Proof Suppose, on the contrary, that there is an irreducible tempered subquotient $\tau $ of $\langle [ \nu ^{a} \rho , \nu ^{b} \rho ] \rangle \rtimes \sigma $ . Inspecting its cuspidal support, we deduce that $\tau $ is not strongly positive, and it is easy to see that $\tau $ can be written as a subrepresentation of an induced representation of the form $\delta ([\nu ^{x} \rho , \nu ^y \rho ]) \rtimes \tau '$ , where $x \leq 0, y \geq 0$ , $x+y \geq 0$ , and $\tau '$ is irreducible tempered. Using the Frobenius reciprocity, we obtain $\mu ^{\ast }(\langle [ \nu ^{a} \rho , \nu ^{b} \rho ] \rangle \rtimes \sigma ) \geq \delta ([\nu ^{x} \rho , \nu ^y \rho ]) \otimes \tau '$ , and Theorem 2.1 implies that there are $0 \leq j \leq i \leq b-a+1$ and an irreducible constituent $\pi \otimes \sigma '$ of $\mu ^{\ast }(\sigma )$ such that

$$ \begin{align*} \delta([\nu^{x} \rho, \nu^y \rho]) \leq \langle [\nu^{-b} \rho, \nu^{-i-a} \rho] \rangle \times \langle [\nu^{a} \rho, \nu^{j+a-1} \rho] \rangle \times \pi. \end{align*} $$

Using the same reasoning as in the proof of [Reference Matić11, Theorem 3.9], we conclude that $\rho $ appears in the cuspidal support of $\pi $ , which contradicts [Reference Matić9, Theorem 5.3] (or [Reference Matić and Tadić12, Section 7]).

Proof The proof closely follows that of [Reference Matić11, Proposition 4.2]. We restrict our attention to the main differences.

Let $i \in \{ 1, 2, \ldots , k \}$ be such that $b_i - a_i = 1$ . Following the same lines as in the proof of [Reference Matić11, Proposition 4.2], we deduce that there is a c such that $1 \leq c \leq -a+1$ , $a_i = -c$ , $L(\delta _{i}, \ldots , \delta _k; \tau )$ is an irreducible subquotient of $\langle [ \nu ^{-c+1} \rho , \nu ^{c} \rho ] \rangle \rtimes \sigma $ , and $L(\delta _{i+1}, \ldots , \delta _k; \tau )$ is an irreducible subquotient of $\langle [ \nu ^{-c+2} \rho , \nu ^{c-1} \rho ] \rangle \rtimes \sigma $ .

If for all $j \in \{ i+1, \ldots , k \}$ , we have $b_j - a_j = 1$ , a repeated application of [Reference Matić10, Lemma 2.6] shows that either $\tau \cong \sigma $ or there is a $d \geq 1$ such that $\tau $ is an irreducible subquotient of $\langle [ \nu ^{-d+1} \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma $ . If $\tau \cong \sigma $ , it follows that $a_k = -1, b_k = 0$ , and $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma )$ is an irreducible subquotient of $\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma $ . If $\tau $ is an irreducible subquotient of $\langle [ \nu ^{-d+1} \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma $ , with $d \geq 1$ , Lemma 3.1 implies $d = 1$ .

Suppose that there is a $j \in \{ i+1, \ldots , k \}$ such that $a_j = b_j$ . In a completely analogous way as in the proof of [Reference Matić11, Proposition 4.2], we deduce that $a_t = b_t$ holds for all $t \in \{ j, \ldots , k\}$ , and there are x and y, $0 \leq -y < x \leq -a$ , such that $\tau $ is an irreducible subquotient of $\langle [ \nu ^{y} \rho , \nu ^{x} \rho ] \rangle \rtimes \sigma $ . Lemma 3.1 gives $y = 0$ and ends the proof.

