Is there a nonunimodular group for which Wiener's Tauberian theorem is true?
Is a locally compact topological group whose volume grows polynomially with radius always unimodular?
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$\begingroup$ To provide extra context, could you please state the version of Wiener's Tauberian theorem that you mean in a non-abelian context? (The classical notion for $L^1$ group algebras is very reliant on the group being abelian; perhaps you have in mind one of the versions in terms of the Fourier algebra?) $\endgroup$– Yemon ChoiCommented Jan 4, 2012 at 15:04
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$\begingroup$ Note also that all discrete groups are unimodular, even the Type II ones with nasty unitary duals; so the classical version of WTT in terms of the Fourier transform of functions on the group will be problematic. $\endgroup$– Yemon ChoiCommented Jan 4, 2012 at 15:10
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1$\begingroup$ By WTT I meant the classical version. When the group $G$ is nonabelian, we state WTT as: if for $f\in L^1(G)$, the Fourier transform of $f$ is nonvanishing on every point of the unitary dual then the (both-sided) ideal generated by $f$ is dense in $L^1(G)$. Both-sided ideal can be changed to linear span of both-sided translations also. Even if the group is non-unimodular, $L^1(G)$ is a Banach algebra and invariant under left and right translations, though norm of $f$ and that of its right translation change. Am I right? I am not aware of other version of WTT. What is a good source? $\endgroup$– sprCommented Jan 5, 2012 at 6:30
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$\begingroup$ The ax+b group is non-unimodular. But perhaps WTT is true here. Is there other examples? $\endgroup$– sprCommented Jan 5, 2012 at 9:41
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1$\begingroup$ Thank you for clarifying. It seems that Yulia Kuznetsova has answered your original question. If you want to find out about the Fourier algebra formulation of Wiener's theorem, I recommend the book of Reiter and Stegemann, which is itself based on some older notes/book of Reiter. $\endgroup$– Yemon ChoiCommented Jan 5, 2012 at 18:03
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2 Answers
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The answer to the second question is yes. Let us show that a non-unimodular, locally compact group $G$ cannot have polynomial growth. Let $\mu$ be left Haar measure, $\Delta$ be the modular function, so that $\mu(Ag)=\mu(A)\Delta(g)$ for $A$ a Borel subset in $G$. Now take for $A$ a compact neighborhood of identity, and $g\in G$ such that $\Delta(g)>1$. Then $Ag^n\subset (Ag)^n$, so $\mu(A)\Delta(g)^n\leq\mu((Ag)^n)$, hence if $\Omega$ is a compact neighborhood of identity containing $Ag$, we have $\mu(A)\Delta(g)^n\leq\mu(\Omega^n)$, and the sequence $(\mu(\Omega^n))_{n\geq 1}$ has exponential growth.
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To the first is yes also. The example is already given above: ax+b group. It is the semidirect product $\mathbb R^\times \ltimes \mathbb R$. Leptin has proved in Leptin, H., Ideal theory in group algebras of locally compact groups. Invent.Math. 31 (1975/76), no. 3, 259-278 that every semidirect product of separable abelian l.c. groups has the Wiener property (in the sense that every proper closed two-sided ideal is annihilated by a nonzero *-representation).
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$\begingroup$ I wanted to accept both yours and Prof Valette's answers as "answers". But is it true that I can accept only one answer? If I click the right-symbol in yours, it goes away from Valette's. $\endgroup$– sprCommented Jan 6, 2012 at 12:00
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1$\begingroup$ I don't know, I am new here. I'm happy my answer helped. $\endgroup$ Commented Jan 6, 2012 at 13:22