Intuition behind the failure of unimodularity

Question

If $G$ is a locally compact group then up to normalization it admits a unique Haar measure: a left invariant measure defined on all Borel subsets of $G$, which assigns every compact set a finite measure. The measure can be chosen to be right invariant, but usually not both simultaneously: if a locally compact group admits a measure which is both left and right invariant then it is said to be unimodular. Examples of unimodular groups are compact groups and abelian groups.

I am interested in an intuitive characterization of unimodularity. The following will do:

1. Something in the vein of "simply connected" = "no holes" or "compact" = "bounded and with no holes"
2. An example of a group that fails to be unimodular with a convincing argument that this is the case, but without direct computations. I know the example of orientation preserving affine transformations on the real line, but the proof that this group is not unimodular uses the fact that this group is homeomorphic to a manifold. I would like to see an example that's more conceptual, such as a proof for the automorphism group of a locally finite graph.

Answer 1

Here is an example involving automorphisms of locally finite graphs. Let $T$ be the infinite $r$-regular for $r \geq 3$ tree. Let $(v_0, v_1, v_2, \ldots)$ be some infinite non-backtracking path through it. Let $K$ be the group of all symmetries $g$ of $T$ so that $g(v_i)=v_i$. Let $G$ be the group of all symmetries $g$ of $T$ for which there exist $N \geq 0$ and $d \in \mathbb{Z}$ so that $g(v_i) = v_{i+d}$ for $i \geq N$, topologized in the usual way.

I now sketch proofs of three things:

$K$ is compact Let $U(i,d)$ be the set of all vertices of $T$ at distance $\leq d$ from $v_i$. Then $T = \bigcup_{i \geq 0, d \geq 0} U(i,d)$ and $K$ takes each $U(i,d)$ to itself, so $K$ embeds into the product of symmetric group $\prod_{i,d} S(U(i,d))$. With a little thought you can see that $K$ is closed in the product topology, so $K$ is compact. $\square$

$K$ is an open neighborhood of the identity in $G$ Define a continuous homomorphism $G \to \mathbb{Z}$ by $g \mapsto d$ where $g(v_i) = v_{i+d}$ for $d$ sufficiently large. Let $G_0$ be the kernel; since $\mathbb{Z}$ is discrete, $G_0$ is an open neighborhood of the identity in $G$. Now let $L$ be the set of all vertices $x$ of $T$ so that $x$ is at distance $i$ from $v_i$ for some $i$. We have a map $\phi: G_0 \to L$ sending $g$ to $g(v_0)$, and this is continuous for the discrete topology on $L$. So $\phi^{-1}(v_0)$ is an open neighborhood of the identity in $G_0$, and we claim $\phi^{-1}(v_0)=K$. By definition, if $g \in K$, then $g(v_0)=v_0$. Conversely, if $g \in G_0$ and $g(v_0)=v_0$, then $g$ fixes $(v_0, v_i, v_{i+1}, v_{i+2}, \ldots, )$ for some $i$. But then it also fixes all points on the path from $v_0$ to $v_i$, so $g$ is in $K$. $\square$

$G$ is not unimodular Suppose there were a left and right invariant measure $\mu$. Since $K$ is open and compact, we would have $0 < \mu(K) < \infty$. Now, let $g \in G$ be some map with $g(v_i) = v_{i+1}$ for all $i$. If $\mu$ were left and right invariant, we would have $\mu(g K g^{-1}) = \mu(K)$. But it is easy to see that $g K g^{-1}$ is the set of symmetries of the tree which fix $(v_1, v_2, \ldots, )$. So $K$, being those symmetries which also fix $v_0$, is an index $r-1$ subgroup of $g K g^{-1}$, so $\mu(g K g^{-1}) = (r-1) \mu(K)$.

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Here is an example which interpolates between the above one and the more standard $\mathbb{R}^{\times} \ltimes \mathbb{R}$. Look at $G:= \mathbb{Q}_p^{\times} \ltimes \mathbb{Q}_p$. The standard computation with matrices (see my other answer) shows that conjugation by $\left( \begin{smallmatrix} p & 1 \\ 0 & 1 \end{smallmatrix} \right)$ dilates the tangent space by $p$, so there can't be a Haar measure. I can make this really explicit in terms of neighborhoods of the identity: If $K = \left( \begin{smallmatrix} \mathbb{Z}_p^{\times} & \mathbb{Z}_p \\ 0 & 1 \end{smallmatrix} \right)$ and $g = \left( \begin{smallmatrix} p & 1 \\ 0 & 1 \end{smallmatrix} \right)$ , then $g K g^{-1} = \left( \begin{smallmatrix} \mathbb{Z}_p^{\times} & p \mathbb{Z}_p \\ 0 & 1 \end{smallmatrix} \right)$. So $gK g^{-1}$ is index $p$ in $K$, and we can't have $\mu(g K g^{-1}) = \mu(K)$.

