\section{Introduction}

A locally recoverable code is a linear code where a single erased position may be recovered from a small set of coordinates. Tamo and Barg constructed a family of optimal LRCs using subcodes of Reed--Solomon codes. Guo et. al. \cite{GKS13} used the point--line geometry of affine space over $\F_q$ to make LRCs.  Later L\'opez et. al \cite{LMMPW21} used the point--line incidence of the affine part of the Hermitian curve in order to define LRCs.  We begin introducing the basic notions of codes with locality.

    \begin{definition}[Locality of a Linear Code]\cite{GHSY12}
    A code $C$ has \emph{locality} $r$ if for every $i \in [n]$ there exists a subset $R_{i} \subset [n] \setminus i,\| R_{i} \| \leq r$ and a function $\phi_{i}$ such that for every codeword  $c \in C$:
    \begin{equation}
        c_{i} = \phi_{i} (\{c_{j}, j \in R_{i}\})
    \end{equation}
    
    A code $C$ with locality $r$ has \emph{availability} $s$ if for any $i \in [n]$ there exists $s$ subsets $R_1, R_2, \ldots, R_s$ such that $i \in R_u$ and $R_{s_1} \cap R_{s_2} = \{i\}$.
    \end{definition}
    A linear code with locality $r$ and availability $s$ is a linear code that any erased position may be recovered by querying any of $s$ disjoint sets of $r$ positions. 

\begin{definition}[Affine Points of the Hermitian curve]
Let $q$ be a prime power. The \emph{affine points of the Hermitian curve} over $\Fqd$ are the solutions to the polynomial equation $$X^{q+1} = Y^q+Y$$ over $\Fqd$. That is, the points are defined by $$ \mathcal{H} := \{ (\alpha, \beta) \in \Fqd^2 \ \| \ \alpha^{q+1} = \beta^q+\beta  \} $$

\end{definition}

Hermitian codes may be defined as evaluation codes of polynomials over $\mathcal{H}$. The set of affine points $\mathcal{H}$ is finite. One only needs a finite set of monomials in order to describe any function on $\mathcal{H}$. We introduce the following notation.

\begin{definition}
Denote by $\mathcal{M}$ the set $\mathcal{M}  := \{ X^iY^j \ \| \ 0 \leq i < q^2, 0 \leq j < q\}$.

Let $f \in \F_{q^2}[X,Y]$. Let $P_1, P_2, \ldots, P_n$ be some ordering of the elements of the set $L$. We denote the evaluation of $f$ on $L$ by $$ev_L(f) = (f(P_1),f(P_2), \ldots, f(P_n)) $$
\end{definition} The following proposition can be established with Gr\"obner bases, normal bases and quotient rings.

\begin{proposition}\cite{CLO07}
Let $q$ be a prime power. Let $\mathcal{H}$ denote the set of all points over $\F_{q^2}$ of the Hermitian curve. Let $f \in \F_{q^2}[X,Y]$ be any polynomial. Then there exists $g \in \langle \mathcal{M} \rangle_{F_{q^2}}$ such that $$ev_\mathcal{H}(f) = ev_\mathcal{H}(g).$$


\end{proposition}

In fact, the evaluation functions from $\langle \mathcal{M} \rangle_{F_{q^2}}$ are linearly independent.

\begin{proposition}\cite{CLO07}
Let $q$ be a prime power. Let $\mathcal{H}$ denote the set of all points over $\F_{q^2}$ of the Hermitian curve. Let $f, g\in \langle \mathcal{M} \rangle_{F_{q^2}}$.  Then there $$f = g \makebox{ if and only if } ev_\mathcal{H}(f) = ev_\mathcal{H}(g).$$


\end{proposition}

Codes over the Hermitian curve may be defined as follows.

