Lifting representations of finite reductive groups: a character relation

Given a connected reductive group $\tilde{G}$ over a finite field $k$, and a semisimple $k$-automorphism $\varepsilon$ of $\tilde{G}$ of finite order, let $G$ denote the connected part of the group of $\varepsilon$-fixed points. Then there exists a lifting from packets of representations of $G(k)$ to packets for $\tilde{G}(k)$. In the case of Deligne-Lusztig representations, we show that this lifting satisfies a character relation analogous to that of Shintani.

Introduction {#sec:intro}
============

Suppose $k$ is a finite field, $\tilde{G}$ is a connected reductive $k$-group, and $\varepsilon$ is a semisimple $k$-automorphism of $\tilde{G}$ of finite order $\ell$. From @steinberg:endomorphisms\*<span>Theorem 7.5</span>, $\varepsilon$ must preserve some pair $(\tilde{B},\tilde{T})$ consisting of a Borel subgroup of $\tilde{B}\subseteq \tilde{G}$ and a maximal torus $\tilde{T} \subseteq \tilde{B}$. Call such a pair a *Borel-torus pair* for $\tilde{G}$. Let $G$ be the connected part of the group $\tilde{G}^\varepsilon$ of $\varepsilon$-fixed points of $\tilde{G}$. Then we have the following from @adler-lansky:lifting\*<span>Proposition 3.5</span>.

\[prop:intro\]  

-   $G$ is a connected reductive $k$-group.

-   For every $\varepsilon$-invariant Borel-torus pair $(\tilde{B},\tilde{T})$ for $\tilde{G}$, one has a Borel-torus pair $(\tilde{B}^\varepsilon,(\tilde{T}^\varepsilon){^\circ})$ for $G$. Moreover, $(\tilde{T}^\varepsilon){^\circ}= \tilde{T} \cap G$.

-   For every Borel-torus pair $(B,T)$ for $G$, one has an $\varepsilon$-invariant Borel-torus pair $(\tilde{B},\tilde{T})$ where $\tilde{T} = C_{\tilde{G}}(T)$.

Note that by choosing $T$ to be defined over $k$, we can show that $\tilde{G}$ has an $\varepsilon$-invariant Borel-torus pair $(\tilde{B},\tilde{T})$ whose torus $\tilde{T}$ is defined over $k$.

Let $\tilde{G}^*$ and $G^*$ denote the duals of $\tilde{G}$ and $G$. For each semisimple element $s\in G$, one obtains a collection ${{\mathcal{E}}}_s(G(k))$ of irreducible representations of $G(k)$, and these collections, known as *Lusztig series*, partition the set ${{\mathcal{E}}}(G(k))$ of (equivalence classes of) irreducible representations of $G(k)$ @lusztig:chars-finite\*<span>§14.1</span>. Suppose that $s$ is regular, and let $T^*\subseteq G^*$ be the unique maximal $k$-torus containing $s$. Then the pair $(T^*,s)$ corresponds to a pair $(T,\theta)$, where $T\subseteq G$ is a maximal $k$-torus, and $\theta$ is a character of $T(k)$. This latter pair is uniquely determined up to $G(k)$-conjugacy. The Lusztig series ${{\mathcal{E}}}_s(G(k))$ corresponding to $s$ is the set of irreducible components of the Deligne-Lusztig virtual representation whose character is $\operatorname{R}_T^G \theta$.

