### Archive

Archive for the ‘representation theory’ Category

## Grothendieck’s functions-sheaves correspondence

Today begins a shift in perspective for this series of posts. Whereas before we took as our basic objects of study the (abelian) étale covers of a smooth projective curve $X$, from now on we will study (rank $1$) $\ell$-adic local systems on $X$, which can be considered as a “linearized” version of the problem. More precisely, since understanding the structure of the fundamental group $\pi_1(X,\overline{\eta})$ (here $\overline{\eta}$ is the geometric generic point, not that it really matters) is tantamount to understanding all étale covers of $X$, the Tannakian philosophy says that we should consider the monoidal category of finite-dimensional representations of $\pi_1(X,\overline{\eta})$, from which the group can be recovered. Just as in topology, representations of the fundamental group are equivalent to local systems.

In fact, the other side of the Artin reciprocity map also has a useful interpretation in terms of $\ell$-adic local systems: characters of the Picard group $\text{Pic}(X)$ (more precisely, its profinite completion) correspond to “multiplicative” rank $1$ local systems on the Picard scheme $\text{Pic}_X$. This situation is very special to finite fields! A multiplicative local system on $\text{Pic}_X$ can be pulled back along the Abel-Jacobi map $X \to \text{Pic}_X$, given by $x \mapsto \mathscr{O}(x)$ on points, and global unramified class field theory says that this pullback establishes a bijection between multiplicative local systems on $\text{Pic}_X$ and rank $1$ local systems on $X$. This reformulation of class field theory is true over any field whatsoever, and has a beautiful geometric proof due to Deligne which we will hopefully get to next time.

Fix a prime $\ell$ (traditionally one assumes $\ell$ is not equal to $p$, the characteristic of our finite ground field $\mathbb{F}_q$, but for us $\ell = p$ is fine). Let $\overline{\mathbb{Q}}_{\ell}$ be an algebraic closure of the $\ell$-adic numbers $\mathbb{Q}_{\ell}$. The totally disconnected topology on $\overline{\mathbb{Q}}_{\ell}$ makes it better suited to our purposes than the complex numbers, although it is worth mentioning that they are isomorphic as discrete fields. For the most part we will only care that $\overline{\mathbb{Q}}_{\ell}$ is an algebraically closed field of characteristic zero.

Let $S$ be a connected scheme and $\overline{s} : \text{Spec } \Omega \to S$ a geometric point.

Definition An $\ell$-adic local system $\mathscr{F}$ on $S$ is a finite-dimensional continuous $\overline{\mathbb{Q}}_{\ell}$-representation of $\pi_1(S,\overline{s})$. The dimension of the representation is called the rank of $\mathscr{F}$.

As our terminology and notation suggest, $\ell$-adic local systems can be thought of as locally constant sheaves of $\overline{\mathbb{Q}}_{\ell}$-vector spaces. This is not literally true, however: in order to get nontrivial local systems one must first consider locally constant étale sheaves over finite coefficient rings, pass to pro-systems of these sheaves, then localize (or kill torsion) to obtain a $\overline{\mathbb{Q}}_{\ell}$-linear category, which one can prove is monoidally equivalent to representations of the fundamental group. It seems clear that there is no such procedure for complex coefficients. Although these beasts are not literally sheaves, most of our sheafy intuition and technique applies, whence the power of this approach.

We will choose our notation accordingly: for instance, if $f : T \to S$ and $\overline{t} \in f^{-1}(\overline{s})$, we will write $f^*\mathscr{F}$ for the local system on $T$ obtained by restricting $\mathscr{F}$ along the homomorphism $\pi_1(T,\overline{t}) \to \pi_1(S,\overline{s})$ induced by $f$. The underlying $\overline{\mathbb{Q}}_{\ell}$-vector space of the representation $\mathscr{F}$ is denoted by $\mathscr{F}_{\overline{s}}$.

Now we come to the namesake of this post, for which we should assume that $S$ is defined over $\mathbb{F}_q$. This construction takes an $\ell$-adic local system $\mathscr{F}$ on $S$ and produces from it a function $t_{\mathscr{F}} : S(\mathbb{F}_q) \to \overline{\mathbb{Q}}_{\ell}$. Given $x : \text{Spec } \mathbb{F}_q \to S$,  there is an isomorphism $\pi_1(S,\overline{x}) \cong \pi_1(S,\overline{s})$ well-defined up to conjugation, so that after making this choice we can form the pullback $x^*\mathscr{F}$, a local system on $\text{Spec } \mathbb{F}_q$. Then we set

$t_{\mathscr{F}}(x) = \text{tr}(\text{Frob};x^*\mathscr{F})$,

the trace of the action of the Frobenius $a \mapsto a^q$, which does not depend on our choice because the trace is conjugation-invariant.

We will be interested in the case  where $\mathscr{F}$ has rank $1$, and then it is easier to describe $t_{\mathscr{F}}$. Namely, $x$ determines a canonical map $\widehat{\mathbb{Z}} = \pi_1(\mathbb{F}_q) \to \pi_1(S,\overline{s})^{\text{ab}}$, which we compose with $\mathscr{F}$ (now thought of a one-dimensional representation) to obtain a homomorphism $\widehat{\mathbb{Z}} \to \overline{\mathbb{Q}}_{\ell}^{\times}$. Evaluate this map at $1$ to obtain $t_{\mathscr{F}}(x)$.

Sometimes interesting classes of functions and sheaves match up under this correspondence. Let $G$ be a commutative algebraic group over $\mathbb{F}_q$: then one such class of functions is the group of characters of $G(\mathbb{F}_q)$, meaning one-dimensional representations $G(\mathbb{F}_q) \to \overline{\mathbb{Q}}_{\ell}^{\times}$. What sort of local system $\mathscr{F}$ on $G$ has the property that $t_{\mathscr{F}}$ is a character?

Definition A rank $1$ local system $\mathscr{F}$ on $G$ is called a character sheaf provided that $\mu^*\mathscr{F} \cong \mathscr{F} \boxtimes \mathscr{F}$, where $\mu : G \times G \to G$ is the multiplication map.

(Notation for those who haven’t seen it: if $\mathscr{F}$ is a sheaf on $S$ and $\mathscr{G}$ is a sheaf on $T$, then their external tensor product is $\mathscr{F} \boxtimes \mathscr{G} = p_S^*\mathscr{F} \otimes p_T^*\mathscr{G}$ where $p_S : S \times T \to S$ and $p_T : S \times T \to T$ are the projections.)

Sometimes character sheaves are called multiplicative local systems. The latter terminology is arguably better, since “character sheaf” has other meanings. This is analogous to how “character” can refer not only to a one-dimensional representation but also to the trace function associated to a higher-dimensional representation.

Before we prove the main result of this post, we need a lemma.

Lemma Let $\mathscr{F}$ be an $\ell$-adic local system on $S$. Then there is a canonical isomorphism $\mathscr{F} \to \text{Fr}_S^*\mathscr{F}$.

