Mixing
Supercuspidal representation of $mathrm{GL}_{2}$ over finite field
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I'm reading Bump's automorphic form, chapter 4.1, and this note written by Garrett. The later note said that there are $q(q-1)/2$ different supercuspidal representations of $mathrm{GL}_{2}(mathbb{F}_{q})$, which comes from anisotropic torus $T_{a}(F) = E^{times}$ ($E$ is an extension field of $F$ with $[E:F]=2$.) In Bump, it said that for each character $chi:E^{times} to mathbb{C}^{times}$ that does not factor through the norm map $N:E^{times}to F^{times}$, we have a Weil representation $pi = (pi(chi), W(chi))$ of $mathrm{GL}_{2}(F)$. However, there are $(q-1)^{2} - (q-1) = (q-1)(q-2)$ such characters, so I think there are some characters $chi, chi'$ that gives same representation $pi(chi)simeq pi(chi')$.
However, I don't know how to decide whether the representations are isomorphic or not, and I think both articles don't explain about this case.
Do we have to compute character (trace) of the representation $W(chi)$ and compare it, or is there any other methods?
I found that there's one way to do it by using a Whittaker model: Theorem 4.1.2 of Bump claims that the induced representation $mathrm{Ind}_{N}^{mathrm{GL}_{2}}psi_{N}$ contains every irreducible representation of dimension $>1$ exactly once, and dimension counting gives an answer. But I'm sure that there's a more elementary way to do this.
number-theory finite-groups representation-theory automorphic-forms
asked Jan 25 at 5:56
Seewoo LeeSeewoo Lee
7,109927
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add a comment |
$begingroup$
I'm reading Bump's automorphic form, chapter 4.1, and this note written by Garrett. The later note said that there are $q(q-1)/2$ different supercuspidal representations of $mathrm{GL}_{2}(mathbb{F}_{q})$, which comes from anisotropic torus $T_{a}(F) = E^{times}$ ($E$ is an extension field of $F$ with $[E:F]=2$.) In Bump, it said that for each character $chi:E^{times} to mathbb{C}^{times}$ that does not factor through the norm map $N:E^{times}to F^{times}$, we have a Weil representation $pi = (pi(chi), W(chi))$ of $mathrm{GL}_{2}(F)$. However, there are $(q-1)^{2} - (q-1) = (q-1)(q-2)$ such characters, so I think there are some characters $chi, chi'$ that gives same representation $pi(chi)simeq pi(chi')$.
However, I don't know how to decide whether the representations are isomorphic or not, and I think both articles don't explain about this case.
Do we have to compute character (trace) of the representation $W(chi)$ and compare it, or is there any other methods?
I found that there's one way to do it by using a Whittaker model: Theorem 4.1.2 of Bump claims that the induced representation $mathrm{Ind}_{N}^{mathrm{GL}_{2}}psi_{N}$ contains every irreducible representation of dimension $>1$ exactly once, and dimension counting gives an answer. But I'm sure that there's a more elementary way to do this.
number-theory finite-groups representation-theory automorphic-forms
asked Jan 25 at 5:56
Seewoo LeeSeewoo Lee
7,109927
$endgroup$
$begingroup$
If $chi'=chi^q$, the conjugate of $chi$ under the Frobenius, then surely $pi(chi')$ must be isomorphic to $pi(chi)$?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 6:00
$begingroup$
@LordSharktheUnknown Could you elaborate more? Also, is it true that $pi(chi)simeq pi(chi')$ if and only if $chi' = chi^{p^{k}}$ for some $k$?
$endgroup$
– Seewoo Lee
Jan 25 at 6:06
add a comment |
$begingroup$
I'm reading Bump's automorphic form, chapter 4.1, and this note written by Garrett. The later note said that there are $q(q-1)/2$ different supercuspidal representations of $mathrm{GL}_{2}(mathbb{F}_{q})$, which comes from anisotropic torus $T_{a}(F) = E^{times}$ ($E$ is an extension field of $F$ with $[E:F]=2$.) In Bump, it said that for each character $chi:E^{times} to mathbb{C}^{times}$ that does not factor through the norm map $N:E^{times}to F^{times}$, we have a Weil representation $pi = (pi(chi), W(chi))$ of $mathrm{GL}_{2}(F)$. However, there are $(q-1)^{2} - (q-1) = (q-1)(q-2)$ such characters, so I think there are some characters $chi, chi'$ that gives same representation $pi(chi)simeq pi(chi')$.
However, I don't know how to decide whether the representations are isomorphic or not, and I think both articles don't explain about this case.
Do we have to compute character (trace) of the representation $W(chi)$ and compare it, or is there any other methods?
