension of $\mathfrak{g}$ up to isomorphism (c.f. \cite{Neher2003,ScheunertZhang1998}). They play crucial rules in the theory of Lie (super)algebras.

A remarkable work about the second homology of a Lie algebra is the nice connection between the second homology of a matrix Lie algebra and the first cyclic homology of its coordinates associative algebra established in \cite{KasselLoday1982}. Concretely, let $A$ be a unital associative algebra over a commutative ring with $2$ invertible, one denotes $\mathfrak{gl}_n(A)$ the Lie algebra of all $n\times n$-matrices with entries in $A$ under commutator operation and $\mathfrak{sl}_n(A)$ the derived Lie subalgebra of $\mathfrak{gl}_n(A)$. It is shown in \cite{KasselLoday1982} that the second homology $\mathrm{H}_2(\mathfrak{sl}_n(A))$ with $n\geqslant2$ is isomorphic to the first cyclic homology $\mathrm{HC}_1(A)$.

Such an isomorphism has been extended to many other classes of Lie (super)algebras. For instance, Y. Gao showed in \cite{Gao1996} that the second homology of elementary unitary Lie algebra $\mathfrak{eu}_n(R,{}^-)$ with $n\geqslant5$ is identified with the first skew-dihedral homology of $(R,{}^-)$ that is a unital associative algebra with anti-involution. The super analogue of C. Kassel and J. L. Loday's work was obtained in \cite{ChenGuay2013,ChenSun2015}.  The isomorphism from the second homology of the Lie superalgebra $\mathfrak{sl}_{m|n}(S)$ coordinated by a unital associative superalgebra $S$ with $m+n\geqslant5$ to the first $\mathbb{Z}/2\mathbb{Z}$-graded cyclic homology $\mathrm{HC}_1(S)$ was established. Recent investigation \cite{ChangWang2016} further gave the identification between the second homology of the orthosymplectic Lie superalgebra $\mathfrak{osp}_{m|2n}(R,{}^-)$ with positive integer pair $(m,n)\neq(1,1)$ or $(2,1)$ and the first $\mathbb{Z}/2\mathbb{Z}$-grade skew-dihedral homology of $(R,{}^-)$, where $(R,{}^-)$ is a unital associative superalgebra with superinvolution (see (\ref{eq:supinv}) for definition). A series of deeep investigations on the relationship between the homology theory of Lie algebras and the homology theory of associative algebras have been made in \cite{LodayQuillen1984,LodayProcesi1988}.

Inspired by above developments, we aim to establish an isomorphism that is analogous to C. Kassel and J. L. Loday's isomorphism for the generalized periplectic Lie superalgebra $\mathfrak{p}_m(R,{}^-)$ coordinatized by a unital associative superalgebra $(R,{}^-)$ with superinvolution. As in Section~\ref{sec:palg}, a generalized periplectic Lie superalgebra is defined as the derived sub-supalgebra of the Lie superalgebra of all skew-symmetric matrices with respect to certain superinvolution. It could be understood as a super analogue of a unitary Lie algebra introduced in \cite{AllisonFaulkner1993}. This family of Lie superalgebras provides us with a realization of an arbitrary generalized root graded Lie superalgebra of type $P(m-1)$ for $m\neq4$ up to central isogeneous (c.f. \cite{ChengGao2015}), which is a complement to the realization of a root graded Lie superalgebra of type $P(m-1)$ given in \cite{MartinezZelmanov2003}.

A primary result of this paper is Theorem~\ref{thm:hml_pm} which states that the second homology of the Lie superalgebra $\mathfrak{p}_m(R,{}^-)$ with $m\geqslant5$ is isomorphic to the first $\mathbb{Z}/2\mathbb{Z}$-graded dihedral homology of $(R,{}^-\circ\rho)$, where ${}^-\circ\rho$ is the superinvolution on $R$ obtained by twisting the superinvolution ${}^-$ with the sign map $\rho$ (see (\ref{eq:rho}) in Section~\ref{sec:palg}). In the special case where $R$ is super-commutative, the isomorphism indicates that the second homology of $\mathfrak{p}_m(\Bbbk)\otimes_{\Bbbk}R$ for a super-commutative superalgebra $R$ is trivial, which was obtained by K. Iohara and Y. Koga in \cite{IoharaKoga2001, IoharaKoga2005}. While the isomorphism also reveals that the second homology  of $\mathfrak{p}_m(R,{}^-)$ is not necessarily trivial if $R$ is not super-commutative. 

The methods used in this paper non-suprisingly involve explicitly construction of the universal central extension of $\mathfrak{p}_m(R,{}^-)$, which will be achieved via introducing the notion of Steinberg periplectic Lie superalgebra $\mathfrak{stp}_m(R,{}^-)$ in Section~\ref{sec:stp}.

The isomorphism between the second homology of $\mathfrak{p}_m(R,{}^-)$ and the first $\mathbb{Z}/2\mathbb{Z}$-graded dihedral homology of $(R,{}^-\circ\rho)$ fails when $m=3$ or $4$. Nonetheless, the second homology of $\mathfrak{p}_4(R,{}^-)$ and $\mathfrak{p}_3(R,{}^-)$ will also been explicitly computed in Theorems~\ref{thm:p4_hml} and~\ref{thm:p3_hml}.
 
 \section{Basics on generalized periplectic Lie superalgebras}
 \label{sec:palg}
 
We briefly review the definition of a generalized periplectic Lie superalgebra and prove a few properties in this section.

Throughout the paper, we always assume $\Bbbk$ is a commutative base ring with $2$ invertible. All modules, associative superalgebras and Lie superalgebras are assumed to be over $\Bbbk$.
Let $R$ be a unital associative superalgebra, in which the parity of $a\in R$ is denoted by $|a|$. Then the associative superalgebra $\mathrm{M}_{m|m}(R)$ of all $2m\times 2m$-matrices is also equipped with a $\mathbb{Z}/2\mathbb{Z}$-gradation by setting 
\begin{equation}
|e_{ij}(a)|:=|i|+|j|+|a|,\quad a\in R,\quad 1\leqslant i,j\leqslant 2m,
\label{eq:parity_matrix}
\end{equation}
where $e_{ij}(a)$ is the matrix unit with $a$ at the $(i,j)$-position and $0$ elsewhere,
\begin{equation}
|i|=\begin{cases}0,&\text{if }i\leqslant m,\\1,&\text{if }i>m.\end{cases}
\label{eq:parity_index}
\end{equation}
This makes $\mathrm{M}_{m|m}(R)$ an associative superalgebra. 

We assume in addition that $R$ is equipped with a superinvolution\footnote{The superinvolutions on the matrix superalgebra $\mathrm{M}_{m|n}(\Bbbk)$ over a field $\Bbbk$ of characteristic not $2$ were classified in \cite{Racine1998}. A superinvolution on $\mathrm{M}_{m|n}(\Bbbk)$ may not exist. Whenever it exists, a superinvolution on $\mathrm{M}_{m|n}(\Bbbk)$ is equivalent to either a periplectic superinvolution or an orthosymplectic superinvolution. This motivates us to define the periplectic superinvolution on $\mathrm{M}_{m|m}(R)$ here.} ${}^-:R\rightarrow R$ that is a $\Bbbk$-linear map satisfying
\begin{equation}
\overline{ab}=(-1)^{|a||b|}\bar{b}\bar{a},\text{ and }\bar{\bar{a}}=a,
\label{eq:supinv}
\end{equation}
for homogeneous $a,b\in R$.  This further gives rise to a periplectic superinvolution on the associative superalgebra $\mathrm{M}_{m|m}(R)$ defined by
\begin{equation}
\begin{pmatrix}A&B\\C&D\end{pmatrix}^{\mathrm{prp}}:=
\begin{pmatrix}\overline{D}^t&-\overline{\rho(B)}^t\\\overline{\rho(C)}^t&\overline{A}^t\end{pmatrix},
\label{eq:inv_pR}\end{equation}
where $A,B,C,D$ are $m\times m$-matrices with entries in $R$,  $\rho:R\rightarrow R$ is the $\Bbbk$-linear map defined by
\begin{equation}
\rho(a)=(-1)^{|a|}a,\label{eq:rho}
\end{equation}
for homogeneous $a\in R$, $\rho(A)$ denotes the matrix $(\rho(a_{ij}))$ and $\overline{A}=(\overline{a_{ij}})$ for $A=(a_{ij})$. In this situation, one defines a Lie superalgebra 
\begin{equation}
\widetilde{\mathfrak{p}}_m(R,{}^-):=\{X\in\mathrm{M}_{m|m}(R)|X^{\mathrm{prp}}=-X\},
\label{eq:ptilde}
\end{equation}
with the standard super-commutator as the super-bracket. Its derived Lie sub-superalgebra
\begin{equation}
\mathfrak{p}_m(R,{}^-):=[\widetilde{\mathfrak{p}}_m(R,{}^-),\widetilde{\mathfrak{p}}_m(R,{}^-)]
\label{eq:palg}
\end{equation}
is called \textit{the generalized periplectic Lie superalgebra coordinatized by the associative superalgebra $(R,{}^-)$ with superinvolution}.
\medskip

As an example, we consider $R=\Bbbk$ on which the identity map is a superinvolution. The Lie superalgebra $\mathfrak{p}_m(\Bbbk,\mathrm{id})$ coincides with the simple Lie superalgebra of type $P(m-1)$ as defined in \cite{Kac1977}. We simply write $\mathfrak{p}_m(\Bbbk):=\mathfrak{p}_m(\Bbbk,\mathrm{id})$.

If $R$ is super-commutative, there is a natural Lie superalgebra structure on $\mathfrak{p}_m(\Bbbk)\otimes_{\Bbbk}R$ with the super-bracket
\begin{equation}
[x\otimes a,y\otimes b]=(-1)^{|a||y|}[x,y]\otimes ab,
\label{eq:ploop}
\end{equation}
for homogeneous $x,y\in\mathfrak{p}_m(\Bbbk)$ and $a,b\in R$. The Lie superalgebra $\mathfrak{p}_m(\Bbbk)\otimes_{\Bbbk}R$ is actually isomorphic to the generalized periplectic Lie superalgebra $\mathfrak{p}_m(R,\rho)$, where the $\Bbbk$-linear map $\rho:R\rightarrow R$ (\ref{eq:rho}) is a superinvolution on $R$ when $R$ is super-commutative.
\medskip

Before going into the discussion on the properties of generalized periplectic Lie superalgebras, we exhibit another example here:
\begin{example}
\label{ex:iso_p_SS}
Let $S$ be an arbitrary unital associative superalgebra and $S^{\mathrm{op}}$ denote its \textit{opposite superalgebra} with the multiplication
\begin{equation}
a\overset{\mathrm{op}}{\cdot}b=(-1)^{|a||b|}b\cdot a,\label{eq:opalg}
\end{equation}
for homogeneous $a,b\in S$. Then the $\Bbbk$-linear map
\begin{equation}
\mathrm{ex}:S\oplus S^{\mathrm{op}}\rightarrow S\oplus S^{\mathrm{op}}, \quad a\oplus b\mapsto b\oplus a.
\label{eq:inv_ex}
\end{equation}
is a superinvolution on $S\oplus S^{\mathrm{op}}$. In this situation, we have an isomorphism of Lie superalgebras
\begin{equation}
\mathfrak{p}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})\cong\mathfrak{sl}_{m|m}(S)
:=[\mathfrak{gl}_{m|m}(S),\mathfrak{gl}_{m|m}(S)],\quad m\geqslant1,
\label{eq:iso_p_SS}
\end{equation}
where $\mathfrak{gl}_{m|m}(S)$ is the Lie superalgebra of $2m\times 2m$-matrices with entrices in $S$.
\end{example}

\begin{proof}
In fact, the Lie superalgebra $\widetilde{\mathfrak{p}}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})$ is isomorphic to the Lie superalgebra $\mathfrak{gl}_{m|m}(S)$, where an explicit isomorphism $\mathfrak{gl}_{m|m}(S)\rightarrow\tilde{\mathfrak{p}}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})$ is given as follows:
\begin{align*}
e_{i,j}(a)&\mapsto e_{i,j}(a\oplus 0)-e_{m+j,m+i}(0\oplus a),\\
e_{i,m+j}(a)&\mapsto e_{i,m+j}(a\oplus 0)+e_{j,m+i}(0\oplus \rho(a)),\\
e_{m+i,j}(a)&\mapsto e_{m+i,j}(a\oplus 0)-e_{m+j,i}(0\oplus \rho(a)),\\
e_{m+i,m+j}(a)&\mapsto -e_{j,i}(0\oplus a)+e_{m+i,m+j}(a\oplus0),
\end{align*}
for $a\in S$ and $1\leqslant i,j\leqslant m$. Taking their derived Lie sub-superalgebras, we conclude that the Lie superalgebra $\mathfrak{p}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})$ is isomorphic to the Lie superalgebra $\mathfrak{sl}_{m|m}(S)$.
\end{proof}
\bigskip

It is known from \cite{Neher2003} that a Lie superalgebra admits a universal central extension if and only if it is perfect. In order to discuss the universal central extension of the generalized periplectic Lie superalgebra $\mathfrak{p}_m(R,{}^-)$, we explore the perfectness of $\mathfrak{p}_m(R,{}^-)$. We will use the following notation:
\begin{align}
t_{ij}(a):&=e_{ij}(a)-e_{m+j,m+i}(\bar{a}),\label{eq:p_gtor_t}\\
f_{ij}(a):&=e_{i,m+j}(a)+e_{j,m+i}(\rho(\bar{a})),\label{eq:p_gtor_f}\\
g_{ij}(a):&=e_{m+i,j}(a)-e_{m+j,i}(\rho(\bar{a})).\label{eq:p_gtor_g}
\end{align}
We always denote $R_{(\pm)}:=\{a\in R|\bar{a}=\pm \rho(a)\}$.

\begin{lemma}
\label{lem:p_elm}
For $m\geqslant2$, every element of $x\in\tilde{\mathfrak{p}}_m(R,{}^-)$ is written as
\begin{equation}
\begin{aligned}
x=&{t}_{11}(a)+\sum\limits_{i=2}^m({t}_{ii}(a_i)-{t}_{11}(a_i))+\sum\limits_{1\leqslant i\neq j\leqslant m}{t}_{ij}(a_{ij})\\
&+\sum\limits_{i=1}^m(e_{i,m+i}(b_i)+e_{m+i,i}(c_i))+\sum\limits_{1\leqslant i<j\leqslant m}({f}_{ij}(b_{ij})+{g}_{ij}(c_{ij})),
\end{aligned}\label{eq:p_ele}
\end{equation}
where $a, a_i, a_{ij}, b_{ij}, c_{ij}\in R$, $b_i\in R_{(+)}$ and $c_i\in R_{(-)}$ are uniquely determined by $x$. Moreover, such an element $x$ is contained in 
$\mathfrak{p}_m(R,{}^-)
=[\widetilde{\mathfrak{p}}_m(R,{}^-),\widetilde{\mathfrak{p}}_m(R,{}^-)]
$ 
if and only if $a\in [R,R]+R_{(-)}$.
\end{lemma}
\begin{proof}
The first statement follows from the definition of $\tilde{\mathfrak{p}}_m(R,{}^-)$. We show that $x\in\mathfrak{p}_m(R,{}^-)$ if and only if $a\in [R,R]+R_{(-)}$.

We observe that each term on the right hand side of (\ref{eq:p_ele}) except $t_{11}(a)$ is a super-commutator of two elements in $\widetilde{\mathfrak{p}}_m(R,{}^-)$, i.e., they are contained in $\mathfrak{p}_m(R,{}^-)=[\widetilde{p}_m(R,{}^-),\widetilde{p}_m(R,{}^-)]$.
Hence, it suffices to show that ${t}_{11}(a)\in\mathfrak{p}_m(R,{}^-)$ if and only if $a\in [R,R]+R_{(-)}$.

If $a\in[R,R]$, we write $a=\sum [a'_i,a''_i]$ with $a'_i, a''_i\in R$, then
$${t}_{11}(a)=\sum[{t}_{11}(a'_i),{t}_{11}(a''_i)]\in \mathfrak{p}_m(R,{}^-).$$
While an element $a\in R_{(-)}$ satisfies that $${t}_{11}(a)+{t}_{22}(a)={t}_{11}(a)-{t}_{22}(\rho(\bar{a}))=[{f}_{12}(a),{g}_{21}(1)]\in\mathfrak{p}_m(R,{}^-).$$
Combining with ${t}_{22}(a)-{t}_{11}(a)\in\mathfrak{p}_m(R,{}^-)$ and $\frac{1}{2}\in\Bbbk$, we conclude that ${t}_{11}(a)\in\mathfrak{p}_m(R,{}^-)$. This shows that ${t}_{11}(a)\in\mathfrak{p}_m(R,{}^-)$ if $a\in [R,R]+R_{(-)}$.