In an earlier work @adler-lansky:lifting\*<span>Corollary 11.3</span>, two of the authors constructed a map from semisimple classes in $G^*(k)$ to semisimple classes in $\tilde{G}^*(k)$, thus lifting each Lusztig series for $G(k)$ to one for $\tilde{G}(k)$. The series of representations coming from $\pm \operatorname{R}_T^G\theta$ lifts to that coming from $\pm \operatorname{R}_{\tilde{T}}^{\tilde{G}} \tilde\theta$, where $\tilde{T} = C_{\tilde{G}}(T)$ and $\tilde\theta = \theta \circ {{\mathcal{N}}}$, and ${{{\mathcal{N}}}\colon \tilde{T} \longrightarrow T}$ is the norm map defined by $${{\mathcal{N}}}(t) = t \varepsilon(t) \cdots \varepsilon^{\ell-1}(t).$$ (Of course, one could define a similar map ${{\mathcal{N}}}$ on any $\varepsilon$-invariant torus in $\tilde{G}$.) In order to understand better this lifting of representations, one would like to have a relation between the character $\operatorname{R}_T^G\theta$ (for $\theta$ an arbitrary character of $T(k)$, not necessarily associated to a regular element of $G^*(k)$) and the $\varepsilon$-twisted character $(\operatorname{R}_{\tilde{T}}^{\tilde{G}}\tilde\theta)_\varepsilon$ associated to the $\varepsilon$-invariant character $\operatorname{R}_{\tilde{T}}^{\tilde{G}}\tilde\theta$. The purpose of the present paper is to prove that such a relation holds at sufficiently regular points.

Suppose $\tilde{s} \in \tilde{G}(k)$ belongs to an $\varepsilon$-invariant, maximal $k$-torus and that ${{\mathcal{N}}}(\tilde{s})$ is regular in $\tilde{G}$. Let $S$ denote the unique maximal torus in $G$ containing ${{\mathcal{N}}}(\tilde{s})$. Then $$(\operatorname{R}_{\tilde{T}}^{\tilde{G}}\tilde{\theta})_{\varepsilon}(\tilde{s})
=
\sum_{w\in W_k(G,T)\backslash W_k(\tilde{G},S,T)} (\operatorname{R}_T^G\theta)(w{{\mathcal{N}}}(\tilde{s})w{^{-1}}).$$

Here $W_k(\tilde{G},S,T)$ is defined to be the quotient $\tilde T(k)\backslash\{ \tilde g\in \tilde G(k)\mid \tilde g S \tilde g{^{-1}}= T\}$ and $W_k(G,T) = N_{G(k)}(T)/T(k)$. We see that $W_k(\tilde{G},S,T)$ is a union of right $W_k(G,T)$-cosets from the fact that $N_{G(k)}(T)  \cap \tilde T(k) = T(k)$, so the index set for the summation makes sense. We define $w {{\mathcal{N}}}(\tilde s) w{^{-1}}$ to be $n {{\mathcal{N}}}(\tilde s) n{^{-1}}$ for any lift $n$ of $w$ to $\tilde{G}(k)$, and claim that the right-hand side does not depend on our choices of lifts.

Here is what we mean by $(\operatorname{R}_{\tilde{T}}^{\tilde{G}}\tilde{\theta})_{\varepsilon}$. Since $\tilde{T}\subseteq \tilde{G}$ is an $\varepsilon$-invariant maximal $k$-torus, and $\tilde{\theta}$ is an $\varepsilon$-invariant character of $\tilde{T}(k)$, we have that $\varepsilon$ preserves $(\tilde T, \tilde \theta)$, so it acts on the corresponding Deligne-Lusztig variety, and thus on the virtual representation whose character is $\operatorname{R}_{\tilde T}^{\tilde G} \tilde\theta$. That is, even if this representation is reducible, we can form its $\varepsilon$-twisted character. One can construct this character as follows. Extend $\tilde\theta$ to a character of $\tilde{T}(k)\rtimes{\Gamma}$ by setting $\tilde\theta(\varepsilon) = 1$. Define the $\varepsilon$-twisted Deligne-Lusztig character $(\operatorname{R}_{\tilde{T}}^{\tilde{G}}\tilde{\theta})_{\varepsilon}$ induced from $\tilde{\theta}$ by $(\operatorname{R}_{\tilde{T}}^{\tilde{G}}\tilde{\theta})_{\varepsilon}(g) = 
(\operatorname{R}_{\tilde{T}\rtimes{\Gamma}}
    ^{\tilde{G}\rtimes{\Gamma}}\tilde\theta)(g\varepsilon)$ for $g\in\tilde{G}(k)$. (See @digne-michel:non-connected for the definition of Deligne-Lusztig induction for nonconnected groups.)

