In this post, I mainly focus on introducing Gelfand pairs of finite groups.

Representations of finite groups

Let’s recall some concepts. Let \(G\) be a finite group. A (complex) representation of \(G\) is a pair \((\rho, V)\) where \(V\) is a vector space over \(\C\) and \(\rho: G \to GL(V)\) is a group homomorphism. Let \((\rho_1, V_1)\) and \((\rho_2, V_2)\) be representations of \(G\), a \(G\)-morphism, or simply a morphism, between these two representations is a linear map \(\varphi: V_1 \to V_2\) such that for any \(g \in G\) and \(v \in V_1\), \[\varphi(\rho_1(g)v) = \rho_2(g)\varphi(v).\] Then \(\mathrm{Hom}_G(\rho_1, \rho_2)\), the set of morphisms between \(\rho_1\) and \(\rho_2\), is a vector space over \(\C\). A representation \((\rho, V)\) is irreducible, if there is no proper subspace \(W \subset V\), such that \(\rho(g)w \in W\) for any \(g \in G\) and \(w \in W\). Between irreducible representations, we have Schur’s lemma.

Theorem 1 (Schur’s lemma). Let \(\rho_1\) and \(\rho_2\) be irreducible representations of \(G\). Then \[\begin{equation*} \mathrm{Hom}_G(\rho_1, \rho_2) \cong \left\{ \begin{array}{ll} \C, & \text{ if } \rho_1 \cong \rho_2; \\ 0, & \text{ otherwise.} \end{array} \right. \end{equation*}\]

It’s well known that \(G\) is semisimple, i.e., any representation of \(G\) decomposes into a direct sum of irreducible representations. A representation \((\rho, V)\) of \(G\) is called multiplicity free, if any irreducible representation appearing in the decomposition of \((\rho, V)\) appears exactly once, up to isomorphism.

Let \(H\) be a subgroup of \(G\). If \((\rho, V)\) is a representation of \(G\), we can certainly restrict the group homomorphism \(\rho: G \to GL(V)\) to \(H\) so that we can regard \((\rho, V)\) as a representation of \(H\). This is called the restriction of representations from \(G\) to \(H\), and the restricted representation of \(\rho\) will be denoted as \(r^G_H(\rho)\). Conversely, if \((\pi, W)\) is a representation of \(H\), we can build a representation \((\rho, V)\) of \(G\) out of \(\pi\), which is the induced representation, as follow.

\(\newcommand{\slash}{{\bf /}}\)

Let \(V\) be a space of functions defined by \[V = \{f: G \to W: f(hg) = \pi(h)f(g)\}.\] The group homomorphism \(\rho: G \to GL(V)\) is given by \[(\rho(g)f)(g^\prime) = f(g^\prime g).\] We use the notations \(\rho = i_H^G(\pi)\) and \(V = i_H^G(W)\) for induced representations. Restrictions and inductions of representations are related by Frobenius reciprocity theorem.

Theorem 2 (Frobenius reciprocity). Let \((\rho, V)\) be representation of \(G\) and let \((\pi, W)\) be representation of \(H\). Then there is a canonical linear isomorphism \(L: \mathrm{Hom}_G(\rho, i_H^G \pi) \to \mathrm{Hom}_H(r^G_H \rho, \pi)\) given by \[L(\varphi)(v) = \varphi(v)(e),\] where \(\varphi \in \mathrm{Hom}_G(\rho, i_H^G \pi), v \in V\) and \(e\) is the identity of \(G\). Moreover, the inverse \(L^{-1}: \mathrm{Hom}_H(r^G_H \rho, \pi) \to \mathrm{Hom}_G(\rho, i_H^G \pi)\) is given by \[L^{-1}(\psi)(v)(g) = \psi(\rho(g)v),\] where \(\psi \in \mathrm{Hom}_H(r^G_H \rho, \pi), v \in V\) and \(g \in G\).

Also, we have Mackey’s theorem relating two representations inducing from different representations. Let \(H_1\) and \(H_2\) be subgroups of \(G\). Let \((\pi_1, W_1)\) and \((\pi_2, W_2)\) be representations of \(H_1\) and \(H_2\), respectively. Consider the space \(\mathcal{D}\) of functions \(\Delta: G \to \mathrm{Hom}_\C(W_1, W_2)\) such that \[\Delta(h_2 g h_1) = \pi_2(h_2) \Delta(g) \pi_1(h_1),\] for any \(h_1 \in H_1, h_2 \in H_2\) and \(g \in G\).