For the inverse implication, we observe that every element
$$\begin{pmatrix}A&B\\C&-\overline{A}^t\end{pmatrix}\in\mathfrak{p}_m(R,{}^-)=[\tilde{\mathfrak{p}}_m(R,{}^-),\tilde{\mathfrak{p}}_m(R,{}^-)]$$
satisfies $\mathrm{Tr}(A)\in[R,R]+R_{(-)}$.  Hence, $a\in [R,R]+R_{(-)}$ if ${t}_{11}(a)\in\mathfrak{p}_m(R,{}^-)$.
\end{proof}

\begin{proposition}
\label{prop:p_pft}
Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution and $m\geqslant2$.
\begin{enumerate}
\item There is an exact sequence of Lie superalgebras
$$0\rightarrow\mathfrak{p}_m(R,{}^-)\rightarrow\tilde{\mathfrak{p}}_m(R,{}^-)\rightarrow \frac{R}{[R,R]+R_{(-)}}\rightarrow0.$$
\item The Lie superalgebra $\mathfrak{p}_m(R,{}^-)$ is generated by ${t}_{ij}(a)$, ${f}_{ij}(a)$, ${g}_{ij}(a)$ for $a\in R$, $1\leqslant i\neq j\leqslant m$.
\item If $m\geqslant3$, then the Lie superalgebra $\mathfrak{p}_m(R,{}^-)$ is perfect, i.e.,
$$\mathfrak{p}_m(R,{}^-)=[\mathfrak{p}_m(R,{}^-),\mathfrak{p}_m(R,{}^-)].$$
\end{enumerate}
\end{proposition}
\begin{proof}
(i) We define a surjective $\Bbbk$--linear map
$$\eta:\tilde{\mathfrak{p}}_m(R,{}^-)\rightarrow\frac{R}{[R,R]+R_{(-)}},\quad\begin{pmatrix}A&B\\C&-\overline{A}^t\end{pmatrix}\mapsto \mathrm{Tr}(A)+([R,R]+R_{(-)}).$$
By Lemma~\ref{lem:p_elm}, $\ker\eta=\mathfrak{p}_m(R,{}^-)$. Hence, we obtain an exact sequence of $\Bbbk$--modules:
$$0\rightarrow \mathfrak{p}_m(R,{}^-)\rightarrow\tilde{\mathfrak{p}}_m(R,{}^-)\rightarrow\frac{R}{[R,R]+R_{(-)}}\rightarrow0.$$
Note that $R/([R,R]+R_{(-)})$ is a super-commutative Lie superalgebra, we obtain that all $\Bbbk$--linear maps appearing in this exact sequence are homomorphisms of Lie superalgebras. 
\medskip

(ii) By Lemma~\ref{lem:p_elm}, it suffices to show $t_{11}(a)$ with $a\in [R,R]+R_{(-)}$, $t_{ii}(a)-t_{11}(a)$ with $a\in R, 2\leqslant i\leqslant m$, $e_{i,m+i}(a)$ with $a\in R_{(+)}, 1\leqslant i\leqslant m$, and $e_{m+i,i}(a)$ with $a\in R, 1\leqslant i\leqslant m$ can be generated by $t_{ij}(b), f_{ij}(b)$ and $g_{ij}(b)$ for $b\in R$ and $1\leqslant i\neq j\leqslant m$. Indeed,
$$t_{ii}(a)-t_{11}(a)=[t_{i1}(a),t_{1i}(1)],
\quad e_{i,m+i}(a)=\frac{1}{2}[t_{ij}(1),f_{ji}(a)],
\text{ and } 
e_{m+i,i}(a)=\frac{1}{2}[g_{ij}(a),t_{ji}(1)],$$
where $1\leqslant j\leqslant m$ is chosen such that $i\neq j$. Furthermore, for $a\in [R,R]+R_{(-)}$, we also have 
\begin{align*}
{t}_{11}([a',a''])&=[{t}_{12}(a'),{t}_{21}(a'')]-(-1)^{|a'||a''|}[{t}_{12}(1),{t}_{21}(a''a')],&& a', a''\in R\\
{t}_{11}(a)&=\frac{1}{2}[{t}_{12}(1),{t}_{21}(a)]+\frac{1}{2}[{f}_{12}(1),{g}_{21}(a)],
&&a\in R_{(-)}.
\end{align*}
This proves (ii).
\medskip

(iii) By (ii), $\mathfrak{p}_m(R,{}^-)$ is generated by ${t}_{ij}(a),{f}_{ij}(a),{g}_{ij}(a)$ with $a\in R$ and $1\leqslant i\neq j\leqslant m$. We shall show that
these elements are also contained in $[\mathfrak{p}_m(R,{}^-),\mathfrak{p}_m(R,{}^-)]$. Note that $m\geqslant3$, for $1\leqslant i\neq j\leqslant m$, we may choose $1\leqslant k\leqslant m$ such that $i,j,k$ are distinct. Then the equalities
\begin{equation*}
{t}_{ij}(a)=[{t}_{ik}(1),{t}_{kj}(a)],\quad
{f}_{ij}(a)=[{t}_{ik}(1),{f}_{kj}(a)],\quad
{g}_{ij}(a)=[{g}_{ik}(a),{t}_{kj}(1)]
\end{equation*}
imply that ${t}_{ij}(a),{f}_{ij}(a),{g}_{ij}(a)\in[\mathfrak{p}_m(R,{}^-),\mathfrak{p}_m(R,{}^-)]$. Hence, $\mathfrak{p}_m(R,{}^-)$ is perfect for $m\geqslant3$.
\end{proof}

\begin{remark}
\label{rmk:p1_impft}
The Lie superalgebra $\mathfrak{p}_1(R,{}^-)$ is not necessarily perfect. For instance, if $R$ is  super-commutative, then
$$\tilde{\mathfrak{p}}_1(R,\rho):=\left\{\begin{pmatrix} a&b\\0&-a\end{pmatrix}\middle| a,b,c\in R\right\},\text{ and }\mathfrak{p}_1(R,\rho)=\left\{\begin{pmatrix} 0&b\\0&0\end{pmatrix}\middle| a,b,c\in R\right\}.$$
The Lie superalgebra $\mathfrak{p}_1(R,\rho)$ is not perfect since $[\mathfrak{p}_1(R,\rho),\mathfrak{p}_1(R,\rho)]=0$. In general, the condition for the perfectness of $\mathfrak{p}_1(R,{}^-)$ is unknown yet.

Similarly, the Lie superalgebra $\mathfrak{p}_2(R,{}^-)$ is also not necessarily perfect. Hence, the existence of a universal central extension of $\mathfrak{p}_1(R,{}^-)$ or $\mathfrak{p}_2(R,{}^-)$ is not ensured. We only consider the second homology of $\mathfrak{p}_m(R,{}^-)$ for $m\geqslant3$.
\end{remark}

\section{Steinberg periplectic Lie superalgebras}
\label{sec:stp}
In the previous section, we have shown that the Lie superalgebra $\mathfrak{p}_m(R,{}^-)$ is perfect for $m\geqslant3$. The perfectness allows us to further study its universal central extension, whose kernel will finally provides us with the second homology of $\mathfrak{p}_m(R,{}^-)$.

In this section, we will introduce the notion of Steinberg periplectic Lie superalgebra $\mathfrak{stp}_m(R,{}^-)$ and prove that it is a central extension of $\mathfrak{p}_m(R,{}^-)$. Its universality will be dicussed in the sequent sections.

\begin{definition}
\label{def:stp}
Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution and $m\geqslant3$. \textit{The Steinberg periplectic Lie superalgebra coordinatized by $(R,{}^-)$,} denoted by $\mathfrak{stp}_m(R,{}^-)$, is defined to be the abstract Lie superalgebra generated by homogenous elements $\mathbf{t}_{ij}(a)$, $\mathbf{f}_{ij}(a)$, $\mathbf{g}_{ij}(a)$ with parity $|a|$, $|a|+1$, $|a|+1$ respectively, for homogeneous $a\in R$ and $1\leqslant i\neq j\leqslant m$, subjecting to the relations:
\begin{align}
&\mathbf{t}_{ij},\mathbf{f}_{ij},\mathbf{g}_{ij}\text{ are all }\Bbbk\text{-linear},
&&\text{for }i\neq j,\tag{STP00}\label{STP00}\\
&\mathbf{f}_{ij}(\bar{a})=\mathbf{f}_{ji}(\rho(a)),
&&\text{for }i\neq j,\tag{STP01}\label{STP01}\\
&\mathbf{g}_{ij}(\bar{a})=-\mathbf{g}_{ji}(\rho(a)),
&&\text{for }i\neq j,\tag{STP02}\label{STP02}\\
&[\mathbf{t}_{ij}(a),\mathbf{t}_{jk}(b)]=\mathbf{t}_{ik}(ab),
&&\text{for distinct }i,j,k,\tag{STP03}\label{STP03}\\
&[\mathbf{t}_{ij}(a),\mathbf{t}_{kl}(b)]=0,
&&\text{for }i\neq j\neq k\neq l\neq i,\tag{STP04}\label{STP04}\\
&[\mathbf{t}_{ij}(a),\mathbf{f}_{jk}(b)]=\mathbf{f}_{ik}(ab),
&&\text{for distinct }i,j,k,\tag{STP05}\label{STP05}\\
&[\mathbf{t}_{ij}(a),\mathbf{f}_{kl}(b)]=0,
&&\text{for }i\neq j\neq k\neq l\neq j,\tag{STP06}\label{STP06}\\
&[\mathbf{g}_{ij}(a),\mathbf{t}_{jk}(b)]=\mathbf{g}_{ik}(ab),
&&\text{for distinct } i,j,k,\tag{STP07}\label{STP07}\\
&[\mathbf{g}_{ij}(a),\mathbf{t}_{kl}(b)]=0,
&&\text{for }l\neq k\neq j\neq i\neq k,\tag{STP08}\label{STP08}\\
&[\mathbf{f}_{ij}(a),\mathbf{f}_{kl}(b)]=0,
&&\text{for }i\neq j,\text{ and }k\neq l,\tag{STP09}\label{STP09}\\
&[\mathbf{g}_{ij}(a),\mathbf{g}_{kl}(b)]=0,
&&\text{for }i\neq j,\text{ and }k\neq l,\tag{STP10}\label{STP10}\\
&[\mathbf{f}_{ij}(a),\mathbf{g}_{jk}(b)]=\mathbf{t}_{ik}(ab),
&&\text{for distinct }i,j,k,\tag{STP11}\label{STP11}\\
&[\mathbf{f}_{ij}(a),\mathbf{g}_{kl}(b)]=0,
&&\text{for distinct }i,j,k,l,\tag{STP12}\label{STP12}
\end{align}
where $a,b\in R$ and $1\leqslant i,j,k,l\leqslant m$.
\end{definition}

Recall Proposition~\ref{prop:p_pft} that $\mathfrak{p}_m(R,{}^-)$ is generated by $t_{ij}(a), f_{ij}(a)$ and $g_{ij}(a)$ for $a\in R$ and $1\leqslant i\neq j\leqslant m$. These generators satisfy all relations (\ref{STP00})-(\ref{STP12}). Hence, there is a canonical homomorphism of Lie superalgebras:
\begin{equation}
\psi:\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{p}_m(R,{}^-),\label{eq:can_hom}
\end{equation}
such that $\psi(\mathbf{t}_{ij}(a))=t_{ij}(a),\psi(\mathbf{f}_{ij}(a))=f_{ij}(a)$ and $\psi(\mathbf{g}_{ij}(a))=g_{ij}(a)$, which will be demonstrated to be a central extension, i.e., the kernel of $\psi$ is included in the center of $\mathfrak{stp}_m(R,{}^-)$.
\bigskip

It is easy to observe that all diagonal, upper triangular and lower triangular matrices in $\mathfrak{p}_m(R,{}^-)$ form three Lie sub-superalgebras of $\mathfrak{p}_m(R,{}^-)$, respectively. Their direct sum gives a decomposition of $\mathfrak{p}_m(R,{}^-)$. We first show that the Steinberg periplectic Lie superalgebra $\mathfrak{stp}_m(R,{}^-)$ also possesses a similar decomposition. 

\begin{lemma}
\label{lem:munui_wd}
In the Lie superalgebra $\mathfrak{stp}_m(R,{}^-)$, the following equalities hold:
\begin{align*}
[\mathbf{t}_{ij}(a),\mathbf{f}_{ji}(b)]=[\mathbf{t}_{ik}(a),\mathbf{f}_{ki}(b)],\text{ and }
[\mathbf{g}_{ij}(a),\mathbf{t}_{ji}(b)]=[\mathbf{g}_{ik}(a),\mathbf{t}_{ki}(b)],
\end{align*}
for $a,b\in R$ and $1\leqslant i,j,k\leqslant m$ with $i\neq j,k$.
\end{lemma}
\begin{proof}
We assume $i, j, k$ are distinct and deduce from (\ref{STP03}), (\ref{STP05}) and (\ref{STP06}) that
\begin{align*}
[\mathbf{t}_{ik}(a),\mathbf{f}_{ki}(b)]&=[[\mathbf{t}_{ij}(a),\mathbf{t}_{jk}(1)],\mathbf{f}_{ki}(b)]\\
&=[[\mathbf{t}_{ij}(a),\mathbf{f}_{ki}(b)],\mathbf{t}_{jk}(1)]+[\mathbf{t}_{ij}(a),
[\mathbf{t}_{jk}(1),\mathbf{f}_{ki}(b)]]\\
&=0+[\mathbf{t}_{ij}(a),\mathbf{f}_{ji}(b)]\\
&=[\mathbf{t}_{ij}(a),\mathbf{f}_{ji}(b)].
\end{align*}
Similarly, $[\mathbf{g}_{ij}(a),\mathbf{t}_{ji}(b)]=[\mathbf{g}_{ik}(a),\mathbf{t}_{ki}(b)]$ follows from (\ref{STP03}), (\ref{STP07}) and (\ref{STP08}).
\end{proof}

Lemma~\ref{lem:munui_wd} permits us to introduce the following well-defined elements of $\mathfrak{stp}_m(R,{}^-)$:
\begin{align}
\mathbf{f}_{i}(a):&=[\mathbf{t}_{ij}(1),\mathbf{f}_{ji}(a)],&&\text{for some }j\neq i,\label{eq:mui}\\
\mathbf{g}_{i}(a):&=[\mathbf{g}_{ij}(a),\mathbf{t}_{ji}(1)],&&\text{for some }j\neq i,\label{eq:nui}\\
\mathbf{h}_{ij}(a,b):&=[\mathbf{f}_{ij}(a),\mathbf{g}_{ji}(b)],&&\text{for }i\neq j,\label{eq:kp}
\end{align}
where $a,b\in R$ and $1\leqslant i,j\leqslant m$. One easily deduce that
\begin{equation}
\boldsymbol{f}_i(\bar{a})=\boldsymbol{f}_i(\rho(a)),\text{ and }
\boldsymbol{g}_i(\bar{a})=-\boldsymbol{g}_i(\rho(a)),
\end{equation}
for $1\leqslant i\leqslant m$ and $a\in R$.

\begin{proposition}
\label{prop:stp_tridec}
The Lie superalgebra $\mathfrak{stp}_m(R,{}^-)$ is decomposed as a direct sum of $\Bbbk$-modules:
\begin{equation}
\mathfrak{stp}_m(R,{}^-)=\mathfrak{stp}_m^-(R,{}^-)\oplus\mathfrak{stp}_m^0(R,{}^-)\oplus\mathfrak{stp}_m^+(R,{}^-),
\label{eq:stp_tridec}
\end{equation}
where
\begin{align*}
\mathfrak{stp}_m^0(R,{}^-):&=\mathrm{span}_{\Bbbk}\{\mathbf{h}_{ij}(a,b)|a,b\in R, 1\leqslant i\neq j\leqslant m\},\\
\mathfrak{stp}_m^+(R,{}^-):&=\mathrm{span}_{\Bbbk}\{\mathbf{t}_{ij}(a),\mathbf{f}_{ij}(a),\mathbf{f}_k(a)|a\in R, 1\leqslant i,j,k\leqslant m\text{ and }i<j\},\\
\mathfrak{stp}_m^-(R,{}^-):&=\mathrm{span}_{\Bbbk}\{\mathbf{t}_{ij}(a),\mathbf{g}_{ij}(a),\mathbf{g}_k(a)|a\in R, 1\leqslant i,j,k\leqslant m \text{ and }i>j\},
\end{align*}
are all Lie sub-superalgebras of $\mathfrak{stp}_m(R,{}^-)$ and $[\mathfrak{stp}_m^0(R,{}^-),\mathfrak{stp}_m^{\pm}(R,{}^-)]\subseteq\mathfrak{stp}_m^{\pm}(R,{}^-)$.
\end{proposition}
\begin{proof}
We first deduce from (\ref{STP00})-(\ref{STP12}) that $\mathfrak{stp}_m^0(R,{}^-)$, $\mathfrak{stp}_m^-(R,{}^-)$ and $\mathfrak{stp}_m^+(R,{}^-)$ are all Lie sub-superalgebras of $\mathfrak{stp}_m(R,{}^-)$ and $$[\mathfrak{stp}_m^0(R,{}^-),\mathfrak{stp}_m^{\pm}(R,{}^-)]\subseteq\mathfrak{stp}_m^{\pm}(R,{}^-).$$

Next, we denote $\mathfrak{g}:=\mathfrak{stp}_m^-(R,{}^-)+\mathfrak{stp}_m^0(R,{}^-)+\mathfrak{stp}_m^+(R,{}^-)$ and show that $\mathfrak{stp}_m(R,{}^-)=\mathfrak{g}$. The $\Bbbk$--module $\mathfrak{g}$ is invariant under $\mathrm{ad}(\mathbf{t}_{ij}(a))$, $\mathrm{ad}(\mathbf{f}_{ij}(a))$, and $\mathrm{ad}(\mathbf{g}_{ij}(a))$. Note that $\mathbf{t}_{ij}(a)$, $\mathbf{f}_{ij}(a)$, and $\mathbf{g}_{ij}(a)$ with $a\in R$ and $1\leqslant i\neq j\leqslant m$ generate the Lie superalgebra $\mathfrak{stp}_m(R,{}^-)$, we obtain that $\mathfrak{g}$ is an ideal of the Lie superalgebra $\mathfrak{stp}_m(R,{}^-)$. It follows that $\mathfrak{stp}_m(R,{}^-)=\mathfrak{g}$ since $\mathfrak{g}$ contains a complete family of generators of $\mathfrak{stp}_m(R,{}^-)$.