Consider the automorphism of $\operatorname{GL}(2)$ given by $\varepsilon(g) = {}^t g {^{-1}}$. Then an analogous relation holds for all irreducible representations, not just those of Deligne-Lusztig type. On the other hand, while the relation can hold for unipotent elements, it fails if $\tilde{s}$ is regular but ${{\mathcal{N}}}(\tilde s)$ is singular.

Consider the special case where $E/k$ is a finite extension, and $\tilde{G}$ arises from $G$ via restriction of scalars: $\tilde{G} = R_{E/k} G$. Suppose that $\varepsilon$ is the $k$-automorphism of $\tilde{G}$ associated to the action of a generator of the Galois group $\operatorname{Gal}(E/k)$. Given a representation $\pi$ of $G(k)$, one often has an associated representation $\tilde\pi$ of $\tilde{G}(k)$, known as the *Shintani lift* of $\pi$. (See @kawanaka:shintani for a discussion.) The character of $\pi$ and the $\varepsilon$-twisted character of $\tilde\pi$ are related by the Shintani relation: $\Theta_\pi({{\mathcal{N}}}(g)) = \Theta_{\tilde\pi,\varepsilon}(g)$. From work of Digne @digne:shintani\*<span>Cor. 3.6</span>, one already knows that if $\operatorname{R}_T^G \theta$ has a Shintani lift, then it must be $\operatorname{R}_{\tilde T}^{\tilde G} \tilde\theta$. Thus, our character relation is a generalization of Shintani's, if one ignores the fact that the former only holds at sufficiently regular elements of $\tilde{G}(k)$, while that latter holds at all elements, provided that one interprets the norm map ${{\mathcal{N}}}$ correctly.

<span style="font-variant:small-caps;">Acknowledgements.</span> The authors were partially supported by a National Science Foundation Focused Research Group grant (DMS-0854844). The fourth-named author was partially supported by the Frank Cox Jones Scholarship Fund and a Dean's Undergraduate Research Award from the College of Arts and Sciences of American University. We have benefited from conversations with Prof. Jeffrey Hakim.

Consequences of regularity
==========================

From now on, we let $k$, $\tilde{G}$, $\varepsilon$, $\ell$, $G$, and ${{\mathcal{N}}}$ be as in §\[sec:intro\]. Let $\Gamma$ be the group generated by $\varepsilon$.

\[prop:regularity\] Suppose $\tilde{s}$ belongs to an $\varepsilon$-invariant maximal $k$-torus $\tilde{S}$ in $\tilde{G}$, and $s:= {{\mathcal{N}}}(\tilde{s})$ is regular in $\tilde{G}$. Let $S:=(\tilde{S}\cap G)$. Then:

(a) \[item:torus-is-relevant\] $\tilde{S}$ belongs to an $\varepsilon$-invariant Borel-torus pair for $\tilde{G}$;

(b) \[item:se-is-regular\] $\tilde{s}\varepsilon$ is regular in $\tilde{G}\rtimes{\Gamma}$; and

(c) \[item:yifei\] $N_{\tilde{G}(k)}(S)
    =
    {     {\bigl\{\bigl.
        g\in \tilde{G}(k)\vphantom{gsg{^{-1}}\in \operatorname{Im}({{\mathcal{N}}}|_{\tilde{S}(k)})          }\,\bigr|
        \,gsg{^{-1}}\in \operatorname{Im}({{\mathcal{N}}}|_{\tilde{S}(k)})\bigr\}}}$.

Since $\tilde{S}$ is $\varepsilon$-invariant, we have that $s \in S$. Let $S'$ be a maximal $k$-torus in $G$ containing $S$, and let $\tilde{S}' = C_{\tilde{G}}(S')$. From Proposition \[prop:intro\], $\tilde{S}'$ is a maximal torus in $\tilde{G}$, and belongs to an $\varepsilon$-invariant Borel-torus pair. But $$\tilde{S}' = 
C_{\tilde{G}}(S') \subseteq 
C_{\tilde{G}}(S) \subseteq 
C_{\tilde{G}}(s) = \tilde{S},$$ proving statement .

