My friend here in Goettingen, Daniel Pape, has told me about his very short paper in which he proves an interesting lemma about Von Neumann dimension. AFAIU it was earlier proved by Gabor Elek in a bit less general context. The lemma concerns discrete groups and it has immediate interesting applications for amenable groups.


Let G be a discrete group, let \theta be some element of \mathbb{C}G, and let F be some finite subset of G. Furthermore, let l^2 G denote a Hilbert space of square-integrable function on G. We are interested in \dim_G (\ker\theta ), where \theta is now seen as an operator on l^2 G, and dim_G is the Von Neumann dimension defined as follows: given V\subset l^2 G, take orthogonal projection P onto V and put

dim_G V: = (P(e),e),

where e is a neutral element of G and (\cdot,\cdot) is a standard scalar product on l^2 G.

Computing \dim_G (\ker\theta ) is rather difficult. The lemma relates \dim_G to a (much more easily computable) standard linear dimensions of certain finite-dimensional spaces.


Before giving the lemma, few words on the Von Neumann dimension. It is an interesting number mostly for G-equivariant spaces (right equivariant, because we assume that elements of the group act on the left, and images of such operators are right equivariant).

Suppose for a second that G = \{g_1,\ldots, g_n\} is a finite group. What is a (standard) dimension of a given V\subset l^2 G? To compute it you can take a projection P onto V and take the trace of this projection, so the (standard) dimension of V is

\dim V: = \text{tr} P = \sum_{i=1}^n (Pg_i,g_i).

But when V is right invariant then so is projection P (these statements are equivalent) and so every summand above is equal to (Pe,e). Therefore we get that

\dim_G V := \frac{\dim V}{|G|},

or in other words: dim_G is just a normalized dimension (normalized in such a way so that \dim_G l^2 G = 1).

The point is the following: in the case when G is finite, the formula gives a normalized dimension. However, in the case of infinite G the formula also makes sense, although the “normalized dimension” is \frac{\infty}{\infty}!

So the Von Neumann dimension is potentially useful in situations where we want to say how big certain infinitely dimensional equivariant set is, when taken into account the group symmetries.

As an example consider G=Z, the group of integers. By Fourier transform we have that l^2 Z = l^2 S^1, and the action of Z on the latter is realized by a pointwise multiplication by the function f(z)=z. Example of an equivariant subspace of l^2 S^1 is a subspace V of functions which have the support contained in some set X. What is \dim_G V? It is very easy to see that it is a measure of the set X.


One of the big conjectures in the geometric group theory is, however, that when the group is torsion free then \dim_G (\ker \theta) is 0. More generally, one considers not only \mathbb{C}G acting on l^2 G, but also M_n(\mathbb{C}G) acting on (l^2 G)^n. In this case the Von Neumann dimension is defined as a sum of standard von Neumann dimensions of diagonal elements of a matrix, and the conjecture says that (for a torsion-free G) so defined dimension of the kernel of an element of M_n(\mathbb{C}G) is an integer.

Similarly, everything I write further can also be generalized to matrices over a group ring in a straightforward way, but to simplify the notation I will consider only \mathbb{C}G.


One more definition: when F =\{f_1,\ldots,f_n\} is a finite subset of G then we define dim_F of a given subspace V of l^2 G to be

\dim_F V := \frac{1}{|F|}\sum_{i=1}^n (Pf_i,f_i),

where P is, as usual, an orthogonal projection onto V.

In general it is rather difficult to compute \dim_G \ker\theta because it involves a projection onto potentially infinite dimensional space. However, it turns out that sometimes we can hope to approximate this situation by a finite dimensional one. This is what the lemma is about.

Lemma: Let G be a discrete group, \theta be an element of \mathbb{C}G, and F be a finite subset of G. Then we have that

0 \le \dim_G \ker \theta - \dim_F \ker \theta_F \le \frac{|\partial_\theta F|}{|F|},

where \partial_\theta Fis a subset of F consisting of those element which are mapped by \theta outside of \text{span} ( F ); and \theta_F is an operator \theta restricted to \text{span} (F-\partial_\theta F)

Proof: First ingredient is the following standard equality:

\dim_G \ker \theta + \dim_G \text{im } \theta = 1,

where \text{im } denotes the closed image. I don’t know how to prove it, but you can find it in Wolfgan Lueck’s book on L^2-invariants. The second ingredient is another equality:

\dim_F \ker \theta_F + \dim_F \text{im } \theta_F = 1 - \frac{|\partial_\theta F|}{|F|},

which is just a standard linear algebra of finitely dimensional spaces.

Now you plug the first equation to the second one and use the following simple observation:

\dim_F\text{im }\theta_F \le \dim_G\text{im }\theta,

which you prove by noticing that \dim_F V \le \dim_F W for V\subset W (this alone proves the first inequality), and that \dim_F V = \dim_G V for a G-equivariant V. This gives the second inequality.


The lemma is valid in a general context of discrete groups. However, it’s clearly “designed” to work well for amenable groups. In the case of amenable groups, for every \theta there exists a finite set F such that \frac{|\partial_\theta F|}{|F|} is arbitrarily small. Therefore for amenable groups we get the following corollary:

Cory: Let G be amenable, and let \theta\in \mathbb{C}G. Then the following are equivalent:

  • there exists an element x\in L^2 G such that \theta x =0
  • there exists an element x\in \mathbb{C}G such that \theta x = 0.

This follows from the above lemma and the fact that for G-equivariant space V, \dim_G V is 0 iff V = \{0\}. For general discrete groups above cory is an open conjecture (maybe one should restrict atttention to torsion free groups).

In as similar (slightly more involved) fashion one gets so called strong approximation conjecture for amenable groups:

Cory: Let \{G_i\}_{i\in I} be a directed system of normal subgroups of G, such that \cap G_i = \{e\}. Then

\dim_G \theta = \lim \dim_{G/G_i}\theta.


Two things catch my attention: does the condition
For every \theta\in\mathbb{C}G, and \epsilon>0 there exists a finite set F such that \frac{|\partial_\theta F|}{|F|} < 0 characterize amenable groups? Most likely so, but I have to think about it a trifle more.

And second, can you take some more general subspace of l^2G instead of \text{span}F? The main problem here seems to be the equality
\dim_F V = \dim_G V, which no longer holds (after suitable deinifition od \dim_F as a normalized trace of a projection from F to V) in general.