\section{Introduction}
In 1964, H.\ Bass \cite{bass} showed that if $R$ is a ring and $H$ a subgroup of the general linear group $\operatorname{GL}_n(R)$, then
\begin{equation*}
H\text{ is normalised by }\operatorname{E}_n(R)~\Leftrightarrow~ \exists \text{ ideal }I:~\operatorname{E}_n(R,I)\subseteq H\subseteq \operatorname{C}_n(R,I)
\end{equation*}
provided $n$ is large enough with respect to the stable rank of $R$. Here $\operatorname{E}_n(R)$ denotes the elementary subgroup, $\operatorname{E}_n(R,I)$ the relative elementary subgroup of level $I$ and $\operatorname{C}_n(R,I)$ the full congruence subgroup of level $I$ (cf. \cite{preusser2}). Bass's result, which is one of the central points in the structure theory of general linear groups, is known as {\it Sandwich Classification Theorem}. In the 1970's and 80's, the validity of this theorem was extended by J.\ Wilson \cite{wilson}, I.\ Golubchik \cite{golubchik}, L.\ Vaserstein \cite{vaserstein, vaserstein_ban, vaserstein_neum} and others. It holds true, for example, if $R$ is almost commutative (i.e. finitely generated as a module over its center) and $n\geq 3$.

It follows from the Sandwich Classification Theorem that if $R$ is a commutative ring, $n\geq 3$, $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$ and $k\neq l$, then the elementary transvection $t_{kl}(\sigma_{ij})$ can be expressed as a finite product of $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
In 1960, J.\ Brenner \cite{brenner} showed that if $R=\mathbb{Z}$, then there is a bound for the number of factors needed for such an expression of $t_{kl}(\sigma_{ij})$, but until recently there was not much hope to prove such a result for arbitrary commutative rings $R$.

However, in 2018 the author found an explicit expression of $t_{kl}(\sigma_{ij})$ (where $\sigma\in \operatorname{GL}_n(R)$, $R$ is a commutative ring, $n\geq 3$, $i\neq j$, $k\neq l$) as a product of $8$ elementary conjugates of $\sigma$ and $\sigma^{-1}$, yielding a very short proof of the Sandwich Classification Theorem for commutative rings \cite{preusser2}. Similar results were obtained for the even- and odd-dimensional orthogonal groups $\operatorname{O}_{2n}(R)$ and $\operatorname{O}_{2n+1}(R)$, and the even- and odd-dimensional unitary groups $\operatorname{U}_{2n}(R,\Lambda)$ and $\operatorname{U}_{2n+1}(R,\Delta)$ where $R$ is a commutative ring and $n\geq 3$ \cite{preusser2, preusser3}.

In order to show the normality of the elementary subgroup $\operatorname{E}_n(R)$ in $\operatorname{GL}_n(R)$, one has to prove that $\operatorname{E}_n(R)^\sigma\subseteq \operatorname{E}_n(R)$ for any $\sigma\in\operatorname{GL}_n(R)$. {\it Decomposition of unipotents} provides explicit formulae expressing the generators $t_{kl}(x)^\sigma~(k\neq l, x\in R)$ of $\operatorname{E}_n(R)^{\sigma}$ as products of elementary transvections, see \cite{stepanov-vavilov}. In order to show the Sandwich Classification Theorem for commutative rings, one has to prove that $\operatorname{E}_n(I)\subseteq \sigma^{E_n(R)}$ for any $\sigma\in\operatorname{GL}_n(R)$ where $I$ denotes the ideal generated by the nondiagonal entries of $\sigma$ and all differences of diagonal entries. The paper \cite{preusser2} provides explicit formulae expressing the generators $t_{kl}(x\sigma_{ij}),t_{kl}(x(\sigma_{ii}-\sigma_{jj}))~(k\neq l,i\neq j,x\in R)$ of $\operatorname{E}_n(I)$, as products of $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. N.\ Vavilov considered the papers \cite{preusser2, preusser3} ``the first major advance in the direction of what can be dubbed the {\it reverse decomposition of unipotents}'', see \cite{vavilov}.

In this paper we obtain bounds for the reverse decomposition of unipotents over different classes of noncommutative rings.
Pick a $\sigma\in \operatorname{GL}_n(R)$. Our goal is to express the matrix $t_{kl}(\sigma_{ij})$ where $k\neq l$ and $i\neq j$ as a product of $E_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. An idea is to start with $\sigma$ and then apply successively operations of types (a) and (b) below until one arrives at $t_{kl}(\sigma_{ij})$.
\begin{enumerate}[(a)]
\item $\operatorname{GL}_n(R)\rightarrow \operatorname{GL}_n(R),\tau\mapsto \tau^\xi$ where $\xi\in \operatorname{E}_n(R)$.
\item $\operatorname{GL}_n(R)\rightarrow \operatorname{GL}_n(R),\tau\mapsto [\tau,\xi]$ where $\xi\in \operatorname{E}_n(R)$.
\end{enumerate}
If $\tau$ lies in $\sigma^{\operatorname{E}_n(R)}$ and $\rho$ is obtained from $\tau$ by applying an operation of type (a) or (b), then clearly $\rho$ again lies in $\sigma^{\operatorname{E}_n(R)}$.

In this paper we also use operations of type (c) below.
\begin{enumerate}[(c)]
\setcounter{enumi}{3}
\item $\operatorname{GL}_n(R)\times\operatorname{GL}_n(R)\rightarrow \operatorname{GL}_n(R)\times\operatorname{GL}_n(R),(\tau_1,\tau_2)\mapsto ([\tau_1^{-1},\xi],[\xi,\tau_2])$ where $\xi\in \operatorname{E}_n(R)$.
\end{enumerate}
We show in Section 4 that if the product $\tau_1\tau_2$ lies in $\sigma^{\operatorname{E}_n(R)}$ where $\tau_1\in\operatorname{E}_n(R)$, and $(\rho_1, \rho_2)$ is obtained from $(\tau_1,\tau_2)$ by applying an operation of type (c), then the product $\rho_1\rho_2$ again lies in $\sigma^{\operatorname{E}_n(R)}$.

The rest of the paper is organised as follows. In Section 2 we recall some standard notation which is used throughout the paper. In Section 3 we recall the definitions of the general linear group $\operatorname{GL}_n(R)$ and its elementary subgroup $\operatorname{E}_n(R)$. In Section 4, we introduce the operations $(c)$ above in the abstract context of groups. Then we use these operations to obtain results for general linear groups which are crucial for the following sections. In Section 5 we show how \cite[Theorem 12]{preusser2} can be derived from the results of Section 4. In Sections 6-10 we get bounds for the reverse decomposition of unipotents over von Neumann regular rings, Banach algebras, rings satisfying a stable range condition, rings with Euclidean algorithm and almost commutative rings. In the last section we list some open problems.

\section{Notation}
$\mathbb{N}$ denotes the set of positive integers. If $G$ is a group and $g,h\in G$, we let $g^h:=h^{-1}gh$, and $[g,h]:=ghg^{-1}h^{-1}$. If $g\in G$ and $H$ is a subgroup of $G$, we denote by $g^{H}$ the subgroup of $G$ generated by the set $\{g^h\mid h\in H\}$. By a ring  we mean an associative ring with $1\neq 0$. By an ideal of a ring we mean a twosided ideal.

Throughout the paper $R$ denotes a ring and $n$ a positive integer greater than $2$. We denote by $^n\!R$ the set of all rows $u=(u_1,\dots,u_n)$ with entries in $R$ and by $R^n$ the set of all columns $v=(v_1,\dots,v_n)^t$ with entries in $R$. Furthermore, the set of all $n\times n$ matrices over $R$ is denoted by $\operatorname{M}_n(R)$. The identity matrix in $\operatorname{M}_n(R)$ is denoted by $e$ or $e_{n\times n}$ and the matrix with a one at position $(i,j)$ and zeros elsewhere is denoted by $e^{ij}$. If $\sigma\in \operatorname{M}_{n
}(R)$, we denote the
the entry of $\sigma$ at position $(i,j)$ by $\sigma_{ij}$. We denote the $i$-th row of $\sigma$ by $\sigma_{i*}$ and its $j$-th column by $\sigma_{*j}$. If $\sigma\in \operatorname{M}_n(R)$ is invertible, we denote the entry
of $\sigma^{-1}$ at position $(i,j)$ by $\sigma'_{ij}$, the $i$-th row of $\sigma^{-1}$ by $\sigma'_{i*}$ and the $j$-th column of $\sigma^{-1}$ by $\sigma'_{*j}$.

\section{The general linear group and its elementary subgroup}

\begin{definition}
The group $\operatorname{GL}_{n}(R)$ consisting of all invertible elements of $\operatorname{M}_n(R)$ is called the {\it general linear group} of degree $n$ over $R$.
\end{definition}

\begin{definition}
Let $x\in R$ and $i,j\in\{1,\dots,n\}$ such that $i\neq j$. Then the matrix $t_{ij}(x):=e+xe^{ij}$ is called an {\it elementary transvection}. The subgroup $\operatorname{E}_n(R)$ of $\operatorname{GL}_n(R)$ generated by the elementary transvections is called the {\it elementary subgroup}.
\end{definition}

\begin{lemma}\label{lemelrel}
The relations
\begin{align*}
t_{ij}(x)t_{ij}(y)&=t_{ij}(x+y), \tag{R1}\\
[t_{ij}(x),t_{hk}(y)]&=e \tag{R2}\text{ and}\\
[t_{ij}(x),t_{jk}(y)]&=t_{ik}(xy) \tag{R3}\\
\end{align*}
hold where $i\neq k, j\neq h$ in $(R2)$ and $i\neq k$ in $(R3)$.
\end{lemma}
\begin{proof}
Straightforward computation.
\end{proof}

\begin{definition}\label{defp}
Let $i,j\in\{1,\dots,n\}$ such that $i\neq j$. Then the matrix $p_{ij}:=e+e^{ij}-e^{ji}-e^{ii}-e^{jj}=t_{ij}(1)t_{ji}(-1)t_{ij}(1)\in \operatorname{E}_{n}(R)$ is called a {\it generalised permutation matrix}. It is easy to show that $p_{ij}^{-1}=p_{ji}$. The subgroup of $\operatorname{E}_n(R)$ generated by the generalised permutation matrices is denoted by $\operatorname{P}_n(R)$.
\end{definition}

\begin{lemma}\label{lemp}
Let $\sigma\in \operatorname{GL}_n(R)$. Further let $x\in R$ and $i,j,k,l\in\{1,\dots,n\}$ such that $i\neq j$ and $k\neq l$. Then there are $\tau,\rho\in \operatorname{P}_n(R)$ such that $(\sigma^{\tau})_{kl}=\sigma_{ij}$ and $t_{kl}(x)^{\rho}=t_{ij}(x)$.
\end{lemma}
\begin{proof}
Easy exercise.
\end{proof}

\section{The key results}
\subsection{Simultaneous reduction in groups}

In this subsection $G$ denotes a group.
\begin{definition}
Let $(a_1,b_1), (a_2,b_2)\in G\times G$. If there is an $g\in G$ such that
\[a_2=[a_1^{-1},g]\text{ and }b_2=[g,b_1],\]
then we write $(a_1,b_1)\xrightarrow{g} (a_2,b_2)$. In this case $(a_1,b_1)$ is called {\it reducible to $(a_2,b_2)$ by $g$}.
\end{definition}

\begin{definition}
If $(a_1,b_1),\dots, (a_{n+1},b_{n+1})\in G\times G$ and $g_1,\dots,g_{n}\in G$  such that
\[(a_1,b_1) \xrightarrow{g_1} (a_2,b_2)\xrightarrow{g_2}\dots \xrightarrow{g_{n}} (a_{n+1},b_{n+1}),\]
then we write $(a_1,b_1) \xrightarrow{g_1,\dots,g_{n}}(a_{n+1},b_{n+1})$. In this case $(a_1,b_1)$ is called {\it reducible to $(a_{n+1},b_{n+1})$ by $g_1,\dots,g_n$}.
\end{definition}

Let $H$ be a subgroup of $G$. If $g\in G$ and $h\in H$, then we call $g^h$ an {\it $H$-conjugate} of $g$.
\begin{lemma}\label{lemredux}
Let $(a_1,b_1),(a_2,b_2)\in G\times G$. If $(a_1,b_1)\xrightarrow{g_1,\dots,g_n}(a_2,b_2)$ for some $g_1,\dots,g_n\in G$, then $a_2b_2$ is a product of $2^n$ $H$-conjugates of $a_1b_1$ and $(a_1b_1)^{-1}$ where $H$ is the subgroup of $G$ generated by $\{a_1,g_1,\dots, g_n\}$.
\end{lemma}
\begin{proof}
Assume that $n=1$. Then
\[a_2b_2=[a_1^{-1},g_1][g_1,b_1]=(a_1b_1)^{g_1^{-1}a_1}\cdot((a_1b_1)^{-1})^{a_1}.\]
Hence $a_2b_2$ is a product of two $H$-conjugates of $a_1b_1$ and $(a_1b_1)^{-1}$. The general case follows by induction (note that $a_1,\dots,a_n\in H$).
\end{proof}

