# $E_n(\ell^\infty)=SL_n(\ell^\infty)$?

Let $R$ be a commutative unital ring $R$ with unit element $1$.

For $n\in \mathbb{N}=\{1,2,3,\cdots\}$, let $SL_n(R)$ be the group of all $n\times n$ matrices with entries from $R$ having determinant $1$.

An element $E\in SL_n(R)$ is called an elementary matrix if $E=I+a E_{ij}$, with $i\neq j$, where

• $a\in R$,
• $I$ denotes the identity matrix, and
• $E_{ij}$ the matrix with entry in $i$th row and $j$th column equal to $a$ and all other entries $0$.

The subgroup of $SL_n(R)$ generated by the elementary matrices is denoted by $E_n(R)$.

We know that when $R=\mathbb{C}$, then for all $n$, $SL_n(\mathbb{C})=E_n(\mathbb{C})$.

Now let $R=\ell^\infty$, the algebra of all complex valued bounded sequences with pointwise operations.

Question: Is it true that $SL_n(\ell^\infty)=E_n(\ell^\infty)$ for all $n$?

• Do you have a nice reference that elementary matrices of determinant one generate $SL_n(\mathbb C)$? – Matthew Daws Mar 14 '17 at 18:46
• @MatthewDaws Elementary matrices have determinant 1 by definition. And this fact is basic linear algebra. References are given in math.stackexchange.com/questions/654206 – YCor Mar 14 '17 at 18:49
• @YCor I had thought that it was at least reasonably standard to say that an "elementary matrix" is any matrix you can get from the identity by applying an elementary row operator. So "of determinant 1" was my, probably clumsy, way of singling out those "of the first type" (as Artin seems to say). Thank you very much for the link! – Matthew Daws Mar 14 '17 at 19:55
• @MatthewDaws Also depends on what you call elementary row operator. In coherence with this convention (which is well-admitted in algebraic K-theory), elementary row/column operators mean $C_i\leftarrow C_i+tC_j$ for $i\neq j$. – YCor Mar 15 '17 at 18:02

The answer is yes. The identity $SL_n(\ell^{\infty}) = E_n(\ell^{\infty})$ holds for every $n \ge 2$ and $\mathbb{C}$ can be replaced by any normed field. This can be inferred using the following lemma and the two observations below.

Lemma. Let $R = \prod_i R_i$ be a direct product of unital and commutative rings $R_i$. Let $n \ge 2$ and $A = (A_i )_i\in SL_n(R) \simeq \prod_i SL_n(R_i)$. If for every $i$ the matrix $A_i$ is the product of at most $k$ elementary matrices over $R_i$, then $A$ is the product of at most $k(n^2 - n)^k$ elementary matrices over $R$. In particular, we have $A \in E_n(R)$.

Proof. Consider each matrix $A_i$ as a word $w_i$ over an alphabet of elementary matrix types (there are $n(n -1)$ types). Since there are at most $(n^2 - n)^k$ such words $w_i$, they all fit in a word of length at most $k(n^2 - n)^k$.

Let $A \in SL_n(\mathbb{C})$. Then we have:

• The largest complex modulus of a coefficient of $A$ is greater or equal to $1/\sqrt[n]{n!}$ (expend $\det(A)$ and use the triangle inequality).
• If the coefficients of $A$ are bounded above by $C > 1$, then $A$ is the product of $\nu(n)$ elementary matrices with coefficients bounded above by $2\sqrt[n]{n!} C$ where $\nu(n)$ depends on $n$ only (use the above observation as a basis for an induction on $n$).

Combining the last observation with the lemma yields the result.

Let us detail the case of $SL_2(\ell^{\infty})$.

Proof for $n = 2$. Multiplying $A = (a_{ij})_{1 \le i, j \le 2}$ by at most four elementary matrices whose coefficients are bounded above by $1$, we obtain a matrix $A' = (a'_{ij})_{1 \le i, j \le 2}$ whose coefficients are those of $A$, up to some permutation and some sign changes, and such that $\vert a'_{11} \vert$ is maximal. Using $a'_{11}$ as a pivot, we can turn $A$ into $\begin{pmatrix} a'_{11} & 0 \\ 0 & 1/a'_{11} \end{pmatrix}$ by means of two elementary matrices whose coefficients are bounded above by $1$. Using four more matrices whose coefficients are bounded above by $2\sqrt{2}C$, such as those used in Whitehead's lemma, we obtain the identity matrix. All in all, we used at most ten elementary matrices to perform our reduction and the coefficients of all these matrices are bounded above by $2\sqrt{2}C$ (hence we can set $\nu(2) = 10$).

• The argument seems to work in any normed field, right? – YCor Mar 14 '17 at 18:13
• @YCor: Yes, you are right. I edited my answer so that it incorporates your comment. – Luc Guyot Mar 14 '17 at 20:30
• Additional remark: in the ultrametric setting the bound $1/\sqrt[n]{n}$ can be replaced by 1, which provides a better bound in that case. – YCor Mar 14 '17 at 21:36