$\textrm{Spin}(7)$ Is Unacceptable

Abstract

We classify to some extent the pairs of group morphisms $\Gamma \rightarrow \textrm{Spin}(7)$ which are element-conjugate but not globally conjugate. As an application, we study the case where $\Gamma$ is the Weil group of a p-adic local field, which is relevant to the recent approach to the local Langlands correspondence for $\textrm{G}_2$ and $\textrm{PGSp}_6$ in Gan and Savin (Forum Math Pi 11:e28, 2023). As a second application, we improve some result in Kret and Shin (J Eur Math Soc 25(1):75–152, 2023) about $\textrm{GSpin}_7$-valued Galois representations.

Appendix A

We gather in this appendix a few lemmas that we used. The first is the following folklore variant of the acceptability of $\textrm{O}(n)$ and $\textrm{U}(n)$.

Lemma A.1

Let G be either $\textrm{U}(n)$ or $\textrm{O}(n)$ for $n\ge 1$, $\rho : G \rightarrow \textrm{GL}_n({\mathbb {C}})$ its tautological representation, and $r,r': \Gamma \rightarrow G$ two group morphisms. Then, r and $r'$ are conjugate in G if, and only if, the representations $\rho \circ r$ and $\rho \circ r'$ of $\Gamma$ are isomorphic. Moreover, the same result holds if G is $\textrm{SU}(n)$, or $\textrm{SO}(n)$ with n odd.

Proof

Two elements $g,g' \in G$ are conjugate in G if, and only if, $\rho (g)$ and $\rho (g')$ have the same characteristic polynomial. The non-trivial implication of the statement is then equivalent to the acceptability of G (proved e.g. in [11]). $\square$

The next lemma is about the transfer morphism to an index 2 subgroup.

Lemma A.2

Let $\Gamma$ be a group, $\chi : \Gamma \rightarrow \{ \pm 1\}$ an order $\textrm{2}$ character, $\Gamma _0 \subset \Gamma$ the kernel of $\chi$, c a character of $\Gamma _0$ and t the transfer of c to $\Gamma$. Then:

• (i) For all $\gamma \in \Gamma _0$ and $z \in \Gamma \setminus \Gamma _0$, we have $t(z) = c(z^2)$ and $t(\gamma )= c(\gamma z^{-1} \gamma z)$.

• (ii) If U is a finite-dimensional representation of $\Gamma _0$ with determinant c, then $\det \textrm{Ind}_{\Gamma _0}^\Gamma U = \chi ^{\dim U} t$.

Proof

Part (i) is straightforward and part (ii) is due to Gallagher [6]. $\square$

The second is about a notion of orthogonal induction.

Lemma A.3

Let V be a Euclidean space, $\Gamma$ a group, $\rho : \Gamma \rightarrow \textrm{O}(V)$ a representation, $\Gamma _0 \subset \Gamma$ an index 2 subgroup and $z \in \Gamma \setminus \Gamma _0$. Assume that there is a $\Gamma _0$-stable subspace $V_0 \subset V$, such that:

• (i) $V_0$ is a direct sum of absolutely irreducible representations of $\Gamma _0$.

• (ii) $V = V_0 \oplus zV_0$, or equivalently, the natural morphism $\textrm{Ind}_{\Gamma _0}^\Gamma V_0 \rightarrow V$ is an isomorphism.

Then, there is a $\Gamma _0$-stable subspace $U_0 \subset V$ which is isomorphic to $V_0$ as $\Gamma _0$-module, and satisfying $V = U_0 \perp zU_0$.

Proof

Consider first the case where $V_0$ is an absolutely irreducible ${\mathbb {R}}[\Gamma _0]$-module. We have the $\Gamma _0$-stable decompositions $V = V_0 \oplus z V_0$ and $V = V_0 \perp V_0^\perp$. If $zV_0$ is not isomorphic to $V_0$, the orthogonal projection $z V_0 \rightarrow V_0$ is zero, so we have $zV_0 = V_0^\perp$ and we are done. Otherwise, the ${\mathbb {R}}[\Gamma _0]$-module V is isotypical. As we have $\textrm{End}_{{\mathbb {R}}[\Gamma _0]}(V_0)={\mathbb {R}}$ by assumption, and by the theory of isotypic components, we may assume that V is the tensor product of $V_0$ and of some Euclidean plane $P \simeq {\mathbb {R}}^2$, and that the action of $\Gamma _0$ on $V= V_0 \otimes P$ is the given one on the first factor, and trivial on the second.

