Timeline for What are the cohomological dimensions of ${\rm Aut}(F_n)$, ${\rm Out}(F_n)$, ${\rm SL}_n(\mathbb{Z})$ over the rationals ℚ and integers ℤ?
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18 events
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| Jan 9, 2023 at 2:35 | comment | added | LSpice | Name of @MattZaremsky's reference: Church, Farb, and Putman - The rational cohomology of the mapping class group vanishes in its virtual cohomological dimension. | |
| Jan 9, 2023 at 2:29 | history | edited | LSpice | CC BY-SA 4.0 |
While this is on the front page, fixed typo in title; `\mathsf` -> `\rm` (`\operatorname` won't fit) for consistency with body; deleted "thanks"
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| Jan 8, 2023 at 20:27 | answer | added | Matt Zaremsky | timeline score: 11 | |
| Jan 8, 2023 at 14:41 | comment | added | Matt Zaremsky | @AndyPutman That is pretty curious, maybe indicative of mapping class groups containing more "weird" subgroups than the others, or something. Well, I suppose I should officially answer the question now, once the kids give me time to actually think.... | |
| Jan 8, 2023 at 13:26 | comment | added | Andy Putman | @MattZaremsky: That’s right! It’s curious that it’s so hard to find the correct lower bound on the cd for the mapping class groups, but trivial for automorphism groups of free groups and nearly trivial for SL(n,Z). | |
| Jan 8, 2023 at 12:38 | comment | added | Matt Zaremsky | @AndyPutman Ah, good point. So actually, right, thinking a little more about achieving bounds using subgroups, the easiest possible case is if vcd_Z=d and your group contains Z^d, then right away we get cd_Q=d. I guess it hadn't explicitly occurred to me that MCG(S_g) and SL_n(Z) don't satisfy this, but Out(F_n) and Aut(F_n) do satisfy this (it's easy to embed Z^{2n-2} into Aut(F_n)), so they're "easier" and I guess this answers the original question (I think). | |
| Jan 8, 2023 at 4:58 | comment | added | Andy Putman | @MattZaremsky: The easiest way to see they are equal is to note that the subgroups realizing the virtual cohomological dimension (the “Mess subgroups”) have nontrivial rational cohomology in their top degree. In fact, I think you need these subgroups to prove Harer’s theorem. | |
| Dec 31, 2022 at 12:11 | comment | added | Matt Zaremsky | @archipelago Ah, yes, I misread what cd_Q was. Actually, I guess the result of Harer they state in the first paragraph, using the Steinberg module, shows cd_Q = vcd_Z for mapping class groups. | |
| Dec 31, 2022 at 2:45 | comment | added | Matt Zaremsky | @IJL For mapping class groups, it seems $\textrm{cd}_{\mathbb{Q}}$ actually is less than $\textrm{vcd}_{\mathbb{Z}}$, see arxiv.org/abs/1108.0622. By analogy I would guess then that it's also true for $Out(F_n)$, but I'm not finding this by googling, so maybe it's open. | |
| Dec 30, 2022 at 15:27 | comment | added | IJL | It is not so easy to find groups of finite vcd for which $\mathrm{cd}_{\mathbb{Q}}$ is not equal to $\mathrm{vcd}_{\mathbb{Z}}$: I don't know that this is the case for the groups asked about, but it would be surprising if it wasn't. | |
| S Dec 28, 2022 at 18:33 | history | suggested | Pedro | CC BY-SA 4.0 |
Title was breaking on main site.
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| Dec 28, 2022 at 16:39 | review | Suggested edits | |||
| S Dec 28, 2022 at 18:33 | |||||
| Dec 28, 2022 at 16:00 | comment | added | YCor | But for $\mathrm{cd}_\mathbf{Q}$ one also needs, in these cases, to know that it is at least equal to $\mathrm{vcd}_\mathbf{Z}$ (this is probably false in general?) and probably the way the lower bound for $\mathrm{vcd}_\mathbf{Z}$ is obtained also provides a lower bound for $\mathrm{cd}_\mathbf{Q}$. I'm not sure about Aut/Out($F_n$), but for $\mathrm{SL}_n(\mathbf{Z})$ this is indeed $n(n-1)/2$ (lower bound being given by the inclusion of the upper unipotent group). | |
| Dec 28, 2022 at 15:55 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Dec 28, 2022 at 15:08 | comment | added | Peter Kropholler | The groups you list always have torsion when $n\ge2$. The virtual cohomological dimension is defined to be the cohomological dimension of a torsion-free subgroup of finite index (if there is one), or $\infty$ (if not). One always has $$\mathrm{cd}_{\mathbb Q}(G)\le \mathrm{vcd}_{\mathbb Z}(G)\le\mathrm{cd}_{\mathbb Z}(G).$$ These facts may help you extract the information you need from the results about vcd that you have already found. | |
| Dec 28, 2022 at 12:48 | comment | added | Benjamin Steinberg | If a group has non-trivial torsion it has infinite cohomological dimension over Z and so virtual cohomological dimension is a better way to go. | |
| Dec 28, 2022 at 12:28 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| Dec 28, 2022 at 12:24 | history | asked | John Depp | CC BY-SA 4.0 |