Proof By [Reference Matić9, Theorem 5.3] or [Reference Matić and Tadić12, Section 7], we have $c> 0$ , $\sigma '$ is strongly positive, and $\delta ([\nu ^c \rho ', \nu ^d \rho ']) \otimes \sigma '$ is a unique irreducible constituent of $\mu ^{\ast }(\sigma )$ of the form $\delta ([\nu ^c \rho ', \nu ^d \rho ']) \otimes \pi $ .

If $\rho ' \cong \rho $ , then $c \geq 2$ . This leads to

$$ \begin{align*} \delta([\nu^{-1} \rho, \rho]) \times \delta([\nu^c \rho', \nu^d \rho']) \cong \delta([\nu^c \rho', \nu^d \rho']) \times \delta([\nu^{-1} \rho, \rho]). \end{align*} $$

Thus, $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma )$ is a subrepresentation of

$$ \begin{align*} \delta([\nu^c \rho', \nu^d \rho']) \times \delta([\nu^{-1} \rho, \rho]) \rtimes \sigma'. \end{align*} $$

By [Reference Jantzen6, Lemma 5.5], there is an irreducible subquotient $\pi $ of $\delta ([\nu ^{-1} \rho , \rho ]) \rtimes \sigma '$ such that $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma )$ is a subrepresentation of $\delta ([\nu ^c \rho ', \nu ^d \rho ']) \rtimes \pi $ . By the Frobenius reciprocity, $\mu ^{\ast }(L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma ))$ contains $\delta ([\nu ^{-1} \rho , \rho ]) \otimes \sigma $ , and it follows that $\mu ^{\ast }(\delta ([\nu ^c \rho ', \nu ^d \rho ']) \rtimes \pi )$ also contains $\delta ([\nu ^{-1} \rho , \rho ]) \otimes \sigma $ . Now, Theorem 2.1 implies that $\mu ^{\ast }(\pi )$ contains an irreducible constituent of the form $\delta ([\nu ^{-1} \rho , \rho ]) \otimes \pi '$ . Strong positivity of $\sigma '$ implies $\pi \cong L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma ')$ , so

$$ \begin{align*} \mu^{\ast}(L(\delta([\nu^{-1} \rho, \rho]); \sigma)) \geq \delta([\nu^c \rho', \nu^d \rho']) \otimes L(\delta([\nu^{-1} \rho, \rho]); \sigma'). \end{align*} $$

From

$$ \begin{align*} L(\delta([\nu^{-1} \rho, \rho]); \sigma) \leq \langle [ \rho, \nu \rho] \rangle \rtimes \sigma \end{align*} $$

follows that $\mu ^{\ast }(\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma )$ contains $\delta ([\nu ^c \rho ', \nu ^d \rho ']) \otimes L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma ')$ .

It is now a direct consequence of Theorem 2.1 that $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma ')$ is an irreducible subquotient of $\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma '$ .

Proof We use a case-by-case consideration. Let us first consider the case $1 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ and $3 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ . By [Reference Muić16, Theorem 2.1], in $R(G)$ , we have

$$ \begin{align*} \delta [ \rho, \nu \rho] \rangle \rtimes \sigma = L(\delta([\nu^{-1} \rho, \rho]); \sigma) + \sigma^{(1)}_{ds} + \sigma^{(2)}_{ds}, \end{align*} $$

where $\sigma ^{(1)}_{ds}$ and $\sigma ^{(2)}_{ds}$ are discrete series subrepresentations of $\delta [ \rho , \nu \rho ] \rangle \rtimes \sigma $ .

Since $\mu ^{\ast }(\sigma )$ does not contain an irreducible constituent of the form $\nu \rho \otimes \pi $ , and $\rho \rtimes \sigma $ is a length-two representation, Theorem 2.1 can be used to show that $\mu ^{\ast }(L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma ))$ does not contain an irreducible constituent of the form $\nu \rho \otimes \pi $ .

