Timeline for Some question about polynomial representations of $GL(V)$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 15, 2014 at 8:30 | vote | accept | Alex | ||
| May 15, 2014 at 8:30 | comment | added | Alex | Ok, now I understand. For some reason I didn't make the distinction between rational and polynomial. As I looked more in the literature, there seems to be quite a number of sources making the same confusion. For instance math.mit.edu/~etingof/replect.pdf, even in Fulton and Harris it's not very clear. | |
| May 14, 2014 at 22:32 | comment | added | Allen Knutson | While this is a justification, it doesn't explain why people would do it. The answer is that every polynomial irrep of $GL(r)$ extends to a functor ${\bf Vec}\to {\bf Vec}$ (where $GL(r)$ is considered as a one-object subcategory of $\bf Vec$). | |
| May 14, 2014 at 16:33 | comment | added | Sam Hopkins | Yes thanks, I was not trying to say your account of the situation was wrong, just thought that this close relationship between rational and polynomial representations is a kind of justification for only studying the polynomial representations. | |
| May 14, 2014 at 16:18 | history | edited | Steven Sam | CC BY-SA 3.0 |
added 587 characters in body
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| May 14, 2014 at 16:15 | comment | added | Steven Sam | Yes, but when you tensor with determinant, you have to add 1 to all of the entries of $\lambda$. | |
| May 14, 2014 at 16:15 | comment | added | Sam Hopkins | But I seem to remember some statement along the lines that any rational representation of $GL(n,\mathbb{C})$ is a product of the determinant to some nonpositive power and a polynomial representation, so there is only a slight distinction between these two classes of representations. | |
| May 14, 2014 at 16:11 | history | answered | Steven Sam | CC BY-SA 3.0 |