text:
\begin{document}
\title{ Lie Nilpotency Index of a Modular Group Algebra}
\author{Meena Sahai}
\address{Department of Mathematics and Astronomy, University of Lucknow,
Lucknow, U.P. 226007, India}
\email{meena\_sahai@hotmail.com}
\author{Bhagwat Sharan}
\address{Department of Mathematics and Astronomy, University of Lucknow,
Lucknow, U.P. 226007, India}
\email{bhagwat\_sharan@hotmail.com}
\subjclass[2010]{Primary 16S34 ; Secondary 17B30.}
\keywords{ Modular group algebras; Nilpotency index; Lie dimension subgroups.}
\begin{abstract}
 In this paper, we classify the modular group algebra $KG$ of a group $G$ over a field $K$ of characteristic $p>0$ having upper Lie nilpotency index $t^{L}(KG)= \vert G^{\prime}\vert - k(p-1) + 1$ for $k=14$ and $15$. Group algebras of upper Lie nilpotency index $\vert G^{\prime}\vert - k(p-1) + 1$ for $k\leq 13$, have already been characterized completely.
\end{abstract}
\maketitle
\section{\textbf{Introduction and Preliminaries }}
Let $G$ be a group, not necessarily finite, and let $KG$ be its group algebra over a field $K$ of characteristic $p>0$. Let $KG^{[n]}$ and $KG^{(n)}$ denote the $n$th Lie power  and  upper Lie power of $KG$, respectively. The lower and the upper Lie nilpotency indices of $KG$  are  $t_{L}{(KG)} =  \min \{{n: KG^{[n]} = 0\}}$ and $t^{L}{(KG)} =  \min \{{n: KG^{(n)} = 0\}}$, respectively. In \cite[pp. 46, 48]{Pa}, for every $m\geq 1$, the $m$th Lie dimension subgroup of $G$ over $K$  is defined as
\begin{equation*}
	D_{(m),K}(G) = G\cap(1+KG^{(m)})=\underset{\left( i-1\right) p^{j}\geq m-1}{\Pi }\gamma_{i}\left( G\right)
	^{p^{j}}.
\end{equation*}

If $KG$ is Lie nilpotent such that $|G'|=p^{n}$, then according to Jennings' theory (see \cite{Sh1}), the upper Lie nilpotency index $t^{L}(KG)$ $=$ $2+$ $(p-1)\sum_{m\geq1}md_{(m+1)}$, where $p^{d_{(m)}} = | D_{(m),K}(G) : D_{(m+1),K}(G)|$ for every $m\geq 2$. It is clear that $\sum_{m\geq2} d_{(m)}=n$. A detailed study of Lie nilpotent group algebras and their unit groups is given in \cite{BK}.   In \cite{SB}, it is proved that  if $G$ is a  nilpotent group with $|G'|=p^{n}$, then  $p+1\leq t_{L}(KG)\leq t^{L}(KG)\leq |G'|+1$. A complete description of Lie nilpotent modular group algebras $KG$ of upper Lie nilpotency index $t^{L}(KG)\leq 9p-7$  is given in \cite{Sa,msbs1,msbs2,msbs3,Sh3}. On the other hand, group algebras with $t^{L}(KG)= \vert G^{\prime}\vert - k(p-1) + 1$ for $k\leq 13$ are classified in \cite{Bo,BJS,BSp,BS,msrs2,msrs1,SB}. In this paper, we classify the group algebra $KG$ with $t^{L}(KG)= \vert G^{\prime}\vert - k(p-1) + 1$ for $k=14$ and $15$. our terminology and notations  are same as in \cite{msbs1}. Throughout this paper, $S(n,m)$ denotes the group number $m$ of order $n$ from the Small Groups Library-GAP \cite{GAP}. We have freely used the following lemma in our work:

\begin{lemma} \cite{BP, Sh2, Sh3}{\label{thm1}}
 Let $K$ be a field of characteristic $p>0$ and let $G$ be a nilpotent group such that $ |G^{\prime}| =p^{n}$ and $exp(G^{\prime})= p ^{e}$.
\begin{enumerate}
\item If  $d_{(m+1)}=0$ and $m$ is a power of $p$, then $d_{( s+1) }=0$ for all $s\geq $ $m$ $($i.e. $D_{( m+1),k}(G)=1)$.
\item If $d_{(m+1)}=0$ and $p^{e-1}$ divides $m$, then $D_{( m+1),k}(G)=1$.
\item If $p\geq 5$ and $t_{L}(KG) < p^{n} + 1$, then $t_{L}(KG)=t^{L}(KG)\leq p^{n-1} + 2p - 1$.
\item If $d_{(l+1)}=0$ for some $l < pm$, then $d_(pm + 1)\leq d_{(m+1)}$.
\item If $\ d_{(m+1)}=0$, then $d_{(s+1)}=0$ for all $s\geq m$ with $\upsilon _{p'}(s)\geq \upsilon_{p'}(m)$, where $\upsilon _{p'}(x)$ is the maximal divisor of $x$ which is relatively prime to $p$.
\end{enumerate}
\end{lemma}
\section {Main Results}
\begin{lemma} {\label{l1}}
	Let $G$ be a group and let $K$ be a field of characteristic $p>0$ such that $KG$ is Lie nilpotent. Then  $t^{L}(KG) = \vert G^{\prime}\vert - 14p + 15$ if and only if  $p=2$, $d_{(2)} = d_{(5)} = d_{(9)}=1$, $d_{(3)}=2$.
\end{lemma}
\begin{proof}
Let $t^{L}(KG) = \vert G^{\prime}\vert - 14p + 15$ and let $\vert G'\vert = p^{n}$. If $n=1$, then $G'$ is cyclic and hence by \cite[Theorem 1]{BSp}, $t^{L}(KG) = \vert G^{\prime}\vert  + 1$. If $p=2$, then $t^{L}(KG) = 2^{n} -13$ and so we must have $n\geq 4$. Now if $n=4$, then $3 = t^{L}(KG)\geq 6$. Let $n=5$. So $t^{L}(KG) = 19$,  $\sum\limits_{m=1}^{17}md_{(m+1)}=17$ and $d_{(k)}=0$ for all $k\geq 19$. Also $\sum\limits_{m\geq2}d_{(m)}=5$ and $d_{(2)}\neq 0$. If $d_{(3)}=0$, then $D_{(3), K}(G)=1$ and  $d_{(2)}=17$, which is not possible. Similarly if $d_{(5)}=0$, then $D_{(5), K}(G)=1$ and $d_{(2)} + 2d_{(3)} +3 d_{(4)}=17$. Since $d_{(3)}\neq 0$, so $d_{(4)}\leq 3$ and  $d_{(2)} + 2d_{(3)} +3 d_{(4)} < 17$. If $d_{(9)}=0$, then  $D_{(9), K}(G)=1$ and  $\sum\limits_{m=1}^{7}md_{(m+1)}=17$. Now it is easy to see that $d_{(2)}, d_{(3)}, d_{(5)}\notin \{ 3, 4, 5\} $.  If $d_{(5)}=2$, then the only possible solution is $d_{(2)}= d_{(3)}= d_{(7)}=1$. But then  $\upsilon_{2'}(6)=3=\upsilon_{2'}(3)$, which yields $d_{(7)}=0$, by Lemma \ref{thm1}. Let $d_{(5)}=1$. Then the possible solutions are $d_{(2)}= d_{(3)}= 1$, $ d_{(6)}=2$ and $d_{(2)}= d_{(3)} = d_{(4)} = d_{(8)} = 1  $. Again these cases are not possible by Lemma \ref{thm1}(v). Hence $d_{(9)}\neq 0$ and we have $d_{(2)}=  d_{(5)} = d_{(9)} = 1 $, $d_{(3)} =2$.

Let $p=2$ and $n\geq 6$. We claim that $d_{(2^{i}+1)}>0$ for $0\leq i\leq  n-2$. Let, if possible, there exist some $0\leq s\leq  n-2$ such that $d_{(2^{s}+1)}=0$, then $s\neq 0$ and by Lemma \ref{thm1}, $D_{(2^{s}+1), K}(G)=1$. Thus $d_{(r)}=0$ for all $r\geq 2^{s}+1$. Moreover, if $d_{(q+1)}\neq 0$, then $q < 2^{s}$. Therefore we get
\begin{align*}
 t^{L}(KG)&= 2+ \sum\limits_{i=0}^{s-1} 2^{i}d_{(2^{i}+1)} + \sum\limits_{q\neq 2^{i}}^{}qd_{(q+1)}\\
&  = 2+  \sum\limits_{i=0}^{s-1} 2^{i} + \sum\limits_{i=0}^{s-1} 2^{i}(d_{(2^{i}+1)}-1) + \sum\limits_{q\neq 2^{i}}^{}qd_{(q+1)}\\
& < 2+\sum\limits_{i=0}^{s-1} 2^{i} + \left (\sum\limits_{i=0}^{s-1} (d_{(2^{i}+1)}-1) + \sum\limits_{q\neq 2^{i}}^{}d_{(q+1)}\right ) 2^{s}\\
& = 1 + 2^{s} + (n-s)2^{s}\\
&  < 1 + 2^{n-2} +  \left\lbrace n- (n-2)   \right\rbrace  2^{n-2}\\
&  = 1 +  3.2^{n-2} < 2^{n}-13.
\end{align*}
which is a contradiction to the assumption that $t^{L}(KG) = 2^{n}-13$. Hence $d_{(2^{i}+1)}>0$ for $0\leq i\leq  n-2$ and there exists $\alpha\geq2$ such that $d_{(\alpha +1)}=1$. Thus
\begin{align*}
2^{n}-13 = t^{L}(KG) & = 2+ \sum\limits_{i=0}^{n-2} 2^{i} + \alpha d_{(\alpha +1)}\\
& = 1+ 2^{n-1} + \alpha.
\end{align*}
So $\alpha = 2^{n-1}-14$,  $d_{(2^{i}+1)}= d_{(2^{n-1} -14 )+1}=1$ and $d_{(j)}=0$ where $0\leq i\leq  n-2$, $j\neq 2^{i}+1$, $j\neq 2^{n-1}- 13$ and $j>1$. Let $ m = 2^{n-2}-7$ and $l= \alpha = 2^{n-1}-14$. Since $n\geq 6$, $\upsilon _{2'}(l)= \upsilon_{2'}(m)$ and $d_{(m+1)}=0$, so by Lemma \ref{thm1}(v), $d_{(\alpha +1)}=0$, a contradiction.

Let $p=3$. Then $t^{L}(KG) = 3^{n}-27$. So we must have $n\geq4$. If there exists some $0\leq s\leq n-2$ such that $d_{(3^{s}+1)}=0$, then
\begin{align*}
t^{L}(KG)&= 2+ 2\sum\limits_{i=0}^{s-1} 3^{i}d_{(3^{i}+1)} + 2\sum\limits_{q\neq 3^{i}}qd_{(q+1)}\\
&  = 1 +  5.3^{n-2} < 3^{n}-27.
\end{align*}
Thus $d_{(3^{i}+1)}>0$ for $0\leq i\leq n-2$ and hence $d_{(\alpha +1)}=1$ for $\alpha = 3^{n-1}-14$. For $n=4$, $\upsilon_{3'}(\alpha)=13\geq\upsilon_{3'}(4)$ and so $d_{(\alpha +1)}=0$ by Lemma \ref{thm1}(v). For $n\geq5$, since $exp(G')\leq 3^{n-1}$,
and since $d_{(2.3^{n-2}+1)}=0$, so by Lemma \ref{thm1}(ii), $D_{(2.3^{n-2}+1),K}(G)=1$. But then $d_{(r)}=0$ for all $r\geq 2.3^{n-2}+1$ and so $d_{(3^{n-1}-13) = 0}$.

If $p\in \{5, 7, 11, 13 \}$, then by Lemma \ref{thm1}(iii), $n\leq 2$ and so $t^{L}(KG)<0$. If $p\geq 17$, then $n=1$, by Lemma \ref{thm1}(iii).
\end{proof}
\begin{theorem}
Let $K$ be a field of characteristic $p>0$ and let $G$ be a nilpotent group such that $\vert G'\vert = p^{n}$. Then $t^{L}(KG)= |G'|- 14p+ 15$ if and only if one of the following conditions holds:
\begin{enumerate}
\item $G'\cong C_{16}\times C_{2}$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$ and $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}$;
\item \begin{enumerate}
\item $G'\cong C_{16}\times C_{2}$, $\gamma_{3}(G)\cong C_{8}$ and $\gamma_{4}(G)\subseteq G'^{4}=\gamma_{3}(G)^{2}=G'^{2}\cap\gamma_{3}(G)\cong C_{4}$;

\item $G'\cong C_{16}\times C_{2}$, $\gamma_{3}(G)\cong C_{4}\times C_{2}$, $G'^{2}\cap\gamma_{3}(G)\cong C_{4}$ and $\gamma_{4}(G)$, $\gamma_{3}(G)^{2}\subseteq G'^{4}$;
\end{enumerate}
\item $G'\cong C_{16}\times C_{2}$, $|\gamma_{3}(G)|=4$, $\vert G'^{2}\cap\gamma_{3}(G)\vert= 2$ and $\gamma_{4}(G)$, $\gamma_{3}(G)^{2}\subseteq G'^{4}$;

\item $G'\cong C_{16}\times C_{2}$, $\gamma_{3}(G)\cong C_{2}$ and $G'^{2}\cap\gamma_{3}(G)= 1$;

\item $G'\cong C_{8}\times C_{4}$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$ and $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}$;

\item $G'\cong C_{8}\times C_{4}$, $\gamma_{3}(G)\cong C_{8}$, $\vert G'^{2}\cap\gamma_{3}(G)\vert= 2$ and $\gamma_{4}(G)$, $G'^{4}\subseteq\gamma_{3}(G)^{2}$;

\item $G'\cong C_{8}\times C_{2}\times C_{2}$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$ and $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}$;

\item $G'\cong C_{8}\times C_{2}\times C_{2}$, $\gamma_{3}(G)\cong C_{8}$, $\vert G'^{2}\cap\gamma_{3}(G)\vert= 2$ and $\gamma_{4}(G)$, $G'^{4}\subseteq\gamma_{3}(G)^{2}$;