By [Reference Muić16, Proposition 3.1], the induced representation $\nu \rho \rtimes \sigma $ is irreducible, so we have

$$ \begin{align*} L(\delta([\nu^{-1} \rho, \rho]); \sigma) \hookrightarrow \rho \times \nu^{-1} \rho \rtimes \sigma \cong \rho \times \nu \rho \rtimes \sigma. \end{align*} $$

By [Reference Jantzen6, Lemma 5.5], there is an irreducible subquotient $\pi $ of $\rho \times \nu \rho $ such that $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma )$ is a subrepresentation of $\pi \rtimes \sigma $ . Since, in $R(GL)$ , we have

$$ \begin{align*} \rho \times \nu \rho = \delta([\rho, \nu \rho]) + \langle [ \rho, \nu \rho] \rangle, \end{align*} $$

and $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma )$ is not a subrepresentation of $\nu \rho \times \rho \rtimes \sigma $ , we get $\pi \cong \langle [ \rho , \nu \rho ] \rangle $ .

Now, we consider the case $1 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ and $3 \in \mathop {\text {Jord}}_{\rho }(\sigma )$ . We denote by $\sigma '$ a strongly positive discrete series such that $\sigma $ is a subrepresentation of $\nu \rho \rtimes \sigma '$ . It follows from [Reference Muić16, Theorem 4.1(iii)] that, in $R(G)$ , we have

$$ \begin{align*} \delta ([ \rho, \nu \rho] ) \rtimes \sigma = L(\delta([\nu^{-1} \rho, \rho]); \sigma) + \tau, \end{align*} $$

where $\tau $ is an irreducible tempered representation, which is a subrepresentation of both $\delta ([\nu ^{-1} \rho , \nu \rho ]) \rtimes \sigma '$ and $\nu \rho \times \nu \rho \times \rho \rtimes \sigma '$ . We note that $\rho \rtimes \sigma '$ is irreducible. It can be seen using Theorem 2.1 and the description of the Jacquet modules of strongly positive discrete series ([Reference Matić9, Theorem 5.3] or [Reference Matić and Tadić12, Section 7]) that $\nu \rho \times \nu \rho \otimes \rho \rtimes \sigma '$ is a unique irreducible constituent of the form $\nu \rho \times \nu \rho \otimes \pi $ appearing in $\mu ^{\ast }(\delta [ \rho , \nu \rho ] \rangle \rtimes \sigma )$ . Consequently, $\mu ^{\ast }(L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma ))$ does not contain an irreducible constituent of the form $\nu \rho \times \nu \rho \otimes \pi $ .

Since $\nu \rho \rtimes \sigma $ is irreducible, we have

$$ \begin{align*} L(\delta([\nu^{-1} \rho, \rho]); \sigma) \hookrightarrow \rho \times \nu \rho \rtimes \sigma. \end{align*} $$

We denote by $\pi $ an irreducible subquotient of $\rho \times \nu \rho $ such that $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma )$ is a subrepresentation of $\pi \rtimes \sigma $ .

If $\pi \cong \delta ([\rho , \nu \rho ])$ , we have

$$ \begin{align*} L(\delta([\nu^{-1} \rho, \rho]); \sigma) &\hookrightarrow \delta([\rho, \nu \rho]) \rtimes \sigma \\ &\hookrightarrow \delta([\rho, \nu \rho]) \times \nu \rho \rtimes \sigma' \\ &\cong \nu \rho \times \delta([\rho, \nu \rho]) \rtimes \sigma' \\ &\hookrightarrow \nu \rho \times \nu \rho \times \rho \rtimes \sigma', \end{align*} $$

which is impossible. Thus, $\pi \cong \langle [ \rho , \nu \rho ] \rangle $ .

