Timeline for Representations of products of symmetric groups
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2021 at 5:27 | history | edited | LSpice | CC BY-SA 4.0 |
Forgot the order of the group
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| Apr 14, 2021 at 3:23 | comment | added | LSpice | I originally thought I had an argument that $\mathbb Z$-equivalence was the same as $\mathbb Q$-equivalence for symmetric groups, but @JeffAdler pointed out to me that, as @JohnBaez suspected, it's not true: consider the obvious representations of $\operatorname S_3$ on $\{(a, b, c) \in \mathbb Z^3 : a + b + c = 0\}$ and on $\mathbb Z^3/\mathbb Z(1, 1, 1)$, which are $\mathbb Q$- but not $\mathbb Z$-equivalent. | |
| Apr 11, 2021 at 19:58 | comment | added | LSpice | I have written a re-phrased version that works only with representations, not with characters, and so hopefully makes it more explicit when we're changing base, and why it's OK to do so (namely, flatness of field extensions—I think is the jargon?). | |
| Apr 11, 2021 at 19:57 | history | edited | LSpice | CC BY-SA 4.0 |
Representation-theoretic re-phrasing
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| Apr 11, 2021 at 19:19 | comment | added | LSpice | The equality is just a statement about certain functions; it doesn't even know what the ground field is. If you regard those functions in taking values in (the copy of the integers in) $k$, then you get a statement about characters of representations over $k$. If you regard them instead as taking values in (the copy of the integers in) $\mathbb C$, then you get a statement about characters of representations over $\mathbb C$, which is known to be true. But the underlying $\mathbb Z$-valued functions are the same, regardless of where we consider their values to lie. | |
| Apr 11, 2021 at 19:02 | comment | added | John Baez | I would find $\mathbb{Q}$ less scary. Btw, I'm a bit nervous about the step where you say something works over an algebraically closed field of characteristic zero because it works that way over $\mathbb{C}$. There's probably some general principles that let one make arguments like that, but what's the easiest explanation in this case? | |
| Apr 11, 2021 at 18:17 | history | edited | LSpice | CC BY-SA 4.0 |
Positive -> non-0; what's an inner product?
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| Apr 11, 2021 at 18:15 | comment | added | LSpice | I'm not sure why it would imply that, but I'm certainly willing to believe that the representation theory over $\mathbb Z$ is harder than that over $\mathbb C$. Perhaps I should just work over $\mathbb Q$, which is good enough since we're passing to a field anyway, to avoid ambiguities. | |
| Apr 11, 2021 at 18:07 | comment | added | John Baez | Okay, for some reason I imagined the $\rho_j$ as held fixed at the start of your argument. Btw, classifying irreps of symmetric groups over finite fields is supposed to be hard, they're not just classified by Young diagrams as they are over $\mathbb{Q}$ or $\mathbb{C}$. Doesn't that imply the representation theory of $S_n$ over $\mathbb{Z}$ must be quite different than over $\mathbb{C}$? | |
| Apr 11, 2021 at 17:43 | comment | added | LSpice | It is a sum over the equivalence classes of irreducible representations $\rho_j$ of $S_{n_j}$ (of which there are finitely many). I don't know off the top of my head if equivalence over $\mathbb Z$ is stronger than equivalence over $\mathbb C$ in this special case, so let's say that I mean equivalence over $\mathbb C$ (which is enough to give that the characters are the same, so you could view it as a sum over characters if that's more palatable). | |
| Apr 11, 2021 at 17:42 | comment | added | John Baez | What does summation over $\rho_1, \dots, \rho_p$ mean? | |
| Apr 11, 2021 at 17:32 | history | answered | LSpice | CC BY-SA 4.0 |