\item $G'\cong S(32,4)$ or $S(32,5)$ or $S(32,12)$,  $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)= \gamma_{3}(G)^{2}$ and $\gamma_{5}(G)\cong C_{2}$;

\item $G'\cong S(32,4)$, $\gamma_{3}(G)\cong C_{8}$, $G'^{4}\subseteq G'^{2}\cap\gamma_{3}(G)= \gamma_{4}(G)= \gamma_{3}(G)^{2}$ and  $\gamma_{5}(G)\cong C_{2}$;

\item $G'\cong S(32,37)$  and  $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)= \gamma_{3}(G)^{2}$, $\gamma_{5}(G)\cong C_{2}$;

\item $G'\cong S(32,17)$, $\gamma_{5}(G)\cong C_{2}$  and  $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)= \gamma_{3}(G)^{2}$  or $\gamma_{3}(G)\cong C_{8}$, $ G'^{2}\cap\gamma_{3}(G)= \gamma_{4}(G)= \gamma_{3}(G)^{2}= G'^{4}$ or $\gamma_{3}(G)\cong C_{4}\times C_{2}$, $\gamma_{3}(G)^{2}\subseteq G'^{2}\cap\gamma_{3}(G)= \gamma_{4}(G)=G'^{4}\cong C_{4}$;

\item $G'\cong S(32,38)$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)= \gamma_{3}(G)^{2}$ and $\gamma_{5}(G)\cong C_{2}$.
\end{enumerate}
\end{theorem}

\begin{proof}
Let $t^{L}(KG)= |G'|- 14p+ 15$. Then by Lemma \ref*{l1}, $p=2$ and $d_{(2)} = d_{(5)} = d_{(9)}=1$, $d_{(3)}=2$. Therefore $|G'|= 2^{5}$, $|D_{(3),K}(G)|=2^{4}$, $|D_{(4),K}(G)|=|D_{(5),K}(G)|=2^{2}$, $|D_{(6),K}(G)|= |D_{(7),K}(G)|= |D_{(8),K}(G)|$ $ = |D_{(9),K}(G)|=2$ and $D_{(10),K}(G)= 1$. Since $|D_{(4),K}(G)|= 4$, so $|\gamma_{4}(G)|\leq 4$ and  $\gamma_{6}(G)=1$. Thus  $G'$ is nilpotent of class at most $2$, $\gamma_{3}(G)$ is abelian and $\gamma_{4}(G)\subseteq\zeta(G')$. Also $\vert\gamma_{5}(G)\vert\leq 2$, so $\gamma_{4}(G)^{2}\subseteq\gamma_{3}(G)^{4}G'^{8}\cong C_{2}$ and $G'^{4}\neq 1$. Hence exponent of $\gamma_{3}(G)^{2}G'^{4}$ is not equal to $2$ and $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}G'^{4}\cong C_{4}$.

Let $G'$ be an abelian group. Then  $G'\cong C_{16}\times C_{2}$ or $C_{8}\times C_{4}$ or $ C_{8}\times C_{2}\times C_{2}$. If $G'\cong C_{16}\times C_{2}$, then $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}$ or $\gamma_{3}(G)\cong C_{8}$, $\gamma_{4}(G)\subseteq G'^{4}=\gamma_{3}(G)^{2} = G'^{2}\cap\gamma_{3}(G)\cong C_{4}$  or  $\gamma_{3}(G)\cong C_{4}\times C_{2}$, $G'^{2}\cap\gamma_{3}(G)\cong C_{4}$, $\gamma_{3}(G)^{2}\subseteq G'^{4}$, $\gamma_{4}(G)\subseteq G'^{4}$ or  $|\gamma_{3}(G)|=4$, $G'^{2}\cap\gamma_{3}(G)\cong C_{2}$, $\gamma_{3}(G)^{2}\subseteq G'^{4}$, $\gamma_{4}(G)\subseteq G'^{4}$ or $|\gamma_{3}(G)|=2$, $G'^{2}\cap\gamma_{3}(G)=1$. If $G'^{8}=1$, then $\gamma_{3}(G)^{4}\cong C_{2}$, so $|\gamma_{3}(G)|=8$ or $16$. Thus $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}$ or $\gamma_{3}(G)\cong C_{8}$, $G'^{2}\cap\gamma_{3}(G)\cong C_{2}$, $G'^{4}\subseteq\gamma_{3}(G)^{2}$, $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}$.

Let $G'$ be a nonabelian group. If $\gamma_{5}(G)\cong C_{2}$, then $\gamma_{4}(G)=\gamma_{3}(G)^{2}G'^{4}$ $\cong C_{4}$. Thus $|\zeta(G')|=4 $ or $8$. If $|\zeta(G')|=4$, then $\gamma_{4}(G)=\zeta(G')\cong C_{4}$ and we see from Table \ref{tab:table1} that $cl(G') = 3$. Thus $\zeta(G')\cong C_{8}$ or $C_{4}\times C_{2}$. If $\zeta(G')\cong C_{8}$, then from Table \ref{tab:table1}, the possible choices for $G'$ are $S(32,17)$ or $S(32,38)$. If $G'\cong S(32,17)$, then $G'^{2}= \zeta(G')\cong C_{8}$ and so $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)= \gamma_{3}(G)^{2}$ or $\gamma_{3}(G)\cong C_{8}$, $G'^{2}\cap\gamma_{3}(G)= \gamma_{4}(G)= G'^{4}=\gamma_{3}(G)^{2}\cong C_{4}$ or $\gamma_{3}(G)\cong C_{4}\times C_{2}$, $G'^{2}\cap\gamma_{3}(G)= \gamma_{4}(G)=G'^{4}\cong C_{4}$, $\gamma_{3}(G)^{2}\subseteq G'^{4}$. If $G'\cong S(32,38)$, then $G'^{2}\cong C_{4}$ and so  $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)= \gamma_{3}(G)^{2}$. If $\zeta(G')\cong C_{4}\times C_{2}$, then the possible  choices for $G'$  are  $S(32,4)$ or $S(32,5)$ or $S(32,12)$ or $S(32,37)$. If $G'\cong S(32,4)$ or $S(32,5)$ or $S(32,12)$, then $G'^{2}\cong C_{4}\times C_{2}$  so either  $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)= \gamma_{3}(G)^{2}$ or $\gamma_{3}(G)\cong C_{8}$, $\gamma_{4}(G)= G'^{2}\cap\gamma_{3}(G)\cong C_{4}$, $G'^{4}\subseteq\gamma_{4}(G)=\gamma_{3}(G)^{2}$.
But $S(32,5)$ or $S(32,12)$ do not contain a normal subgroup of the type $C_{8}$.  If $G'\cong S(32,37)$, then $G'^{2}\cong C_{4}$ and so $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}\times C_{2}$, $\gamma_{4}(G)= \gamma_{3}(G)^{2}$.

If $\gamma_{5}(G)=1$, then $\gamma_{3}(G)\subseteq\zeta(G')$ and so $|\gamma_{3}(G)|\neq 16$. Also $G'^{4}\neq1$. Thus $|\gamma_{4}(G)|\geq2$ and $|\gamma_{3}(G)|= 4$ or $8$. If  $\gamma_{3}^{4}(G)=1$, then  $G'^{8}\cong C_{2}$. Therefore $G'\cong S(32,17)$. But then $\gamma_{3}(G)\subseteq\zeta(G')= G'^{2}$ and $|D_{(3),K}(G)|=8$. So $\gamma_{3}(G)=\zeta(G')\cong C_{8}$ and the possible choices for $G'$ are $S(32,17)$ or $S(32,38)$. If $G'\cong S(32,17)$ or $S(32,38)$ , then $G'^{2}\subseteq\zeta(G') \cong C_{8}$ and so $|D_{(3),K}(G)|\neq16$.\\
Conversely, if conditions (i) to (xiii) holds, then  $d_{(2)} = d_{(5)} = d_{(9)}=1$, $d_{(3)}=2$. So $n=5$ and $t^{L}(KG)=19= |G'|- 14p+ 15$.

\end{proof}

\begin{lemma}{\label{l2}}
Let $G$ be a group and let $K$ be a field of characteristic $p>0$ such that $KG$ Lie nilpotent. Then  $t^{L}(KG) = \vert G^{\prime}\vert - 15p + 16$ if and only if one of the following conditions holds :
\begin{enumerate}
\item   $p=2$, $n=5$ and $d_{(2)} = d_{(3)} = d_{(4)} = d_{(5)}= d_{(7)}=1$;
\item   $p=2$, $n=5$ and $d_{(2)}=2$, $ d_{(3)} = d_{(5)}= d_{(9)}=1$;
\item $p=17$, $n=2$ and $d_{(2)} = d_{(3)} =1$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $t^{L}(KG) = \vert G^{\prime}\vert - 15p + 16$ and let $\vert G^{\prime}\vert = p^{n}$. As in Lemma \ref{l1}, we can not have $n=1$. It is easy to see that if $p=2$, then $n\geq 5$, if $p=3$, then $n\geq 4$, if $p\in \{5, 7, 11, 13\}$, then $n\geq 3$ and if $p\geq 17$, then $n\geq 2$.

 If $p=2$ and $n=5$, then we have $\sum\limits_{m=1}^{16}md_{(m+1)}=16$ and $d_{(k)}=0$ for all $k\geq 17$. Also $\sum_{m\geq 2} d_{(m)}=5$ and $d_{(2)}$, $d_{(3)}$, $d_{(5)}\neq 0$. If $d_{(9)}=0$, then $\sum\limits_{m=1}^{7}md_{(m+1)}=16$. Clearly $d_{(2)}, d_{(3)}, d_{(5)} \notin\{ 3, 4, 5 \}$. If $d_{(5)}=2$, then  $d_{(2)} = d_{(3)} = d_{(6)}=1$. But then $2= d_{(2.2+1)}\leq d_{(2+1)}=1$ by Lemma \ref{thm1}(iv). So $d_{(5)}=1$, and either $d_{(2)}= d_{(8)}= 1$, $ d_{(3)}=2$ or $d_{(2)}= d_{(3)} = d_{(4)} = d_{(7)} = 1$. If  $d_{(2)}= d_{(5)}= d_{(8)}= 1$, $ d_{(3)}=2$, then  $\upsilon _{2'}(7)\geq \upsilon_{2'}(5)$ and  $d_{(8)}=0$ by Lemma \ref{thm1}(v), so this is not possible. Clearly, if $d_{(9)}\neq 0$, then $d_{(2)}=2$, $d_{(3)}= d_{(5)}= d_{(9)} =1$.

Let $p=2$ and $n\geq 6$, then proceeding in the same way as we have done in Lemma \ref*{l1}, we obtain  $d_{(2^{i}+1)}= d_{(2^{n-1} -15 )+1}=1$ and $d_{(j)}=0$ where $0\leq i\leq  n-2$, $j\neq 2^{i}+1$, $j\neq 2^{n-1}- 14$ and $j>1$. Let $ m = 2^{n-2}+1$ and $l=  2^{n-1}-15$. Since $n\geq 6$, $\upsilon_{2'}(l)\geq \upsilon_{2'}(m)$ and by Lemma \ref{thm1}(v) we get $d_{(2^{n-1}-14)}=0$,  which is a contradiction.

If $p=3$ and $n\geq 4$, then similarly  we get $d_{(3^{i}+1)}= d_{(3^{n-1} -15 )+1}=1$ and $d_{(j)}=0$ where $0\leq i\leq n-2$, $j\neq 3^{i}+1$, $j\neq 3^{n-1}- 14$ and $j>1$. Let $ m = 3^{n-2}-5$ and $l=  3^{n-1}-15$. Since $n\geq 4$, $\upsilon_{3'}(l)= \upsilon_{3'}(m)$ and so $d_{(l+1)}=0$,  a contradiction.

Let $p\in \{5, 7, 11, 13 \}$. Then by Lemma \ref{thm1}(iii), $n=2$ which is not possible. Also if $p=17$, then $n=2$ and $d_{(2)}= d_{(3)}=1$. We can not have $p\geq 19$  by Lemma \ref{thm1}(iii).
\end{proof}

\begin{lemma}{\label{l3}}
Let $K$ be a field with characteristic $p=2$ and let $G$ be a nilpotent group such that $\vert G'\vert = 2^{n}$. Then $t^{L}(KG)= 2^{n}-14$ if and only if one of the following conditions holds:

\begin{enumerate}
\item $G'\cong C_{16}\times C_{2}$, $\gamma_{3}(G)\subseteq G'^{2}$ and $\gamma_{4}(G)\subseteq G'^{4}$;
\item $G'\cong C_{8}\times C_{2}\times C_{2}$, $G'^{2}\subseteq \gamma_{3}(G)\cong C_{8}$ and $\gamma_{4}(G)\subseteq \gamma_{3}(G)^{2}$;
\item \begin{enumerate} \item $G'\cong S(32,17)$ or $S(32,38)$, $G'^{2}\subseteq \gamma_{3}(G)\cong C_{8}$, $\gamma_{4}(G)\subseteq \gamma_{3}(G)^{2}$ and $\gamma_{5}(G)=1$;

\item $G'\cong S(32,17)$ or $S(32,38)$, $G'^{2}\subseteq \gamma_{3}(G)\cong C_{8}$, $\gamma_{4}(G)=\gamma_{3}(G)^{2}$ and $\gamma_{5}(G)\cong C_{2}$;
\end{enumerate}
\item $G'\cong S(32,37)$, $G'^{2}\subseteq \gamma_{3}(G)\cong C_{8}$, $\gamma_{4}(G)=\gamma_{3}(G)^{2}$ and $\gamma_{5}(G)\cong C_{2}$;

\item $G'\cong S(32,17)$, $\gamma_{3}(G)\subseteq G'^{2}\cong C_{8}$, $\gamma_{3}(G)\cong C_{4}$;

\item $G'\cong C_{8}\times C_{2}\times C_{2}$ or $C_{4}\times C_{4}\times C_{2}$ or  $C_{4}\times C_{2}\times C_{2}\times C_{2}$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{2}\times C_{2}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$ and $\gamma_{6}(G)\subseteq\gamma_{3}(G)^{2}= \gamma_{4}(G)^{2}\subseteq\gamma_{5}(G)\cong C_{2}\times C_{2}$;
\item $G'\cong S(32,5)$ or $S(32,24)$ or $S(32,25)$ or $S(32,37)$ or $S(32,48)$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{2}\times C_{2}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$, $\gamma_{3}(G)^{2}\subseteq\gamma_{5}(G)\cong C_{2}\times C_{2}$ and $\gamma_{6}(G)=1$;