Let us now assume $1 \in \mathop {\text {Jord}}_{\rho }(\sigma )$ and $3 \in \mathop {\text {Jord}}_{\rho }(\sigma )$ . Then, $\mu ^{\ast }(\sigma )$ does not contain an irreducible constituent of the form $\nu \rho \otimes \pi $ . Also, in this case, $\delta ([ \rho , \nu \rho ] ) \rtimes \sigma $ is irreducible, so

$$ \begin{align*} L(\delta([\nu^{-1} \rho, \rho]); \sigma) \cong \delta ([ \rho, \nu \rho]) \rtimes \sigma. \end{align*} $$

Thus, $\mu ^{\ast }(L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma ))$ contains an irreducible constituent of the form $\nu \rho \otimes \pi '$ , and it follows that $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma )$ is not an irreducible subquotient of $\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma $ .

It remains to consider the case $1 \in \mathop {\text {Jord}}_{\rho }(\sigma )$ and $3 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ . Note that, in this case, $\nu \rho $ does not appear in the cuspidal support of $\sigma $ , and there is a positive integer d, $d \geq 2$ , such that $\sigma $ is a subrepresentation of $\delta ([\nu ^{2} \rho , \nu ^d \rho ]) \rtimes \sigma '$ , for a strongly positive discrete series $\sigma '$ .

Suppose that $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma )$ is an irreducible subquotient of $\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma $ . Using a repeated application of Proposition 2.2 and Lemma 3.4, we obtain that then $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _c)$ is an irreducible subquotient of $\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma _c$ . Thus, it is enough to prove that in this case $\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma _c$ does not contain $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _c)$ .

Note that $\alpha \geq 1$ . The case $\alpha = 1$ follows directly from [Reference Tadić19, Proposition 3.8], so we comment only case $\alpha> 1$ .

For $\alpha> 1$ , $\delta ([\rho , \nu \rho ]) \rtimes \sigma _c$ is irreducible by [Reference Muić16, Theorem 4.1(i)], so we have $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _c) \cong \delta ([\rho , \nu \rho ]) \rtimes \sigma _c$ and $\mu ^{\ast }(L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _c)) \geq \nu \rho \otimes \rho \rtimes \sigma _c$ . Since, by Theorem 2.1, $\mu ^{\ast }(\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma _c)$ does not contain $\nu \rho \otimes \rho \rtimes \sigma _c$ , we obtain that $\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma _c$ does not contain $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _c)$ .

Proof Let us first show that there are no irreducible tempered subquotients in the case $1 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ . Suppose, on the contrary, that $1 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ and there is an irreducible tempered subquotient $\tau $ of $\langle [ \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma $ . If $2d+1 \in \mathop {\text {Jord}}_{\rho }(\sigma )$ , using cuspidal support considerations, similarly as in [Reference Hanzer4, Section 4.2.1], we deduce that $\tau $ is a subrepresentation of an induced representation of the form $\delta ([\nu ^{-d} \rho , \nu ^d \rho ]) \rtimes \sigma _1$ , where $\sigma _1$ is a strongly positive discrete series.

The Frobenius reciprocity implies $\mu ^{\ast }(\langle [ \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma ) \geq \delta ([\nu ^{-d} \rho , \nu ^d \rho ]) \otimes \sigma _1$ . Using Theorem 2.1 and [Reference Matić9, Theorem 5.3], we easily get $d = 1$ and $\mu ^{\ast }(\sigma ) \geq \nu \rho \otimes \sigma _1$ . Consequently, $1 \in \mathop {\text {Jord}}_{\rho }(\sigma _1)$ .

Thus,

$$ \begin{align*} \tau \hookrightarrow \nu \rho \times \delta([\nu^{-1} \rho, \rho]) \rtimes \sigma_1, \end{align*} $$

and [Reference Jantzen6, Lemma 5.5] implies that there is an irreducible subquotient $\pi $ of $\delta ([\rho , \nu \rho ]) \rtimes \sigma _1 = \delta ([\nu ^{-1} \rho , \rho ]) \rtimes \sigma _1$ such that $\tau $ is a subrepresentation of $\nu \rho \rtimes \pi $ .