\begin{definition}[Evaluation codes over the Hermitian curve]\cite{G08}
Let $L$ be a subset of $\F_{q^2}[X,Y]$. Let $\mathcal{H}$ denote the set of affine points of the Hermitian curve. We define an evaluation code over $\mathcal{H}$ as:

$$C(L, \mathcal{H}) := \{ ev_{\mathcal{H}}(f) \ \| \ f \in \langle L \rangle_{\F_{q^2}} \} $$
\end{definition}

An evaluation code is defined as the linear code generated by the evaluation of a set of functions on a given set of points. In the case of Hermitian codes the set $L$ is taken as the set of all monomials of the form $X^iY^j$ where $iq+j(q+1) \leq s$ for a given $s$. Due to the kernel of the evaluation map, we may assume all function $f(X,Y)$ are contained in  $\mathcal{M}$ without loss of generality. Any linear code may be defined as an evaluation code using multivariate Lagrange interpolation polynomials. Now we shall define the lines of the Hermitian curve. These lines will be the locally recovery sets for the Hermitian Lifted codes.


\begin{definition}[Lines of the Hermitian curve]\cite{LMMPW21}
Let $q$ be a prime power. Let $a,b \in \Fqd$. A \emph{line of the Hermitian curve} is a set $L_{a,b}$ of the form $$L_{a,b} := \{(x,y) \in \mathcal{H} \ \| y = ax+b \} \makebox{ and } \#(L_{a,b}) = q+1.$$

\end{definition} 

Not all lines of the affine space are lines of the Hermitian curve. We exclude vertical lines (those of the form $X = c$) because they have only $q$ points in common with the Hermitian curve. We also omit tangent lines with the Hermitian curve because those have only one point in common with the Hermitian curve. The following lemma states the polynomial equation that the $x$--coordinates of a line of the Hermitian curve must satisfy.

\begin{definition}
Let $q$ be a prime power. Let $a,b \in \Fqd$. Define by $L_{a,b,x}$ the set of $x$--coordinates of the line $L_{a,b}$. That is: $$L_{a,b,x} = \{ x \ \| \ (x,y) \in L_{a,b} \}.$$

\end{definition}
\begin{lemma}

Let $a,b \in \Fqd$ be such that $L_{a,b}$ is a line of the Hermitian curve. Then $$\prod\limits_{\gamma \in L_{a,b,x}} (X - \gamma) = (X-a^q)^{q+1} -(a^{q+1}-b^q-b).$$

\end{lemma} \begin{proof} We need to determine the common points to $Y = aX+b$ and $X^{q+1} = Y^q +Y$ over $\Fqd$. Substitute $Y = aX+b$ on the equation of the Hermitian curve to obtain:
 
 $$X^{q+1} = (aX+b)^q+(aX+b).$$ We rearrange terms and obtain:
  $$X^{q+1}-a^qX^q -aX^q = b^q+b.$$ We add $a^{q+1}$ to both sides.
  $$X^{q+1}-a^qX^q -aX^q +a^{q+1} = b^q+b + a^{q+1}.$$ The right hand side factors as: $$(X^q-a)(X-a^q) = b^q+b + a^{q+1}.$$
  Therefore $$(X-a^q)^{q+1} = b^q+b + a^{q+1}.$$ This implies the elements of $L_{a,b,x}$ satisfy  $(X-a^q)^{q+1} -(a^{q+1}-b^q-b) = 0$ \end{proof}


Now we state the condition on $a,b$ such that $L_{a,b}$ is a line of the Hermitian curve.
\begin{lemma}\cite{LMMPW21} Let $a,b \in \Fqd$ be such that $L_{a,b}$ is a line of the Hermitian curve. Then $a^{q+1} + b^q+b \neq 0$.

