Nonarchimedean components of non-endoscopic automorphic representations for quasisplit Sp(N) and O(N)

Abstract

Arthur classified the discrete automorphic representations of symplectic and orthogonal groups over a number field by that of the general linear groups. In this classification, those that are not from endoscopic lifting correspond to pairs $(\phi , b)$, where $\phi$ is an irreducible unitary cuspidal automorphic representation of some general linear group and b is an integer. In this paper, we study the local components of these automorphic representations at a nonarchimedean place, and we give a complete description of them in terms of their Langlands parameters.

Appendices

Appendix A: A nonvanishing result

In this appendix, we will prove the following nonvanishing result. Let $\psi$ be an Arthur parameter of G(F) (cf. (1.2)) under the Assumption (1.5). Let $>_{\psi }$ be an admissible order and we index the Jordan blocks in $Jord(\psi )$ such that

$$\begin{aligned} (\rho , A_{i+1}, B_{i+1}, \zeta _{i+1}) >_{\psi } (\rho , A_{i}, B_{i}, \zeta _{i}). \end{aligned}$$

Let

$$\begin{aligned} J {:=} \cup _{i = 1}^{n}\{(\rho , A_{i}, B_{i}, \zeta _{i})\} \subseteq Jord(\psi ) \end{aligned}$$

Suppose

$$\begin{aligned} A_{i+1} \geqslant A_{i}, \quad B_{i+1} \geqslant B_{i}, \quad \zeta _{i+1} = \zeta _{i} \quad \text { for } i < n \end{aligned}$$

and

$$\begin{aligned} J^{c} \gg J, \quad J^{c} \text { has discrete diagonal restriction}, \end{aligned}$$

where $J^{c} {:=} Jord(\psi ) \backslash J$. Then we have the following theorem.

Theorem A.1

$\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }) \ne 0$ if and only if the following condition are satisfied for all $i < n$:

$$\begin{aligned} {\left\{ \begin{array}{ll} \eta _{i+1} = (-1)^{A_{i} - B_{i}}\eta _{i} &{} \Rightarrow A_{i+1} - l_{i+1} \geqslant A_{i} - l_{i}, \quad B_{i+1} + l_{i+1} \geqslant B_{i} + l_{i}, \\ \eta _{i+1} \ne (-1)^{A_{i} - B_{i}}\eta _{i} &{} \Rightarrow B_{i+1} + l_{i+1} > A_{i} - l_{i} \end{array}\right. } \end{aligned}$$

(A.1)

Proof

The necessity of the condition follows from [10, Lemma 5.5]. So it remains to prove its sufficiency. We will proceed by induction on |J|. If $|J| = 2$, this has been proved in [10, Proposition 5.2].

Suppose $|J| = m+1$. We first “expand” $[B_{m+1}, A_{m+1}]$ to $[B^{*}_{m+1}, A^{*}_{m+1}]$ (cf. [10, Section 7.2]), so that $B_{m+1}^{*} = B_{m}$. By [10, Proposition 7.4], we know $\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }) \ne 0$ if and only if

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M, >_{\psi }} \Big (\psi _{-}, \underline{l}_{-}, \underline{\eta }_{-}; (\rho , A^{*}_{m+1}, B^{*}_{m+1}, l^{*}_{m+1}, \eta _{m+1}, \zeta _{m+1}) \Big ) \ne 0 \end{aligned}$$

(A.2)

where $\psi _{-}$ is defined by

$$\begin{aligned} Jord(\psi _{-}) = Jord(\psi ) \backslash \{(\rho , A_{m+1}, B_{m+1}, \zeta _{m+1})\} \end{aligned}$$

and

$$\begin{aligned} l^{*}_{m+1} = l_{m+1} + (B_{m+1} - B_{m}). \end{aligned}$$

It is easy to check that the condition (A.1) holds for $\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta })$ if and only if it holds for the representation in (A.2). So we will assume $B_{m+1} = B_{m}$ from now on.

Next we can “pull” $[B_{m+1}, A_{m+1}], [B_{m}, A_{m}]$ (cf. [10, 7.1]), so that they are far away from $\cup _{i < m}\{(\rho , A_{i}, B_{i}, \zeta _{i})\}$. By [10, Proposition 7.1, 7.3], we know $\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }) \ne 0$ if the following representations are all nonzero. So it suffices to show each of them is nonzero by our induction assumption. Let $\psi _{-}$ be defined by

$$\begin{aligned} Jord(\psi _{-}) = Jord(\psi ) \backslash \{(\rho , A_{m+1}, B_{m+1}, \zeta _{m+1}), (\rho , A_{m}, B_{m}, \zeta _{m})\}. \end{aligned}$$