\item $G'\cong S(32,28)$ to $S(32,30)$,  $G'^{2} \subseteq \gamma_{3}(G)\cong C_{4}\times C_{2}\times C_{2}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$ and $\gamma_{6}(G)=\gamma_{3}(G)^{2}\subseteq\gamma_{5}(G)\cong C_{2}\times C_{2}$;

\item $G'\cong S(32,2)$ or  $S(32,5)$  or $S(32,22)$ to $S(32,25)$ or $S(32,37)$ or $S(32,46)$ to $S(32,48)$,  $G'^{2} \subseteq \gamma_{3}(G)\cong C_{4}\times C_{2}\times C_{2}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$ and $\gamma_{6}(G)=\gamma_{3}(G)^{2}\subseteq\gamma_{5}(G)\cong C_{2}\times C_{2}$;

\item $G'\cong C_{8}\times C_{2}\times C_{2}$ or $C_{4}\times C_{4}\times C_{2}$ or  $C_{4}\times C_{2}\times C_{2}\times C_{2}$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{2}\times C_{2}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$, $\gamma_{3}(G)^{2}\subseteq\gamma_{4}(G)$, $\gamma_{5}(G)\cong C_{2}$ and	 $\gamma_{5}(G)\cap\gamma_{3}(G)^{2}=1$;

\item $G'\cong S(32,5)$ or $S(32,24)$, $S(32,25)$ or $S(32,37)$ or $S(32,48)$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{2}\times C_{2}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$, $\gamma_{3}(G)^{2}\subseteq\gamma_{4}(G)$, $\gamma_{5}(G)\cong C_{2}$ and	 $\gamma_{5}(G)\cap\gamma_{3}(G)^{2}=1$;

\item $G'\cong C_{8}\times C_{4}$ or $C_{4}\times C_{4}\times C_{2}$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{4}$, $\gamma_{5}(G)\subseteq\gamma_{3}(G)^{2}\subseteq\gamma_{4}(G)\cong C_{4}\times C_{2}$;

\item $G'\cong S(32,4)$  or $S(32,12)$ or $S(32,24)$ to $S(32,26)$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{4}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$, $\gamma_{5}(G)=\gamma_{3}(G)^{2}$ and $\gamma_{6}(G)=1$;

\item $G'\cong S(32,4)$ or $S(32,12)$ or $S(32,24)$ to $S(32,26)$ or $S(32,31)$ to $S(32,35)$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{4}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$, $\gamma_{5}(G)=\gamma_{3}(G)^{2}$ and $\gamma_{6}(G)\cong C_{2}$;

\item $G'\cong S(32,4)$ or $S(32,12)$ or $S(32,24)$ to $S(32,26)$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{4}$, $\gamma_{5}(G)\subseteq\gamma_{3}(G)^{2}\subseteq\gamma_{4}(G)\cong C_{4}\times C_{2}$ and $\gamma_{5}(G)\cong C_{2}$;

\item $G'\cong C_{8}\times C_{4}$ or $C_{4}\times C_{4}\times C_{2}$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{4}$, $\gamma_{4}(G)\cong C_{4}$, $|\gamma_{3}(G)^{2}\cap \gamma_{4}(G)|=2$ and $\gamma_{5}(G)\subseteq\gamma_{3}(G)^{2}$;

\item $G'\cong S(32,4)$ or $S(32,12)$ or $S(32,24)$ to $S(32,26)$, $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{4}$, $\gamma_{4}(G)\cong C_{4}$, $|\gamma_{3}(G)^{2}\cap \gamma_{4}(G)|=2$, $\gamma_{5}(G)\subseteq\gamma_{3}(G)^{2}$ and $\gamma_{5}(G)\cong C_{2}$;

\item $G'\cong S(32,4)$ or $S(32,5)$ or $S(32,12)$ or $S(32,17)$, $\gamma_{3}(G)\cong C_{4}\times C_{2}$, $|G'^{2}\cap \gamma_{3}(G)|=4$, $\gamma_{4}(G)\cong C_{4}$, $|\gamma_{3}(G)^{2}G^{\prime 4}\cap \gamma_{4}(G)|=2$,
$\gamma_{5}(G)\cong C_{2}$ and $\gamma_{5}(G)\subseteq \gamma_{3}(G)^{2}G^{\prime 4}\cong C_{2}\times C_{2}$;

\item $G'\cong C_{8}\times C_{2}\times C_{2}$ or $C_{4}\times C_{4}\times C_{2}$, $\gamma_{3}(G)\cong C_{4}\times C_{2}$, $|G'^{2}\cap\gamma_{3}(G)|=2$, $\gamma_{4}(G)\cong C_{4}$, $|\gamma_{3}(G)^{2}G^{\prime 4}\cap\gamma_{4}(G)|=2$
and $\gamma_{5}(G)\subseteq \gamma_{3}(G)^{2}G^{\prime 4}\cong C_{2}\times C_{2}$;

\item $G'\cong S(32,24)$ or $S(32,25)$ or $S(32,26)$ or $S(32,37)$ or $S(32,38)$, $\gamma_{3}(G)\cong C_{4}\times C_{2}$, $|G'^{2}\cap \gamma_{3}(G)|=2$, $\gamma_{4}(G)\cong C_{4}$, $|\gamma_{3}(G)^{2}G^{\prime 4}\cap \gamma_{4}(G)|=2$,
$\gamma_{5}(G)\cong C_{2}$ and $\gamma_{5}(G)\subseteq \gamma_{3}(G)^{2}G^{\prime 4}\cong C_{2}\times C_{2}$.

\end{enumerate}
\end{lemma}

\begin{proof}
Let $t^{L}(KG)= 2^{n}-14$, then by Lemma \ref{l2}, we have either  $d_{(2)} = d_{(3)}= d_{(4)} = d_{(5)} = d_{(7)} = 1$ or $d_{(2)}=2$, $d_{(3)} = d_{(5)} = d_{(9)}=1$. We consider these two cases separately.\\
{\bf Case (I).} $d_{(2)}=2$, $d_{(3)} = d_{(5)} = d_{(9)}=1$.
Thus $|G'|= 2^{5}$, $|D_{(3),K}(G)|=2^{3}$, $|D_{(4),K}(G)|=|D_{(5),K}(G)|=2^{2}$, $D_{(6),K}(G)=D_{(7),K}(G)=D_{(8),K}(G)= D_{(9),K}(G) \cong C_2$ and $D_{(10),K}(G)= 1$. Since $|D_{(4),K}(G)|$ $= 4$, so $\gamma_{6}(G)=1$ and nilpotency class of $G'$ is at most $2$. Also $|\gamma_{5}(G)^{2}\gamma_{3}(G)^{4}G'^{8}|= 2$ leads to $|\gamma_{3}(G)^{4}G'^{8}|=2$.
Hence $G'^{4}\neq1$, $exp(\gamma_{3}(G)^{2}G'^{4})= 4$ and $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}G'^{4}\cong C_{4}$. Let $G'$ be an abelian group. Then $G'\cong C_{16}\times C_{2}$ or $C_{8}\times C_{4}$ or $ C_{8}\times C_{2}\times C_{2}$. If $G'\cong C_{16}\times C_{2}$, then $\gamma_{3}(G)\subseteq G'^{2}$ and $\gamma_{4}(G)\subseteq G'^{4}$. If $G'\cong C_{8}\times C_{4}$, then again  $\gamma_{3}(G)\subseteq G'^{2}$. But then $\gamma_{3}(G)^{4}G'^{8} = G'^{8}=1$. If $G'\cong C_{8}\times C_{2}\times C_{2}$, then $|\gamma_{3}(G)^{4}G'^{8}|=2$ yields that
$G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}$ and $\gamma_{4}(G)\subseteq\gamma_{3}(G)^{2}$.

Let $G'$ be a nonabelian group. Then $\gamma_{4}(G)\neq1$. If $|\gamma_{3}(G)|=8$, then $G'^{2}\subseteq\gamma_{3}(G)\cong C_{8}$. Thus if $\gamma_{5}(G)=1$, then $\gamma_{3}(G)=\zeta(G')$ and so possible choices for  $G'$ are $S(32,17)$ or $S(32,38)$. On the other hand, if $|\gamma_{5}(G)|=2$, then $G'^{4}\gamma_{3}(G)^{2}=\gamma_{4}(G)\subseteq\zeta(G')$ and hence $\zeta(G')\cong C_{4}$ or $C_{4}\times C_{2}$ or $C_{8}$. From Table \ref{tab:table1}, we see that we can not have $\zeta(G')\cong C_{4}$. Thus the possible choices for $G'$ are  $S(32,4)$ or $S(32,5)$ or $S(32,12)$ or $S(32,17)$ or $S(32,37)$ and $S(32,38)$. But if $G'\cong S(32,4)$ or $S(32,5)$ or $S(32,12)$, then $G'^{2}\cong C_{4}\times C_{2}$ whereas $\gamma_{3}(G)\cong C_{8}$. If $|\gamma_{3}(G)|= 4$, then $G'^{8}\cong C_{2}$, so $exp (G')=16$, $G'\cong S(32,17)$, $\gamma_{3}(G)\subseteq G'^{2}=\zeta(G')\cong C_{8}$ and
$\gamma_{4}(G)\subseteq G'^{4}$.\\
{\bf Case (II).} $d_{(2)} = d_{(3)}= d_{(4)} = d_{(5)} = d_{(7)}=1$. Thus $|G'|= 2^{5}$, $|D_{(3),K}(G)|=2^{4}$, $|D_{(4),K}(G)|= 2^{3}$, $|D_{(5),K}(G)|=2^{2}$, $D_{(6),K}(G)= D_{(7),K}(G)\cong C_{2}$ and $D_{(8),K}(G)=\gamma_{8}(G)\gamma_{5}(G)^{2}\gamma_{3}(G)^{4}G^{\prime 8}=1$. Hence $|\gamma_{5}(G)|\leq4$, $\gamma_{6}(G)\subseteq\gamma_{4}(G)^{2}\cong C_{2}$, $\gamma_{7}(G)=1$ and  $|\gamma_{3}(G)^{2}G^{\prime 4}|=2$ or $4$.\\
{\bf{(a)}}  $|\gamma_{3}(G)^{2}G^{\prime 4}|=2$. Clearly $\gamma_{4}(G)^{2}= \gamma_{3}(G)^{2}=\gamma_{3}(G)^{2}G^{\prime 4}\cong C_{2}$.

If $|\gamma_{5}(G)|=4$, then $\gamma_{3}(G)^{2}G^{\prime 4}\subseteq \gamma_{5}(G)\cong C_{2}\times C_{2}$, $\gamma_{4}(G)\cong C_{4}\times C_{2}$ and  $G'^{2}\subseteq \gamma_{3}(G)$. Since $\gamma_{7}(G)=1$, we have $\gamma_{4}(G)\subseteq\zeta(\gamma_{3}(G))$ and hence $\gamma_{3}(G)$ must be abelian and isomorphic to $C_{4}\times C_{2}\times C_{2}$. Thus if $G'$ is abelian, then $G'\cong C_{8}\times C_{2}\times C_{2}$ or $C_{4}\times C_{4}\times C_{2}$ or $C_{4}\times C_{2}\times C_{2}\times C_{2}$. Now let $G'$ be nonabelian. If $\gamma_{6}(G)=1$, then $cl(G')=2$ and $\gamma_{4}(G)=\zeta(G')\cong C_{4}\times C_{2}$.
So possible choices for $G'$ are $S(32,4)$ or $S(32,5)$ or $S(32,12)$ or  $S(32,24)$ or $S(32,25)$ or $S(32,26)$ or $S(32,37)$ or $S(32,48)$. Since $S(32,4)$ or $S(32,12)$ or $S(32,26)$ do not contain a normal subgroup of the type $C_{4}\times C_{2}\times C_{2}$, so these groups are not possible. Hence $G'\cong S(32,5)$ or  $S(32,24)$ or $S(32,25)$  or $S(32,37)$ or $S(32,48)$. If $\gamma_{6}(G)\cong C_{2}$, then $\gamma_{5}(G)\subseteq\zeta(G')$. Thus $|\zeta(G')|= 4$ or $8$. If $|\zeta(G')|= 4$, then $\gamma_{5}(G)=\zeta(G')\cong C_{2}\times C_{2}$ and hence possible choices for $G'$ are $S(32,9)$ or $S(32,10)$ or $S(32,13)$ or $S(32,14)$ or $S(32,27)$ to  $S(32,35)$ or $S(32,39)$ to $S(32,41)$. But the groups $S(32,9)$ or $S(32,10)$ or $S(32,13)$ or $S(32,14)$ or $S(32,27)$ or $S(32,31)$ to  $S(32,35)$ or $S(32,39)$ to $S(32,41)$, do not contain a normal subgroup of the type $C_{4}\times C_{2}\times C_{2}$. Hence $G'\cong S(32,28)$ to  $S(32,30)$. If $|\zeta(G')|= 8$, then either $\zeta(G')\cong C_{2}\times C_{2}\times C_{2}$ or $C_{4}\times C_{2}$. So possible choices for $G'$ are $S(32,2)$ or $S(32,4)$ or $S(32,5)$ or $S(32,12)$ or $S(32,22)$ to $S(32,26)$ or $S(32,37)$ or $S(32,46)$ to $S(32,48)$. But a normal subgroup of the type $C_{4}\times C_{2}\times C_{2}$ is not possible in  $S(32,4)$, $S(32,12)$ and $S(32,26)$. Hence $G'\cong S(32,2)$ or  $S(32,5)$ or $S(32,22)$ to $S(32,25)$ or $S(32,37)$ or $S(32,46)$ to $S(32,48)$.

