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Mar 15, 2016 at 8:23 comment added Jay Taylor @user61318 Well it was useful because I wouldn't have thought about the general case without it, so thanks for sharing it! For clarity, I should say that we are talking about the paper here arxiv.org/abs/1603.03914 which uses the techniques in the answer below to give a new proof of the results in the above mentioned paper of Regev.
Mar 15, 2016 at 1:07 comment added user35313 @JayTaylor: Funny I thought the Regev paper would be useful. Instead you ended up deriving results therein. Nice!
Mar 8, 2016 at 12:38 vote accept Marcel
Mar 8, 2016 at 9:40 answer added Jay Taylor timeline score: 5
Mar 7, 2016 at 20:15 comment added Marcel @JayTaylor Funny, huh? So the proof for $m> 0$ cannot rely on the $m=0$ case; it has to be a different proof...
Mar 7, 2016 at 19:48 comment added Jay Taylor Your formula is independent of $\mu$ so long as $m > 0$. I think there is a difference when $m=0$ due to the way that hooks are computed in the skew diagram versus a proper Young diagram. For instance a hook Young diagram of a partition of $n$ has a hook of length $n$ but a hook skew diagram whose shape is a partition of $n$ has no hook of length $n$. The Murnaghan–Nakayama formula indicates that these situations are therefore different.
Mar 7, 2016 at 19:04 history edited Marcel CC BY-SA 3.0
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Mar 7, 2016 at 17:05 comment added user35313 Hooks and characters... the following might be useful: arxiv.org/pdf/1108.3170.pdf
Mar 7, 2016 at 16:27 comment added Jay Taylor Yeah, sorry, I forgot about the sum.
Mar 7, 2016 at 13:38 history asked Marcel CC BY-SA 3.0