Timeline for Existence of a finite group with a given decomposition for a tensor square of one irreducible complex representation
Current License: CC BY-SA 4.0
19 events
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| Jan 16, 2020 at 12:41 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
deleted paragraph, plus minor amendment
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| Jan 16, 2020 at 12:34 | comment | added | user6976 | @Geoff: Thabk you for the explanation. | |
| Jan 16, 2020 at 10:20 | comment | added | Geoff Robinson | @MarkSapir: As I explained above, it is because the normal subgroup $K$ is the kernel of $5_{1}$, the statement would not be true in general for an arbitrary normal subgroup. Once $K$ is in thhe kernek of $5_{1}$, $K$ is also in the kernel of everything in sight in the original equation, so, as I said, $G/K$ has the same property (regarding $5_{1}$ as a representation of $G/K)$. | |
| Jan 16, 2020 at 9:49 | comment | added | Sebastien Palcoux | @MarkSapir: Dietmar and Vaughan know all that, so if you and (one of) them are currently at Vanderbilt, it could be a good opportunity to discuss. | |
| Jan 16, 2020 at 9:40 | comment | added | Sebastien Palcoux | @MarkSapir: the isomorphic classes of irreps form a ring under the tensor product (it has the strucutre of a fusion ring, and it is the Grothendieck ring of the fusion category $\mathrm{Rep}(G)$, but it does not matter). Now let consider $5_1$ as an element of this fusion ring, it generates a fusion subring which, as every fusion subring, is isomorphic to the Grothendieck ring of $\mathrm{Rep}(G/N)$ with $N$ a normal subgroup depending on the subring (here $N=ker(5_1)$). | |
| Jan 16, 2020 at 9:03 | comment | added | user6976 | @SebastianPalcoux: I do not know what fusion sucategory is. But whatever it is it cannot be the same for $G$ and for $G/N$ if $N=G$. An easier question is (in Geoff's notation) why $G/K$ has an irrep of degree 7? If this is true it should have something to do with the fact that $K$ is the kernel of nn irrep of degree 5. It can't be true for arbitrqry normal $K$. | |
| Jan 16, 2020 at 8:55 | comment | added | Geoff Robinson | @MarkSapir : Because if $5_{1}$ has $K$ in its kernel, so does every representation on the right side, because $5_{1} \otimes 5_{1}$ has $K$ in its kernel. | |
| Jan 16, 2020 at 8:37 | comment | added | Sebastien Palcoux | @MarkSapir: $N=G$ corresponds to the trivial subcategory (i.e. given by the trivial irrep). | |
| Jan 16, 2020 at 8:35 | comment | added | user6976 | @SebastienPalcoux: What if $N=G$? | |
| Jan 16, 2020 at 8:30 | comment | added | Sebastien Palcoux | @MarkSapir: a fusion subcategory of $\mathrm{Rep}(G)$ is isomorphic to $\mathrm{Rep}(G/N)$ with $N \unlhd G$ (and reciprocally). | |
| Jan 16, 2020 at 2:12 | comment | added | user6976 | @Geoff: Why is it true (in your notation) that $G/K$ has the same property as $G$? | |
| Jan 15, 2020 at 21:22 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Some typoss corrected
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| Jan 15, 2020 at 18:39 | vote | accept | Sebastien Palcoux | ||
| Jan 15, 2020 at 18:27 | comment | added | Geoff Robinson | I think it is probably true that if $p >3$ is a prime such that $q = p+2$ is also prime, then there is no finite group $G$ with a complex irreducible characters $\chi,\mu$ of respective degrees $p$ and $q$ such that $\chi^{2} = 1 + \chi + \mu + \theta$, where $\theta$ is a character ( or $0$). The argument (using Feit's Theorem rather than Brauer's) goes through more or less unchanged, after noting that $(3,5)$ is the only prime pair $(p,q)$ such that $p+1 = q-1$ is power of $2$. But this seems very specialized. | |
| Jan 15, 2020 at 16:46 | comment | added | Sebastien Palcoux | I meant a weaker assumption on the decomposition: - Firstly, perhaps something like: the existence of (at least) one irrep of dim $5$ and of dim $7$ and $$5_1 \otimes 5_1 \ge 1 \oplus 5_1 \oplus 7_1$$ would be enough for your argument, correct? - And secondly, perhaps the couple $(5,7)$ can be generalized to a class of couples $(n,m)$, isn't it? | |
| Jan 15, 2020 at 16:27 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Added remark
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| Jan 15, 2020 at 16:24 | comment | added | Geoff Robinson | Yes, this argument shows that there is no such finite $G$, simple or not. As for a more general statement, I am not sure: arguments about complex linear groups of low dimension tend to be rather ad hoc, and some arguments above are very specific to your hypotheses. | |
| Jan 15, 2020 at 16:11 | comment | added | Sebastien Palcoux | My comment about Hiss-Malle is only for the simple groups. Your answer works for every group, right? Does your answer prove a more general statement than what expected? (because you seem to use just partially the assumption) If so, what is it? and to what could it be extended? | |
| Jan 15, 2020 at 15:46 | history | answered | Geoff Robinson | CC BY-SA 4.0 |