\end{lemma}
 \begin{proof} Let $L_{a,b}$ be a line of the Hermitian curve. The $x$--coordinates satisfy the polynomial equation  $(X-a^q)^{q+1} -(a^{q+1}-b^q-b) = 0$. Note that since $b^q+b + a^{q+1} \in \F_q$, the equation has $1$ solution if $b^q+b + a^{q+1} = 0$ and $q+1$ solutions if $b^q+b + a^{q+1} \neq 0$. For each solution in $X$ there is one point in $L_{a,b}$. Therefore if $L_{a,b}$ is a line of the Hermitian curve, it has $q+1$ points. This implies $(X-a^q)^{q+1} = b^q+b + a^{q+1}$ has $q+1$ solutions and therefore $b^q+b + a^{q+1} \neq 0$. \end{proof}

\begin{definition}\cite{LMMPW21} Let $f(X,Y)$ be a bivariate polynomial. Let $L_{a,b}$ be a line of the Hermitian curve. The \emph{restriction of $f$ onto $L_{a,b}$} is the function obtained by evaluating $f$ on the points of the line $L$. We denote the restriction by $f_{L_{a,b}}$. \end{definition}

We must make the distinction between a polynomial and its evaluation. If the the line $L_{a,b}$ is represented by the equations $X = T, Y = aX+b$ then the polynomial restricted to the line $L_{a,b}$ is the univariate polynomial $f(T, aT+b)$. An evaluation code is defined via the evaluation function $ev_{\mathcal{H}}(f)$. The restriction of the evaluation code to the line $L_{a,b}$ is simply the evaluation vector $ev_{L_{a,b}}(f)$. Even if $f(T, aT+b)$ has an arbitrarily high degree, the evaluation vector $ev_{L_{a,b}}(f)$ may be represented by evaluating a univariate polynomial of degree $q$ or less on the $q+1$ points in $L_{a,b,x}$. In order to construct an LRC we need functions which restrict "nicely" on each line. 

\begin{definition}[Good functions]\cite{LMMPW21}
Let $f(X,Y)$ be a bivariate polynomial. Let $a,b \in \Fqd$ such that $L_{a,b}$ is a line of the Hermitian curve. We say $f$ is a \emph{good polynomial} if and only the evaluation $$ev_{L_{a,b}}(f) = ev_{L_{a,b,x}}(g(T))$$ where $g$ is a univariate polynomial of degree  less than $q$ on $L_{a,b,x}$ for each line $L_{a,b}$ of the Hermitian curve.

We denote the set of good functions as $\mathcal{G}_f$.

\end{definition}

\begin{definition}[Hermitian Lifted Codes]\cite{LMMPW21}

Let $q = 2^k$. We define the \emph{Hermitian Lifted code} over the Hermitian curve as $$\mathcal{C}  := C(\mathcal{G}_f, \mathcal{H}).$$


\end{definition}

We denote the dimension of $\mathcal{C}$ by $K$, its rate by $R$ and its distance by $D$. L\'opez et. al \cite{LMMPW21} claim the following.

\begin{proposition} \cite[Claim 12]{LMMPW21}
Let $q = 2^k$. Then the monomial set $\mathcal{M}$ contains at least $$\sum\limits_{r=0}^{k-1}(4^r-3^r)4^{k-r-2}2^{k-r-1}$$ good monomials.  Consequently $$K \geq \sum\limits_{r=0}^{k-1}(4^r-3^r)4^{k-r-2}2^{k-r-1}.$$
\end{proposition}

As a corollary it was proven that the rate of Hermitian Lifted codes is at least $R \geq 0.007$. Our aim is to find more monomials in $\mathcal{M}$ which are also good functions. Our proof is based in univariate polynomial division. To simplify our arguments, we introduce the following notation.

\begin{definition}
Let $\gamma \neq 0 $. We shall denote by $$P_{a,\gamma} := (T-a^q)^{q+1} - \gamma.$$
\end{definition}

The nonzero positions of the parity check equations for $\mathcal{C}$ are the point sets of lines of the Hermitian unital. The binary code generated by the lines of the Hermitian unital is a code of dimension $q^3+1$. This means that its dual code and the LRC defined by the lines of the Hermitian unital has dimension $0$ (see \cite[Theorem 8.3.1]{AK92}). We find it remarkable that even though the parity check equations for $\mathcal{C}$ have the same nonzero positions than the Hermitian unital code, the dimension of $\mathcal{C}$ is relatively large. 

