# Using Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ring

Just recently I've stumbled across Warren Dicks' book Groups, trees and projective modules (1980) and I was pretty stunned. I know nothing of group cohomology, but I gather the "tree" component is a special case of space studied in general group cohomology.

The results that caught my attention were:

The augmentation ideal of $R[G]$ is projective iff $G$ has cohomological dimension at most $1$, and it has cohomological dimension exactly $1$ iff it is an infinite group of cohomological dimension $1$.

The augmentation ideal of $R[G]$ is projective iff $G$ is the fundamental group of a graph of finite groups having order invertible in $R$.

For me this resonated with two other results I know well about group rings:

(Renault) $R[G]$ is right self-injective iff $R$ is right self-injective and $G$ is finite.

and

(Connell) $R[G]$ is von Neumann regular iff 1) $R$ is von Neumann regular; 2) $G$ is locally finite; and 3) the order of every finite subgroup of $G$ is invertible in $R$.

Suppose from here on that $R[G]$ is von Neumann regular, $G$ is (at most) countable, and you don't know Connell's proof about regular group rings.

Then the augmentation ideal is countably generated, and by a result of Kaplansky the augmentation ideal must be projective. Apparently now $G$ is "the fundamental group of a graph of finite groups having order invertible in $R$."

First question: is it obvious somehow that $G$ is locally finite and that the groups which are vertices of the graph of groups tell us about the finite subgroups of $G$ and the invertibility of their orders? (I.e., can you recover Connell's theorem from Dunwoody's theorem?)

Second question: now additionally assume $R[G]$ is right self-injective and you don't know Renault's result. Can right self-injectivity be interpreted in this context to explain why $G$ is finite, i.e. $G$ has cohomological dimension $0$? (I.e., can you recover Renault's theorem in the special case of VNR right self-injective rings from Dunwoody's theorem by explaining why injectivity of $R[G]$ decides that $G$ is finite?)

I'm just probing around here seeing if I can get some connection between the two disciplines.

• If $G$ is countable, then the augmentation ideal is clearly countably generated as an $R$-module, but is it necessarily countably generated as an $R[G]$-module? Or does this use the assumption of von Neumann regularity (I'm sorry that I've not yet looked up the definition)? – Mark Grant Apr 5 '18 at 10:54
• @MarkGrant Any set that $R$-generates an $R[G]$ module will $R[G]$-generate a module... Of course, being finitely $R[G]$ generated does not imply finitely $R$ generated, which I think must be what you're thinking of... – rschwieb Apr 5 '18 at 11:10
• I believe that you need the order of each finite subgroup to be invertible in R in your first statement. Also you say at most 1 and the switch to infinite group of cohomological dimension one at the end. Any finite group has cohomological dimension 0 over $\mathbb C$. – Benjamin Steinberg Apr 5 '18 at 15:02
• I’m confused about question 2. To get G to be cohomological dimension $0$ you need the order of $G$ invertible $R$. But $QG$ is always self injective for $G$ finite. – Benjamin Steinberg Apr 5 '18 at 16:20
• Ok. I didn't know you were assuming von Neumann regularity. – Benjamin Steinberg Apr 5 '18 at 16:51

Question 1:

I can `simplify' Connell's argument using Dunwoody-Dicks. Assume that $RG$ is von Neumann regular. Note that if $g\in G$, then $(1-g)r(1-g)=1-g$ for some $r\in R$. So $(1-g)(1- r(1-g))=0$. As $r(1-g)$ is in the augmentation ideal, it follows that $1-r(1-g)\neq 0$ and so $1-g$ is a left zero divisor. This implies $g$ has finite order by standard group ring arguments (in order for $a=ga$ we must have $g$ permutes the finite support of $a$ and hence has finite order). So $G$ is torsion.

If $G$ is countable, it has projective augmentation ideal (since countably generated left ideals in a von Neumann regular ring are projective) and so $G$ acts on a tree with finite stabilizers with order invertible $R$ by Dunwoody-Dicks and because the augmentation ideal is projective. Thus each finitely generated subgroup of $H$ also acts on a tree with finite stabilizers of order invertible in $R$ and we prove that $H$ is finite. But any finitely generated torsion group acing on a tree has a global fixed point (see Serre's book, where it is shown any finitely generated group of elliptic automorphisms has a global fixed point) and so $H$ is finite with order invertible in $R$. Alternatively, a finitely generated group acting on a tree with finite stabilizers has a finite index free subgroup by Bass-Serre theory, which must be trivial in our case since $H$ is a torsion group. So $H$ is finite and hence is a subgroup of a vertex stabilizer so its order is invertible in $R$. Thus $G$ is locally finite.