Now choose an $\varepsilon$-invariant Borel subgroup of $\tilde{G}$ containing $\tilde{S}$, and let $\Phi^+$ denote the corresponding system of positive roots. For each $\alpha\in \Phi^+$, consider the following sum of root spaces in the Lie algebra $\tilde{{\mathfrak{g}}}$ of $\tilde{G}$: $V_\alpha:= \bigoplus_{\beta \in {\Gamma}\cdot \alpha} \tilde{{\mathfrak{g}}}_\beta$. Via the adjoint action, the element $\tilde{s}\varepsilon$ induces a linear transformation on $\tilde{{\mathfrak{g}}}$, preserving each $V_\alpha$. To see that $\tilde{s}\varepsilon$ is regular, it is enough to choose a positive root $\alpha\in\Phi^+$ and show that the action of $\tilde{s}\varepsilon$ on $V_\alpha$ does not have $1$ as an eigenvalue.

Let $m = m_\alpha =  | {\Gamma}\cdot \alpha|$. There is a nonzero scalar $\mu = \mu_\alpha$, not necessarily rational, such that $\varepsilon^m$ acts on $V_\alpha$ via multiplication by $\mu$. By necessity, $\mu$ is a root of unity of order dividing $\ell /  m$. Let $\chi' = \chi'_\alpha = \sum_{\beta \in {\Gamma}\cdot \alpha}\beta$, and let $\chi = \chi_\alpha = \sum_{\gamma \in {\Gamma}} \gamma(\alpha)$. Thus, $\chi = \frac{\ell}{m} \chi'$.

It is not hard to see that the characteristic polynomial (in the variable $X$) of our transformation of $V_\alpha$ is $X^m - \mu\cdot \chi'(\tilde{s})$, so it is enough to show that $\mu \cdot \chi'(\tilde{s}) \neq 1$.

Since $s$ is regular, we have that $$1 \neq \alpha(s)  = \chi(\tilde{s})
= \chi'(\tilde{s})^{\ell/m}
= (\mu \cdot \chi'(\tilde{s}))^{\ell/m},$$ and therefore, $\mu \cdot \chi'(\tilde{s}) \neq 1$, proving statement .

Now consider statement . From statement and Proposition \[prop:intro\], we have that $S$ is a maximal $k$-torus in $G$, and $\tilde{S} = C_{\tilde{G}}(S)$. To show that the right-hand-side is included in the left, take any $g\in\tilde{G}$ such that $gsg{^{-1}}\in \operatorname{Im}({{\mathcal{N}}}|_{\tilde{S}(k)}) \subseteq S(k)$. Since $s$ and $gsg{^{-1}}$ are in $S(k)$, they are both $\varepsilon$-invariant, so we have $\varepsilon(g)s\varepsilon(g){^{-1}}=gsg{^{-1}}$. This implies $g{^{-1}}\varepsilon(g)\in C_{\tilde{G}}(s) = \tilde{S}$. Therefore $\varepsilon(g)=g\tilde{t}$ for some $\tilde{t}\in \tilde{S}(k)$.

Since $g$ conjugates a regular element of $\tilde{S}$ back into $\tilde{S}$, we must have that $g$ normalizes $\tilde{S}$. Let $t\in S(k)$. Since $$\varepsilon (gtg{^{-1}})
=
g\tilde{t}t\tilde{t}{^{-1}}g{^{-1}}= 
gtg{^{-1}},$$ we have that $gtg{^{-1}}\in \tilde{S}^\varepsilon$. But since $t \in (\tilde{S}^\varepsilon){^\circ}= S$, so is $gtg{^{-1}}$. Thus, $g\in N_{\tilde{G}}(S)$.