Finally, we prove that the summation in the decomposition (\ref{eq:stp_tridec}) is a direct sum. We claim that the restriction $\psi|\mathfrak{stp}_m^{\pm}(R,{}^-)$ of the canonical homomorphism (\ref{eq:can_hom}) is injective. Suppose that $\boldsymbol{x}^+\in\mathfrak{stp}_m^+(R,{}^-)$ satisfying $\psi(\boldsymbol{x}^+)=0$. Write
$$\boldsymbol{x}^+=\sum\limits_{1\leqslant i<j\leqslant m}\mathbf{t}_{ij}(a_{ij})+\sum\limits_{1\leqslant i<j\leqslant m}\mathbf{f}_{ij}(b_{ij})+\sum\limits_{i}\mathbf{f}_i(c_i),$$
where $a_{ij}, b_{ij}, c_i\in R$. Applying $\psi$, we obtain
$$0=\psi(\boldsymbol{x}^+)=\sum\limits_{1\leqslant i<j\leqslant m}({t}_{ij}(a_{ij})+{f}_{ij}(b_{ij}))
+\sum\limits_{i=1}^me_{i,m+i}(c_i+\rho(\bar{c}_i))\in\mathfrak{p}_m(R,{}^-).$$
It follows that $a_{ij}=b_{ij}=0$ for $1\leqslant i<j\leqslant m$ and $c_i+\rho(\bar{c}_i)=0$ for $i=1,\ldots,m$. Now,
\begin{align*}
\boldsymbol{x}^+&=\sum\limits_{1\leqslant i<j\leqslant m}\mathbf{t}_{ij}(a_{ij})+\sum\limits_{1\leqslant i<j\leqslant m}\mathbf{f}_{ij}(b_{ij})+\sum\limits_{i}\mathbf{f}_i(c_i)\\
&=\sum\limits_{1\leqslant i<j\leqslant m}\mathbf{t}_{ij}(a_{ij})+\sum\limits_{1\leqslant i<j\leqslant m}\mathbf{f}_{ij}(b_{ij})+\sum\limits_{i}\frac{1}{2}(\mathbf{f}_i(c_i)+\mathbf{f}_i(\rho(c_i)))=0,
\end{align*}
since $\boldsymbol{f}_i(\bar{c})=\boldsymbol{f}_i(\rho(c))$. Hence, $\psi|\mathfrak{stp}_m^+(R,{}^-)$ is injective. Similarly, $\psi|\mathfrak{stp}_m^-(R,{}^-)$ is injective.

Now, if $0=\boldsymbol{x}^-+\boldsymbol{x}^0+\boldsymbol{x}^+$ for $\boldsymbol{x}^0\in\mathfrak{stp}_m^0(R,{}^-)$ and $\boldsymbol{x}^{\pm}\in\mathfrak{stp}_m^{\pm}(R,{}^-)$, then
$$0=\psi(\boldsymbol{x}^-)+\psi(\boldsymbol{x}^0)+\psi(\boldsymbol{x}^+)$$
in $\mathfrak{p}_m(R,{}^-)$, where $\psi(\boldsymbol{x}^-)$ (resp. $\psi(\boldsymbol{x}^0)$ or $\psi(\boldsymbol{x}^+)$) is a lower-triangular (resp. diagonal or upper-triangular) matrix. It follows that $\psi(\boldsymbol{x}^-)=\psi(\boldsymbol{x}^0)=\psi(\boldsymbol{x}^+)=0$ and yields $\boldsymbol{x}^-=\boldsymbol{x}^+=0$ since $\psi|\mathfrak{stp}_m^{\pm}(R,{}^-)$ is injective. Hence, the summation (\ref{eq:stp_tridec}) is a direct sum.
\end{proof}

\begin{proposition}
\label{prop:stp_ce}
Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution and $m\geqslant3$. Then $\psi:\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{p}_m(R,{}^-)$ is a central extension and $\ker\psi\subseteq\mathfrak{stp}_m^0(R,{}^-)$.
\end{proposition}
\begin{proof}
Let $\boldsymbol{x}\in\ker\psi$. We write $\boldsymbol{x}=\boldsymbol{x}^-+\boldsymbol{x}^0+\boldsymbol{x}^+$ with respect to the decomposition~(\ref{eq:stp_tridec}). Then
$$0=\psi(\boldsymbol{x})=\psi(\boldsymbol{x}^-)+\psi(\boldsymbol{x}^0)+\psi(\boldsymbol{x}^+)\in\mathfrak{p}_m(R,{}^-),$$
where $\psi(\boldsymbol{x}^-)$ (resp. $\psi(\boldsymbol{x}^0)$ or $\psi(\boldsymbol{x}^+)$) is a lower-triangular (resp. diagonal or upper-triangular) matrix. Hence, $\psi(\boldsymbol{x}^-)=\psi(\boldsymbol{x}^0)=\psi(\boldsymbol{x}^+)=0$. Recall from the proof of Proposition~\ref{prop:stp_tridec} that $\psi|\mathfrak{stp}_m^{\pm}(R,{}^-)$ are injective. It follows that $\boldsymbol{x}^+=\boldsymbol{x}^-=0$. Hence, $\boldsymbol{x}=\boldsymbol{x}^0\in\mathfrak{stp}_m^0(R,{}^-)$.

It remains to show that every element $\boldsymbol{x}\in\ker\psi$ commutes with the generators $\mathbf{t}_{ij}(a)$, $\mathbf{f}_{ij}(a)$ and $\mathbf{g}_{ij}(a)$. For $\boldsymbol{x}\in\ker\psi$, we have
$$\psi([\boldsymbol{x},\mathbf{t}_{ij}(a)])=
\psi([\boldsymbol{x},\mathbf{f}_{ij}(a)])=
\psi([\boldsymbol{x},\mathbf{g}_{ij}(a)])=0.$$
Note that $\boldsymbol{x}\in\ker\psi\subseteq\mathfrak{stp}_m^0(R,{}^-)$, it follows from Proposition~\ref{prop:stp_tridec} that $[\boldsymbol{x},\mathbf{t}_{ij}(a)]$, $[\boldsymbol{x},\mathbf{f}_{ij}(a)]$ and $[\boldsymbol{x},\mathbf{g}_{ij}(a)]$ are all contained in either $\mathfrak{stp}_m^+(R,{}^-)$ or $\mathfrak{stp}_m^-(R,{}^-)$. Hence,
$$[\boldsymbol{x},\mathbf{t}_{ij}(a)]=[\boldsymbol{x},\mathbf{f}_{ij}(a)]=[\boldsymbol{x},\mathbf{g}_{ij}(a)]=0$$
since $\psi|\mathfrak{stp}_m^{\pm}(R,{}^-)$ are injective. 
\end{proof}
\bigskip

In the special case where $(R,{}^-)=(S\oplus S^{\mathrm{op}},\mathrm{ex})$ for a unital associative superalgebra $S$, Example~\ref{ex:iso_p_SS} implies that the Lie superalgebra $\mathfrak{p}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})$ is isomorphic to $\mathfrak{sl}_{m|m}(S)$. According to Proposition~\ref{prop:stp_ce}, a central extension of $\mathfrak{p}_{m}(S\oplus S^{\mathrm{op}},\mathrm{ex})$ is given by $\mathfrak{stp}_m(S\oplus S^{\mathrm{op}}, \mathrm{ex})$, which is isomorphic to \textit{the Steinberg Lie superalgebra} $\mathfrak{st}_{m|m}(S)$ defined in \cite{ChenSun2015}:
\begin{proposition}
\label{prop:stpSS}
Let $S$ be an arbitrary unital associative superalgebra and $m\geqslant3$. Then
$$\mathfrak{stp}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})\cong\mathfrak{st}_{m|m}(S)$$
as Lie superalgebras over $\Bbbk$.
\end{proposition}
\begin{proof}
The Steinberg Lie superalgebra $\mathfrak{st}_{m|m}(S)$ as defined in \cite{ChenSun2015} is the abstract Lie superalgebra generated by homogeneous elements $\boldsymbol{e}_{ij}(a)$ of degree $|i|+|j|+|a|$ for $a\in R$ and $1\leqslant i\neq j\leqslant m+n$, subjecting to the relations:
\begin{align}
&a\mapsto\boldsymbol{e}_{ij}(a)\text{ is }\Bbbk\text{-linear},\tag{ST0}\label{ST0}\\
&[\boldsymbol{e}_{ij}(a),\boldsymbol{e}_{jk}(b)]=\boldsymbol{e}_{ik}(ab),
&&\text{ for distinct }i,j,k,\tag{ST1}\label{ST1}\\
&[\boldsymbol{e}_{ij}(a),\boldsymbol{e}_{kl}(b)]=0,
&&\text{ for }i\neq j\neq k\neq l\neq i,\tag{ST2}\label{ST2}
\end{align}
where $a,b\in R$ and $1\leqslant i,j,k,l\leqslant m+n$. 

According to the relations (\ref{STP00})-(\ref{STP12}), there is a homomorphism of Lie superalgebras $\phi:\mathfrak{st}_{m|m}(S)\rightarrow\mathfrak{stp}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})$ such that
\begin{align*}
\phi(\boldsymbol{e}_{ij}(a)):&=\mathbf{t}_{ij}(a\oplus 0),
&\phi(\boldsymbol{e}_{i,m+j}(a)):&=\mathbf{f}_{ij}(a\oplus 0),\\
\phi(\boldsymbol{e}_{m+i,j}(a)):&=\mathbf{g}_{ij}(a\oplus 0),
&\phi(\boldsymbol{e}_{m+i,m+j}(a)):&=-\mathbf{t}_{ji}(0\oplus a),\\
\phi(\boldsymbol{e}_{i,m+i}(a)):&=[\mathbf{t}_{ij}(1\oplus0),\mathbf{f}_{ji}(a\oplus 0)],
&\phi(\boldsymbol{e}_{m+i,i}(a)):&=[\mathbf{g}_{ij}(a\oplus0),\mathbf{t}_{ji}(1\oplus0)],
\end{align*}
for $a\in S$ and $1\leqslant i\neq j\leqslant m$. It has an inverse $\tilde{\phi}:\mathfrak{stp}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})\rightarrow\mathfrak{st}_{m|m}(S)$ given by
\begin{align*}
\tilde{\phi}(\boldsymbol{t}_{ij}(a\oplus b))&=\boldsymbol{e}_{ij}(a)-\boldsymbol{e}_{m+j,m+i}(b),\\
\tilde{\phi}(\boldsymbol{f}_{ij}(a\oplus b))&=\boldsymbol{e}_{i,m+j}(a)+\boldsymbol{e}_{j,m+i}(\rho(b)),\\
\tilde{\phi}(\boldsymbol{g}_{ij}(a\oplus b)&=\boldsymbol{e}_{m+i,j}(a)-\boldsymbol{e}_{m+j,i}(\rho(b)),
\end{align*}
for $a,b\in S$ and $1\leqslant i\neq j\leqslant m$. Hence, we obtain the desired isomorphism.
\end{proof}

\section{Characterization of the kernel}
\label{sec:kernel}
We have shown that the canonical epimorphism $\psi:\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{p}_m(R,{}^-)$ is a central extension. This section is devoted to explicitly characterizing the kernel of $\psi$. Here, we need the notion of the first $\mathbb{Z}/2\mathbb{Z}$-graded dihedral homology $\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-)$ for a unital associative superalgebra $(R,{}^-)$ with superinvolution.

The $\mathbb{Z}/2\mathbb{Z}$-graded dihedral homology of $(R,{}^-)$ is a natrual $\mathbb{Z}/2\mathbb{Z}$-graded analogue of the dihedral homology of a unital associative algebra with anti-involution. It can be defined through the coinvariant complex of the Hoschild complex under certain action of the dihedral group as in \cite{Loday1998}. For the use in this paper, we only describe its degree one term here:

Let $I$ be the $\Bbbk$-submodule of $R\otimes_{\Bbbk}R$ spanned by
\begin{equation*}
a\otimes b+(-1)^{|a||b|}b\otimes a, \quad
a\otimes b+\bar{a}\otimes\bar{b},\text{ and }
(-1)^{|a||c|}ab\otimes c+(-1)^{|b||a|}bc\otimes a+(-1)^{|c||b|}ca\otimes b,
\end{equation*}
for homogenous $a,b,c\in R$. Let $\langle R,R\rangle:=(R\otimes_{\Bbbk}R)/I$ and $\langle a,b\rangle=a\otimes b+I$, then \textit{the first $\mathbb{Z}/2\mathbb{Z}$-graded dihedral homology of $(R,{}^-)$} is 
\begin{equation}
\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-):=\left\{\sum\limits_{i}\langle a_i,b_i\rangle\middle|\sum\limits_{i}\overline{[a_i,b_i]}=-\sum\limits_{i}[a_i,b_i] \right\}.
\label{eq:HD1}
\end{equation}

\begin{proposition}
\label{prop:stp_ct}
Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution, $m\geqslant3$, and $\psi:\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{p}_m(R,{}^-)$ the canonical epimorphism (\ref{eq:can_hom}). Then
\begin{equation}
\ker\psi\cong\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)\label{eq:ker_psi}
\end{equation}
as $\Bbbk$-modules, where $\rho$ is the $\Bbbk$-linear map (\ref{eq:rho}) and ${}^-\circ\rho$ is also a superinvolution on $R$.
\end{proposition}

In order to prove this proposition, we need a few lemmas:

\begin{lemma}
\label{lem:ht_wd}
The elements $\mathbf{h}_{ij}(a,b)=[\mathbf{f}_{ij}(a),\mathbf{g}_{ji}(b)]\in\mathfrak{stp}_m(R,{}^-)$ satisfy
\begin{align}
\mathbf{h}_{1i}(a,b)-(-1)^{|a||b|}\mathbf{h}_{1i}(1,ba)
&=\mathbf{h}_{1k}(a,b)-(-1)^{|a||b|}\mathbf{h}_{1k}(1,ba),
\label{eq:h_wd}\\
\mathbf{h}_{i1}(1,a)+\mathbf{h}_{1j}(1,a)-\mathbf{h}_{ij}(1,a)
&=\mathbf{h}_{k1}(1,a)+\mathbf{h}_{1l}(1,a)-\mathbf{h}_{kl}(1,a).
\label{eq:t_wd}
\end{align}
for homogenous $a,b\in R$ and $2\leqslant i,j,k,l\leqslant m$ with $i\neq j$ and $k\neq l$.
\end{lemma}
\begin{proof}
Observing that the equality (\ref{eq:h_wd}) is trivial when $i=k$, we assume that $2\leqslant i\neq k\leqslant m$.
\begin{align*}
&\mathbf{h}_{1k}(a,b)-(-1)^{|a||b|}\mathbf{h}_{1k}(1,ba)\\
&=[\mathbf{f}_{1k}(a),\mathbf{g}_{k1}(b)]
-(-1)^{|a||b|}[\mathbf{f}_{1k}(1),\mathbf{g}_{k1}(ba)]\\
&=-[[\mathbf{f}_{1i}(a),\mathbf{t}_{ki}(1)],\mathbf{g}_{k1}(b)]
+(-1)^{|a||b|}[[\mathbf{f}_{1i}(1),\mathbf{t}_{ki}(1)],\mathbf{g}_{k1}(ba)]\\
&=-[[\mathbf{f}_{1i}(a),\mathbf{g}_{k1}(b)],\mathbf{t}_{ki}(1)]
-[\mathbf{f}_{1i}(a),[\mathbf{t}_{ki}(1),\mathbf{g}_{k1}(b)]]\\
&\quad+(-1)^{|a||b|}[[\mathbf{f}_{1i}(1),\mathbf{g}_{k1}(ba)],\mathbf{t}_{ki}(1)]
+(-1)^{|a||b|}[\mathbf{f}_{1i}(1),[\mathbf{t}_{ki}(1),\mathbf{g}_{k1}(ba)]]\\
&=(-1)^{|a|+|b|}[\mathbf{t}_{ik}(\bar{a}\bar{b}),\mathbf{t}_{ki}(1)]
+[\mathbf{f}_{1i}(a),\mathbf{g}_{i1}(b)]\\
&\quad-(-1)^{|a|+|b|}[\mathbf{t}_{ik}(\bar{a}\bar{b}),\mathbf{t}_{ki}(1)]
-(-1)^{|a||b|}[\mathbf{f}_{1i}(1),\mathbf{g}_{i1}(ba)]\\
&=[\mathbf{f}_{1i}(a),\mathbf{g}_{i1}(b)]-(-1)^{|a||b|}[\mathbf{f}_{1i}(1),\mathbf{g}_{i1}(ba)]\\
&=\mathbf{h}_{1i}(a,b)-(-1)^{|a||b|}\mathbf{h}_{1i}(1,ba).
\end{align*}
It yields the equality (\ref{eq:h_wd}).

