Commutator ideal of $\mathfrak{gl}(V)$ for $V$ infinite-dimensional It is well-known that the commutator ideal of $\mathfrak{gl}(V)$ is $\mathfrak{sl}(V)$, the Lie algebra of traceless endomorphisms, if $V$ is a finite-dimensional vector space. I am now wondering what is known about $[\mathfrak{gl}(V), \mathfrak{gl}(V)]$ if the vector space $V$ is infinite-dimensional. I am in particular wondering about the the codimension of $[\mathfrak{gl}(V), \mathfrak{gl}(V)]$ in $\mathfrak{gl}(V)$.
 A: The codimension is zero. That is, $\mathfrak{gl}(V)$ is a perfect Lie algebra.
Indeed, first use that $V\simeq W\oplus W$ for some vector space $W$. Under this "block" decomposition, every element of $\mathfrak{gl}(V)$ can be written as a square block matrix
$$\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}A&0\\0&0\end{pmatrix}+\begin{pmatrix}0&B\\0&0\end{pmatrix}+\begin{pmatrix}0&0\\C&0\end{pmatrix}+\begin{pmatrix}0&0\\0&D\end{pmatrix}.$$
It is enough to prove that each of these matrices is a commutator (thus every element of $\mathfrak{gl}(V)$ is a sum of 4 commutators; 4 is probably not optimal). Indeed,
$$\begin{pmatrix}0&B\\0&0\end{pmatrix}=\Big[\begin{pmatrix}I&0\\0&0\end{pmatrix},\begin{pmatrix}I&B\\0&0\end{pmatrix}\Big],$$
and similarly $\begin{pmatrix}0&0\\C&0\end{pmatrix}$ is a commutator.
Next, to deal with $f=\begin{pmatrix}A&0\\0&0\end{pmatrix}$, we use that $W$ is isomorphic to $W^{(\mathbf{N})}$ and thus write $V$ as $\bigoplus_{n\in\mathbf{Z}} W_n$, where each $W_n$ is given with an isomorphism with $W$, and $W_0$ is identified to the left-hand $W$ in the previous $W\oplus W$ decomposition. Hence, in this new decomposition, $f$ is the block-diagonal matrix whose diagonal is $(\dots 0,0,A,0,0\dots)$ where $A$ is in position zero. For $w\in W$, write $w[n]$ as the element $w$ viewed in the $n$-th copy $W_n$ (through the given identification). Define $g,h\in\mathfrak{gl}(V)$ as follows: $g(w[n])=0$ if $n\le 0$, $g(w[n])=w[n-1]$ if $n>0$; $h(w[n])=(Aw)[n+1]$ for all $n\in\mathbf{Z}$. Then
$$[g,h](w[0])=gh(w[0])-hg(w[0])=g((Aw)[1])=(Aw)[0];$$
$$[g,h](w[n])=gh(w[n])-hg(w[n])=0-0=0 \quad (n<0);$$
$$[g,h](w[n])=gh(w[n])-hg(w[n])=g((Aw)[n+1])-h(w[n-1])=$$
$$= (Aw)[n]-(Aw)[n])=0  \quad (n\ge 0).$$
That is, $[g,h]=f$, so $f$ is a commutator. Similarly, $\begin{pmatrix}0&0\\0&D\end{pmatrix}$ is a commutator.
A: Motivated by YCor’s answer I came up with the following argumentation, which shows that every element of $\mathfrak{gl}(V)$ is a commutator.
The vector space $V$ is isomorphic to the direct sum $V^{\oplus \mathbb{N}_{\geq 1}}$.
We may therefore assume that $V = W^{\oplus \mathbb{N}_{\geq 1}}$ for some vector space $W$.
We can thus represent every endomorphism $f$ of $V$ as a matrix
$$
  f
  \equiv
  \begingroup
  \renewcommand{\arraystretch}{1.3}
  \begin{pmatrix}
    f_{11}  & f_{12}  & f_{13}  & f_{14}  & \cdots  \\
    f_{21}  & f_{22}  & f_{23}  & f_{24}  & \cdots  \\
    f_{31}  & f_{32}  & f_{33}  & f_{34}  & \cdots  \\
    f_{41}  & f_{42}  & f_{43}  & f_{44}  & \cdots  \\
    \vdots  & \vdots  & \vdots  & \vdots  & \ddots
  \end{pmatrix} \,.