• (1) Show

  $$\begin{aligned}&\pi ^{\Sigma _{0}}_{M, >_{\psi }} \Big (\psi _{-}, \underline{l}_{-}, \underline{\eta }_{-}; (\rho , A_{m+1} + T, B_{m+1} + T, l_{m+1}, \eta _{m+1}, \zeta _{m+1}), (\rho , A_{m} \nonumber \\&\quad +T, B_{m} + T, l_{m}, \eta _{m}, \zeta _{m})\Big ) \ne 0 \end{aligned}$$

  (A.3)

  for some T. Let $J_{-} = Jord(\psi _{-})$. Then we will choose T so that $J^{c}_{-} \gg J_{-}$. To make $J^{c}_{-}$ having discrete diagonal restriction, we will shift $[B_{m+1} + T, A_{m+1} + T]$ further to $[B_{m+1} + T', A_{m+1} + T']$ such that $B_{m+1} + T' > A_{m} +T$. Then by our induction assumption,

  $$\begin{aligned}&\pi ^{\Sigma _{0}}_{M, >_{\psi }} \Big (\psi _{-}, \underline{l}_{-}, \underline{\eta }_{-}; (\rho , A_{m+1} + T', B_{m+1} + T', l_{m+1}, \eta _{m+1}, \zeta _{m+1}), (\rho , A_{m} \\&\quad +T, B_{m} + T, l_{m}, \eta _{m}, \zeta _{m}) \Big ) \ne 0 \end{aligned}$$

  Let $\psi _{\gg }$ be the dominating parameter with discrete diagonal restriction, obtained by shifting $[B_{i}, A_{i}]$ to $[B_{i} + T_{i}, A_{i} + T_{i}]$ with $A_{i} + T_{i} < B_{m} + T$ for all $1 \leqslant i \leqslant m - 1$. Then

  $$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi _{\gg }, \underline{l}, \underline{\eta }) \hookrightarrow \times _{i < m} \begin{pmatrix} \zeta _{i}(B_{i} + T_{i}) &{} \cdots &{} \zeta _{i}(B_{i} + 1) \\ \vdots &{} &{} \vdots \\ \zeta _{i}(A_{i} + T_{i}) &{} \cdots &{} \zeta _{i}(A_{i} + 1) \end{pmatrix} \rtimes \pi ^{\Sigma _{0}}_{M, >_{\psi }}\Big (\psi _{-}, \underline{l}_{-}, \underline{\eta }_{-}; \\ (\rho , A_{m+1} + T', B_{m+1} + T', l_{m+1}, \eta _{m+1}, \zeta _{m+1}), (\rho , A_{m} +T, B_{m} + T, l_{m}, \eta _{m}, \zeta _{m}) \Big ) \end{aligned}$$

  By [10, Proposition 5.2],

  $$\begin{aligned} \text {Jac}_{(\rho , A_{m+1} + T', B_{m+1} + T', \zeta _{m+1}) \mapsto (\rho , A_{m+1} + T, B_{m+1} + T, \zeta _{m+1})} \pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi _{\gg }, \underline{l}, \underline{\eta }) \ne 0. \end{aligned}$$

  So after we apply the same Jacquet functor to the full induced representation above, we should get something nonzero. Since $B_{m+1} + T + 1 > A_{i} + T_{i}$ for $i < m$, the result is

  $$\begin{aligned}&\times _{i < m} \begin{pmatrix} \zeta _{i}(B_{i} + T_{i}) &{} \cdots &{} \zeta _{i}(B_{i} + 1) \\ \vdots &{} &{} \vdots \\ \zeta _{i}(A_{i} + T_{i}) &{} \cdots &{} \zeta _{i}(A_{i} + 1) \end{pmatrix}\\&\quad \rtimes \text {Jac}_{(\rho , A_{m+1} + T', B_{m+1} + T', \zeta _{m+1}) \mapsto (\rho , A_{m+1} + T, B_{m+1} + T, \zeta _{m+1})} \\&\quad \pi ^{\Sigma _{0}}_{M, >_{\psi }} \Big (\psi _{-}, \underline{l}_{-}, \underline{\eta }_{-}; (\rho , A_{m+1} + T', B_{m+1} + T', l_{m+1}, \eta _{m+1}, \zeta _{m+1}),\\&\quad (\rho , A_{m} +T, B_{m} + T, l_{m}, \eta _{m}, \zeta _{m}) \Big ) \ne 0 \end{aligned}$$

  This shows (A.3).