The centralizer of $\rho (\Gamma _0)$ in $\textrm{O}(V)$ is $1 \otimes \textrm{O}(P)$, and that of $1 \otimes \textrm{O}(P)$ is $\textrm{O}(V_0) \otimes 1$. The element $\rho (z)$ acts on V by normalizing $\rho (\Gamma _0)$, hence by normalizing $1 \otimes \textrm{O}(P)$ as well. As each automorphism of $\textrm{O}(P)$ is inner, we may thus write $\rho (z) = \gamma \otimes \delta$ for some $\gamma \in \textrm{O}(V_0)$ and $\delta \in \textrm{O}(P)$. As $z^2 \in \Gamma _0$, we have

$$\begin{aligned} \delta ^2 \in (\textrm{O}(V_0) \otimes 1) \cap (1 \otimes \textrm{O}(P)) = \{\pm \textrm{id}_V\}. \end{aligned}$$

The proper $\Gamma _0$-stable subspaces of V are the $V_0 \otimes v$ for $v \in P$ nonzero. By assumption (ii), there is $v \in P$, such that $\delta (v) \notin {\mathbb {R}}v$, i.e., $\delta$ is not a homothety. It follows that either $\delta$ is an orthogonal symmetry (case $\delta ^2=1$), or a rotation of angle $\pi /2$ (case $\delta ^2=-1$). In both cases, there is a nonzero $v_0 \in P$, such that $v_0$ and $\delta (v_0)$ are orthogonal. The ${\mathbb {R}}[\Gamma _0]$-module $U_0 = V_0 \otimes v_0$ does the trick.

Consider now the general case. Let A be an irreducible ${\mathbb {R}}[\Gamma _0]$-submodule of $V_0$. We have $zA \cap A =\{0\}$ by (ii). By applying the first paragraph to $V_1=A \oplus zA$, we may find a $\Gamma _0$-stable $A' \subset V_1$ isomorphic to A and with $V_1 = A' \perp z A'$. Write $V=V_1 \perp V_2$; both $V_i$ are $\Gamma$-stable. Let B be an ${\mathbb {R}}[\Gamma _0]$-module, such that $V_0 \simeq A \oplus B$. We must have $V_2 \simeq \textrm{Ind}_{\Gamma _0}^\Gamma B$ by semi-simplicity. By induction on $\dim V$, we may write $V_2 = B' \perp zB'$ with $B'$ a $\Gamma _0$-stable subspace of $V_2$ isomorphic to B as $\Gamma _0$-module. The subspace $U_0 = A' \perp B'$ concludes the proof. $\square$

We also used the more specific:

Lemma A.4

Assume we are in the situation of Proposition 7.3. Then:

• (i) the ${\mathbb {R}}[\Gamma _0]$-module F does not contain 1 nor $\eta _{|\Gamma _0}$;

• (ii) the ${\mathbb {R}}[\Gamma _0]$-module $V_0$ is a direct sum of absolutely irreducible representations.

Proof

Assume that the trivial representation $1_0$ of $\Gamma _0$ appears in F. Recall that the trivial representation 1 of $\Gamma$ does not appear in F by definition in types II or III. If $1_0$ appears in $V_0$ (or equivalently, in its outer conjugate by $\Gamma /\Gamma _0$), then $\textrm{Ind}_{\Gamma _0}^\Gamma 1_0 \simeq 1 \oplus \chi$ embeds in F, a contradiction. Therefore, we are in type IIIa and $1_0$ appears in $Q_{|\Gamma _0}$. However, $\det Q$ is 1 on $\Gamma _0$, so we have $Q_{|\Gamma _0} \simeq 1_0 \oplus 1_0$ and again $Q \simeq 1 \oplus \chi$. For similar reasons, $\eta _{|\Gamma _0}$ does not occur in F: we have

$$\begin{aligned} \textrm{Ind}_{\Gamma _0}^\Gamma \eta _{|\Gamma _0} \simeq \eta \otimes \textrm{Ind}_{\Gamma _0}^\Gamma 1 \simeq \eta \oplus \eta \chi , \end{aligned}$$

and neither $\eta$ nor $\eta \chi$ appears in F as $(r,\eta )$ is unacceptable. This proves (i). If S is an irreducible ${\mathbb {R}}[\Gamma _0]$-submodule of $V_0$, we have $1 \le \dim S \le 3$. If S is not absolutely irreducible, we necessarily have $\dim S=2$ and $\textrm{End}_{{\mathbb {R}}[\Gamma _0]} S = {\mathbb {C}}$. However, in this case, we have $\det S = 1$. As $\det V_0 = \eta _{|\Gamma _0} \ne 1$, we have $\dim V_0=3$, and so, $V_0 \simeq S \oplus \eta _{|\Gamma _0}$, in contradiction with (i). $\square$