For the equality (\ref{eq:t_wd}), we first consider the case where $2\leqslant i\neq k\leqslant m$. 
We have
\begin{align*}
[\mathbf{t}_{1i}(1),\mathbf{t}_{i1}(a)]
&=[[\mathbf{t}_{1k}(1),\mathbf{t}_{ki}(1)],\mathbf{t}_{i1}(a)]
=[\mathbf{t}_{ki}(1),\mathbf{t}_{ik}(a)]+[\mathbf{t}_{1k}(1),\mathbf{t}_{k1}(a)].
\end{align*}
On the other hand, it follows from $\boldsymbol{t}_{ij}(a)=[\boldsymbol{f}_{ik}(a),\boldsymbol{g}_{kj}(1)]$ for distinct $i,j,k$ that
\begin{equation}
[\boldsymbol{t}_{ij}(a),\boldsymbol{t}_{ji}(b)]
=\boldsymbol{h}_{ik}(a,b)
-(-1)^{|a||b|}\boldsymbol{h}_{ik}(1,ba).
\label{eq:stp_tth}
\end{equation}
Hence,
$$\mathbf{h}_{1j}(1,a)-\mathbf{h}_{ij}(1,a)
=\mathbf{h}_{k1}(1,a)-\mathbf{h}_{i1}(1,a)
+\mathbf{h}_{1l}(1,a)-\mathbf{h}_{kl}(1,a),$$
i.e., (\ref{eq:t_wd}) holds when $2\leqslant i\neq k\leqslant m$.

For $i=k$, the equality (\ref{eq:t_wd}) is reduced to
$$\mathbf{h}_{1j}(1,a)-\mathbf{h}_{ij}(1,a)=\mathbf{h}_{1l}(1,a)-\mathbf{h}_{il}(1,a),$$
for distinct $i,j,l\in\{2,\ldots,m\}$, whose both sides are equal to $[\mathbf{t}_{1i}(1),\mathbf{t}_{i1}(a)]$ by (\ref{eq:stp_tth}). 
\end{proof}

Lemma~\ref{lem:ht_wd} ensures that
\begin{equation}
\boldsymbol{\lambda}(a,b):=\mathbf{h}_{1i}(a,b)-(-1)^{|a||b|}\mathbf{h}_{1i}(1,ba)\in\mathfrak{stp}_m(R,{}^-)\label{eq:h_def}
\end{equation}
is independent of $2\leqslant i\leqslant m$, and
\begin{equation}
\boldsymbol{\mu}(a):=\mathbf{h}_{i1}(1,a)+\mathbf{h}_{1j}(1,a)-\mathbf{h}_{ij}(1,a)\in\mathfrak{stp}_m(R,{}^-)\label{eq:t_def}
\end{equation}
is independent of $2\leqslant i\neq j\leqslant m$. Moreover, they satisfy the following properties:


\begin{lemma}
\label{lem:ht_rln}
For homogeneous $a,b,c\in R$, we have
\begin{enumerate}
\item $(-1)^{|a||c|}\boldsymbol{\lambda}(ab,c)+(-1)^{|b||a|}\boldsymbol{\lambda}(bc,a)+(-1)^{|c||b|}\boldsymbol{\lambda}(ca,b)=0$,
\item $\boldsymbol{\lambda}(a,1)=\boldsymbol{\lambda}(1,b)=0$,
\item $\boldsymbol{\lambda}(a,b)=-(-1)^{|a||b|}\boldsymbol{\lambda}(b,a)$,
\item $\boldsymbol{\mu}(\bar{a})=-\boldsymbol{\mu}(\rho(a))$.
\end{enumerate}
\end{lemma}
\begin{proof}
We claim that
\begin{equation}
\boldsymbol{\lambda}(a,b)=[\mathbf{t}_{1j}(a),\mathbf{t}_{j1}(b)]
-(-1)^{|a||b|}[\mathbf{t}_{1j}(1),\mathbf{t}_{j1}(ba)],\label{eq:h_lmd}
\end{equation}
for $a,b\in R$ and $j\neq 1$. Indeed, we deduce from (\ref{eq:stp_tth}) that
\begin{align*}
[\mathbf{t}_{1j}(a),\mathbf{t}_{j1}(b)]
&=\mathbf{h}_{1k}(a,b)-(-1)^{|a||b|}\mathbf{h}_{jk}(1,ba),\\
[\mathbf{t}_{1j}(1),\mathbf{t}_{j1}(ba)]
&=\mathbf{h}_{1k}(1,ba)-\mathbf{h}_{jk}(1,ba),
\end{align*}
for some $k\neq 1,j$. Hence, (\ref{eq:h_lmd}) holds.
\medskip

Secondly, (\ref{STP03}) and the Jacobi identity yield that
\begin{equation}
(-1)^{|a||c|}[\mathbf{t}_{ij}(ab),\mathbf{t}_{ji}(c)]
+(-1)^{|a||b|}[\mathbf{t}_{ki}(bc),\mathbf{t}_{ik}(a)]
+(-1)^{|b||c|}[\mathbf{t}_{jk}(ca),\mathbf{t}_{kj}(b)]
=0,
\label{eq:lmd_Jac}
\end{equation}
for distinct $i,j,k$.
\medskip

(i) We deduce from (\ref{eq:h_lmd}) and (\ref{eq:lmd_Jac}) that
\begin{align*}
&(-1)^{|c||a|}\boldsymbol{\lambda}(ab,c)\\
&=(-1)^{|c||a|}[\mathbf{t}_{1j}(ab),\mathbf{t}_{j1}(c)]
-(-1)^{|b||c|}[\mathbf{t}_{1j}(1),\mathbf{t}_{j1}(cab)]\\
&=(-1)^{|c||a|}[\mathbf{t}_{1j}(ab),\mathbf{t}_{j1}(c)]
-(-1)^{|b||c|}[\mathbf{t}_{1j}(1),[\mathbf{t}_{ji}(ca),\mathbf{t}_{i1}(b)]]\\
&=(-1)^{|c||a|}[\mathbf{t}_{1j}(ab),\mathbf{t}_{j1}(c)]
+(-1)^{|b||c|}[\mathbf{t}_{ji}(ca),\mathbf{t}_{ij}(b)]
-(-1)^{|b||c|}[\mathbf{t}_{1i}(ca),\mathbf{t}_{i1}(b)]\\
&=-(-1)^{|a||b|}[\mathbf{t}_{i1}(bc),\mathbf{t}_{1i}(a)]
-(-1)^{|b||c|}[\mathbf{t}_{1i}(ca),\mathbf{t}_{i1}(b)].
\end{align*}
On the other hand, we also compute that
\begin{align*}
\boldsymbol{\lambda}(ab,c)&=
[\mathbf{t}_{1j}(ab),\mathbf{t}_{j1}(c)]
-(-1)^{(|a|+|b|)|c|}[\mathbf{t}_{1j}(1),\mathbf{t}_{j1}(cab)]\\
&=[\mathbf{t}_{1j}(ab),\mathbf{t}_{j1}(c)]
-(-1)^{(|a|+|b|)|c|}[\mathbf{t}_{1j}(1),[\mathbf{t}_{ji}(c),\mathbf{t}_{i1}(ab)]]\\
&=[\mathbf{t}_{1j}(ab),\mathbf{t}_{j1}(c)]
+[\mathbf{t}_{i1}(ab),\mathbf{t}_{1i}(c)]
-[\mathbf{t}_{ij}(ab),\mathbf{t}_{ji}(c)].
\end{align*}
It follows from (\ref{eq:lmd_Jac}) again that
\begin{align*}
&(-1)^{|a||c|}\boldsymbol{\lambda}(ab,c)+(-1)^{|a||b|}\boldsymbol{\lambda}(bc,a)\\
&=-(-1)^{|a||b|}[\mathbf{t}_{i1}(bc),\mathbf{t}_{1i}(a)]
-(-1)^{|b||c|}[\mathbf{t}_{1i}(ca),\mathbf{t}_{i1}(b)]\\
&\quad-(-1)^{|b||c|}[\mathbf{t}_{j1}(ca),\mathbf{t}_{1j}(b)]
-(-1)^{|c||a|}[\mathbf{t}_{1j}(ab),\mathbf{t}_{j1}(c)]\\
&=(-1)^{|b||c|}([\mathbf{t}_{ji}(ca),\mathbf{t}_{ij}(b)]
-[\mathbf{t}_{1i}(ca),\mathbf{t}_{i1}(b)]
-[\mathbf{t}_{j1}(ca),\mathbf{t}_{1j}(b)])\\
&=-(-1)^{|b||c|}\boldsymbol{\lambda}(ca,b).
\end{align*}
This proves (i).
\medskip

(ii) $\boldsymbol{\lambda}(1,b)=0$ is obvious. Taking $b=c=1$ in (i), we obtain
$$\boldsymbol{\lambda}(a,1)+\boldsymbol{\lambda}(1,a)+\boldsymbol{\lambda}(a,1)=0$$
which implies $\boldsymbol{\lambda}(a,1)=0$ since $\frac{1}{2}\in\Bbbk$ and $\boldsymbol{\lambda}(1,a)=0$.
\medskip

(iii) follows from (i) by taking $c=1$.
\medskip

(iv) follows from the equality $\mathbf{h}_{ij}(a,b)=-\mathbf{h}_{ji}(\rho(\bar{a}),\rho(\bar{b}))$.
\end{proof}

\begin{lemma}
\label{lem:diag}
Every element $x\in\mathfrak{stp}_{m}^0(R,{}^-)$ can be written as
\begin{equation}
x=\sum\limits_{i\in I_x}\boldsymbol{\lambda}(a_i,b_i)+\boldsymbol{\mu}(c)+\sum\limits_{j=2}^m\mathbf{h}_{1j}(1,d_j),
\label{eq:diag_ele}
\end{equation}
where $I_x$ is a finite index set, $a_i, b_i, c, d_j\in R$ for $i\in I_x$ and $j=2,\ldots,m$. Moreover,
\begin{equation}
\boldsymbol{\mu}([a,b])=\boldsymbol{\lambda}(a,b)+\boldsymbol{\lambda}(\rho(\bar{a}),\rho(\bar{b}))
\label{eq:t_h}
\end{equation}
for homogeneous $a,b\in R$.
\end{lemma}
\begin{proof}
Recall that $\mathfrak{stp}_{m}^0(R,{}^-)$ is spanned by $\mathbf{h}_{ij}(a,b)$ for homogeneous $a,b\in R$ and $1\leqslant i\neq j\leqslant m$. It suffices to show that every $\mathbf{h}_{ij}(a,b)$ can be written in the form of (\ref{eq:diag_ele}).

We first observe that
$$-\mathbf{h}_{i1}(\rho(\bar{a}),\rho(\bar{b}))=\mathbf{h}_{1i}(a,b)=\mathbf{h}(a,b)+(-1)^{|a||b|}\mathbf{h}_{1i}(1,ba)$$
for $a,b\in R$ and $i=2,\ldots,m$.

If $2\leqslant i\neq j\leqslant m$, then
\begin{align*}
\mathbf{h}_{ij}(a,b)&=[\mathbf{f}_{ij}(a),\mathbf{g}_{ji}(b)]\\
&=[[\mathbf{t}_{i1}(1),\mathbf{f}_{1j}(a)],\mathbf{g}_{ji}(b)]\\
&=[\mathbf{f}_{1j}(a),\mathbf{g}_{j1}(b)]+[\mathbf{t}_{i1}(1),\mathbf{t}_{1i}(ab)]\\
&=\mathbf{h}_{1j}(a,b)+[\mathbf{t}_{i1}(1),\mathbf{t}_{1i}(ab)]\\
&=\mathbf{h}_{1j}(a,b)+\mathbf{h}_{ij}(1,ab)-\mathbf{h}_{1j}(1,ab)\\
&=\mathbf{h}_{1j}(a,b)-\boldsymbol{\mu}(ab)+\mathbf{h}_{i1}(1,ab),
\end{align*}
which is of the form (\ref{eq:diag_ele}) since $\mathbf{h}_{1j}(a,b)$ and $\mathbf{h}_{i1}(1,ab)$ have already been of the form (\ref{eq:diag_ele}).
\medskip

Now, we prove the equality (\ref{eq:t_h}). For $2\leqslant i\neq j\leqslant m$, we have already obtained that
\begin{align*}
\boldsymbol{\mu}(ab)&=\mathbf{h}_{1j}(a,b)+\mathbf{h}_{i1}(1,ab)-\mathbf{h}_{ij}(a,b)\\
&=\mathbf{h}_{1j}(a,b)-(-1)^{|a||b|}\mathbf{h}_{1i}(1,\rho(\bar{b})\rho(\bar{a}))-\mathbf{h}_{ij}(a,b).
\end{align*}
It follows from Lemma~\ref{lem:ht_rln} that
\begin{align*}
\boldsymbol{\mu}(ba)
&=-(-1)^{|a||b|}\boldsymbol{\mu}(\rho(\bar{a})\rho(\bar{b}))\\
&=-(-1)^{|a||b|}
(\mathbf{h}_{1i}(\rho(\bar{a}),\rho(\bar{b}))-(-1)^{|a||b|}\mathbf{h}_{1j}(1,ba)-\mathbf{h}_{ji}(\rho(\bar{a}),\rho(\bar{b})))\\
&=-(-1)^{|a||b|}\mathbf{h}_{1i}(\rho(\bar{a}),\rho(\bar{b}))+\mathbf{h}_{1j}(1,ba)-(-1)^{|a||b|}\mathbf{h}_{ij}(a,b).
\end{align*}
Hence,
\begin{align*}
\boldsymbol{\mu}([a,b])&=\boldsymbol{\mu}(ab)-(-1)^{|a||b|}\boldsymbol{\mu}(ba)\\
&=\mathbf{h}_{1j}(a,b)-(-1)^{|a||b|}\mathbf{h}_{1j}(1,ba)\\
&\quad+\mathbf{h}_{1i}(\rho(\bar{a}),\rho(\bar{b}))-(-1)^{|a||b|}\mathbf{h}_{1i}(1,\rho(\bar{b})\rho(\bar{a}))\\
&=\boldsymbol{\lambda}(a,b)+\boldsymbol{\lambda}(\rho(\bar{a}),\rho(\bar{b})).
\end{align*}
This completes the proof.
\end{proof}

Now, we may proceed to prove Proposition~\ref{prop:stp_ct},

\begin{proof}[Proof of Proposition~\ref{prop:stp_ct}]
Recall (\ref{eq:HD1}) that
$$\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)=\left\{\sum_i\langle a_i,b_i\rangle\in\langle R,R\rangle\middle|\sum_i \overline{[a_i,b_i]}=-\sum_i[\rho(a_i),\rho(b_i)]\right\},$$
where $\langle R,R\rangle=(R\otimes_{\Bbbk}R)/I$ and $I$ is the $\Bbbk$-submodule of $R\otimes_{\Bbbk}R$ spanned by $a\otimes b+(-1)^{|a||b|}b\otimes a$, $a\otimes b+\rho(\bar{a})\otimes\rho(\bar{b})$ and $(-1)^{|a||c|}ab\otimes c+(-1)^{|b||a|}bc\otimes a+(-1)^{|c||b|}ca\otimes b$ for homogeneous $a,b,c\in R$.