\subsection{Application to general linear groups}

\begin{proposition}\label{propkey}
Let $\sigma\in \operatorname{GL}_n(R)$. Let $x_1,\dots,x_{n},y\in R$ such that $y\sum\limits_{p=1}^{n}\sigma_{1p}x_p=0$. Then the following holds.
\begin{enumerate}[(i)]
\itemsep0pt
\item Suppose that $x_n=0$, $y=1$. Then for any $k\neq l$ and $a,b\in R$, the elementary transvection $t_{kl}(ax_1b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\item Suppose that $x_n=1$, $y=1$. Then for any $k\neq l$ and $a,b\in R$, the elementary transvection $t_{kl}(ax_1b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\item Suppose that $x_1=1$. Then for any $k\neq l$ and $a,b\in R$, the elementary transvection $t_{kl}(ayb)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{proposition}
\begin{proof}
\begin{enumerate}[(i)]
\item Set $\tau:=\prod\limits_{p=1}^{n-1}t_{pn}(x_p)$. Clearly $(\sigma\tau^{-1})_{1*}=\sigma_{1*}$ and hence $(\sigma\tau^{-1}\sigma^{-1})_{1*}=e_{1*}$. A straightforward computation shows
\[(\tau,\sigma\tau^{-1}\sigma^{-1})\xrightarrow{t_{21}(a),t_{n1}(-b)}(t_{21}(ax_1b), e).\]
It follows from Lemma \ref{lemredux} that $t_{21}(ax_1b)$ is a product of $4$ $\operatorname{E}_n(R)$-conjugates of $[\tau,\sigma]$ and $[\tau,\sigma]^{-1}$. Thus $t_{21}(ax_1b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (i) follows now from Lemma \ref{lemp}.
\item Set $\tau:=\prod\limits_{p=1}^{n-1}t_{pn}(x_p)$. Clearly $(\sigma\tau)_{1n}=0$. A straightforward computation shows that
\[(\tau,\tau^{-1}\sigma^{-1})\xrightarrow{t_{21}(a),t_{n1}(-b),t_{n2}(1)}(t_{n1}(ax_1b), e).\]
It follows from Lemma \ref{lemredux} that $t_{n1}(ax_1b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (ii) follows now from Lemma \ref{lemp}.
\item Clearly $t_{n1}(y)^{\sigma}=e+\sigma'_{*n}y\sigma_{1*}$. Set $\tau:=\prod\limits_{p=2}^{n}t_{p1}(x_p)$. One checks easily that $(t_{n1}(y)^{\sigma\tau})_{*1}=e_{*1}$. A straightforward computation shows that
\[(t_{n1}(-y),t_{n1}(y)^{\sigma\tau})\xrightarrow{t_{12}(b),t_{1n}(-a)}(t_{12}(ayb), e).\]
It follows from Lemma \ref{lemredux} that $t_{12}(ayb)$ is a product of $4$ $\operatorname{E}_n(R)$-conjugates of $t_{n1}(-y)t_{n1}(y)^{\sigma\tau}=(\sigma^{-1})^{\tau t_{n1}(y)}\cdot \sigma^\tau$ and its inverse. Thus $t_{12}(ayb)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (iii) follows now from Lemma \ref{lemp}.
\end{enumerate}
\end{proof}

\begin{corollary}\label{corkeyB}
Let $\sigma\in \operatorname{GL}_n(R)$ such that $\sigma_{1n}=0$. Then for any $2\leq j\leq n$, $k\neq l$ and $a,b\in R$, the elementary transvection $t_{kl}(a\sigma'_{1j}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{corollary}
\begin{proof}
Clearly we have $\sum\limits_{p=1}^{n-1}\sigma_{1p}\sigma'_{pj}=0$ for any $2\leq j\leq n$. Hence, by Proposition \ref{propkey}(i), the elementary transvection $t_{kl}(a\sigma'_{1j}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{proof}

\begin{corollary}\label{corkeyA}
Let $\sigma\in \operatorname{GL}_n(R)$ such that $\sigma_{11}$ is right invertible. Then for any $k\neq l$ and $a,b\in R$, the elementary transvection $t_{kl}(a\sigma_{1n}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{corollary}
\begin{proof}
Let $z$ be a right inverse of $\sigma_{11}$. Then $-\sigma_{11}z\sigma_{1n}+\sigma_{1n}=0$. Hence, by Proposition \ref{propkey}(ii), the elementary transvection $t_{kl}(a\sigma_{1n}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{proof}

\begin{corollary}\label{corkeyC}
Let $\sigma\in \operatorname{GL}_n(R)$ such that $\sigma_{1n}$ is an idempotent. Then for any $k\neq l$ and $a,b\in R$, the elementary transvection $t_{kl}(a\sigma_{1n}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. In particular, if $\sigma_{1n}=1$, then  $\operatorname{E}_n(R)\subseteq \sigma^{\operatorname{E}_n(R)}$.
\end{corollary}
\begin{proof}
Clearly we have $\sigma_{1n}(\sigma_{11}-\sigma_{1n}\sigma_{11})=0$. Hence, by Proposition \ref{propkey}(iii), the elementary transvection $t_{kl}(a\sigma_{1n}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{proof}

\begin{corollary}\label{corkeyD}
Let $\sigma\in \operatorname{GL}_n(R)$. Let $x_1,\dots,x_{n-1}\in R$ such that $\sum\limits_{p=1}^{n-1}\sigma_{1p}x_p+\sigma_{1n}=0$. Then for any $2\leq j\leq n$, $k\neq l$ and $a,b\in R$, the elementary transvection $t_{kl}(a\sigma'_{1j}b)$ is a product of $16$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{corollary}
\begin{proof}
One checks easily that $\sum\limits_{p=1}^{n-1}\sigma_{1p}(\sigma'_{pj}-x_p\sigma'_{nj})=0$. Hence, by Proposition \ref{propkey}(i), the elementary transvection $t_{kl}(a(\sigma'_{1j}-x_1\sigma'_{nj})b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. By Proposition \ref{propkey}(ii), $t_{kl}(ax_1\sigma'_{nj}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Thus $t_{kl}(a\sigma'_{1j}b)$ is a product of $16$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{proof}

\section{RDU over commutative rings}
\begin{theorem}[cf. {\cite[Theorem 12]{preusser2}}]\label{thmcomm}
Suppose that $R$ is commutative. Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij})$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(\sigma_{ii}-\sigma_{jj}))$ is a product of $24$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item Clearly $\sigma_{11}\sigma_{12}-\sigma_{12}\sigma_{11}=0$. Hence, by Proposition \ref{propkey}(i), $t_{kl}(a\sigma_{12})$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (i) now follows from Lemma \ref{lemp}.
\item Clearly the entry of $\sigma^{t_{ji}(-1)}$ at position $(j,i)$ equals $\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij}$. Applying (i) to $\sigma^{t_{ji}(-1)}$ we get that $t_{kl}(a(\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij}))$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Applying (i) to $\sigma$ we get that $t_{kl}(a(\sigma_{ij}-\sigma_{ji}))=t_{kl}(a\sigma_{ij})t_{kl}(-a\sigma_{ji})$ is a product of $8+8=16$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. It follows that $t_{kl}(a(\sigma_{ii}-\sigma_{jj}))=t_{kl}(a(\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij}))t_{kl}(a(\sigma_{ij}-\sigma_{ji}))$ is a product of $24$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{proof}

\section{RDU over von Neumann regular rings}

Recall that $R$ is called {\it von Neumann regular}, if for any $x\in R$ there is a $y\in R$ such that $xyx=x$.
\begin{theorem}\label{thmNeum}
Suppose that $R$ is von Neumann regular. Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a,b,c\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)$ is a product of $24$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item Choose a $z$ such that $\sigma_{12}z\sigma_{12}=\sigma_{12}$. Then $\sigma_{12}z(\sigma_{11}-\sigma_{12}z\sigma_{11})=0$. By Proposition \ref{propkey}(iii) (applied with $y=\sigma_{12}z$, $x_1=1$, $x_2=-z\sigma_{11}$ and $x_3,\dots,x_n=0$), we get that $t_{kl}(a\sigma_{12}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (i) now follows from Lemma \ref{lemp}.
\item Clearly the entry of $\sigma^{t_{ji}(-c)}$ at position $(j,i)$ equals $c\sigma_{ii}-\sigma_{jj}c+\sigma_{ji}-c\sigma_{ij}c$. Applying (i) to $\sigma^{t_{ji}(-c)}$ we get that $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c+\sigma_{ji}-c\sigma_{ij}c)b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Applying (i) to $\sigma$ we get that $t_{kl}(a(c\sigma_{ij}c-\sigma_{ji})b)=t_{kl}(ac\sigma_{ij}cb)t_{kl}(-a\sigma_{ji}b)$ is a product of $8+8=16$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. It follows that $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)=t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c+\sigma_{ji}-c\sigma_{ij}c)b)t_{kl}(a(c\sigma_{ij}c-\sigma_{ji})b)$ is a product of $24$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{proof}

\section{RDU over Banach algebras}
Recall that a {\it Banach algebra} is an algebra $R$ over the real or the complex numbers that at the same time is also a {\it Banach space}, i.e. a normed vector space that is complete with respect to the metric induced by the norm. The norm is required to satisfy $\|x\,y\|\ \leq \|x\|\,\|y\|$ for any $x,y\in R$.

It follows from \cite[Chapter VII, Lemma 2.1]{conway_book} that any Banach algebra $R$ has the property (1) below where $R^*$ denotes the set of all right invertible elements of $R$.
\begin{equation}
\text{For any }x,z\in R\text{ there is a }y\in R^*\text{ such that } 1+xyz\in R^*.
\end{equation}

\begin{theorem}\label{thmBan}
Suppose that $R$ is a ring satisfying (1) (which is true e.g. if $R$ is a Banach algebra). Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a,b,c\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij}b)$ is a product of $160$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)$ is a product of $480$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item {\bf Step 1} Let $x\in R$ and set $\tau:=[\sigma,t_{1n}(x)]$. Then $\tau_{11}=1+\sigma_{11}x\sigma'_{n1}\in R^*$ for an appropriate $x\in R^*$. Let $x^{-1}$ denote a right inverse of $x$. It follows from Corollary \ref{corkeyA} that for any $k'\neq l'$ and $a',b'\in R$, the elementary transvection
\[t_{k'l'}(a'\tau_{1n}b')=t_{k'l'}(a'(\sigma_{11}x\sigma'_{nn}-(1+\sigma_{11}x\sigma'_{n1})x)b')=t_{k'l'}(a'(\sigma_{11}\alpha-1)xb'),\]
where $\alpha=x\sigma'_{nn}x^{-1}-x\sigma'_{n1}$, is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\tau$ and $\tau^{-1}$. Hence
\begin{equation}
t_{k'l'}(a'(\sigma_{11}\alpha-1)b')\text{ is a product of }16~\operatorname{E}_n(R)\text{-conjugates of }\sigma \text{ and }\sigma^{-1}.
\end{equation}
{\bf Step 2} Set $\zeta:=\sigma t_{1n}(-\alpha\sigma_{1n})$ where $\alpha$ is defined as in Step 1. Clearly
\[(t_{1n}(\alpha\sigma_{1n}),\zeta)\xrightarrow{t_{n2}(y)}(t_{12}(-\alpha\sigma_{1n}y),\xi)\]
for any $y\in R$, where $\xi=[t_{n2}(y),\zeta]$. It follows from Lemma \ref{lemredux} that
\begin{equation}
t_{12}(-\alpha\sigma_{1n}y)\xi\text{ is a product of }2~\operatorname{E}_n(R)\text{-conjugates of }\sigma \text{ and }\sigma^{-1}.
\end{equation}
Choose a $y\in R^*$ such that $\xi_{11}=1-\zeta_{1n}y\zeta'_{21}\in R^*$. Let $\xi_{11}^{-1}$ denote a right inverse of $\xi_{11}$. Then $\rho:=\xi t_{12}(-\xi_{11}^{-1}\xi_{12})$ has the property that $\rho_{12}=0$. Clearly $\xi_{12}=-\zeta_{1n}y\zeta'_{22}=-(1-\sigma_{11}\alpha)\sigma_{1n}y\zeta'_{22}$. Hence, by (2),
\begin{equation}
t_{12}(-\xi_{11}^{-1}\xi_{12})\text{ is a product of }16~\operatorname{E}_n(R)\text{-conjugates of }\sigma \text{ and }\sigma^{-1}.
\end{equation}
It follows from (3) and (4) that
\begin{equation}
t_{12}(-\alpha\sigma_{1n}y)\rho\text{ is a product of }18~\operatorname{E}_n(R)\text{-conjugates of }\sigma \text{ and }\sigma^{-1}.
\end{equation}
{\bf Step 3} Let $y^{-1}$ denote a right inverse of $y$. One checks easily that for any $a'',b''\in R$ we have
\[(t_{12}(-\alpha\sigma_{1n}y),\rho)\xrightarrow{t_{23}(y^{-1}b''),t_{21}(a''),t_{31}(-1)}(t_{21}(a''\alpha\sigma_{1n}b''), e)\]
(recall from Step 2 that $\rho_{12}=0$). It follows from Lemma \ref{lemredux} that $t_{21}(a''\alpha\sigma_{1n}b'')$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $t_{12}(-\alpha\sigma_{1n}y)\rho$ and its inverse. Hence, by (5),
\begin{equation}
t_{21}(a''\alpha\sigma_{1n}b'')\text{ is a product of }8\cdot 18=144~\operatorname{E}_n(R)\text{-conjugates of }\sigma \text{ and }\sigma^{-1}.
\end{equation}
It follows from (2) and (6) that $t_{21}(a\sigma_{1n}b)=t_{21}(a(1-\sigma_{11}\alpha)\sigma_{1n}b)t_{21}(a\sigma_{11}\alpha\sigma_{1n}b)$ is a product of $144+16=160$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (i) now follows from Lemma \ref{lemp}.
\item See the proof of Theorem \ref{thmNeum}.
\end{enumerate}
\end{proof}

\section{RDU in the stable range}

Recall that a row vector $u\in {}^m\!R$ is called {\it unimodular} if there is a column vector $v\in R^m$ such that $uv=1$. The {\it stable rank $\operatorname{sr}(R)$} is the least $m\in\mathbb{N}$ such that for any $p\geq m$ and unimodular row $(u_1,\dots,u_{p+1})\in {}^{p+1}\!R$ there are elements $x_1,\dots,x_p\in R$ such that $(u_1+u_{p+1}x_1,\dots,u_p+u_{p+1}x_p)\in {}^p\!R$ is unimodular (if no such $m$ exists, then $\operatorname{sr}(R)=\infty$).