• (2) Show

  $$\begin{aligned} \pi ^{\Sigma _{0}}_{M, >_{\psi }} \Big (\psi _{-}, \underline{l}_{-}, \underline{\eta }_{-}; (\rho , A_{m+1} + T, B_{m+1} + T, l_{m+1}, \eta _{m+1}, \zeta _{m+1}), (\rho , A_{m}, B_{m}, l_{m}, \eta _{m}, \zeta _{m})\Big ) \ne 0 \end{aligned}$$

  (A.4)

  for some T. Let $J_{-} = Jord(\psi _{-}) \sqcup \{(\rho , A_{m}, B_{m}, \zeta _{m})\}$. We can choose T so that $J_{-}^{c} \gg J_{-}$. Then the statement follows from our induction assumption immediately.

• (3) Show

  $$\begin{aligned}&\pi ^{\Sigma _{0}}_{M, >'_{\psi }}\Big (\psi _{-}, \underline{l}'_{-}, \underline{\eta }'_{-}; (\rho , A_{m+1}, B_{m+1}, l'_{m+1}, \eta '_{m+1}, \zeta _{m+1}), \nonumber \\&\quad (\rho , A_{m} + T, B_{m} +T , l'_{m}, \eta '_{m}, \zeta _{n-1})\Big ) \ne 0 \end{aligned}$$

  (A.5)

  for some T, where $>'_{\psi }$ is obtained by switching $(\rho , A_{m+1}, B_{m+1}, \zeta _{m+1})$ with $(\rho , A_{m}, B_{m}, \zeta _{m})$, and $(\underline{l}', \underline{\eta }') = S^{+}_{m+1}(\underline{l}, \underline{\eta })$ (cf. [10, Section 6.1]) given by the change of order formula. Let $J_{-} = Jord(\psi _{-}) \sqcup \{(\rho , A_{m+1}, B_{m+1}, \zeta _{m+1})\}$. We can choose T so that $J_{-}^{c} \gg J_{-}$. Then the statement follows from our induction assumption again, provided we can verify the representation in (A.5) satisfies (A.1). Indeed, we only need to show

  $$\begin{aligned} {\left\{ \begin{array}{ll} \eta '_{m+1} = (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1} &{} \Rightarrow A_{m +1} - l'_{m +1} \geqslant A_{m-1} - l_{m-1}, \quad B_{m +1} + l'_{ m+1} \geqslant B_{m-1} + l_{m-1}, \\ \eta '_{m+1} \ne (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1} &{} \Rightarrow B_{m +1} + l'_{m +1} > A_{m-1} - l_{m-1}. \end{array}\right. } \end{aligned}$$

  (A.6)

  We leave it to the next lemma.$\square$

Lemma A.2

(A.6) holds.

Proof

We divide into three cases according to the change of order formula.

• (1) If $\eta _{m+1} \ne (-1)^{A_{m} - B_{m}} \eta _{m}$, then

  $$\begin{aligned} {\left\{ \begin{array}{ll} \eta '_{m + 1} = \eta _{m} \\ l'_{m+1} = (B_{m} + l_{m}) - (A_{m} - l_{m}) + l_{m+1} - 1 \end{array}\right. } \end{aligned}$$

  We get

  $$\begin{aligned} B_{m+1} + l'_{m+1} = (B_{m+1} + l_{m+1}) + (B_{m} + l_{m}) - (A_{m} - l_{m}) - 1 \\ A_{m+1} - l'_{m+1} = (A_{m+1} - l_{m+1}) + (A_{m} - l_{m}) - (B_{m} + l_{m}) + 1 \end{aligned}$$

  By (A.1), we have

  $$\begin{aligned} B_{m+1} + l_{m+1} > A_{m} - l_{m}. \end{aligned}$$

  • (a) When $\eta _{m} \ne (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$, then $\eta '_{m + 1} \ne (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$. We need to show

    $$\begin{aligned} B_{m+1} + l'_{m+1} > A_{m-1} - l_{m-1}. \end{aligned}$$

    By (A.1), we have

    $$\begin{aligned} B_{m} + l_{m} > A_{m-1} - l_{m-1}. \end{aligned}$$

    Then

    $$\begin{aligned} B_{m+1} + l'_{m+1} \geqslant B_{m} + l_{m} > A_{m-1} - l_{m-1}. \end{aligned}$$

  • (b) When $\eta _{m} = (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$, then $\eta '_{m + 1} = (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$. We need to show

    $$\begin{aligned} {\left\{ \begin{array}{ll} B_{m+1} + l'_{m+1} \geqslant B_{m-1} + l_{m-1} \\ A_{m+1} - l'_{m+1} \geqslant A_{m-1} - l_{m-1} \end{array}\right. } \end{aligned}$$

