2
$\begingroup$

Let $G$ be a locally compact group, $K$ a compact subgroup in $G$, and $\mu_K$ the normalized Haar measure on $K$: $$ \mu_K(K)=1. $$ Let us denote by $\widetilde{\mu_K}$ the measure on $G$ defined as the functional on ${\mathcal C}(G)$ by the formula $$ \widetilde{\mu_K}(u)=\int_K u(t) \ \mu_K(dt), \qquad u\in{\mathcal C}(G) $$ Suppose now that $K$ and $L$ are two normal compact subgroups in $G$, and $H$ is the closed normal subgroup in $G$ generated by $K$ and $L$: $$ H=\operatorname{Gr}(K\cup L) $$ I have two questions:

  1. Is it true that $H$ is always compact?

and

  1. Suppose $H$ is compact, does the following equality hold: $$ \widetilde{\mu_K}*\widetilde{\mu_L}=\widetilde{\mu_H} $$ ?
$\endgroup$
2
  • 2
    $\begingroup$ Can you do 1. in case of discrete groups? If two normal subgroups $K,L$ are finite, then the group generated by $K \cup L$ is finite ?? $\endgroup$ Apr 10 at 10:14
  • $\begingroup$ @GeraldEdgar you know my imagination only suggests that if $G$ is generated by two elements $a$ and $b$ with the identities $a^2=1$ and $b^2=1$, then we can take $K=\{1,a\}$ and $L=\{1,b\}$, and the answer will be "no" if we remove the requirement that $K$ and $L$ are normal. It seems to me the answer anyway is "no", but my intuition is not good here. $\endgroup$ Apr 10 at 10:27

1 Answer 1

1
$\begingroup$

The answer to both questions seems yes to me.

First, only assume that $K$ is a normal subgroup of $G$. Then, $(\text{Ad } g)_{g \in L}$ defines a continuous action of $L$ by automorphisms of $K$ and we can define the semidirect product group $K \rtimes L$, which is just the set $K \times L$ with product $$(x,g) \cdot (y,h) = (x (gyg^{-1}),gh) \; .$$ Then $K \times L$ is a compact group. Because the Haar measure $\mu_K$ of $K$ is invariant under every continuous automorphism of $K$, we get that $\mu_K \times \mu_L$ is the Haar measure of $K \times L$.

The product map $\theta : K \times L \to G : \theta(x,g) = xg$ is a continuous group homomorphism. The image $\theta(K \times L)$ is compact and equals the group considered in the question. The Haar measure on this group is given by $\theta_*(\mu_K \times \mu_L)$, which is equal to $\widetilde{\mu_K} * \widetilde{\mu_L}$.

Note that $\theta(K \times L) = K \cdot L$ equals the subgroup of $G$ generated by $K$ and $L$.

Finally, if both $K$ and $L$ are normal in $G$, then for every $g \in G$, we have that $$g \cdot (K \cdot L) \cdot g^{-1} = (gKg^{-1}) \cdot (gLg^{-1}) = K \cdot L$$ so that $K \cdot L$ is normal in $G$. This means that $K \cdot L$ equals the group $H$ in the question.

$\endgroup$
7
  • $\begingroup$ Stefaan, I don't understand, where is $H$ in this reasoning? $\endgroup$ Apr 10 at 12:51
  • $\begingroup$ I have updated the answer to make this more clear. $\endgroup$ Apr 10 at 13:06
  • $\begingroup$ Ah, yes, if $K$ and $L$ are normal subgroups, then $K\cdot L$ is a normal subgroup... I was wondering, why I couldn't prove non-compactness... :} $\endgroup$ Apr 10 at 13:28
  • $\begingroup$ Stefaan, OK, this answers the first question, but your proof of the second one looks complicated. I need some time to analyse this. $\endgroup$ Apr 10 at 13:33
  • $\begingroup$ Stefaan, yes, actually, this obsevration, that $H=K\cdot L$ implies everything immediately: first, for each $x\in K$ and $y\in L$ we have $$ (\widetilde{\mu_K}*\widetilde{\mu_L})*\delta^{x\cdot y}=\widetilde{\mu_K}*\widetilde{\mu_L}*\delta^x*\delta^y=\widetilde{\mu_K}*\delta^x*\widetilde{\mu_L}*\delta^y=\widetilde{\mu_K}*\widetilde{\mu_L} $$ - and this means that $\widetilde{\mu_K}*\widetilde{\mu_L}$ is the Haar measure on $K\cdot L$, and second, $$ (\widetilde{\mu_K}*\widetilde{\mu_L})(1)=1 $$ - and this means that it is normalized. $\endgroup$ Apr 10 at 14:01

Your Answer

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

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