Hello,
if $G$ is a compact Lie group. Let $\mu$ be an infinitely divisible measure on $G$, such that $e$, the neutral element of $G$, is in the support of $\mu$. Is that true that the support of $\mu $ is a group ? Why ?
Thanks a lot.
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$\begingroup$ Why do you ask? $\endgroup$– Anthony QuasCommented Dec 18, 2012 at 16:56
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1$\begingroup$ Let's understand what "infinitely divisible" means when $G$ is not abelian . If $n$ is a positive integer, then there exist indpendent identically distributed random elements $Y_1,\dots,Y_n$ of $G$ such that $Y_1\cdot \dots\cdot Y_n$ has distribution $\mu$. $\endgroup$– Gerald EdgarCommented Dec 18, 2012 at 18:15
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3 Answers
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Have you tried to look up
Heyer, Herbert, Probability measures on locally compact groups. Ergebnisse der Mathematik und ihrer Grenzgebiete. 94. Berlin-Heidelberg-New York: Springer-Verlag. (1977)?
I don't have access to this book right now, but I expect that you can find the answer there.
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Actually Martin the support in your exemple would be the closure of $\{\exp(ix), x\in \mathbb{N}\}$, which is equal to $U(1)$, which is a group.
In a finite group the proof would be just that if $x$ is in the support of $\mu$, looking at $\mu$ like more or less the measure of $X_1$ with $X$ a Levy process going from $e$, then it is obvious that we can "speed" up the jumps and so $x$ is in $X_{\frac{1}{2}}$, and so as the support of $X_1$ is equal to the closure of the product of the support of $X_{\frac{1}{2}}$ with himself then $\text{Support}(X_1)$ is stable by multiplication. Thus it is a group as a subset of a finite group which is stable by multiplication is a group. But in a Lie group can we "speed up" the jumps ? If $X$ is a Levy process beginning from $e$ having at any time $e$ in his support, is it true that : $Support(X_{1}) \subset Support(X_{\frac{1}{2}})$ ?
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Hello, Let consider a pure jump process on the unitary group on the complex plane : the jumps $exp(i)$ arrive randomly according to a Poisson process ; the support is composed of $exp(ix)$ with $x\in \mathbb{N}$. Is it a counter-example or I forget something ?
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1$\begingroup$ I believe "support" should be a closed set, so you should take the closure of your countable set. Does it include also $\exp(ix)$ with $x<0$? $\endgroup$ Commented Dec 19, 2012 at 14:18