Timeline for Kronecker product preserves the conjugacy relation?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Nov 14, 2022 at 7:42 | comment | added | Sean Eberhard | No. It's redundant, because any finite abelian subgroup of $\mathrm{GL}_n(\mathbf{C})$ is diagonalizable, i.e., conjugate to a subgroup of $T$. | |
| Nov 12, 2022 at 23:11 | comment | added | user488802 | I'm sorry, I just realized this: has the condition that $A,B\leqslant T$ been used anywhere in the argument? | |
| Nov 12, 2022 at 21:01 | vote | accept | user488802 | ||
| Nov 12, 2022 at 14:58 | comment | added | Sean Eberhard | Yes $A$ lifts uniquely to $\dot A$ and conversely $A$ is just the image of $\dot A$ in $\mathrm{PGL}$, so the correspondence is bijective, and $A \sim B$ iff $\dot A \sim \dot B$. | |
| Nov 12, 2022 at 8:17 | comment | added | user488802 | I think this is the last question: My start point is two nonconjugate elementary abelian 2-subgroups $A,B$ in $PGL_{n}$, they lift uniquely to $\dot{A},\dot{B}$ in $GL_{n}$, then $\dot{A}⊗I_{2},\dot{B}⊗I_{2}$ in $GL_{2n}$, at last, $A⊗I_{2},B⊗I_{2}$ in $PGL_{2n}$. So it boils down to $A\sim B$ iff $\dot{A}∼\dot{B}$, it I'm correct? But I am not sure how to see from the right to the left. Intuitively, it seems correct. $\sim$ means conjugacy relation. Thx! | |
| Nov 11, 2022 at 23:34 | comment | added | user488802 | Yes! I've just worked out a similar example to see this! And I've finally got most of your answer. I'll keep digesting. Not sure what time it is where you are, but really appreciate your patience! | |
| Nov 11, 2022 at 23:27 | comment | added | Sean Eberhard | Consider two maps $C_2^2 \to T \le \mathrm{GL}_2(\mathbf C)$. Here let us identify $C_2$ with $\{\pm1\}$. The first map is $(x, y) \mapsto \mathrm{diag}(x, y)$. The second is $(x, y) \mapsto \mathrm{diag}(x, xy)$. The characters are $x+y$ and $x+xy$, so the representations are not equivalent. But they have the same image. | |
| Nov 11, 2022 at 22:13 | comment | added | user488802 | Sorry, I am still digesting. For the second, you are saying, before quotienting, there could be two different parameterizations of faithful reps yielding a same embedded image? Would you please give a short example to make it concrete? Like a Klein 4 in $\operatorname{GL}_{4}$? I'd appreciate your help. | |
| Nov 11, 2022 at 11:07 | comment | added | Sean Eberhard | That's what I mean, yes. | |
| Nov 11, 2022 at 10:39 | comment | added | user488802 | Thank you for your patience. For the first, do you mean the intersection of the kernels of the reps is trivial instead of zero? Thx. | |
| Nov 11, 2022 at 10:06 | comment | added | Sean Eberhard | For the second: without quotienting we have a parameterization of faithful embeddings $\rho : C_2^r \to \mathrm{GL}_n(\mathbf{C})$ up to conjugacy. But we want to consider two of these to be the same if they have the same image. Therefore we need to quotient by the action of the automorphism group of $C_2^r$. This quotienting has nothing to do with $\mathrm{PGL}$. | |
| Nov 11, 2022 at 10:04 | comment | added | Sean Eberhard | For the first: if $\chi = w_1 \chi_1 + \cdots + w_m \chi_m$ then $\rho$ is equivalent up to conjugation to a diagonal representation with $\chi_1, \dots, \chi_m$ entries (and the appropriate multiplicities). Clearly this is faithful iff $\ker \chi_1 \cap \cdots \cap \ker \chi_m = 0$, which is equivalent to $\mathrm{supp}(w)$ generating $D$. | |
| Nov 11, 2022 at 9:16 | comment | added | user488802 | Also, I don't quite get this: to get the conjugacy classes in $\operatorname{PGL}_{n}(\textbf{C})$, we quotient $\{w:D \to \mathbf N \mid \langle \mathrm{supp}~w\rangle = D\}$ by the action of $\mathrm{Aut}(C_2^r) \cong \mathrm{GL}_r(2)$. Could you please elaborate a bit more? Thank you very much! | |
| Nov 11, 2022 at 9:12 | comment | added | user488802 | Thank you for the added words which helped me a lot. Could you please point me to a proof of "the representation is faithful iff the support of $w$ generates $D$"? | |
| Nov 10, 2022 at 11:48 | comment | added | Sean Eberhard | I added some words. I hope it helps. | |
| Nov 10, 2022 at 11:46 | history | edited | Sean Eberhard | CC BY-SA 4.0 |
some clarification
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| Nov 10, 2022 at 10:50 | comment | added | user488802 | I really appreciate your input. May I ask about the definition of the dual of a group and the support of a map? Also, I'm not quite sure how $w$ maps elements... And how do we sum up the $w$ so the sum is $n$? Sorry for the trouble. Thank you! | |
| Nov 10, 2022 at 10:26 | history | answered | Sean Eberhard | CC BY-SA 4.0 |