If  $|\gamma_{5}(G)|=2$,, then $\gamma_{4}(G)\subseteq\zeta(G')$. Now if  $|\gamma_{4}(G)|=4$, then $|D_{(4),K}(G)|=8$ leads to $G'^{4}\gamma_{3}(G)^{2}\cap \gamma_{4}(G)=1$, but $\gamma_{4}(G)^{2}\neq1$. So $\gamma_{4}(G)\cong C_{4}\times C_{2}$ and $G'^{2}\subseteq\gamma_{3}(G)\cong C_{4}\times C_{2}\times C_{2}$. If $G'$ is abelian, then $G'\cong C_{8}\times C_{2}\times C_{2}$  or $C_{4}\times C_{4}\times C_{2}$ or $C_{4}\times C_{2}\times C_{2}\times C_{2}$. If $G'$ is nonabelian, then $\gamma_{4}(G)=\zeta(G')\cong C_{4}\times C_{2}$. Thus possible choices for $G'$ are $S(32,4)$ or $S(32,5)$ or $S(32,12)$ or  $S(32,24)$ or $S(32,25)$ or $S(32,26)$ or $S(32,37)$ or $S(32,48)$. Again we observe that $G'\ncong S(32,4)$, $S(32,12)$ and $S(32,26)$ as these groups do not contain a normal subgroup of the type $C_{4}\times C_{2}\times C_{2}$. Hence $G'\cong S(32,5)$ or  $S(32,24)$ or $S(32,25)$  or $S(32,37)$ or $S(32,48)$.\\~\\
{\bf(b)} $|\gamma_{3}(G)^{2}G^{\prime 4}|=4$. Then $\gamma_{5}(G)\subseteq \gamma_{3}(G)^{2}G^{\prime 4}\cong C_{2}\times C_{2}$. Also $\gamma_{5}(G)\subseteq\zeta(G')$ and $|\gamma_{4}(G)|=8$ or $4$. We consider these two cases separately :

 {\bf(i)} If $|\gamma_{4}(G)|=8$, then $G'^{2}\subseteq\gamma_{3}(G)$ and $\gamma_{3}(G)^{2}G^{\prime 4}\subseteq \gamma_{4}(G)\cong C_{4}\times C_{2}$. Also $\gamma_{4}(G)\subseteq\zeta(\gamma_{3}(G))$, so $\gamma_{3}(G)$ is  abelian. Since $\gamma_{3}(G)^{2}=\gamma_{3}(G)^{2}G^{\prime 4}\cong C_{2}\times C_{2}$, so $\gamma_{3}(G)\cong C_{4}\times C_{4}$. Thus if $G'$ is abelian, then $G'\cong C_{8}\times C_{4}$ or $C_{4}\times C_{4}\times C_{2}$. Let $G'$ be nonabelian. If $\gamma_{5}(G)=1$, then $\gamma_{3}(G)\subseteq\zeta(G')$ and $G'$ is abelian. So $\gamma_{5}(G)\neq1$.

If $|\gamma_{5}(G)|=4$, then $\gamma_{5}(G)=\gamma_{3}(G)^{2}G^{\prime 4}\cong C_{2}\times C_{2}$. Now if $\gamma_{6}(G)=1$, then $\gamma_{4}(G)\subseteq\zeta(G')$ and hence possible choices for $G'$ are $S(32,4)$ or $S(32,5)$ or $S(32,12)$ or $S(32,24)$ to $S(32,26)$  or $S(32,37)$ or $S(32,48)$. But a normal subgroup of the type $C_{4}\times C_{4}$ is not available in $S(32,5)$, $S(32,37)$ and $S(32,48)$, so these groups are not possible. Hence $G'\cong S(32,4)$ or $S(32,12)$ or $S(32,24)$ to $S(32,26)$. If $\gamma_{6}(G)\neq1$, then $C_{2}\times C_{2}\subseteq\zeta(G')$. So $\zeta(G')\cong C_{2}\times C_{2}$ or $C_{4}\times C_{2}$ or $C_{2}\times C_{2}\times C_{2}$. Therefore possible choices for $G'$ are $S(32,2)$ or $S(32,4)$ or $S(32,5)$ or $S(32,9)$ or $S(32,10)$ or $S(32,12)$ or $S(32,13)$ or $S(32,14)$ or $S(32,22)$ to  $S(32,35)$ or  $S(32,37)$ or $S(32,39)$ to $S(32,41)$ or $S(32,46)$ to  $S(32,48)$. But the groups $S(32,2)$, $S(32,5)$, $S(32,9)$, $S(32,10)$, $S(32,13)$, $S(32,14)$, $S(32,22)$, $S(32,23)$, $S(32,27) - S(32,30)$, $S(32,37)$, $S(32,39)-S(32,41)$ and $S(32,46)- S(32,48)$ do not have a normal subgroup of the type $ C_{4}\times C_{4}$. Hence $G'\cong S(32,4)$ or $S(32,12)$ or $S(32,24)$ to  $S(32,26)$ or  $S(32,31)$ to $S(32,35)$.

If $|\gamma_{5}(G)|=2$, then $\gamma_{4}(G)=\zeta(G')\cong C_{4}\times C_{2}$. So possible choices for $G'$ are  $S(32,4)$ or $S(32,5)$ or $S(32,12)$ or $S(32,24)$ to $S(32,26)$ or $S(32,37)$ or $S(32,48)$. But $\gamma_{3}(G)\cong C_{4}\times C_{4}$ is not possible in $S(32,5)$, $S(32,37)$, $S(32,38)$ and $S(32,48)$. Hence $G'\cong S(32,4)$ or $S(32,12)$ or $S(32,24)$ or $S(32,25)$ or $S(32,26)$.

{\bf(ii)} If $|\gamma_{4}(G)|=4$, then $\gamma_{4}(G)\cong C_{4}$, $|\gamma_{3}(G)^{2}G^{\prime 4}\cap \gamma_{4}(G)|=2$, $\gamma_{4}(G)\subseteq\zeta(G')$ and $|\gamma_{3}(G)|=16$ or $8$.

If $|\gamma_{3}(G)|=16$, then $G'^{2}\subseteq\gamma_{3}(G)$. Therefore $\gamma_{3}(G)^{2}\cong C_{2}\times C_{2}$ and hence $\gamma_{3}(G)\cong C_{4}\times C_{4}$. If $G'$ is abelian, then $G'\cong C_{8}\times C_{4}$ or  $C_{4}\times C_{4}\times C_{2}$. If $G'$ is nonabelian, then $\gamma_{5}(G)\cong C_{2}$, $cl(G')= 2$ and $\zeta(G')\cong C_{4}$ or $C_{8}$ or $C_{4}\times C_{2}$. So possible choices for $G'$ are $S(32,4)$ or $S(32,5)$ or $S(32,12)$ or $S(32,24)$ or $S(32,25)$ or $S(32,26)$ or $S(32,37)$ or  $S(32,38)$ or $S(32,48)$. Since a normal subgroup of the type $C_{4}\times C_{4}$ is not available in $S(32,5)$, $S(32,37)$, $S(32,38)$ and $S(32,48)$, so these groups are not possible. Hence $G'\cong S(32,4)$ or $S(32,12)$ or $S(32,24)$ or $S(32,25)$ or $S(32,26)$.

Let $|\gamma_{3}(G)|=8$. Then $\gamma_{3}(G)\cong C_{4}\times C_{2}$. If $|G'^{2}|=2$, then $G'^{2}\cap \gamma_{3}(G)\neq 1$, as $|\gamma_{3}(G)^{2}G^{\prime 4}\cap \gamma_{4}(G)|=2$. So $|G'^{2}|=8$ or $4$. If $|G'^{2}|=8$, then $|G'^{2}\cap \gamma_{3}(G)|= 4$. Suppose that $G'$ is abelian. Then $G'\cong C_{8}\times C_{4}$. But then $\gamma_{3}(G)\subseteq G^{\prime 2}$, which is not possible. So  $G'$ is nonabelian. If $\gamma_{5}(G)=1$, then $\gamma_{3}(G)=\zeta(G')\cong C_{4}\times C_{2}$. So the possible choices for $G'$ are $S(32,4)$ or $S(32,5)$ or $S(32,12)$. But in all these cases $G'^{2}=\zeta(G')=\gamma_{3}(G)$. If $\gamma_{5}(G)\cong C_{2}$, then $\zeta(G')\cong C_{4}$ or $C_{8}$ or $C_{4}\times C_{2}$. Hence $G'\cong S(32,4)$ or $S(32,5)$ or $S(32,12)$ or $S(32,17)$. If $|G'^{2}|=4$, then $|G'^{2}\cap \gamma_{3}(G)|=2$. Suppose that $G'$ is abelian. Then $G'\cong C_{8}\times C_{2}\times C_{2}$ or $C_{4}\times C_{4}\times C_{2}$. Let $G'$ be nonabelian. If $\gamma_{5}(G)=1$, then $\gamma_{3}(G)=\zeta(G')\cong C_{4}\times C_{2}$. So the possible choices for $G'$ are $S(32,24)$ or $S(32,25)$ or $S(32,26)$ or $S(32,37)$. But in all these cases $G'^{2}\subseteq\zeta(G')=\gamma_{3}(G)$. If $\gamma_{5}(G)\cong C_{2}$, then $\zeta(G')\cong C_{4}$ or $C_{8}$ or $C_{4}\times C_{2}$. Hence $G'\cong S(32,24)$ or $S(32,25)$ or $S(32,26)$ or $S(32,37)$ or $S(32,38)$.

Conversely, if conditions (i) to (v) hold, then $d_{(2)}=2$, $d_{(3)} = d_{(5)} = d_{(9)}=1$ and for the remaining conditions $d_{(2)} = d_{(3)}= d_{(4)} = d_{(5)} = d_{(7)} = 1$. Thus $n=5$ and $t^L(KG)=18=2^n-14$.

\end{proof}
\begin{lemma}{\label{l4}}
	Let $K$ be a field of characteristic $p=17$ and let $G$ be a nilpotent group such that $\vert G'\vert = 17^{n}$. Then $t^{L}(KG)= 17^{n}-239$ if and only if $G'\cong C_{17}\times C_{17}$, $\gamma_{3}(G)\cong C_{17}$
\end{lemma}
\begin{proof}
Let $t^{L}(KG)= 17^{n}-239$, then by Lemma \ref{l2},  $d_{(2)} = d_{(3)}=1$. Thus $|G'|= 17^{2}$, $|D_{(3),K}(G)|=17$ . So  $t^{L}(KG)= 50 = 3p-1$. Rest follows by \cite[Theorem 3.5]{Sa}.
\end{proof}
\begin{theorem}
	Let $K$ be a field of characteristic $p>0$ and let $G$ be a nilpotent group such that $\vert G'\vert = p^{n}$. Then $t^{L}(KG)= |G'|- 15p+ 16$ if and only if one of the following conditions holds:

\begin{enumerate}
\item $p=2$ and $G'$ is a group of one of the types in Lemma~\ref{l3};

\item $p=17$, $G'\cong C_{17}\times C_{17}$, $\gamma_{3}(G)\cong C_{17}$
\end{enumerate}
\end{theorem}
\begin{proof} Lemma~\ref{l3} and Lemma~\ref{l4} together imply the complete proof of the Theorem.

\end{proof}

\begin{table}[htp]
\caption{ Nonabelian Groups of order $2^{5}$}
\label{tab:table1}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$G$ & $\exp(G)$ & $\zeta(G)$  & $G^{2}$ & $G^{4}$  & $G^{2}\cap \zeta(G)$ & $G^{4}\cap \zeta(G)$   & $cl(G)$ \\
\hline	$S(32, 2)$ & 4 & $(C_{2})^{3}$  & $(C_{2})^{3}$ & 1 & $ (C_{2})^{3}$ & 1  & 2 \\
\hline	$S(32, 4)$ & 8 & $C_{4}\times C_{2}$  & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{4}\times C_{2}$ & $C_{2}$ & 2 \\
\hline	$S(32, 5)$ & 8 & $C_{4}\times C_{2}$  & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{4}\times C_{2}$ & $C_{2}$ & 2 \\
\hline	$S(32, 6)$ & 4 & $C_{2}$  & $ (C_{2})^{3}$ & 1 &  $C_{2}$ & 1   & 3 \\
\hline	$S(32, 7)$ & 8 & $C_{2}$  & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{2}$ & $C_{2}$  & 3 \\
\hline	$S(32, 8)$ & 8 & $C_{2}$  & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{2}$ & $C_{2}$  & 3 \\
\hline	$S(32, 9)$ & 8 & $C_{2}\times C_{2}$  & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{2}\times C_{2}$ & $C_{2}$   & 3 \\
\hline	$S(32, 10)$ & 8 & $C_{2}\times C_{2}$  & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{2}\times C_{2}$ & $C_{2}$   & 3 \\
\hline	$S(32, 11)$ & 8 & $C_{4}$  & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{4}$ & $C_{2}$  & 3 \\
\hline	$S(32, 12)$ & 8 & $C_{4}\times C_{2}$ & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{4}\times C_{2}$ & $C_{2}$   & 2 \\
\hline	$S(32, 13)$ & 8 & $C_{2}\times C_{2}$  & $C_{4}\times C_{2}$ & $C_{2}$ &  $C_{2}\times C_{2}$ & $C_{2}$  & 3 \\
\hline	$S(32, 14)$ & 8 & $C_{2}\times C_{2}$  & $C_{4}\times C_{2}$ & $C_{2}$ & $C_{2}\times C_{2}$ & $C_{2}$  & 3 \\
\hline	$S(32, 15)$ & 8 & $C_{4}$  & $C_{4}\times C_{2}$ &  1& $C_{4}$ & $C_{2}$   & 3 \\
\hline	$S(32, 17)$ & 16 & $C_{8}$  & $C_{8}$ & $C_{4}$ &  $C_{8}$ & $C_{4}$ & 2 \\
\hline	$S(32, 18)$ & 16 & $C_{2}$  & $C_{8}$ & $C_{4}$ & $C_{2}$ & $C_{2}$   & 4 \\
\hline	$S(32, 19)$ & 16 & $C_{2}$ & $C_{8}$ & $C_{4}$  & $C_{2}$ & $C_{2}$ &  4 \\
\hline	$S(32, 20)$ & 16 & $C_{2}$ & $C_{8}$ & $C_{4}$  & $C_{2}$ & $C_{2}$  & 4 \\
\hline	$S(32, 22)$ & 4 & $(C_{2})^{3}$  & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 23)$ & 4 & $(C_{2})^{3}$  & $C_{2}\times C_{2}$ & 1  & $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 24)$ & 4 & $C_{4}\times C_{2}$  & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 25)$ & 4 & $C_{4}\times C_{2}$ & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 & 2 \\
\hline	$S(32, 26)$ & 4 & $C_{4}\times C_{2}$ & $C_{2}\times C_{2}$ & 1  & $C_{2}\times C_{2}$ & 1 & 2 \\
\hline	$S(32, 27)$ & 4 & $C_{2}\times C_{2}$ & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 28)$ & 4 & $C_{2}\times C_{2}$ & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 & 2 \\
\hline	$S(32, 29)$ & 4 & $C_{2}\times C_{2}$  & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 30)$ & 4 & $C_{2}\times C_{2}$ & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 31)$ & 4 & $C_{2}\times C_{2}$ & $C_{2}\times C_{2}$ & 1  & $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 32)$ & 4 & $C_{2}\times C_{2}$ & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 33)$ & 4 & $C_{2}\times C_{2}$ & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 & 2 \\
\hline	$S(32, 34)$ & 4 & $C_{2}\times C_{2}$  & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 &  2 \\
\hline	$S(32, 35)$ & 4 & $C_{2}\times C_{2}$  & $C_{2}\times C_{2}$ & 1 &  $C_{2}\times C_{2}$ & 1 &  2 \\

