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Kevin Buzzard
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EDIT: I assumed the OP was asking about reductive groups over non-arch local fields. Emerton raises the possibility that the question is actually about groups over R or C, and he's probably right. So the answer below is probably irrelevant.

Do correct me if I'm wrong; I'm not an expert. But I thought that if $V$ was any supercuspidal representation of, say, $GL_2(\mathbf{Q}_p)$, then $V/uV=0$. So in fact you know very little about $V$ if you only know $V/uV$. Can't $V$ basically be recovered from $V/uV$ in the $GL_2$ case when it's principal series or Steinberg, and in the supercuspidal case you have nothing? Actually, even in the Steinberg case you might have trouble distinguishing $V$ from a 1-dimensional representation...

Do correct me if I'm wrong; I'm not an expert. But I thought that if $V$ was any supercuspidal representation of, say, $GL_2(\mathbf{Q}_p)$, then $V/uV=0$. So in fact you know very little about $V$ if you only know $V/uV$. Can't $V$ basically be recovered from $V/uV$ in the $GL_2$ case when it's principal series or Steinberg, and in the supercuspidal case you have nothing? Actually, even in the Steinberg case you might have trouble distinguishing $V$ from a 1-dimensional representation...

EDIT: I assumed the OP was asking about reductive groups over non-arch local fields. Emerton raises the possibility that the question is actually about groups over R or C, and he's probably right. So the answer below is probably irrelevant.

Do correct me if I'm wrong; I'm not an expert. But I thought that if $V$ was any supercuspidal representation of, say, $GL_2(\mathbf{Q}_p)$, then $V/uV=0$. So in fact you know very little about $V$ if you only know $V/uV$. Can't $V$ basically be recovered from $V/uV$ in the $GL_2$ case when it's principal series or Steinberg, and in the supercuspidal case you have nothing? Actually, even in the Steinberg case you might have trouble distinguishing $V$ from a 1-dimensional representation...

Source Link
Kevin Buzzard
  • 41.4k
  • 13
  • 166
  • 245

Do correct me if I'm wrong; I'm not an expert. But I thought that if $V$ was any supercuspidal representation of, say, $GL_2(\mathbf{Q}_p)$, then $V/uV=0$. So in fact you know very little about $V$ if you only know $V/uV$. Can't $V$ basically be recovered from $V/uV$ in the $GL_2$ case when it's principal series or Steinberg, and in the supercuspidal case you have nothing? Actually, even in the Steinberg case you might have trouble distinguishing $V$ from a 1-dimensional representation...