Timeline for Irreducible representations and Jaccard Kernel for Groups?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2020 at 10:11 | comment | added | user6671 | @DanielCopeland: Thanks for your comment. You might be interested in the following question: mathoverflow.net/questions/362406/… | |
| Jun 6, 2020 at 21:35 | comment | added | Daniel Copeland | The matrix $M$ can be interpreted as a $G×G$-invariant map from the group algebra of $G$ to itself. Since the group algebra has a multiplicity free decomposition, $M$ acts by a scalar times the identity on each isotypic component. Hence the characteristic polynomial of $M$ has the form $(x−a_1)^{d_1^2}...(x−a_r)^{d_r^2}$ where $r$ is the number of irreps of $G$ and $d_i$ is the dim of the $i$th irrep. However, as in the $D_8$ case, there may be coincidences among the eigenvalues $a_i$. It looks like a nice problem to figure out when these coincidences happen. | |
| Jun 6, 2020 at 21:16 | comment | added | Daniel Copeland | @orgesleka the code used above doesn't use the cayley embedding, since sage implements DihedralGroup(4) as a permutation group acting on 4 elements (subgroup of $S_4$), while the cayley embedding expresses $D_8$ as subgroup of $S_8$ | |
| Jun 6, 2020 at 20:47 | history | edited | user6671 | CC BY-SA 4.0 |
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| Jun 6, 2020 at 20:45 | comment | added | user6671 | @DanielCopeland: I don't understand why in your eyes it is not the Calyley embedding? | |
| Jun 6, 2020 at 20:44 | comment | added | user6671 | @LSpice: Do you have to make a choice for each given group concerning the embedding? | |
| Jun 6, 2020 at 19:59 | comment | added | Daniel Copeland | The matrix in the question differs from the Cayley embedding, no? In the Cayley embedding the row corresponding to the identity should be all zeros off the diagonal, if I understand correctly (since only the identity has any fixed points in this action). It seems the matrix above comes from the action of $D_4$ on vertices of a square. | |
| Jun 6, 2020 at 19:57 | comment | added | LSpice | It can't literally be independent, because you get different-sized matrices; but there might still be something interesting. For example, what happens if you regard $\operatorname D_8$ as a subgroup not of $\operatorname{Sym}_{\operatorname D_8}$ but of $\operatorname{Sym}_4$? | |
| Jun 6, 2020 at 19:37 | comment | added | user6671 | @LSpice. I mean the Cayley embedding: en.wikipedia.org/wiki/Cayley%27s_theorem. Which you consider the canonical realisation. I am not sure if it is independent for different embeddings. | |
| Jun 6, 2020 at 19:29 | comment | added | LSpice | Is there any reasonable sense in which your notion is independent of the choice of embedding in a symmetric group? Of course there is the canonical realisation of a finite group as a group of permutations of its underlying set. | |
| Jun 6, 2020 at 19:18 | history | asked | user6671 | CC BY-SA 4.0 |