A function $f(X,Y)$ is good or bad depending on the residue $f(T, aT+b) \mod (T-a^q)^{q+1}-\gamma$. The polynomial $P_{a,\gamma}$ will determine the degree of $f(T,aT+b) \mod P_{a, \gamma}$. In this section we prove key properties of this reduction. For the remainder of this article we assume we are working over characteristic $2$.


We remark that if $0 \leq i,j \leq \frac{q}{2^r}-1$ then $ 0 \leq i 2^r \leq q-2^r$. Now we shall prove a lemma on the degree of certain $2^r$--powers of $T$.
\begin{lemma}\label{lem:little2r pow}
Let $q$ be a power of $2$, let $0 < 2^r < q$. Let $k$ be odd and let $j$ be such that $0 \leq k,j < \frac{q}{2^r}$. Then 

$$T^{k2^rq}T^{j2^r} \mod P_{a, \gamma} $$ has degree at most $q+1-2^r$.
\end{lemma}
\begin{proof}
Recall that $P_{a, \gamma} = (T-a^q)^{q+1}-\gamma$, where $\gamma \neq 0$. We shall apply the change of variables $T \mapsto S+a^q$ and compute the degree of $(S+a^q)^{k2^rq}(S+a^q)^{j2^r} \mod S^{q+1}-\gamma$. instead. Since we are working over  characteristic $2$:

$$(S+a^q)^{k2^rq}(S+a^q)^{j2^r} = (S^{2^rq}+a^{2^r})^{k}(S^{2^r}+a^{2^rq})^{j}. $$ We expand the binomial and obtain
$$  = \left(\sum\limits_{i_1=0}^k\binom{k}{i_1}S^{i_12^rq}\left(a^{(k-i_1)2^r}\right)\right)\left (\sum\limits_{i_2=0}^j \binom{j}{i_2}S^{i_2 2^r}\left(a^{(j-i_2)2^rq}\right)\right)$$
We multiply both binomial expansions and obtain:
$$  = \sum\limits_{i_1=0}^k\sum\limits_{i_2=0}^j \binom{k}{i_1}\binom{j}{i_2}S^{(i_1q+i_2)2^r}a^{(j-i_2)q+(k-i_1)2^r}$$
The degree of  $S^{(i_12^r q+i_2 2^r}\mod S^{q+1}-\gamma$ depends on $(i_1q+i_2)2^r$.
\begin{itemize} \item  \textbf{Case 1: $i_1 = 0$} If $i_1 = 0$, then $(i_1q+i_2)2^r = i_2 2^r$ and $i_2 2^r \leq q-2^r$. Therefore $$S^{(i_12^r)q+i_22^r} = S^{i_2 2^r} \equiv S^{i_2 2^r} \mod S^{q+1}-\gamma.$$
\item \textbf{Case 2: $i_1 \geq 1, i_1 < i_2$}. If $i_1 \geq 1$ and $i_1 < i_2$, then let $i_2 = i_1 +\delta$ where $\delta > 0$. Write $(i_1q+i_2)2^r = i_1 2^r q + i_12^r +\delta 2^r =  (i_1 2^r)(q+1) +\delta 2^r$.   In this case $$S^{(i_12^r)q+i_22^r} = S^{ (i_1 2^r)(q+1) +\delta 2^r} \equiv \gamma^{i_1 2^r}S^{\delta 2^r} \mod S^{q+1}-\gamma.$$ Clearly the reduction has degree at most $\delta 2^r \leq q-2^r$.
\item \textbf{Case 3: $i_1 \geq 1, i_1 =i_2$ } If $i_1 \geq 1$ and $i_1  = i_2$ then $(i_1q+i_2)2^r = (i_2+\delta)2^r q = (i_2 2^r)(q+1)$. Reducing $\mod S^{q+1}-\gamma$ we obtain $$S^{(i_12^r q + i_2 2^r)} = S^{((q+1)i_2 2^r)} \equiv \gamma^{i_2 2^r}. $$ Therefore $S^{(i_12^r q + i_2 2^r)} \mod S^{q+1}-\gamma$ is a constant.
\item \textbf{Case 4: $i_1 \geq 1, i_1 > i_2$ } If $i_1 \geq 1$ and $i_1 \geq i_2$ then $i_1 = i_2 + \delta$ where $\delta >  0$. Write $(i_1q+i_2)2^r = (i_2+\delta)2^r q + i_22^r = (i_2 2^r)(q+1) + \delta 2^r q$. This implies that  $$S^{(i_2 2^r)(q+1) + \delta 2^r} = S^{((i_2+\delta)2^r-1)(q+1) +(q+1-\delta 2^r)}.$$
  Reducing $S$ modulo $S^{q+1}- \gamma$ we obtain  $$S^{(i_2 2^r)(q+1) + \delta 2^r} \equiv \gamma^{(i_2+\delta)2^r-1} S^{q+1-\delta2^r}.$$ Since $1 \leq \delta \leq \frac{q}{2^r}-1$, it follows that the degree of the reduction is at most $q+1-2^r$.\end{itemize}

