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
- 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 $X\in \mathfrak{g}$).
- 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 $\rho$...
Help please :)