Proof. We reduce immediately to the case that $\mathscr{F}$ is a locally constant étale sheaf of finite sets, so there is some finite étale map $f : T \to S$ whose sheaf of local sections is $\mathscr{F}$. Write $\text{Fr}_S^*T$ for the fiber product of $\text{Fr}_S : S \to S$ and $f$, so it suffices to produce an isomorphism $T \to \text{Fr}_S^*T$ of $S$-schemes. Using the relation $f \circ \text{Fr}_T = \text{Fr}_S \circ f$, we obtain the desired map $g : T \to \text{Fr}_S^*T$ from $\text{Fr}_T$ and $f$. Since $f$ and $\text{Fr}_S^*T \to S$ are finite étale, so is $g$, and similarly $g$ is radicial (i.e. universally injective) because $\text{Fr}_T$ is. But a map which is both étale and radicial must be an open embedding, and $g$ is also finite, hence an isomorphism.

$\Box$

Now we come to the really interesting part. As usual $G$ is a commutative algebraic group over $\mathbb{F}_q$, which we now assume to be smooth and connected (the smoothness hypothesis is really not necessary).

Proposition Under the above assumptions, $\mathscr{F} \to t_{\mathscr{F}}$ is a bijection from character sheaves on $G$ to characters of $G(\mathbb{F}_q)$.

Proof. That $t_{\mathscr{F}}$ is a character follows from the easy identities $t_{f^*\mathscr{F}} = f^*t_{\mathscr{F}}$ and $t_{\mathscr{F} \otimes \mathscr{G}} = t_{\mathscr{F}} \cdot t_{\mathscr{G}}$. Suppose that we are given a character $\chi : G(\mathbb{F}_q) \to \overline{\mathbb{Q}}_{\ell}^{\times}$. The Lang isogeny $L : G \to G$ is a pointed finite Galois covering with group $G(\mathbb{F}_q)$, hence gives rise to a map $\pi_1(G,\overline{1}) \to G(\mathbb{F}_q)$, which we can restrict $\chi$ along to obtain a rank $1$ local system $\mathscr{F}(\chi)$ on $G$. The same identities show that $\mathscr{F}(\chi)$ is a character sheaf if one argues in the opposite direction.

It remains to show that these constructions are mutually inverse, and first we’ll check that $t_{\mathscr{F}(\chi)} = \chi$. Given $x \in G(\mathbb{F}_q)$, we obtain a canonical map $\widehat{\mathbb{Z}} = \pi_1(\mathbb{F}_q) \to \pi_1(G,\overline{1})^{\text{ab}}$, whose value on $1 = \text{Frob}$ we will call $\text{Frob}_x$. By definition $t_{\mathscr{F}(\chi)}(x)$ is the value of $\mathscr{F}(\chi)$ on $\text{Frob}_x$, but $\mathscr{F}(\chi)$ factors through $\chi : G(\mathbb{F}_q) \to \overline{\mathbb{Q}}_{\ell}^{\times}$ by construction, so it suffices to show that the map $\pi_1(G,\overline{1}) \to G(\mathbb{F}_q)$ induced by $L$ sends $\text{Frob}_x$ to $x$. This means that when $\text{Frob}_x$ acts on the fiber $L^{-1}(\overline{1}) = G(\mathbb{F}_q)$, it just translates by $x$. Fixing $\overline{y} \in L^{-1}(\overline{x})$, we obtain an identification $L^{-1}(\overline{1}) \cong L^{-1}(\overline{x})$ since $L^{-1}(\overline{x}) = \overline{y}G(\mathbb{F}_q)$. Tracing through definitions, our claim follows from the calculation

$\text{Frob}_x \cdot \overline{y} = \text{Fr}_G(\overline{y}) = \overline{x} \overline{y}$.

Finally, we must prove that $\mathscr{F}(t_{\mathscr{F}}) = \mathscr{F}$. We claim that $L^*\mathscr{F}$ is trivial, or equivalently that $\mathscr{F}$ factors through the homomorphism $\pi_1(G,\overline{1}) \to G(\mathbb{F}_q)$ determined by $L$. Since this map sends $\text{Frob}_x \mapsto x$ the claim implies that $\mathscr{F}$ is determined by its values on $\text{Frob}_x$ for all $x \in G(\mathbb{F}_q)$, and we just proved that $t_{\mathscr{F}(t_{\mathscr{\chi}})} = t_{\mathscr{F}}$, from which it follows that $\mathscr{F}(t_{\mathscr{F}}) = \mathscr{F}$. As for the claim, observe that

$L^*\mathscr{F} = (i,\text{Fr}_G)^*\mu^*\mathscr{F} = (i,\text{Fr}_G)^*(\mathscr{F} \boxtimes \mathscr{F}) = i^*\mathscr{F} \otimes \text{Fr}_G^*\mathscr{F}$.

By the lemma $\text{Fr}_G^*\mathscr{F} = \mathscr{F}$, so we just have to check that $i^*\mathscr{F} \otimes \mathscr{F}$ is trivial. The latter sheaf is the pullback of $\mathscr{F}$ along $\mu \circ (i,\text{id}_G) : G \to G$, which is the trivial homomorphism, and since $t_{\mathscr{F}}(1) = 1$ we are done.

$\Box$

Next time we will consider $G = \text{Pic}_X$, a disconnected group for which the proposition is almost true.

## Conjugacy classes in the finite general and special linear groups

Now that I’m finally done with school for the summer, I’d like to get back into the routine of blogging regularly. If you were following last summer: I never completed my project of understanding the Weil representation, so I probably won’t be continuing that series of posts. I may be helping some people complete that project this summer, in which case I can hopefully link to some further information eventually.

This week I’m going to give a detailed description of the conjugacy classes in $\text{GL}_2(\mathbb{F}_q)$ and $\text{SL}_2(\mathbb{F}_q)$, where $\mathbb{F}_q$ is the finite field with $q$ elements. This is relevant to representation theory because the conjugacy classes in a finite group correspond bijectively to irreducible representations, and in particular we will find out how many irreducible representations these groups have. A quick Google search reveals that it is easy to find the final answers, but somewhat harder to find a careful explanation, which is what I will attempt now.

First, the general linear group: for any $A \in \text{GL}_2(\mathbb{F}_q)$, consider the $\mathbb{F}_q[t]$-module $\mathbb{F}_q^2$ where $t$ acts by $A$. Two matrices are conjugate if and only if the corresponding modules are isomorphic, and it is easy to analyze these isomorphism classes using the structure theorem for principal ideal domains. Note that since we are counting invertible matrices, we need only consider polynomials with nonzero constant term.

