# Lie group representation $\rho$ completely reducible iff $d\rho$ is completely reducible

What I want to show (with your help)

Show that $\rho$ is completely reducible iff $d\rho$ is completely reducible.

Level: a first course in Lie groups and algebras.

Relevant definitions

1. For a representation $\rho$ of a Lie group $G$ (Lie algebra $\mathfrak{g}$) on a vector space $V$, a vector subspace $W\subset V$ is said to be $\rho$ invariant if $\rho(g)W \subset W$ for all $g\in G$ (respectively $\rho(X)W \subset W$ for all $X\in \mathfrak{g}$).
2. A representation $\rho$ of a lie group or lie algebra on a vector space $V$ is completely reducible if every $\rho$ invariant subspace $W\subset V$ has a $\rho$ invariant complement $W'\subset V$, so that $V = W \oplus W'$.

Ramblings

So firstly I have to convince myself that $d\rho$ is indeed a representation on the corresponding Lie algebra - this follows from the chain rule.

Next, what does it really mean for either of these representations to be completely reducible... it seems like this condition is just requesting that the matrices $\rho(g)$ or $d\rho(X)$ are simultaneously block diagonalisable... or something like that...

I guess the 'hardest' bit is that I don't have much intuition about what $d\rho$ ought to look like as the differential of $\rho$...

• @TobiasKildetoft Well i guess one direction is easy... if $W$ is an invariant subspace of the group then we can view $\rho$ as a representation of $G$ on $W$, by restriction. Then $d\rho$ must also be a representation acting on $W$ for the same reason it was a representation acting on $V$, so $W$ must be $d\rho$ invariant. Would you agree with this argument? Is there something as simple for the converse? – abc Nov 11 '16 at 13:07
• ok how about this: we have that $\rho \circ \exp = \exp \circ d\rho = \sum (d\rho)^k/k!$. Applying this to $X\in \mathfrak{g}$ yields that anything of the form $\rho \circ \exp (X)$ invariates $W$, and since the image of exp generates $G$ we are done... sound reasonable? (Also, is it a fact that the image of of exp generates $G$... it certainly seems true for a connected Lie group... what if the Lie group is not connected?) – abc Nov 11 '16 at 13:48