    By (A.1), we have

    $$\begin{aligned} {\left\{ \begin{array}{ll} B_{m} + l_{m} \geqslant B_{m-1} + l_{m-1} \\ A_{m} - l_{m} \geqslant A_{m-1} - l_{m-1} \end{array}\right. } \end{aligned}$$

    Then

    $$\begin{aligned} B_{m+1} + l'_{m+1}&\geqslant B_{m} + l_{m} \geqslant B_{m-1} + l_{m-1} \\ A_{m+1} - l'_{m+1}&\geqslant A_{m+1} - l_{m+1} \geqslant A_{m} - l_{m} \geqslant A_{m-1} - l_{m-1} \end{aligned}$$

• (2) If $\eta _{m+1} = (-1)^{A_{m} - B_{m}} \eta _{m}$ and

  $$\begin{aligned} l_{m+1} - l_{m} < (A_{m+1} - B_{m+1})/2 - (A_{m} - B_{m}) + l_{m}, \end{aligned}$$

  then

  $$\begin{aligned} {\left\{ \begin{array}{ll} \eta '_{m + 1} \ne \eta _{m} \\ l'_{m+1} = (A_{m} - l_{m}) - (B_{m} + l_{m}) + l_{m+1} - 1 \end{array}\right. } \end{aligned}$$

  We get

  $$\begin{aligned} B_{m+1} + l'_{m+1} = (B_{m+1} + l_{m+1}) - (B_{m} + l_{m}) + (A_{m} - l_{m}) + 1 \\ A_{m+1} - l'_{m+1} = (A_{m+1} - l_{m+1}) - (A_{m} - l_{m}) + (B_{m} + l_{m}) - 1 \end{aligned}$$

  By (A.1), we have

  $$\begin{aligned} {\left\{ \begin{array}{ll} B_{m+1} + l_{m+1} \geqslant B_{m} + l_{m} \\ A_{m+1} - l_{m+1} \geqslant A_{m} - l_{m} \end{array}\right. } \end{aligned}$$

  • (a) When $\eta _{m} \ne (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$, then $\eta '_{m + 1} = (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$. We need to show

    $$\begin{aligned} {\left\{ \begin{array}{ll} B_{m+1} + l'_{m+1} \geqslant B_{m-1} + l_{m-1} \\ A_{m+1} - l'_{m+1} \geqslant A_{m-1} - l_{m-1} \end{array}\right. } \end{aligned}$$

    By (A.1), we have

    $$\begin{aligned} B_{m} + l_{m} > A_{m-1} - l_{m-1}. \end{aligned}$$

    Then

    $$\begin{aligned} B_{m+1} + l'_{m+1}&\geqslant (A_{m} - l_{m}) + 1 \geqslant (A_{m-1} - l_{m-1}) + 1 \geqslant B_{m-1} + l_{m-1} \\ A_{m+1} - l'_{m+1}&\geqslant (B_{m} + l_{m}) - 1 \geqslant A_{m-1} - l_{m-1} \end{aligned}$$

  • (b) When $\eta _{m} = (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$, then $\eta '_{m + 1} \ne (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$. We need to show

    $$\begin{aligned} B_{m+1} + l'_{m+1} > A_{m-1} - l_{m-1} \end{aligned}$$

    By (A.1), we have

    $$\begin{aligned} {\left\{ \begin{array}{ll} B_{m} + l_{m} \geqslant B_{m-1} + l_{m-1} \\ A_{m} - l_{m} \geqslant A_{m-1} - l_{m-1} \end{array}\right. } \end{aligned}$$

    Then

    $$\begin{aligned} B_{m+1} + l'_{m+1} \geqslant (A_{m} - l_{m}) + 1 \geqslant (A_{m-1} - l_{m-1}) + 1 > A_{m-1} - l_{m-1} \end{aligned}$$

• (3) If $\eta _{m+1} = (-1)^{A_{m} - B_{m}} \eta _{m}$ and

  $$\begin{aligned} l_{m+1} - l_{m} \geqslant (A_{m+1} - B_{m+1})/2 - (A_{m} - B_{m}) + l_{m}, \end{aligned}$$

  then

  $$\begin{aligned} {\left\{ \begin{array}{ll} \eta '_{m + 1} = \eta _{m} \\ l'_{m+1} = (A_{m+1} - B_{m+1}) - l_{m+1} - (A_{m} - l_{m}) + (B_{m} + l_{m}) \end{array}\right. } \end{aligned}$$

  We get

  $$\begin{aligned} B_{m+1} + l'_{m+1} = (A_{m+1} - l_{m+1}) - (A_{m} - l_{m}) + (B_{m} + l_{m}) \\ A_{m+1} - l'_{m+1} = (B_{m+1} + l_{m+1}) - (B_{m} + l_{m}) + (A_{m} - l_{m}) \end{aligned}$$