Note that on all cases the degree of each power of $S$ in the expansion of $(S+a^q)^{k2^rq}(S+a^q)^{j2^r}$ reduces $\mod S^{q+1}-\gamma$ to something of degree at most $q+1-2^r$. Since the change of variables $T \mapsto S+a^q$ preserves the degree, the result follows. \end{proof} 

It is worth mentioning that because $k$ is odd, $\binom{k}{1} = 1$. Therefore $S^{2^rq} \equiv  \gamma^{2^r-1}S^{q+1-2^r} \mod S^{q+1}-\gamma $. If $k$ was even, (say $k = k'2^s$) then the same argument holds with $2^{r+s}$ instead of $2^r$ and obtain better bounds on the degree.





\section{Finding good monomials}
In the previous section we have found that certain powers of $T$, namely $T^{k2^rq+j2^r}$ have degree $q+1-2^r$ when reduced modulo $P_{a,\gamma}$. As a good monomial has a reduction $\mod P_{a,\gamma}$ with degree is less than $q$. Therefore if $2^r \geq 2$, then those monomials are good. Now we count the monomials of the form $X^iY^j$ where $0 \leq i < q^2, 0 \leq j < q$ which are known to give good reductions.

\begin{definition}\cite{GKS13}

Let $0 \leq i, j \leq 2^r-1$. Suppose that the expansion of $i$ in base $2$ is $i = i_{r-1}2^{r-1}+i_{r-2}2^{r-2}+  \cdots+ i_12+i_0$ where $0 \leq i_s \leq 1$. and that the expansion of $j$ in base $2$ is $j_{r-1}2^{r-1}+j_{r-2}2^{r-2}+  \cdots+ j_12+j_0$ where $0 \leq j_s \leq 1$. We say $i$ \emph{lies in the $2$--shadow of $j$} if $i_s \leq j_s$ for all $s$, We denote it by $i \leq_2 j$.
\end{definition}

\begin{lemma}[Lucas' Theorem]\cite{L78}
Let $0 \leq i,j \leq 2^r-1$.
$$\binom{j}{i} = 0 \mod 2 \makebox{ if and only if } i \not \leq_2 j.$$ 
\end{lemma} 

With an elementary counting argument with $0 \leq i,j < 2^r$ it may be proven that there are $4^r-3^r$ pairs $(i,j)$ such that $i \not\leq_2 j$.