Define $E^*_n(R):=\langle t_{kl}(x)\mid x\in R,k\neq l, k\neq 1, l\neq n\rangle$.

\begin{lemma}\label{lemsr}
Suppose that $\operatorname{sr}(R)<n$ and let $\sigma\in \operatorname{GL}_n(R)$. Then there is a $\rho\in E^*_n(R)$ such that the row vector $((\sigma^{\rho})_{11},\dots,(\sigma^{\rho})_{1,\operatorname{sr}(R)})$ is unimodular.
\end{lemma}
\begin{proof}
Since $\operatorname{sr}(R)<n$, there is a $\rho$ of the form
\[\rho=(\prod\limits_{j=1}^{n-1}t_{nj}(*))(\prod\limits_{j=1}^{n-2}t_{n-1,j}(*))\dots (\prod\limits_{j=1}^{\operatorname{sr}(R)}t_{\operatorname{sr}(R)+1,j}(*))\in E^*_n(R)\]
such that $((\sigma\rho)_{11},\dots,(\sigma\rho)_{1,\operatorname{sr}(R)})$ is unimodular. Clearly $((\sigma^{\rho})_{11},\dots,(\sigma^{\rho})_{1,\operatorname{sr}(R)})=((\sigma\rho)_{11},\dots,(\sigma\rho)_{1,\operatorname{sr}(R)})$ since $\rho\in E^*_n(R)$.
\end{proof}

\begin{theorem}\label{thmsr=1}
Suppose that $\operatorname{sr}(R)=1$. Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a,b,c\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)$ is a product of $24$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item By Lemma \ref{lemsr} there there is a $\rho\in E^*_n(R)$ such that $\hat\sigma_{11}$ is right invertible where $\hat\sigma=\sigma^{\rho}$. It follows from Corollary \ref{corkeyA} that $t_{kl}(a\hat\sigma_{1n}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\hat\sigma$ and $\hat\sigma^{-1}$. Clearly $\hat\sigma_{1n}=\sigma_{1n}$ since $\rho\in E^*_n(R)$. Moreover, any $\operatorname{E}_n(R)$-conjugate of $\hat\sigma$ or $\hat\sigma^{-1}$ is an $\operatorname{E}_n(R)$-conjugate of $\sigma$ or $\sigma^{-1}$. Hence $t_{kl}(a\sigma_{1n}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (i) now follows from Lemma \ref{lemp}.
\item See the proof of Theorem \ref{thmNeum}.
\end{enumerate}
\end{proof}

\begin{theorem}
Suppose that $1<\operatorname{sr}(R)<n$. Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a,b,c\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij}b)$ is a product of $16$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)$ is a product of $48$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item By Lemma \ref{lemsr} there there is a $\rho\in E^*_n(R)$ such that $(\hat\sigma_{11},\dots,\hat\sigma_{1,\operatorname{sr}(R)})$ is unimodular where $\hat\sigma=\sigma^{\rho}$.  Clearly there are $x_1,\dots,x_{\operatorname{sr}(R)}\in R$ such that $\sum\limits_{p=1}^{\operatorname{sr}(R)}\hat\sigma_{1p}x_p+\hat\sigma_{1n}=0$. It follows from Corollary \ref{corkeyD} that $t_{kl}(a\hat\sigma'_{1n}b)$ is a product of $16$ $\operatorname{E}_n(R)$-conjugates of $\hat\sigma$ and $\hat\sigma^{-1}$. Clearly $\hat\sigma'_{1n}=\sigma'_{1n}$ since $\rho\in E^*_n(R)$. Moreover, any $\operatorname{E}_n(R)$-conjugate of $\hat\sigma$ or $\hat\sigma^{-1}$ is an $\operatorname{E}_n(R)$-conjugate of $\sigma$ or $\sigma^{-1}$. Hence $t_{kl}(a\sigma'_{1n}b)$ is a product of $16$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (i) now follows from Lemma \ref{lemp} (after swapping the roles of $\sigma$ and $\sigma^{-1}$).
\item See the proof of Theorem \ref{thmNeum}.
\end{enumerate}
\end{proof}

\section{RDU over rings with Euclidean algorithm}

We call $R$ a {\it ring with $m$-term Euclidean algorithm} if for any row vector $v\in {}^m\!R$ there is a $\tau\in \operatorname{E}_m(R)$ such that $v\tau$ has a zero entry. Note that the rings with $2$-term Euclidean algorithm are precisely the right quasi-Euclidean rings defined in \cite{alahmadi} (follows from \cite[Theorem 11]{alahmadi}).

If $\tau\in \operatorname{GL}_m(R)$ for some $1\leq m<n$, then we identify $\tau$ with its image in $\operatorname{GL}_n(R)$ under the embedding
\begin{align*}
\operatorname{GL}_m(R)&\hookrightarrow \operatorname{GL}_n(R)\\
\sigma&\mapsto\begin{pmatrix}e_{(n-m)\times(n-m)}&0\\0&\sigma\end{pmatrix}
\end{align*}

\begin{lemma}\label{lemeucl}
Let $\sigma\in \operatorname{GL}_n(R)$ and suppose $(\sigma\tau)_{1n}=0$ for some $\tau\in E_m(R)$ where $1\leq m<n$. Then for any $k\neq l$ and $a,b\in R$,
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma'_{1j}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ if $j\in \{2,\dots,n-m\}$ and
\item $t_{kl}(a\sigma'_{1j}b)$ is a product of $8m$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ if $j\in \{n-m+1,\dots,n\}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Set $\xi:=\sigma^{\tau}$. Then $\xi_{1n}=0$. It follows from Corollary \ref{corkeyB} that for any $2\leq j\leq n$, $k\neq l$ and $a,b\in R$,
\begin{equation}
t_{kl}(a\xi'_{1j}b)\text{ is a product of }8~ \operatorname{E}_n(R)\text{-conjugates of }\sigma\text{ and }\sigma^{-1}.
\end{equation}
Clearly $\sigma^{-1}=(\xi^{-1})^{\tau^{-1}}$ and hence
\begin{equation}
\sigma'_{1j}=\begin{cases}
\xi'_{1j},&\text{ if } j\in \{2,\dots,n-m\},\\
\sum\limits_{p=n-m+1}^{n}\xi'_{1p}\tau'_{pj},&\text{ if } j\in \{n-m+1,\dots,n\}.
\end{cases}
\end{equation}
The assertion of the theorem follows from (7) and (8).
\end{proof}

\begin{theorem}\label{thmeucl_1}
Suppose that $R$ is a ring with $m$-term Euclidean algorithm for some $1\leq m\leq n-2$. Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a,b,c\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)$ is a product of $24$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item Since $R$ is a ring with $m$-term Euclidean algorithm, there is a $\tau\in \operatorname{E}_m(R)$ such that $(\sigma\tau)_{1n}=0$. It follows from Lemma \ref{lemeucl} that $t_{kl}(a\sigma'_{12}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ (note that $n-m\geq 2$). Assertion (i) now follows from Lemma \ref{lemp} (after swapping the roles of $\sigma$ and $\sigma^{-1}$).
\item See the proof of Theorem \ref{thmNeum}.
\end{enumerate}
\end{proof}

\begin{theorem}\label{thmeucl_2}
Suppose that $R$ is a ring with $n-1$-term Euclidean algorithm. Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a,b,c\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij}b)$ is a product of $8(n-1)$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)$ is a product of $24(n-1)$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item Since $R$ is a ring with $n-1$-term Euclidean algorithm, there is a $\tau\in \operatorname{E}_{n-1}(R)$ such that $(\sigma\tau)_{1n}=0$. It follows from Lemma \ref{lemeucl} that $t_{kl}(a\sigma'_{12}b)$ is a product of $8(n-1)$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (i) now follows from Lemma \ref{lemp} (after swapping the roles of $\sigma$ and $\sigma^{-1}$).
\item See the proof of Theorem \ref{thmNeum}.
\end{enumerate}
\end{proof}

Theorems \ref{thmeucl_1} and \ref{thmeucl_2} show that if $\sigma\in \operatorname{GL}_n(R)$ where $R$ is a ring with $m$-term Euclidean algorithm for some $1\leq m\leq n-1$, then the nondiagonal entries of $\sigma$ can be ``extracted'' (i.e. the matrices $t_{kl}(\sigma_{ij})~(k\neq l, i\neq j)$ lie in $\sigma^{\operatorname{E}_n(R)}$). Does this also hold for $m=n$? The author does not know the answer to this question. However, if $R$ is a ring with {\it strong} $n$-term Euclidean algorithm (see Definition \ref{defstrong} below), then the nondiagonal entries of $\sigma$ can be extracted, as Theorem \ref{thmeucl_3} shows.

\begin{definition}\label{defstrong}
We call $R$ a {\it ring with strong $m$-term Euclidean algorithm} if for any row vector $v\in {}^m\!R$ there is a $\tau\in \operatorname{E}_m(R)$ with $\tau_{11}=1$ such that $(v\tau)_{n}=0$.
\end{definition}

\begin{remark}
If $R$ is a ring with strong $m$-term Euclidean algorithm, then obviously $R$ is also a ring with $m$-term Euclidean algorithm.
\end{remark}

\begin{theorem}\label{thmeucl_3}
Suppose that $R$ is a ring with strong $n$-term Euclidean algorithm. Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a,b,c\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij}b)$ is a product of $80(n-1)$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)$ is a product of $240(n-1)$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item Since $R$ is a ring with strong $n$-term Euclidean algorithm, there is a $\tau\in \operatorname{E}_{n}(R)$ with $\tau_{11}=1$ such that $(\sigma\tau)_{1n}=0$. Clearly the matrix $\tau\prod\limits_{j=2}^{n-1}t_{1j}(-\tau_{1j})$ has the same properties as $\tau$. Hence we may assume that $\tau_{1j}=0~(j=2,\dots,n-1)$. \\
{\bf Step 1}
Clearly $\tau=\rho t_{1n}(\tau_{1n})$ for some $\rho\in \operatorname{E}_n(R)$ with trivial first row. Set $\xi:=t_{1n}(\tau_{1n})\sigma^{\tau}=\rho^{-1}\sigma\tau$. Then $\xi_{1n}=0$. One checks easily that
\[(t_{1n}(-\tau_{1n}),\xi)\xrightarrow{t_{n2}(b'),t_{31}(a'),t_{21}(-1)}(t_{31}(a'\tau_{1n}b'), e)\]
for any $a',b'\in R$. It follows from Lemma \ref{lemp} and Lemma \ref{lemredux} that
\begin{equation}
t_{k'l'}(a'\tau_{1n}b')\text{ is a product of }8~\operatorname{E}_n(R)\text{-conjugates of }\sigma\text{ and }\sigma^{-1}
\end{equation}
for any $k'\neq l'$ and $a',b'\in R$.\\
{\bf Step 2} Let $\xi$ be defined as in Step 1. Corollary \ref{corkeyB} implies that for any $2\leq j\leq n$, $k''\neq l''$ and $a'',b''\in R$, the matrix $t_{k''l''}(a''\xi'_{1j}b'')$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\xi$ and $\xi^{-1}$. Hence, by (9),
\begin{equation}
t_{k''l''}(a''\xi'_{1j}b'')\text{ is a product of }8\cdot 9=72~ \operatorname{E}_n(R)\text{-conjugates of }\sigma\text{ and }\sigma^{-1}.
\end{equation}
{\bf Step 3} Clearly $\sigma^{-1}=\tau\xi^{-1}\rho^{-1}$ and hence $\sigma'_{12}=\sum\limits_{j=2}^{n}(\xi'_{1j}+\tau_{1n}\xi'_{nj})\rho'_{j2}$. It follows from (9) and (10) that $t_{kl}(a\sigma'_{12}b)$ is a product of $(72+8)(n-1)=80(n-1)$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. Assertion (i) now follows from Lemma \ref{lemp} (after swapping the roles of $\sigma$ and $\sigma^{-1}$).
\item See the proof of Theorem \ref{thmNeum}.
\end{enumerate}
\end{proof}

\section{RDU over almost commutative rings}
In this subsection $C$ denotes the center of $R$.

\begin{definition}
Let $x\in R$. An element $z\in C$ is called a {\it central multiple} of $x$ if $z=xy=yx$ for some $y\in R$. The ideal of $C$ consisting of all central multiples of $x$ is denoted by $\mathcal{O}(x)$.
\end{definition}

\begin{lemma}\label{lemalmcom1}
Let $\sigma\in \operatorname{GL}_n(R)$ and $z\in \mathcal{O}(\sigma_{11})$. Then for any $k\neq l$ and $a,b\in R$, $t_{kl}(az\sigma_{12}b)$ is a product of of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{lemma}
\begin{proof}
Since $z\in \mathcal{O}(\sigma_{11})$, there is a $y\in R$ such that $z=\sigma_{11}y=y\sigma_{11}$. Clearly $\sigma_{11}y\sigma_{12}-\sigma_{12}\sigma_{11}y=0$. It follows from Proposition \ref{propkey}(i) that $t_{kl}(az\sigma_{12}b)$ is a product of $8$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{proof}

Set $\operatorname{E}^{**}_n(R):=\langle t_{k1}(x)\mid x\in R,k\neq 1\rangle$.

\begin{lemma}\label{lemalmcom2}
Let $\sigma\in \operatorname{GL}_n(R)$. Suppose that there are $\tau_p\in E^{**}_n(R)~(1\leq p\leq q)$ and $z_p\in \mathcal{O}((\sigma^{\tau_{_p}})_{11})~(1\leq p\leq q)$ such that $\sum\limits_{p=1}^{q}z_p=1$. Then for any $k\neq l$ and $a,b\in R$, $t_{kl}(a\sigma_{12}b)$ is a product of of $8q$ $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{lemma}
\begin{proof}
The previous lemma implies that for any $1\leq p \leq q$, the elementary transvection $t_{kl}(az_p(\sigma^{\tau_{p}})_{12}b)=t_{kl}(az_p\sigma_{12}b)$ (for this equation we have used that $\tau_{p}\in E^{**}_n(R)$) is a product of $8$ elementary $\sigma$-conjugates (note that any elementary $\sigma^{\tau_{p}}$-conjugate is also an elementary $\sigma$-conjugate). It follows that $t_{kl}(a\sigma_{12}b)=t_{kl}(az_1\sigma_{12}b)\dots t_{kl}(az_q\sigma_{12}b)$ is a product of $8q$ elementary $\sigma$-conjugates.
\end{proof}

Recall that $R$ is called {\it almost commutative} if it is finitely generated as a $C$-module. We will show that if $R$ is almost commutative, then the requirements of Lemma \ref{lemalmcom2} are always satisfied.