• The nonzero scalar matrices are precisely the center of $\text{GL}_2(\mathbb{F}_q)$, so these account for $q-1$ conjugacy classes with one element each.
• For each $\lambda,\mu \in \mathbb{F}_q^{\times}$ such that $\lambda \neq \mu$, there is the semisimple conjugacy class of matrices with minimal polynomial $(x-\lambda)(x-\mu)$: the centralizer of such a matrix is a split maximal torus $\mathbb{F}_q^{\times} \times \mathbb{F}_q^{\times} \subset \text{GL}_2(\mathbb{F}_q)$, so each of these $\frac12 (q-1)(q-2)$ conjugacy classes has $q^2+q$ elements.
•  For each $\lambda \in \mathbb{F}_q^{\times}$ there is a conjugacy class of matrices with minimal polynomial $(x-\lambda)^2$ which are not semisimple, and hence conjugate to a Jordan block. If we write a Jordan block as $\lambda I + N$, where $N$ is the nilpotent matrix defined by $Ne_1 = 0$ and $Ne_2 = e_1$, it is easy to see that the centralizer consists of matrices of the form $a I + bN$ where $a,b \in \mathbb{F}_q$ and $a \neq 0$. Thus each of these $q-1$ conjugacy classes has $q^2 - 1$ elements.
•  Finally, there are the matrices which have no eigenvalue in $\mathbb{F}_q$, and therefore have a conjugate pair of eigenvalues $\alpha,\alpha^q \in \mathbb{F}_{q^2} \setminus \mathbb{F}_q$. Such matrices are semisimple because $\mathbb{F}_q$ is perfect, so their conjugacy class is determined by their eigenvalues, and in particular we see that there are $\frac12 (q^2 - q)$ conjugacy classes of these matrices. If $A \in \text{GL}_2(\mathbb{F}_q)$ has eigenvalue $\alpha \in \mathbb{F}_{q^2} \setminus \mathbb{F}_q$, then the subalgebra $\mathbb{F}_q[A] \subset \text{Mat}_2(\mathbb{F}_q)$ is isomorphic to $\mathbb{F}_{q^2}$. If we use the basis $1,\alpha$ to identify $\mathbb{F}_{q^2}$ with $\mathbb{F}_q \oplus \mathbb{F}_q$, we get an isomorphism $\text{Mat}_2(\mathbb{F}_q) \cong \text{End}_{\mathbb{F}_q}(\mathbb{F}_{q^2})$, and here $\mathbb{F}_q[A]$ corresponds to $\text{End}_{\mathbb{F}_{q^2}}(\mathbb{F}_{q^2}) \cong \mathbb{F}_{q^2}$. The centralizer of this subalgebra is $\mathbb{F}_{q^2}^{\times}$, so we see that the centralizer of $A$ in $\text{GL}_2(\mathbb{F}_q)$ is isomorphic to the non-split torus $\mathbb{F}_{q^2}^{\times}$ and in particular the conjugacy class of $A$ has $q^2 - q$ elements.

Note that the total number of conjugacy classes of $\text{GL}_2(\mathbb{F}_q)$ is

$q-1 + \frac12 (q-1)(q-2) + q-1 + \frac12 (q^2 - q) = q^2 - 1.$

As for $\text{SL}_2(\mathbb{F}_q)$, we first find the $\text{GL}_2(\mathbb{F}_q)$-conjugacy classes in $\text{SL}_2(\mathbb{F}_q)$ and then determine how they split into $\text{SL}_2(\mathbb{F}_q)$-conjugacy classes. Unfortunately, we must now keep track of whether $q$ is even or odd.

• The center of $\text{SL}_2(\mathbb{F}_q)$ is trivial if $q$ is even or $\{ \pm I \}$ if $q$ is odd. Hence this accounts for one conjugacy class if $q$ is even or two if $q$ is odd, with one element each in either case.
• For each $\lambda \in \mathbb{F}_q^{\times}$ with $\lambda \neq \pm 1$, there is the semisimple conjugacy class of matrices with minimal polynomial $(x-\lambda)(x-\lambda^{-1})$. If $q$ is even then there are $\frac12 (q-2)$ of these conjugacy classes, and if $q$ is odd then there are $\frac12 (q-3)$. We already saw that the stabilizer of such a matrix in $\text{GL}_2(\mathbb{F}_q)$ is a split maximal torus, so each conjugacy class has $q^2+1$ elements.
• There are matrices with minimal polynomial $(x \pm 1)^2$ which are not semisimple, and hence conjugate to a Jordan block. If $q$ is even then there is only one such$\text{GL}_2(\mathbb{F}_q)$-conjugacy class, and if $q$ is odd then there are two. We saw that the stabilizer in $\text{GL}_2(\mathbb{F}_q)$ of such a matrix has $q(q-1)$ elements, so these $\text{GL}_2(\mathbb{F}_q)$-conjugacy classes contain $q^2-1$ matrices each.
• The conjugacy classes of matrices which have no eigenvalue in $\mathbb{F}_q$ are parameterized by conjugate pairs $\alpha,\alpha^q \in \mathbb{F}_{q^2} \setminus \mathbb{F}_q$ where $\alpha^{q+1} = 1$. The latter equation has $q+1$ solutions in the cyclic group $\mathbb{F}_{q^2}^{\times}$, and if $q$ is even only one of those solutions comes from $\mathbb{F}_q$, while if $q$ is odd then two do. Thus there are $\frac12 q$ such conjugacy classes if $q$ is even and $\frac12 (q-1)$ if $q$ is odd. As we saw, the stabilizers in $\text{GL}_2(\mathbb{F}_q)$ of these matrices are non-split maximal tori, so each of these conjugacy classes has $q^2-q$ elements.

So we have described the  $\text{GL}_2(\mathbb{F}_q)$-conjugacy classes in  $\text{SL}_2(\mathbb{F}_q)$, but it remains to see how these split as $\text{SL}_2(\mathbb{F}_q)$-conjugacy classes. We will show momentarily that semisimple $\text{GL}_2(\mathbb{F}_q)$-conjugacy classes in $\text{SL}_2(\mathbb{F}_q)$ do not split further as $\text{SL}_2(\mathbb{F}_q)$-conjugacy classes, and here the only non-semisimple matrices are conjugate to one of the Jordan blocks $\pm I + N$ (where $N$ is the nilpotent matrix mentioned earlier). Let’s write $G = \text{GL}_2(\mathbb{F}_q)$ and $H = \text{SL}_2(\mathbb{F}_q)$ for the moment to improve the notation. Now if $X$ is the $G$-conjugacy class in $H$ of $A \in H$, then $X \cong G/C_G(A)$ as $G$-sets and in particular as $H$-sets. In particular we get a bijection $X/H \cong G/HC_G(A) \cong \mathbb{F}_q^{\times}/\text{det} C_G(A)$. We saw earlier that if $A$ is a Jordan block then $C_G(A)$ consists of matrices of the form $aI + bN$ with $a,b \in \mathbb{F}_q$ and $a \neq 0$, so $\text{det} C_G(A) = \mathbb{F}_q^{\times 2}$ is the subgroup of squares. Thus if $q$ is odd then the two $G$-conjugacy classes of $\pm I + N$ split into two $H$-conjugacy classes with $\frac12 (q^2-1)$ elements each, and if $q$ is even then the $G$-conjugacy class of $I + N$ does not split further as an $H$-conjugacy class. We see now that if $q$ is odd then $\text{SL}_2(\mathbb{F}_q)$ has