Theorem 3 (Mackey’s theorem). As vector spaces, \[\mathcal{D} \cong \mathrm{Hom}_G(i_{H_1}^G \pi_1, i_{H_2}^G \pi_2).\]

**Remark. **If \((\pi_1, W_1) = (\pi_2, W_2)\), implicitly, this implies \(H_1 = H_2\). Let \(H=H_1=H_2\) and let \(\pi=\pi_1=\pi_2\). In this case, for any \(\Delta_1, \Delta_2 \in \mathcal{D}\), we can define convolution by \[\begin{equation*} \Delta_1 \ast \Delta_2(g) = \sum_{g^\prime \in G} \Delta_1({g^\prime}) \circ \Delta_2({g^\prime}^{-1}g). \end{equation*}\] Then \(\mathcal{D}\) becomes a ring. On the other hand, \(\mathrm{Hom}_G(i_H^G \pi, i_H^G \pi)\) is naturally a ring by compositions. Under this special situation, the isomorphism in Theorem 3 is actually a ring isomorphism.

Gelfand pairs with respect to trivial character

Let \(C(H \backslash G \slash H)\) be a set of functions \(\Delta: G \to \C\) such that \[\Delta(h_1 g h_2) = \Delta(g).\] Given \(\Delta_1, \Delta_2 \in C(H \backslash G \slash H)\), define the convolution \[\begin{equation} \Delta_1 \ast \Delta_2(g) = \sum_{{g^\prime} \in G} \Delta_1({g^\prime})\Delta_2({g^\prime}^{-1}g). \label{20150928-1} \end{equation}\]

It’s quite straightforward to check that \(\Delta_1 \ast \Delta_2 \in C(H \backslash G \slash H)\). Then under point-wise addition and convolution, \(C(H \backslash G \slash H)\) becomes a ring.

Theorem 4.Let \(G\) be a finite group and let \(H\) be a subgroup. Let \[C(H \backslash G) = \{f:G \to \C: f(hg)=f(g) \, \forall h \in H, g \in G\}.\] The right translation \(\rho\) of \(G\) on \(C(H \backslash G)\) is a representation. The following are equivalent.

(1) Every irreducible representation \(V\) of \(G\) has at most one \(H\)-fixed vector, up to scalar multiplication.

(2) \((\rho, C(H \backslash G))\) is multiplicity free.

(3) \(C(H \backslash G \slash H)\) is commutative under convolution.

Proof. Suppose statement \(1\) and statement \(2\) are equivalent. Let \((\chi_0, \C)\) be the trivial representation of \(H\), i.e., \(\chi_0(h) = id_\C\) for any \(h \in H\). Then by the definition of induction, \(\rho = i_H^G \chi_0\). \(\rho\) is multiplicity free, if and only if for any irreducible representation \(\pi\) of \(G\), \(\pi\) appears in \(\rho\) at most one. By Theorem 1, the number of times of \(\pi\) in \(\rho\) can be calculated by \(\dim \mathrm{Hom}_G(\pi, \rho)\). Using Theorem 2, this is the same as \(\dim \mathrm{Hom}_H(i_H^G \pi, \chi_0)\), i.e., the dimension of \(H\)-fixed vectors in \(\pi\).

Suppose statement \(2\) is equivalent to statement \(3\). By Theorem 3 (See also Remark after the theorem), \[C(H \backslash G \slash H) \cong \mathrm{Hom}_G(\rho, \rho) \text{ as rings}.\] Using Theorem 1, we can see that \(\rho\) is multiplicity free if and only if \(\mathrm{Hom}_G(\rho, \rho)\) is commutative, therefore if and only if \(C(H \backslash G \slash H)\) is commutative. \(\square\)

Definition 5.Let \(G\) be a finite groups and let \(H\) be a subgroup. Then \((G, H)\) is called a Gelfand pair (with respect to \(\chi_0\)), if \(G\) and \(H\) satisfy the equivalent statements in Theorem 4.

Gelfand pairs with respect to general characters

Let’s look back to Theorem 4 and see how it relates to the trivial character \(\chi_0\) of \(H\). As seen in the proof of Theorem 4, the first statement is equivalent to \(\dim \mathrm{Hom}_H(r^G_H \pi, \chi_0) \le 1\) for any irreducible representation \(\pi\) of \(G\). In the second statement, \[(\rho, C(H \backslash G)) = (i_H^G \chi_0, i_H^G \C).\] And in the third statement, we know by Theorem 3 that \[C(H \backslash G \slash H) \cong \mathrm{Hom}_G(\rho, \rho) = \mathrm{Hom}_G(i_H^G \chi_0, i_H^G \chi_0).\] Therefore, all statements are actually coming from \(\chi_0\) and it’s not surprising that Theorem 4 holds with \(\chi_0\) replaced by any character \(\chi\) of \(H\).

