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Let $G$ be a finite group and let $\chi$ be the character of an irreducible complex representation $\rho$ of $G$ on $V$.

Let $x$ be an involution in $G$.

I'd like to ask the following

Question 1:

What is known about $\chi(x)?$

1a) Are there criteria when $\chi(x)$ is positive / negative / zero ?

Of course, $1_{\text{Aut}(V)}=\rho(x^2)=\rho(x)^2$, such that the only possible eigenvalues of $\rho(x)$ are $\pm 1$.

Moreover, there is an article written by P.X. Gallagher with the title "Character values at involutions" (DOI: https://doi.org/10.1090/S0002-9939-1994-1185260-1) dealing with the case that $\int_G\chi_1\chi_2\chi_3 \neq 0$, where the integral is in the sense of the Haar measure.

EDIT: The main parts of Gallagher's results are the following ones:

For an involution $\sigma$ of a finite group $G$ and an irreducible complex representation $R$ of $G$, denote by $q$ the proportion of $-1$'s among the eigenvalues of $R(\sigma)$. Then:

$(*)$ $\frac{1}{h}\leq q \leq 1-\frac{1}{h}$, unless $q = 0$ or $1$, where $h$ is the index of the centralizer $C$ of $\sigma$.

Moreover, if $\int_G\chi_1\chi_2\chi_3 \neq 0$, then $(*)$ is refined to prove that the proportions of $-1$'s among the eigenvalues of $\rho_1, \rho_2$ and $\rho_3$ (i.e., the corresponding representations) at $\sigma$ are the sides of a triangle on a sphere of circumference 2.

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1b) When does $\int_G\chi_1\chi_2\chi_3 \neq 0$ happen (necessary / sufficient criteria)?

1c) When does $\int_G\chi_1\chi_2\chi_3 = 0$ happen (necessary / sufficient criteria)?

1d) Are there results apart from Gallagher's result?

1e) Can one deduce additional information, if all considered characters lie in the same 2-block?

Thanks for the help.

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    $\begingroup$ Since $\int_G \chi_1\chi_2\chi_3 = \langle \chi_1\chi_2, \overline{\chi}_3 \rangle$, the questions (1b) and (1c) are equivalent to decomposing arbitrary tensor products of representations. This is a tough problem on which there are many interesting results but no general theory. For instance, Saxl has conjectured that for the symmetric group if $\chi = \chi^{(m,m-1,\ldots, 1)}$ is the 'staircase' character then every irreducible character appears in $\chi^2$. $\endgroup$ – Mark Wildon May 31 at 17:22
  • $\begingroup$ I think your question would be slightly improved if you could briefly indicate why Gallagher's result is related to involutions. $\endgroup$ – Mark Wildon May 31 at 17:28
  • $\begingroup$ Thank you very much for the comment. I've edited the question. $\endgroup$ – Bernhard Boehmler Jun 1 at 11:09
  • $\begingroup$ Just a small remark: Theorem 3.1 on page 456 of Feit's book is also related, see books.google.de/… $\endgroup$ – Bernhard Boehmler Jun 1 at 13:48
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There are many results about the values of $\chi(t)$ when $t$ is an involution of a finite group $G$ and $\chi$ is an irreducible character: Isaacs' book on Character Theory has many such results collected from the literature, but there are many others scattered around:

For example, if $G = O^{2}(G)$ (equivalently, if $G/G^{\prime}$ has odd order), then $\chi(1) \equiv \chi(t)$ (mod $4$).

(Knörr): We have $\chi(t) = 0$ for every involution $t$ if and only if $|S|$ divides $\chi(1)$, where $S$ is a Sylow $2$-subgroup of $G.$

Regarding block theory, whenever $t $ is an involution of $G$ and $\chi$ is an irreducible character in the principal $2$-block of $G$, we have $\chi(tuv) = \chi(tu)$ whenever $u,v \in C_{G}(t)$ have odd order and $v \in O_{2^{\prime}}(C_{G}(t)),$ which is a consequence of Brauer's Second and Third Main Theorems.

I could probably give several more examples if you gave further clues as to what you are looking for.