$2 + \frac12 (q-3) + 4 + \frac12 (q-1) = q+4$

conjugacy classes, and if $q$ is even then the number is

$1 + \frac12 (q-2) + 1 + \frac12 q = q+1.$

It remains to show that if $X \subset H$ is a semisimple $G$-conjugacy class, then $X$ does not split further as an $H$-conjugacy class. This is true for $G = \text{GL}_n(F)$ and $H = \text{SL}_n(F)$ where $n \geq 1$ is arbitrary and $F$ is any field with the property that the norm map $N_{E/F} : E^{\times} \to F^{\times}$ is surjective for any finite extension $E/F$. Even more generally, suppose $R$ is a finite-dimensional commutative semisimple algebra over such a field $F$, and $M$ a finite-dimensional $R$-module. Then we have the determinant map $\text{Aut}_F(M) \to F^{\times}$, and we claim the subgroup $\text{Aut}_R(M)$ surjects onto $F^{\times}$. Now $R \cong E_1 \times \cdots \times E_r$ by the semisimplicity hypothesis, where each $E_i$ is a finite field extension of $F$, so $M \cong M_1 \times \cdots \times M_r$ where each $M_i$ is an $E_i$-vector space and $R$ acts diagonally. Thus the automorphism group splits as well:

$\text{Aut}_R(M) \cong \text{Aut}_{E_1}(M_1) \times \cdots \times \text{Aut}_{E_r}(M_r)$.

It is enough to show that $\text{det} : \text{Aut}_{E_i}(M_i) \to F^{\times}$ is surjective for some $1 \leq i \leq r$, so we have reduced to the case that $R = E$ is a finite field extension of $F$. But now we can see from the definitions that the determinant $\text{Aut}_E(M) \to F^{\times}$ factors into the determinant $\text{Aut}_E(M) \to E^{\times}$ followed by the norm $E^{\times} \to F^{\times}$, and the latter is surjective by assumption. Applying this to the case when $A \in H$ is semisimple, $R = F[A] \subset \text{Mat}_n(F)$, and $M = F^n$, we have $\text{Aut}_R(M) = C_G(A)$ and the claim follows.

Categories: representation theory

## A connection with the Hilbert symbol

This time we will talk about an interesting arithmetical feature of the Weil index, namely its relationship with the Hilbert symbol from quadratic local class field theory. The proof of the theorem is based on the proof in Daniel Bump’s Automorphic Forms and Representations, and the ideas are due to Weil. For the moment, let $F$ be any local field of characteristic not equal to $2$.

Given $a,b \in F^{\times}$ we can define an $F$-algebra $Q(a,b)$ to be the $F$-vector space with basis $1,i,j,k$ and multiplication given by

$i^2 = a, j^2 = b, k^2 = -ab, ij = -ji = k, jk = -kj = -bi, \text{ and } ik = -ki = aj$.

Proposition Either $Q(a,b) \cong \text{Mat}_2(F)$ is the $2 \times 2$ matrix algebra over $F$ or $Q(a,b)$ is a division algebra.

Proof. It is easy to see by direct calculation that $Q(a,b)$ is a central simple $F$-algebra, meaning it has center $F$ and the only two-sided ideals are $0$ and $Q(a,b)$. Thus by the structure theory for central simple algebras (see, for instance, Noncommutative Algebra by R. Keith Dennis and Benton Farb) there is a central division algebra $D$ over $F$ and $n \geq 0$ such that $Q(a,b) \cong \text{Mat}_n(D)$. Clearly the only possibilities are $D = Q(a,b)$ and $n = 1$ or $D = F$ and $n = 2$.

$\Box$

Definition Let $a,b \in F^{\times}$ and define the Hilbert symbol $(a,b)$ to be $1$ if $Q(a,b) \cong \text{Mat}_2(F)$ or $-1$ if $Q(a,b)$ is a division algebra.

It is not hard to see that $(a,b) = 1$ if and only if $a$ is in the image of the norm $F[\sqrt{b}] \to F$, or equivalently $z^2 = ax^2 + by^2$ has a solution with $x,y,z \in F$ not all zero, whence the Hilbert symbol is trivial when $F$ is the field of complex numbers or finite. If $F$ is non-Archimedean, recall that we defined the Weil index $\gamma : W(F) \to \mathbb{C}^{\times}$. For $a \in F^{\times}$ we will abuse notation and write $\gamma( \langle a \rangle ) = \gamma(a)$.

Theorem (Weil) Let $F$ be a non-Archimedean local field of characteristic not equal to $2$. For all $a,b \in F^{\times}$ we have

$(a,b) = \frac{\gamma(1) \gamma(ab)}{\gamma(a)\gamma(b)}$.

Before we go on to the proof, which is somewhat difficult, let us give an easy but surprising application of this result.

Corollary The Weil index $\gamma : W(F) \to \mathbb{C}^{\times}$ takes values in $8$th roots of unity.

Proof. Taking $a = b = -1$ in the equation from the theorem, we get

$\gamma(1)^8 = \big( \frac{\gamma(1)^2}{\gamma(-1)^2} \big)^2 = (1,-1)^2 = 1$.

Then we can deduce that for any $a \in F^{\times}$, we have

$\gamma(a)^8 = (\gamma(a)^2)^4 = (\gamma(1)^2 (a,a))^4 = 1$.

The result follows for general quadratic spaces by diagonalization.

$\Box$

The algebra $Q(a,b)$ has an antiautomorphism called conjugation, which sends $\xi = x + yi + zj + wk$ to $\overline{\xi} = x - yi - zj - wk$. Then the reduced norm $\nu : Q(a,b) \to F$ is given by $\nu(\xi) = \xi \overline{\xi}$, and it is not hard to see that $\nu$ is a nondegenerate quadratic form on $Q(a,b)$. In fact, $Q(a,b) \cong \langle 1 \rangle \oplus \langle -a \rangle \oplus \langle -b \rangle \oplus \langle ab \rangle$ with respect to the form $nu$.

Proof of the theorem. Note that if $Q(a,b) \cong \text{Mat}_2(F)$ then $\nu$ coincides with the determinant, and from this one deduces that $Q(a,b) \cong \mathfrak{H} \oplus \mathfrak{H}$ as quadratic spaces and hence $\gamma(\nu) = 1$.

