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Let $G$ be a finite group, $H$ a subgroup of $G$ and $g \in G$. Define $$ [gH] = \sum_{h \in H} gh, $$ viewed an element in the group algebra $\mathbb{C}[G]$. Given an irreducible character $\chi$ of $G$, one can compute $$ \chi([gH]) = \sum_{h \in H} \chi(gh). $$ Note this is a generalization of character evaluations for group elements (let $H$ be the trivial group). Also, these computations are much easier to understand when $g$ is the identity.

Question: Are there results studying computations of this form? Are there applications of them to representation theory?

I am especially interested in results of this form and applications when $G$ is the symmetric group. My motivation is that recent work of mine (with a coauthor) can be interpreted as performing precisely this computation for $G = S_n$ and $H = S_{n-k}$ for some fixed $k$. In this case, it's not so hard to see that all but finitely many irreducible characters will vanish on $[gH]$. For $\chi$ one of the remaining irreducible characters, naively one would be evaluating $\chi$ on $(n-k)!$ group elements. Our formula shows this computation can be computed in $O(n^k)$ evaluations (I think I have this right, but there is certainly a huge savings).

This seems like a very natural questions in representation theory, but we were unable to find similar results in the literature. We'd be very interested in knowing if there is prior work on this problem or others of a similar flavor.

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    $\begingroup$ Summing over $h\in H$ and dividing by $|H|$ gives the idempotent projector onto the $H$-invariants, so this is the trace of the composition of $g$ with the idempotent projector onto the $H$-invariants, which can be computed in an orthonormal basis where the first $k$ vectors are a basis for the $H$-invariants as the sum of the first $k$ diagonal entries of the matrix for $g$. $\endgroup$
    – Will Sawin
    Oct 16, 2022 at 2:12
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    $\begingroup$ This does show up in existing literature. As Will Sawin says, $\chi([gH])$ will be zero unless ${\rm Res}^{G}_{H}(\chi) > 0.$ Using Frobenius reciprocity, it will be zero unless $\chi$ is an irreducible constituent of ${\rm Ind}^{G}_{H}(1).$This sort of thing comes up a lot in the study of endomorphism rings of permutation modules, for example. Also in work on Gelfand-Graev representations. $\endgroup$ Oct 16, 2022 at 13:07

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Will Sawin and Geoff Robinson have already hinted at what is going on, but I would like to add some detail (and some formulas in the case of symmetric groups).

First of all, let's do as Will Sawin suggests, and divide by $|H|$, so that you are asking for character values on $g e_H$, where $e_H = \frac{1}{|H|}\sum_{h\in H}h$ is the idempotent projecting on to $H$-invariants. Of course, $e_H^2 = e_H$, which tells us that $$ \chi(ge_H) = \chi(g e_H e_H) = \chi(e_H g e_H), $$ where in the last step we have used the cyclicity of trace: if $M$ is the representation with character $\chi$, then $\chi(AB) = tr_M(AB) = tr_M(BA) = \chi(BA)$ (here we take $A = ge_H$ and $B=e_H$). There are two things to point out about this.

First of all $e_H g e_H$ is, up to rescaling, the sum of all elements in an $(H,H)$-double coset. So I would contend that the most natural formulation of these character sums is for double cosets rather than cosets themselves.

Secondly, if $e$ is an idempotent in an associative algebra $A$, then $Ae$ is a left $A$-module, and $$ \mathrm{End}_{A-mod}(Ae) = eA^{op}e, $$ where the action of $eA^{op}e$ on an element of $Ae$ is by multiplication on the right (which is why the $op$ appears). Our situation is similar, but not exactly identical to this. One difference is that we are considering an irreducible module $M$ which is different from $Ae = \mathbb{C}G e_H$ (which is a permutation module). Another difference is that we are computing the trace of the module action (roughly, left multiplication) rather than a module endomorphism (roughly, right multiplication). But we can do some mathematics to reconcile the two settings.

Let us continue to work in the general setting where $A$ is an algebra and $e$ is an idempotent, and let's further assume that $A$ is semisimple and finite dimensional over $\mathbb{C}$ (which is certainly true in the case of group algebras). Then we have the following direct sum decomposition of $A$-modules: $$ Ae = \bigoplus_{V} V \otimes \hom(V, Ae), $$ where $V$ ranges over all simple $A$-modules, and the action of $A$ is on the left tensor factor. The character $\chi$ (corresponding to the irreducible $M$) evaluated at $exe \in eAe$ may be found by taking the trace on $Ae$ of $$ \frac{1}{\dim(\hom(V, Ae))} exe p_{M} $$ where $p_M$ is the central idempotent projecting onto the isotypic component of the irreducible $M$. But as Will Sawin pointed out, we can restrict to just taking the trace over the invariants, i.e. on $$ eAe = \bigoplus_{V} eV \otimes \hom(V, Ae) $$ rather than all of $Ae$.