We denote by $\operatorname{Max}(C)$ the set of all maximal ideals of $C$. If $\mathfrak{m}\in \operatorname{Max}(C)$, then we denote by $R_{\mathfrak{m}}$ the localisation of $R$ with respect to the multiplicative set $S_{\mathfrak{m}}:= C\setminus \mathfrak{m}$. We denote by $Q_{\mathfrak{m}}$ the quotient $R_{\mathfrak{m}}/\operatorname{Rad}(R_{\mathfrak{m}})$ where $\operatorname{Rad}(R_{\mathfrak{m}})$ is the Jacobson radical of $R_{\mathfrak{m}}$. Moreover, we denote the canonical ring homomorphism $R\rightarrow Q_{\mathfrak{m}}$ by $\phi_{\mathfrak{m}}$.

\begin{lemma}\label{lemalmcom3}
If $R$ is almost commutative, then for any $\mathfrak{m}\in \operatorname{Max}(C)$, $\phi_{\mathfrak{m}}$ is surjective and $\operatorname{sr}(Q_{\mathfrak{m}})=1$.
\end{lemma}
\begin{proof}
First we show that $\phi_\mathfrak{m}$ is surjective. Set $C_\mathfrak{m}:=S_\mathfrak{m}^{-1}C$. Clearly $R_\mathfrak{m}$ is finitely generated as a $C_\mathfrak{m}$-module. Hence, by \cite[Corollary 5.9]{lam_book}, we have $\mathfrak{m}=\operatorname{Rad}(C_\mathfrak{m})\subseteq \operatorname{Rad} (R_\mathfrak{m})$. Let now $\frac{r}{s}\in R_\mathfrak{m}$. Since $sC+\mathfrak{m}=C$, there is a $c\in C$ and an $m\in\mathfrak{m}$ such that $sc+m=1$. Multiplying this equality by $\frac{r}{s}$ we get $\frac{r}{s}=rc+\frac{rm}{s}\in rc+\operatorname{Rad}(R_\mathfrak{m})$. Thus $\phi_{\mathfrak{m}}$ is surjective.\\
Next we show that $\operatorname{sr}(Q_{\mathfrak{m}})=1$. Clearly $Q_\mathfrak{m}=R_\mathfrak{m}/\operatorname{Rad}(R_{\mathfrak{m}})$ is finitely generated as a $C_\mathfrak{m}/\operatorname{Rad}(C_{\mathfrak{m}})$-module. But $C_\mathfrak{m}/\operatorname{Rad}(C_{\mathfrak{m}})$ is a field. Hence $Q_\mathfrak{m}$ is a semisimple ring (see \cite[\S 7]{lam_book}) and therefore $\operatorname{sr}(Q_{\mathfrak{m}})=1$ (see \cite[\S6]{bass}).
\end{proof}

\begin{lemma}\label{lemalmcom4}
Suppose that $R$ is almost commutative. Let $u=(u_1,\dots,u_n)\in {}^n\!R$ be a unimodular row. Then there are matrices $\tau_{\mathfrak{m}}\in \operatorname{E}^{**}_n(R)~(\mathfrak{m}\in \operatorname{Max}(C))$ such that $C=\sum\limits_{\mathfrak{m}\in \operatorname{Max}(C)}\mathcal{O}((u\tau_{\mathfrak{m}})_1)$.
\end{lemma}
\begin{proof}
Let $\mathfrak{m}\in\operatorname{Max}(C)$ and set $\hat u:=\phi_\mathfrak{m}(u)\in {}^n\!(Q_m)$. Since $\operatorname{sr}(Q_{\mathfrak{m}})=1$ and $\hat u$ is unimodular, there is a $\hat\tau_{\mathfrak{m}}\in \operatorname{E}_n(Q_{\mathfrak{m}})$ such that $(\hat u\hat\tau_{\mathfrak{m}})_1$ is invertible (note that rings of stable rank $1$ are Dedekind finite, see e.g. \cite[Lemma 1.7]{lam}). Clearly $\hat\tau_{\mathfrak{m}}$ can be chosen to be in $E_n^{**}(Q_\mathfrak{m})$ (follows from \cite[Theorem 1]{vaserstein_sr}). Since $\phi_{\mathfrak{m}}$ is surjective, it induces a surjective homomorphism $\operatorname{E}^{**}_n(R)\rightarrow\operatorname{E}^{**}_n(Q_{\mathfrak{m}})$ which we also denote by $\phi_\mathfrak{m}$. Choose a $\tau_\mathfrak{m}\in E^{**}_n(R)$ such that $\phi_\mathfrak{m}(\tau_\mathfrak{m})=\hat\tau_\mathfrak{m}$.\\
Since $\phi_{\mathfrak{m}}((u\tau_{\mathfrak{m}})_1)=(\hat u\hat\tau_{\mathfrak{m}})_1$ is invertible in $Q_\mathfrak{m}$, $(u\tau_{\mathfrak{m}})_1$ is invertible in $R_{\mathfrak{m}}$. Write $((u\tau_{\mathfrak{m}})_1)^{-1}=\frac{x}{s}$ where $x\in R$ and $s\in S_{\mathfrak{m}}$. Then
\begin{align*}
\frac{(u\tau_{\mathfrak{m}})_1}{1}\frac{x}{s}=\frac{1}{1}~\Leftrightarrow~\exists t\in S_{\mathfrak{m}}:t(u\tau_{\mathfrak{m}})_1x=ts
\end{align*}
and
\begin{align*}
\frac{x}{s}\frac{(u\tau_{\mathfrak{m}})_1}{1}=\frac{1}{1}~\Leftrightarrow~\exists t'\in S_{\mathfrak{m}}:t'x(u\tau_{\mathfrak{m}})_1=t's.
\end{align*}
Clearly $z_{\mathfrak{m}}:=(u\tau_{\mathfrak{m}})_1tt'x=tt's=tt'x(u\tau_{\mathfrak{m}})_1\in \mathcal{O}((u\tau_{\mathfrak{m}})_1)\cap S_{\mathfrak{m}}$. We have shown that for any $\mathfrak{m}\in\operatorname{Max}(C)$ there is a $\tau_{\mathfrak{m}}\in \operatorname{E}^{**}_n(R)$ and a $z_{\mathfrak{m}}\in \mathcal{O}((u\tau_{\mathfrak{m}})_1)$ such that $z_{\mathfrak{m}}\not\in \mathfrak{m}$. The assertion of the lemma follows.
\end{proof}

\begin{corollary}\label{coralmcom}
Suppose that $R$ is almost commutative and let $\sigma\in \operatorname{GL}_n(R)$. Then there is a $q\leq |\operatorname{Max}(C)|$, $\tau_p\in E^{**}_n(R)~(1\leq p\leq q)$ and $z_p\in \mathcal{O}((\sigma^{\tau_{_p}})_{11})~(1\leq p\leq q)$ such that $\sum\limits_{p=1}^{q}z_p=1$.
\end{corollary}
\begin{proof}
By Lemma \ref{lemalmcom4} there are matrices $\tau_{\mathfrak{m}}\in E^{**}_n(R)~(\mathfrak{m}\in \operatorname{Max}(C))$ such that
\[C=\sum\limits_{\mathfrak{m}\in \operatorname{Max}(C)}\mathcal{O}((\sigma_{1*}\tau_{\mathfrak{m}})_1)=\sum\limits_{\mathfrak{m}\in \operatorname{Max}(C)}\mathcal{O}((\sigma\tau_{\mathfrak{m}})_{11})=\sum\limits_{\mathfrak{m}\in \operatorname{Max}(C)}\mathcal{O}((\sigma^{\tau_{\mathfrak{m}}})_{11})\]
(for the last equation we have used that $\tau_{\mathfrak{m}}\in E^{**}_n(R)$). The assertion of the corollary follows.
\end{proof}

\begin{theorem}\label{thmalmcom}
Suppose that $R$ is almost commutative. Let $\sigma\in \operatorname{GL}_n(R)$, $i\neq j$, $k\neq l$ and $a,b,c\in R$. Then
\begin{enumerate}[(i)]
\itemsep0pt
\item $t_{kl}(a\sigma_{ij}b)$ is a finite product of $8|\operatorname{Max}(C)|$ or less $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$ and
\item $t_{kl}(a(c\sigma_{ii}-\sigma_{jj}c)b)$ is a finite product of $24|\operatorname{Max}(C)|$ or less $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item It follows from Lemma \ref{lemalmcom2} and Corollary \ref{coralmcom} that that $t_{kl}(a\sigma_{12}b)$ is a product of $8q$ elementary $\sigma$-conjugates for some $q\leq|\operatorname{Max}(C)|$. Assertion (i) now follows from Lemma \ref{lemp}.
\item See the proof of Theorem \ref{thmNeum}.
\end{enumerate}
\end{proof}

\section{Open problems}
\begin{definition}
Let $m\in\mathbb{N}$. Then $m$ is called an {\it RDU bound for $\operatorname{GL}_n(R)$} if for any $\sigma\in\operatorname{GL}_n(R)$, $k\neq l$ and $i\neq j$ the matrix $t_{kl}(\sigma_{ij})$ is a product of $m$ or less $\operatorname{E}_n(R)$-conjugates of $\sigma$ and $\sigma^{-1}$. If $m$ is an RDU bound for $\operatorname{GL}_n(R)$ for any $n\geq 3$, then $m$ is called an {\it RDU bound for $R$}. If $\mathfrak{C}$ is a class of rings and $m$ is an RDU bound for $R$ for any $R\in \mathfrak{C}$, then $m$ is called an {\it RDU bound for $\mathfrak{C}$}.
\end{definition}

It follows from Sections 5, 6 and 8 that $8$ is an RDU bound for the class of commutative rings, the class of von Neumann regular rings, and the class of rings of stable rank $1$. It follows from Sections 7 and 9 that $160$ is an RDU bound for the class of Banach algebras and $16$ is an RDU bound for the class of quasi-Euclidean rings. Note that Section 10 does not yield an RDU bound for the class of almost commutative rings since the maximal spectrum of an almost commutative ring might be infinite.

It is natural to ask the following two question.

\begin{question}\label{Q1}
Which rings have an RDU bound?
\end{question}

\begin{question}\label{Q2}
What is the optimal (i.e. smallest) RDU bound for the class of commutative rings (resp. von Neumann regular rings, rings of stable rank $1$ etc.)?
\end{question}

Using a computer program the author found out that $2$ is the optimal RDU bound for $\operatorname{GL}_3(\mathbb{F}_2)$ and $\operatorname{GL}_3(\mathbb{F}_3)$ where $\mathbb{F}_2$ and $\mathbb{F}_3$ denote the fields of $2$ and $3$ elements, respectively. He does not know the optimal bounds in any other cases.

\begin{lemma}
Suppose that $1$ is an RDU bound for a ring $R$. Then $\operatorname{GL}_n(R)=\operatorname{E}_n(R)$ for any $n\geq 3$.
\end{lemma}
\begin{proof}
If $n\geq 3$ and $\sigma\in \operatorname{GL}_n(R)$, then, by the definition of an RDU bound, $\sigma=t_{12}(\sigma_{12})^{\tau}$ or $\sigma^{-1}=t_{12}(\sigma_{12})^{\tau}$ for some $\tau\in \operatorname{E}_n(R)$. Hence $\sigma\in \operatorname{E}_n(R)$.
\end{proof}

Since $\operatorname{E}_n(R)\subseteq\operatorname{SL}_n(R)\neq \operatorname{GL}_n(R)$ if $R$ is a field with more than two elements, we get the following result.

\begin{theorem}
The optimal RDU bound for the class of commutative rings (resp. von Neumann regular rings, rings of stable rank $1$) lies between $2$ and $8$.
\end{theorem}

The author thinks that the following question is interesting too (compare Lemma \ref{lemeucl}).

\begin{question}\label{Q3}
Let $\sigma\in \operatorname{GL}_n(R)$ and suppose $(\sigma\tau)_{1n}=0$ for some $\tau\in E_n(R)$. Can one extract some nondiagonal entry of $\sigma$ or $\sigma^{-1}$? (I.e., is it true that $t_{kl}(\sigma_{ij})\in \sigma^{\operatorname{E}_n(R)}$ or $t_{kl}(\sigma'_{ij})\in \sigma^{\operatorname{E}_n(R)}$ for some $k\neq l$ and $i\neq j$?)
\end{question}

A positive answer to Question \ref{Q3} would have the interesting consequence that for any $\sigma$ lying in the elementary subgroup $\operatorname{E}_n(R)$, one could extract some nondiagonal entry of $\sigma$ or $\sigma^{-1}$ (since $(\sigma\sigma^{-1})_{1n}=0$).