So we need to treat the case where $Q(a,b) = D$ is a division algebra. First observe that the absolute value on $F$ extends to a non-Archimedean absolute value on $D$ as follows: for any $\alpha \in D \setminus F$, we can consider the degree two extension $F[\alpha] / F$ of local fields, which is automatically separable because $F$ has odd (or zero) characteristic. It is well-known that in this situation the absolute value on $F$ extends to $F[\alpha]$ via the formula $| \alpha | = |N(\alpha)|^{1/2} = |\nu(\alpha)|^{1/2}$, where $N : F[\alpha] \to F$ is the field norm. Thus by letting $\alpha$ vary we get the desired non-Archimedean absolute value on $D$. Moreover, $\nu^{-1}(\mathcal{O}_F) = \{ x \in D \ | \ |x| \leq 1 \}$ is a maximal $\mathcal{O}_F$-order in $D$, and in particular is an $\mathcal{O}_F$-lattice.

Let $\mu$ be the canonical additive Haar measure on $D$ which is self-dual with respect to $\nu$. Recall that we proved last time that

$\gamma(\nu) = \int_{\Lambda} \psi(\frac12 \nu(x)) \ d\mu(x)$

for any sufficiently large lattice $\Lambda \subset D$ (we fixed a continuous additive character $F : \psi \to \mathbb{T}$ a long time ago). Choose a uniformizer $\pi \in \mathcal{O}_F$: if we make $n \geq 0$ large enough, then the lattice $\Lambda = \nu^{-1}(\pi^{-n}\mathcal{O}_F)$ will work.

We would like to rewrite the integral above as an integral in $F$ by pushing forward along $\nu$, but first we must pass to the multiplicative Haar measure $d\mu^{\times}(x) = | \nu(x) |^{-2} d\mu(x)$, so the integral becomes

$\int_{\nu^{-1}(\pi^{-n}\mathcal{O}_F)} \psi(\frac12 \nu(x) ) |\nu(x)|^2 \ d\mu^{\times}(x)$.

Since the reduced norm $\nu$ is a proper (meaning preimages of compact sets are compact) homomorphism, it follows that $\nu_*\mu^{\times}$ is a multiplicative Haar measure on $F^{\times}$. Now if we write $\alpha$ for the additive Haar measure on $F$ satisfying $\alpha(\mathcal{O}_F) = 1$, then $d\alpha^{\times}(z) = |z|^{-1} d\alpha(z)$ defines a multiplicative Haar measure on $F^{\times}$, so it follows that $\nu_*\mu^{\times}$ is a scalar multiple of $\alpha^{\times}$, which we will write as $\nu_*\mu^{\times} \sim \alpha^{\times}$. Thus

$\gamma(\nu) \sim\int_{\pi^{-n}\mathcal{O}_F} \psi(\frac12 z) | z |^2 \ d\alpha^{\times}(z) \sim \int_{\pi^{-n}\mathcal{O}_F} \psi(z) | z | \ d\alpha(z)$.

Decompose $\pi^{-n}\mathcal{O}_F = \coprod_{k = -n}^{\infty} \pi^k\mathcal{O}_F^{\times}$, so the integral can be rewritten as

$\sum_{k = -n}^{\infty} q^{-k} \int_{|z| = q^{-k}} \psi(z) \ d\alpha(z)$.

Suppose $\pi^r\mathcal{O}_F$ is the conductor of $\psi$, meaning the largest $\mathcal{O}_F$-submodule of $F$ on which $\psi$ is trivial. Then an easy computation shows that $\int_{|x| = q^{-k}} \psi(x) \ dx = q^{-k}(1-q^{-1})$ if $k \geq r$, $-q^{-r}$ if $k = r-1$, and $0$ if $k < r-1$, so for sufficiently large $n$ we see that

$\sum_{k = -n}^{\infty} q^{-k} \int_{|z| = q^{-k}} \psi(z) \ d\alpha(z) = -q^{1-2r} + (1-q^{-1})\sum_{k = r}^{\infty} q^{-2k} = \frac{-q^{1-2r}}{1+q^{-1}}$

is negative. But we know that $|\gamma(\nu)| = 1$, so $\gamma(\nu) < 0$ implies $\gamma(\nu) = -1$.

$\Box$

## Proving the existence of the Weil index

Before going on to the proof of the theorem from last time, we need to prove the basic result about Gauss sums on finite abelian groups. A quadratic form on a finite abelian group $A$ is a map $q : A \to \mathbb{C}^{\times}$ satisfying $q(n \cdot a) = q(a)^{n^2}$ for $a \in A$ and $n \in \mathbb{Z}$, and if we define $B : A \times A \to \mathbb{C}^{\times}$ by the formula $B(x,y) = q(x+y)q(x)^{-1}q(y)^{-1}$, we also require that $y \mapsto B(x,y)$ is a group homomorphism $A \to \mathbb{C}^{\times}$ for all $x \in A$. The dual group of $A$ is $A^* = \text{Hom}(A,\mathbb{C}^{\times})$, and we call $q$ nondegenerate if the homomorphism $A \to A^*$ which sends $x \mapsto B(x,-)$ is an isomorphism.

The Gauss sum associated with $q$ is the number

$S(q) = \sum_{a \in A} q(a)$.

Lemma If $q : A \to \mathbb{C}^{\times}$ is a nondegenerate quadratic form, then $|S(q)| = \sqrt{\# A}$.

Proof. We begin with the calculation

$|S(q)|^2 = S(q) \cdot \overline{S(q)} = \sum_{a,b \in A} q(a)\overline{q(b)} = \sum_{a,b \in A} q(a-b)B(a-b,b)$.

Next make the substitutions $x = a-b$ and $y = b$, so the above expression becomes

$\sum_{x \in A} q(x) \sum_{y \in A} B(x,y)$.

Now if $x \neq 0$ then $y \mapsto B(x,y)$ is a nontrivial character of $A$, whence $\sum_{y \in A} B(x,y) = 0$. So the sum simplifies to

$q(0) \sum_{y \in A} B(0,y) = \# A$.

$\Box$

Now let us prove the theorem. Recall our situation: $F$ is a non-Archimedean local field of characteristic not equal to $2$, $V$ is a finite-dimensional vector space over $F$, and $q : V \to F$ is a nondegenerate quadratic form. Fix a nontrivial continuous character $\psi : F \to \mathbb{T}$, and let $\mu$ be the Haar measure on $V$ which is self-dual with respect to $\psi \circ q$. Consider the function $f_q : V \to \mathbb{C}$, which is locally constant and therefore may be regarded as a distribution on $V$. Then the statement of the theorem is

Theorem (Weil) There exists $\gamma(q) \in \mathbb{C}^{\times}$, called the Weil index of $q$, such that

$\mathcal{F}(f_q) = \gamma(q) \cdot f_{-q}$.

Furthermore, we have $|\gamma(q)| = 1$, and for any sufficiently large lattice $\Lambda \subset V$,

$\gamma(q) = \int_{\Lambda} f_q \ d\mu$.

Proof. A straightforward calculation shows that $T_af_q = f_q(a)M_{a}f_q$ for $a \in V$, and then the first part of the lemma from last time tells us that $M_{-a}\mathcal{F}(f_q) = f_q(a)T_a\mathcal{F}(f_q)$.

