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Let $V$ be a (possibly infinite-dimensional) vector space over a division ring $d$, and consider the projective special linear group $\mathbf{PSL}(V)$. We suppose that if the dimension of $V$ would be $2$, then $d$ is not isomorphic to $\mathbb{F}_2$ or $\mathbb{F}_3$.

My questions are:

  • Is it known that $\mathbf{PSL}(V)$ is simple (so even in infinite dimension) ? (In finite dimension, the answer is positive, I have read.) If I am not mistaken, it is simple if and only if it is perfect (due to Iwasawa's criterion).

  • What is the automorphism group of $\mathbf{PSL}(V)$ ? Is it isomorphic to the projective semilinear $\mathbf{P\Gamma L}(V)=\mathbf{PGL}(V)\rtimes\mathrm{Aut}(d)$ ?

I would greatly appreciate, besides an answer, good references !

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    $\begingroup$ When ${\rm dim}(V) = n\ge 3$ we also have the duality/graph automorphism, even when $d$ is a field, which is the automorphism induced by $A \mapsto (A^{\mathsf T})^{-1}$ for matrices $A$. When $d$ is a field, ${\rm P \Gamma L}(V)$ has index $2$ in ${\rm Aut}({\rm PSL}(V))$ when $n \ge 3$. $\endgroup$
    – Derek Holt
    Commented Oct 14, 2022 at 10:11
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    $\begingroup$ What is your definition of $\mathbf{PSL}(V)$ when $V$ is infinite-dimensional? $\endgroup$
    – Tom De Medts
    Commented Oct 14, 2022 at 10:50
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    $\begingroup$ Most likely in infinite dimension $\mathrm{PSL}(V)$ should be defined as $\mathrm{PGL}(V)$. In any case it has a normal subgroup $N$ consisting of those $f$ such that $f-\mathrm{Id}$ has finite rank. When $V$ has countable dimension this might be "essentially" the only nontrivial proper normal subgroup. "Essentially" because there is the determinant map on $N$ — say when $d$ is a field, and hence $\det^{-1}$ of any subgroup of $d^*$ yields a normal subgroup of $\mathrm{PGL}(V)$ contained in $N$. In uncountable dimension one also has the set $N_\alpha$ of $f$ such that $f-I$ has rank $<\alpha$. $\endgroup$
    – YCor
    Commented Oct 14, 2022 at 11:56
  • $\begingroup$ @TomDeMedts: it is the subgroup of $\mathbf{PGL}(V)$ generated by the transvections (in its action on the underlying projective space). $\endgroup$
    – THC
    Commented Oct 14, 2022 at 13:04
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    $\begingroup$ Then for $d$ commutative, in infinite dim., $\mathrm{PSL}(V)=\mathrm{SL}(V)$ is the kernel of the determinant map, in the group of linear aut. $f$ of $V$ such that $f-\mathrm{Id}$ has finite rank. It is indeed a simple group (easy consequence of the finite-dimensional case). Proof: let $f\neq 1$, $g$ nontrivial elements, hence they are in $\mathrm{SL}(W)$ for some finite-dim subspace $W$. Enlarging $W$ if necessary, we can suppose that $f$ is not scalar. All proper normal subgroups of $\mathrm{SL}(W)$ are central. Hence $g$ is in the normal subgroup generated by $f$. Whence simplicity. $\endgroup$
    – YCor
    Commented Oct 15, 2022 at 11:21

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