\section{appendix}
\label{app}
Here we give details about the static Bogoliubov transformation, the classical model, the Floquet theory and Mathieu functions.
\subsection{Static Bogoliubov transformation}
For the time-independent case, Hamiltonian (2) in the main text can be expressed as
\begin{equation}
H_q=  v_F q\left[(1+g_4)\left(2 J_{0,q}-1\right)+ g_2 \left(J_{+,q}+J_{-,q}\right)\right]
\end{equation}
where  $2J_{0,q}=b_{Rq}^{\dagger}b_{Rq}+b_{Lq}b_{Lq}^{\dagger}$, $J_{+,q}=b_{Rq}^{\dagger}b_{Lq}^{\dagger}$, $J_{-,q}=J_{+,q}^{\dagger}$. For the sake of simplicity, in the following we will drop the index $q$. We observe that $J_i (i=0,\pm)$ form a $su(1,1)$ algebra; therefore the Tomonaga-Luttinger Hamiltonian can be diagonalized by the Schrieffer-Wolff transformation obtained through the unitary operator
\begin{equation}
\label{eq:Schrieffer_Wolff}
U=e^{z}\qquad z=\vartheta\left(J_+-J_-\right)
\end{equation}
\begin{equation}
\begin{split}
 H & \longrightarrow \tilde{H}=UHU^{-1}\\
|\rm{GS}\rangle & \longrightarrow |\widetilde{\rm{GS}}\rangle=U|\rm{GS}\rangle \hspace{2pt}.
\end{split}
\end{equation}
The transformed Hamiltonian $\tilde{H}$ is diagonal in the old bosonic basis
\begin{equation}
\begin{split}
\tilde{H}&=\Delta\left(b_R^{\dagger}b_R+b_L^{\dagger}b_L\right)+\Delta- v_F q (1+g_4)\\
&=2\Delta J_0- v_F q (1+g_4) \hspace{10pt}
\end{split}
\end{equation}
with $\Delta= v_F q\sqrt{(1+g_4)^2-g_2^2}$ if $\tanh{(2\vartheta)}=\tanh{(2\bar{\vartheta})}\equiv g_2/(1+g_4)$, implying $\cosh(2\bar{\vartheta})=v_Fq(1+g_4)/\Delta$, $\sinh(2\bar{\vartheta})=v_Fqg_2/\Delta$.\\
Notice that we have used a passive transformation, where the operators are rotated, while the states defined by the original creation and annihilation operators stay the same.
  \subsection{Classical model}
  The Hamiltonian
  \begin{equation}
  H(t)=(A+B(t))\left(a^{\dagger}a+b^{\dagger}b\right)+C(t)\left(ab+a^{\dagger}b^{\dagger}\right)
  \end{equation}
  with real coefficients $A, B(t), C(t)$ can be mapped to the classical model
  \begin{equation}\label{eq:Ham_cl}
  H(t)=\frac{1}{2}(A+B(t))\left(x^2+y^2+p_x^2+p_y^2\right)+C(t)\left(xy-p_xp_y\right)
  \end{equation}
  with the substitution
  \begin{equation}
  a=\frac{1}{\sqrt{2}}\left(x+\imath p_x\right)\qquad b=\frac{1}{\sqrt{2}}\left(y+\imath p_y\right) \hspace{2pt}.
  \end{equation}
  The Hamilton's equations for Hamiltonian (\ref{eq:Ham_cl}) are
  \begin{eqnarray}
  &\dot{x}=\frac{\partial H}{\partial p_x}=\left(A+B(t)\right)p_x-C(t)p_y \label{eq:Hamilton_x}\\
  &\dot{y}=\frac{\partial H}{\partial p_y}=\left(A+B(t)\right)p_x-C(t)p_x \label{eq:Hamilton_y}\\
  &\dot{p_x}=-\frac{\partial H}{\partial x}=-\left(A+B(t)\right)x-C(t)y \label{eq:Hamilton_px}\\
  &\dot{p_y}=-\frac{\partial H}{\partial y}=-\left(A+B(t)\right)y-C(t)x \label{eq:Hamilton_py} \hspace{2pt}.
  \end{eqnarray}
  By summing equations (\ref{eq:Hamilton_x}) and (\ref{eq:Hamilton_y}), and equations (\ref{eq:Hamilton_px}) and (\ref{eq:Hamilton_py}), we get the system
  \begin{eqnarray}
  &\dot{v}=\left(A+B(t)-C(t)\right)p_v\\
  &\dot{p_v}=-\left(A+B(t)+C(t)\right)v
  \end{eqnarray}
  for $v=x+y$ and $p_v=p_x+p_y$, which yields
  \begin{equation}
  \ddot{v}-\frac{\dot{B}(t)-\dot{C}(t)}{A+B(t)-C(t)}\dot{v}+\left[\left(A+B(t)\right)^2-C^2(t)\right]v=0 \hspace{2pt}.
  \end{equation}
  In the special case $C(t)=B(t)=\bar{\rho}+\rho\cos(\omega t)$, one recovers the Mathieu equation
  \begin{equation}
  \ddot{v}+\left[A^2+2A\bar{\rho}+2A\rho\cos(\omega t)\right]v=0 \hspace{2pt}.
  \end{equation}
The classical equations of motions also give insight to the role of damping,
  which generally reduces the regions of instability of the Mathieu equation.
In fact, in Ref.~\cite{StabilityChart} it is shown that the presence of  linear damping
  pushes the instability zone upwards in Fig.~2, so that below a critical value of amplitude
 $\rho$
stable solution are always possible and no instabilities occur.
In Ref.~\cite{StabilityChart} it is also shown that nonlinear effects can generate subharmonic stable motions. These cubic terms correspond to band curvature in the original band structure, so

   in a real system the nonlinearity of the band further stabilizes the
system.  While a large number of density waves is still expected to occur at the critical $q-$values, the sum over all momenta becomes well defined leading to the predicted
density order of scenario 2) in the main paper.
 \subsection{Floquet theory}
 Analogously to the Bloch theorem, the Floquet theorem asserts that the Schr\"odinger equation for a time-periodic Hamiltonian admits steady-state solutions of the form
 \begin{equation}
 |\Psi(t)\rangle=e^{-\imath\epsilon t}|u(t)\rangle  \hspace{2pt},
 \end{equation}
 where the modes $|u(t)\rangle=|u(t+T)\rangle $ inherit the periodicity from the Hamiltonian, and the quantity $\epsilon$ is the so-called Floquet quasienergy. Indeed the steady-state Schr\"odinger equation can be recasted in the form of an eigenvalue equation for the quasienergy operator $\mathscr{H}=H(t)-\imath\partial_t$ in the extended Hilbert space generated by the product of the state space of the quantum system and the space of square-integrable $T$-periodic functions:
 \begin{equation}
 \label{eq:eig2}
 \mathscr{H}|u(t)\rangle=\epsilon|u(t)\rangle \hspace{2pt}.
 \end{equation}
 By expanding both the Hamiltonian and the Floquet mode in Fourier series
 \begin{gather}
 H(t)=\sum_m e^{\imath m \omega t}H^{(m)} \hspace{2pt},\\
 |u(t)\rangle=\sum_m e^{\imath m \omega t}|u_m\rangle \hspace{2pt},
 \end{gather}
 Eq.~(\ref{eq:eig2}) yields
 \begin{equation}
 \label{eq:Fl_components}
 \left(H^{(0)}+m\omega\right)|u_m\rangle +H^{(1)}\left(|u_{m-1}\rangle+|u_{m+1}\rangle\right)=\epsilon|u_m\rangle \hspace{2pt},
 \end{equation}
 which turns out to be an eigenvalue equation for the infinite tridiagonal matrix
 \begin{equation}
 \label{eq:matr1}
 {\cal M}_{\cal F}=\left(\begin{array}{ccccc}
   \vdots & \vdots  & \vdots  & \vdots  & \\
 \hdots & H^{(0)}-\omega & H^{(1)}  & 0 &  \hdots \\
 \hdots & H^{(1)} & H^{(0)} & H^{(1)}&  \hdots\\
 \hdots  & 0 & H^{(1)} & H^{(0)}+\omega  & \hdots\\
  & \vdots  & \vdots  & \vdots  &
 \end{array}
 \right) \hspace{2pt}.
 \end{equation}
\subsection{Floquet solution in terms of Mathieu functions}
The solution of the Mathieu equation
\begin{equation}
\ddot{y}(t)+(a-2p\cos{\omega t})y(t)=0
\end{equation}
is usually discussed in terms of even and odd solutions, known respectively as Mathieu cosine ${\cal C}$ and Mathieu sine ${\cal S}$ functions. A general solution can be therefore written as
\begin{equation}
\label{SM:Mat_Sol}
y(t)=c_1 {\cal C}(a,p,\tau)+c_2{\cal S}(a,p,\tau) \hspace{2pt},
\end{equation}
with $\tau=\omega t/2$.
However, being the equation time-periodic, it always admits a Floquet solution
\begin{equation}
\label{SM:Fl_Sol}
y(t)=e^{\imath\nu\tau}P_{\nu}(\tau)
\end{equation}
with $P_{\nu}(\tau)=P_{\nu}(\tau\pm\pi)$. We want to use the quantum number $\nu$, which is commonly referred to as Mathieu characteristic exponent. Therefore, in this section we clarify the relation between the latter and the Mathieu functions.
Comparing Eqs.~(\ref{SM:Mat_Sol}) and (\ref{SM:Fl_Sol}) and employing the periodicity of $P_{\nu}(\tau)$, we get the following relation
\begin{equation}
\begin{split}
c_1{\cal C}(a,p,\tau)+c_2&{\cal S}(a,p,\tau)=\\
&e^{\mp\imath \nu\pi}\left(c_1{\cal C}(a,p,\tau\pm\pi)+c_2{\cal S}(a,p,\tau\pm\pi)\right)\hspace{2pt}.
\end{split}
\end{equation}
Evaluating this expression in $\tau=0$ and normalizing the Mathieu functions such that ${\cal C}(a,p,0)={\cal S}(a,p,\pi)=1$, we obtain
\begin{equation}
c_1(e^{\pm\imath\pi\nu}-{\cal C}(a,p,\pi))=\pm c_2 {\cal S}(a,p,\pi) \hspace{2pt},
\end{equation}
from which we finally get
\begin{equation}
\begin{split}
&\cos{\pi\nu}={\cal C}(a,p,\pi)\hspace{2pt},\qquad\text{and}\\
& c_2=\imath c_1\sin{\pi\nu} \hspace{2pt}.
\end{split}
\end{equation}


\section{Introduction}
The entanglement entropy is an important quantity for quantum field theories. The entanglement entropy is non-vanishing even at zero temperature and it is proportional to the degrees of freedom. Moreover,  it can classify the topological phases described by the ground state of the FQHE or the Chern-Simons effective theory~\cite{wenx,kita,DFLN}. In the context of the AdS/CFT correspondence~\cite{juan,anom}, the formula of holographic Entanglement entropy (holographic EE)~\cite{RT} for CFTs has been proposed as the area of minimal surface $\gamma_A$ which ends on the boundary of the region $A$ as follows:
\begin{eqnarray}
S_A=\dfrac{\text{Area}(\gamma_A)}{4G_N}, \label{HOL1}
\end{eqnarray}
where $G_N$ is the Newton constant in the Einstein gravity. Here, $\gamma_A$ is required to be homologous to $A$. It has been shown~\cite{RT,Sol,LNST,CaHu,Solog} that the log term of the entanglement entropy agrees with the computation in the gravity side. In~\cite{FRWT}, the topological entanglement entropy is clarified via the AdS/CFT correspondence.
 Moreover, it is shown~\cite{HHM} that using holographic EE,  the correlation in the holographic system arises from the quantum entanglement rather than the classical correlation. The review of the holographic EE are given in~\cite{NRT} and in the review part of~\cite{Tonni:2010pv}.

While the holographic EE of~\cite{RT} can be used for the Einstein gravity with no higher derivative terms, it is shown~\cite{Myers:2010xs,Myers:2010tj,Casini:2011kv,Hung:2011ta,Hung:2011nu} that for the spherical surface of the boundary of $A$, the proposal for holographic EE \eqref{HOL1} can be extended to certain higher curvature theories such as Gauss-Bonnet gravities.
It is proposed that using the conformal mapping, EE across the spherical surface $S^2$ with the radius $L$ can be related with the thermal entropy on $\mathbb{R}\times H^{3}$ with $T=1/(2\pi L)$ and this thermal entropy is given by the black hole entropy in the gravity side. It is also shown that the log term of the holographic EE is then proportional to the $A$-type anomaly.

Thus, it is considered to be interesting to apply these results for higher derivative corrections in the holographic EE in some string models.
An interesting model is the 5-dimensional curvature gravity instead of beginning with the 10-dimensional theory, where the curvature-squared terms arise from the D-branes~\cite{bng,ode,Buchel:2008vz,kp}. Here, the higher curvature terms correct also the cosmological constant which should be determined to reproduce the central charges $a$ and $c$ of the 4-dimensional theory. Using the holographic Weyl anomaly~\cite{bng,ode,holo4}, these terms affect the two central charges $a$ and $c$ in the subleading order of $N$. Moreover, this curvature-squared terms  appear in the context of perturbative expansion of the string theory and it is known that they can be modified by field redefinitions.

A purpose of this paper is to compute the holographic EE and $A$-type anomaly in the gravity dual to $d=4$ $\mathcal{N}=2$ SCFTs in F-theory. Here, this gravity dual is described by the curvature gravity theory of above the string model.
These SCFTs include the theory with $G=D_4,E_n,...$ flavor symmetries. Except for $D_4$ all of them are isolated theories, i.e. their gauge couplings are frozen and a perturbative description is not available.  There are previous works for holographic computation of the central charges in these SCFTs~\cite{anom2,at} in which they analyzed Chern-Simons terms in the dual string theory. Their result implies that the AdS/CFT correspondence can be applied even for the finite string coupling. On the other hand, the curvature term of our gravity theory makes the analysis difficult since for $d=5$ these terms depend on the metric of the spacetime. According to~\cite{Buchel:2008vz}, we choose the cosmological constant to reproduce the central charges $a,c$ of the dual SCFT in the context of the holographic Weyl anomaly. As stated above, we can also transform the effective $d=5$ curvature gravity obtained from the F-theory compactifications to the $d=5$ Gauss-Bonnet gravity in the subleading order of $l_s^2/L^2$ expansion using the field redefinition, which is important in the context of the holographic c-theorem.
 And we derive the holographic EE computing the black hole entropy.

This paper is organized as follows: In section 2, we briefly review the Wald's entropy formula which is important for computing the black hole entropy in the presence of the higher curvature terms. In section 3, we review the construction of the Type IIB background after including the backreaction of the $N$ D3-branes at singularities in F-theory.
In the beginning of section 4, we review the construction of the effective curvature gravity from F-theory. In subsection 4.1, we analyze what terms contribute to the central charges in SCFTs.  In subsection 4.2, by using the holographic Weyl anomaly,
we determine the parameters of the gravity to realize the central charges in the field theory side. In subsection 4.3, we generalize the argument in section 4 for the general SCFTs in F-theory.
 In section 5, we compute the holographic EE and the $A$-type anomaly using the $d=5$ effective curvature gravity. We also consider the transformation of the effective $d=5$ curvature gravity to the $d=5$ Gauss-Bonnet gravity using the field redefinition.