I found that there's one way to do it by using a Whittaker model: Theorem 4.1.2 of Bump claims that the induced representation $mathrm{Ind}_{N}^{mathrm{GL}_{2}}psi_{N}$ contains every irreducible representation of dimension $>1$ exactly once, and dimension counting gives an answer. But I'm sure that there's a more elementary way to do this.
number-theory finite-groups representation-theory automorphic-forms
asked Jan 25 at 5:56
Seewoo LeeSeewoo Lee
7,109927
$endgroup$
I'm reading Bump's automorphic form, chapter 4.1, and this note written by Garrett. The later note said that there are $q(q-1)/2$ different supercuspidal representations of $mathrm{GL}_{2}(mathbb{F}_{q})$, which comes from anisotropic torus $T_{a}(F) = E^{times}$ ($E$ is an extension field of $F$ with $[E:F]=2$.) In Bump, it said that for each character $chi:E^{times} to mathbb{C}^{times}$ that does not factor through the norm map $N:E^{times}to F^{times}$, we have a Weil representation $pi = (pi(chi), W(chi))$ of $mathrm{GL}_{2}(F)$. However, there are $(q-1)^{2} - (q-1) = (q-1)(q-2)$ such characters, so I think there are some characters $chi, chi'$ that gives same representation $pi(chi)simeq pi(chi')$.
However, I don't know how to decide whether the representations are isomorphic or not, and I think both articles don't explain about this case.
Do we have to compute character (trace) of the representation $W(chi)$ and compare it, or is there any other methods?
I found that there's one way to do it by using a Whittaker model: Theorem 4.1.2 of Bump claims that the induced representation $mathrm{Ind}_{N}^{mathrm{GL}_{2}}psi_{N}$ contains every irreducible representation of dimension $>1$ exactly once, and dimension counting gives an answer. But I'm sure that there's a more elementary way to do this.
number-theory finite-groups representation-theory automorphic-forms
number-theory finite-groups representation-theory automorphic-forms
asked Jan 25 at 5:56
Seewoo LeeSeewoo Lee
7,109927
asked Jan 25 at 5:56
Seewoo LeeSeewoo Lee
7,109927
edited Jan 25 at 6:08
Seewoo Lee
asked Jan 25 at 5:56
Seewoo LeeSeewoo Lee
7,109927
asked Jan 25 at 5:56
Seewoo LeeSeewoo Lee
7,109927
asked Jan 25 at 5:56
Seewoo LeeSeewoo Lee
7,109927
7,109927
$begingroup$
If $chi'=chi^q$, the conjugate of $chi$ under the Frobenius, then surely $pi(chi')$ must be isomorphic to $pi(chi)$?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 6:00
$begingroup$
@LordSharktheUnknown Could you elaborate more? Also, is it true that $pi(chi)simeq pi(chi')$ if and only if $chi' = chi^{p^{k}}$ for some $k$?
$endgroup$
– Seewoo Lee
Jan 25 at 6:06
add a comment |
$begingroup$
If $chi'=chi^q$, the conjugate of $chi$ under the Frobenius, then surely $pi(chi')$ must be isomorphic to $pi(chi)$?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 6:00
$begingroup$
@LordSharktheUnknown Could you elaborate more? Also, is it true that $pi(chi)simeq pi(chi')$ if and only if $chi' = chi^{p^{k}}$ for some $k$?
$endgroup$
– Seewoo Lee
Jan 25 at 6:06
$begingroup$
If $chi'=chi^q$, the conjugate of $chi$ under the Frobenius, then surely $pi(chi')$ must be isomorphic to $pi(chi)$?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 6:00
$begingroup$
If $chi'=chi^q$, the conjugate of $chi$ under the Frobenius, then surely $pi(chi')$ must be isomorphic to $pi(chi)$?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 6:00
$begingroup$
@LordSharktheUnknown Could you elaborate more? Also, is it true that $pi(chi)simeq pi(chi')$ if and only if $chi' = chi^{p^{k}}$ for some $k$?
$endgroup$
– Seewoo Lee
Jan 25 at 6:06
$begingroup$
@LordSharktheUnknown Could you elaborate more? Also, is it true that $pi(chi)simeq pi(chi')$ if and only if $chi' = chi^{p^{k}}$ for some $k$?
$endgroup$
– Seewoo Lee
Jan 25 at 6:06
add a comment |
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$begingroup$
If $chi'=chi^q$, the conjugate of $chi$ under the Frobenius, then surely $pi(chi')$ must be isomorphic to $pi(chi)$?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 6:00
$begingroup$
@LordSharktheUnknown Could you elaborate more? Also, is it true that $pi(chi)simeq pi(chi')$ if and only if $chi' = chi^{p^{k}}$ for some $k$?
$endgroup$
– Seewoo Lee
Jan 25 at 6:06