Timeline for Number of permutations in $S_{a+b}$ with $\operatorname{maj}(\pi)=a$ and $\operatorname{maj}(\pi^{-1})=b$
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2021 at 21:23 | comment | added | Per Alexandersson | The Cauchy identity can be proved using RSK, and the Garsia-Gessel reference probably does basically the combinatorics you are after. | |
| Feb 22, 2021 at 19:31 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| Feb 22, 2021 at 19:31 | comment | added | Sam Hopkins | @DavidESpeyer: the $\mu[n]$ construction is very common in representation stability; maybe these computations have some meaning in that context. | |
| Feb 22, 2021 at 19:22 | comment | added | David E Speyer | You're welcome. I actually think there is more going on here; see the second proof I added below the horizontal line. | |
| Feb 22, 2021 at 19:22 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| Feb 22, 2021 at 17:53 | comment | added | Per Alexandersson | Great answer and comment, SamHopkins and @DavidESpeyer ! This now gives a nice record of this identity! I need this for some background in a paper, so I'll probably include this argument, with proper attributions | |
| Feb 22, 2021 at 17:50 | vote | accept | Per Alexandersson | ||
| Feb 22, 2021 at 17:04 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
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| Feb 22, 2021 at 16:58 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
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| Feb 22, 2021 at 16:57 | comment | added | David E Speyer | Thanks to both of you, you are both right. | |
| Feb 22, 2021 at 16:56 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| Feb 22, 2021 at 16:36 | comment | added | Will Sawin | Should $\neq$ in the second equation be $\geq$? | |
| Feb 22, 2021 at 16:36 | comment | added | Sam Hopkins | FWIW, I think the stabilization is easy to see if you work with maj and inv instead of maj and imaj. Let $\pi = \pi_1\pi_2\ldots\pi_m \in S_m$ be a permutation, where $m \geq a+b$, and with $\mathrm{maj}(\pi)=a$ and $\mathrm{inv}(\pi)=b$. Since $\mathrm{maj}(\pi)=a$, no descents can occur after the first $a$ positions; since $\mathrm{inv}(\pi)=b$, no letter of $a+b+i$ for $i\geq 1$ can occur in the first $a$ positions. Together these force all $a+b+i$ for $i\geq 1$ to be fixed points. So $\pi$ is really only a permutation of $\{1,\ldots,a+b\}$. | |
| Feb 22, 2021 at 16:14 | history | answered | David E Speyer | CC BY-SA 4.0 |