Return to Answer

EDIT: Having written up this answer, I realized Qiaochu was probably more interested in the case of $G$ not a Lie group. Oh well, I'll let the answer stand, even if it doesn't fully address the question.

A compact Lie group has finitely many simple representations if and only if its Lie algebra is semi-simple.

$(\Leftarrow)$: First check that it holds for a group iff it holds for the connected component of the identity (every irrep is in the induction of an irrep from that connected component). So now, assume the group is connected.

Then, note that if you have a quotient which is a torus (which is true iff the Lie algebra is not semi-simple), you're shot, because you can pull back all the irreps of the torus.

This shows you must have semi-simple Lie algebra.

$(\Rightarrow)$: Now, assume your group has semi-simple Lie algebra. You might as well pass to the universal cover, since this just makes more irreps. So really, you just have to prove the result for a semi-simple Lie algebra.

Now, use the Weyl dimension formula

$d_{\lambda}=\frac{\prod_{\alpha}(\lambda+\rho,\alpha)}{\prod_{\alpha}(\rho,\alpha)}$

to see that any rep with dimension below $n$ has its inner product with any simple root $\leq n$, and so is confined to a compact box.

EDIT: Having written up this answer, I realized Qiaochu was probably more interested in the case of $G$ not a Lie group. Oh well, I'll let the answer stand, even if it doesn't fully address the question.

A compact Lie group has finitely many simple representations of any given dimension if and only if its Lie algebra is semi-simple.

$(\Leftarrow)$: First check that it holds for a group iff it holds for the connected component of the identity (every irrep is in the induction of an irrep from that connected component). So now, assume the group is connected.

Then, note that if you have a quotient which is a torus (which is true iff the Lie algebra is not semi-simple), you're shot, because you can pull back all the irreps of the torus.

This shows you must have semi-simple Lie algebra.

$(\Rightarrow)$: Now, assume your group has semi-simple Lie algebra. You might as well pass to the universal cover, since this just makes more irreps. So really, you just have to prove the result for a semi-simple Lie algebra.

Now, use the Weyl dimension formula

$d_{\lambda}=\frac{\prod_{\alpha}(\lambda+\rho,\alpha)}{\prod_{\alpha}(\rho,\alpha)}$

to see that any rep with dimension below $n$ has its inner product with any simple root $\leq n$, and so is confined to a compact box.

EDIT: Having written up this answer, I realized Qiaochu was probably more interested in the case of $G$ not a Lie group. Oh well, I'll let the answer stand, even if it doesn't fully address the question.

A compact Lie group has finitely many simple representations if and only if its Lie algebra is semi-simple.

$(\Leftarrow)$: First check that it holds for a group iff it holds for the connected component of the identity (every irrep is in the induction of an irrep from that connected component). So now, assume the group is connected.

Then, note that if you have a quotient which is a torus (which is true iff the Lie algebra is not semi-simple), you're shot, because you can pull back all the irreps of the torus.

This shows you must have semi-simple Lie algebra.

$(\Rightarrow)$: Now, assume your group has semi-simple Lie algebra. You might as well pass to the universal cover, since this just makes more irreps. So really, you just have to prove the result for a semi-simple Lie algebra.

Now, use the Weyl dimension formula

$d_{\lambda}=\frac{\prod_{\alpha}(\lambda+\rho,\alpha)}{\prod_{\alpha}(\rho,\alpha)}$

to see that any rep with dimension below $n$ has its inner product with any simple root $\leq n$, and so is confined to a compact box.

EDIT: Having written up this answer, I realized Qiaochu was probably more interested in the case of $G$ not a Lie group. Oh well, I'll let the answer stand, even if it doesn't fully address the question.

A compact Lie group has finitely many simple representations if and only if its Lie algebra is semi-simple.

$(\Leftarrow)$: First check that it holds for a group iff it holds for the connected component of the identity (every irrep is in the induction of an irrep from that connected component). So now, assume the group is connected.

Then, note that if you have a quotient which is a torus (which is true iff the Lie algebra is not semi-simple), you're shot, because you can pull back all the irreps of the torus.

This shows you must have semi-simple Lie algebra.

$(\Rightarrow)$: Now, assume your group has semi-simple Lie algebra. You might as well pass to the universal cover, since this just makes more irreps. So really, you just have to prove the result for a semi-simple Lie algebra.

Now, use the Weyl dimension formula

$d_{\lambda}=\frac{\prod_{\alpha}(\lambda+\rho,\alpha)}{\prod_{\alpha}(\rho,\alpha)}$

to see that any rep with dimension below $n$ has its inner product with any simple root $\leq n$, and so is confined to a compact box.

A compact Lie group has finitely many simple representations of each dimension if and only if its Lie algebra is semi-simple.

$(\Leftarrow)$: First check that it holds for a group iff it holds for the connected component of the identity (every irrep is in the induction of an irrep from that connected component). So now, assume the group is connected.

Then, note that if you have a quotient which is a torus (which is true iff the Lie algebra is not semi-simple), you're shot, because you can pull back all the irreps of the torus.

This shows you must have semi-simple Lie algebra.

$(\Rightarrow)$: Now, assume your group has semi-simple Lie algebra. You might as well pass to the universal cover, since this just makes more irreps. So really, you just have to prove the result for a semi-simple Lie algebra.

Now, use the Weyl dimension formula

$d_{\lambda}=\frac{\prod_{\alpha}(\lambda+\rho,\alpha)}{\prod_{\alpha}(\rho,\alpha)}$

to see that any rep with dimension below $n$ has its inner product with any simple root $\leq n$, and so is confined to a compact box.

EDIT: Having written up this answer, I realized Qiaochu was probably more interested in the case of $G$ not a Lie group. Oh well, I'll let the answer stand, even if it doesn't fully address the question.

A compact Lie group has finitely many simple representations if and only if its Lie algebra is semi-simple.

$(\Leftarrow)$: First check that it holds for a group iff it holds for the connected component of the identity (every irrep is in the induction of an irrep from that connected component). So now, assume the group is connected.

Then, note that if you have a quotient which is a torus (which is true iff the Lie algebra is not semi-simple), you're shot, because you can pull back all the irreps of the torus.

This shows you must have semi-simple Lie algebra.

$(\Rightarrow)$: Now, assume your group has semi-simple Lie algebra. You might as well pass to the universal cover, since this just makes more irreps. So really, you just have to prove the result for a semi-simple Lie algebra.

Now, use the Weyl dimension formula

$d_{\lambda}=\frac{\prod_{\alpha}(\lambda+\rho,\alpha)}{\prod_{\alpha}(\rho,\alpha)}$

to see that any rep with dimension below $n$ has its inner product with any simple root $\leq n$, and so is confined to a compact box.