Now we apply the second part of that lemma to see that

$T_a(f_q \cdot \mathcal{F}(f_q)) = T_{-a}f_q \cdot T_a\mathcal{F}(f_q) = f_q(a)f_q(a)^{-1}M_{a}f_q \cdot M_{-a}\mathcal{F}(f_q) = f_q \cdot \mathcal{F}(f_q)$,

so $f_q \cdot \mathcal{F}(f_q)$ is a translation-invariant distribution. If we consider the action of $V$ on $C^{\infty}_c(V)$ by translations, then this means $f_q \cdot \mathcal{F}(f_q)$ factors through the coinvariants for this representation, which are one-dimensional. Since integration against $d\mu$, i.e. the constant distribution $1$, also has this property, we see that $f_q \cdot \mathcal{F}(f_q)$ is a constant distribution $\gamma(q)$ and $\mathcal{F}(f_q) = \gamma(q) \cdot f_{-q}$ as desired.

Before we prove $|\gamma(q)| = 1$, let us pass to the more concrete formula for $\gamma(q)$. It is easy to see that for any $g \in C^{\infty}_c(V)$ we have

$(f_q * g)(x) = f_q(x)\mathcal{F}(f_q \cdot g)(-x) = f_q(x)\mathcal{F}^{-1}(f_q \cdot g)(x)$,

and since the latter is compactly supported we see that $f_q * g$ is also. In fact, we can even conclude that

$\int_V f_q * g \ d\mu = \int_V f_q \mathcal{F}^{-1}(f_q \cdot g) \ d\mu = (f_q \cdot \mathcal{F}(f_q))(g) = \gamma(q)\int_V g \ d\mu$.

We can simplify this even further: first find a sufficiently large lattice $\Lambda$ so that $f_q|_{\Lambda^{\perp}}$ is trivial. This can be done by starting with any lattice $X \subset V$ and taking $\Lambda = (2 \cdot (X \cap X^{\perp}))^{\perp}$. Then let $g = e_{\Lambda^{\perp}}$ be the indicator function, so

$(f_q * e_{\Lambda^{\perp}})(x) = \int_{\Lambda^{\perp}} f_q(x+y) \ d\mu(y) = f_q(x) \int_{\Lambda^{\perp}} \psi(B(x,y)) \ d\mu(y)$.

Now if $x \notin \Lambda$, then $y \mapsto \psi(B(x,y))$ is a nontrivial smooth character of $\Lambda^{\perp}$, whence its integral is zero. On the other hand, if $x \in \Lambda$ then by definition $\psi(B(x,y)) = 1$ for all $y \in \Lambda^{\perp}$. Thus

$(f_q * e_{\Lambda^{\perp}})(x) = \mu(\Lambda^{\perp})f_q(x)e_{\Lambda}(x)$,

so by substituting into the last formula for $\gamma(q)$ we get

$\gamma(q)\mu(\Lambda^{\perp}) = \gamma(q) \int_V e_{\Lambda^{\perp}} \ d\mu = \int_V f_q * e_{\Lambda^{\perp}} \ d\mu = \mu(\Lambda^{\perp})\int_{\Lambda} f_q \ d\mu$,

which then yields $\gamma(q) = \int_{\Lambda} f_q \ d\mu$.

Now by our choice of $\Lambda$, the last integral factors through $\Lambda/\Lambda^{\perp}$ and becomes

$\gamma(q) = \mu(\Lambda^{\perp}) \sum_{x \in \Lambda/\Lambda^{\perp}} f_q(x)$.

It is not hard to see that $f_q$ induces a nondegenerate quadratic form on $\Lambda/\Lambda^{\perp}$ in the sense of the discussion above, so it follows that

$|\gamma(q)| = \mu(\Lambda^{\perp}) \sqrt{\# (\Lambda/\Lambda^{\perp})} = \sqrt{\mu(\Lambda^{\perp})^2\mu(\Lambda)/\mu(\Lambda^{\perp})} = 1$,

where we used the fact that $\mu(\Lambda^{\perp})\mu(\Lambda) = 1$ because $\mu$ is self-dual.

$\Box$

Categories: representation theory

## Existence of the Weil index

From now on, $F$ will be a non-Archimedean local field, with characteristic not equal to 2 as always, and $\psi : F \to \mathbb{T}$ a fixed nontrivial continuous character. For a finite-dimensional vector space $V$ over $F$, we will denote by $C^{\infty}_c(V)$ the space of locally constant, compactly supported $\mathbb{C}$-valued functions on $V$. If we are given a nondegenerate symmetric bilinear form $B : V \times V \to F$, there is a unique self-dual Haar measure $\mu$ with respect to the pairing $\psi \circ B : V \times V \to \mathbb{T}$. Concretely, given a subset $X \subset V$, we can define

$X^{\perp} = \{ v \in V \ | \ \psi(B(v,w)) = 1 \ \text{for all } w \in X \},$

and then $\mu$ is the unique Haar measure on $V$ such that $\mu(\Lambda) \cdot \mu(\Lambda^{\perp}) = 1$ for some (equivalently, for any) lattice $\Lambda \subset V$. So we obtain a canonical Fourier transform $\mathcal{F} : C^{\infty}_c(V) \to C^{\infty}_c(V)$ defined by the formula

$\mathcal{F}(f)(y) = \int_V f(x) \psi(B(x,y)) \ d\mu(x).$

Define the space of distributions $\mathcal{D}(V) = C^{\infty}_c(V)^*$ on $V$ to be the dual vector space to $C^{\infty}_c(V)$. There is a natural injection $C^{\infty}(V) \to \mathcal{D}(V)$, where $C^{\infty}(V)$ denotes the space of all locally constant $\mathbb{C}$-valued functions on $V$, which sends $f \in C^{\infty}(V)$ to the linear functional

$g \mapsto \int_V fg \ d\mu.$

Any invertible operator $T : C^{\infty}_c(V) \to C^{\infty}_c(V)$ induces an adjoint operator $(T^*)^{-1} : \mathcal{D}(V) \to \mathcal{D}(V)$ defined by the formula $T^*(\varphi)(f) = \varphi(T^{-1}(f))$. We will usually abuse notation and denote $(T^*)^{-1}$ by $T$ also.

Let $q(v) = B(v,v)$ be the associated quadratic form and consider the function $f_q : V \to \mathbb{T}$ given by $f_q(v) = \psi(\tfrac12 q(v))$. Obviously $f_q$ does not lie in $C^{\infty}_c(V)$, since $\psi$ never vanishes. However, since $q$ is continuous and $\psi$ is locally constant it follows that $f_q$ is locally constant, so we can think of $f_q$ as a distribution and take its Fourier transform.

