Return to Answer

This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem.

General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is a nontrivial irreducible rep and $k$ has a trivial action. The argument for Weyl's theorem goes as follows: the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a splitting via an element in its kernel.

The reason we should expect this to be so is that the Casimir element is the Laplacian of the compact Lie group $G$. By the Peter-Weyl theorem, every irreducible representation of $G$ embeds into $L^2(G)$ with the translation action.

On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constants since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, the Casimir element acts by a nonzero scalar, hence the argument for Weyl's theorem should work.

This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem.

General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is a nontrivial irreducible rep and $k$ has a trivial action. The argument for Weyl's theorem goes as follows: the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a splitting via an element in its kernel.

The reason we should expect this to be so is that the Casimir element is the Laplacian of the compact Lie group $G$. Every irreducible representation of $G$ embeds into $L^2(G)$ with the translation action via matrix coefficients.

On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constants since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, the Casimir element acts by a nonzero scalar, hence the argument for Weyl's theorem should work.

This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem.

General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is an irreducible rep and $k$ has a trivial action. The argument for Weyl's theorem goes as follows: the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a splitting via an element in its kernel.

The reason we should expect this to be so is that the Casimir element is the Laplacian of the compact Lie group $G$. By the Peter-Weyl theorem, every irreducible representation of $G$ embeds into $L^2(G)$ with the translation action.

On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constants since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, the Casimir element acts by a nonzero scalar, hence the argument for Weyl's theorem should work.

This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem.

General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is a nontrivial irreducible rep and $k$ has a trivial action. The argument for Weyl's theorem goes as follows: the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a splitting via an element in its kernel.

The reason we should expect this to be so is that the Casimir element is the Laplacian of the compact Lie group $G$. By the Peter-Weyl theorem, every irreducible representation of $G$ embeds into $L^2(G)$ with the translation action.

On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constants since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, the Casimir element acts by a nonzero scalar, hence the argument for Weyl's theorem should work.

This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem.

General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is an irreducible rep and $k$ has a trivial action. The argument for Weyl's theorem goes as follows: the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a splitting via an element in its kernel.

The reason we should expect this to be so is that the Casimir element is the Laplacian of the compact Lie group $G$. By the Peter-Weyl theorem, every irreducible representation of $G$ embeds into $L^2(G)$ with the translation action.

On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constants since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, the Casamir element acts by a nonzero scalar, hence the argument for Weyl's theorem should work.

This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem.

General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is an irreducible rep and $k$ has a trivial action. The argument for Weyl's theorem goes as follows: the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a splitting via an element in its kernel.

The reason we should expect this to be so is that the Casimir element is the Laplacian of the compact Lie group $G$. By the Peter-Weyl theorem, every irreducible representation of $G$ embeds into $L^2(G)$ with the translation action.

On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constants since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, the Casimir element acts by a nonzero scalar, hence the argument for Weyl's theorem should work.