\section{Holographic EE as the Wald's entropy formula}
In this section, we briefly review the holographic EE formula in the presence of the higher curvature terms.
It was conjectured~\cite{Myers:2010xs,Myers:2010tj} that EE across the sphere $S^2$ with the radius $L$ can be given by the black hole entropy in the gravity side and the contribution to the EE is then proportional to the coefficient of $A$-type anomaly. This is since when we write the pure AdS by the topological BH with the horizon of radius $L$, there appear two hyperbolic spaces inside and outside the horizon, respectively. By the  conformal mapping, the thermal entropy with temperature $T=1/(2\pi L)$ in a hyperbolic space is shown to be equal to the EE across the sphere $S^2$ as follows:
\begin{eqnarray}
S_{EE}=S_{thermal}.
\end{eqnarray}
Thus, the horizon entropy of the topological BH should be considered as EE of the dual theory using the AdS/CFT correspondence.
Moreover, two hyperbolic spaces are considered as two separated space when we compute the entanglement entropy.

We start with the general gravity theory in $AdS$ with higher derivative terms as follows:
\begin{eqnarray}
S=\dfrac{1}{2\kappa^2_5}\int d^5x\sqrt{-g}F(R^{\mu\nu}{}_{\rho\lambda})+S_{matter}, \label{ACT03}
\end{eqnarray}
where $F(R^{\mu\nu}{}_{\rho\lambda})$ is constructed from the Riemann tensors and $\mu,\nu,\rho,\lambda=0,1,2,3,4$.
The Einstein equation for \eqref{ACT03} is given by
\begin{eqnarray}
F'_{(\mu}{}^{\rho\lambda\alpha}R_{\nu)\rho\lambda\alpha}-\dfrac{F(R^{\mu\nu}{}_{\rho\lambda})}{2}g_{\mu\nu}+2\nabla^{\rho}\nabla^{\lambda}F'_{\mu\rho\nu\lambda}=\kappa^2_5T_{\mu\nu},
\end{eqnarray}
where $F'_{\mu\nu\rho\lambda}=\delta F(R^{\alpha\beta}{}_{\gamma\delta})/\delta R^{\mu\nu\rho\lambda}$, round brackets describe the symmetrization, and $\nabla_{\mu}$ is the  covariant derivative. Note that the tensor $F'_{\mu\nu\rho\lambda}$ is antisymmetrized and symmetrized as seen in the Riemann tensor.
To compute the holographic EE, we should choose the $AdS_5$ topological black hole solution with its radius $L$ where the terms including the covariant derivatives vanish since Riemann curvature of maximally symmetric space like $AdS_5$ can be described by the metric (see appedix A):
\begin{equation}
ds^2=\dfrac{d\rho^2}{\frac{\rho^2}{L^2}-1}-\Big(\frac{\rho^2}{L^2}-1\Big)d\tau^2+\rho^2(du^2+\sinh^2ud\Omega_{2}^2), \label{MET05}
\end{equation}
where $d\Omega_2^2$ represents the metric of $S^2$ and the BH horizon is at $\rho ={L}$.
Then, the holographic EE of the topological black hole is given by extremizing Wald's entropy formula as follows:
\begin{eqnarray}
S_{entropy}=-2\pi \int_{horizon}d^3x\sqrt{h}\dfrac{\delta S}{\delta R^{\mu\nu}{}_{\rho\lambda}}\epsilon^{\mu\nu}\epsilon_{\rho\lambda}.
\end{eqnarray}
where $\epsilon^{\mu\nu}$ is the binormal to the horizon satisfying $\epsilon_{\mu\nu}\epsilon^{\mu\nu}=-2$ for \eqref{MET05}.
According to~~\cite{Sinha:2010ai,Myers:2010xs,Myers:2010tj,Casini:2011kv},
the integrand can be related with the $A$-type anomaly in the SCFT side as follows:
\begin{eqnarray}
a_4=-\pi^2L^3\dfrac{\delta S}{\delta R^{\mu\nu}{}_{\rho\lambda}}\epsilon^{\mu\nu}\epsilon_{\rho\lambda}\Big|_{horizon}. \label{CEN5}
\end{eqnarray}
This $a_4$ is also given by the $a$-function used in the context of the holographic c-theorem if it is evaluated at an AdS vacua.
For the case of Einstein gravity $F(R^{\mu\nu}{}_{\rho\lambda})=R-\Lambda$, in addition, the $A$-type anomaly is given by
\begin{eqnarray}
a=\dfrac{\pi^2L^3}{\kappa_5^2}.
\end{eqnarray}
We will use \eqref{CEN5} in section 4 to estimate the corrections
 for the $A$-type anomaly and in section 5, use for computing the $A$-type anomaly for the $\mathcal{N}=2$ SCFT in F-theory.
\section{The gravity dual to $d=4$ $\mathcal{N}=2$ SCFTs}

In this section, we review the construction of the Type IIB background dual to  $\mathcal{N}=2$ SCFTs.
It has been known that $\mathcal{N}=2$ SCFTs arise as worldvolume theories of D3-branes moving near 7-branes, namely 3-branes in F-theory~\cite{ft1,ft2,ft3,ft4,ft5}.
The simplest theory is the worldvolume theory of D3-branes in a $\mathbf{Z}_2$ orientifold and becomes a $USp(2N)$ $\mathcal{N}=2$ gauge theory with an antisymmetric and four fundamental hypermultiplets in the low energy limit. Its gravity dual is the type IIB supergravity on an orientifold of $AdS_5\times S^5$~\cite{Fayyazuddin:1998fb,Aharony:1998xz}.

For the gravity side, we consider the type IIB string theory on $AdS_5\times X_{5}$ with $N$ units of $F_5$ flux on $X_5$ taking the near-horizon limit.
The metric is as follows:
\begin{equation}
ds^2=\dfrac{L^2}{z^2}(d\vec{x}^2+dz^2)+L^2ds^2_{X_5},
\end{equation}
where $L=(4\pi g_s\Delta N)^{1/4}l_s$ is the radius of $AdS_5$~\cite{Gubser:1998vd}
satisfying $L^4\text{Vol}(X_5)=L^4_{S^5} \text{Vol}(S^5)$.
Here, $X_{5}$ is the 5-sphere
\begin{equation}
\{|z_1|^2+|z_2|^2+|z_3|^2=const \}, \label{MET6}
\end{equation}
 with the periodicity $2\pi/\Delta$ along the phase
 $\phi\equiv$arg$(z_3)$ and
the metric of $X_{5}$ is given by
\begin{equation}
ds^2_{X_5}=d\theta^2+\sin^2\theta d\phi^2 +\cos^2\theta d\Omega^2_3,
\end{equation}
 where $d\Omega ^2_3$ is the metric of $S^3$, $0\le \theta  \le \pi/2$. At the singularity $\theta =0$, the metric becomes $S^3$.  It is known that the singularities are related with flavor groups of 7-branes, which are called $G$-type 7-branes for the $G$-type singularity ($G=H_0,H_1,H_2,D_4,E_6,..$). This 7-brane then wraps whole $AdS_5\times S^3$. Indeed, the transverse direction $Re(z_3),Im(z_3)$ operated by $\mathbb{Z}_2$ orbifold is the transverse direction of an O7-brane for $\Delta=2$. The number of 7-branes is related with the deficit angle $\Delta$ and
 is given by $n_7=12(1-1/\Delta)$.

The gravity action dimensionally reduced to 5-dimension is given by
\begin{equation}
I\equiv \int d^{5}x\sqrt{-g}\mathcal{L}=\dfrac{1}{2\kappa_5^2}\int d^5x\sqrt{-g}(R+12/L^2+...), \label{AC09}
\end{equation}
where $2\kappa^2_{10} =(2\pi)^7g_s^2l_s^8$ and we defined
\begin{equation}
\dfrac{1}{2\kappa_5^2}\equiv \dfrac{L^5\text{Vol} (X_5)}{2\kappa_{10}^2} \label{DEF10}
\end{equation} Here, dots include the matter fields and the higher derivative corrections. To describe the gravity dual of the $d=4$ $\mathcal{N}=2$ SCFT with flavor, moreover, we should
include the curvature-squared terms of 7-branes since these terms affect the $A$-type anomaly as the $1/N$ correction (see appendix B).
In next section, we start with the 5-dimensional effective curvature gravity from F-theory instead of considering Type IIB D-branes.

\section{Effective curvature gravity dual with the orientifold}

It has been known~\cite{Buchel:2008vz,kp} that if the central charges of the dual CFT satisfy
\begin{eqnarray}
c\sim a \gg 1 \ \text{and} \ |c-a|/c \ll 1,
\end{eqnarray}
 the effective gravity theory is described by the curvature gravity with the higher curvature term coupled to a negative cosmological constant.
Using the constants $a_1$, $a_2$, and $a_3$, the effective curvature gravity is given by
\begin{eqnarray}
S=\dfrac{1}{2\kappa_{5}^2}\int d^5x\sqrt{-g}\Big(R+\dfrac{12}{ \tilde{L}^2}+a_1R^2+a_2R_{\mu\nu}R^{\mu\nu}+a_3R_{\mu\nu\rho\lambda}R^{\mu\nu\rho\lambda} \Big), \label{ACT6}
\end{eqnarray}
where $-12/\tilde{L}$ describes the cosmological constant.
It is known that in the case of $\mathcal{N}=2$ SCFT dual to $AdS_5\times S^5/\mathbb{Z}_2$, moreover, we can construct the effective curvature gravity theory from F-theory~\cite{bng,anom2}. In this section, we review construction of this effective curvature gravity from the F-theory. And in subsection 4.1, we consider the generalization for the case of the curvature
 gravity on $AdS_5\times X_5$ with general $\Delta$.

In order to obtain the higher derivative terms in the action of F-theory compactifications, we use a duality chain
\begin{equation}
{\rm Heterotic\;}SO(32)/T^{2} \rightarrow {\rm Type\; I}/T^{2} \rightarrow {\rm Type IIB}/T^{2}/\mathbb{Z}_{2}, \label{dual7}
\end{equation}
where we break the gauge symmetry of $SO(32)$ into $SO(8)^{4}$ in Heterotic string theory and Type I string theory by turning on Wilson lines.

In the ten dimensional supergravity action of Heterotic string theory, the terms which include Riemann tensors up to the quadratic order is
\begin{equation}
S_{het}=\frac{1}{(2 \pi)^{7}l_{s}^{8}} \int dx^{10}\sqrt{g^{h}}e^{-2\phi^{h}}\left(R+\frac{l_{s}^{2}}{4}R_{MNLP}R^{MNLP}\right),
\label{eq:action_het}
\end{equation}
where $M,N,L,P=1,...,10$.
When one moves to Type I string theory, the duality relations are
\begin{equation}
\phi^{h} = -\phi^{I},\;\;
g_{MN}^{h} = e^{-\phi_{I}} g_{MN}^{I}.
\end{equation}
Then, the action \eqref{eq:action_het} becomes
\begin{equation}
S_{I}=\frac{1}{(2 \pi)^{7}l_{s}^{8}} \int dx^{10}\sqrt{g^{I}}\left(e^{-2\phi_{I}} R + \frac{l_{s}^{2}}{4}e^{-\phi_{I}}R_{MNLP}R^{MNLP} \right),
\label{eq:action_typeI}
\end{equation}
and we have 32 D9-branes and 1 O9-plane. We also consider $N$ coincident 5-branes wrapped on the torus $T^{2}$. From such a configuration, one obtains $USp(2N)$ gauge symmetry with a hypermultiplet in the anti-symmetric representation on 5-branes. Besides we have 32 additional half-hypermultiplets in the {\bf 2N} representation of $USp(2N)$.

As for the second step, we take two T-duality along the torus. Then, we have 16 D7-branes and 4 O-planes which are separated into four groups. The difference of the number of D7-brane comes from the fact that the charges on both sides are related as $\mu_{7}^{I}=\frac{1}{2}\mu_{7}^{IIB}$. Each group locate at the four fixed points of $T^{2}/\mathbb{Z}_{2}$. We consider the case where $T^{2}$ is large and focus on one fixed point. Other fields related to other fixed points are very massive and can be integrated out in the low energy field theory. Furthermore, $N$ 5-branes become $N$ 3-branes after the two T-dualities along $T^{2}$. Hence, the low energy field theory is described by $\mathcal{N}=2$ $USp(2N)$ gauge theory with four fundamental hypermultiplets and an additional hypermultiplet in the anti-symmetric representation. The duality relations between Type I and Type IIB string theory are
\begin{equation}
V_{I}e^{-2\phi^{I}} = V_{IIB} e^{-2\phi^{IIB}},\;\; R_{I}^2 = \frac{l_{s}^{4}}{L^2},
\end{equation}
where $L$ is the radius of the torus $T^2$ in Type IIB theory and $V_I$ and $V_{II}$ are the volume of the torus in Type I and Type IIB theory, respectively.
The action of Type IIB string theory becomes
\begin{equation}
S_{IIB}=\frac{1}{(2 \pi)^{7}l_{s}^{8}} \int dx^{10}\sqrt{g^{IIB}}e^{-2\phi_{IIB}} R + \frac{1}{(2 \pi)^{7}l_{s}^{8}} \frac{(2\pi)^{2}l_{s}^{2}}{2}\int d^{8}x \sqrt{g^{IIB}} \frac{l_{s}^{2}}{4}e^{-\phi_{IIB}}R_{MNLP}R^{MNLP}.
\label{eq:action_typeIIB}
\end{equation}
Note that the second term in \eqref{eq:action_typeIIB} is proportional to $e^{-\phi^{IIB}}$ and the world volume is eight dimension. Hence, this term originates from the D7-brane and O7-plane. We also take into account the difference between the charge of D9-branes in Type I string theory and the charge of D9-branes in Type IIB string theory $\mu_{9}^{I} = \frac{1}{\sqrt{2}}\mu_{9}^{IIB}$. Since we will focus on the one of the four fixed points in $T^{2}/\mathbb{Z}_{2}$, we further multiply a factor $\frac{6}{24}=\frac{1}{4}$ to the second term in \eqref{eq:action_typeIIB}
\begin{equation}
S_{IIB}=\frac{1}{(2 \pi)^{7}l_{s}^{8}} \int dx^{10}\sqrt{g^{IIB}}e^{-2\phi_{IIB}} R + \frac{1}{(2 \pi)^{7}l_{s}^{4}} \frac{(2\pi)^{2}}{2} \frac{1}{4} \frac{1}{4}\int d^{8}x \sqrt{g^{IIB}}e^{-\phi_{IIB}}R_{MNLP}R^{MNLP}.
\label{s1}
\end{equation}

By the dimensional reduction of the 5-dimensional and 3-dimensional internal directions, \eqref{s1} reduces following effective five dimensional action plus cosmological constant $-12/\tilde{L}^2$ which can have the $AdS_5$ vacua as the solution:
\begin{eqnarray}
&S_{IIB}=\frac{L^5vol(X_{\Delta=2})e^{-2\phi_{IIB}}}{(2 \pi)^{7}l_{s}^{8}} \int dx^{5}\sqrt{g^{IIB}} \Big(R+\dfrac{12}{{\tilde{L}}^{2}}\Big) \\ \notag
&+ \frac{L^3vol(S^{3})}{(2 \pi)^{7}l_{s}^{4}} \frac{(2\pi)^{2}}{2} \frac{1}{4} \frac{1}{4}\int d^{5}x \sqrt{g^{IIB}}e^{-\phi_{IIB}}R_{MNLP}R^{MNLP},
\label{s3}
\end{eqnarray}
Note that the higher derivative terms also correct the cosmological constant.
As pointed out in~\cite{bng,Fukuma:2001uf,Fukuma:2002sb},
 the cosmological constant $-12/\tilde{L}^2$, which is different from $-12/L^2$ via $1/N$ correction, is determined later for the gravity dual to $\mathcal{N}=2$ $USp(2N)$ gauge theory and also depends on the field redefinition.