Theorem There exists a scalar $\gamma(q) \in \mathbb{C}^{\times}$, called the Weil index of $q$, such that

$\mathcal{F}(f_q) = \gamma(q) \cdot f_{-q},$

and furthermore $|\gamma(q)| = 1$. In fact, for a sufficiently large lattice $\Lambda \subset V$, we have

$\gamma(q) = \int_{\Lambda} f_q \ d\mu.$

The proof of the theorem we will give is, unlike Weil’s original proof, completely independent from other results on the Weil representation and uses only the theory of Fourier transforms. The fact that $\gamma$ takes values in complex numbers of modulus one will follow from an elementary result on Gauss sums.

Unfortunately, finishing the proof this time would probably make this post unreasonably long, so I’ll content myself with a lemma on Fourier transforms. Let us define translation and multiplication operators on $C^{\infty}(V)$ by the formulas

$(T_af)(x) = f(x + a) \ \text{and} \ (M_af)(x) = \psi(B(a,x))f(x)$

for all $a \in V$. In particular, these operate on $C^{\infty}_c(V)$, so as discussed above we obtain adjoint operators $T_a$ and $M_a$ on $\mathcal{D}(V)$.

Observe also that $C^{\infty}_c(V)$ is a $C^{\infty}(V)$-module, so formally we see that $C^{\infty}(V)$ also operates on $\mathcal{D}(V)$ by the formula $(f \cdot \varphi)(g) = \varphi(f \cdot g)$.

Lemma We have the following equations of operators on $\mathcal{D}(V)$:

$\mathcal{F} \circ T_a = M_{-a} \circ \mathcal{F} \ \text{and} \ \mathcal{F} \circ M_a = T_a \circ \mathcal{F}.$

Also, the translation and multiplication operators interact with the $C^{\infty}(V)$-module structure on $\mathcal{D}(V)$ via

$T_a(f \cdot \varphi) = T_{-a}(f) \cdot T_a(\varphi) \ \text{and} \ M_a(f \cdot \varphi) = M_a(f) \cdot \varphi = f \cdot M_a(\varphi).$

Proof. These are trivial calculations.

$\Box$

Categories: representation theory

## The Witt group and its Weil character

For the next couple of posts I’ll be proving the existence and some properties of the Weil index, which is a numerical invariant of a quadratic space over a local field, and is an important ingredient in the construction of the metaplectic group and its (linear) Weil representation. I will treat only the case of a non-Archimedean local field $F$ with characteristic not equal to $2$. This post will probably consist mostly of setup to put us in a natural context for the proofs.

For now let $k$ be any field of characteristic not equal to $2$.

Definition By a quadratic space over $k$ we mean a finite-dimensional vector space $W$ over $k$ equipped with a nondegenerate quadratic form $q : W \to k$.

Since the characteristic of $k$ is not $2$, quadratic forms correspond bijectively with symmetric bilinear forms. An isometry between quadratic spaces $(W_1,q_1)$ and $(W_2,q_2)$ is a linear isomorphism $T : W_1 \to W_2$ satifying $q_2(T(w)) = q_1(w)$ for all $w \in W_1$. Clearly isometry is an equivalence relation on quadratic spaces.

Example For any $a \in k^{\times}$, the map $x \mapsto ax^2$ is a nondegenerate quadratic form on $k$ which we denote by $\langle a \rangle$. It is not hard to see that any one-dimensional quadratic space has this form, and $\langle a \rangle$ is isometric to $\langle b \rangle$ if and only if $a \equiv b \ (\text{mod } k^{\times 2})$.

Given two quadratic spaces $(W_1,q_1)$ and $(W_2,q_2)$, we can construct their sum in the obvious way: its underlying space is $W_1 \oplus W_2$, and the quadratic form is defined by the formula $(q_1 \oplus q_2)(w_1,w_2) = q_1(w_1) + q_2(w_2)$.

Example The quadratic space $\mathfrak{H} = \langle 1 \rangle \oplus \langle -1 \rangle$ is called the hyperbolic plane. It is isometric to the form on $k^2$ which sends $(x,y) \mapsto xy$.

Thankfully, all quadratic spaces decompose (noncanonically) into a direct sum of spaces of the form $\langle a \rangle$, so understanding arbitrary quadratic forms reduces to understanding $\langle a \rangle$.

Proposition Any quadratic space $(V,q)$ is isometric to one of the form $\langle a_1 \rangle \oplus \cdots \oplus \langle a_n \rangle$ for some $a_1,\cdots,a_n \in k^{\times}$.

Proof. We induct on the dimension $n$ of $V$: the case $n = 0$ is trivial. For the inductive step, certainly there exists $v \in V$ with $q(v) \neq 0$, and if we write $W$ for the subspace of $V$ spanned by $v$ then we have $W \cong \langle q(v) \rangle$. Denote by $W^{\perp}$ the subspace of $V$ orthogonal to $W$ with respect to $q$, so we need only show $V = W \oplus W^{\perp}$ and we will be done by the inductive hypothesis.

Clearly $W \cap W^{\perp} = 0$ since $q$ is nondegerate when restricted to $W$, so it remains to prove $V = W + W^{\perp}$. For this, one observes that the isomorphism $V \to V^*$ induced by $q$ restricts to an isomorphism $W^{\perp} \to (V/W)^*$.

$\Box$

Finally we can introduce the group of which the Weil index is a character.

Definition-Proposition The quotient of the commutative monoid formed by isometry classes of quadratic spaces by the submonoid generated by the hyperbolic plane $\mathfrak{H}$ is a group, called the Witt group of $k$ and denoted by $W(k)$.

Proof. The content of the statement is that for any quadratic space $(V_1,q_1)$, there exists another $(V_2,q_2)$ such that $(V_1 \oplus V_2,q_1 \oplus q_2)$ is isometric to a direct sum of copies of $\mathfrak{H}$. So choose a diagonalization $\langle a_1 \rangle \oplus \cdots \oplus \langle a_n \rangle$ of $(V_1,q_1)$ and take $(V_2,q_2)$ to be $\langle -a_1 \rangle \oplus \cdots \oplus \langle -a_n \rangle$. Then the resulting direct sum $(V_1 \oplus V_2,q_1 \oplus q_2)$ is isometric to $\langle a_1 \rangle \oplus \langle -a_1 \rangle \oplus \cdots \oplus \langle a_n \rangle \oplus \langle -a_n \rangle$, so it suffices to show that for any $a \in k^{\times}$ we have $\langle a \rangle \oplus \langle -a \rangle \cong \mathfrak{H} = \langle 1 \rangle \oplus \langle -1 \rangle$. This isometry is given by $(x,y) \mapsto \tfrac{a+1}{2}(x,y)$.

$\Box$

In fact, the Witt group can be given a ring structure using the tensor product of quadratic spaces, but we will not need this.

Next time we will prove the existence of the Weil character of $W(F)$ for $F$ a non-Archimedean local field of characteristic not equal to $2$ using the theory of distributions and their Fourier transforms.

## Rodier’s Mackey machine

So I’m finally getting around to the proof of that theorem of Rodier’s from several posts back. After this we’ll be done with the Stone-von Neumann theorem (keeping in mind that all this work only applies to the non-Archimedean local case).  Let’s recall the situation of the theorem and state it again for convenience.

