Decomposing a Lie Group representation into irreds using Lie Algebra representations.

Question

I know that a Lie Group representation $\rho: G \rightarrow GL(V)$ gives rise to a Lie Algebra representation: $\rho_*: \operatorname{Lie}(G) \rightarrow \mathfrak{gl}(V)$. I don‘t understand, however, how these $\rho_*$ are useful when it comes to decomposing representations into irreducible ones.

Let’s consider the following example $\rho: U(1) \rightarrow GL(\mathbb{C}^2)$, $e^{i\theta} \rightarrow \begin{pmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{pmatrix}$, we want to decompose this into a direct some of the irreds of $U(1)$, which are given by $\rho_j(e^{i\theta})=e^{ij\theta}$, one for each $\theta$.

We can consider the Lie Algebra representations for the $\rho_j$:

$$
\rho_{j*}(i \theta):=\frac{d}{dt} e^{tij \theta}|_{t=0}=ij\theta
$$

And for $\rho$:

$$
\rho_*(i\theta):= \frac{d}{dt} \begin{pmatrix} \cos (t\theta) & i \sin (t\theta) \\ i \sin (t\theta) & \cos (t\theta) \end{pmatrix}|_{t=0}= \begin{pmatrix}0 & i \theta \\ i \theta & 0 \end{pmatrix}
$$

Now the eigenvalues of $\rho_*$ are $\pm i \theta$ so $\rho_*=\rho_{1*} \oplus \rho_{-1*}$ but how does that help me with $\rho$? I assume it will turn out that $\rho=\rho_1 \oplus \rho_{-1}$, but why is that? I only know that if a represenentation is irreducible, then so is its Lie Algebra representation.

Answer 1

Here is the answer in the form of a two-part exercise.

Exercise 1a: Let $G$ be a connected Lie group, $\rho : G \to GL(V)$ be a finite-dimensional representation of $G$ over $\mathbb{R}$ or $\mathbb{C}$, and $d \rho : \mathfrak{g} \to \mathfrak{gl}(V)$ be its derivative. Let $W \subseteq V$ be a subspace of $V$. Then $W$ is invariant under $G$ iff it's invariant under $\mathfrak{g}$.

Exercise 1b: Deduce that if $V$ is completely reducible (e.g. if $G$ is compact) then the irreducible decomposition of $V$ as a representation of $G$ coincides with its irreducible decomposition as a representation of $\mathfrak{g}$.