By Lemmas~\ref{lem:ht_rln} and~\ref{lem:diag}, there exists a well-defined $\Bbbk$--linear map
\begin{align*}
\eta:\langle R,R\rangle&\rightarrow\mathfrak{stp}_m(R,{}^-),\\
\langle a,b\rangle&\mapsto\boldsymbol{\lambda}(a,b)-\frac{1}{2}\boldsymbol{\mu}([a,b])
=\frac{1}{2}(\boldsymbol{\lambda}(a,b)-\boldsymbol{\lambda}(\rho(\bar{a}),\rho(\bar{b}))).
\end{align*}
We will prove that its restriction on $\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)$ is an isomorphism of $\Bbbk$-modules onto $\ker\psi$.
\medskip

We claim that $\eta\left(\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)\right)\subseteq\ker\psi$. For $\sum_i\langle a_i,b_i\rangle\in\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)$, we have 
$$\sum_i\overline{[a_i,b_i]}=-\sum_i[\rho(a_i),\rho(b_i)].$$ 
Hence,
$$\psi(\eta(\textstyle{\sum}_i\langle a_i,b_i\rangle))
=\frac{1}{2}\textstyle{\sum}_i\psi(\boldsymbol{\lambda}(a_i,b_i)-\boldsymbol{\lambda}(\rho(\bar{a}_i),\rho(\bar{b}_i)))
=\frac{1}{2}\textstyle{\sum}_ie_{11}([a_i,b_i]-[\rho(\bar{a}_i),\rho(\bar{b}_i)])
=0.$$

Conversely, let $\boldsymbol{x}\in\ker{\psi}\subseteq\mathfrak{stp}_m^0(R,{}^-)$ (see Proposition~\ref{prop:stp_ce}). It follows from Lemma~\ref{lem:diag} that
$$\boldsymbol{x}=\sum_{i\in I_{\boldsymbol{x}}}\boldsymbol{\lambda}(a_i,b_i)+\boldsymbol{\mu}(c)+\sum\limits_{j=2}^m\mathbf{h}_{1j}(1,d_j),$$
where $I_{\boldsymbol{x}}$ is a finite index set, $a_i,b_i,c,d_j\in R$ for $i\in I_{\boldsymbol{x}}$ and $j=2,\ldots,m$, and hence,
$$0=\psi(\boldsymbol{x})=\sum_{i\in I_{\boldsymbol{x}}}e_{11}([a_i,b_i])+e_{11}(c_{(-)})+\sum\limits_{j=2}^m(e_{11}(d_j)-e_{jj}(\bar{d}_j)),$$
which implies that $d_j=0$ for $j=2,\ldots,m$ and
$$\sum\limits_{i\in I_{\boldsymbol{x}}}[a_i,b_i]=-c_{(-)}\in R_{(-)}.$$
Since $\frac{1}{2}\in\Bbbk$ and $\boldsymbol{\mu}(\bar{a})=-\boldsymbol{\mu}(\rho(a))$ for homogeneous $a\in R$, we have
$$\boldsymbol{\mu}(c)=\frac{1}{2}\boldsymbol{\mu}(c_{(-)})=-\frac{1}{2}\sum_{i\in I_{\boldsymbol{x}}}\boldsymbol{\mu}([a_i,b_i]).$$
Hence, we conclude that
$$x=\sum_{i\in I_{\boldsymbol{x}}}(\boldsymbol{\lambda}(a_i,b_i)-\frac{1}{2}\boldsymbol{\mu}([a_i,b_i]))=\sum_{i\in I_x}\eta(\langle a_i,b_i\rangle),$$
and $\sum_{i\in I_x}\langle a_i,b_i\rangle\in\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)$.
\medskip

It remains to show the injectivity of $\eta$. Define a $\Bbbk$--bilinear map
$$\alpha:\mathfrak{gl}_{m|m}(R)\times\mathfrak{gl}_{m|m}(R)\rightarrow\langle R, R\rangle$$
by
$$\alpha(e_{ij}(a),e_{kl}(b))=\delta_{jk}\delta_{il}(-1)^{|i|(|i|+|a|+|b|)}\langle a,b\rangle$$
for homogeneous $a,b\in R$ and $1\leqslant i,j\leqslant 2m$, where $|i|$ is the parity of $i$ given by (\ref{eq:parity_index}). It is verified that $\alpha$ is a $2$-cocycle on the Lie superalgebra $\mathfrak{gl}_{m|m}(R)$.

Now, the restriction of $\alpha$ on $\mathfrak{p}_m(R,{}^-)\times\mathfrak{p}_m(R,{}^-)$ is a $2$-cocycle on $\mathfrak{p}_m(R,{}^-)\subseteq\mathfrak{gl}_{m|m}(R)$. Hence, there is a Lie superalgebra structure on $\mathfrak{p}_m(R,{}^-)\oplus\langle R,R\rangle$:
$$[x\oplus c, y\oplus c']=[x,y]\otimes \alpha(x,y),\quad x,y\in\mathfrak{p}_m(R,{}^-)\text{ and }c,c'\in\langle R,R\rangle.$$
Observing that ${t}_{ij}(a)\oplus0$, ${f}_{ij}(a)\oplus0$ and ${g}_{ij}(a)\oplus0\in\mathfrak{p}_m(R,{}^-)\oplus\langle R,R\rangle$ satisfy all relations (\ref{STP00})-(\ref{STP12}), there is a canonical homomorphism of Lie superalgebras
$$\phi:\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{p}_m(R,{}^-)\oplus\langle R,R\rangle$$
such that
$$\phi(\mathbf{t}_{ij}(a))={t}_{ij}(a)\oplus0,
\quad\phi(\mathbf{f}_{ij}(a))={f}_{ij}(a)\oplus0,
\quad\phi(\mathbf{g}_{ij}(a))={g}_{ij}(a)\oplus0,$$
for $a\in R$ and $1\leqslant i\neq j\leqslant m$. We now compute that
\begin{align*}
\phi(\mathbf{h}_{ij}(a,b))
&=\phi([\mathbf{f}_{ij}(a),\mathbf{g}_{ji}(b)])=[{f}_{ij}(a)\oplus0,{g}_{ji}(b)\oplus0]\\
&=[{f}_{ij}(a),{g}_{ji}(b)]\oplus \alpha({f}_{ij}(a),{g}_{ji}(b))\\
&=({t}_{ii}(ab)-{t}_{jj}(\rho(\bar{a})\rho(\bar{b})))\oplus(\langle a,b\rangle-\langle\rho(\bar{a}),\rho(\bar{b})\rangle)\\
&=({t}_{ii}(ab)-{t}_{jj}(\rho(\bar{a})\rho(\bar{b})))\oplus2\langle a,b\rangle,
\end{align*}
which implies that
\begin{align*}
\phi(\boldsymbol{\lambda}(a,b))&=\phi(\mathbf{h}_{1i}(a,b)-(-1)^{|a||b|}\mathbf{h}_{1i}(1,ba))\\
&={t}_{11}([a,b])\oplus(2\langle a,b\rangle-2(-1)^{|a||b|}\langle1,ba\rangle).
\end{align*}
Since $\langle1,a\rangle=-\langle a,1\rangle$ and $\langle a,1\rangle+\langle a,1\rangle+\langle 1,a\rangle=0$, we obtain that $\langle1,a\rangle=0$. Hence,
$$\phi(\boldsymbol{\lambda}(a,b))={t}_{11}([a,b])\oplus2\langle a,b\rangle.$$
Since $\frac{1}{2}\in\Bbbk$, 
\begin{align*}
\phi(\eta(\langle a,b\rangle))&=\frac{1}{2}(\phi(\boldsymbol{\lambda}(a,b))-\phi(\boldsymbol{\lambda}(\rho(\bar{a}),\rho(\bar{b}))))\\
&=\frac{1}{2}({t}_{11}([a,b])\oplus2\langle a,b\rangle_{\mathsf{d}}^{(+)}-{t}_{11}([\rho(\bar{a}),\rho(\bar{b})])\oplus2\langle\rho(\bar{a}),\rho(\bar{b})\rangle_{\mathsf{d}}^{(+)})\\
&=\frac{1}{2}({t}_{11}([a,b]-[\rho(\bar{a}),\rho(\bar{b})])\oplus(\langle a,b\rangle_{\mathsf{d}}^{(+)}-\langle\rho(\bar{a}),\rho(\bar{b})\rangle_{\mathsf{d}}^{(+)}))\\
&=\frac{1}{2}{t}_{11}([a,b]-[\rho(\bar{a}),\rho(\bar{b})])\oplus2\langle a,b\rangle_{\mathsf{d}}^{(+)},
\end{align*}
which shows that $\eta$ is injective and completes the proof.
\end{proof}

\section{The universality of the central extension $\psi$}
\label{sec:uce}
 
 It is shown in Section~\ref{sec:stp} that the canonical homomorphism $\psi:\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{p}_m(R,{}^-)$ is a central extension, whose kernel has been explicitly characterized in Section~\ref{sec:kernel}. In this section, we will prove that the central extension $\psi$ is universal for $m\geqslant5$ and thus obtain a precise description of the second homology $\mathfrak{p}_m(R,{}^-)$.

A necessary condition for the universality of $\psi$ is the perfectness $\mathfrak{stp}_m(R,{}^-)$, which can be easily observed from the defining relations (\ref{STP03}), (\ref{STP05}) and (\ref{STP07}). Next, we proceed to prove the universality of $\psi$.

Let $\varphi:\mathfrak{E}\rightarrow\mathfrak{p}_m(R,{}^-)$ be an arbitrary central extension of $\mathfrak{p}_m(R,{}^-)$ with $m\geqslant3$. For $a\in R$ and $1\leqslant i\neq j\leqslant m$, we pick
$$\hat{t}_{ij}(a)\in\varphi^{-1}({t}_{ij}(a)),
\quad\hat{f}_{ij}(a)\in\varphi^{-1}({f}_{ij}(a)),\text{ and }\hat{g}_{ij}(a)\in\varphi^{-1}({g}_{ij}(a)).$$
Obviously, the element  $[\hat{x},\hat{y}]\in\mathfrak{E}$ is independent the choice of the representatives $\hat{x}\in\varphi^{-1}(x)$ and $\hat{y}\in\varphi^{-1}(y)$ for $x,y\in\mathfrak{p}_m(R,{}^-)$.  Moreover, we have the following lemma:

\begin{lemma}
\label{lem:ele_tilde}
In the Lie superalgebra $\mathfrak{E}$, the following equalities hold:
\begin{enumerate}
\item $[\hat{f}_{ik}(a),\hat{g}_{kj}(b)]=[\hat{f}_{il}(a),\hat{g}_{lj}(b)]$,
\item $[\hat{t}_{ik}(a),\hat{f}_{kj}(b)]=[\hat{t}_{il}(a),\hat{f}_{lj}(b)]$,
\item $[\hat{g}_{ik}(a),\hat{t}_{kj}(b)]=[\hat{g}_{il}(a),\hat{t}_{lj}(b)]$.
\end{enumerate}
for $a,b\in R$ and distinct $i,j,k,l$.
\end{lemma}
\begin{proof}
(i) Since $[\hat{f}_{ik}(a),\hat{t}_{lk}(1)]+\hat{f}_{il}(a)\in\ker\varphi$ that is included in the center of $\mathfrak{E}$, we deduce
\begin{align*}
[\hat{f}_{il}(a),\hat{g}_{lj}(b)]
&=-[[\hat{f}_{ik}(a),\hat{t}_{lk}(1)],\hat{g}_{lj}(b)]\\
&=-[[\hat{f}_{ik}(a),\hat{g}_{lj}(b)],\hat{t}_{lk}(1)]-[\hat{f}_{ik}(a),[\hat{t}_{lk}(1),\hat{g}_{lj}(b)]]\\
&=0+[\hat{f}_{ik}(a),\hat{g}_{kj}(b)],
\end{align*}
which shows (i). The equalities (ii) and (iii) follows similarly.
\end{proof}

According to Lemma~\ref{lem:ele_tilde}, we define for each pair $(i,j)$ with $1\leqslant i\neq j\leqslant m$ that:
\begin{equation}
\tilde{t}_{ij}(a):=[\hat{f}_{ik}(a),\hat{g}_{kj}(1)],\quad
\tilde{f}_{ij}(a):=[\hat{t}_{ik}(1),\hat{f}_{kj}(a)],\text{ and }
\tilde{g}_{ij}(a):=[\hat{g}_{ik}(a),\hat{t}_{kj}(1)],\label{eq:tilde}
\end{equation}
where $a\in R$ and $1\leqslant k\leqslant m$ is an arbitrary integer such that $k\neq i,j$.

\begin{lemma}
\label{lem:tilde_rln}
Suppose $m\geqslant3$. Let $\tilde{t}_{ij}(a),\tilde{f}_{ij}(a)$ and $\tilde{g}_{ij}(a)$ be the elements of $\mathfrak{E}$ given in (\ref{eq:tilde}), where $a\in R$ and $1\leqslant i\neq j\leqslant m$. Then they satisfy all relations (\ref{STP00})-(\ref{STP12}) except (\ref{STP04}) and (\ref{STP10}). Moreover, for $a,b\in R$, we have
\begin{align}
\/[\tilde{t}_{ik}(a),\tilde{t}_{jk}(b)]&=0,&&\text{if }i,j,k\text{ are distinct,}\tag{STP04a}\label{STP04a}\\
\/[\tilde{t}_{ij}(a),\tilde{t}_{kl}(b)]&=0,&&\text{if }i,j,k,l\text{ are distinct,}\tag{STP04b}\label{STP04b}\\
\/[\tilde{g}_{ij}(a),\tilde{g}_{ij}(b)]&=0,&&\text{if }i\neq j,\tag{STP10a}\label{STP10a}\\
\/[\tilde{g}_{ij}(a),\tilde{g}_{jk}(b)]&=0,&&\text{if }i,j,k\text{ are distinct.}\tag{STP10b}\label{STP10b}
\end{align}
\end{lemma}

\begin{proof}
The $\Bbbk$-linearity of $\tilde{t}_{ij}$ is obvious since both $\hat{f}_{ik}(ca)$ and $c\hat{f}_{ik}(a)$ are contained in $\varphi^{-1}(f_{ik}(a))$ for $a\in R$ and $c\in\Bbbk$. Similarly, we have the $\Bbbk$-linearity of $\tilde{f}_{ij}$ and $\tilde{g}_{ij}$, which shows (\ref{STP00}).

For (\ref{STP01}), we aim to show $\tilde{f}_{ij}(\bar{a})=\tilde{f}_{ji}(\rho(a))$ for $a\in R$. Note that $m\geqslant3$, we choose $1\leqslant k\leqslant m$ such that $i,j,k$ are distinct and set
\begin{equation}
\tilde{h}_{ik}=[\hat{f}_{ik}(1),\hat{g}_{ki}(1)],\label{eq:h_tilde}
\end{equation}
which is independent of the choice of $k$ and the representatives $\hat{f}_{ik}(1)\in\varphi^{-1}(f_{ik}(1))$ and $\hat{g}_{ki}(1)\in\varphi^{-1}(g_{ki}(1))$. Then we compute that
\begin{align*}
[\tilde{h}_{ik},\tilde{f}_{ij}(a)]=[\tilde{h}_{ik},[\hat{t}_{ik}(1),\hat{f}_{kj}(a)]]
=[[\tilde{h}_{ik},\hat{t}_{ik}(1)],\tilde{f}_{kj}(a)]+[\hat{t}_{ik}(1),[\tilde{h}_{ik},\hat{f}_{kj}(a)]].
\end{align*}
Now, $\varphi(h_{ik})=e_{ii}(1)-e_{kk}(1)-e_{m+i,m+i}(1)+e_{m+k,m+k}(1), \varphi(\hat{t}_{ik}(1))=t_{ik}(1)$ and $\varphi(\hat{f}_{kj}(a))=f_{kj}(a)$. Hence,
$$[\tilde{h}_{ik},\hat{t}_{ik}(1)]\in\varphi^{-1}(2t_{ik}(1)),\text{ and }[\tilde{h}_{ik},\hat{f}_{kj}(a)]\in\varphi^{-1}(-f_{kj}(a)).$$
It yields that $[\tilde{h}_{ik},\tilde{f}_{ij}(a)]=\tilde{f}_{ij}(a)$.

Observing that $f_{ij}(\bar{a})=f_{ji}(\rho(a))$ in $\mathfrak{p}_m(R,{}^-)$, we have $\tilde{f}_{ij}(\bar{a})-\tilde{f}_{ji}(\rho(a))\in\ker\varphi$ which is contained in the center of $\mathfrak{E}$. Applying $\tilde{h}_{ik}$, we obtain that
$$0=[\tilde{h}_{ik},\tilde{f}_{ij}(\bar{a})-\tilde{f}_{ji}(\rho(a))]=\tilde{f}_{ij}(\bar{a})-\tilde{f}_{ji}(\rho(a)).$$
This shows $\tilde{f}_{ij}(a)$ satisfies the relation (\ref{STP01}). All other relations can be verified similarly.
\end{proof}

\begin{remark}The relation (\ref{STP04}) states that
$$[\mathbf{t}_{ij}(a),\mathbf{t}_{kl}(b)]=0,\text{ if }i\neq j\neq k\neq l\neq i,$$
which is equivalent to (\ref{STP04a}), (\ref{STP04b}) and
\begin{equation}
[\mathbf{t}_{ij}(a),\mathbf{t}_{ik}(b)]=0,\text{ if }i,j,k\text{ are distinct.}\tag{STP04c}\label{STP04c}
\end{equation}
By Lemma~\ref{lem:tilde_rln}, the elements $\tilde{t}_{ij}(a)$'s in an arbitrary central extension $\mathfrak{E}$ of $\mathfrak{p}_m(R,{}^-)$ always satisfy (\ref{STP04a}) and (\ref{STP04b}), but do not necessarily satisfy (\ref{STP04c}). Such examples will appear in central extensions of $\mathfrak{p}_3(R,{}^-)$. The similar phenomenon also occurs for the relation (\ref{STP10}).
\end{remark}

\begin{proposition}
\label{thm:pm_uce}
 Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution and $m\geqslant5$. Then $\psi:\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{p}_m(R,{}^-)$ is a universal central extension.
\end{proposition}
\begin{proof}
Let $\varphi:\mathfrak{E}\rightarrow\mathfrak{p}_m(R,{}^-)$ be an arbitrary central extension. Take $\tilde{t}_{ij}(a),\tilde{f}_{ij}(a),\tilde{g}_{ij}(a)\in\mathfrak{E}$ for $a\in R$ and $1\leqslant i\neq j\leqslant m$ as in (\ref{eq:tilde}). Then we have already known from Lemma~\ref{lem:tilde_rln} that they satisfy all relations (\ref{STP00})-(\ref{STP12}) except (\ref{STP04}) and (\ref{STP10}). Now, under the additional assumption that $m\geqslant5$, we will show that these elements also satisfy (\ref{STP04}) and (\ref{STP10}).