  By (A.1), we have

  $$\begin{aligned} {\left\{ \begin{array}{ll} B_{m+1} + l_{m+1} \geqslant B_{m} + l_{m} \\ A_{m+1} - l_{m+1} \geqslant A_{m} - l_{m} \end{array}\right. } \end{aligned}$$

  • (a) When $\eta _{m} \ne (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$, then $\eta '_{m + 1} \ne (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$. We need to show

    $$\begin{aligned} B_{m+1} + l'_{m+1} > A_{m-1} - l_{m-1} \end{aligned}$$

    By (A.1), we have

    $$\begin{aligned} B_{m} + l_{m} > A_{m-1} - l_{m-1}. \end{aligned}$$

    Then

    $$\begin{aligned} B_{m+1} + l'_{m+1} \geqslant B_{m} + l_{m} > A_{m-1} - l_{m-1} \end{aligned}$$

  • (b) When $\eta _{m} = (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$, then $\eta '_{m + 1} = (-1)^{A_{m-1} - B_{m-1}} \eta _{m-1}$. We need to show

    $$\begin{aligned} {\left\{ \begin{array}{ll} B_{m+1} + l'_{m+1} \geqslant B_{m-1} + l_{m-1} \\ A_{m+1} - l'_{m+1} \geqslant A_{m-1} - l_{m-1} \end{array}\right. } \end{aligned}$$

    By (A.1), we have

    $$\begin{aligned} {\left\{ \begin{array}{ll} B_{m} + l_{m} \geqslant B_{m-1} + l_{m-1} \\ A_{m} - l_{m} \geqslant A_{m-1} - l_{m-1} \end{array}\right. } \end{aligned}$$

    Then

    $$\begin{aligned} B_{m+1} + l'_{m+1} \geqslant B_{m} + l_{m} \geqslant B_{m-1} + l_{m-1} \\ A_{m+1} - l'_{m+1} \geqslant A_{m} - l_{m} \geqslant A_{m-1} - l_{m-1} \end{aligned}$$

$\square$

More generally, we can drop the Assumption (1.5), but only assume $\psi = \psi _{p}$. Suppose for each $\rho$ appearing in $Jord(\psi )$, we have the same setup as in Theorem A.1. Then we have

Theorem A.3

$\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }) \ne 0$ if and only if the condition (A.1) is satisfied for each $\rho$.

Proof

We can apply the arguments of the proof of Theorem A.1 to each $\rho$ one by one, which reduces it to the case that $|J| = 2$ for each $\rho$. Then this case follows from [10, Proposition 5.3]. $\square$

Appendix B. Change sign

In this appendix, we would like to extend [10, Proposition 7.6] as follows. Let $\psi$ be an Arthur parameter of G(F) such that $\psi = \psi _{p}$. We choose an admissible order $>_{\psi }$ and fix an irreducible unitary supercuspidal representation $\rho$ of $GL(d_{\rho }, F)$. Let us index the Jordan blocks in $Jord_{\rho }(\psi )$ such that

$$\begin{aligned} (\rho , A_{i+1}, B_{i+1}, \zeta _{i+1}) >_{\psi } (\rho , A_{i}, B_{i}, \zeta _{i}). \end{aligned}$$

Suppose there exists n such that for $i > n$,

$$\begin{aligned} (\rho , A_{i}, B_{i}, \zeta _{i}) \gg \cup _{j=1}^{n}\{(\rho , A_{j}, B_{j}, \zeta _{j})\}. \end{aligned}$$

Moreover, there exists $1 \leqslant m \leqslant n$ such that

$$\begin{aligned} A_m = \cdots = A_1 \geqslant A_i, \quad B_m = \cdots = B_1 = 1/2, \quad \zeta _{m} = \cdots = \zeta _{1} \ne \zeta _{i} \end{aligned}$$

for $m < i \leqslant n$. We introduce another parameter $\psi ^{*}$ by changing $(\rho , A_{i}, B_{i}, \zeta _{i})$ to $(\rho , A_{i} + 1, B_{i}, -\zeta _{i})$ for $i \leqslant m$. For any $(\underline{l}, \underline{\eta })$, such that

$$\begin{aligned} \quad l_{i+1} = l_{i}, \quad \eta _{i+1} = (-1)^{A_{i} - \frac{1}{2}}\eta _{i} \quad \text { for } i < m, \end{aligned}$$

(B.1)

we can associate it with $(\underline{l}^*, \underline{\eta }^*)$, defined as follows. For $i > m$,

$$\begin{aligned} l^{*}_{i} = l_{i}, \quad \eta ^{*}_{i} = \eta _{i}. \end{aligned}$$

