Is the number of irreducible unitary representations of a finitely-generated group of a given dimension finite?
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Let $Gamma$ be a finitely-generated group. Fix an integer $ngeq 1$. Is it possible that $Gamma$ has infinitely many pairwise non-isomorphic irreducible unitary representations of dimension $n$?
I tried to show that it is not possible:
We can fix a generating set $S$, and think of a unitary representation $rho$ as a function $Srightarrow U(n)$. Since $U(n)^S$ is compact, it would suffice to show that every element of $U(n)^S$ which corresponds to an irreducible representation has an open neighborhood $U$, such that all unitary representations that are mapped into $U$ are isomorphic to $rho$. I don't know if that's true.
representation-theory finitely-generated
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up vote
-3
down vote
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Let $Gamma$ be a finitely-generated group. Fix an integer $ngeq 1$. Is it possible that $Gamma$ has infinitely many pairwise non-isomorphic irreducible unitary representations of dimension $n$?
I tried to show that it is not possible:
We can fix a generating set $S$, and think of a unitary representation $rho$ as a function $Srightarrow U(n)$. Since $U(n)^S$ is compact, it would suffice to show that every element of $U(n)^S$ which corresponds to an irreducible representation has an open neighborhood $U$, such that all unitary representations that are mapped into $U$ are isomorphic to $rho$. I don't know if that's true.
representation-theory finitely-generated
2
Doesn't $Gamma=Bbb Z$, $n=1$ give a counterexample? I'm pretty sure the unit circel in the complex plane is infinite.
â David C. Ullrich
Jul 22 at 15:05
add a comment |Â
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
Let $Gamma$ be a finitely-generated group. Fix an integer $ngeq 1$. Is it possible that $Gamma$ has infinitely many pairwise non-isomorphic irreducible unitary representations of dimension $n$?
I tried to show that it is not possible:
We can fix a generating set $S$, and think of a unitary representation $rho$ as a function $Srightarrow U(n)$. Since $U(n)^S$ is compact, it would suffice to show that every element of $U(n)^S$ which corresponds to an irreducible representation has an open neighborhood $U$, such that all unitary representations that are mapped into $U$ are isomorphic to $rho$. I don't know if that's true.
representation-theory finitely-generated
Let $Gamma$ be a finitely-generated group. Fix an integer $ngeq 1$. Is it possible that $Gamma$ has infinitely many pairwise non-isomorphic irreducible unitary representations of dimension $n$?
I tried to show that it is not possible:
We can fix a generating set $S$, and think of a unitary representation $rho$ as a function $Srightarrow U(n)$. Since $U(n)^S$ is compact, it would suffice to show that every element of $U(n)^S$ which corresponds to an irreducible representation has an open neighborhood $U$, such that all unitary representations that are mapped into $U$ are isomorphic to $rho$. I don't know if that's true.
representation-theory finitely-generated
asked Jul 22 at 15:03
Scott Harris
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Doesn't $Gamma=Bbb Z$, $n=1$ give a counterexample? I'm pretty sure the unit circel in the complex plane is infinite.
â David C. Ullrich
Jul 22 at 15:05
add a comment |Â
2
Doesn't $Gamma=Bbb Z$, $n=1$ give a counterexample? I'm pretty sure the unit circel in the complex plane is infinite.
â David C. Ullrich
Jul 22 at 15:05
2
2
Doesn't $Gamma=Bbb Z$, $n=1$ give a counterexample? I'm pretty sure the unit circel in the complex plane is infinite.
â David C. Ullrich
Jul 22 at 15:05
Doesn't $Gamma=Bbb Z$, $n=1$ give a counterexample? I'm pretty sure the unit circel in the complex plane is infinite.
â David C. Ullrich
Jul 22 at 15:05
add a comment |Â
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2
Doesn't $Gamma=Bbb Z$, $n=1$ give a counterexample? I'm pretty sure the unit circel in the complex plane is infinite.
â David C. Ullrich
Jul 22 at 15:05