To prove the converse, let $n \in N_{\tilde{G}(k)}(S)$. We want to show that $nsn{^{-1}}\in \operatorname{Im}({{\mathcal{N}}}|_{\tilde{S}(k)})$. Since every element in $S$ is $\varepsilon$-fixed, we have $\varepsilon^i (n)s\varepsilon^i (n){^{-1}}=nsn{^{-1}}$ for all $i$, and therefore $n{^{-1}}\varepsilon^i (n)$ belongs to $C_{\tilde{G}(k)}(s) = \tilde{S}$, and thus commutes with every element of $\tilde{S}(k)$. Consider the following product: $$\left(n{^{-1}}\varepsilon(n)\right)\left(\varepsilon(n){^{-1}}\varepsilon^{2}(n)\right)
\left(\varepsilon^{2}(n){^{-1}}\varepsilon^{3}(n)\right)\cdots\left(\varepsilon^{\ell-1}(n){^{-1}}n\right)=1$$ Since each term in the above equation commutes with $\varepsilon^{j}(\tilde{s})$, we have $$\begin{aligned}
nsn{^{-1}}&= n\tilde{s}\varepsilon(\tilde{s})\cdots \varepsilon^{\ell-1}(\tilde{s})n{^{-1}}\\
&= n\tilde{s}\left(n{^{-1}}\varepsilon(n)\right)\varepsilon(\tilde{s})
\left(\varepsilon(n){^{-1}}\varepsilon^2(n)\right) \cdots
\varepsilon^{\ell-1}(\tilde{s})\left(\varepsilon^{\ell-1}(n){^{-1}}n\right) n{^{-1}}\\ 
&=\left(n\tilde{s}n{^{-1}}\right)
\left(\varepsilon(n)\varepsilon(\tilde{s})\varepsilon(n){^{-1}}\right)
\cdots \left(\varepsilon^{\ell-1}(n)\varepsilon^{\ell-1}(\tilde{s})\varepsilon^{l-1}(n){^{-1}}\right) \\ 
&= \left(n\tilde{s}n{^{-1}}\right)\left(\varepsilon(n\tilde{s}n{^{-1}})\right)
\cdots \left(\varepsilon^{\ell-1}(n\tilde{s}n{^{-1}})\right) \\
&= {{\mathcal{N}}}(n\tilde{s}n{^{-1}}).\end{aligned}$$ Since $n$ conjugates a regular element ($s$) of $\tilde{S}$ back into $\tilde{S}$, we must have that $n\in N_{\tilde{G}}(\tilde{S})$, and so $n\tilde{s}n{^{-1}}\in \tilde{S}(k)$. Therefore $$nsn{^{-1}}={{\mathcal{N}}}(n\tilde{s}n{^{-1}})\in \operatorname{Im}({{\mathcal{N}}}|_{\tilde{S}(k)}),$$ as desired.

Proof of the Theorem {#sec:proof}
====================

We start by recalling some facts about Deligne-Lusztig virtual characters.

\[lemma:char-formulas\]  

(a) \[item:char-downstairs\] If $x\in G(k)$ is a regular semisimple element, then $$(\operatorname{R}_{T}^{G}\theta)(x)=
    \begin{cases}
    \displaystyle
    \sum_{v\in W_k(G,T)}\theta(vx'v{^{-1}})
    &\text{
    \parbox{.35\textwidth}{
    \hangindent 3ex
    if $x$ is $G(k)$-conjugate to an element $x'\in T(k)$,
    }
    }
    \\
    0& \text{otherwise}.
    \end{cases}$$

