Assume that $G$ is a group and $R$ is a p.i.d. What can we say about the projective dimension of $R[G]$? For example can we say that this dimension is at most $1$ for reductive groups? (I think if $R$ is a field then this is true.) What about an arithmetic subgroup of a reductive group?
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$\begingroup$ What do you mean by the group ring of a reductive group? $\endgroup$– მამუკა ჯიბლაძეCommented Dec 17, 2022 at 16:23
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$\begingroup$ RG is projective and so has projective dimension 0. Projective dimension is for modules not rings. Do you mean global dimension or cohomological dimension (projective dimension of the trivial module 1)? $\endgroup$– Benjamin SteinbergCommented Dec 17, 2022 at 16:49
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1$\begingroup$ Yes I mean global dimension $\endgroup$– aliCommented Dec 17, 2022 at 16:56
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$\begingroup$ @მამუკაჯიბლაძე it means the formal sum of elements of G with coefficient in R.( by a reductive group I mean the points over $\mathbb{C}$ not the group scheme) $\endgroup$– aliCommented Dec 17, 2022 at 16:59
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2$\begingroup$ @ali If you mean "global dimension" and not "projective dimension" you should edit your problem statement asap to clear things up. $\endgroup$– rschwiebCommented Dec 19, 2022 at 16:26
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1 Answer
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There are the inequalities $$\max\{\textrm{gl.dim}(R),\textrm{cohom.dim}_R(G)\}\le\textrm{gl.dim}(RG)\le\textrm{gl.dim}(R)+\textrm{cohom.dim}_R(G).$$ The right hand inequality is often realized (as an equality). In general I think it would be hard to compute when the right hand inequality is realized.
answered Dec 17, 2022 at 18:08
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1$\begingroup$ The global dimension always is an upper bound on the cohomological dimension and the global dimension of a pid is \leq 1 $\endgroup$ Commented Dec 17, 2022 at 19:41
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1$\begingroup$ Anyway it won't be 1 for a non virtually free group. $\endgroup$ Commented Dec 17, 2022 at 19:42
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$\begingroup$ I am pretty sure $\mathrm{cohom.dim}_{\mathbb Z}(\mathrm{GL}_n(\mathbb C))$ is infinity: cohomologies of $\mathrm{GL}_n(\mathbb C)$ contain lots of stuff coming from algebraic K-theory $\endgroup$ Commented Dec 17, 2022 at 19:50
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1$\begingroup$ That is why I added that the order of the finite group should not be invertible. $\endgroup$ Commented Dec 17, 2022 at 21:15
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1$\begingroup$ In any event the cohomological dimension cannot be at most 1 unless the group acts on a tree with finite stabilizers with order invertible in R and that won't happen for non-virtually free arithmetic subgroups $\endgroup$ Commented Dec 17, 2022 at 21:17