Support of an infinitely divisible measure. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:38:08Z http://mathoverflow.net/feeds/question/116713 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116713/support-of-an-infinitely-divisible-measure Support of an infinitely divisible measure. Gabriel 2012-12-18T15:03:18Z 2013-01-03T03:18:20Z <p>Hello, </p> <p>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 ? </p> <p>Thanks a lot. </p> http://mathoverflow.net/questions/116713/support-of-an-infinitely-divisible-measure/116787#116787 Answer by Martin for Support of an infinitely divisible measure. Martin 2012-12-19T13:40:57Z 2012-12-19T13:40:57Z <p>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 ?</p> http://mathoverflow.net/questions/116713/support-of-an-infinitely-divisible-measure/116876#116876 Answer by Gabriel for Support of an infinitely divisible measure. Gabriel 2012-12-20T15:55:58Z 2012-12-20T15:55:58Z <p>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. </p> <p>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 : $Support(X_{1}) \subset Support(X_{\frac{1}{2}})$ ? </p> http://mathoverflow.net/questions/116713/support-of-an-infinitely-divisible-measure/117925#117925 Answer by Uwe Franz for Support of an infinitely divisible measure. Uwe Franz 2013-01-03T03:18:20Z 2013-01-03T03:18:20Z <p>Have you tried to look up</p> <p>Heyer, Herbert, Probability measures on locally compact groups. Ergebnisse der Mathematik und ihrer Grenzgebiete. 94. Berlin-Heidelberg-New York: Springer-Verlag. (1977)?</p> <p>I don't have access to this book right now, but I expect that you can find the answer there.</p>