Let $G$ be a connected semisimple Lie group and $\mathfrak{g}$ it's Lie algebra. Then $\mathfrak{g}$ is semisimple. Let $V$ be a finite dimensional representation of $G$. Viewing $V$ as a representation of $\mathfrak{g}$, we may decompose $V$ into a direct sum of irreducible representations $V=\oplus V_i$. Does it follow that $\oplus V_i$ is the decomposition of $V$ into irreducible representations of $G$?

I would appreciate a reference or sketch of proof.


1 Answer 1


The answer is yes. This is because $G$-submodule of any $G$-module $V$ are the same as $\mathfrak{g}$-submodules of $V$.

Let me start with the direction I think you already know. Let $V$ be a representation of $G$ and let $\phi : G \rightarrow GL(V)$ be the structure map. If $W$ is a $G$-submodule of $V$, then by restricting the image of $\phi$ to $W$, we get a map $\phi_{W} : G \rightarrow GL(W)$. Differentiating this map at $e$ gives us a map from $\mathfrak{g} \rightarrow \mathfrak{gl}(W)$ compatible with the Lie algebra module structure map $d\phi_{e}$. Namely, $W$ is also a $\mathfrak{g}$-submodule of $V$.

Conversely, if $W$ is a $\mathfrak{g}$-submodule of $V$, then every element in $d\phi_{e}(\mathfrak{g})$ stabilizes $W$ as a set. Since $W$ is a closed subspace of $V$, this means that $\exp(d\phi_{e}(\mathfrak{g}))$ stabilizes $W$ as a set. But

$$\exp \circ d\phi_{e} = \phi \circ \exp$$

and hence this implies that $\exp(\mathfrak{g})$ stabilizes $W$. But $\exp(\mathfrak{g})$ contains an open neighborhood of the identity which generates $G$ as $G$ is connected. Hence, $G$ stabilizes $W$ i.e. $W$ is a $G$-submodule of $V$.

  • $\begingroup$ Thank you, that was exactly what I was looking for. $\endgroup$
    – Seth
    Dec 2, 2014 at 13:30

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