Timeline for Stirling number identity via homology?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 13, 2012 at 0:05 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Jul 13, 2012 at 0:02 | comment | added | David E Speyer | You are right. Which is sad, because now it is less obvious that $\chi=0$. I'll note in my defense that the original post said I only had to give a proof for $m=1$ :). | |
| Jul 12, 2012 at 20:38 | comment | added | Gary Kennedy | Don't we also need the symmetric group on m elements to act? For example, if n=3 and m=2 then we're supposed to have 3 zero-cells rather than 6. | |
| Jul 3, 2012 at 23:02 | comment | added | David E Speyer | Edited because the complex isn't quite regular. | |
| Jul 3, 2012 at 23:00 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Jul 3, 2012 at 20:38 | history | answered | David E Speyer | CC BY-SA 3.0 |