\hline	$S(32, 37)$ & 8 & $C_{4}\times C_{2}$  & $C_{4}$ & $C_{2}$  & $C_{4}$ & $C_{2}$ & 2 \\

\hline	$S(32, 38)$ & 8 & $C_{8}$ & $C_{4}$ & $C_{2}$  & $C_{4}$ & $C_{2}$ &  2 \\

\hline	$S(32, 39)$ & 8 & $C_{2}\times C_{2}$ & $C_{4}$ & $C_{2}$ & $C_{2}$ & $C_{2}$ & 3 \\

\hline	$S(32, 40)$ & 8 & $C_{2}\times C_{2}$ & $C_{4}$ & $C_{2}$  & $C_{2}$ & $C_{2}$ & 3 \\

\hline	$S(32, 41)$ & 8 & $C_{2}\times C_{2}$ & $C_{4}$ & $C_{2}$  & $C_{2}$ & $C_{2}$ &  3 \\

\hline	$S(32, 42)$ & 8 & $C_{4}$ & $C_{4}$ & $C_{2}$ &  $C_{4}$ & $C_{2}$ & 3 \\

\hline	$S(32, 43)$ & 8 & $C_{2}$ & $C_{4}$ & $C_{2}$ &  $C_{2}$ & $C_{2}$ &  3 \\

\hline	$S(32, 44)$ & 8 & $C_{2}$ & $C_{4}$ & $C_{2}$ &  $C_{2}$ & $C_{2}$ & 3 \\

\hline	$S(32, 46)$ & 4 & $(C_{2})^{3}$ & $C_{2}$ & 1 &  $ C_{2}$ & 1 & 2 \\

\hline	$S(32, 47)$ & 4 & $(C_{2})^{3}$ & $C_{2}$ & 1 &  $ C_{2}$ & 1 & 2 \\

\hline	$S(32, 48)$ & 4 & $C_{4}\times C_{2}$ & $C_{2}$ & 1 & $ C_{2}$ & 1 & 2 \\

\hline	$S(32, 49)$ & 4 & $C_{2}$ & $C_{2}$ & 1 &  $ C_{2}$ & 1 & 2 \\

\hline	$S(32, 50)$ & 4 & $C_{2}$ & $C_{2}$ & 1 &  $ C_{2}$ & 1 & 2 \\
\hline
\end{tabular}
\end{table}

{\bf{Acknowledgments}}: The financial assistance provided to the second author in the form of a Senior Research Fellowship from University Grants Commission, India is gratefully acknowledged.

config: arxiv


TITLE: Is there an easy proof for ${\aleph_\omega} ^ {\aleph_1} = {2}^{\aleph_1}\cdot{\aleph_\omega}^{\aleph_0}$?
QUESTION [12 upvotes]: The question contains 2 stages:

Prove that ${\aleph_n} ^ {\aleph_1} = {2}^{\aleph_1}\cdot\aleph_n$

This one is pretty clear by induction and by applying Hausdorff's formula.

Prove  ${\aleph_\omega} ^ {\aleph_1} = {2}^{\aleph_1}\cdot{\aleph_\omega}^{\aleph_0}$

Is there an easy proof for the second one?
Thanks in advance.

REPLY [11 votes]: As you mention, the first equation is a consequence of Hausdorff's formula and induction.
For the second: Clearly the right hand side is at most the left hand side. Now: Either $2^{\aleph_1}\ge\aleph_\omega$, in which case in fact $2^{\aleph_1}\ge{\aleph_\omega}^{\aleph_1}$, and we are done, or $2^{\aleph_1}<\aleph_\omega$.
I claim that in this case we have ${\aleph_\omega}^{\aleph_1}={\aleph_\omega}^{\aleph_0}$. Once we prove this, we are done.
Note that $\aleph_\omega={\rm sup}_n\aleph_n\le\prod_n\aleph_n$, so
$$ {\aleph_\omega}^{\aleph_1}\le\left(\prod_n\aleph_n\right)^{\aleph_1}=\prod_n{\aleph_n}^{\aleph_1}. $$
Now use part 1, to conclude that ${\aleph_n}^{\aleph_1}<\aleph_\omega$ for all $n$, since we are assuming that $2^{\aleph_1}<\aleph_\omega$. This shows that the last product is at most $\prod_n \aleph_\omega={\aleph_\omega}^{\aleph_0}$.
This shows that ${\aleph_\omega}^{\aleph_1}\le {\aleph_\omega}^{\aleph_0}$. But the other inequality is obvious, and we are done.


text:
\section{Well-posedness}
\label{sec:wellposedness}
In this section we prove that ($P_1$) has a weak solution by approximating it with a Galerkin projection.
We show the projection exists and is unique, and proceed by proving it converges to the weak solution of ($P_1$).
First, we introduce the necessary tools for defining the Galerkin approximation.

We use one mesh partition for each of the two spatial scales.
Let $\B_H$ be a mesh partition for $\Omega$ consisting of simplices. We denote the diameter of an element $B \in \B_H$ with $H_B$, and the global mesh size with $H:= \max_{B \in \B_H} H_B$.
We introduce a similar mesh partition $\K_h$ for $Y$ with global mesh size $h:= \max_{K \in \K_h} h_K$.

Our macroscopic and microscopic finite element spaces $V_H$ and $W_h$ are, respectively:
\begin{align*}
    V_H &:= \left\{ \left. v \in \L^2(\Omega)\right|\,v|_B \in H^2(B) \mbox{ for all } B \in \B_H  \right\},\\
        W_h &:= \left\{ \left. w \in \L^2(Y)\right|\,w|_K \in H^2(K) \mbox{ for all } K \in \K_h  \right\}.
\end{align*}
Let $ \langle \xi_B \rangle_{\B_H}:= \operatorname{span}(V_H)$ and $ \langle \eta_K \rangle_{\K_h}:= \operatorname{span}(W_h)$, and let $\alpha_B,\beta_{BK} : S \to \R$ denote the Galerkin projection coefficient for a patch $B$ and $B\times K$, respectively. We introduce the following finite-dimensional Galerkin approximations of the functions $\pi$ and $\rho$:
\begin{equation}
    \begin{split}
        \pi^H(t,x) &:= \sum_{B \in \B_H} \alpha_B(t) \xi_B(x),\\
        \rho^{H,h}(t,x,y) &:= \sum_{B \in \B_H, K \in \K_h} \beta_{BK}(t) \xi_B(x) \eta_K(y),
    \end{split}
    \label{eq:trunc}
\end{equation}
where we clamp $\alpha_B(t)=0$ for all $B\in \B_H$ with $\partial B \cap \Omega\neq \emptyset $ to represent the macroscopic Dirichlet boundary condition.

Reducing the space of test functions to $V^H$ and $W^h$, we obtain the following discrete weak formulation: find a solution pair $(\pi^H(t,x),\rho^{H,h}(t,x,y)) \in \L^2(S;V^H)\times \L^2(S;V^H\times W^h)$ that are solutions to
\begin{equation}
    \label{eq:weak_pi}
    A\int_\Omega\nabla_x\pi^H\cdot\nabla_x\varphi dx=\int_\Omega f(\pi^H,\rho^{H,h})\varphi dx,
\end{equation}
and
\begin{equation}
    \label{eq:weak_rho}
    \begin{split}
    &\int_\Omega\int_Y\partial_t\rho^{H,h}\psi dydx+D\int_\Omega \int_Y\nabla_y \rho^{H,h}\cdot\nabla_y\psi dydx\\
    &\quad= \kappa\int_\Omega\int_{\Gamma_R}(\pi^H+p_F-R\rho^{H,h})\psi d\sigma_ydx,
    \end{split}
\end{equation}
for any $\varphi \in V_H$ and $\psi \in V_H \times W_h$ and almost every $t \in S$.

These concepts lead us to the first proposition.
\ \\
\begin{proposition}[Existence and uniqueness of the Galerkin approximation]
    \label{prop:ex_un}
    There exists a unique solution $(\pi^H,\rho^{H,h})$ to the system in \eqref{eq:weak_pi}-\eqref{eq:weak_rho}.
\end{proposition}
\begin{proof}
    The proof is divided in three steps.  In step 1, the local existence in time is proven. In step 2, global existence in time is proven. Step 3 is concerned with proving the uniqueness of the system.

    We introduce an integer index for $\alpha_B(t)$ and $\beta_{BK}(t)$ to increase the legibility of arguments in this proof.
    Let $N_1 := \{1,\dots, |\B_H|\}$ and $N_2 := \{1,\dots, |\K_h|\}$. We introduce bijective mappings $n_1:N_1\to \B_H$ and $n_2: N_2 \to \K_h$, so that each index $j\in N_1$ corresponds to an element $B \in \B_H$ and each index $k\in N_2$ corresponds to a $K \in \K_h$.

    \textit{Step 1: local existence of solutions to \eqref{eq:weak_pi} - \eqref{eq:weak_rho}}:
    By substituting $\varphi = \xi_i$ and $\psi = \xi_i\eta_k$ for $i\in N_1$ and $k\in N_2$ in \eqref{eq:weak_pi}-\eqref{eq:weak_rho} we obtain the following system of ordinary differential equations coupled with algebraic equations.

    \begin{align}
        \sum_{j\in N_1} P_{ij} \alpha_j(t) &= F_i(\alpha,\beta)\label{eq:ode_alpha} \mbox{ for }i\in N_1,\\
        \beta^\prime_{ik}(t) + \sum_{j\in N_1,l\in N_2} Q_{ijkl} \beta_{jl}(t) &= c_{ik} + \sum_{j\in N_1} E_{ijk}\alpha_j\label{eq:ode_beta}\mbox{ for }i\in N_1\mbox{ and }k \in N_2,
    \end{align}

    with

    \begin{equation}
        \begin{split}
            P_{ij} &:= A \int_\Omega \nabla_x \xi_i \nabla_x \xi_j\,dx,\\
            F_i &:= \int_\Omega f \left( \sum_{j\in N_1}\alpha_j(t) \xi_j,\sum_{j\in N_1,l\in N_2}\beta_{jl}(t)\xi_j\eta_l \right)\xi_i\,dx,\\
            Q_{ijkl} &:= D \int_\Omega \xi_i\xi_jdx\int_{Y} \nabla_y \eta_k \nabla_y \eta_ldy + \kappa R\int_\Omega \xi_i\xi_j dx\int_{\Gamma_R}\eta_k \eta_ld\sigma_y,\\
            E_{ijk} &:= \kappa\int_\Omega\xi_i\xi_j dx\int_{\Gamma_R} \eta_kd\sigma_y,\\
            c_{ik}&:= \kappa p_F\int_\Omega \xi_i dx\int_{\Gamma_R} \eta_k d\sigma_y\,
        \end{split}
        \label{eq:matrices}
    \end{equation}
    Applying \eqref{eq:main_initial} to \eqref{eq:weak_pi} and \eqref{eq:weak_rho} yields:
    \begin{equation}
        \begin{split}
            \alpha_i(0) &= \int_\Omega \xi_i\pi_I\,dx,\\
            \beta_{ik}(0) &= \int_\Omega\int_Y\xi_i\eta_k\rho_Idydx.
        \end{split}
        \label{eq:fem_init}
    \end{equation}

    For all $t \in S$, the coefficients $\alpha_i(t), \beta_{ik}(t)$ of \eqref{eq:trunc} are determined by \eqref{eq:ode_alpha}, \eqref{eq:ode_beta} and \eqref{eq:fem_init}.