For $i < m$,

$$\begin{aligned} l^{*}_{i+1} = l^{*}_{i}, \quad \eta ^{*}_{i+1} = (-1)^{A_{i} + \frac{1}{2}} \eta ^{*}_{i}. \end{aligned}$$

(B.2)

Then it remains to specify $l^{*}_1, \eta ^{*}_1$, which are given by

$$\begin{aligned} \eta ^{*}_{1} = -\eta _{1}, \quad l^{*}_{1} = {\left\{ \begin{array}{ll} l_{1} + 1 &{} \text { if } \eta _{1} = 1 \\ l_{1} &{} \text { if } \eta _{1} = -1 \\ \end{array}\right. } \end{aligned}$$

In case $l_{1} = (A_1 + \frac{1}{2})/2$, we fix $\eta _{1} = -1$.

Proposition B.1

$\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }) \ne 0$ if and only if $\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi ^{*}, \underline{l}^{*}, \underline{\eta }^{*}) \ne 0$. Moreover,

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi ^{*}, \underline{l}^{*}, \underline{\eta }^{*}) \hookrightarrow \times _{i = 1}^{m} \langle -\zeta _{i}1/2, \ldots , -\zeta _{i}(A_{i} + 1) \rangle \rtimes \pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }) \end{aligned}$$

(B.3)

Remark B.2

[10, Proposition 7.6] settles the case when $m = 1$.

Proof

As in the proof of [10, Proposition 7.6], we can reduce it to the case that

$$\begin{aligned} m = n \text { and } Jord(\psi ) \backslash \cup _{i=1}^{n}\{(\rho , A_{i}, B_{i}, \zeta _{i})\} \text { has discrete diagonal restriction}. \end{aligned}$$

Because of the conditions (B.1) and (B.2), we have

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi , \underline{l}, \underline{\eta }) \ne 0 \text { and } \pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi ^{*}, \underline{l}^{*}, \underline{\eta }^{*}) \ne 0 \end{aligned}$$

by Theorem A.3. So we only need to show (B.3), and we will proceed by induction on n.

Let $\psi ^{*}_{>}$ be obtained from $\psi ^{*}$ by changing $(\rho , A_{n} + 1, 1/2, -\zeta _{n})$ to $(\rho , A_{n} + 1 + T_{n}, 1/2 + T_{n}, -\zeta _{n})$ for $T_{n}$ sufficiently large. Then by our induction assumption, we have

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi ^{*}_{>}, \underline{l}^{*}, \underline{\eta }^{*}) \hookrightarrow \times _{i = 1}^{n-1} \langle -\zeta _{i}1/2, \ldots , -\zeta _{i}(A_{i} + 1) \rangle \rtimes \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi ^{(n)}_{>}, \underline{l}^{(n)}, \underline{\eta }^{(n)}) \end{aligned}$$

where $\psi ^{(n)}_{>}$ is obtained from $\psi ^{*}_{>}$ by changing $(\rho , A_{i} + 1, 1/2, -\zeta _{i})$ back to $(\rho , A_{i}, 1/2, \zeta _{i})$ for $1 \leqslant i < n$. Moreover,

$$\begin{aligned} l^{(n)}_{i} = l_{i}, \quad \eta ^{(n)}_{i} = \eta _{i} \quad \text { for } i < n, \end{aligned}$$

and

$$\begin{aligned} l^{(n)}_{i} = l^{*}_{i}, \quad \eta ^{(n)}_{i} = \eta ^{*}_{i} \quad \text { for } i \geqslant n. \end{aligned}$$

Then we claim

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi ^{(n)}_{>}, \underline{l}^{(n)}, \underline{\eta }^{(n)}) \hookrightarrow \underbrace{\begin{pmatrix} -\zeta _{n}(1/2 + T_{n}) &{} \cdots &{} -\zeta _{n}1/2 \\ \vdots &{} &{} \vdots \\ -\zeta _{n}(A_{n} + 1 + T_{n}) &{} \cdots &{} -\zeta _{n}(A_{n} + 1) \end{pmatrix}}_{{\mathcal {C}}_{X_{n}}} \rtimes \pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }). \end{aligned}$$

(B.4)

where

$$\begin{aligned} X_{n} {:=} \begin{bmatrix} -\zeta _{n}(1/2 + T_{n}) &{} \cdots &{} -\zeta _{n}1/2 \\ \vdots &{} &{} \vdots \\ -\zeta _{n}(A_{n} + 1 + T_{n}) &{} \cdots &{} -\zeta _{n}(A_{n} + 1) \end{bmatrix} \end{aligned}$$

