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May 10, 2012 at 18:17 answer added Alexander Chervov timeline score: 3
Dec 16, 2011 at 21:32 answer added Alexander Chervov timeline score: 2
Dec 16, 2011 at 2:57 answer added David E Speyer timeline score: 5
Dec 15, 2011 at 3:52 answer added Igor Makhlin timeline score: 13
Dec 11, 2011 at 15:55 comment added Igor Makhlin Well, actually I have been reading this article simply with the purpose of educating myself on the subject of representation theory, which (I guess) is my main field of interest. But there's a lot of combinatorics to it, yes.
Dec 11, 2011 at 15:44 vote accept Igor Makhlin
Dec 11, 2011 at 15:03 comment added darij grinberg Oh hi! So we meet again. I've just started studying for a PhD at MIT. As you see by the comment I'm doing some algebraic combinatorics, at least as a pastime. Are you, too, or do you need this for some kind of infinite symmetric groups / probability theory?
Dec 11, 2011 at 2:05 history edited Ben Webster CC BY-SA 3.0
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Dec 11, 2011 at 1:41 answer added Florian Eisele timeline score: 13
Dec 10, 2011 at 23:23 comment added Igor Makhlin I'm sorry for the off topic, but I had to say this. Don't know if you remember, but I know you, me and my twin brother Anton visited you at your home with our parents some 8-10 years ago when we (and you) lived in Karlsruhe. I'm in my 5th year at MSU now =) Quite a coincidence that you were the first person to respond to my first question here.
Dec 10, 2011 at 23:00 comment added darij grinberg I myself have asked this question to several people to no avail. Not only are the eigenvalues integers; we also have $\prod\limits_{i=-n+1}^{n-1}\left(X-i\right)=0$, where $X=\left(1,n\right)+\left(2,n\right)+...+\left(n-1,n\right)\in\mathbb Z\left[S_n\right]$ is the $n$-th YJM element. I am sure this has a combinatorial proof, probably even a smart elementary induction one - but I had no success whatsoever in finding one over several weeks.
Dec 10, 2011 at 22:53 history asked Igor Makhlin CC BY-SA 3.0