For (\ref{STP04}), since (\ref{STP04a}) and (\ref{STP04b}) have already been verified in Lemma~\ref{lem:tilde_rln}, it suffices to show
\begin{equation}
[\tilde{t}_{ij}(a),\tilde{t}_{ik}(b)]=0,\text{ if }i,j,k\text{ are distinct.}\tag{\ref{STP04c}}
\end{equation}
Indeed, we observe that $[\tilde{t}_{ij}(a),\tilde{t}_{ik}(b)]\in\ker\varphi$. Since $m\geqslant5$, we are allowed to choose $1\leqslant l\leqslant m$ such that $l\neq i,j,k$. Applying $\tilde{h}_{lj}$ defined in (\ref{eq:h_tilde}), we obtain that
\begin{align*}
0&=[\tilde{h}_{lj},[\tilde{t}_{ij}(a),\tilde{t}_{ik}(b)]]\\
&=[[\tilde{h}_{lj},\tilde{t}_{ij}(a)],\tilde{t}_{ik}(b)]
+[\tilde{t}_{ij}(a),[\tilde{h}_{lj},\tilde{t}_{ik}(b)]]\\
&=[\tilde{t}_{ij}(a),\tilde{t}_{ik}(b)].
\end{align*}
Then (\ref{STP04c}) follows.

For (\ref{STP10}), we have obtained (\ref{STP10a}) and (\ref{STP10b}) in Lemma~\ref{lem:tilde_rln}. It suffices to show
\begin{equation}
[\tilde{g}_{ij}(a),\tilde{g}_{kl}(b)]=0,\text{ if }i,j,k,l\text{ are distinct}.\tag{STP10c}\label{STP10c}
\end{equation}
Since $m\geqslant5$, we are permitted to choose $k'$ such $k'\neq i,j,k,l$. Hence,
\begin{align*}
[\tilde{g}_{ij}(a),\tilde{g}_{kl}(b)]
&=[\tilde{g}_{ij}(a),[\hat{g}_{kk'}(b),\hat{t}_{k'l}(1)]]\\
&=[[\tilde{g}_{ij}(a),\hat{g}_{kk'}(b)],\hat{t}_{k'l}(1)]+(-1)^{(1+|a|)(1+|b|)}[\hat{g}_{kk'}(b),[\tilde{g}_{ij}(a),\hat{t}_{k'l}(1)]]
=0.
\end{align*}
This proves (\ref{STP10}).

In summary, we have shown that the elements $\tilde{t}_{ij}(a),\tilde{f}_{ij}(a),\tilde{g}_{ij}(a)\in\mathfrak{E}$ with $a\in R$ and $1\leqslant i\neq j\leqslant m$ satisfy all relations (\ref{STP00})-(\ref{STP12}). Hence, there is a homomorphism of Lie superalgebras
$$\varphi':\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{E}$$
such that
$$\varphi'(\mathbf{t}_{ij}(a))=\tilde{t}_{ij}(a),\quad
\varphi'(\mathbf{f}_{ij}(a))=\tilde{f}_{ij}(a),\quad
\varphi'(\mathbf{g}_{ij}(a))=\tilde{g}_{ij}(a),$$
for $a\in R$ and $1\leqslant i\neq j\leqslant m$, i.e., $\varphi\circ\varphi'=\psi$.

To show the uniqueness of $\varphi'$, we let $\tilde{\varphi}':\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{E}$ be another homomorphism of Lie superalgebras such that $\varphi\circ\tilde{\varphi}'=\psi$. Then
$$\tilde{\varphi}'(\mathbf{t}_{ij}(a))\in\varphi^{-1}({t}_{ij}(a)),\quad
\tilde{\varphi}'(\mathbf{f}_{ij}(a))\in\varphi^{-1}({f}_{ij}(a)),\text{ and }
\tilde{\varphi}'(\mathbf{g}_{ij}(a))\in\varphi^{-1}({g}_{ij}(a)),$$
for $1\leqslant i\neq j\leqslant m$ and homogeneous $a\in R$. Note that
$$\mathbf{t}_{ij}(a)=[\mathbf{t}_{ik}(a),\mathbf{t}_{kj}(1)],\quad
\mathbf{f}_{ij}(a)=[\mathbf{t}_{ik}(1),\mathbf{g}_{kj}(a)],\text{ and }
\mathbf{g}_{ij}(a)=[\mathbf{g}_{ik}(a),\mathbf{t}_{kj}(1)],$$
for $a\in R$ and distinct $i,j,k$, we deduce that
$$\tilde{\varphi}'(\mathbf{t}_{ij}(a))=\tilde{t}_{ij}(a),\quad
\tilde{\varphi}'(\mathbf{f}_{ij}(a))=\tilde{f}_{ij}(a),\text{ and }
\tilde{\varphi}'(\mathbf{g}_{ij}(a))=\tilde{g}_{ij}(a).$$
It yields that $\tilde{\varphi}'=\varphi'$ since $\mathfrak{stp}_m(R,{}^-)$ is generated by $\mathbf{t}_{ij}(a),\mathbf{f}_{ij}(a),\mathbf{g}_{ij}(a)$ with $a\in R$ and  $1\leqslant i\neq j\leqslant m$. Thus, there is a unique homomorphism $\varphi':\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{E}$ such that $\varphi\circ\varphi'=\psi$. Therefore, the central extension $\psi:\mathfrak{stp}_m(R,{}^-)\rightarrow\mathfrak{p}_m(R,{}^-)$ is universal.
\end{proof}

\begin{theorem}
\label{thm:hml_pm}
Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution and $m\geqslant5$. Then
$$\mathrm{H}_2(\mathfrak{p}_m(R,{}^-))=\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)$$
where $\rho$ is the $\Bbbk$-linear map given in (\ref{eq:rho}).
\end{theorem}

\begin{proof}
The second homology of $\mathfrak{p}_m(R,{}^-)$ can be identified with the kernel of its universal central extension $\psi$, which has been shown in Proposition~\ref{prop:stp_ct} to be $\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)$.
\end{proof}

\begin{remark}
\label{rmk:hml_com}
If $R$ is super-commutative, then one deduce from the definition that $\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,\mathrm{id})=0$. Hence,
$$\mathrm{H}_2(\mathfrak{p}_m(\Bbbk)\otimes_{\Bbbk}R)
\cong\mathrm{H}_2(\mathfrak{p}_m(R,\rho))
\cong\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,\mathrm{id})
=0.$$
This recovers the results about the second homology of $\mathfrak{p}_m(\Bbbk)\otimes_{\Bbbk}R$ given in \cite{IoharaKoga2005} and \cite{MartinezZelmanov2003}.
\end{remark}
\medskip

In the special case where $(R,{}^-)=(S\oplus S^{\mathrm{op}},\mathrm{ex})$ for a unital associative superalgebra $S$, we have

\begin{corollary}
\label{cor:slmm}
Let $S$ be an arbitrary unital associative superalgebra and $m\geqslant5$. Then
$$\mathrm{H}_2(\mathfrak{sl}_{m|m}(S))\cong\fourIdx{}{+}{}{1}{\mathrm{HD}}(S\oplus S^{\mathrm{op}},\mathrm{ex}\circ\rho)\cong\mathrm{HC}_1(S),$$
where $\mathrm{HC}_1(S)$ is the first $\mathbb{Z}/2\mathbb{Z}$-graded cyclic homology of $S$ as defined in \cite{ChenSun2015}.
\end{corollary}
\begin{proof}
By Example~\ref{ex:iso_p_SS}, the Lie superalgebra $\mathfrak{sl}_{m|m}(S)$ is isomorphic to $\mathfrak{p}_m(S\oplus S^{\mathrm{op}},\mathrm{ex})$. Theorem~\ref{thm:hml_pm} insures that 
$$\mathrm{H}_2(\mathfrak{sl}_{m|m}(S))
\cong
\mathrm{H}_2(\mathfrak{p}_m(S\oplus S^{\mathrm{op}},\mathrm{ex}))
\cong
\fourIdx{}{+}{}{1}{\mathrm{HD}}(S\oplus S^{\mathrm{op}},\mathrm{ex}\circ\rho),$$
for $m\geqslant 5$.

Next, we identify $\fourIdx{}{+}{}{1}{\mathrm{HD}}(S\oplus S^{\mathrm{op}},\mathrm{ex}\circ\rho)$ with the first $\mathbb{Z}/2\mathbb{Z}$-graded cyclic homology $\mathrm{HC}_1(S)$ defined in \cite{ChenSun2015}. In the $\Bbbk$-module $\langle S\oplus S^{\mathrm{op}},S\oplus S^{\mathrm{op}}\rangle$ defined in (\ref{eq:HD1}), we have
\begin{align*}
\langle a\oplus0,0\oplus b\rangle
=&\langle (a\oplus0)(1\oplus0),0\oplus b\rangle+0+0\\
=&\langle (a\oplus0)(1\oplus0),0\oplus b\rangle
+(-1)^{|a||b|}\langle (1\oplus0)(0\oplus b),a\oplus0\rangle\\
&+(-1)^{|a||b|}\langle (0\oplus b)(a\oplus 0),1\oplus0\rangle\\
=&0.
\end{align*}
This shows that 
$$\langle a_1\oplus a_2,b_1\oplus b_2\rangle=\langle a_1\oplus0,b_1\oplus0\rangle
+\langle a_2\oplus0,b_2\oplus0\rangle,$$
for $a_1,a_2,b_1,b_2\in S$.

Let $I_{\mathsf{c}}$ be the $\Bbbk$-submodule of $S$ generated by $a\otimes b-(-1)^{|a||b|}b\otimes a$ and $(-1)^{|a||c|}ab\otimes c+(-1)^{|b||a|}bc\otimes a+(-1)^{|c||b|}ca\otimes b$ for homogeneous $a,b,c\in S$ and $\langle S,S\rangle_{\mathsf{c}}=(S\otimes S)/I_{\mathsf{c}}$. Then one observes that
$$\langle a_1\oplus a_2,b_1\oplus b_2\rangle\mapsto\langle a_1,b_1\rangle_{\mathsf{c}}
+\langle a_2,b_2\rangle_{\mathsf{c}}$$
defines an isomorphism $\langle S\oplus S^{\mathrm{op}},S\oplus S^{\mathrm{op}}\rangle\rightarrow\langle S,S\rangle_{\mathsf{c}}$. Its restriction on $\fourIdx{}{+}{}{1}{\mathrm{HD}}(S\oplus S^{\mathrm{op}},\mathrm{ex}\circ\rho)$ gives an isomorphism onto 
$$\mathrm{HC}_1(S):=\left\{\sum\limits_i\langle a_i,b_i\rangle_{\mathsf{c}}
\in\langle S,S\rangle_{\mathsf{c}}\middle|\sum\limits_i[a_i,b_i]=0\right\}.$$
This completes the proof.
\end{proof}

\begin{remark}
The above corollary recovers the second homology of $\mathfrak{sl}_{m|m}(S)$ for $m\geqslant5$ obtained in \cite{ChenSun2015}. As a byproduct, we obtain the isomorphism
$$\fourIdx{}{+}{}{1}{\mathrm{HD}}(S\oplus S^{\mathrm{op}},\mathrm{ex}\circ\rho)\cong\mathrm{HC}_1(S),$$
which indicates that the first $\mathbb{Z}/2\mathbb{Z}$-graded cyclic homology can be regarded as a special case of the first $\mathbb{Z}/2\mathbb{Z}$-graded dihedral homology. However, it is unknown yet whether such an isomorphism exists for higher degree cyclic homology and higher degree dihedral homology.
\end{remark}

\section{The second homology of $\mathfrak{p}_4(R,{}^-)$}
\label{sec:p4}

The second homology of $\mathfrak{p}_m(R,{}^-)$ for $m\geqslant5$ have been explicitly characterized in the previous section. However, the central extension $\psi:\mathfrak{stp}_4(R,{}^-)\rightarrow\mathfrak{p}_4(R,{}^-)$ is not necessarily universal. This section is devoted to explicitly constructing a universal central extension of $\mathfrak{stp}_4(R,{}^-)$, which can be accomplished by creating a $2$-cocycle on $\mathfrak{stp}_4(R,{}^-)$.

Such a $2$-cocycle takes values in the $\Bbbk$-module $R/(R_{(-)}\cdot R)$, where $R_{(-)}\cdot R$ is the right ideal of $R$ generated by $\bar{a}-\rho(a)$ for $a\in R$. In fact, the $\Bbbk$-module $R/(R_{(-)}\cdot R)$ is  a super-commutative $\Bbbk$-superalgebra since $[R,R]\cdot R\subseteq R_{(-)}\cdot R$. We denote $\boldsymbol{\pi}:R\rightarrow R/(R_{(-)}\cdot R)$ the canonical quotient map of $\Bbbk$--modules. It satisfies
\begin{equation}
\boldsymbol{\pi}(\bar{a}b)=\boldsymbol{\pi}(\rho(a)b), \quad a,b\in R.
\end{equation} 

Similar to Proposition~\ref{prop:stp_tridec}, $\mathfrak{stp}_4(R,{}^-)$ is decomposed as a direct sum of Lie sub-superalgebras:
$$\mathfrak{stp}_4(R,{}^-)=\mathfrak{a}\oplus\mathfrak{b}$$
where
\begin{align*}
\mathfrak{a}:&=\mathrm{span}_{\Bbbk}\{\mathbf{h}_{ij}(a,b), \mathbf{t}_{ij}(a),\mathbf{f}_i(a),\mathbf{f}_{ij}(a)|a,b\in R, 1\leqslant i\neq j\leqslant 4\},\\
\mathfrak{b}:&=\mathrm{span}_{\Bbbk}\{\mathbf{g}_i(a),\mathbf{g}_{ij}(a)|a\in R, 1\leqslant i\neq j\leqslant 4\}.
\end{align*}
Then, we define a $\Bbbk$-linear map $\beta_0:\mathfrak{b}\times\mathfrak{b}\rightarrow R/(R_{(-)}\cdot R)$ by
\begin{eqnarray*}
&\beta_0(\mathbf{g}_{ij}(a),\mathbf{g}_{kl}(b))=\epsilon(ijkl)\boldsymbol{\pi}(a\cdot\rho(b)),&\\
&\beta_0(\mathbf{g}_{i}(a),\mathfrak{b})=\beta_0(\mathfrak{b},\mathbf{g}_i(a))=0,&
\end{eqnarray*}
for $a,b\in R$, $1\leqslant i\neq j\leqslant 4$ and $1\leqslant k\neq l\leqslant 4$, where $\epsilon(ijkl)$ denotes the sign of the permutation $(ijkl)$ if $(ijkl)$ is a permutation of $\{1,2,3,4\}$ and denotes $0$ if $(ijkl)$ is not a permutation. Such a $\Bbbk$-linear map $\beta_0$ is well-defined since $\mathbf{g}_{ij}(\bar{a})=-\mathbf{g}_{ji}(\rho(a))$ while $\boldsymbol{\pi}(ab)=\boldsymbol{\pi}(\rho(a)b)$.

Furthermore, the $\Bbbk$--bilinear map $\beta_0:\mathfrak{b}\times\mathfrak{b}\rightarrow R/(R_{(-)}\cdot R)$ is extended to a $\Bbbk$--bilinear map
$$\beta:\mathfrak{stp}_4(R,{}^-)\times\mathfrak{stp}_4(R,{}^-)\rightarrow R/(R_{(-)}\cdot R)$$
such that $\mathfrak{a}$ lies in the radical of $\beta$, i.e.,
$$\beta(\mathfrak{a},\mathfrak{stp}_4(R,{}^-))=
\beta(\mathfrak{stp}_4(R,{}^-),\mathfrak{a})=0.$$
Now, we can show that:

\begin{lemma}
\label{lem:stp4_2cy2}
The $\Bbbk$-bilinear map $\beta$ is a 2-cocycle on $\mathfrak{stp}_4(R,{}^-)$ with values in $R/(R_{(-)}\cdot R)$.
\end{lemma}
\begin{proof}
We have to show $\beta$ satisfies
\begin{eqnarray}
&\beta(y,x)=-(-1)^{|x||y|}\beta(x,y),&\label{stp4:eq:2cyl1}\\
&(-1)^{|x||z|}\beta([x,z],y)+(-1)^{|y||x|}\beta([y,z],x)+(-1)^{|z||y|}\beta([z,x],y)=0,&\label{stp4:eq:2cyl2}
\end{eqnarray}
for homogeneous $x,y,z\in\mathfrak{stp}_4(R,{}^-)$.

