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

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.

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 $$\mathfrak{g}$$.

• I wrote an answer based on your suggestions, thank you:) Aug 13, 2022 at 21:00

If someone is interested here an answer based on Qiaochu Yuan‘s suggestions and definitions:

„Exercise 1a“:

$$\forall g \in G \ \ \forall w \in W \ \ \forall X \in \mathfrak{g}$$

$$\rho(g)w \in W \Rightarrow \ \frac{d}{dt}\rho(\exp(tX))|_{t=0}w \in W$$, because $$W$$ is topologically closed.

$$\rho_*(g)w \in W \Rightarrow \rho(g)w=\rho(\exp(X_1))\dots\rho(\exp(X_k))w=\exp(\rho_*(X_1))\dots \exp(\rho_*(X_k))w \in W$$

Where I used that every element $$g \in G$$ can be written as $$\prod_i\exp(X_i)$$, because $$G$$ is connected. And $$\rho(\exp(X))=\exp(\rho_*(X))$$, which follows from $$\forall X \in \mathfrak{gl}(V) \ \exists Y \in \mathfrak{g}: \rho(\exp(tX))=\exp(tY) \Rightarrow \rho_*(X)=Y \Rightarrow \exp(\rho_*(X))=\rho(\exp(X)))$$.

„Exercise 1b“:

So if $$V = \bigoplus_i V_i$$ is a decomposition of $$V$$ into irreds of $$\rho$$, then it is also a decomposition into irreds of $$\rho_*$$, furthermore if $$\rho_i$$ acts irreducibly on $$V_i$$ so does $$\rho_{i*}$$ and vice versa.