Further edit to address a question from comments: if $t$ is an involution of $G$ and $B$ is a $2$-block of $G$, then results of Brauer imply the following facts (among others):

If $B$ has defect group $D$ and $t$ is not $G$-conjugate to an element of $D$, then we have $\chi(t) = 0 $ for every complex irreducible character $\chi \in B$.

If $B$ has dfect group $D$ and some conjugate of $t$ lies in $D$, then there is an irreducible character $\chi \in B$ with $\chi(t) \neq 0$, and we have $\sum_{ \chi \in {\rm Irr}(B)} \chi(1)\chi(t) =0,$ so there are irreducible characters in $B$ taking both positive and negative values at $t$. Later edit: Another theorem of Brauer is that if $B$ is a $2$-block of defect $d >1$, then the number of irreducible characters in $B$ of degree exactly divisible by $2^{a-d}$ is divisible by $4$. In particular, this implies that if $|G|$ is divisible by $4$ and $t$ is an involution of $G$, then the number of irreducible characters $\chi$ in the principal $2$-block such that $\chi(t)$ is odd is a multiple of $4$.

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  • $\begingroup$ Thank you very much for the answer. I would be interested in results in the literature concerning the case that all ordinary irreducible characters in question are lying in a non-principal $2$-block which has a non-cyclic defect group...Are there criteria known when they have negative values at involutions / an involution? $\endgroup$ – Bernhard Boehmler Jun 1 at 11:12
  • $\begingroup$ I will make a small edit . $\endgroup$ – Geoff Robinson Jun 1 at 11:48
  • $\begingroup$ Thank you very much. $\endgroup$ – Bernhard Boehmler Jun 1 at 12:08
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For the symmetric group, let $\chi^\lambda$ be the symmetric group charater canonically labelled by the partition $\lambda$. Then $\chi^\lambda(x) = 0$ whenever $\lambda = \lambda'$ is a self-conjugate partition and the involution $x$ has an odd number of disjoint transpositions. This isn't very deep: in fact $\chi^\lambda(x) = 0$ for any odd permutation $x$. The converse does not hold: for example $\chi^{(6,3,2,2,2)}(1,2) = 0$.

As a very weak sufficient condition, it follows easily from the Murnaghan–Nakayama rule that if all the $2$-hooks in the partition $\lambda$ are horizontal (i.e. two boxes in the same row) then $\chi^\lambda(1,2) \ge 0$, with strict inequality unless $\lambda$ is a $2$-core (i.e. a staircase partition as in Saxl's Conjecture).

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  • $\begingroup$ Thank you very much for the answer. $\endgroup$ – Bernhard Boehmler Jun 1 at 11:12
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If the involution has no fixed-points, then the Murnaghan-Nakayama rule is cancellation-free. Hence, the character value is (up to a sign) the number of domino tableaux of the shape $\lambda$. The number of such tableaux can be computed via a hook-formula (Fomin-Lulov / James-Kerber).

You can extend this to all involutions, but you need to sum over all possible ways to distribute single boxes in $\lambda$, so that they occupy some skew shape $\lambda/\mu$, then apply the above argument for each shape $\mu$.

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  • $\begingroup$ Thank you very much for the answer. $\endgroup$ – Bernhard Boehmler Jun 1 at 11:12
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Suppose that $\chi$ is an irreducible character of a finite group $G$ and $t$ is an involution in $G$. It is well known that $\chi(1)\equiv\chi(t)$ (mod $2$). Stephen Gagola and Sidney Garrison showed (J. Algebra 74 (1982) 20--51) that if $\chi$ is a faithful orthogonal character of $G$, and $\chi(1)-\chi(t)\equiv4$ (mod $8$), then $G$ has a non-trivial double cover. Moreover if $t$ lies in the commutator subgroup of $G$, then $H^2(G,{\mathbb C}^\times)$ has even order. They also have results related to the restriction of a real-valued character to a Klein-four subgroup of $G$. They used these results to verify that $M_{22}$ has a four-fold cover.

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  • $\begingroup$ Thank you very much for the answer. $\endgroup$ – Bernhard Boehmler Jun 2 at 15:41

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