8
$\begingroup$

Let $(G,+)$ be a locally compact (abelian) group endowed with its natural Haar measure. The doubling-map $T_2 : g \in G \mapsto g + g $ defines an endomorphism of $G$.

Is there some (natural) condition on $G$ which ensures that the map $T_2$ is measure-preserving?

When $G$ is finite, it is of course the same as asking $G$ to be of odd order (i.e. that there's no $2$-torsion). But $T_2$ may fail even when $G$ has no $2$-torsion, as shows $G = \mathbb{R}$. The group of $2$-adic integers $\mathbb{Z}_2$ (with its natural topology) shows that lack of 2-torsion and compactness are not sufficient either.

It is well-known that $T_2$ is measure-preserving when $G = \mathbb{T} = \mathbb{R}/ \mathbb{Z}$, or even when $G = \mathbb{T}^d$ (despite the $2$-torsion). I believe this is still true for any compact connected (abelian) group (I think connectedness would prevent pathologies such as $\mathbb{Z}_2$) but I have no convincing argument so far.

Improve this question
$\endgroup$
2
  • $\begingroup$ How is it measure-preserving for $\mathbb T$? More generally, for any Lie group $T_2$ increases Haar measure on small neighbourhoods of the identity. $\endgroup$
    – Robert Israel
    Feb 27, 2012 at 19:34
  • 2
    $\begingroup$ @Robert: to preserve measure means that $\lambda(T_2^{-1}(A))=\lambda(A)$ for all Borel $A$. This holds in the circle (and thus it's not a local property). $\endgroup$
    – YCor
    Feb 27, 2012 at 19:41

1 Answer 1

Reset to default
6
$\begingroup$

For an endomorphism $\rho$ to preserve Haar measure on a compact group $G$, it is necessary and sufficient that $\rho$ be surjective. One can easily check that the image $\rho^*(m)$ of Haar measure $m$ under $\rho$ is a translation-invariant probability measure as long as $\rho$ is surjective. If $\rho$ is not surjective, the image $H=\rho(G)$ is a compact subgroup, either of positive finite index or infinite index. In either case, $\rho^{-1}(H)$ will not have the same Haar measure as $H.$

When $G$ is compact and connected, the doubling map is surjective (one way to see this is to observe that the dual $\widehat{G}$ is torsion-free, so the doubling map induces an injective map on the dual).

More generally, if the dual of the compact group $G$ lacks $2$-torsion, then the doubling map will be surjective, and so will preserve Haar measure.

Edit: The following argument and conclusion are in error, as Yves' comment shows.

The structure theorem for locally compact abelian groups reduces the general question to two cases: (1) $G$ is compact, and (2) $G$ is discrete. Fix a compact group $G$ where the doubling map preserves Haar measure, and let $\rho:G\to G$ be the doubling map. By the structure theorem, $G$ has an open subgroup $W$ isomorphic to $K \times \mathbb R^n$ for some nonnegative integer $n,$ where $K$ is a compact group. Since the doubling map does not preserve Haar measure on $\mathbb R$, we must have $n=0.$ If $\rho$ preserves Haar measure on $G$ then the induced map $\tilde{\rho}:G/W\to G/W$ preserves Haar measure on $G/W,$ so $\tilde{\rho}$ is injective.

Since $\tilde{\rho}$ is injective, we have $\rho^{-1}(W)=W.$ Thus $\rho(W)$ is a subgroup of $W$, and by the argument in the first paragraph, we necessarily have $\rho(W)=W.$

We have (edit: not) shown:

The doubling map on G preserves Haar measure if and only if $G$ has a compact open subgroup $W$ such that:

(i) The doubling map on $G/W$ is injective (i.e. $G/W$ lacks 2-torsion).

(ii) $\widehat{W}$ lacks $2$-torsion.

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ Your characterization is correct as long as compact groups are concerned. However, it is not correct in general: if $P_2$ is the 2-Prüfer group (inductive limit of Z/2^nZ) , and $R$ is the group of reals, then the doubling map on $R\times C$ preserves the measure. Other example: $R\times Q_2$. The general answer is: those groups $R^k\times H$, where $H$ is a group with an open compact subgroup so that the doubling of $H$ multiplies the measure by $2^k$. It remains to characterize those groups. $\endgroup$
    – YCor
    Feb 27, 2012 at 22:43
  • $\begingroup$ In the first paragraph, is compactness of G necessary to ensure that $\rho^{\star}(m)$ is proportional to $m$ when $\rho$ is surjective ? If not, the above comment shows that it is sufficient to study surjectivity of $\rho$ (which is in practice easier), and then to check the value of the multiplicative constant between $\rho^{\star} (m)$ and $m$ on a compact subgroup. I need to think about it. $\endgroup$
    – js21
    Feb 28, 2012 at 9:47
  • $\begingroup$ A LCA group in which $\rho$ is a quotient map should be characterized as with a group with a nested sequence of closed subgroups $N_1\le N_2\le N_3$ with $N_1$ connected (i.e. the direct product of $\mathbb{R}^k$ with a compact connected group), $N_2/N_1=\mathbf{Q}_2^n/\mathbf{Z}_2^m$ for some finite $m\le n$, $N_3/N_2$ product of pro-$p$-groups for odd $p$, $N/N_3$ a discrete group in which $\rho$ is bijective. In such a group, multiplication by 2 multiplies the measure by $2^{n-k}$. (Note that $N_1$ and $N_2$ are characteristic.) $\endgroup$
    – YCor
    Feb 29, 2012 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.