\begin{lemma}\label{lem:goodmonomialsnb}

Let $i = i_1q+i_2 2^r+ i_3$ where $0 \leq i_1, i_2, i_3$ where $2^r$ is the highest power of $2$ dividing $i_1$,$0 \leq i_2 < \frac{q}{2^{r}}$ and $0 \leq i_3 < 2^r$ Let $j = j_2 2^r + j_3$ where $0 \leq j < q$, $0 \leq j_2 < \frac{q}{2^{r}}$ and $j_3 < 2^r$. The monomial $X^{i}Y^{j}$ is a good monomial if $i_2 2^r + i_3 +j_2 2^r + j_3 < q$ and $ 2^r-1-i_3 \not \leq_2 j_3 $



\end{lemma}

\begin{proof}

We shall evaluate the monomial $X^iY^j$ on the line given by $Y = aX+b$, $b^q+b \neq a^{q+1}$. Denote $a^{q+1} - b^q-b$ by $\gamma$. In this case, we set $X = T$ and $Y = aT+b$. The powers of $T$ satisfy the polynomial equation $(T-a^q)^{q+1} = \gamma$.

From the conditions of the theorem, $i_1 = k_12^r$ where $k_1$ is odd. We need to determine the degree of $$T^{i}(aT+b)^j \mod (T-\alpha^q)^{q+1}-\gamma $$


Note that  $$T^{i}(aT+b)^j  = T^{k_1 2^r q} T^{i_2 2^r} T^{i_3}(aT+b)^{j_2 2^r} (aT+b)^{j_3}.$$

Therefore we rewrite the product as:

$$T^{i}(aT+b)^j  = T^{k_1 2^r q} (T^{2^r})^{i_2} T^{i_3}(a^{2^r}T^{2^r}+b^{2^r})^{j_2} (aT+b)^{j_3}.$$

Expanding the terms of $(a^{2^r}T^{2^r}+b^{2^r})^{j_2}$ we obtain:

$$T^{i}(aT+b)^j  = T^{k_1 2^r q} (T^{2^r})^{i_2} T^{i_3}(aT+b)^{j_3}\left( \sum\limits_{k_1=0}^{j_2}\binom{j_2}{k_1} a^{k_2 2^r}b^{2^r(j_2-k_2)} T^{k_2 2^r} \right)  .$$

We include the term $(T^{2^r})^{i_2}$ into the sum 
$$T^{i}(aT+b)^j  = T^{k_1 2^r q}  T^{i_3}(aT+b)^{j_3}\left( \sum\limits_{k_1=0}^{j_2}\binom{j_2}{k_1} a^{k_2 2^r}b^{2^r(j_2-k_2)} T^{ (k_2+i_2) 2^r} \right)  .$$


Since $k_2+j_2<\frac{q}{2^r}$ Lemma \ref{lem:little2r pow} implies that $T^{k_1 2^r q}T^{k_2+j_2} \mod P(a^q, \gamma)$ has degree $q+1-2^r$. Let us denote $$T^{k_1 2^r q}  \left( \sum\limits_{k_1=0}^{j_2}\binom{j_2}{k_1} a^{k_2 2^r}b^{2^r(j_2-k_2)} T^{ (k_2+i_2) 2^r} \right) \mod P(a^q, \gamma)$$ by = $A(T)$ where $deg A = q+1-2^r$.

