Timeline for Two questions about $\mathbf{PSL}(V)$ with $V$ a vector space over a division ring
Current License: CC BY-SA 4.0
8 events
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| Oct 15, 2022 at 11:21 | comment | added | YCor | 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. | |
| Oct 14, 2022 at 13:04 | history | edited | THC | CC BY-SA 4.0 |
deleted 6 characters in body
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| Oct 14, 2022 at 13:04 | comment | added | THC | @TomDeMedts: it is the subgroup of $\mathbf{PGL}(V)$ generated by the transvections (in its action on the underlying projective space). | |
| Oct 14, 2022 at 11:56 | comment | added | YCor | 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 $<\alpha$. | |
| Oct 14, 2022 at 11:51 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo, defined P\Gamma L
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| Oct 14, 2022 at 10:50 | comment | added | Tom De Medts | What is your definition of $\mathbf{PSL}(V)$ when $V$ is infinite-dimensional? | |
| Oct 14, 2022 at 10:11 | comment | added | Derek Holt | 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 $n \ge 3$. | |
| Oct 14, 2022 at 10:03 | history | asked | THC | CC BY-SA 4.0 |