Let $G$ be a finite group and $\Lambda = (\lambda_{i,j})$ its character table with $\lambda_{i,1}$ the degree of the ith character.

Consider the following combinatorial property of $\Lambda$: for all triple $(j,k,\ell)$ $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$ It is a consequence of a more general result involving subfactor planar algebra and fusion category (see here Corollary 7.5, see also this answer).

Question: Is this combinatorial property already known to finite group theorists?
If yes: What is a reference?
If no: Is there a group theoretical elementary proof?
In any case: Are there other properties of the same kind?

To avoid any misunderstanding, let us see one example. Take $G=A_5$, its character table is:
$$\left[ \begin{matrix} 1&1&1&1&1 \\ 3&-1&0&\frac{1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2} \\ 3&-1&0&\frac{1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2} \\ 4&0&1&-1&-1 \\ 5&1&-1&0&0 \end{matrix} \right] $$ Take for example $(j,k,\ell) = (2,4,5)$, then $\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} = \frac{5}{3} \ge 0$.

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    $\begingroup$ This would be easier to read replacing $j_1,j_2,j_3$ with $j,k,\ell$ $\endgroup$ – Matt F. Oct 31 '19 at 18:53
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    $\begingroup$ In other words, for all triples of group elements $f$, $g$, $h$, we have $\sum_{\chi} \tfrac{\chi(f) \chi(g) \chi(h)}{\chi(1)} \geq 0$, where the sum is over irreducible characters? $\endgroup$ – David E Speyer Oct 31 '19 at 19:11
  • $\begingroup$ @MattF. Yes, done! $\endgroup$ – Sebastien Palcoux Oct 31 '19 at 21:45
  • $\begingroup$ @DavidESpeyer: Correct! $\endgroup$ – Sebastien Palcoux Oct 31 '19 at 21:45
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    $\begingroup$ Dear Sebastien, I've added a follow-up question here: math.stackexchange.com/questions/3774014/… $\endgroup$ – Ines Institoris Jul 30 at 1:44

By standard manipulations with the group algebra, your sum has a combinatorial/probabilistic interpretation that makes its nonnegativity clear.

The element $ \frac{1}{|G|} \sum_{ g\in G} [g hg^{-1} ]$ in the group algebra is conjugacy invariant, and so acts by scalars on each irreducible representation. Because its trace on a representation with the character $\chi$ is $ \frac{1}{|G|} \sum_{ g\in G} \chi( g hg^{-1} ) = \frac{1}{|G|} \sum_{ g\in G} \chi( h )= \chi(h)$, its unique eigenvalue must be $\frac{\chi(h)}{\chi(1)}$. Hence for $h_1,h_2,h_3$ three elements of the group,

$$ \left( \frac{1}{|G|} \sum_{ g\in G} [g h_1g^{-1} ]\right) \left( \frac{1}{|G|} \sum_{ g\in G} [g h_2g^{-1} ]\right) \left( \frac{1}{|G|} \sum_{ g\in G} [g h_3g^{-1} ]\right) $$

acts on this representation with eigenvalue $\frac{ \chi(h_1) \chi(h_2) \chi(h_3)}{\chi(1)^3}$.

Now the group algebra, as a module over itself, is the sum over irreducible characters $\chi$ of $\chi(1) $ copies of the representation with character $\chi$. Hence the trace of this element on the group algebra is $$\sum_{\chi} \chi(1) \cdot \chi(1) \cdot \frac{ \chi(h_1) \chi(h_2) \chi(h_3)}{\chi(1)^3}= \sum_{\chi} \frac{ \chi(h_1) \chi(h_2) \chi(h_3)}{\chi(1)}.$$

On the other hand, the trace of an element of the group algebra on itself is the order of the group times the coefficient of $[1]$. The coefficient of $[1]$ in this particular element is $\frac{1}{ |G|^3}$ times the number of $g_1,g_2,g_3$ such that $g_1 h_1 g_1^{-1} g_2 h_2 g_2^{-1} g_3 h_3 g_3^{-1} =1$. This gives the combinatorial interpretation

$$\sum_{\chi} \frac{ \chi(h_1) \chi(h_2) \chi(h_3)}{\chi(1)} = \frac{1}{ |G|^2} \left| \{ g_1,g_2,g_3 \in G \mid g_1 h_1 g_1^{-1} g_2 h_2 g_2^{-1} g_3 h_3 g_3^{-1} =1 \}\right|$$

from which non-negativity is clear.

I would guess this is probably in the group theory literature somewhere but I wouldn't know where.

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    $\begingroup$ I think it is a known identity mathoverflow.net/a/226026/4312 $\endgroup$ – Fedor Petrov Oct 31 '19 at 20:26
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    $\begingroup$ Not an exact citation, but close to Serre's Topics in Galois Theory, theorem 7.2.1 (the case $k=3$). $\endgroup$ – MyNinthAccount Oct 31 '19 at 20:43

This is indeed well-known in the character theory literature, and goes back to Frobenius and Burnside. What you are calculating is a positive rational multiple of a class algebra constant, and class algebra constants are clearly non-negative.

Using the notation in David Speyer's comment, it is well-known, and derived in most representation theory texts that $\frac{|G|}{|C_{G}(f)| |C_{G}(g)|} \sum_{\chi} \frac{\chi(f)\chi(g)\chi(h)}{\chi(1)}$ is the number of times $h^{-1}$ is expressible as a product of a conjugate of $f$ and a conjugate of $g$. The character theory formula is easily derived from the expressions of the class sums as linear combinations of primitive central idempotents of the group algebra $\mathbb{C}G$, and can be found in many texts .

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  • $\begingroup$ I look for other properties of the character table to see those which could be extended to Grothendieck ring of (commutative) fusion categories, because they are obstructions for a fusion ring to admit a unitary categorification (see this answer). Serre's Topics in Galois Theory, Theorem 7.2.1 (pointed out by MyNinthAccount) admits such extension for any $k$, but unfortunatly it provides no additional obstruction (than $k=3$ already gives). Then do you know where I can find a list of such properties? What would by your favorite one for such goal? $\endgroup$ – Sebastien Palcoux Nov 1 '19 at 10:41
  • $\begingroup$ Yes this is what is given by Serre's Topics in Galois Theory, Theorem 7.2.1, but as I said, it provides no new obstruction than $k=3$ already gives. I am looking for something else. $\endgroup$ – Sebastien Palcoux Nov 1 '19 at 11:52
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    $\begingroup$ @SebastienPalcoux I think the same is true upon multiplying by any negative even power of $\chi(1)^{-2}$, because you can put commutators in there / count RIemann surfaces, but that presumably will imply no new obstruction, or maybe just the $k=1$ case. $\endgroup$ – Will Sawin Nov 1 '19 at 12:06
  • $\begingroup$ You wrote "call algebra constants". Did you mean to write "class algebra constants", as in the first clause of the sentence? I'm not familiar with the terminology, so may not be understanding your usage. $\endgroup$ – Faheem Mitha Nov 1 '19 at 13:41
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    $\begingroup$ These non-negative numbers satisfy some inequalities. Say, $prob(abcdefg=1)\ge prob(abc=1)prod(defg=1)$ where $a,b,c,d,e,f,g$ are chosen at random in conjugacy classes $\endgroup$ – Fedor Petrov Nov 1 '19 at 15:37

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