If this is the case, then

$$\begin{aligned} \pi ^{\Sigma _{0}}_{>_{\psi }}(\psi ^{*}_{>}, \underline{l}^{*}, \underline{\eta }^{*})&\hookrightarrow \times _{i = l}^{n-1} \langle -\zeta _{i}1/2, \ldots , -\zeta _{i}(A_{i} + 1) \rangle \times {\mathcal {C}}_{X_{n}} \rtimes \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi , \underline{l}, \underline{\eta }) \\&\cong {\mathcal {C}}_{X_{n}} \times \times _{i = l}^{n-1} \langle -\zeta _{i}1/2, \ldots , -\zeta _{i}(A_{i} + 1) \rangle \rtimes \pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }), \end{aligned}$$

from which (B.3) follows.

We still need to show the claim (B.4). Let $\psi ^{(n)}$ be obtained from $\psi ^{(n)}_{>}$ by moving $(\rho , A_{n} + 1 + T_{n}, 1/2 + T_{n}, -\zeta _{n})$ back to $(\rho , A_{n} + 1, 1/2, -\zeta _{n})$. Suppose $\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi ^{(n)}, \underline{l}^{(n)}, \underline{\eta }^{(n)}) \ne 0$, then

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi ^{(n)}_{>}, \underline{l}^{(n)}, \underline{\eta }^{(n)}) \hookrightarrow \begin{pmatrix} -\zeta _{n}(1/2 + T_{n}) &{} \cdots &{} -\zeta _{n}3/2 \\ \vdots &{} &{} \vdots \\ -\zeta _{n}(A_{n} + 1 + T_{n}) &{} \cdots &{} -\zeta _{n}(A_{n} + 2) \end{pmatrix} \rtimes \pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi ^{(n)}, \underline{l}^{(n)}, \underline{\eta }^{(n)}). \end{aligned}$$

(B.5)

To show the nonvanishing of $\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi ^{(n)}, \underline{l}^{(n)}, \underline{\eta }^{(n)})$, we need to switch to a new order $>'_{\psi }$ by moving $(\rho , A_{n} + 1, 1/2, -\zeta _{n})$ to the last position. Then

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi ^{(n)}, \underline{l}^{(n)}, \underline{\eta }^{(n)}) = \pi ^{\Sigma _{0}}_{M, >'_{\psi }}(\psi ^{(n)}, \underline{l}^{'(n)}, \underline{\eta }^{'(n)}), \end{aligned}$$

where

$$\begin{aligned} l^{'(n)}_{i} = l_{i}^{(n)}, \quad \eta ^{'(n)}_{i} = \eta ^{(n)}_{i} \quad \text { for } i > n, \end{aligned}$$

and

$$\begin{aligned} l^{'(n)}_{i} = l_{i}^{(n)}, \quad \eta ^{'(n)}_{i} = (-1)^{A_{n} - 1/2}\eta ^{(n)}_{i} \quad \text { for } i < n, \end{aligned}$$

and

$$\begin{aligned} l^{'(n)}_{n} = l^{*}_{1}, \quad \eta ^{'(n)}_{n} = \eta ^{*}_{1}. \end{aligned}$$

Let $\psi ^{(n)}_{\gg }$ be a dominating parameter for $\psi ^{(n)}$ with respect to $>'_{\psi }$, obtained by changing $(\rho , A_{i}, B_{i}, \zeta _{i})$ to $(\rho , A_{i} + T_{i}, B_{i} + T_{i}, \zeta _{i})$ for $i < n$. We also require that $\psi ^{(n)}_{\gg }$ has discrete diagonal restriction. Then by [10, Proposition 7.6],

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>'_{\psi }}(\psi ^{(n)}_{\gg }, \underline{l}^{'(n)}, \underline{\eta }^{'(n)}) \hookrightarrow \langle -\zeta _{n}1/2, \ldots , -\zeta _{n}(A_{n} + 1) \rangle \rtimes \pi ^{\Sigma _{0}}_{M, >'_{\psi }}(\psi _{\gg }, \underline{l}', \underline{\eta }'), \end{aligned}$$

where $\psi _{\gg }$ is obtained from $\psi ^{(n)}_{\gg }$ by changing $(\rho , A_{n} + 1, 1/2, -\zeta _{n})$ back to $(\rho , A_{n}, 1/2, \zeta _{n})$. Note

$$\begin{aligned} l'_{i} = l^{'(n)}_{i}, \quad \eta '_{i} = \eta ^{'(n)}_{i} \quad \text { for } i \ne n, \end{aligned}$$

and

$$\begin{aligned} l'_{n} = l_{1}, \quad \eta '_{n} = \eta _{1}. \end{aligned}$$

It is easy to check by the change of order formula that

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi , \underline{l}, \underline{\eta }) = \pi ^{\Sigma _{0}}_{M, >'_{\psi }}(\psi , \underline{l}', \underline{\eta }'). \end{aligned}$$