(b) \[item:char-upstairs\] Suppose that $\tilde{x}\in\tilde G(k)$ and that $\tilde x\varepsilon$ is a regular semisimple element of $\tilde G(k)\rtimes{\Gamma}$. Let $\tilde{\theta}$ be an $\varepsilon$-invariant character of $\tilde{T}(k)$, extended to trivially $\tilde{T}(k) \rtimes{\Gamma}$. Then $$(\operatorname{R}_{\tilde{T}}^{\tilde{G}}\tilde{\theta})_{\varepsilon}(\tilde{x})
    =
    \frac{1}{\vert\tilde{T}(k)\rtimes{\Gamma}\vert}
                \sum_
            {{     {\bigl\{\bigl.
        h\in\tilde{G}(k)\rtimes{\Gamma}\vphantom{\tilde{x}\varepsilon\in h(\tilde{T}(k)\rtimes{\Gamma})h{^{-1}}}\,\bigr|
        \,\tilde{x}\varepsilon\in h(\tilde{T}(k)\rtimes{\Gamma})h{^{-1}}\bigr\}}}}
        \tilde\theta (h{^{-1}}(\tilde{x}\varepsilon)h) .$$

For each formula, see @digne-michel:non-connected\*<span>Proposition 2.6</span>. For the second formula, note that $C_{\tilde G}(x)$ contains no nontrivial unipotent elements and that the Green function $Q_{C_{\tilde{G}}(\tilde{x}\varepsilon)^{\circ}(k)}^{C_{\tilde{G}}(\tilde{x}\varepsilon)^{\circ}(k)}$ in this proposition takes the value $|C_{\tilde{G}}(\tilde{x}\varepsilon)^{\circ}(k)|$ at $(1,1)$.

Let $\tilde{S}$ denote an $\varepsilon$-invariant maximal $k$-torus in $\tilde{G}$ containing $\tilde{s}$. From Proposition \[prop:regularity\](\[item:torus-is-relevant\]) and Proposition \[prop:intro\], $S = \tilde{S} \cap G$ and $\tilde{S} = C_{\tilde{G}}(S)$.

Suppose that for all $\tilde{h} \in \tilde{G}(k)\rtimes {\Gamma}$, we have that $\tilde{h} (\tilde{s}\varepsilon) \tilde{h}{^{-1}}\notin \tilde{T}(k) \varepsilon$. Then $\tilde{h} \tilde{s} \varepsilon(\tilde{h}{^{-1}}) \notin \tilde{T}(k)$, so in particular $g \tilde{s} g{^{-1}}\notin \tilde{T}(k)$ for all $g\in G(k)$. This implies that for all $g \in G(k)$, $g \tilde{S} g{^{-1}}\neq \tilde{T}$, and thus from Proposition \[prop:intro\] that $gS g{^{-1}}\neq T$. Then the regularity of ${{\mathcal{N}}}(\tilde{s})$ implies that $g {{\mathcal{N}}}(\tilde{s}) g{^{-1}}\notin T(k)$. For every lift $n$ of $w$ to $\tilde{G}(k)$, we have that the summand corresponding to $w$ in the right-hand side of the equation in the theorem is zero. But our assumption on $\tilde{s}$ implies that the left-hand side is zero, too. Thus, the theorem holds in this case.

Now suppose that $\tilde{h} (\tilde{s}\varepsilon) \tilde{h}{^{-1}}\in \tilde{T}(k) \varepsilon$ for some $\tilde{h} = \tilde{g}\varepsilon^i \in \tilde{G}(k) \rtimes {\Gamma}$. Then $\tilde{g}\cdot \tilde s \cdot \varepsilon(\tilde{g}){^{-1}}\in \tilde T(k)$ so $\tilde{g} {{\mathcal{N}}}(\tilde{s}) \tilde{g}{^{-1}}= {{\mathcal{N}}}(\tilde{g} \cdot \tilde{s} \cdot \varepsilon(\tilde{g}){^{-1}})  \in T(k)$. Thus $\tilde{g}S\tilde{g}^{-1} = T$. Rewriting each index $w$ in the summation in the theorem in the form $w'\tilde{g}$, where $w'$ is a coset representative for $W_k(G,T) \backslash W_k(\tilde{G},T)$ (where $W_k(\tilde{G},T)$ is defined to be $W_k(\tilde{G},T,T)$), we see that the right-hand side is equal to $$\sum_{w\in W_k(G,T)\backslash W_k(\tilde{G},T)} (\operatorname{R}_T^G\theta)(w \tilde{g}{{\mathcal{N}}}(\tilde{s})\tilde{g} {^{-1}}w{^{-1}}).$$ Thus, we may replace $\tilde{s}$ by its twisted conjugate $\tilde{h}\cdot \tilde{s}\cdot \varepsilon(\tilde{h}{^{-1}})$, and ${{\mathcal{N}}}(\tilde{s})$ by its conjugate $\tilde{g} {{\mathcal{N}}}(\tilde{s}) \tilde{g}{^{-1}}$. Having done so, we may now assume that $S = T$ and $\tilde{S} = \tilde{T}$.