• Renault doesn't argue with anything obviously cohomological. Everything he did was purely a ring theoretic or group theoretic reduction, backstopped by a lemma which says that the group ring of a Prufer group can't be self-injective. Maybe someday when I understand these things better though, I'll see a hint of the cohomology – rschwieb Apr 5 '18 at 17:18
• I have to think more about question 2. I am use to finite dimensional algebras where every projective module for a self-injective algebra is also injective. – Benjamin Steinberg Apr 5 '18 at 17:26
• I made the answer to question 1 self-contained except I assumed Bass-Serre theory that a finitely generated torsion group acting on a tree has a fixed point or that a finitely generated group acting on a tree with finite vertex stabilizers has a free subgroup of finite index. – Benjamin Steinberg Apr 5 '18 at 17:57
• This proof only handles the coubtable case. – Benjamin Steinberg Apr 5 '18 at 18:50
• The proof that a finitely generated torsion group has a fixed point on the tree is explained pretty well in Section I 6.4 and 6.5 of Serre's tree book and can be read without looking at the rest of the book. The proof that the fundamental group of a finite graph of groups with finite vertex groups has a free subgroup of finite index is Section II 2.6 of Serre. You will also need that whenever a finitely generated group G acts on a tree it has a minimal invariant subtree which has finite quotient. This is on page 20 of Dunwoody-Dicks. – Benjamin Steinberg Apr 5 '18 at 19:37

I am not sure exactly what constitutes an answer to question 1, but here is a proof using the Dunwoody-Dicks stuff that if $G$ is a countable locally finite group, each of whose finite subgroups has order invertible in $R$, then the augmentation ideal of $RG$ is projective (recovering Kaplansky's result without using Connell).

Since $G$ is countable, we can write $G=\bigcup_{n\geq } G_n$ where $\{1\}=G_1\subset G_2\subset \cdots$ is a countable chain of finite subgroups (an infinite chain if $G$ is infinite, which is the interesting case) with orders invertible in $R$.

Let $T$ be the tree with vertex set $\coprod_{n\geq 1}G/G_n$ and we connect $gG_n$ to $gG_{n+1}$ by an edge for $g\in G$. Then $G$ acts on the tree $T$ in an obvious way, and the vertex stabilizers are conjugates of the $G_n$, and hence finite with orders invertible in $R$. Thus $G$ has a projective augmentation module by the result from Dunwoody-Dicks you cited.

The corresponding graph of groups is just a one-sided infinite ray with $G_i$ at vertex $i$ and edge $i$ and the edge groups include in the natural way ($G_i$ is mapped to $G_i$ by the identity and to $G_{i+1}$ by the inclusion) so the expression as a fundamental group of a graph of groups is the direct limit.

Having said that, this is most likely almost the same as Kaplansky's proof (I've never seen it) written in a geometric way. Going from an action on a tree with finite stabilizers of order invertible in $R$ to projective augmentation module is the easy direction. One just needs that the permutation module $R[G/H]$ is projective when $H$ is finite with order invertible in $R$, which is obvious since it is isomorphic to $RGe$ with $e$ the idempotent $e=\frac{1}{|H|}\sum_{h\in H}h$. Then one can use that the augmented simplicial chain complex of a tree is exact.

Going from projective augmentation module to the action on the tree is the hard direction and uses the Almost Stability Theorem, which is a fancy version of Stallings Ends Theorem.

• The goal of question 1 is to recover Connell's theorem on VNR group rings from Dunwoody's theorem (hopefully.) The goal of question 2 is to recover Renault's theorem restricted to right self-injective VNR rings using Dunwoody. I consider the fact the augmentation ideal is projective to be already established by Dunwoody's theorem. So the big question is "why does injectivity of $R[G]$ tip the scales and make $G$ finite?" I'm not sure if this solution has those aims yet, but I appreciate the explanation all the same, so please don't think of omitting anything already written :) – rschwieb Apr 5 '18 at 16:37
• My first comment is that I think part of the challenge of #1 is proving $G$ must be locally finite, but you've used that as an assumption. However, again, I still appreciate the version you've given. Also, thanks for these pointers to the Almost Stability and Stallings Ends Theorems – rschwieb Apr 5 '18 at 16:47
• I think you've deleted the comment already but Kaplansky's proof appears in the paper On the Dimension of Modules and Algebras, X: A Right Hereditary Ring which is not left Hereditary (1958) It's for arbitrary VNR rings, no relation to group rings afaict – rschwieb Apr 5 '18 at 18:47
• Thanks. I saw in Goodearl that countably generated left ideals in vnr rings are projective. – Benjamin Steinberg Apr 5 '18 at 18:49