# What is known about ordinary character values at involutions?

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.

$$\$$

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.

• 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 $\chi^2$. May 31, 2020 at 17:22
• I think your question would be slightly improved if you could briefly indicate why Gallagher's result is related to involutions. May 31, 2020 at 17:28
• Thank you very much for the comment. I've edited the question. Jun 1, 2020 at 11:09
• Just a small remark: Theorem 3.1 on page 456 of Feit's book is also related, see books.google.de/… Jun 1, 2020 at 13:48

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 $$4$$.

• 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? Jun 1, 2020 at 11:12
• I will make a small edit . Jun 1, 2020 at 11:48
• Thank you very much. Jun 1, 2020 at 12:08

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).

• Thank you very much for the answer. Jun 1, 2020 at 11:12

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 $$\mu$$.

• Thank you very much for the answer. Jun 1, 2020 at 11:12

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.

• Thank you very much for the answer. Jun 2, 2020 at 15:41