Let $G$ be an $\ell$-group and $U \subset G$ a closed normal abelian $\ell_c$-subgroup. Then $G$ acts on $U$ by conjugation, hence acts on $\widehat{U}$ as well. If $M$ is a smooth representation of $G$, then after restricting to $U$ we can construct the corresponding sheaf $\widetilde{M}$ on $\widehat{U}$, and in particular it is easy to see that $\text{Supp } \widetilde{M}$ is $G$-stable. Fix a smooth character $\chi \in \widehat{U}$. Then the stalk $\widetilde{M}_{\chi}$ carries a natural smooth action of the stabilizer $Z_G(\chi)$: this is clear if we recall that $\widetilde{M}_{\chi}$ is identified with the space of $\chi$-coinvariants of $M$.

If you read the earlier post in which I stated this theorem, you might notice that I’ve added an extra hypothesis (although I did go back and change that post also). The map $Z_G(\chi) \backslash G \to \chi^G$ is a homeomorphism under some mild technical assumptions on $G$ such as second countability, but for mostly aesthetic reasons I simply assumed this, and indeed it is obvious in the case where $G$ is the Heisenberg group.

Theorem (Rodier) Suppose that the $G$-orbit $\chi^G$ is locally closed in $\widehat{U}$ and that the natural map $Z_G(\chi) \backslash G \to \chi^G$ is a homeomorphism. Then the functor $M \mapsto \widetilde{M}_{\chi}$ is an equivalence of categories from the category of smooth representations of $G$ whose restriction to $U$ is supported on $\chi^G$ and the category of smooth representations of $Z_G(\chi)$ whose restriction to $U$ is supported at $\chi$.

Proof. First let us construct the quasi-inverse equivalence, which is really just a geometric form of compactly supported induction. Given a smooth representation $N$ of $Z_G(\chi)$, the product space $N \times G$ (where $N$ carries the discrete topology as usual) has a left action of $Z_G(\chi)$ on each factor, so we can form the quotient space $E_N = Z_G(\chi) \backslash (N \times G)$. There is a natural projection $E_N \to Z_G(\chi) \backslash G$, and we will now use the result from the last post to show that this is an étale space. In fact, something a little better is true: it is a “covering space” in the usual sense, although this term is not usually applied in totally disconnected situations.

To simplify notation, write $H = Z_G(\chi)$. It is not hard to see that if $\pi : G \to H \backslash G$ denotes the canonical projection, it suffices to work in a neighborhood of $\pi(1)$. We already know that $\pi(1)$ has a neighborhood $V$ such that there is an $H$-equivariant homeomorphism $H \times V \to \pi^{-1}(V)$. It is also easy to see that if $N_0$ is the space $N$ with the trivial $H$-action, then $(n,h) \mapsto (h \cdot n,h)$ is an $H$-equivariant homeomorphism $N_0 \times H \to N \times H$. Combining these yields an $H$-isomorphism $N_0 \times H \times V \to N \times \pi^{-1}(V)$. If we denote by $\rho : E_N \to H \backslash G$ the projection map, then after descending to the quotient we get a homeomorphism $N_0 \times V \to \rho^{-1}(V)$ as desired.

We need to finish describing the quasi-inverse functor. Write $\mathcal{F}_N$ for the sheaf of continuous sections of the étale space $E_N \to Z_G(\chi) \backslash G$, and note that $G$ acts on global sections $s$ of this sheaf via the formula $(g \cdot s)(Z_G(\chi)h) = s(Z_G(\chi)hg) \cdot g^{-1}$. Clearly this restricts to an action on the compactly supported sections $\Gamma_c(\mathcal{F}_N)$, and this is the desired $G$-representation. Of course we must check that the restriction to $U$ is supported on $\chi^G$. For this, recall that by hypothesis the $G$-equivariant map $i : Z_G(\chi) \backslash G \to \widehat{U}$ which sends $Z_G(\chi)g \mapsto \chi^g$ is a locally closed embedding, so we can form the pushforward with compact supports $i_!\mathcal{F}_N$, and $G$ acts on $\Gamma_c(i_!\mathcal{F}_N) \cong \Gamma_c(\mathcal{F}_N)$ by the same formula. Fix $s \in \Gamma_c(i_!\mathcal{F}_N)$ and $\chi^g \in \chi^G$, so we can write $s(\chi^g) = [n,g]$ for some $n \in N$. Then compute

$(u \cdot s)(\chi^g) = [n,gu^{-1}] = [n,gu^{-1}g^{-1}g] = [\chi^g(u) \cdot n,g] = \chi^g(u) \cdot s(\chi^g)$

for $u \in U$, where we used the fact that $\text{Supp } \widetilde{N} \subset \{ \chi \}$. This shows that $\widetilde{\Gamma_c(i_!\mathcal{F}_N)} \cong i_!\mathcal{F}_N$, whence this $U$-representation has support $\text{Supp } i_!\mathcal{F}_N \subset \chi^G$.

Finally, we check that these functors are actually quasi-inverse equivalences. Since $\widetilde{\Gamma_c(i_!\mathcal{F}_N)} \cong i_!\mathcal{F}_N$, the stalk at $\chi$ of the former sheaf is naturally identified with $(\mathcal{F}_N)_\chi$, which is in turn naturally isomorphic to $N$ because this is the fiber of the étale space $E_N$ over $\chi$. This isomorphism is $Z_G(\chi)$-equivariant because

$[n,1] \cdot z^{-1} = [n,z^{-1}] = [z \cdot n,1]$.

As for the other direction, let $M$ be a smooth representation of $G$ supported on $\chi^G$. Write $P = \widetilde{M}_{\chi}$ for the corresponding $Z_G(\chi)$-representation, so we need to show $M \cong \Gamma_c(\mathcal{F}_P)$ as $G$-representations. This map sends $m \in M$ to the section $s_m : \chi^G \to E_P$ defined by $s_m(\chi^g) = [g \cdot m,g]$, where we have confused $g \cdot m$ with its image in the quotient $M \to P$. Note $s_m$ has compact support: viewing $M$ as a nondegenerate $C^{\infty}_c(\widehat{U})$-module, find a compact open subset $K$ such that $e_K \cdot m = m$, where $e_K$ is the indicator function for $K$. This implies that $\text{Supp } s_m \subset K \cap \chi^G$. It is completely straightforward to check that $M \to \Gamma_c(\mathcal{F}_P)$ is $G$-equivariant, and to see that this is an isomorphism we may work on the level of $U$-representations, or equivalently sheaves on $\widehat{U}$. Passing to stalks, we get the map $\widetilde{M}_{\chi^g} \to (\mathcal{F}_P)_{\chi^g}\cong \widetilde{M}_{\chi}$ which sends $m \mapsto g \cdot m$, and since these are the spaces of $\chi^g$– and $\chi$– coinvariants this is clearly an isomorphism.

$\Box$

Categories: representation theory