By [Reference Muić16, Theorem 4.1(ii)], in $R(G)$ , we have

$$ \begin{align*} \delta([\rho, \nu \rho]) \rtimes \sigma_1 = \tau' + L(\delta([\nu^{-1} \rho, \rho]); \sigma_1), \end{align*} $$

where $\tau '$ is an irreducible tempered subrepresentation of $\delta ([\rho , \nu \rho ]) \rtimes \sigma _1$ . If $\pi \cong \tau '$ , we have

$$ \begin{align*} \tau &\hookrightarrow \nu \rho \rtimes \tau' \hookrightarrow \nu \rho \times \delta([\rho, \nu \rho]) \rtimes \sigma_1 \\ &\hookrightarrow \nu \rho \times \nu \rho \times \rho \rtimes \sigma_1, \end{align*} $$

which is impossible since $\mu ^{\ast }(\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma )$ does not contain an irreducible constituent of the form $\nu \rho \times \nu \rho \otimes \pi '$ . Thus, $\pi \cong L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _1)$ , and $\mu ^{\ast }(\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma )$ contains $\nu \rho \otimes L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _1)$ . Using Theorem 2.1 and [Reference Matić9, Theorem 5.3] again, we obtain $L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _1) \leq \langle [ \rho , \nu \rho ] \rangle \rtimes \sigma _1$ , contradicting Proposition 3.5.

If $2d+1 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ , using cuspidal support considerations, we obtain that $\tau $ is a discrete series representation, such that $\mathop {\text {Jord}}_{\rho }(\tau ) = \mathop {\text {Jord}}_{\rho }(\sigma ) \cup \{ 1, 2d+1 \}$ . Also, $\tau $ is a subrepresentation of an induced representation of the form $\delta ([\nu ^x \rho , \nu ^y \rho ]) \rtimes \sigma _1$ , where $\sigma _1$ is a strongly positive discrete series, $x \leq 0$ , $x + y> 0$ , and in $\mathop {\text {Jord}}_{\rho }(\tau )$ , we have $(2y+1)\_ = -2x+1$ .

Using the Frobenius reciprocity, together with the transitivity of the Jacquet modules, we deduce $\mu ^{\ast }(\langle [ \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma ) \geq \delta ([\nu ^x \rho , \nu ^y \rho ]) \otimes \sigma _1$ . The calculation of Jacquet modules gives $x \in \{ 0, -1 \}$ , $d = 1$ if $x = -1$ , and $\mu ^{\ast }(\sigma ) \geq \delta ([ \nu \rho , \nu ^y \rho ]) \otimes \sigma '$ . Consequently, $2y + 1 = \min (\mathop {\text {Jord}}_{\rho }(\sigma ))$ , and $\sigma '$ is a strongly positive discrete series such that $1 \in \mathop {\text {Jord}}_{\rho }(\sigma ')$ and $2y+1 \notin \mathop {\text {Jord}}_{\rho }(\sigma ')$ .

If $y < d$ , we have $x = 0$ , and the calculation of Jacquet modules leads to $\sigma _1 \leq \langle [ \nu \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma '$ . This contradicts [Reference Matić11, Proposition 3.3], the proof of which covers the integral case without any modification.

If $d < y$ , we have $x = -d$ , so $d = 1$ . From $\mu ^{\ast }(\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma ) \geq \delta ([\nu ^{-1} \rho , \nu ^y \rho ]) \otimes \sigma _1$ , we directly get $\sigma _1 \cong \sigma '$ .

Note that

$$ \begin{align*} \tau \hookrightarrow \delta([\nu^{-1} \rho, \nu^y \rho]) \rtimes \sigma_1 \hookrightarrow \delta([\nu \rho, \nu^y \rho]) \times \delta([\nu^{-1} \rho, \rho]) \rtimes \sigma_1, \end{align*} $$

so there is an irreducible subquotient $\pi $ of $\delta ([\nu ^{-1} \rho , \rho ]) \rtimes \sigma _1$ such that $\tau $ is a subrepresentation of $\delta ([\nu \rho , \nu ^y \rho ]) \rtimes \pi $ .