In particular, the right hand side is nonzero. Therefore,

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>'_{\psi }}(\psi _{\gg }, \underline{l}', \underline{\eta }') \hookrightarrow \times _{i = 1}^{n-1} \begin{pmatrix} \zeta _{i}(1/2 + T_{i}) &{} \cdots &{} \zeta _{i}3/2 \\ \vdots &{} &{} \vdots \\ \zeta _{i}(A_{i} + T_{i}) &{} \cdots &{} \zeta _{i}(A_{i} + 1) \end{pmatrix} \rtimes \pi ^{\Sigma _{0}}_{M, >'_{\psi }}(\psi , \underline{l}', \underline{\eta }') \end{aligned}$$

Combined with the previous inclusion, we get

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>'_{\psi }}(\psi ^{(n)}_{\gg }, \underline{l}^{'(n)}, \underline{\eta }^{'(n)})&\hookrightarrow \langle -\zeta _{n}1/2, \ldots , -\zeta _{n}(A_{n} + 1) \rangle \times \\&\times _{i = 1}^{n-1} \begin{pmatrix} \zeta _{i}(1/2 + T_{i}) &{} \cdots &{} \zeta _{i}3/2 \\ \vdots &{} &{} \vdots \\ \zeta _{i}(A_{i} + T_{i}) &{} \cdots &{} \zeta _{i}(A_{i} + 1) \end{pmatrix} \rtimes \pi ^{\Sigma _{0}}_{M,>'_{\psi }}(\psi , \underline{l}', \underline{\eta }') \\&\cong \times _{i = 1}^{n-1} \begin{pmatrix} \zeta _{i}(1/2 + T_{i}) &{} \cdots &{} \zeta _{i}3/2 \\ \vdots &{} &{} \vdots \\ \zeta _{i}(A_{i} + T_{i}) &{} \cdots &{} \zeta _{i}(A_{i} + 1) \end{pmatrix} \\&\times \langle -\zeta _{n}1/2, \ldots , -\zeta _{n}(A_{n} + 1) \rangle \rtimes \pi ^{\Sigma _{0}}_{M, >'_{\psi }}(\psi , \underline{l}', \underline{\eta }') \end{aligned}$$

Consequently, $\pi ^{\Sigma _{0}}_{M, >'_{\psi }}(\psi ^{(n)}, \underline{l}^{'(n)}, \underline{\eta }^{'(n)}) \ne 0$ and

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>'_{\psi }}(\psi ^{(n)}, \underline{l}^{'(n)}, \underline{\eta }^{'(n)}) \hookrightarrow \langle -\zeta _{n}1/2, \ldots , -\zeta _{n}(A_{n} + 1) \rangle \rtimes \pi ^{\Sigma _{0}}_{M, >'_{\psi }}(\psi , \underline{l}', \underline{\eta }'). \end{aligned}$$

Substitute the above expression into (B.5), we obtain

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi ^{(n)}_{>}, \underline{l}^{(n)}, \underline{\eta }^{(n)})&\hookrightarrow \begin{pmatrix} -\zeta _{n}(1/2 + T_{n}) &{} \cdots &{} -\zeta _{n}3/2 \\ \vdots &{} &{} \vdots \\ -\zeta _{n}(A_{n} + 1 + T_{n}) &{} \cdots &{} -\zeta _{n}(A_{n} + 2) \end{pmatrix} \\&\times \langle -\zeta _{n}1/2, \ldots , -\zeta _{n}(A_{n} + 1) \rangle \rtimes \pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }). \end{aligned}$$

Note the Jordan blocks in $Jord_{\rho }(\psi )$ satisfies $A_{i} < A_{n} + 1$ for $i \leqslant n$, and $B_{i} > A_{n} + 1 + T_{n}$ for $i > n$. If we apply $\text {Jac}_{X_{n}}$ to the right hand side of the above expression, we can only get $\pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta })$. This means the left hand side is the unique irreducible subrepresentation of the right hand side. Therefore,

$$\begin{aligned} \pi ^{\Sigma _{0}}_{M,>_{\psi }}(\psi ^{(n)}_{>}, \underline{l}^{(n)}, \underline{\eta }^{(n)}) \hookrightarrow {\mathcal {C}}_{X_{n}} \rtimes \pi ^{\Sigma _{0}}_{M, >_{\psi }}(\psi , \underline{l}, \underline{\eta }), \end{aligned}$$

which is exactly (B.4). This finishes our proof. $\square$