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I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering the case $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in this case?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp formsextending cusp forms)

I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering the case $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in this case?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp forms)

I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering the case $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in this case?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp forms)

deleted 1 characters in body
Source Link
Jude
  • 263
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  • 6

I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering the case $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in thesethis case?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp forms)

I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering the case $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in these case?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp forms)

I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering the case $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in this case?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp forms)

deleted 188 characters in body
Source Link
Jude
  • 263
  • 1
  • 6

I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering either the case $G=U(1)\times U(1) , \tilde{G}=U(2)$ or $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in these case?

That is, among these two irreducible cuspidal representation of G, is there anything which has extension to representation of $\tilde{G}$?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp forms)

I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering either the case $G=U(1)\times U(1) , \tilde{G}=U(2)$ or $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in these case?

That is, among these two irreducible cuspidal representation of G, is there anything which has extension to representation of $\tilde{G}$?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp forms)

I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.

Let me first invoke his original question.

"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."

The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.

Because, I am considering the case $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in these case?

Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.

(Harris's original post.

extending cusp forms)

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Jude
  • 263
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  • 6
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Source Link
Jude
  • 263
  • 1
  • 6
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