From Proposition \[prop:regularity\](\[item:se-is-regular\]), we have that $\tilde{s}\varepsilon$ is regular, so we can analyze the right-hand side of the equation ($\operatorname{LHS}$) using Lemma \[lemma:char-formulas\](\[item:char-upstairs\]): $$\operatorname{LHS}=
\frac{1}{\ell|\tilde{T}(k)|}\sum \tilde\theta (h{^{-1}}(\tilde{s}\varepsilon) h)
=
\frac{1}{\ell|\tilde{T}(k)|}\sum \tilde\theta \left(h{^{-1}}\tilde{s}\varepsilon (h)\varepsilon\right),$$ where each sum is over the set ${     {\bigl\{\bigl.
    h\in \tilde{G}(k)\rtimes{\Gamma}\vphantom{
\tilde{s}\varepsilon \in h \bigl(\tilde{T}\rtimes {\Gamma}\bigr) h{^{-1}}}\,\bigr|
    \,
\tilde{s}\varepsilon \in h \bigl(\tilde{T}\rtimes {\Gamma}\bigr) h{^{-1}}\bigr\}}}$. If $g\in \tilde{G}(k)$, then $g\varepsilon^i$ belongs to the index set if and only if $g{^{-1}}\tilde{s}\varepsilon(g)\in \tilde{T}(k)$. Thus, $$\begin{aligned}
\operatorname{LHS}&=
\frac{1}{\ell|\tilde{T}(k)|}
\sum_{i=0}^{\ell-1}
\sum_{     {\bigl\{\bigl.
    g \in \tilde{G}(k)\vphantom{g{^{-1}}\tilde{s}\varepsilon(g)\in \tilde{T}(k)          }\,\bigr|
    \,g{^{-1}}\tilde{s}\varepsilon(g)\in \tilde{T}(k)\bigr\}}}
\tilde\theta \left( (g\varepsilon^i){^{-1}}\tilde{s}\varepsilon (g\varepsilon^i) \varepsilon \right) \\
&=
\frac{1}{|\tilde{T}(k)|}
\sum_{     {\bigl\{\bigl.
    g \in \tilde{G}(k)\vphantom{g{^{-1}}\tilde{s}\varepsilon(g)\in \tilde{T}(k)          }\,\bigr|
    \,g{^{-1}}\tilde{s}\varepsilon(g)\in \tilde{T}(k)\bigr\}}}
\tilde\theta (g{^{-1}}\tilde s \varepsilon(g))\end{aligned}$$ Letting $s = {{\mathcal{N}}}(\tilde s)$, we hence have $$\begin{aligned}
\operatorname{LHS}&= \frac{1}{|\tilde{T}(k)|}\sum_{\tilde{t}\in\tilde{T}(k)}
    \sum_{{     {\bigl\{\bigl.
    g\in \tilde{G}(k)\vphantom{g{^{-1}}\tilde{s}\varepsilon (g)= \tilde{t}          }\,\bigr|
    \,g{^{-1}}\tilde{s}\varepsilon (g)= \tilde{t}\bigr\}}}}
    \tilde{\theta} (g{^{-1}}\tilde{s}\varepsilon (g)) \\
\displaybreak[0]
&= \frac{1}{|\tilde{T}(k)|}\sum_{\tilde{t}\in\tilde{T}(k)}
    \sum_{{     {\bigl\{\bigl.
    g\in \tilde{G}(k)\vphantom{g\tilde{s}\varepsilon (g){^{-1}}=\tilde{t}          }\,\bigr|
    \,g\tilde{s}\varepsilon (g){^{-1}}=\tilde{t}\bigr\}}}}