Now, standard double-centraliser theory tells us that $$ eAe = \mathrm{End}_{A-mod}(Ae) = \bigoplus_{V} \mathrm{End}_{\mathbb{C}}(\hom(V, Ae)) $$ is a product of matrix algebras, so in particular is itself semisimple. For semisimple finite-dimensional algebras it turns out that taking the trace of left multiplication on the algebra by an element is equal to the the trace of right multiplication on the algebra by that element. One can check this directly for matrix algebras from which the semisimple case follows. Note though, that this is not usually true for non-semisimple algebras, see here. So let us consider the trace of $$ \frac{1}{\dim(\hom(V, Ae))} exe p_{M} = \frac{1}{\dim(\hom(V, Ae))} p_{M} exe $$ (since $p_M$ is central) on $$ eAe = \bigoplus_{V} eV \otimes \hom(V, Ae), $$ where the action is now by right multiplication. Now, the central idempotent $p_M$ annihilates all summands other than the one with $V=M$, where it acts by the identity. So we are left to compute the trace of $$ \frac{1}{\dim(\hom(V, Ae))} exe $$ on $\hom(V, Ae)$. This amounts to understanding $\hom(V, Ae)$ as an $\mathrm{End}_{A-mod}(Ae)$-module. Translated back to the original setting, we are studying the action of $\mathrm{End}(\mathbb{C}G e_H)$ on $\hom(M, \mathbb{C}G e_H)$, where $\mathbb{C}Ge_H$ and a permutation module; so as Geoff Robinson commented, the character sums are natural quantities that come up when studying permutation modules. But the real reason I went through the above is because it is now in a form where we can give some formulas in the case of the symmetric group (with $H$ a Young subgroup, which in particular subsumes the case $H = S_{n-k} \times S_1^k$).

Before we get into that, let's see some examples that are easy enough to compute directly. Specifically let $H = S_{n-1} \times S_1$ which has two double cosets in $G = S_n$, namely $H$ and $HsH$, where $s$ is the transposition $(n-1,n)$. One easily checks that $$ e_G = \frac{1}{n} e_H + \frac{n-1}{n} e_H s e_H. $$ The actions of $e_G$ and $e_H$ are easy enough to compute, so this equation gives us a way of computing the action of $e_H s e_H$. I wish to consider the action on the Specht module $M = S^{(n-1,1)}$. In that case $e_G$ acts by zero and $e_H$ acts by one, so we find that $e_H s e_H$ acts by $-1/(n-1)$.

By standard facts about representations of $S_n$, the multiplicity of $S^{\lambda}$ in the permutation module $M^{\lambda}$ is 1, so $\hom(S^{\lambda}, M^{\lambda})$ is one dimensional. So to compute the trace is to compute the whole action. In my paper, there is a formula giving the scalar for any $\lambda$ and $(S_\lambda, S_\lambda)$-double coset via a generating function.

The data of a $(S_\lambda, S_\lambda)$-double coset is a contingency table $q_{ij}$ with row and column sums equal to $\lambda$. If $c_{q_{ij}}$ is the scalar by which the double-coset indexed by $q_{ij}$ acts, then we have $$ \sum_{q} c_{(q_{ij})} \prod_{ij} x_{ij}^{q_{ij}} = \det(x_{11})^{\lambda_1 - \lambda_2} \det \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}^{\lambda_2 - \lambda_3} \cdots \det(X)^{\lambda_l}, $$ where $l$ is the length of $\lambda$ and $X$ is an $l \times l$ matrix with entries $x_{ij}$. Let's use this to verify the examples we computed before. In the paper I worked with carefully normalised cosets rather than idempotents because I wanted everything to be defined over the integers. So we have to account for this: it turns out that for the identity double coset $H = H \mathrm{Id} H$ we work with $$ \frac{|H \mathrm{Id} H|}{|H|}e_H = e_H, $$ but for $HsH$ we instead work with $$ \frac{|H s H|}{|H|} e_H s e_H = \frac{n! - (n-1)!}{(n-1)!} e_H s e_H = (n-1) e_H s e_H. $$ In our formula we take $\lambda = (n-1,1)$ so we must expand $$ x_{11}^{n-2}(x_{11}x_{22} - x_{12}x_{21})^{1} = (1)x_{11}^{n-1} x_{22} + (-1) x_{11}^{n-2}x_{12}x_{21}. $$ The first monomial tells us that $e_H$ acts by the scalar $1$. The second monomial tells us that $(n-1)e_H s e_H$ acts by the scalar $(-1)$, so that $e_H s e_H$ acts by $-1/(n-1)$ as we already knew.

One question remains, namely, how to understand $\hom(S^\lambda, M^\mu)$ as an $\mathrm{End}(M^\mu)$-module when $\mu \neq \lambda$. Well, in that case it is explained in the paper that by semisimplicity $$ \hom(S^\lambda, M^\mu) =\hom(M^\lambda, M^\mu) \otimes_{\mathrm{End}(M^\lambda)} \hom(S^\lambda, M^\lambda), $$ and moreover an explicit basis and composition rule is given of all the spaces involved (indexed by contingency tables). The upshot of computing it in this way is that, as the paper explains, we can upgrade the parts of $\lambda$ to polynomial variables, like how we computed the case of $(n-1,1)$ for general $n$ (but we can also interpolate parts of the partition other than the first). I believe that in my normalisation, the traces will be polynomials in the parts (in the case of one varying part, they will be polynomials in $n$), after passing to the idempotent normalisation, they will be rational functions (like $-1/(n-1)$ above).

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  • $\begingroup$ Minor comment: Group algebras are Frobenius and the left and right traces agree because the left regular representation is isomorphic to the dual of the right representation. So for groups this is ok in any characteristic. $\endgroup$ Oct 19, 2022 at 2:12

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