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.


2 Answers 2


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 $\mathfrak{g}$.

  • $\begingroup$ I wrote an answer based on your suggestions, thank you:) $\endgroup$
    – Henry T.
    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.


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