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Aug 3 at 4:29 comment added Qiaochu Yuan @Ewrt: this is incredible overkill, there is no need to discuss either $\ell^2$ or bounded operators here. Every group $G$ already acts by automorphisms on, say, the polynomial ring in $|G|$ generators, by permuting the generators.
Feb 1 at 22:09 answer added Brian Pinsky timeline score: 9
May 3, 2023 at 12:50 comment converted from answer Ewrt Wert (Comment: If you accept subgroup of automorphism groups, then it's easy: Every group can be represented on the ring $B(\ell^2(G))$ by left regular representation $g \mapsto U_g$ and is so subgroup of $Aut(B(\ell^2(G)))$ by inner automorphisms induced by $U_g$.)
May 3, 2023 at 7:47 vote accept A. Bailleul
May 3, 2023 at 7:46 history edited A. Bailleul CC BY-SA 4.0
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May 3, 2023 at 7:39 comment added A. Bailleul @YCor I am mostly interested in the associative unital case.
May 3, 2023 at 3:26 history became hot network question
May 2, 2023 at 20:03 answer added M.G. timeline score: 14
May 2, 2023 at 19:48 history edited YCor
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May 2, 2023 at 19:48 comment added YCor Could you define exactly what you mean by ring, since there are many definitions according to different communities. (Unital? Associative? Commutative?)
May 2, 2023 at 19:48 comment added Dave Benson Ah, true. But some construction based on the graph should work...
May 2, 2023 at 19:47 history edited YCor
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May 2, 2023 at 19:46 comment added Wojowu @DaveBenson I believe that ring will usually have many more automorphisms than those of the graph it comes from.
May 2, 2023 at 19:44 comment added Dave Benson Since every group is the automorphism group of a graph (even a cubic graph), can't you just use the Stanley-Reisner ring of the graph over a field of your choice?
May 2, 2023 at 19:24 history asked A. Bailleul CC BY-SA 4.0