        \tilde{\theta} (g\tilde{s}\varepsilon (g){^{-1}}) \\
\displaybreak[0]
&= \frac{1}{|\tilde{T}(k)|}\sum_{\tilde{t}\in\tilde{T}(k)}
    \sum_{{     {\bigl\{\bigl.
    g\in \tilde{G}(k)\vphantom{ g\tilde{s}\varepsilon (g){^{-1}}=\tilde{t}          }\,\bigr|
    \, g\tilde{s}\varepsilon (g){^{-1}}=\tilde{t}\bigr\}}}}
        \theta ({{\mathcal{N}}}(\tilde{t})) \\
\displaybreak[0]
&= \frac{1}{|\tilde{T}(k)|}\sum_{\tilde{t}\in\tilde{T}(k)}
    \left|{     {\bigl\{\bigl.
    g\in \tilde{G}(k)\vphantom{g\tilde{s}\varepsilon (g){^{-1}}=\tilde{t}          }\,\bigr|
    \,g\tilde{s}\varepsilon (g){^{-1}}=\tilde{t}\bigr\}}}\right|
    \theta ({{\mathcal{N}}}(\tilde{t}))\\
\displaybreak[0]
&= \frac{1}{|\tilde{T}(k)|}\sum_{t\in T(k)}
    \left|{{     {\bigl\{\bigl.
    g\in \tilde{G}(k)\vphantom{\text{$g\tilde{s}\varepsilon (g){^{-1}}\in\tilde{T}$ and $gsg{^{-1}}=t$}          }\,\bigr|
    \,\text{$g\tilde{s}\varepsilon (g){^{-1}}\in\tilde{T}$ and $gsg{^{-1}}=t$}\bigr\}}}}\right|
    \theta (t)\\
\displaybreak[0]
&= \frac{1}{|\tilde{T}(k)|}
    \sum_{t\in \operatorname{Im}({{\mathcal{N}}}|_{\tilde{T}(k)})}\left|{     {\bigl\{\bigl.
    g\in \tilde{G}(k)\vphantom{gsg{^{-1}}=t          }\,\bigr|
    \,gsg{^{-1}}=t\bigr\}}}\right| \theta (t)\\
\displaybreak[0]
&= \frac{1}{|\tilde{T}(k)|}
    \sum_{{     {\bigl\{\bigl.
    g\in \tilde{G}(k)\vphantom{gsg{^{-1}}\in\operatorname{Im}({{\mathcal{N}}}|_{\tilde{T}(k)})          }\,\bigr|
    \,gsg{^{-1}}\in\operatorname{Im}({{\mathcal{N}}}|_{\tilde{T}(k)})\bigr\}}}} \theta (gsg{^{-1}}) \\
\displaybreak[0]
&=\frac{1}{|\tilde{T}(k)|}\sum_{g\in N_{\tilde{G}(k)}(T)} \theta (gsg{^{-1}})
\qquad (\text{by Proposition~\ref{prop:regularity}(\ref{item:yifei})}) \\
\displaybreak[0]
&= \sum_{w\in W_k(\tilde{G},T)} \theta (wsw{^{-1}}) \\
& =\sum_{w\in W_k(G,T)\backslash W_k(\tilde{G},T)}  \sum_{v\in W_k(G,T)} \theta(vwsw{^{-1}}v{^{-1}}),\end{aligned}$$ where the first summation in the final line is over a set of representatives for the right-coset space $W_k(G,T)\backslash W_k(\tilde{G},T)$. But from Lemma \[lemma:char-formulas\](\[item:char-downstairs\]), the final expression above is equal to the right-hand side of the relation in the Theorem.

{'source': 'arXiv_out/1205'}