For (\ref{stp4:eq:2cyl1}), it suffices to show
$$\beta(\mathbf{g}_{ij}(a),\mathbf{g}_{kl}(b))=-(-1)^{(1+|a|)(1+|b|)}\beta(\mathbf{g}_{kl}(b),\mathbf{g}_{ij}(a)),$$
for homogeneous $a,b\in R$, $i\neq j$ and $k\neq l$. Note that $\boldsymbol{\pi}(ab)=(-1)^{|a||b|}\boldsymbol{\pi}(ba)$, we deduce that
\begin{align*}
\beta(\mathbf{g}_{ij}(a),\mathbf{g}_{kl}(b))
&=\epsilon(ijkl)\boldsymbol{\pi}(a\cdot\rho(b))=\epsilon(klij)(-1)^{|a||b|}\boldsymbol{\pi}(\rho(b)\cdot a)\\
&=(-1)^{|a||b|}\beta(\mathbf{g}_{kl}(\rho(b)),\mathbf{g}_{ij}(\rho(a)))\\
&=-(-1)^{(1+|a|)(1+|b|)}\beta(\mathbf{g}_{kl}(b),\mathbf{g}_{ij}(a)).
\end{align*}

Next, we show (\ref{stp4:eq:2cyl2}). Observing that $\mathfrak{a}$ is a Lie sub-superalgebra of $\mathfrak{stp}_4(R,{}^-)$  included in the radical of $\beta$ and $[\mathfrak{b},\mathfrak{b}]=0$, we deduce that $\beta([x,y],z)=\beta([y,z],x)=\beta([z,x],y)=0$ if $(x,y,z)$ is contained in one of the subspaces $\mathfrak{a}\times\mathfrak{a}\times\mathfrak{a}$, $\mathfrak{a}\times\mathfrak{a}\times\mathfrak{b}$, $\mathfrak{b}\times\mathfrak{b}\times\mathfrak{b}$. Note also that (\ref{stp4:eq:2cyl2}) is symmtric with respect to all permutations on $\{x,y,z\}$, the proof is reduced to verify (\ref{stp4:eq:2cyl2}) for $x\in\mathfrak{a}$ and $y,z\in\mathfrak{b}$. In this situation, $[y,z]=0$ and (\ref{stp4:eq:2cyl2}) is equivalent to 
\begin{equation}
\beta([y,x],z)=(-1)^{|x||y|+|y||z|+|z||x|}\beta([z,x],y)\label{eq:stp4_2cy4}
\end{equation}

If $x=\boldsymbol{f}_{ij}(a)$ or $x=\boldsymbol{f}_{i}(a)$, then $[x,\mathfrak{b}]\subseteq\mathfrak{a}$ is included in the radical of $\beta$. It yields that both sides of (\ref{eq:stp4_2cy4}) are zero, and hence, we may assume that $x=\boldsymbol{h}_{ij}(a)$ or $x=\boldsymbol{t}_{ij}(a)$. In this situation, it is also obvious that both sides of (\ref{eq:stp4_2cy4}) are zero if $y=\boldsymbol{g}_i(a)$ or $z=\boldsymbol{g}_i(a)$. Now, it remains to verify the following two equalities
\begin{eqnarray}
&\beta([\mathbf{g}_{ij}(a),\mathbf{t}_{rs}(c)],\mathbf{g}_{kl}(b))
=-(-1)^{(|a|+|b|)(1+|c|)+|a||b|}\beta([\mathbf{g}_{kl}(b),\mathbf{t}_{rs}(c)],\mathbf{g}_{ij}(a)),&
\label{eq:stp4_2cy2}\\
&\beta([\mathbf{g}_{ij}(a),\mathbf{h}_{rs}(c,c')],\mathbf{g}_{kl}(b))
=-(-1)^{(|a|+|b|)(1+|c|+|c'|)+|a||b|}\beta([\mathbf{g}_{kl}(b),\mathbf{h}_{rs}(c,c')],\mathbf{g}_{ij}(a)),\quad&
\label{eq:stp4_2cy1}
\end{eqnarray}
for homogeneous $a,b,c,c'\in R$ and $i\neq j$, $k\neq l$ and $r\neq s$.

For (\ref{eq:stp4_2cy2}), we compute that
\begin{align*}
\beta([\boldsymbol{g}_{ij}(a),\boldsymbol{t}_{rs}(c)],\boldsymbol{g}_{kl}(b))
&=\delta_{jr}\beta(\boldsymbol{g}_{is}(ac),\boldsymbol{g}_{kl}(b))
-\delta_{ir}\beta(\boldsymbol{g}_{js}(\rho(\bar{a})c),\boldsymbol{g}_{kl}(b))\\
&=\delta_{jr}\epsilon(iskl)\boldsymbol{\pi}(ac\rho(b))
-\delta_{ir}\epsilon(jskl)\boldsymbol{\pi}(\rho(\bar{a})c\rho(b))\\
&=(\delta_{jr}\epsilon(iskl)-\delta_{ir}\epsilon(jskl))\boldsymbol{\pi}(ac\rho(b)).
\end{align*}
Then, (\ref{eq:stp4_2cy2}) follows from the facts that $\boldsymbol{\pi}(abc)=(-1)^{|a||b|+|b||c|+|c||a|}\boldsymbol{\pi}(cba)$ and
$$\delta_{jr}\epsilon(iskl)-\delta_{ir}\epsilon(jskl)
=\delta_{lr}\epsilon(ksij)-\delta_{kr}\epsilon(lsij)$$
for all $1\leqslant i,j,k,l\leqslant 4$. The equality (\ref{eq:stp4_2cy1}) follows similarly. 
\end{proof}
\medskip

The $2$-cocycle $\beta:\mathfrak{stp}_4(R,{}^-)\times\mathfrak{stp}_4(R,{}^-)
\rightarrow R/(R_{(-)}\cdot R)$ gives rise to a new Lie superalgebra
$$\widehat{\mathfrak{stp}}_4(R,{}^-):=\mathfrak{stp}_4(R,{}^-)\oplus (R/(R_{(-)}\cdot R)),$$
under the super-bracket
$$[x\oplus c,y\oplus c]:=[x,y]\oplus \beta(x,y)$$
for $x,y\in\mathfrak{stp}_4(R,{}^-)$ and $c,c'\in R/(R_{(-)}\cdot R)$, which is a central extension with the canonical projection $\psi'_4:\widehat{\mathfrak{stp}}_4(R,{}^-)\rightarrow\mathfrak{stp}_4(R,{}^-)$. Furthermore, we may show that

\begin{proposition}
\label{prop:p4_uce}
Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution. Then the central extension $\psi':\widehat{\mathfrak{stp}}_4(R,{}^-)\rightarrow \mathfrak{stp}_4(R,{}^-)$ is universal.
\end{proposition}
\begin{proof}
We have already known from Proposition~\ref{prop:stp_ce} that $\psi:\mathfrak{stp}_4(R,{}^-)\rightarrow\mathfrak{p}_4(R,{}^-)$ is a central extension. Hence, $\psi\circ\psi':\widehat{\mathfrak{stp}}_4(R,{}^-)\rightarrow\mathfrak{p}_4(R,{}^-)$ is a central extension. It suffices to show that $\psi\circ\psi'$ is universal.

Let $\varphi:\mathfrak{E}\rightarrow\mathfrak{p}_4(R,{}^-)$ be an arbitrary central extension of $\mathfrak{p}_4(R,{}^-)$. We also take $\tilde{t}_{ij}(a)$, $\tilde{f}_{ij}(a)$ and $\tilde{g}_{ij}(a)\in\mathfrak{E}$ as in (\ref{eq:tilde}). By Lemma~\ref{lem:tilde_rln}, these elements satisfy (\ref{STP00})-(\ref{STP12}) except (\ref{STP04}) and (\ref{STP10}). While the same argument as in Theorem~\ref{thm:pm_uce} also shows that (\ref{STP04}) holds.

For $a\in R$, we define
$$\tilde{\pi}(a):=[\tilde{g}_{12}(a),\tilde{g}_{34}(1)]\in\mathfrak{E},$$
which is contained in the center of $\mathfrak{E}$ since $\varphi(\tilde{\pi}(a))=0$.
We next prove that $\tilde{\pi}(R_{(-)}\cdot R)=0$.

Let $a,b\in R$ be homogenous. We compute that
\begin{align*}
\tilde{\pi}(ab)&=[\tilde{g}_{12}(ab),\tilde{g}_{34}(1)]
=[[\tilde{g}_{13}(a),\tilde{t}_{32}(b)],\tilde{g}_{34}(1)]\\
&=[\tilde{g}_{13}(a),[\tilde{t}_{32}(b),\tilde{g}_{34}(1)]]
=-[\tilde{g}_{13}(a),\tilde{g}_{24}(\bar{b})]\\
&=-[[\tilde{g}_{14}(1),\tilde{t}_{43}(a)],\tilde{g}_{24}(\bar{b})]
=-[\tilde{g}_{14}(1),[\tilde{t}_{43}(a),\tilde{g}_{24}(\bar{b})]]\\
&=(-1)^{|a|(1+|b|)}[\tilde{g}_{14}(1),\tilde{g}_{23}(\bar{b}a)]
=(-1)^{|a|(1+|b|)}[\tilde{g}_{14}(1),[\tilde{g}_{21}(\bar{b}a),\tilde{t}_{13}(1)]]\\
&=(-1)^{|a|(1+|b|)}(-1)^{1+|a|+|b|}[\tilde{g}_{21}(\bar{b}a),[\tilde{g}_{14}(1),\tilde{t}_{13}(1)]]
=-(-1)^{|b|+|a||b|}[\tilde{g}_{21}(\bar{b}a),\tilde{g}_{34}(1)]\\
&=(-1)^{|a|}[\tilde{g}_{12}(\bar{a}b),\tilde{g}_{34}(1)]
=(-1)^{|a|}\tilde{\pi}(\bar{a}b).
\end{align*}
It follows that $(a-(-1)^{|a|}\bar{a})b\in\ker{\tilde{\pi}}$ for homogeneous $a,b\in R$. Hence, $\tilde{\pi}(R_{(-)}\cdot R)=0$. We obtain a $\Bbbk$--linear map
$$R/(R_{(-)}\cdot R)\rightarrow\ker\varphi, \quad\boldsymbol{\pi}(a)\mapsto\tilde{\pi}(a).$$

Since $\tilde{\pi}(a)=[\tilde{g}_{12}(a),\tilde{g}_{34}(1)]$ and $\tilde{\pi}(\bar{a}b)=\tilde{\pi}(\rho(a)b)$, we deduce that
$$[\tilde{g}_{ij}(a),\tilde{g}_{kl}(b)]=\epsilon(ijkl)\tilde{\pi}(a\cdot\rho(b)),$$
for distinct $i,j,k,l$. Combining with (\ref{STP10a}) and (\ref{STP10b}), we have
$$[\tilde{g}_{ij}(a),\tilde{g}_{kl}(b)]=\epsilon(ijkl)\tilde{\pi}(a\cdot\rho(b)),$$
for $i\neq j$ and $k\neq l$.

Hence, there is a homomorphism
$\varphi':\widehat{\mathfrak{stp}}_4(R,{}^-)\rightarrow\mathfrak{E}$ such that
\begin{align*}
\varphi'(\mathbf{t}_{ij}(a)\oplus0)=\tilde{t}_{ij}(a),\quad
\varphi'(\mathbf{f}_{ij}(a)\oplus0)=\tilde{f}_{ij}(a),\quad
\varphi'(\mathbf{g}_{ij}(a)\oplus0)=\tilde{g}_{ij}(a),\quad
\varphi'(0\oplus \boldsymbol{\pi}(a))=\tilde{\pi}(a),
\end{align*}
where $a\in R$ and $1\leqslant i\neq j\leqslant4$, i.e., $\psi\circ\psi'=\varphi\circ\varphi'$. The uniqueness of $\varphi'$ follows from a similar argument as the proof of Proposition~\ref{thm:pm_uce}. Hence, we conclude that $\psi\circ\psi':\widehat{\mathfrak{stp}}_4(R,{}^-)\rightarrow\mathfrak{p}_4(R,{}^-)$ is a universal central extension.
\end{proof}

Using Propositions~\ref{prop:stp_ce} and~\ref{prop:p4_uce}, we conclude that

\begin{theorem}
\label{thm:p4_hml}
Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution. Then
$$\mathrm{H}_2(\mathfrak{p}_4(R,{}^-))=\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)\oplus R/(R_{(-)}\cdot R).\qed$$
\end{theorem}

\begin{remark}
\label{rmk:p4_hml_com}
If $R$ is super-commutative, then $R_{(-)}=0$. In this situation,
$$\mathrm{H}_2(\mathfrak{p}_4(\Bbbk)\otimes_{\Bbbk}R)\cong
\mathrm{H}_2(\mathfrak{p}_4(R,\rho))
\cong \fourIdx{}{+}{}{1}{\mathrm{HD}}(R,\mathrm{id})\oplus R\cong
R,$$
which recovers the second homology of $\mathfrak{p}_4(\Bbbk)\otimes_{\Bbbk}R$ obtained in \cite{IoharaKoga2005}.
\end{remark}
\medskip

In the special case where $(R,{}^-)=(S\oplus S^{\mathrm{op}},\mathrm{ex})$, Theorem~\ref{thm:p4_hml} recovers the result about the universal central extension of $\mathfrak{sl}_{4|4}(S)$ given in \cite{ChenSun2015}.

\begin{corollary}
\label{cor:sl44}
Let $S$ be an arbitrary unital associative superalgebra. Then
$$\mathrm{H}_2(\mathfrak{sl}_{4|4}(S))=\mathrm{HC}_1(S).$$
\end{corollary}
\begin{proof}
Recall from Example~\ref{ex:iso_p_SS} that the Lie superalgebra $\mathfrak{sl}_{4|4}(S)$ is isomorphic to $\mathfrak{p}_4(S\oplus S^{\mathrm{op}},\mathrm{ex})$. Hence,
$$\mathrm{H}_2(\mathfrak{sl}_{4|4}(S))
\cong\mathrm{H}_2(\mathfrak{p}_4(S\oplus S^{\mathrm{op}},\mathrm{ex}))
\cong\fourIdx{}{+}{}{1}{\mathrm{HD}}(S\oplus S^{\mathrm{op}},\mathrm{ex}\circ\rho)\oplus R/(R_{(-)}\cdot R),$$
where $(R,{}^-)=(S\oplus S^{\mathrm{op}},\mathrm{ex})$. Now, $R_{(-)}$ contains a unit element $1\oplus(-1)$, which yields that $R/(R_{(-)}\cdot R)=0$. Hence,
$$\mathrm{H}_2(\mathfrak{sl}_{4|4}(S))
\cong\fourIdx{}{+}{}{1}{\mathrm{HD}}(S\oplus S^{\mathrm{op}},\mathrm{ex}\circ\rho)
\cong\mathrm{HC}_1(S),$$
where the last isomorphism follows from Corollary~\ref{cor:slmm}.
\end{proof}

\section{The second homology of $\mathfrak{p}_3(R,{}^-)$}
\label{sec:p3}
Analogous to Section~\ref{sec:p4}, we will calculate the second homology of $\mathfrak{p}_3(R,{}^-)$ via explicitly creating the universal central extension of $\mathfrak{stp}_3(R,{}^-)$.
This will be accomplished by introducing a $2$-cocycle on $\mathfrak{stp}_3(R,{}^-)$ with values in the $\Bbbk$-module:
$$\mathfrak{z}:=\frac{R}{3R+R_{(-)}\cdot R}\oplus\frac{R}{3R+R_{(-)}\cdot R}\oplus\frac{R}{3R+R_{(-)}\cdot R},$$
where $R_{(-)}\cdot R$ is the right-ideal of $R$ generated by $\bar{a}-\rho(a)$ for $a\in R$.

Let $\boldsymbol{\pi}_i(a), i=1,2,3$ denote the canonical image of $a$ in one of the three direct summands, respectively. For distinct $i,j,k$, we will also use $\epsilon(ijk)$ to denote the sign of the permutation $(ijk)$.