Note that $\frac{(2\pi)^{2}}{2}=vol(T^{2}/\mathbb{Z}_{2})/L^{2}$ and we can formally write the second term with a coefficient $vol(X_{5,\Delta=2})$. Hence this factor can be factored out and the action \eqref{s3} becomes
\begin{equation}
S_{IIB}=\frac{L^5vol(X_{5,\Delta=2})e^{-2\phi_{IIB}}}{(2 \pi)^{7}l_{s}^{8}} \int dx^{5}\sqrt{g^{IIB}}\left( R+\dfrac{12}{\tilde{L}^2} +  \frac{L^2}{16N}R_{MNLP}R^{MNLP} \right),
\label{eq:action_typeIIB_4}
\end{equation}
where we used the relation $L=(4\pi g_s\Delta N)^{1/4}l_s$ with $\Delta =2$.

Thus, the coefficient $a_3=L^2/(16N)$ for the gravity dual on $AdS_5\times S^5/\mathbb{Z}_2$ is determined using the duality chain \eqref{dual7}.

\subsection{Power counting}

So far, we have only focused on the two terms, namely the Einstein-Hilbert action in the bulk and the quadratic term of the Riemann tensor on 7-branes. Let us see which terms will contribute to the $A$-type anomaly from the entanglement entropy. For simplicity, we call $R$ as either $R_{MNKL}, R_{MN}, R$. Since the $A$-type anomaly depends on the parameters $N, \Delta$ and $n_{7}$, we will focus on the $N, \Delta$ and $n_{7}$ dependence in the action. In general, we have the terms
\begin{equation}
\frac{e^{-2\phi_{IIB}}}{l_{s}^{8}} \int d^{10}x \; \sqrt{g^{IIB}} l_{s}^{2(k-1)} R^{k} + \frac{n_{7}e^{-\phi_{IIB}}}{l_{s}^{8}} \int d^{8}x \; \sqrt{g^{IIB}} l_{s}^{2k} R^{k}.
\label{eq:action_counting1}
\end{equation}
The first term of \eqref{eq:action_counting1} is the one in the bulk action and the second term of \eqref{eq:action_counting1} is the one in the 7-brane action. Hence, the second term is proportional to the number of 7-branes. Remember that supersymmetry protects string loop corrections up to the lowest order of the DBI terms and then the second term only depending on
 $e^{-\phi_{IIB}}$ arises from 7-branes~\cite{Tseytlin:1999dj}. By the dimensional reduction on $X_{5}$, \eqref{eq:action_counting1} becomes
\begin{equation}
\frac{e^{-2\phi_{IIB}}L^{5}}{l_{s}^{10-2k} \Delta} \int d^{5}x \; \sqrt{g^{IIB}} R^{k} + \frac{e^{-\phi_{IIB}}n_{7} L^{3}}{l_{s}^{8-2k}} \int d^{5}x \; \sqrt{g^{IIB}}R^{k}.
\end{equation}
The factor $\frac{1}{\Delta}$ appears from the dimensional reduction of $vol(X_{5}) \propto \frac{1}{\Delta}$. By applying the Wald formula \eqref{CEN5} for the entanglement entropy and the $A$-type anomaly, the $A$-type anomaly is roughly expressed as
\begin{equation}
a \sim \frac{e^{-2\phi_{IIB}}L^{10-2k}}{l_{s}^{10-2k} \Delta} + \frac{e^{-\phi_{IIB}}n_{7} L^{8-2k}}{l_{s}^{8-2k}}.
\label{eq:action_counting3}
\end{equation}
The first term originates from the term in the bulk action and the second term originates from the term in the 7-brane action. Because of the relation $L \propto N^{\frac{1}{4}}$, $\mathcal{O}(N^{2})$ contribution only comes from the $k=1$ term in the bulk action, namely the Einstein-Hilbert term. The $\mathcal{O}(N)$ can come from the $k=3$ terms in the bulk action and $k=2$ term in the 7-brane action. However, there are no $k=3$ terms in the five dimensional action since higher derivative correction for the Type IIB theory starts with $\mathcal{O}(R^4)$~\cite{Gris,Ogawa:2011fw}. Here, $k=3$ term may be generated in terms of the dimensional reduction of $k=4$ term. However, it has the role of $k=4$ term since the dimensional reduction does not change the result. In the dual SCFT side, moreover, the coefficient of $k=3$ term corresponding to the three point function of the stress tensor, which would vanish for any SCFTs~\cite{Hofman:2008ar}.
 Hence $\mathcal{O}(N)$ contribution only comes from the $k=2$ term in the 7-brane action. Therefore, the first term in \eqref{eq:action_typeIIB} will generate a $\mathcal{O}(N^{2})$ contribution to the $A$-type anomaly and the second term in \eqref{eq:action_typeIIB} will make a $\mathcal{O}(N)$ contribution. From the general expression of \eqref{eq:action_counting3} and concentrating only on the $\mathcal(N^{2})$ term and the $\mathcal{O}(N)$ in the $A$-type anomaly, we can also determine the $N, \Delta$ and $n_{7}$ dependence of the $A$-type anomaly. The $A$-type anomaly should behave as
\begin{equation}
a = \alpha_{1}N^{2}\Delta + \alpha_{2} N \Delta n_{7},
\label{eq:a-anomaly_general}
\end{equation}
where $\alpha_{1}$ and $\alpha_{2}$ are numerical coefficients.

Finally, it is interesting to investigate the $1/N^2$ correction. Since the bulk $R^3$ term is not allowed for any SCFTs and since the more higher derivative terms seem not to contribute the central charges, the $1/N^2$ correction should be included in the gravity theory with curvature squared terms. In other words, this correction can be included not in more higher derivative terms but in coefficients of the gravity theory such as the cosmological constant.

\subsection{Determination of $\tilde{L}^2$}
Next, we determine the cosmological constant $-12/\tilde{L}^2$ by comparing the central charges $a,c$ in SCFT side with that in the holographic Weyl anomaly~\cite{bng,ode,holo4}. This method
 includes the review of~\cite{deBoer:2011wk,Hung:2011xb} up to the scaling transformation of the coefficients of the gravity theory.
In the field theory side, we can identify $a,c$-charges from the coefficients of the trace anomaly in 4-dimension as follows:
\begin{eqnarray}
\langle T^{a}{}_{a}\rangle =\dfrac{c}{16\pi^2}(R_{abcd}R^{abcd}-2R_{ab}R^{ab}+R^2/3)-\dfrac{a}{16\pi^2}(R_{abcd}R^{abcd}-4R_{ab}R^{ab}+R^2),
\end{eqnarray}
where $a,b=0,..,3$ and $R_{abcd}$ is the 4-dimensional Riemann tensor. This coefficients $a$ and $c$ can be derived using holographic Weyl anomaly formula. We start with the curvature gravity \eqref{ACT6}.
We first derive the AdS vacua for \eqref{ACT6} by solving the EOM.
In the EOM for the action \eqref{ACT6}, the terms including the covariant derivatives vanish in the critical points describing $AdS_5$. Leaving $a_1$, $a_2$, and $a_3$,
 the EOM becomes
\begin{eqnarray}
&-\dfrac{1}{2}g_{\mu\nu}\Big(a_1R^2+a_2R_{\alpha\beta}R^{\alpha\beta}+a_3R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}+R+\dfrac{12}{\tilde{L}^2}\Big) \nonumber \\
&+2a_1RR_{\mu\nu}+2a_2R_{\mu\alpha}R_{\nu}{}^{\alpha}+2cR_{\mu\alpha\beta\gamma}R_{\nu}{}^{\alpha\beta\gamma}+R_{\mu\nu}=0. \label{EOM11}
\end{eqnarray}
Substituting the formula in the appendix A,
 \eqref{EOM11} is rewritten as
\begin{eqnarray}
&1 l^4-12 \tilde{L}^2l^2+(80a_1+16a_2+8a_3)\tilde{L}^2=0. \label{EQ26}
\end{eqnarray}
Substituting $a_3=L^2/16N$, $a_2=a_1=0$ into above the equation, \eqref{EQ26} is solved by
\begin{eqnarray}
&l^2=\tilde{L}^2\dfrac{1+\sqrt{1-\dfrac{4}{3\tilde{L}^2}(20a_1+4a_2+2a_3)}}{2}, \\
&=\dfrac{\tilde{L}^2}{2}\Big({1+\sqrt{1-\dfrac{L^2}{6N\tilde{L}^2}}}\Big).
\label{CUR12}
\end{eqnarray}
where $l$ is the AdS radius in an AdS vacua obtained from \eqref{EOM11}.

According to~\cite{ode} and using \eqref{CUR12}, $a,c$-charges are given by
 \begin{align}
&a=\dfrac{\pi^2 l^3}{\kappa_5^2}(1-4a_3/l^2), \label{ACA26} \\
& c=\dfrac{\pi^2 l^3}{\kappa^2_5}(1+4a_3/l^2). \nonumber
\end{align}
It can be shown that these central charges are invariant under the scaling transformation of the metric $g_{\mu\nu}\to e^{2a}g_{\mu\nu}$.
Under this scaling transformation, the coefficients in the action \eqref{ACT6} are transformed as $\kappa_5^2\to \kappa_5^2e^{-3a}$, $\tilde{L}\to \tilde{L}e^{-a}$, $l\to  le^{-a}$, and $a_i\to a_ie^{-2a}$. Thus, central charges $a,c$ are invariant under above scaling transformation. For later purpose, we leave this dependence of $e^a$ in the coefficient of parameters.

To recover the central charges in the field theory side~\cite{at} with $\Delta=2$
\begin{align}
&a=\dfrac{\Delta N^2}{4}+\dfrac{(\Delta -1)N}{2}-\dfrac{1}{24},\quad c=\dfrac{\Delta N^2}{4}+\dfrac{(\Delta -1)3N}{4}-\dfrac{1}{12}, \label{CEN47}
\end{align}
we should solve the simultaneous equations~\eqref{ACA26}.
 $l$ and $a_3$ are then given by
\begin{eqnarray}
&l^2=L^2e^{-2a}\Big(1+\dfrac{5}{4N}-\dfrac{1}{8N^2}\Big)^{{2}/{3}}, \label{LLL40} \\
&a_3=\dfrac{L^3e^{-2a}}{16l}\Big(\dfrac{1}{N}-\dfrac{1}{6 N^2} \Big).
\end{eqnarray}
The radius $\tilde{L}$ is determined by solving \eqref{CUR12} and including the factor $e^{-a}$ as follows:
\begin{eqnarray}
\tilde{L}^2=\dfrac{4l^4N(144Nl^2-6L^2)e^{-2a}}{(L^2-24Nl^2)^2},
\end{eqnarray}
where we used $l$ in \eqref{LLL40}.
To recover $a_3=L^2/16N$ obtained from F-theory, moreover, we should set
\begin{eqnarray}
e^{-2a}=\dfrac{l}{L\Big(1-\dfrac{1}{6N}\Big)}.
\end{eqnarray}
It is concluded that the Taylor-series of $\tilde{L}$ and $l$ in terms of $1/N$ are given by
\begin{eqnarray}
l^2=  {L}^{2}+\dfrac{17L^2}{12N}+\dfrac{L^2}{9N^2}+O \left( {N}^{-3} \right),
\end{eqnarray}
\begin{eqnarray}
\tilde{L}^2=  {L}^{2}+\dfrac{35L^2}{24N}+\dfrac{79L^2}{576N^2}+O \left( {N}^{-3} \right).
\end{eqnarray}

\subsection{General cases}
For general cases with the gravity dual on $AdS_5\times X_5$, moreover, the coefficient $a_3$ should be determined using the normalization of $a_3=L^2/(16N)$ for $\Delta =2$ since the origin of this squared curvature term is 7-brane in the Type IIB theory~\cite{at} as follows:
\begin{equation}
a_3=\dfrac{(1-1/\Delta)L^2}{8N}.
\end{equation}
This coefficient has the information of the number of 7-branes $n_7=12(1-1/\Delta)$.
This formula is also obtained in~\cite{Buchel:2008vz} requiring that the difference of two central charges $a-c$ in the field theory side is reproduced using the effective curvature gravity.

