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.

  • $\begingroup$ 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 '12 at 19:34
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    $\begingroup$ @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 '12 at 19:41

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.

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    $\begingroup$ 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 '12 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 '12 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 '12 at 5:10

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