    Since the system of ordinary differential equations in \eqref{eq:ode_beta} is linear, we are able to explicitly formulate the solution representation for $\beta_{ik}$ with respect to $\alpha_i$. Let $\alpha_i$ be given, and let $Q$ and $E$ denote matrices given by:
    \begin{align}
        Q\beta &= \sum_{j\in N_1,l\in N_2} Q_{ijkl} \beta_{jl},\\
        E\alpha &= \sum_{j\in N_1} E_{ijk}\alpha_j.
    \end{align}
    Then $\beta_{ik}$ can be expressed as
    \begin{equation}
        \beta_{ik}(t) = \beta_{ik}(0)e^{-Qt} + Q^{-1}(c+E\alpha_i)(I - e^{-Qt}).
        \label{eq:matrix_exponential}
    \end{equation}
    \label{lem:b_from_a}
    Substituting \eqref{eq:matrix_exponential} in \eqref{eq:ode_beta} results in the expression:
    \begin{equation}
      -Q\beta_{ik}(0)e^{-Qt} + (c+E\alpha_j)e^{-Qt} + Q\beta_{ik}(0)e^{-Qt} + (c+E\alpha_j)(I-e^{-Qt}) = (c+E\alpha_j).
    \end{equation}
    \ref{as:rhs} implies that for all $i \in N_1$, $F_i$ are contractions. Pick $\beta_1(t),\beta_2(t)$ that satisfy \eqref{eq:matrix_exponential} given some $\alpha_1(t), \alpha_2(t)$. Then it holds that:
    \begin{equation}
      \begin{split}
          &|F_i(\alpha_1,\beta_1) - F_i(\alpha_2,\beta_2)|,\\
          &\leq |F_i(\alpha_1,\beta_1) - F_i(\alpha_1,\beta_2) + F_i(\alpha_1,\beta_2) - F_i(\alpha_2,\beta_2)|, \\
          &= \left|\int_\Omega f \left( \sum_{j\in N_1}\alpha_{j,1}(t) \xi_j,\sum_{j\in N_1,l\in N_2}\beta_{jl,1}(t)\xi_j\eta_l \right)\xi_i\right.\\
          &\quad- \left. f \left( \sum_{j\in N_1}\alpha_{j,1}(t) \xi_j,\sum_{j\in N_1,l\in N_2}\beta_{jl,2}(t)\xi_j\eta_l \right)\xi_i\,dx\right|\\
          &+ \left|\int_\Omega f \left( \sum_{j\in N_1}\alpha_{j,1}(t) \xi_j,\sum_{j\in N_1,l\in N_2}\beta_{jl,2}(t)\xi_j\eta_l \right)\xi_i\right.\\
          &\quad- \left. f \left( \sum_{j\in N_1}\alpha_{j,2}(t) \xi_j,\sum_{j\in N_1,l\in N_2}\beta_{jl,2}(t)\xi_j\eta_l \right)\xi_i\,dx\right|,\\
          &\leq \left|\int_\Omega c_\rho \sum_{j\in N_1,l\in N_2}(\beta_{jl,1}(t) - \beta_{jl,2}(t))\xi_j\eta_l
          + c_\pi \sum_{j\in N_1}(\alpha_{j,1}(t) - \alpha_{j,2}(t)) \xi_j \,dx\right|,\\
          &\leq c_\beta \left|\sum_{j\in N_1}\beta_{jl,1}(t) - \beta_{jl,2}(t)\right| +  c_\alpha \left|\sum_{j\in N_1}\alpha_{j,1}(t) - \alpha_{j,2}(t)\right|,
      \end{split}
      \label{eq:lipschitz_carry}
    \end{equation}
    with $c_\alpha, c_\beta$ defined as
    \begin{align}
        c_\alpha:= c_\pi\max_{j\in N_1}\int_\Omega \xi_j\,dx\leq c_\pi,& &c_\beta := c_\rho\max_{j\in N_1,l\in N_2}\int_\Omega \xi_j\eta_ldx\leq c_\rho.
    \end{align}
    Now, we derive a time-dependent continuity estimate for sufficiently small $t$. Again picking $\beta_1(t)$ and $\beta_2(t)$ (not necessarily the same as in \eqref{eq:lipschitz_carry}):
    \begin{equation}
        \begin{split}
            ||\beta_1(t) - \beta_2(t)|| &= ||I-e^{Qt}||\cdot||Q^{-1}D||\cdot||\alpha_1(t) - \alpha_2(t)||,\\
            & = ||Qt + \bigo{t^2}||\cdot||Q^{-1}D||\cdot||\alpha_1(t) - \alpha_2(t)||,\\
            &\leq tC||\alpha_1 - \alpha_2|| \mbox{ for small $t$}.
        \end{split}
        \label{eq:time_lp}
    \end{equation}
    Using \eqref{eq:time_lp} we obtain a Lipschitz bound on all $F_i$  in the interval $[0,\tau]$ for any choice of $\tau < T$:
    \begin{equation}
      \begin{split}
          &||F_i(\alpha_1(t),\beta_1(t)) - F_i(\alpha_2(t),\beta_2(t))||\\ &\leq ||F_i(\alpha_1(t),\beta_1(t)) - F_i(\alpha_1(t),\beta_2(t))|| + ||F_i(\alpha_1(t),\beta_2(t)) - F_i(\alpha_2(t),\beta_2(t))||,\\
          &\leq c_\alpha ||\alpha_1(t) - \alpha_2(t)|| + c_\beta || \beta_1(t) - \beta_2(t)||,\\
          &\leq \left(c_\alpha + c_\beta C\tau  \right) ||\alpha_1(t) - \alpha_2(t)||.
      \end{split}
    \end{equation}

    Choosing $\tau$ small enough to satisfy $c_\alpha + c_\beta C\tau<1$ makes $F$ a contraction on $[0,\tau]$.
    By Banach's fixed point theorem, it follows that the equation $F(\alpha(t),\beta(t)) = \alpha(t)$ has a solution for $\alpha$ in $\L^2(S)$.
    Substitution of $\alpha(t)$ into \eqref{eq:matrix_exponential} leads to the corresponding $\beta$.
    Existence of $\pi^H$ and $\rho^{H,h}$ follows directly.

    \textit{Step 2: global existence of solutions to \eqref{eq:weak_pi} - \eqref{eq:weak_rho}}:
    We cover time interval $S$ into $N$ intervals of length at most $\tau$ such that $S \subseteq \bigcup_n((n-1)\tau,n\tau]$.
    From the arguments in the previous paragraph it is clear a solution exists on the first interval $[0,\tau]$. This allows us to provide an induction argument for the existence of a solution on interval $n$:

    Given that interval $n$ has local solution $\beta \left( \left[ (n-1)\tau,n\tau \right] \right)$, we can obtain values $\beta(n\tau), \beta'(n\tau), \alpha(n\tau)$ as initial values to the local system on interval $n+1$, and show existence of a solution. This way, we are able to construct a solution satisfying \eqref{eq:weak_pi} - \eqref{eq:weak_rho} everywhere on $S$.

    \textit{Step 3: uniqueness of solutions to \eqref{eq:weak_pi} - \eqref{eq:weak_rho}}:
    We decouple the system and use a fixed point argument to show that this system has a globally unique solution in time.

    Let $(\alpha_1, \beta_1)$ and $(\alpha_2,\beta_2)$ be two solutions satisfying \eqref{eq:weak_pi} - \eqref{eq:weak_rho} with the same initial data.
    Let $\bar{\beta}(t) := \beta_1(t) - \beta_2(t)$ and $\bar{\alpha}(t) := \alpha_1(t) - \alpha_2(t)$.
    By starting from \eqref{eq:ode_beta} and multiplying both equations with $\bar{\beta}(t)$, we obtain

    \begin{equation}
        \begin{split}
            \langle\bar{\beta}(t),\bar{\beta}\prime(t)\rangle &= \langle Q\bar{\beta}(t),\bar{\beta}(t)\rangle + \langle E\bar{\alpha}(t),\bar{\beta}(t)\rangle,\\
            \frac{1}{2}\frac{d}{dt} || \bar{\beta}(t) ||^2 &\leq ||Q|||| \bar{\beta}(t) ||^2 + ||E|| || \bar{\alpha}(t)||\ ||\bar{\beta}(t)||.

        \end{split}
    \end{equation}

    Since $\bar{\beta}(0)=0$, by applying Gr\"onwall's differential inequality, we know that $\bar\beta(t)\equiv0$. Combined with \eqref{eq:matrix_exponential}, it immediately follows that $\bar\alpha(t)\equiv0$, and therefore, $(\alpha_1, \beta_1)=(\alpha_2,\beta_2)$.

\end{proof}
Note that showing the stability of the finite element approximation with respect to data and initial conditions follows an analogous argument. The proof is omitted here.

The remaining part of this section is devoted to proving that the system in \eqref{eq:weak_pi}-\eqref{eq:weak_rho} converges to the solution of the Galerkin projection converges to the weak solution of ($P_1$) (as stated in Proposition~\ref{prop:conv}).
To this end, we first formulate the lemmata that help us prove this statement.

\begin{lemma}[Aubin-Lions lemma]
    \label{lem:aubinlions}
Let $B_0 \embed B \subset\subset B_1$ be Banach spaces, i.e. $B_0$ be compactly embedded in $B$ and $B$ be continuously embedded in $B_1$. Let
\begin{equation}
    W := \left\{ u\in \L^2\left( (0,T);B_0 \right) | \partial_t u \in \L^2\left( (0,T); B_1 \right) \right\}.
    \label{eq:aubinlions}
\end{equation}
Then the embedding of $W$ into $\L^2\left( (0,T);B \right)$ is compact.
\end{lemma}

\begin{lemma}[Weak maximum principle]
    \label{lem:max}
    Assume \ref{as:lips_boundary} -- \ref{as:rhs}:
    Then $\pi^H(t,\cdot) \in \L^\infty(\Omega)$ and $\rho^{H,h}(t,\cdot) \in \L^\infty(\Omega\times Y)$ for all $t\in(0,T)$.
\end{lemma}
We refer the reader to \cite{aubin1963} for the original proof of the statement.
\begin{proof}
    We use a weak maximum principle according to Stampacchia (\cite{stampacchia1965}). Consequently, we test the weak formulation with $\varphi = (\pi^H-M_1)^+$ and $\psi = (\rho^{H,h} - M_2)^+$ for suitable $M_1$ and $M_2$.
    Assumptions \ref{as:diff_condition} and \ref{as:rhs} are used in this context.
    From \eqref{eq:weak_pi} we obtain
    \begin{equation}
      A\int_\Omega \left| \nabla_x \pi^H\nabla_x(\pi^H - M_1)^+ \right|\,dx = \int_\Omega \left|f(\pi^H,\rho^{H,h})(\pi^H - M_1)^+\right| \,dx,\\
      \label{eq:stampacchia_formulation}
    \end{equation}
    The left hand side of \eqref{eq:stampacchia_formulation} can be manipulated as:
    \begin{equation}
        \begin{split}
            A\int_\Omega \left| \nabla_x \pi^H\nabla_x(\pi^H - M_1)^+ \right|\,dx &= A\int_\Omega \left| \nabla_x (\pi^H - M_1)\nabla_x(\pi^H - M_1)^+ \right|\,dx,\\
            & = A\int_\Omega \left| \nabla_x (\pi^H - M_1)^+\right|^2,
        \end{split}
    \end{equation}
    which can be bounded with the right hand side of \eqref{eq:stampacchia_formulation}:
  \begin{equation}
      \begin{split}
          &A\int_\Omega \left| \nabla_x (\pi^H - M_1)^+\right|^2\\
          &\leq  \int_\Omega \left| f(\pi^H,\rho^{H,h}) - f(M_1,\rho^{H,h}) + f(M_1,\rho^{H,h})\right|(\pi^H - M_1)^+ \,dx,\\
          &\leq \int_\Omega |c_\pi^H| |\pi^H-M_1|(\pi^H - M_1)^++ |f(M_1,\rho^{H,h})|(\pi^H - M_1)^+ \,dx,\\
          &= \int_\Omega \left((\pi^H - M_1)^{+}\right)^2\,dx \leq c_p \int_\Omega\left(\nabla_x(\pi^H - M_1)^{+}\right)^2\,dx.
      \end{split}
      \label{eq:second_estimate}
    \end{equation}

    Proceeding similarly with \eqref{eq:weak_rho}:
    \begin{equation}
      \begin{split}
          &\int_\Omega\int_Y\partial_t\rho^{H,h}(\rho^{H,h}-M_2)^+ dydx+\int_\Omega \int_YD\nabla_y\rho^{H,h}\cdot\nabla_y (\rho^{H,h}-M_2)^+ dydx,\\
          &= \int_\Omega\int_{\Gamma_R}k(\pi^H+p_F-R\rho^{H,h})(\rho^{H,h}-M_2)^+ d\sigma_ydx,\\
          &= \kappa\int_\Omega\int_{\Gamma_R}\left(\pi^H-M_1 + p_F + M_1 - RM_2 + R(\rho^{H,h} - M_2)\right)(\rho^{H,h} - M_2)^+d\sigma_ydx,\\
          &=\kappa\int_\Omega\int_{\Gamma_R}\left( (\pi^H-M_1)^+ - (\pi^H-M_1)^-d\sigma_ydx(\rho^{H,h} - M_2)^+\right)(\rho^{H,h} - M_2)^+ \\
          &+ \kappa\int_\Omega\int_{\Gamma_R} (p_F + M_1 - RM_2)(\rho^{H,h}-M_2)^+d\sigma_ydx\\
          &+ \kappa R\int_\Omega\int_{\Gamma_R}\left((\rho^{H,h}-M_2)^+ - (\rho^{H,h}-M_2)^-(\rho^{H,h} - M_2)^+\right)d\sigma_ydx,\\
          &\leq \kappa\int_\Omega\int_{\Gamma_R} (\pi^H-M_1)^+ (\rho^{H,h}-M_2)^+ - (\pi^H-M_1)^- (\rho^{H,h}-M_2)^+d\sigma_ydx\\
          &+\kappa \left( p_F + M_1 - RM_2 \right)\int_\Omega\int_{\Gamma_R} (\rho^{H,h}-M_2)^+ d\sigma_ydx + \kappa R\int_\Omega\int_{\Gamma_R}(\rho^{H,h} - M_2)^{+2}d\sigma_ydx\\
          &- \kappa R\int_\Omega\int_{\Gamma_R}(\rho^{H,h}-M_2)^-(\rho^{H,h}-M_2)^+d\sigma_ydx\\
          &\leq \varepsilon\int_\Omega\int_{\Gamma_R} (\pi^H-M_1)^{+2}d\sigma_ydx + \kappa(R+c_\varepsilon)\int_\Omega\int_{\Gamma_R}(\rho^{H,h} - M_2)^{+2}d\sigma_ydx.
      \end{split}
      \label{eq:third_estimate}
    \end{equation}
    Add \eqref{eq:second_estimate} and \eqref{eq:third_estimate} with $\varepsilon>0$ small. Applying the trace inequality twice in \eqref{eq:third_estimate} ensures that $\pi^H$ and $\rho^{H,h}$ are uniformly bounded if the pair $(M_1,M_2)$ is chosen such that
    \begin{equation}
        \begin{cases}
            M_2 = \theta, \\
            M_1 < R\theta -p_F,\\
            M_1 \geq ||\rho_I||_{\L^2(\Omega\times Y)},\\
            \theta > \frac{p_f}{R}.
        \end{cases}
    \end{equation}
\end{proof}