  \endgroup
$$
Each matrix entry $f_{ij}$ is an endomorphism of $W$, namely the composite $\pi_i \circ f \circ \iota_j$, where $\iota_j$ the inclusion from $W$ into the $j$-th direct summand of $V$, and $\pi_i$ the projection from $V$ onto its $i$-th direct summand $W$.
This matrix is pointwise column-finite, in the sense that for every element $w$ of $W$ and every column index $j$ the images $f_{ij}(w)$ vanish for all but finitely many row indices $i$.
Suppose conversely that we have given a family of endomorphisms $g_{ij}$ of $W$ with $i, j \geq 1$ such that the resulting matrix is pointwise column-finite.
Then this matrix describes an endomorphism $g$ of $V$.
Let $s$ be the shift endomorphism of $V$ given by
$$
  s( (w_1, w_2, w_3, \dotsc) )
  =
  ( w_2, w_3, w_4, \dotsc )
$$
for all $(w_1, w_2, w_3, \dotsc) \in V$.
As a matrix, this endomorphism is given by
$$
  s
  \equiv
  \begin{pmatrix}
    0 & 1 &   &   &         \\
      & 0 & 1 &   &         \\
      &   & 0 & 1 &         \\
      &   &   & 0 & \ddots  \\
      &   &   &   & \ddots
  \end{pmatrix}
$$
We therefore have
\begin{align*}
  [s, f]
  &=
  s f - f s
  \\[1em]
  &\equiv
  \begin{pmatrix}
    f_{21}  & f_{22}  & f_{23}  & f_{24}  & \cdots  \\
    f_{31}  & f_{32}  & f_{33}  & f_{34}  & \cdots  \\
    f_{41}  & f_{42}  & f_{43}  & f_{44}  & \cdots  \\
    f_{51}  & f_{52}  & f_{53}  & f_{54}  & \cdots  \\
    \vdots  & \vdots  & \vdots  & \vdots  & \ddots
  \end{pmatrix}
  -
  \begin{pmatrix}
    0       & f_{11}  & f_{12}  & f_{13}  & \cdots  \\
    0       & f_{21}  & f_{22}  & f_{23}  & \cdots  \\
    0       & f_{31}  & f_{32}  & f_{33}  & \cdots  \\
    0       & f_{41}  & f_{42}  & f_{43}  & \cdots  \\
    \vdots  & \vdots  & \vdots  & \vdots  & \ddots
  \end{pmatrix}
  \\[1em]
  &=
  \begingroup
  \renewcommand{\arraystretch}{1.5}
  \begin{pmatrix}
    f_{21}  & f_{22} - f_{11} & f_{23} - f_{12} & f_{24} - f_{13} & \cdots  \\
    f_{31}  & f_{32} - f_{21} & f_{33} - f_{22} & f_{34} - f_{23} & \cdots  \\
    f_{41}  & f_{42} - f_{31} & f_{43} - f_{32} & f_{44} - f_{33} & \cdots  \\
    f_{51}  & f_{52} - f_{41} & f_{53} - f_{42} & f_{54} - f_{43} & \cdots  \\
    \vdots  & \vdots          & \vdots          & \vdots          & \ddots
  \end{pmatrix} \,.
  \endgroup
\end{align*}
Let now $g$ be an endomorphism of $V$.
We can use the above explicit calculation of the commutator $[s, f]$ to construct an endomorphism $f$ with $[s, f] = g$.
Indeed:
Starting with the first column we choose the entry $f_{11}$ arbitrary and $f_{i1}$ as $g_{i-1,1}$ for every $i \geq 2$.
Once the entries $f_{ij}$ are constucted for all $i \geq 1$ and $j = 1, \dotsc, k$, we choose the entries $f_{1,k+1}$ arbitrary and the entry $f_{i,k+1}$ as $g_{i-1, k+1} + f_{i-1, k}$ for every $i \geq 2$.
The matrix described by these entries $f_{ij}$ is pointwise column-finite because the matrix associated to $g$ is pointwise column-finite.
It follows that the entries $f_{ij}$ describe an endomorphism $f$ of $V$.
By construction, this endomorphism satisfies $[s, f] = g$.
We have thus shown that every endomorphism of $V$ is a commutator (namely a commutator of the form $[s, f]$ for some endomorphism $f$ of $V$).