 For the gravity dual on $AdS_5\times X_5$, thus, we should substitute $a_1=0$, $a_2=0$, and $a_3=(1-1/\Delta)L^2/8N$ into \eqref{ACT6} and the
 effective curvature gravity action becomes
\begin{eqnarray}
S=\dfrac{1}{2\kappa_{5}^2}\int d^5x\sqrt{-g}\Big(R+\dfrac{12}{\tilde{L}^2}+\dfrac{(1-1/\Delta)L^2}{8N}R_{\mu\nu\rho\lambda}R^{\mu\nu\rho\lambda} \Big),
\label{ACT7} \end{eqnarray}
where $\tilde{L}$ can be determined by using the method in previous section and by using \eqref{CEN47}. In terms of the $1/N$ expansion, the $AdS$ radius $l$ and $\tilde{L}$ are given by
\begin{eqnarray}
l^2=  {L}^{2}+{\frac {{L}^{2} \left( 15{\Delta}^{
2}-29\Delta+15 \right) }{ 6N\left( \Delta-1 \right) \Delta}}+{
\frac {{L}^{2} \left( 6{\Delta}^{2}-11\Delta+6 \right) }{
 36N^2\left( \Delta-1 \right) ^{2}\Delta}}+O \left( {N}^{-3} \right),
\end{eqnarray}
\begin{eqnarray}
\tilde{L}^2=  {L}^{2}+{\frac {{L}^{2} \left( 31{\Delta}^
{2}-60\Delta+31 \right) }{12N \left( \Delta-1 \right) \Delta}}+{\frac {{L}^{2} \left( 11{\Delta}^{4}-18{\Delta}^{3}+18
{\Delta}^{2}-18\Delta+11 \right) }{144N^2 \left( \Delta-1
 \right) ^{2}{\Delta}^{2}}}+O \left( {N}^{-3} \right).
\end{eqnarray}

\section{Computation of the $A$-type anomaly from the entanglement entropy}

In this section, we compute the holographic EE for $d=4$ $\mathcal{N}=2$ SCFT with its gravity dual using holographic EE formula.

 We first consider $O(N^2)$ part of the anomaly. Since the curvature correction and $1/N$ correction do not contribute the central charge of this order,  we can ignore the $1/N$ corrections for $\tilde{L}$, $\kappa_5$, $a_3$ and the curvature corrections ($a_3=0$). Thus, central charges $a$ equals $c$ since we consider only Einstein-Hilbert term \eqref{AC09}.

An useful metric of $AdS_5$ is a hyperbolic foliation of $AdS_5$ and is given by \eqref{MET05}.
 The horizon entropy can be calculated using Wald's entropy
\begin{eqnarray}
S=-2\pi \int_{horizon}d^{d-1}x\sqrt{h}\dfrac{\partial \mathcal{L}}{\partial R^{\mu\nu}{}_{\rho\lambda}}\epsilon^{\mu\nu}\epsilon_{\rho\lambda},
\end{eqnarray}
where $\epsilon_{ab}$ denotes the binormal to the horizon. Using the wedge product, the binormal is represented by $d\tau\wedge d\rho$.
 The integrand for the action \eqref{AC09} can be rewritten as
\begin{eqnarray}
\dfrac{\partial \mathcal{L}}{\partial R^{\mu\nu}{}_{\rho\lambda}}\epsilon^{\mu\nu}\epsilon_{\rho\lambda}\Big|_{AdS}=-\dfrac{vol(X_5)}{\kappa_{10}^2}L^5,
\end{eqnarray}
where we substituted the only non-zero component of the binormal $\epsilon_{t\rho}=1$ or the relation $\epsilon_{\mu\nu}\epsilon^{\mu\nu}=-2$.
Thus, using the relation proven in~\cite{Myers:2010xs,Myers:2010tj}
\begin{eqnarray}
\dfrac{\partial \mathcal{L}}{\partial R^{\mu\nu}{}_{\rho\lambda}}\epsilon^{\mu\nu}\epsilon_{\rho\lambda}\Big|_{AdS}=-\dfrac{a_4}{\pi^2 L^3}, \label{ACA8} \end{eqnarray}
the central charge $a_4^{(2)}$, namely the $O(N^2)$ contribution for the $A$-type anomaly is computed as follows:
\begin{eqnarray}
a_4^{(2)}=\dfrac{\pi^2L^8vol(X_5)}{\kappa_{10}^2}=\dfrac{\Delta N^2}{4}. \label{CEN32}
\end{eqnarray}
This central charge correctly reproduces the $A$-anomaly of dual $\mathcal{N}=2$ SCFT. Remember that \eqref{CEN32} agrees with the central charges $a_4$ using the GKP-W relation~\cite{anom}.

\subsection{$1/N$ correction of $A$-type anomaly}

 For the action \eqref{ACT7}, the horizon entropy becomes
\begin{align}
&S^E=\dfrac{2\pi}{\kappa_5^2}\Big(1-4a_3  \dfrac{1}{r_h^2}\Big)\oint d^{d-1}x\sqrt{h(r_h)} \nonumber \\
&=\dfrac{2\pi}{\kappa_5^2}\Big(1-\dfrac{(1-1/\Delta)L^2}{2Nl^2} \Big)\oint d^{3}x\sqrt{h(r_h)}, \label{ENT22}
\end{align}
where in the last equality, we used $l=r_h$.

According to~\eqref{ACA8}, the $A$-type anomaly is computed as follows:
\begin{eqnarray}
a_4=\dfrac{\pi^2{l}^3}{\kappa_5^2}\Big(1-\dfrac{(1-1/\Delta)L^2}{2Nl^2}\Big)=\dfrac{\Delta N^2}{4}+\dfrac{(\Delta-1) N}{2}-\dfrac{1}{24}, \label{ATY36}
\end{eqnarray}
where we used \eqref{DEF10} and \eqref{CEN32}. The central charge \eqref{ATY36} is the same as the holographic Weyl anomaly \eqref{ACA26}.  It is concluded that central charge \eqref{ATY36} agrees with the $A$-type anomaly in the CFT side.

Then, we consider the field redefinition of the curvature theory \eqref{ACT7} to avoid the graviton ghost in the subleading order of $l_s^2/L^2$ expansion.
As pointed out in~\cite{Myers:2010tj}, the linearized gravition equation of \eqref{ACT7} is not second order but is 4-th order. So, there appear gravition ghosts for 4th order equations, which lead to the non-unitary operator in the dual SCFTs.
 Using the field redefinition~\cite{Buchel:2008vz,kp}, however, we show that in the subleading order of $l_s^2/L^2$ expansion, we can transform the curvature gravity \eqref{ACT7} to the Gauss-Bonnet gravity, which has second order linearized graviton equations and preserves unitarity.
We transform the metric as
\begin{eqnarray}
g_{\mu\nu}\to g_{\mu\nu}+b_1g_{\mu\nu}R+b_2R_{\mu\nu},
\end{eqnarray}
where $b_1=-(1-1/\Delta)L^2/12N$ and $b_2=(1-1/\Delta)L^2/2N$.
Then, the gravity action \eqref{ACT7} can be transformed to
\begin{eqnarray}
S=\dfrac{1}{2\kappa_{5}^2\beta}\int d^5x\sqrt{-g}\Big(R+\dfrac{12\beta}{ \tilde{L}^2}+\dfrac{(1-1/\Delta)L^2}{8N}(R_{\mu\nu\rho\lambda}R^{\mu\nu\rho\lambda}+R^2-4R_{\mu\nu}R^{\mu\nu})+... \Big),
\label{ACT19} \end{eqnarray}
where $\beta^{ -1}=1+(1-1/\Delta)/(2N)$ and dots describe the terms of $\mathcal{O}(R^3)$ and terms of $O(N^{-2})$ corrections.

Using \eqref{EOM11}, the $AdS$ radius of the $AdS$ solution obtained from \eqref{ACT19} is given by
\begin{eqnarray}
&\tilde{l}^2=\tilde{L}^2\dfrac{1+\sqrt{1-\dfrac{4\beta^2}{3\tilde{L}^2}(20a_1+4a_2+2a_3)}}{2\beta}, \\
&=\dfrac{\tilde{L}^2}{2\beta}\Big({1+\sqrt{1-\dfrac{1-1/\Delta}{N}}}+O(N^{-2})\Big).
\label{CUR2}
\end{eqnarray}

 For the action \eqref{ACT19}, the horizon entropy\footnote{In the context of Kerr/CFT correspondence~\cite{Guica:2008mu}, the Wald's entropy of a $d=5$ rotating black hole in the presence of Gauss-Bonnet terms coincides with the Cardy's formula~\cite{Hayashi:2011uf}.} becomes
\begin{align}
&S^E=\dfrac{2\pi}{\kappa_5^2\beta}\Big(1-12a_3\beta  \dfrac{1}{r_h^2}\Big)\oint d^{d-1}x\sqrt{h(r_h)} \nonumber \\
&=\dfrac{2\pi}{\kappa_5^2\beta}\Big(1-\dfrac{3(1-1/\Delta)\beta}{2N} \Big)\oint d^{3}x\sqrt{h(r_h)}, \label{ENT22x}
\end{align}
where in the last equality, we used ${L}=r_h$ of the leading order of $N$.

According to~\eqref{ACA8}, the $A$-type anomaly is computed as follows:
\begin{eqnarray}
a_4=\dfrac{\pi^2\tilde{l}^3}{\kappa_5^2\beta}\Big(1-\dfrac{3(1-1/\Delta)\beta}{2N}\Big)=\dfrac{\Delta N^2}{4}+\dfrac{(\Delta-1) N}{2}+O(1), \label{ATY36x}
\end{eqnarray}
where we used \eqref{DEF10} and \eqref{CEN32}. Since \eqref{ATY36x} is the same form as \eqref{ACA26}, this central charge is invariant under the scaling transformation $g_{\mu\nu}\to e^{2a}g_{\mu\nu}$.
Central charge \eqref{ATY36x} agrees with the $A$-type anomaly in the CFT side up to order $N$~\eqref{CEN47}.

Lastly, we discuss the holographic c-theorem proposed in~\cite{Myers:2010tj,Liu:2011ii}, where the usual $c$-theorem in quantum field theories~\cite{Zam,Cardy:1988cwa,Osborn:1989td,Jack:1990eb,Komargodski:2011vj} states that central charges decrease in the IR. To describe the holographic $c$-theorem, the bulk matter should satisfy two conditions, namely, the null energy condition $-(T^t_t-T^r_r)\ge 0$ and $\sigma'\ge 0$ where
\begin{eqnarray}
\sigma =\dfrac{16a_1+5a_2+4a_3}{16}R', \label{NUL21}
\end{eqnarray}
where $R'$ is the derivative of the Ricci scalar along the AdS radial direction. Here, we used the convention of the metric in~\cite{Liu:2011ii}.
 The Gauss-Bonnet gravity \eqref{ACT19} clearly satisfies the second condition.
So, it is concluded that if we set the coefficients of the curvature square terms properly using the field-redefinition, we may only consider the bulk null energy condition $-(T^t_t-T^r_r)\ge 0$ to be consistent with the holographic c-theorem.

\section{Discussion}
In this paper, we computed the holographic EE in the effective curvature gravity dual to the $\mathcal{N}=2$ SCFTs in the F-theory using the Wald's entropy~\cite{Myers:2010tj} and confirmed that it is consistent with the proposal in~\cite{Myers:2010tj}. We realized the $A$-type anomaly including the $1/N^2$ corrections in the SCFT side from the log term of the holographic EE not using the Gauss-Bonnet type but using the general curvature squared gravity, where it was also discussed in~\cite{deBoer:2011wk,Hung:2011xb}. Here, the curvature term of this theory can be transformed into the Gauss-Bonnet term \eqref{ACT19} in the subleading order of $l_s^2/L^2$ expansion. Our new analysis in section 4.1 also shows that
 the $1/N^2$ correction is included not in the higher derivative terms of $O(R^3)$ but in the gravity theory with general curvature squared terms. While the $A$-type anomaly can be computed by using the $a$-maximalization in supersymmetric theories, the analysis of this paper gives the interesting method to compute the entanglement entropy and the $A$-type anomaly.
 Precisely, we chose the cosmological constant $-12/\tilde{L}^2$ to reproduce the Weyl anomaly in the dual SCFTs using the holography since  this cosmological constant has not been determined from the 10-dimensional theory yet.

In appendix B, we considered the $1/N$ correction from the higher derivative term of the probe $D_4$-type 7-brane, where we can take the $g_s\to 0$ limit. This $1/N$ correction in our approximation agreed with the result in the CFT side up to a factor 15/16. We leave the similar analysis for more complicated cases $G=E_6,E_7,..$ with a constant coupling.

To obtain consistent results, we also discussed a holographic c-theorem~\cite{Myers:2010tj,Liu:2011ii}.  Using the field redefinition properly, we showed that the Gauss-Bonnet gravity \eqref{ACT19} satisfies the null energy condition in the subleading order of the $l_s^2/L^2$ expansion. This means that using the field redefinition, a theory with non-trivial $\sigma$ \eqref{ACT7} may be equal to the Gauss-Bonnet gravity with $\sigma=0$ \eqref{ACT19} of the finite order of $N$.\footnote{ Note that as another direction, the action \eqref{ACT19} is the Gauss-Bonnet action and we can apply the holographic EE on the $AdS_5$ soliton~\cite{Ogawa:2011fw} using the action \eqref{ACT19}. We then notice an instability of the holographic EE for $a_3>0$.  However, a mechanism of the supersymmetry seems to work for our theory for $\mathcal{N}=2$ SCFTs and we do not mind this instability.  }
 See also~\cite{Gubser:2002zh,Fujita:2008rs} for the paper discussing the holographic c-theorem in the context of double trace deformation.

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\noindent {\bf Acknowledgments:}  We would like to thank A. Buchel, C. Hoyos, M. Kaminski, A. Karch, T. Nishioka, N. Ogawa, H. Ooguri, S. Sugimoto, Y. Tachikawa and T. Takayanagi for discussions and comments. I would like to thank H. Hayashi and T. S. Tai for collaboration in the initial stage of this project and for helpful discussions.  MF is supported by the postdoctoral fellowship program of the Japan Society for the Promotion of Science
(JSPS), and partly by JSPS Grant-in-Aid for JSPS Fellows No. 22-1028.