Now

$$T^i(aT+b)^j \equiv  T^{i_3}(aT+b)^{j_3}A(T) \mod (T-\alpha^q)^{q+1}-\gamma $$


In the binomial expansion of $T^{i_3}(aT+b)^{j_3}$ the coefficient of the term $T^{q-2^r}$ is $\binom{j_3}{2^r-1-i_3}a^{2^r-1-i_3}b^{j_3+i_3+1-2^r}$. The condition on $i_3$ and $j_3$ implies that $\binom{j_3}{2^r-1-i_3} = 0$. Therefore $$T^i(aT+b)^j \equiv  T^{i_3}(aT+b)^{j_3}A(T) \mod (T-\alpha^q)^{q+1}-\gamma $$ has no coefficients of degree $T^{q}$, which implies the reduction of $T^i(aT+b)^j \mod P(a^q, \gamma)$ is good.\end{proof}









\begin{corollary}


There are at least $$\sum\limits_{r = 1}^{k} \frac{q}{2^{r+1}}\left(      \binom{\frac{q}{2^r}}{2}(4^r-3^r)  +  \frac{q}{2^r} \binom{2^r+1}{2}\right)  $$ good monomials in $\mathcal{M}$.



\end{corollary}
\begin{proof}
Lemma \ref{lem:goodmonomialsnb} implies that monomials $X^iY^j$ satisfying: $i = i_1q+i_2 2^r+ i_3$ where $0 \leq i_1, i_2, i_3$ where $2^r$ is the highest power of $2$ dividing $i_1$,$0 \leq i_2 < \frac{q}{2^{r}}$ and $0 \leq i_3 < 2^r$ and $j = j_2 2^r + j_3$ where $0 \leq j < q$, $0 \leq j_2 < \frac{q}{2^{r}}$ and $j_3 < 2^r$ are good. Now we count the number of such $(i,j)$ pairs. If $i < q$ there are $\binom{q+1}{2}$ pairs such that $0 \leq i+j \leq q-1.$

There are $\frac{q}{2^{r+1}}$ numbers $0 \leq i_1 \leq q-1$ such that $2^r$ is the highest power of $2$ dividing $i_1$. For each such $i_1$, we count the possible values of $i_2 2^r+i_3$ and $j_2 2^r + j_3$ satisfying the conditions of the corollary. There are $\frac{q}{2^{r}}$ values of $i_2$ and $j_2$ satisfying $i_2+j_2 = \frac{q}{2^r}-1$. For those values, the conditions of the Lemma imply that $i_3+ j_3 \leq 2^r-1$. There are $\binom{2^r+1}{2}$ pairs of $i_3, j_3$ satisfying that condition. There are $\binom{\frac{q}{2^r}}{2}$ pairs of $i_2, j_2$ which satisfy $0 \leq i_2+j_2 \leq \frac{q}{2^r}-2$. Of those there are $4^r-3^r$ pairs of $i_3, j_3$ such that $2^r-1-j_3 \not\leq_2 i_3$. Adding the cases together we get that there are at least $$\sum\limits_{r = 1}^{k} \frac{q}{2^{r+1}}\left(      \binom{\frac{q}{2^r}}{2}(4^r-3^r)  +  \frac{q}{2^r} \binom{2^r+1}{2}\right)  $$ good independent monomials in the footprint. 



\end{proof}

Since the monomials in $\mathcal{M}$ give linearly independent evaluations over the affine points of the Hermitian curve, we have the following corollary.

\begin{corollary} Let $q = 2^k$.  Then $$K \geq \sum\limits_{r = 1}^{k} \frac{q}{2^{r+1}}\left(      \binom{\frac{q}{2^r}}{2}(4^r-3^r)  +  \frac{q}{2^r} \binom{2^r+1}{2}\right)$$  
\end{corollary}


We can use the improved dimension estimate of the Hermitian Lifted code to improve their asymptotic rate bound.


\begin{corollary} Let $q = 2^k$.  Then the rate of $\mathcal{C}$ is at least $0.1$. 
\end{corollary}
\begin{proof}
There are at least $$\sum\limits_{r = 1}^{k} \frac{q}{2^{r+1}}\left(      \binom{\frac{q}{2^r}}{2}(4^r-3^r)  +  \frac{q}{2^r} \binom{2^r+1}{2}\right)  $$ good monomials in $\mathcal{M}$. 

