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Consider a finite group $G$, a subgroup $H\leq G$, and a transversal $G/H = \{t_1H, t_2H,\ldots,t_rH\}$. Given three characters $\chi_1,\chi_2$ and $\chi_3$ of $G$, I'd like to compute:

$$ \sum_{i}^r \left[\left(\sum_{\sigma_1\in t_iH}\chi_{1}(\sigma_1)\right)\left(\sum_{\sigma_2\in t_iH}\chi_{2}(\sigma_2)\right)\left(\sum_{\sigma_3\in t_iH}\chi_{3}(\sigma_3)\right)\right] $$

Actually, I just need to know if it is zero or not.

In the case where $H=\{e\}$ (the trivial group) we get

$$ \sum_{\sigma\in G}\chi_1(\sigma)\chi_2(\sigma)\chi_3(\sigma), $$

and we can reinterpret this as the multiplicity of the trivial rep of $G$ in the (tensor) product rep $\chi_1\otimes \chi_2\otimes \chi_3$ of $G\times G\times G$ restricted to $G$ via $g \mapsto (g,g,g )$.

In the case where $H=G$ we get

$$ \left(\sum_{\sigma_1\in G}\chi_{1}(\sigma_1)\right)\left(\sum_{\sigma_1\in G}\chi_{2}(\sigma_2)\right)\left(\sum_{\sigma_3\in G}\chi_{3}(\sigma_3)\right), $$

and we can reinterpret this as the product of the multiplicities of the trivial rep of $G$ in each of $\chi_1$, $\chi_2$ and $\chi_3$. Hence if $\chi_1$, $\chi_2$ and $\chi_3$ are irreducible we get a non-zero answer iff each character is trivial.

In the intermediate cases with $H$ a proper subgroup, I was hoping it would also just be a matter of reinterpreting using inner products of induced characters $H\uparrow G$, but this doesn't seem to work out (unless I am missing something).

As per my question title you see it comes down to average values of the characters on each coset, but I can't see a nice way of understanding what this is telling me.

Failing a theoretical answer, I've started coding up in GAP to compute specific examples. Any suggestions on how this could be achieved efficiently would also be very much appreciated.

Thanks!

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    $\begingroup$ One of the common pitfalls in calculating with GAP character tables is not being aware that the ordering of conjugacy classes may not be the same in different calls of CharacterTable since probabilistic methods may be involved. The GAP code should be ordering-agnostic. $\endgroup$ – Alexander Konovalov Apr 7 '14 at 13:48
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    $\begingroup$ Another tip is to use the GAP character table library. This will give you the same table each call and will save the time because there will be no need to recompute it from scratch. If you need to connect the group and the character table for classes identification, an important function is CharacterTableWithStoredGroup. $\endgroup$ – Alexander Konovalov Apr 7 '14 at 13:52
  • $\begingroup$ Thanks for the tips. I'm actually having good luck with coding it up in GAP (something I have never done properly before). $\endgroup$ – Jeremy Sumner Apr 8 '14 at 0:54
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The product will be zero (for any choice of coset representative $t_{i}$ and for each coset) if any of $\chi_{1},\chi_{2},\chi_{3}$ contain the trivial character with multiplicity zero on restriction to $H.$ However, I do not think the converse ( for a particular choice of $t_{i}$) need be true. It suffices to note that if the complex (matrix) representation $\sigma$ of the finite group $H$ does not contain the trivial representation, then $\sum_{h \in H} (h \sigma) = 0.$ It suffices to check this for each irreducible constituent of $\sigma,$ by Maschke's Theorem. So it suffices to check the case that $\sigma$ is irreducible, but non-trivial. Then $\sum_{h \in H} h\sigma$ commutes with $x\sigma$ for each $x \in H.$ Hence it is a scalar matrix by Schur's Lemma. But it has trace zero by the orthogonality relations, so it is the zero matrix.

Note then, that for general $\sigma$ not contaig the trivial representation on restriction to $H,$ and any $t \in G,$ we have $\sum_{h \in H}(th\sigma) = (t\sigma)\sum_{h \in H}h\sigma = [0].$

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  • $\begingroup$ I can't see that that is true. I see it works for the first triple product in the sum over i (where we take $t_1=e$) since we can interpret this as the product of multiplicities of the trivial characters of $\chi_1,\chi_2,\chi_3$ under restriction to $H$. But you need one of $\chi_1,\chi_2,\chi_3$ to be one-dimensional for the argument to extend to the other cosets, don't you? i.e. If $\chi_1$ is 1D then $\sum_{\sigma_1\in t_iH}\chi_1(\sigma_1)=\sum_{\sigma\in H}\chi_1(t_1\sigma)=\chi_1(t_1)\sum_{\sigma\in H}\chi(\sigma)=0 $ if $\chi$ does not contain trivial rep. $\endgroup$ – Jeremy Sumner Apr 8 '14 at 0:33
  • $\begingroup$ Or does it work more generally? Also, when you say $t$ do you mean a choice of transversal? I don't think the answer depends on the choice of transversal. $\endgroup$ – Jeremy Sumner Apr 8 '14 at 0:39
  • $\begingroup$ Example calculation should have taken $t_i$ (to avoid confusion with $t_1=e$) and should read: $\sum_{\sigma_1\in t_iH}\chi_1(\sigma_1)=\sum_{\sigma\in H}\chi_1(t_i\sigma)=\chi_1(t_i)\sum_{\sigma\in H}\chi_1(\sigma)=0$ if $\chi_1$ is 1D and does not contain trivial character. $\endgroup$ – Jeremy Sumner Apr 8 '14 at 0:52
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    $\begingroup$ It is true for general representations, see the expanded edit. $\endgroup$ – Geoff Robinson Apr 8 '14 at 6:29
  • $\begingroup$ I see the argument now. That's awesome, thanks. $\endgroup$ – Jeremy Sumner Apr 8 '14 at 9:14

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