\begin{lemma}[Regularity lift]
    Let $(\pi^H,\rho^{H,h})$ be a solution to \eqref{eq:weak_pi}-\eqref{eq:weak_rho}. Then it must hold that
    \begin{equation}
        \begin{cases}
            \pi^H \in \L^\infty((0,T) \times \Omega) \cap \L^\infty((0,T); H^1_0(\Omega)),\\
            \rho^{H,h} \in \L^2((0,T); \L^2(\Omega; \H^1(Y))) \cap \L^\infty((0,T); \L^\infty(\Omega; \L^\infty(Y)).
        \end{cases}
    \end{equation}
\end{lemma}
\begin{proof}
    Testing \eqref{eq:weak_pi} with $\varphi = \pi^H$ and \eqref{eq:weak_rho} with $\psi = \rho^{H,h}$ yields identities
	\begin{equation}
	      \label{eq:energy_pi}A \left|\left| \nabla_x \pi^H \right|\right|^2_{\L^2(\Omega)} = \int_\Omega f(\pi^H,\rho^{H,h})\pi^H\,dx,
	\end{equation}
	and
	\begin{equation}
	    \begin{split}
	        &\label{eq:energy_rho}\frac{1}{2} \frac{d}{dt} \left|\left|\rho^{H,h}\right|\right|^2_{\L^2(\Omega\times Y)}
	      + D \left|\left|\nabla_y \rho^{H,h}\right|\right|^2_{\L^2(\Omega\times Y)}\\
	      &\quad = \int_\Omega \int_{\Gamma_R} \kappa(\pi^H + p_F)\rho^{H,h}\,d\sigma_y \,dx - \kappa R \left|\left| \rho^{H,h}\right|\right|^2_{\L^2(\Omega \times Y)}.
	    \end{split}
	\end{equation}
	We recall the embedding
	\begin{equation}
	    H^1(Y) \embed \L^2(\Gamma_R),
	\end{equation}
	which implies that there exists a $c_E$ such that
	\begin{equation}
	    ||u||_{\L^2(\Omega; \L^2(\Gamma_R))} \leq c_E ||u||_{\L^2(\Omega; \H^1(Y))},
	    \label{eq:embed}
	\end{equation}
	for all $u \in \L^2(\Omega;\H^1(Y))$.
     Using Cauchy-Schwarz' inequality and \eqref{eq:embed}, we can bound the right hand side of \eqref{eq:energy_rho} as
	\begin{equation}
	    \begin{split}
	      \int_\Omega \int_{\Gamma_R} \kappa(\pi^H + p_F)\rho^{H,h}\,d\sigma_y \,dx
	      &\leq \kappa|\Gamma_R|\left(||\pi^H||_{\L^2(\Omega)}+|\Omega|p_F\right) ||\rho^{H,h}||_{\L^2(\Omega\times\Gamma_R)}.\\
	      &\leq \kappa c_E|\Gamma_R|\left(|\pi^H||_{\L^2(\Omega)} + p_F|\Omega|\right) ||\rho^{H,h}||_{\L^2(\Omega;H^1(Y))}.
	  \end{split}
	    \label{eq:first_est}
	\end{equation}
	Then, we add to both sides of \eqref{eq:energy_rho} a term $D||\rho^{H,h}||^2_{\L^2(\Omega\times Y)}$ to get
	\begin{equation}
	  \begin{split}
	      &\frac{1}{2}\frac{d}{dt}||\rho^{H,h}||^2_{\L^2(\Omega\times Y)} + D||\rho^{H,h}||^2_{ \L^2(\Omega;H^1(Y))},\\
	      &\leq D||\rho^{H,h}||^2_{\L^2(\Omega\times Y)} + \kappa c_E|\Gamma_R|\left(|\pi^H||_{\L^2(\Omega)} + p_F|\Omega|\right) ||\rho^{H,h}||_{\L^2(\Omega;H^1(Y))}.
	  \end{split}
	\end{equation}
	After applying Young's inequality with the small parameter $\varepsilon>0$, we get
	\begin{equation}
	  \begin{split}
	      &\frac{1}{2}\frac{d}{dt}||\rho^{H,h}||^2_{\L^2(\Omega\times Y)} + (D-\varepsilon) ||\rho^{H,h}||^2_{\L^2(\Omega;H^1(Y))},\\
	      &\leq D||\rho^{H,h}||^2_{\L^2(\Omega\times Y)} + \kappa^2c_E^2c_\varepsilon |\Gamma_R|^2\left( ||\pi^H||^2_{\L^2(\Omega)} + p_F^2|\Omega|^2 \right).
	  \end{split}
	\end{equation}
	By applying Gr\"onwall's inequality we obtain the desired estimates:
	\begin{align}
	    ||\rho^{H,h}||^2_{\L^2(\Omega\times Y)} &\leq C_\rho e^{Dt},\\
	    ||\nabla\rho^{H,h}||^2_{\L^2(\Omega\times Y)} &\leq C_\rho + \varepsilon||\rho^{H,h}||_{\L^2(\Omega\times Y)},
	\end{align}
	with
	\[C_\rho = \kappa^2c_E^2c_\varepsilon |\Gamma_R|^2\left( ||\pi^H||^2_{\L^2(\Omega)} + p_F^2|\Omega|^2 \right).\]
    To obtain a bound on $\pi^H$, we test \eqref{eq:weak_pi} with $\pi^H$ and then use \ref{as:diff_condition} and \ref{as:rhs}:
	\begin{equation}
	  \begin{split}
	      A||\nabla_x\pi^H||^2 &= \int_\Omega f(\pi^H,\rho^{H,h})\pi^H\,dx \\
	      &= \int_\Omega \left(f(\pi^H,\rho^{H,h})-f(0,\rho^{H,h})+f(0,\rho^{H,h})\right)\pi^Hdx,\\
          & \leq \sqrt{\int_\Omega (c_\rho\pi^H)^2\,dx}||\pi^H||_{\L^2(\Omega)} \leq c_\pi||\pi^H||_{\L^\infty(\Omega)}||\nabla_x\pi^H||_{\L^2(\Omega)}.
	  \end{split}
	\end{equation}
	This yields the upper bound
	\begin{equation}
	    ||\nabla_x\pi^H||_{\L^2(\Omega)} \leq \frac{c_\pi}{A}||\pi^H||_{\L^\infty(\Omega)}.
	\end{equation}
\end{proof}
With these lemmata we are ready to state and prove a first convergence result.
\begin{proposition}[Convergence of the Galerkin approximation]
    \label{prop:conv}
    Let $(\pi^H,\rho^{H,h}) \in\L^2(S;H_0^1(\Omega))\times \L^2(S;\L^2(\Omega;H^1(Y)))$ be a solution to \eqref{eq:weak_pi}-\eqref{eq:weak_rho} and let $(\pi,\rho)$ be the weak solution to ($P_1$). Then
    \begin{equation}
        \begin{split}
            \pi^H \to \pi\\
            \rho^{H,h} \to \rho
        \end{split}
    \end{equation}
    for $H,h \to 0$.
\end{proposition}

\begin{proof}
	To apply Lemma~\ref{lem:aubinlions}, we first need to show
	\begin{equation}
	     \partial_t \pi^H \in \L^2((0,T); \L^2(\Omega)).
	    \label{eq:e1}
	\end{equation}
	Then, by choosing
	\begin{equation}
	    \begin{split}
		B_0= \H^1_0(\Omega),\\
		B_1=B = \L^2(\Omega),
	    \end{split}
	\end{equation}
    we satisfy the requirements of Aubin-Lions' lemma to get convergence of $\pi^H$ in $\L^2(\Omega)$.

	Concerning $\rho^{H,h}$, again we use Aubin-Lions' lemma to prove convergence, this time for the following spaces:
	\begin{equation}
	    \begin{split}
		B_0= \H^1_0(\Omega; \H^1(Y)),\\
		B_1=B = \L^2(\Omega; \L^2(Y)),
	    \end{split}
	    \label{eq:la-spaces}
	\end{equation}
	Note that $\rho^{H,h} \in \L^2((0,T);B)$. To conclude the argument, what remains to show are the following steps:
	\begin{equation}
	    \partial_t\rho^{H,h}  \in \L^2((0,T); \L^2(\Omega; \L^2(Y)),
	    \label{eq:e2}
	\end{equation}
	and
	\begin{equation}
	    \nabla_x\rho^{H,h} \in \L^2(((0,T); \L^2(\Omega; \H^1(Y))).
	    \label{eq:e3}
	\end{equation}
	We start with the estimates that provide us \eqref{eq:e1} and \eqref{eq:e2} and then we handle \eqref{eq:e3}.

	Let $f_\pi$ and $f_\rho$ denote the partial derivatives of $f$ to respectively $\pi$ and $\rho$.
    We introduce $u^H:=\partial_t \pi^H$ and $v^h := \partial_t \rho^{H,h} - \rho_I$.  We differentiate ($P_1$) with respect to $t$ and obtain the following system
	\begin{align}
        \label{eq:pi_int}-A\Delta_x u^H &= f_\rho v^h + f_\pi u^H\mbox{ in }S\times\Omega,\\
		\label{eq:rho_int}\partial_t v^h &= D\Delta_y v^h \mbox{ in }S\times \Omega \times Y,\\
		 D\nabla_y v^h &= \kappa(u^H - Rv)\mbox{ on }S\times \Omega \times\Gamma_R,\\
         D\nabla_y v^h &= 0\mbox{ on }S\times \Omega \times\Gamma_N,\\
         u^H&=0\mbox{ on }S\times \partial\Omega\\
         v(t=0) &= 0\mbox { on } S\times  \Omega \times Y
	\end{align}
	By multiplying \eqref{eq:rho_int} with $v^h$ and integrating the result over $\Omega\times Y$, we obtain an equation that can be bounded similarly to \eqref{eq:first_est}. This gives:
	\begin{equation}
	    \begin{split}
		&\frac{1}{2}\frac{d}{dt}||v^h||^2_{\L^2(\Omega\times Y)} + D||\nabla_y v^h||^2_{\L^2(\Omega\times Y)}\\
		&= \kappa \int_{\Omega\times \Gamma_R} (u^H - Rv^h)v^h,\\
		&\leq \kappa \int_{\Omega\times \Gamma_R} u^Hv^h - \kappa R\int_{\Omega\times \Gamma_R}(v^h)^2 ,\\
		&\leq \kappa c_E||u^H||_{\L^2(\Omega;H^1(Y))}||v^h||_{\L^2(\Omega;H^1(Y))},\\
		&\leq  \kappa c_E\left(c_\varepsilon|\Gamma_R|||u^H||^2_{\L^2(\Omega)}+\varepsilon||v^h||^2_{\L^2(\Omega;H^1(Y))}\right),\\
		&\leq \kappa c_E\left(c_\varepsilon|\Gamma_R|||u^H||^2_{\L^2(\Omega)}+\varepsilon (c_p+1)||\nabla_y v^h||^2_{\L^2(\Omega\times Y)}\right).
	    \end{split}
	  \label{eq:diff_rho}
	\end{equation}
	From \eqref{eq:diff_rho}, by integration in time we obtain
	\begin{equation}
	    ||\partial_t\rho^{H,h}||^2_{\L^2(\Omega\times Y)} \leq c_\varepsilon|\Gamma_R|||\partial_t\pi^H||^2_{\L^2(\Omega)}.
	    \label{eq:partial_t_rho}
	\end{equation}
	By multiplying \eqref{eq:pi_int} with $v^h$ and integrating the result over $\Omega$, we obtain
	\begin{align*}
	    A||\nabla_x u^H||^2_{\L^2(\Omega)} &= \int_\Omega \left( f_\pi u^H + f_\rho v^h \right)u^H \leq c_1||u^H||^2_{\L^2(\Omega)} + c_2||v^h||^2_{\L^2(\Omega)},\\
	\end{align*}
	which, by combining this inequality with Poincar\'e's inequality for $u^H$, yields
	\begin{equation}
	   ||u^H||^2_{\L^2(\Omega)} \leq c_p ||\nabla u^H||^2_{\L^2(\Omega)}.
	\end{equation}
	Thus, we obtain an upper bound that holds in the interior of $\Omega$, say $\Omega_\delta$.
	\begin{equation}
	    \left( \frac{A}{c_p} - c_1 \right)||\partial_t \pi^H||^2_{\L^2(\Omega_\delta)}\leq c_2 || \partial_t \rho^{H,h}||^2_{\L^2(\Omega_\delta\times Y)}.
	    \label{eq:interior_bound}
	\end{equation}
    Here, $\Omega_\delta$ is defined as any subset of $\Omega$ such that $d(\partial\Omega,\Omega_\delta) \geq \delta$, ($d(\cdot, \cdot)$ measures the distance between two sets) and where $\delta>0$.