We expand the binomial coefficients $$\sum\limits_{r = 1}^{k} \frac{q}{2^{r+1}}\left(      \frac{\frac{q}{2^r}(\frac{q}{2^r}-1)}{2}(4^r-3^r)  +  \frac{q}{2^r} \frac{(2^r+1)2^r}{2}\right) .$$

We take out $\frac{1}{4}$ as a common term from the sums and rewrrite $\frac{q}{2^r} = 2^{k-r}$ $$\frac{1}{4}\sum\limits_{r = 1}^{k} 2^{k-r}\left(      2^{k-r}(2^{k-r}-1)(4^r-3^r)  +  2^{k-r} (2^r+1)2^r\right) .$$

We collect powers $$\frac{1}{4}\sum\limits_{r = 1}^{k} 2^{k-r}\left(      (4^{k-r}-2^{k-r})(4^r-3^r)  +  2^{k+r}+2^k\right) .$$
$$\frac{1}{4}\sum\limits_{r = 1}^{k} 2^{k-r}\left(         q^2 -q^2\left(\frac{3}{4}\right)^r -q2^r + q\left(\frac{3}{2}\right)^r   +  q2^r+q\right) .$$
$$\frac{1}{4}\sum\limits_{r = 1}^{k} 2^{k-r}\left(         q^2 -q^2\left(\frac{3}{4}\right)^r + q\left(\frac{3}{2}\right)^r +q\right) .$$

We multiply everything by $2^{k-r} = \frac{q}{2^r}$ $$\frac{1}{4}\sum\limits_{r = 1}^{k}         q^3\left(\frac{1}{2}\right)^r -q^3\left(\frac{3}{8}\right)^r + q^2\left(\frac{3}{4}\right)^r +q^2\left(\frac{1}{2}\right)^r .$$

We rewrite the geometric series as:


$$\frac{1}{4} \left( q^3 \frac{1}{2}\frac{1-\left( \frac{1}{2}\right)^k}{1- \frac{1}{2}} -  q^3 \frac{3}{8}\frac{1-\left( \frac{3}{8}\right)^k}{1- \frac{3}{8}} +  q^2 \frac{3}{4}\frac{1-\left( \frac{3}{4}\right)^k}{1- \frac{3}{4}} +  q^2 \frac{1}{2}\frac{1-\left( \frac{1}{2}\right)^k}{1- \frac{1}{2}} \right) .$$

$$\frac{1}{4} \left( q^3 \left(1-\left( \frac{1}{2}\right)^k\right) -  q^3 \frac{3}{5} \left(1-\left( \frac{3}{8}\right)^k\right)+  q^2 3 \left(1-\left( \frac{3}{4}\right)^k\right) +  q^2 \left(1-\left( \frac{1}{2}\right)^k\right) \right) .$$

If we divide by $q^3$ and take the limit as $k$ the only two terms that survive are ones multiplied by $q^3.$ The limit of the first one is $1$ and the limit of the second one is $\frac{3}{5}$. After multiplying by $\frac{1}{4}$ the asymptotic limit is $$r = \frac{1}{4} - \frac{3}{20} = \frac{1}{10}.$$\end{proof}
This rate bound is a vast improvement from the previously known  bound of $0.007$.

\section*{Conclusion}

We have improved the dimension bound of a class of Locally Recoverable codes defined from the Hermitian curve using functions which behave like a low--degree polynomial on each line of the Hermitian curve. The rate of Hermitian--Lifted codes is much  smaller than the rate of the Reed--Muller lifted codes. However, it is remarkable that we can find codes of a positive rate using parity check equations over a field $\F_q$ when the code from the corresponding binary vector would give a code of dimension $0$. We hope that this leads to codes with better rates from the Hermitian unital and other designs.

\section*{Acknowledgements}

The authors would like to thank Beth Malmskog for her insightful comments and thoughtful conversations on the article.

