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Let $G$ be a (possibly infinite) group. Let $\mathbb{Z}[G]$ be its integral group ring and let $P$ be a finitely generated projective module over $\mathbb{Z}[G]$. Suppose that the coinvariants of $P$ vanish, i.e. $$ P_G=P/I_G P=0 $$

where $I_G$ is the kernel of the augmentation map $\mathbb{Z}[G]\to\mathbb{Z}$. Does it follow that $P=0$?

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This is true when $G$ is finite. This follows from a theorem of Swan [1], which asserts that all projective $\mathbb{Z}[G]$-modules are locally free, i.e isomorphic to a free module over $\mathbb{Z}_p[G]$ for every prime $p$. And clearly, locally free modules have non-trivial coinvariants.

I don't know what happens for infinite groups.

[1] R. G. Swan, Induced representations and projective modules, Ann. of Math. 71 (1960), 552-578.

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  • $\begingroup$ This is good news, but actually I am interested in the infinite (finitely presented) case. $\endgroup$ – KotelKanim Jul 23 '15 at 20:31
  • $\begingroup$ Presumably your argument also establishes the case when $G$ is residually finite. Of course there are finitely presented, infinite, simple groups . . . $\endgroup$ – Jason Starr Jul 24 '15 at 13:33
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This isn't an area that I'm expert on, and it's quite possible there's a much more elementary and/or more general answer.

But if the Bass Conjecture on Hattori-Stallings ranks for group rings is true for $G$ then I think the answer is yes. I don't think the conjecture has been proved for finitely presented groups in general, but there are large classes of groups for which it is known to be true.

If $A$ is a ring, and $P$ a finitely generated projective $A$-module, then $P$ is the image of an idempotent endomorphism of a finitely generated free module, given by a square matrix $M$ over $A$, and the Hattori-Stallings rank of $P$ is the image in $A/[A,A]$ of the trace of $M$, which depends only on $P$ and not the choice of $M$.

If $A=\mathbb{Z}[G]$ for a group $G$, then $A/[A,A]$ is a free abelian group on the images of a set of conjugacy class representatives $[g]$, and Bass conjectured that the Hattori-Stallings rank of any finitely generated projective is a multiple of $[1]$. Previously (I think), Kaplansky proved that the coefficient of $[1]$ in the rank is strictly positive if $P\not\cong0$, which, if we also assume the Bass conjecture for $G$, proves that the image of $M$ under the augmentation map must have non-zero trace, so that the image of $M$ under the augmentation map is non-zero, which is equivalent to $P$ having a non-zero map to the trivial module $\mathbb{Z}$.

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  • $\begingroup$ Yes, it follows from the conjecture of Bass which is open in general, but isn't it still possible that the answer to my question is known nevertheless? $\endgroup$ – KotelKanim Jul 29 '15 at 9:25
  • $\begingroup$ @KotelKanim Yes, that's pretty much what I said in the first paragraph of my answer. $\endgroup$ – Jeremy Rickard Jul 29 '15 at 9:30

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