Recall from Lemma~\ref{prop:stp_tridec} that $\mathfrak{stp}_3(R,{}^-)$ is spanned as a $\Bbbk$--module by
$$\mathfrak{B}:=\{\mathbf{h}_{ij}(a,b),\mathbf{t}_{ij}(a),\mathbf{f}_{ij}(a),\mathbf{g}_{ij}(a),
\mathbf{f}_k(a),
\mathbf{g}_k(a)|a,b\in R, 1\leqslant i,j,k\leqslant 3\text{ with }i\neq j\}.$$
We define a $\Bbbk$--bilinear map $\beta:\mathfrak{stp}_3(R,{}^-)\times\mathfrak{stp}_3(R,{}^-)\rightarrow\mathfrak{z}$ as follows:
\begin{eqnarray*}
&\beta(\mathbf{t}_{ij}(a),\mathbf{t}_{ik}(b))
=\epsilon(ijk)\boldsymbol{\pi}_i(ab),&\\
&\beta(\mathbf{f}_i(a),\mathbf{g}_{jk}(b))
=-(-1)^{(1+|a|)(1+|b|)}\beta(\mathbf{g}_{jk}(b),\mathbf{f}_i(a))
=\epsilon(ijk)\boldsymbol{\pi}_i(ab),&
\end{eqnarray*}
where $a,b\in R$ are homogeneous and $\{i,j,k\}=\{1,2,3\}$. For other pairs $(x,y)\in\mathfrak{B}\times\mathfrak{B}$, we set $\beta(x,y)=0$. The $\Bbbk$-blinear map $\beta$ is well-defined since 
$$\mathbf{f}_i(\bar{a})=\mathbf{f}_i(\rho(a)),\quad
\mathbf{f}_{ij}(\bar{a})=\mathbf{f}_{ji}(\rho(a)),\quad
\mathbf{g}_{i}(\bar{a})=-\mathbf{g}_i(\rho(a)),\quad
\mathbf{g}_{ij}(\bar{a})=-\mathbf{g}_{ji}(\rho(a)),$$ 
while 
$\boldsymbol{\pi}_i(\bar{a}b)=\boldsymbol{\pi}_i(\rho(a)b)$.

\begin{lemma}
\label{lem:stp3_2cyl}
The $\Bbbk$--bilinear map $\beta$ is a $2$-cocycle on $\mathfrak{stp}_3(R,{}^-)$ with values in $\mathfrak{z}$.
\end{lemma}
\begin{proof}
Since $[R,R]\cdot R\subseteq R_{(-)}\cdot R$ implies that $\boldsymbol{\pi}_i(ab)=(-1)^{|a||b|}\boldsymbol{\pi}_i(ba)$, the $\Bbbk$-bilinear map $\beta$ satisfies
\begin{equation}
\beta(x,y)=-(-1)^{|x||y|}\beta(y,x),\label{eq:stp3_2cyl1}
\end{equation}
for homogeneous elements $x,y\in\mathfrak{stp}_3(R,{}^-)$. It suffices to show
\begin{equation}
J(x,y,z):=(-1)^{|x||z|}\beta([x,y],z)
+(-1)^{|y||x|}\beta([y,z],x)
+(-1)^{|z||x|}\beta([z,x],y)
=0,\label{eq:stp3_2cyl2}
\end{equation}
for $x,y,z\in\mathfrak{stp}_3(R,{}^-)$.  Since (\ref{eq:stp3_2cyl2}) is symmetric under all permutations of $\{x,y,z\}$, we may assume $\beta([x,y],z)\neq0$, which only occurs when $z=\mathbf{t}_{ik}(a)$, $z=\mathbf{g}_{jk}(a)$, or $z=\mathbf{f}_i(a)$.

If $z=\mathbf{t}_{ik}(a)$ for $a\in R$ and $1\leqslant i\neq k\leqslant 3$, we pick $j$ to be the unique element of $\{1,2,3\}$ such that $\{i,j,k\}=\{1,2,3\}$. Then we directly verifed  that $J(x,y,z)=0$ for all possible choices of $(x,y)\in\mathfrak{B}\times\mathfrak{B}$ such that $\beta([x,y],z)\neq0$. The pair $(x,y)$ might be one of the following pairs
$$(\mathbf{h}_{ij}(a,a'), \mathbf{t}_{ij}(b)), 
(\mathbf{h}_{ik}(a,a'), \mathbf{t}_{ij}(b)),
(\mathbf{t}_{ik}(a), \mathbf{t}_{kj}(b)),
(\mathbf{f}_{ik}(a), \mathbf{g}_{kj}(b)),
(\mathbf{f}_i(a), \mathbf{g}_{ij}(b)),
(\mathbf{f}_{ij}(a), \mathbf{g}_{j}(b)),$$
for homogeneous $a,a', b\in R$. Similarly, $J(x,y,z)=0$ when $z=\mathbf{g}_{jk}(a)$ or $z=\mathbf{f}_i(a)$.
\end{proof}
\bigskip

The $2$-cocycle $\beta:\mathfrak{stp}_3(R,{}^-)\times\mathfrak{stp}_3(R,{}^-)\rightarrow\mathfrak{z}$ determines a centeral extension
$$\psi'_3:\mathfrak{stp}_3(R,{}^-)\oplus\mathfrak{z}\rightarrow\mathfrak{stp}_3(R,{}^-),$$
where $\psi'_3$ is the canonical projection and the super-bracket on $\mathfrak{stp}_3(R,{}^-)\oplus\mathfrak{z}$ is given by
$$[x\oplus c,y\oplus c']=[x,y]\oplus \beta(x,y),\quad x,y\in\mathfrak{stp}_3(R,{}^-),\text{ and }c,c'\in\mathfrak{z}.$$

\begin{proposition}
\label{prop:p3_uce}
The central extension $\psi'_3:\mathfrak{stp}_3(R,{}^-)\oplus\mathfrak{z}\rightarrow\mathfrak{stp}_3(R,{}^-)$ is universal.
\end{proposition}

\begin{proof}
It suffices to show the central extension $\psi\circ\psi'_3:\mathfrak{stp}_3(R,{}^-)\oplus\mathfrak{z}\rightarrow\mathfrak{p}_3(R,{}^-)$ is universal.

Let $\varphi:\mathfrak{E}\rightarrow\mathfrak{p}_3(R,{}^-)$ be an arbitrary central extension of $\mathfrak{p}_3(R,{}^-)$. Pick elements $\tilde{t}_{ij}(a)$, $\tilde{f}_{ij}(a)$, $\tilde{g}_{ij}(a)\in\mathfrak{E}$ with $a\in R$ and  $1\leqslant i\neq j\leqslant3$ as in (\ref{eq:tilde}). By Lemma~\ref{lem:tilde_rln}, they satisfying all relations (\ref{STP00})-(\ref{STP11}) except (\ref{STP04}) and (\ref{STP10}). Moreover, since $m=3$, there are no four distinct indices $1\leqslant i,j,k,l\leqslant3$. Hence, (\ref{STP10a}) and (\ref{STP10b}) imply (\ref{STP10}).

For $i\in\{1,2,3\}$, there are unique $j$ and $k$ such that $(i,j,k)$ is an even permutation of $\{1,2,3\}$. We define
\begin{equation}
\tilde{\pi}_i(a):=[\tilde{t}_{ij}(1),\tilde{t}_{ik}(a)]\in\ker\varphi,\label{eq:stp3_01}
\end{equation}
for $i=1,2,3$ and $a\in R$.

We next show that $\tilde{\pi}_i(3R+R_{(-)}\cdot R)=0$ for $i=1,2,3$. First, we take $\tilde{h}_{ij}=[\hat{f}_{ij}(1),\hat{g}_{ji}(1)]$ and deduce that
\begin{align*}
0&=[\tilde{h}_{ij},\tilde{\pi}_i(a)]\\
&=[\tilde{h}_{ij},[\tilde{t}_{ij}(1),\tilde{t}_{ik}(a)]]\\
&=[[\tilde{h}_{ij},\tilde{t}_{ij}(1)],\tilde{t}_{ik}(a)]
+[\tilde{t}_{ij}(1),[\tilde{h}_{ij},\tilde{t}_{ik}(a)]]\\
&=2[\tilde{t}_{ij}(1),\tilde{t}_{ik}(a)]+[\tilde{t}_{ij}(1),\tilde{t}_{ik}(a)]\\
&=\tilde{\pi}_i(3a),
\end{align*}
where $a\in R$ and $\{i,j,k\}=\{1,2,3\}$ are chosen as in (\ref{eq:stp3_01}). It follows that  $\tilde{\pi}_i(3R)=0$.

Now, we claim that
$\tilde{\pi}_i(ab)=(-1)^{|a|}\tilde{\pi}_i(\bar{a}b)$
for $i=1,2,3$ and homogeneous $a,b\in R$. Indeed, for $i\in\{1,2,3\}$, we may pick $j,k$ as in (\ref{eq:stp3_01}). Then
\begin{align*}
\tilde{\pi}_i(ab)&=[\tilde{t}_{ij}(1),\tilde{t}_{ik}(ab)]\\
&=[\tilde{t}_{ij}(1),[\tilde{f}_{ij}(a),\tilde{g}_{jk}(b)]]\\
&=[[\tilde{t}_{ij}(1),\tilde{f}_{ij}(a)],\tilde{g}_{jk}(b)]
+[\tilde{f}_{ij}(a),[\tilde{t}_{ij}(1),\tilde{g}_{jk}(b)]]\\
&=(-1)^{|a|}[[\tilde{t}_{ij}(1),\tilde{f}_{ji}(\bar{a})],\tilde{g}_{jk}(b)].
\end{align*}
Since $[\tilde{t}_{ij}(1),\tilde{f}_{ji}(\bar{a})]=[\tilde{t}_{ik}(1),\tilde{f}_{kj}(\bar{a})]+c$ for some $c\in\ker\varphi$, we further deduce that
\begin{align*}
\tilde{\pi}_i(ab)&=(-1)^{|a|}[[\tilde{t}_{ik}(1),\tilde{f}_{ki}(\bar{a})],\tilde{g}_{jk}(b)]\\
&=(-1)^{|a|+(1+|a|)(1+|b|)}[[\tilde{t}_{ik}(1),\tilde{g}_{jk}(b)],\tilde{f}_{ki}(\bar{a})]
+(-1)^{|a|}[\tilde{t}_{ik}(1),[\tilde{f}_{ki}(\bar{a}),\tilde{g}_{jk}(b)]]\\
&=0-(-1)^{|b|}[\tilde{t}_{ik}(1),\tilde{t}_{ij}(a\bar{b})]\\
&=-(-1)^{|b|}[[\tilde{t}_{ij}(1),\tilde{t}_{jk}(1)],\tilde{t}_{ij}(a\bar{b})]\\
&=-(-1)^{|b|}[[\tilde{t}_{ij}(1),\tilde{t}_{ij}(a\bar{b})],\tilde{t}_{jk}(1)]
-(-1)^{|b|}[\tilde{t}_{ij}(1),[\tilde{t}_{jk}(1),\tilde{t}_{ij}(a\bar{b})]]\\
&=(-1)^{|b|}[\tilde{t}_{ij}(1),\tilde{t}_{ik}(a\bar{b})]\\
&=(-1)^{|b|}\tilde{\pi}_i(a\bar{b}).
\end{align*}
It follows that $\tilde{\pi}_i(b)=(-1)^{|b|}\tilde{\pi}^i(\bar{b})$ and
\begin{align*}
\tilde{\pi}_i(ab)
&=(-1)^{|a|+|b|}\tilde{\pi}_i(\overline{ab})
=(-1)^{|a|+|b|+|a||b|}\tilde{\pi}_i(\bar{b}\bar{a})
=(-1)^{|b|+|a||b|}\tilde{\pi}_i(\bar{b}a)\\
&=(-1)^{|a|+|a||b|}\tilde{\pi}_i(\overline{\bar{b}a})
=(-1)^{|a|}\tilde{\pi}_i(\bar{a}b).
\end{align*}
Therefore, we conclude that $\tilde{\pi}_i(3R+R_{(-)}\cdot R)=0$ for $i=1,2,3$.

Next, we show that $[\tilde{t}_{ij}(a),\tilde{t}_{ik}(b)]=\epsilon(ijk)\tilde{\pi}_i(ab)$
for $\{i,j,k\}=\{1,2,3\}$. We first assume that the permutation taking $1$ to $i$, $2$ to $j$, and $3$ to $k$ has positive sign. Then
\begin{align*}
[\tilde{t}_{ij}(a),\tilde{t}_{ik}(b)]
&=[\tilde{t}_{ij}(a),[\tilde{t}_{ij}(1),\tilde{t}_{jk}(b)]]\\
&=[[\tilde{t}_{ij}(a),\tilde{t}_{ij}(1)],\tilde{t}_{jk}(b)]
+[\tilde{t}_{ij}(1),[\tilde{t}_{ij}(a),\tilde{t}_{jk}(b)]]\\
&=[\tilde{t}_{ij}(1),\tilde{t}_{ik}(ab)]\\
&=\tilde{\pi}_i(ab).
\end{align*}
If the permutation $(ijk)$ has negative sign, then $(ikj)$ has positive sign. We have
\begin{align*}
[\tilde{t}_{ij}(a),\tilde{t}_{ik}(b)]
=-(-1)^{|a||b|}[\tilde{t}_{ik}(b),\tilde{t}_{ij}(a)]
=-(-1)^{|a||b|}\tilde{\pi}_i(ba)
=-\tilde{\pi}_i(ab).
\end{align*}
Hence, we conclude that $[\tilde{t}_{ij}(a),\tilde{t}_{ik}(b)]=\epsilon(ijk)\tilde{\pi}_i(ab)$.

Therefore, there is a homomorphism of Lie superalgebras
$$\varphi':\widehat{\mathfrak{stp}}_3(R,{}^-)\rightarrow\mathfrak{E}$$
such that
$$\varphi'(\mathbf{t}_{ij}(a))=\tilde{t}_{ij}(a),
\quad\varphi'(\mathbf{f}_{ij}(a))=\tilde{f}_{ij}(a),
\quad\varphi'(\mathbf{g}_{ij}(a))=\tilde{g}_{ij}(a),$$
for $1\leqslant i\neq j\leqslant 3$ and $a\in R$. Hence, $\varphi\circ\varphi'=\psi\circ\psi'$. The uniqueness of $\varphi'$ follows from the same argument as in the proof of Proposition~\ref{thm:pm_uce}.
\end{proof}

Now, we conclude from Propositions~\ref{prop:stp_ce} and~\ref{prop:p3_uce} that:

\begin{theorem}
\label{thm:p3_hml}
Let $(R,{}^-)$ be a unital associative superalgebra with superinvolution. Then
$$\mathrm{H}_2(\mathfrak{p}_3(R,{}^-))=\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,{}^-\circ\rho)\oplus\frac{R}{3R+R_{(-)}\cdot R}\oplus\frac{R}{3R+R_{(-)}\cdot R}\oplus\frac{R}{3R+R_{(-)}\cdot R}.\qed$$
\end{theorem}

\begin{remark}
\label{rmk:p3_hml_com}
If $R$ is super-commutative, then $\mathfrak{p}_3(R,\rho)\cong\mathfrak{p}_3(\Bbbk)\otimes_{\Bbbk}R$, $\fourIdx{}{+}{}{1}{\mathrm{HD}}(R,\rho\circ\rho)=0$ and $R_{(-)}=0$. Hence,
$$\mathrm{H}_2(\mathfrak{p}_3(\Bbbk)\otimes_{\Bbbk}R)=(R/3R)\oplus(R/3R)\oplus(R/3R),$$
which equals $0$ whenever $3$ is invertible in $R$. When $\Bbbk$ is a field of characteristic zero, this coincides with the second homology group of $\mathfrak{p}_3(\Bbbk)\otimes_{\Bbbk}R$ given in \cite{IoharaKoga2005}.
\end{remark}
\medskip

In the special case where $(R,{}^-)=(S\oplus S^{\mathrm{op}},\mathrm{ex})$, Theorem~\ref{thm:p3_hml} recovers the second homology of $\mathfrak{sl}_{3|3}(S)$ obtained in \cite{ChenSun2015}.

\begin{corollary}
\label{cor:uce_sl33}
Let $S$ be an arbitrary unital associative superalgebra. Then
$$\mathrm{H}_2(\mathfrak{sl}_{3|3}(S))\cong\mathrm{HC}_1(S).$$
\end{corollary}
\begin{proof}
It is known from Example~\ref{ex:iso_p_SS} that $\mathfrak{sl}_{3|3}(S)$ is isomorphic to $\mathfrak{p}_3(S\oplus S^{\mathrm{op}},\mathrm{ex})$. On the other hand, $\mathfrak{z}=0$ since $1\oplus(-1)$ is invertible and is contained in the right-ideal of $S\oplus S^{\mathrm{op}}$ generated by $(S\oplus S^{\mathrm{op}})_{(-)}$. Hence, $\mathrm{H}_2(\mathfrak{sl}_{3|3}(S))\cong\fourIdx{}{+}{}{1}{\mathrm{HD}}(S\oplus S^{\mathrm{op}},\mathrm{ex})\cong\mathrm{HC}_1(S)$.
\end{proof}

\section*{Acknowledgements}
The authors thank Prof. Yun Gao and Prof. Hongjia Chen for useful suggestions. Zhihua Chang was supported by the National Natural Science Foundation of China (Grant No. 11501213), the China Postdoctoral Science Foundation (Grant No. 2015M570705) and the Fundamental Research Funds for the Central Universities (Grant No. 2015ZM085). Yongjie Wang was supported by the China Postdoctoral Science Foundation (Grant No. 2015M571928) and the Fundamental Research Funds for the Central Universities.