By [Reference Muić16, Theorem 4.1(ii)], in $R(G)$ , we have

$$ \begin{align*} \delta([\rho, \nu \rho]) \rtimes \sigma_1 = \tau" + L(\delta([\nu^{-1} \rho, \rho]); \sigma_1), \end{align*} $$

where $\tau "$ is an irreducible tempered subrepresentation of $\delta ([\rho , \nu \rho ]) \rtimes \sigma _1$ . If $\pi \cong \tau "$ , we have

$$ \begin{align*} \tau &\hookrightarrow \delta([\nu \rho, \nu^y \rho]) \times \delta([\rho, \nu \rho]) \rtimes \sigma_1 \\ &\hookrightarrow \delta([\nu \rho, \nu^y \rho]) \times \nu \rho \times \rho \rtimes \sigma_1 \\ &\cong \nu \rho \times \delta([\nu \rho, \nu^y \rho]) \times \rho \rtimes \sigma_1. \end{align*} $$

Since $y> 1$ and $2y + 1 = \min (\mathop {\text {Jord}}_{\rho }(\sigma ))$ , it follows that $\mu ^{\ast }(\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma )$ does not contain an irreducible constituent of the form $\nu \rho \otimes \pi '$ . Thus, $\pi \cong L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _1)$ so $\mu ^{\ast }(\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma )$ contains $\delta ([\nu \rho , \nu ^y \rho ]) \otimes L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _1)$ . The calculation of the Jacquet modules implies $\langle [ \rho , \nu \rho ] \rangle \rtimes \sigma _1 \geq L(\delta ([\nu ^{-1} \rho , \rho ]); \sigma _1)$ , which contradicts Proposition 3.5, since $\sigma _1 \cong \sigma '$ and $1 \in \mathop {\text {Jord}}_{\rho }(\sigma ')$ .

Let us now assume $1 \in \mathop {\text {Jord}}_{\rho }(\sigma )$ . Using cuspidal support considerations, we obtain that an irreducible tempered subquotient $\tau $ of $\langle [ \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma $ is a subrepresentation of an induced representation of the form $\rho \rtimes \sigma '$ , for a strongly positive discrete series $\sigma '$ . Thus, $\mu ^{\ast }(\langle [ \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma )$ contains $\rho \otimes \sigma '$ . Theorem 2.1 and [Reference Matić9, Theorem 5.3] imply $\sigma ' \leq \langle [ \nu \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma $ . It can now be seen, following the same lines as in the proof of [Reference Matić11, Proposition 3.3], that $x \in \mathop {\text {Jord}}_{\rho }(\sigma )$ for all $x \in \{ 1, 3, \ldots , 2d-1 \}$ and $2d+1 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ .

If $x \in \mathop {\text {Jord}}_{\rho }(\sigma )$ for all $x \in \{ 1, 3, \ldots , 2d-1 \}$ and $2d+1 \notin \mathop {\text {Jord}}_{\rho }(\sigma )$ , [Reference Matić10, Lemma 5.3] shows that the induced representation $\langle [ \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma $ contains an irreducible tempered subrepresentation. Also, by [Reference Matić11, Proposition 3.3], the induced representation $\langle [\nu \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma $ contains a unique discrete series subquotient, which we denote by $\sigma _{ds}$ . Using Theorem 2.1, together with [Reference Matić9, Theorem 5.3], we deduce that $\rho \otimes \sigma _{ds}$ is a unique irreducible constituent of $\mu ^{\ast }(\langle [ \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma )$ of the form $\rho \otimes \sigma '$ , for a discrete series $\sigma '$ , and appears with multiplicity one.

Consequently, if $\langle [ \rho , \nu ^{d} \rho ] \rangle \rtimes \sigma $ contains an irreducible tempered subquotient, then it contains a unique irreducible tempered subquotient, which is a subrepresentation.