	To handle the integral estimates on the boundary layer $\Omega_\delta\setminus\Omega$, we recall \eqref{eq:weak_rho}. For any $\Omega_\delta \subset \Omega$, by testing with $\psi=\partial_t\rho^{H,h}$ and integrating over the time domain, we obtain the identity
	\begin{equation}
	    \begin{split}
		&2\int_0^T||\partial_t\rho^{H,h}||^2_{\L^2(\Omega_\delta\times Y)} + ||\nabla_y\rho^{H,h}||^2_{\L^2(\Omega_\delta\times Y)} - ||\nabla_y\rho^{H,h}_I||^2_{\L^2(\Omega_\delta\times Y)}\\
		&=2\kappa \left[ \int_{\Omega_\delta\times\Gamma_R}\pi^H\rho^{H,h}   \right]^T_0 - 2\kappa\int_0^T\int_{\Omega_\delta\times\Gamma_R} \rho^{H,h}\partial_t \pi^H \\
		&+ \kappa p_f\int_{\Omega_\delta\times \Gamma_R}\rho^{H,h}_I + \kappa ||\rho^{H,h}||^2_{\L^2(\Omega_\delta\times \Gamma_R)}.
	    \end{split}
        \label{eq:boundary_layer_integration}
	\end{equation}

    Conveniently rearranging the terms of \eqref{eq:boundary_layer_integration} yields:
	\begin{align}
	    \label{eq:cilinder_ineq1}&2\int_0^T ||\partial_t\rho^{H,h}||^2_{\L^2(\Omega\setminus\Omega_\delta)}
	    + ||\nabla_y\rho^{H,h}||^2_{\L^2(\Omega\setminus\Omega_\delta \times Y)}\\
	    &\leq 2\kappa||\pi^H\partial_t\rho^{H,h}||_{\L^1(\Omega\setminus\Omega_\delta \times\Gamma_R)} + \tilde{c}||\rho^{H,h}_I||^2_{\L^2(\Omega\setminus\Omega_\delta \times \gamma_R)} + \hat{c}||\rho^{H,h}||^2_{\L^2(\Omega\setminus\Omega_\delta \times \gamma_R)}\\
	    &+||\nabla_y\rho^{H,h}_I||^2_{\L^2(\Omega\setminus\Omega_\delta \times Y)}+2\kappa\int_0^T\int_{\Omega\setminus\Omega_\delta \times \Gamma_R} |\rho^{H,h}\partial_t \pi^H | \leq C\delta.
	    \label{eq:cilinder_ineq2}
	\end{align}

	Now we can extend the bound in \eqref{eq:interior_bound} to hold on the entire domain $\Omega$, i.e.:
	\begin{equation}
	    \left( \frac{A}{c_p} - c_1 \right)||\partial_t \pi^H||^2_{\L^2(\Omega)} = \sup_{\delta>0} \left( \frac{A}{c_p} - c_1 \right)||\partial_t \pi^H||^2_{\L^2(\Omega_\delta)} \leq C_2 || \partial_t \rho^{H,h}||^2_{\L^2(\Omega)}.
        \label{eq:partial_t_pi}
	\end{equation}

	Finally, to obtain \eqref{eq:e3} we adapt an interior regularity argument from \cite{evans10}, Chapter 6.
	We let $\Omega_\delta \subset\subset W \subset\subset \Omega$ and define a smooth cutoff function $\zeta:\Omega \to [0,1]$ satisfying
	\begin{equation}
	    \begin{cases}
            \zeta(x) =&\mbox{ for }x \in \Omega_\delta,\\
            \zeta(x) =&\mbox{ for }x \in \Omega\setminus W.
	    \end{cases}
	\end{equation}

	We introduce the directional finite difference
	\[D_i^\lambda\rho^{H,h}:= \frac{\rho^{H,h}(t,x+\lambda e_i,y) - \rho^{H,h}(t,x,y)}{\lambda }\mbox{ for }\lambda >0.\]
    for $i\in\{1,...,d_2\}$. We let $\lambda$ be small and we test \eqref{eq:weak_rho} with
	\[\psi = -D_i^{-\lambda}\zeta^2D_i^\lambda \rho,\] which gives us:
	\begin{equation}
	     -\int\partial_t\rho D_i^{-\lambda}\zeta^2D_i^\lambda \rho-D\int\nabla_y\rho\cdot\nabla_y D_i^{-\lambda }\zeta^2D_i^h\rho
	     = -\kappa\int(\pi + p_F - R\rho)D_i^{-\lambda}\zeta^2D_i^\lambda \rho.
	    \label{eq:weak_cutoff}
	\end{equation}

	Because of the properties of the support of $\zeta$, it holds that for any $f\in \Omega $
	\begin{equation}
	    \int_{\Omega}\psi D^{-\lambda}_if = -\int_{\Omega}f D^{\lambda}_i\psi.
        \label{eq:diffness}
	\end{equation}

    Applying the property in \eqref{eq:diffness} to \eqref{eq:weak_cutoff} yields
	\begin{equation}
	\begin{split}
	    &\int_{\Omega\times Y}\zeta^2D_i^\lambda \partial_t\rho^{H,h} D_i^\lambda \rho^{H,h}+ D\int_{\Omega\times Y} \zeta^2D_i^{\lambda }\nabla_y\rho^{H,h}\cdot D_i^\lambda\nabla_y \rho^{H,h}\\
	    &\quad = \kappa\int_{\Omega\times \Gamma_R}\zeta^2D_i^\lambda(\pi^H + p_F - R\rho^{H,h})D_i^\lambda\rho^{H,h},
	\end{split}
	\end{equation}
	leading to
	\begin{equation}
    	\begin{split}
    	    &\frac{1}{2}\frac{d}{dt}\int_{\Omega\times Y} \left|\zeta D_i^\lambda\rho^{H,h}\right|^2+ D\int_{\Omega\times Y} \left|\zeta D_i^{\lambda}\nabla_y\rho^{H,h}\right|^2\\
    	    &\quad = \kappa\int_{\Omega\times \Gamma_R}\zeta^2D_i^\lambda\pi^H D_i^\lambda\rho^{H,h}- \kappa R\int_{\Omega\times \Gamma_R}\left|\zeta D_i^\lambda\rho^{H,h}\right|^2.
    	\end{split}
	    \label{eq:compact_cutoff}
	\end{equation}
	Using Young's inequality combined with the inequality, we estimate the third term of \eqref{eq:compact_cutoff} as follows:
	\begin{equation}
	    \begin{split}
		 &\kappa\int_{\Omega\times \Gamma_R}\zeta^2D_i^\lambda\pi^H D_i^\lambda\rho^{H,h}\\
		&\leq \kappa|\Gamma_R|\ ||D_i^\lambda\pi^H||_{\L^2(\Omega)}\ ||\zeta D_i^\lambda\rho^{H,h}||_{\L^2(\Omega\times \Gamma_R)},\\
		&\leq c_\varepsilon\kappa|\Gamma_R|\ ||D_i^\lambda\pi^H||^2_{\L^2(\Omega)}+ \varepsilon||\zeta D_i^\lambda\rho^{H,h}||^2_{\L^2(\Omega\times \Gamma_R)},\\
		&\leq C_\varepsilon\kappa|\Gamma_R|\ ||\nabla_x \pi^H||^2_{\L^2(\Omega)} + \varepsilon||\zeta D_i^\lambda\rho^{H,h}||_{\Omega\times Y}||\zeta D_i^\lambda\nabla_y\rho^{H,h}||_{\L^2(\Omega\times Y)},\\
		&\leq C_\varepsilon\kappa|\Gamma_R|\ ||\nabla_x \pi^H||^2_{\L^2(\Omega)} + \frac{\varepsilon^2}{2}||\zeta D_i^\lambda\rho^{H,h}||_{\Omega\times Y}\\
		&\quad + \frac{\varepsilon^2}{2}||\zeta D_i^\lambda\nabla_y\rho^{H,h}||_{\L^2(\Omega\times Y)}.
	    \end{split}
	    \label{eq:cutoff_ineq}
	\end{equation}

	Now, combining \eqref{eq:cutoff_ineq} with \eqref{eq:compact_cutoff}, we obtain the required estimate for \eqref{eq:e3}:
	\begin{equation}
    	\begin{split}
    	    &\frac{1}{2}\frac{d}{dt}\int_{\Omega\times Y} \left|\zeta D_i^\lambda\rho^{H,h}\right|^2+ \left(D-\frac{\varepsilon^2}{2}\right)\int_{\Omega\times Y} \left|\zeta D_i^{\lambda}\nabla_y\rho^{H,h}\right|^2 \\
    	    &\quad \leq C_\varepsilon\kappa|\Gamma_R|\ ||\nabla_x \pi^H||^2_\Omega + \frac{\varepsilon^2}{2}||\zeta D_i^\lambda\rho^{H,h}||_{\Omega\times Y}.
    	\end{split}
	\end{equation}
	Using Gr\"onwall's inequality, we conclude that $D_i^\lambda\rho^{H,h} \in L^2(\Omega \times Y)$, and by letting $\lambda\to0$, we obtain
	\begin{equation}
	    \nabla_x\rho^{H,h},\ \nabla_x\nabla_y\rho^{H,h} \in \L^2((0,T)\times\Omega\times Y).
        \label{eq:bound_grads}
	\end{equation}
    With the newly found estimates \eqref{eq:partial_t_pi}, \eqref{eq:bound_grads} and \eqref{eq:partial_t_rho}, we are able to apply Lemma~\ref{lem:aubinlions} and we obtain that
	\[W \embed\embed \L^2((0,T)\times\Omega\times Y),\]
	which proves:
	\[(\pi^H,\rho^{H,h}) \to (\pi,\rho).\]
for $h,H\to 0$.
\end{proof}
The preliminary work allows us to state the first main result of this paper.
\ \\
\begin{theorem}[Well-posedness of the system]
    The system in \eqref{eq:weak_pi}-\eqref{eq:weak_rho} has a unique solution $\pi^H \in \L^2(S; V^H)$ and $\rho^{H,h}\in \L^2(S; V^H\times W^h)$.
    \label{thm:well_posedness}
\end{theorem}
\begin{proof}
    The proof of this theorem is a direct result of Proposition~\ref{prop:ex_un} and Proposition~\ref{prop:conv}.
\end{proof}


TITLE: Show that there is a natural one-to-one correspondence
QUESTION [3 upvotes]: This example is the book Functional Analysis by Walter Rudin in page 288 Exercise 3.

If $X$ is a compact Hausdorff space, show that there is a natural one-to-one correspondence between closed subset $X$ and closed ideals of $C(X)$.

REPLY [1 votes]: I guess $C(X)$ is endowed with the natural norm, that is $\lVert  f\rVert:=\sup_{x\in X}|f(x)|$. Let $\mathcal F$ the collection of closed subsets of $X$ and $\mathcal I$ the collection of the closed ideals of $C(X)$. We can define
$$\phi(F):=\{f\in C(X),\forall x\in F,\, f(x)=0\}.$$
What we have to show it that

$\phi(F)$ is an ideal of $C(X)$;
$\phi(F)$ is closed;
given two distinct closed sets $F_1$ and $F_2$, if $F_1$ is not contained in $F_2$ we can find a continuous function which vanishes on $F_1$ but not on $F_2$;
given a closed ideal $I$, we can find a closed set $F$ such that $\phi(F)=I$.


TITLE: Dimension of vector spaces under surjection
QUESTION [2 upvotes]: Definition. Let $V$ be a vector space and $X\subseteq V$ be a definable subset in the language of vector spaces. $X$ is independent if for all $a\in X$, $a\notin span(A\setminus\{a\})$.
Definition. Let $V$ be a vector space and $Y\subseteq V$  be a definable subset (in the language of vector spaces). We say that $B$ is a basis for $Y$ if $B \subseteq Y$ is
independent and $span(A) = span(Y)$. The dimension of Y is the cardinality
of a basis for Y .
My question is this:
Suppose $V$ is a vector space and $X,Y\subseteq V$ are definable subsets. Let $f:X\rightarrow Y$ be a definable surjection. Also, let $\dim(f^{-1}(y))=d$ for all $y\in Y$. Is the following statement always true? why?
$$\dim(X)= \dim(Y) + d$$

REPLY [4 votes]: With your definition of dimension, almost any definable map you try will be a counterexample.
For example, let $V$ be a $1$-dimensional vector space. Then any set $X\subseteq V$ has dimension $1$, except for $\emptyset$ and $\{0\}$ (which both have dimension $0$).
Let $X = Y = V\setminus \{0\}$, and let $f\colon X\to Y$ be the identity map. Then for all $y\in Y$, $\dim(f^{-1}(y)) = \dim(\{y\}) = 1$. But $\dim(X) = 1$,  while $\dim(Y) +1 = 2$.

Here is an alternative (and more sensible) definition of dimension for definable sets: Let $X\subseteq V^n$ be a definable set, and let's assume $V$ is $\omega$-saturated (if not, take a suitable elementary extension, and evaluate $X$ there). An element of $X$ is an $n$-tuple $a$, and we define $\dim(a)$ to be the size of a maximal independent sub-tuple of $a$, equivalently the dimension of $\text{span}(a)$ in your sense. Then we define $\dim(X)$ to be $\max_{a\in X} \dim(a)$, the maximal dimension of an element of $X$.
This definition of dimension agrees with the Morley rank relative to the theory of infinite $K$-vector spaces, and it satisfies the additivity condition in your question.

In the comments, you asked some follow-up questions.

(1) Is my definition of dimension coincide to Morley rank? (I used Definitions 6.1.5 and 6.1.8 and 6.1.10 of Marker's book).

No - as I said in my answer, my definition of dimension coincides with Morley rank.
There's a confusing thing about strongly minimal theories, which is that there are two reasonable meanings of "dimension", which are very different. Let $T$ be a strongly minimal theory.

Let $M\models T$, and let $A\subseteq M$ (not necessarily definable!). Define $\dim_1(A)$ to be the cardinality of a basis for $\text{acl}(A)$.
Let $X$ be a definable set, defined by $\varphi(x)$. Let $\mathbb{M}\models T$ be a monster model of $T$ (or at least $\omega$-saturated). Define $\dim_2(X)$ to be the maximal value of $\dim_1(a)$ over all tuples $a\in X(\mathbb{M})$.

The fact that $\dim_2$ coincides with Morley rank is Theorem 6.2.19, together with Lemma 6.2.16(i), in Marker.
In vector spaces, $\dim_1$ coincides with the ordinary notion of linear dimension, for closed sets. But it's unsatisfactory for definable sets, since it isn't invariant under elementary extensions. For example, let $T$ be the theory of $\mathbb{R}$-vector spaces, and consider the formula $\phi(x): x = x$. We have $\dim_1(\varphi(\mathbb{R}^2)) = 2$, but $\dim_1(\varphi(\mathbb{R}^{42})) = 42$, despite the fact that $\mathbb{R}^2\preceq \mathbb{R}^{42}$. So $\dim_1$ is really assigning a dimension to a particular subset of a particular model.
If you're familiar with field theory or algebraic geometry, it might be morehelpful to think about the theory of algebraically closed fields, where both notions of dimension are familiar mathematical notions.

If $A\subseteq K\models \text{ACF}_0$, then $\dim_1(A)$ is the transcendence degree of the field $\overline{\mathbb{Q}(A)}$ over $\mathbb{Q}$.
If $X$ is a definable set (which is a Boolean combination of zero-sets of polynomials), then $\dim_2(X)$ is the Krull dimension of the Zariski closure of $X$.

(2) Does acl(X)=span(X) in vector spaces?

Yes, in any infinite vector space $V$, $\text{acl}(X) = \text{span}(X)$ for all sets $X$. The inclusion $\text{span}(X)\subseteq \text{acl}(X)$ is clear, and for the other, note that if $b\notin \text{span}(X)$, then $b$ has infinitely many images under automorphisms of $V$, so it's not in $\text{acl}(X)$.
Of course, if $V$ is finite (which can only happen if the base field is finite), then $\text{acl}(X) = V$ for any $X$.

(3) Does the additivity condition in my question hold for Morley rank in any strongly minimal structure (or we need to have extra assumptions)?

Yes, it does. Basically, this comes down to the fact that in any strongly minimal structure, for any tuples $a$ and $b$, we have $$\dim(ab) = \dim(a/b) + \dim(b).$$
Here $\dim$ is what I wrote as $\dim_1$ above. For a proof, see here.