Let \(\chi\) be a character of \(H\). Define \(C(H \backslash G \slash H; \chi)\) to be a space of functions \(\Delta: G \to C\) such that \[\Delta(h_1 g h_2) = \chi(h_1)\Delta(g)\chi(h_2),\] for any \(h_1, h_2 \in H\) and \(g \in G\). Then \(C(H \backslash G \slash H; \chi)\) becomes a ring under the convolution defined in \(\eqref{20150928-1}\). Using Theorem 3, \[C(H \backslash G \slash H; \chi) \cong \mathrm{Hom}_G(i_H^G \chi, i_H^G \chi).\]

Now following the same proof in Theorem 4, we can prove a similar theorem. Henceforth, we get a more general definition for Gelfand pairs.

Theorem 6.The following are equivalent.

(1) \(\dim \mathrm{Hom}_H(r^G_H \pi, \chi) \le 1\), for any irreducible representation \(\pi\) of \(G\).

(2) \(i_H^G \chi\) is multiplicity free.

(3) \(C(H \backslash G \slash H; \chi)\) is commutative under the convolution.

Definition 7.Let \(G\) be a finite groups and let \(H\) be a subgroup. Then \((G, H)\) is called a Gelfand pair with respect to \(\chi\), if \(G\) and \(H\) satisfy the equivalent statements in Theorem 6.

Gelfand’s Lemma

We use notations as in previous sections. A set bijection \(\iota: G \to G\) is called an anti-involution if \(\iota(gh) = \iota(h)\iota(g)\), for any \(h, g \in G\). Then \(\iota\) induces an action on the double coset \(H \backslash G \slash H\), which sends \(HgH\) to \(H\iota(g)H\).

Theorem 8 (Gelfand’s lemma). Let \(\iota: G \to G\) be an anti-involution. The ring \(C(H \backslash G \slash H; \chi)\) is commutative under convolution if \(\iota\) acts trivially on \(H \backslash G \slash H\) and \(\chi\), i.e., \(\iota(r_i)=r_i\) for a set of representatives \(\{r_i\}\) of \(H \backslash G \slash H\) and \(\iota(h) \in H\) with \(\chi(h) = \chi(\iota(h))\) for \(h \in H\).

Proof. Define an action of \(\iota\) on \(C(H \backslash G \slash H; \chi)\) by \[\iota(\Delta)(g) = \Delta(\iota(g)).\] The action is well defined, because for \(h_1, h_2 \in H\), \[\begin{equation} \begin{split} \iota(\Delta)(h_1 g h_2) &= \Delta(\iota(h_2) \iota(g) \iota(h_1)) \\ &= \chi(\iota(h_2)) \Delta(\iota(g)) \chi(\iota(h_1)) \\ &= \chi(h_2) \iota(\Delta)(g) \chi(h_1), \end{split} \label{20150928-2} \end{equation}\]

so \(\iota(\Delta) \in C(H \backslash G \slash H; \chi)\). By an easy calculation, we can see that \(\iota\) is an anti-involution on \(C(H \backslash G \slash H; \chi)\), \[\iota(\Delta_1 \ast \Delta_2) = \iota(\Delta_2) \ast \iota(\Delta_1).\] Moreover, since \(\iota\) acts trivial on \(H \backslash G \slash H\), a calculation similar to \(\eqref{20150928-2}\) shows that \(\iota\) is actually the identity map on \(C(H \backslash G \slash H; \chi)\). If the identity map is anti-involution, then the ring \(C(H \backslash G \slash H; \chi)\) is commutative. \(\square\)

Remark.Actually, we don’t need \(\iota\) to act trivially on all \(r_i\). We just need \(\iota\) act trivially on those \(r_i\) having some \(\Delta\) such that \(\Delta(r_i) \ne 0\).

The condition that \(\iota\) acts trivially on \(\chi\) is obtained under either of the following situations.

(1) \(\iota\) acts trivially on \(H\), i.e., \(\iota(h)=h\) for \(h \in H\).

(2) \(\chi=\chi_0\) and \(\iota\) stabilizes \(H\), i.e., \(\iota(h) \in H\) for \(h \in H\).

