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Let $G$ be a locally compact Hausdorff (LCH) topological group with left Haar measure $\mu$. Given a compact unit neighborhood $U$, consider the function $$ \Phi: \quad G \to (0,\infty), \quad x \mapsto \mu(U x U). $$ My question is:

Can one give a natural characterization of the groups for which this function is bounded?

One conjecture (see below): This happens exactly for IN groups, which are groups for which there exists a compact unit neighborhood $U$ satisfying $x U x^{-1} = U$ for all $x \in G$.

Thoughts/observations:

  1. The question is independent of the choice of $U$. Indeed, if $U,V$ are both compact unit neighborhoods, then $U \subset \bigcup_{i=1}^n x_i V$ and $U \subset \bigcup_{j=1}^m V y_j$ for suitable $x_i,y_j \in G$, and this easily allows to bound $\mu(U x U)$ in terms of $\mu(V x V)$.

  2. If $G$ is not unimodular, then $\Phi$ is not bounded, since $\Phi(x) \geq \mu(U x) = \Delta(x) \cdot \mu(U)$, so that $\Phi$ is bounded from below (up to a constant) by the modular function, which is unbounded for non-unimodular groups.

  3. If $G$ is an IN group, then $\Phi$ is bounded. Indeed, by the first observation from above we can choose $U$ to satisfy $x U x^{-1} = U$ for all $x$, and then $\Phi(x) = \mu(U x U) = \mu(x U U) = \mu(U U)$ for all $x \in G$.

What I have not been able to show is that if $\Phi$ is bounded, then $G$ needs to be IN. Of course, it could be that this simply does not hold.

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    $\begingroup$ What is your typical unimodular group with unbounded $\Phi$? $\endgroup$ Aug 17, 2021 at 14:34
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    $\begingroup$ @ მამუკა ჯიბლაძე: The Heisenberg group is such an example. I can add the details later today if you are interested. $\endgroup$
    – PhoemueX
    Aug 17, 2021 at 15:22
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    $\begingroup$ I had also checked by hand (in an unsuccessful attempt to produce a counterexample) that the (positive) isometry group of the Euclidean plane has unbounded $\Phi$. $\endgroup$
    – YCor
    Aug 17, 2021 at 17:07
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    $\begingroup$ Beyond understanding its boundedness, I'd be curious about understanding the meaning of the growth of this function $x\mapsto \mu(UxU)$. $\endgroup$
    – YCor
    Aug 17, 2021 at 19:33
  • $\begingroup$ @YCor: I cannot tell you much about the meaning related to the geometry/properties of the group. I got interested in this question through certain properties of so-called (two-sided) Wiener Amalgam spaces, where the norm of $f$ is essentially the $L^1$ norm of the maximal function $M f(x) = \| f\|_{L^\infty (QxQ)}$. I then wanted to understand the norm of $1_{x Q}$, in particular compared to the norm in the one-sided Amalgam spaces. $\endgroup$
    – PhoemueX
    Aug 18, 2021 at 5:09

1 Answer 1

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Your conjecture is correct.

Suppose that we have a compact unit neighbourhood $U$ such that $\mu(UxU) \ll 1$ for all $x$. As you have already noted, we can take $\mu$ to be unimodular, and the choice of neighbourhood is not relevant, so we may assume without loss of generality that $U$ is symmetric: $U^{-1} = U$. (This makes $U$ what is called an approximate group in arithmetic combinatorics, and the intuition that $U$ should behave like a subgroup of $G$ is underlying the arguments below.) We allow implied constants in asymptotic notation to depend on $U$, thus for instance $\mu(U) \asymp 1$.

Note that the conjugate $x U x^{-1}$ of $U$ is commensurate with $U$ in the sense that $\mu( U \cdot x U x^{-1} ) = \mu( U x U ) \ll 1$. (In the language of arithmetic combinatorics, $U$ stays close to its conjugates $xUx^{-1}$ in Ruzsa distance.) In the spirit of the isomorphism theorems (or the Ruzsa covering lemma in arithmetic combinatorics), one now expects $U$ and $xUx^{-1}$ to have large intersection, and this can be accomplished (at the cost of enlarging $U$ to $U^2$) by the following convolution argument (cf. the double counting argument that shows that $|H \cdot K| = |H| |K| / |H \cap K|$ for finite subgroups $H,K$ of $G$). Observe that the convolution $1_U * 1_{xUx^{-1}}$ has an $L^1(G,\mu)$ norm of $\mu(U) \mu(x U x^{-1}) \asymp 1$ and is supported on $U x U x^{-1}$, which has measure $O(1)$. Thus there must exist a point $y \in G$ where $1_U * 1_{xUx^{-1}}(y) \gg 1$, thus $$ \mu( yU \cap x U x^{-1} ) \gg 1$$ which implies $$ \mu( (yU \cap x U x^{-1})^{-1} \cdot (yU \cap x U x^{-1}) ) \gg 1$$ and hence $$ \mu( U^2 \cap x U^2 x^{-1} ) \gg 1.$$ To put it another way, the inner products of the functions $1_{x U^2 x^{-1}}$ with $1_{U^2}$ are uniformly bounded from below.

We can now use an "ergodic" argument to extract an invariant object (in the spirit of the Alaoglu--Birkhoff ergodic theorem). Let $S$ be the closed convex hull in $L^2(G,\mu)$ of the functions $1_{x U^2 x^{-1}}$. By the Hilbert projection theorem, this set has a unique element $f$ of minimal norm. All elements of $S$ have inner product with $1_{U^2}$ uniformly bounded from below, so $f$ does also; in particular, $f$ is non-trivial. Since $S$ is conjugation-invariant, symmetric, bounded and consists of non-negative functions, $f$ must be non-negative, symmetric, and conjugation-invariant.

At this point one could already extract a conjugation-invariant set of positive finite measure by taking level sets of $f$, but this is not quite regular enough for the conjecture, so we take a convolution to achieve an additional smoothing (in the spirit of the Steinhaus theorem). The convolution $f*f$ is then in $C_0(G)$ (this follows from a standard limiting argument, approximating $f$ in $L^2(G,\mu)$ by $C_c(G)$ functions and using Young's inequality), conjugation-invariant, and strictly positive at the origin; taking level sets, we obtain a non-trivial conjugation-invariant compact unit neighbourhood, as required.

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  • $\begingroup$ This is really incredible! Thanks! I still need to check the fine print and understand parts of the intuitive explanations that you provide, but I could follow the main argument. One thing you did not stress: the uniqueness statement of the Riesz projection theorem is what allows to deduce the conjugation invariance, right? $\endgroup$
    – PhoemueX
    Aug 17, 2021 at 18:16
  • $\begingroup$ Yes, this is correct; I have now added a mention of uniqueness to the post. $\endgroup$
    – Terry Tao
    Aug 17, 2021 at 19:24

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