I found the above example by taking the action of this one on the Bruhat-Tits tree, and abstracting away all properties couldn't be stated using trees.

Answer 2

This is probably the explanation you don't like, since it involves using the Lie structure, but it seems incredibly simple to me and I don't see what you could ask for that is simpler.

Let $G$ be a Lie group. $G$ acts on itself by conjugation, fixing the identity, and thus $G$ acts on $T_e(G)$, the tangent space at the identity. Write $Ad(g)$ for the action of $g \in G$ on $T_e(G)$. The group is unimodular if and only if $\det(g)= \pm 1$ for all $g \in G$.

Explanation On a Lie group, a measure is basically a volume form. (There are some subtleties about sign, but let's ignore them for now.) If we choose a volume form $\omega$ on the tangent space to the identity (known as $\mathfrak{g}$), then we get a volume form on $T_g(G)$ by $(Lg)_{\ast} \omega$ where $Lg$ is left multiplication by $g$. The form on $G$ which is $(Lg)_{\ast} \omega$ on $g$ is left invariant, so it must be the Haar measure. (Again, up to sign nuisances due to measures versus volume forms.) Simultaneously, we get a right invariant volume form by $(Rg)_{\ast} \omega$. So we want to know whether $(Lg)_{\ast} \omega = (Rg)_{\ast} \omega$. Equivalent, we want $(Rg^{-1})_{\ast} (Lg)_{\ast} \omega = \omega$. The left hand side is precisely the action of conjugation by $g$ on volume forms at the identity. Explicitly, $(R g^{-1})_{\ast} (Lg)_{\ast} \omega = \det Ad(g) \omega$.

Unwinding the issues about signs, we actually want $\det Ad(g) = \pm 1$.

Example Let $G$ be the group $\left( \begin{smallmatrix} u & v \\ 0 & 1 \end{smallmatrix} \right)$, where $u$ is a nonzero real and $v$ is any real. Then $$ \begin{pmatrix} u & v \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1+x & y \\ 0 & 1 \end{pmatrix} \begin{pmatrix} u & v \\ 0 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1+x & -vx + uy \\ 0 & 1 \end{pmatrix}.$$ So $G$ acts on the tangent space by $$ \begin{pmatrix} u & v \\ 0 & 1 \end{pmatrix} : \begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x \\ -vx + uy \end{pmatrix}$$ which is described by the matrix $$\begin{pmatrix} 1 & 0 \\ -v & u \end{pmatrix}.$$ Since the determinant is $u \neq 1$, the group is not unimodular.

Technical note Usually, we pull back differential forms, not push them forward. If $h$ is an automorphism of a manifold $X$, I write $h_{\ast}$ for $(h^{-1})^{\ast}$; this reduces the number of inverses I need, and I find it leads to a cleaner presentation.

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Here is a way to make the above more concrete and geometric. Let $S$ be the square $$\left\{ \begin{pmatrix} 1+x & y \\ 0 & 1 \end{pmatrix} : |x|, |y| \leq 1/2 \right\}$$ and let $p$ be an odd number. If I didn't screw up the details, then $$\bigcup_{i=-(p-1)/2}^{(p-1)/2} \ \begin{pmatrix} \frac{1}{p} & \frac{i}{p} \\ 0 & 1 \end{pmatrix}\cdot S \cdot \begin{pmatrix} \vphantom{\frac{1}{p}} p & 0 \\ 0 & 1 \end{pmatrix} \ = \ \bigcup_{i=-(p-1)/2}^{(p-1)/2} \left\{ \left( x, \frac{y+i}{p} \right) : (x,y) \in S \right\} = S.$$

So, if we had a left and right invariant measure on $G$, we would have $\mu(S) = p \mu(S)$ and thus $\mu(S) = 0$ or $\infty$. But $S$ is compact and contains an open set, so it should have positive finite measure.