An application: Uniqueness of Whittaker models

Let \(F\) be a finite field. Let \(G=GL(2, F)\) be the invertible \(2 \times 2\) matrices, and let \(N\) be a subgroup of \(G\) defined by \[\begin{equation*} N = \left\{ \left(\begin{array}{cc} 1 & x \\ 0 & 1 \end{array}\right) : x \in F \right\}. \end{equation*}\] \(N\) is so called the standard nilpotent subgroup of \(G\). Let \(\psi\) be a nontrivial additive character of \(F\), then it can be extended to a character of \(N\) given by \[\begin{equation*} \psi_N\left(\begin{array}{cc} 1 & x \\ 0 & 1 \end{array}\right) =\psi(x). \end{equation*}\] Then we can induce \(\psi_N\) to form a representation \((\rho, \mathcal{W}_\psi) = (i_N^G \psi_N, i_N^G \C)\) of \(G\). Given an irreducible representation \((\pi, V)\) of \(G\), we say \(\pi\) has a Whittaker model with respect to \(\psi_N\) if \(\pi\) can be embedded into \(\mathcal{W}_\psi\). We call the image of this embedding the Whittaker model of \(\pi\).

The motivation of Whittaker models is clear: We try to realize an abstract irreducible representation as a concrete space of functions \(f: G \to \C\) such that \[\begin{equation*} f\left(\left(\begin{array}{cc} 1 & x \\ 0 & 1 \end{array}\right)g\right) =\psi(x)f(g). \end{equation*}\]

Given an abstract irreducible representation \((\pi, V)\), there are two immediate questions should be ask:

(1) Does \(\pi\) have a Whittaker model with respect to some character \(\psi\)?

(2) If \(\pi\) has a Whittaker model, is it unique?

As for the first question, we need to understand what kinds of irreducible representation of \(G\) have. This would be another long story and not related to this post, so I would rather ignore it. But the upshot is that not every irreducible representation has a Whittaker model.

For the second question, it’s the same to ask whether or not \(\mathcal{W}_\psi\) is multiplicity free. Therefore, based on Theorem 6, it suffices to answer whether or not \((G, N)\) is a Gelfand pair with respect to \(\psi\).

Theorem 9 (Uniqueness of Whittaker models). \((G, N)\) is a Gelfand pairs with respect to any non-trivial character \(\psi\). Equivalently, \(\mathcal{W}_\psi\) is multiplicity free.

Proof. Under the light of Theorem 8, we just need to find an anti-involution \(\iota: G \to G\) such that \(\iota\) acts trivially on \(N \backslash G \slash N\) and \(N\) (see Remark after Theorem 8).

It’s from Bruhat decomposition that the representatives of \(N \backslash G \slash N\) are of forms \[\begin{equation*} \left(\begin{array}{cc} a & \\ & d \end{array}\right), \left(\begin{array}{cc} & b\\ c & \end{array}\right). \end{equation*}\] Recall that we want to prove \(C(N \backslash G \slash N; \psi)\) is commutative as in Theorem 8. Let \(\Delta \in C(N \backslash G \slash N; \psi)\), then \[\begin{equation*} \begin{split} &\Delta \left(\begin{array}{cc} a & \\ & d \end{array}\right) \\ &= \Delta \left( \left(\begin{array}{cc} 1 & -a\\ & 1 \end{array}\right) \left(\begin{array}{cc} a & \\ & d \end{array}\right) \left(\begin{array}{cc} 1 & d\\ & 1 \end{array}\right) \right) \\ &= \psi(d-a) \Delta \left(\begin{array}{cc} a & \\ & d \end{array}\right) \end{split} \end{equation*}\] Since \(\psi\) is non-trivial, \(\Delta\) is support on \(\left(\begin{array}{cc}a &\\& d \end{array}\right)\) if and only if \(a=d\). Therefore, we just need to find an anti-involution \(\iota: G \to G\) such that \(\iota\) acts trivially on \(N\) and on matrices of forms \[\begin{equation*} \left(\begin{array}{cc} a & \\ & a \end{array}\right), \left(\begin{array}{cc} & b\\ c & \end{array}\right). \end{equation*}\] Define \(\iota: G \to G\) by \[\begin{equation*} \iota(g) = \left(\begin{array}{cc} & 1\\ 1 & \end{array}\right) g^T \left(\begin{array}{cc} & 1\\ 1 & \end{array}\right), \end{equation*}\] where \(g^T\) is the transpose of \(g\). It’s easy to verify that \(\iota\) is the desired anti-involution. \(\square\)