Monster coefficients

Question

For three irreducible characters $\phi,\psi,\rho$ of a finite group $G$ define the Kronecker multiplicities as: $$g(\phi,\psi,\rho) = \langle \phi,\psi\cdot\rho\rangle $$ where $$\langle \chi,\eta\rangle = \frac{1}{|G|}\sum_{x\in G} \chi(x)\, \overline{\eta(x)} $$ and $[\psi\cdot\rho] (x) = \psi(x) \rho(x)$ is the usual product.

I am interested in Kronecker multiplicities for the Monster group $M$. While the group is large, there are only 194 conjugacy classes.

$$(1) \qquad \max_{\phi,\psi,\rho} g(\phi,\psi,\rho)$$

$$(2) \qquad \sum_{\phi,\psi,\rho} g(\phi,\psi,\rho)$$ $$(3) \qquad \sum_{\phi,\psi,\rho} g(\phi,\psi,\rho)^2$$

These sums are over all triples of irreducible characters, but because of the symmetries only about 1/6 of them need to be computed to get the answer. If you can do this, I would also be curious about the specific characters maximizing (1).

The computation is beyond my computer algebra skills, but I know that GAP has the whole character table of $M$ ready to use.

Answer 1

In GAP, you could simply iterate in a triple loop over $\phi,\psi,\rho$, calculate the $g$-values and find maximum and sum values:

ct:=CharacterTable("M");
irrs:=Irr(ct);

m:=0; s:=0;q:=0; # max, sum, sumsquare

for rho in irrs do
  for psi in irrs do
    ten:=rho*psi; # tensor product
    for phi in irrs do
      g:=ScalarProduct(phi,ten);
      if g>m then m:=g;fi;
      s:=s+g;
      q:=q+g^2;
    od;
  od;
od;

Afterwards look at the values of m, s, and q. Unless I have mistyped something, the results I get are (for the monster group):

1. Maximum is $21458051228477513179513856=2^{10}\cdot281\cdot443\cdot599\cdot6571\cdot42768299767$

2. Sum is $247017097351847432984363535932$ (Thank you, @James for the correction)

3. Sum-Squares is $808017424794512875894769468067441075690144312450960558$ (ditto corrected, also typo fixed)

Answer 2

I'll answer the other part of the question, namely "I would also be curious about the specific characters maximizing (1)."

I found that the maximum Kronecker multiplicity for the Monster is produced when $\phi$, $\psi$, and $\rho$ are all the last character in GAP's character table, which is the character whose degree is the maximum degree 258823477531055064045234375.

I did the computations for the other sporadic groups as well, and this pattern continues except for the McLaughlin and O'Nan groups. These have two characters of maximum degree, and taking each of $\phi$, $\psi$, and $\rho$ to be either of these two characters produces the maximum Kronecker multiplicity.

I don't know whether "three maximum-degree characters produce the maximum Kronecker multiplicity" is a general result or not because I'm not a group theorist.