Irreducible finite dimensional complex representation of $GL_2(\Bbb C)$

Question

I know the basic theory of representation theory of compact Lie groups and I want to understand a non-compact example:

How to find all irreducible finite dimensional complex representations of $GL_2(\Bbb C)$? Are its finite dimensional complex representations always completely irreducible?

I think there cannot exist an invariant inner product on it's representations, is there some way to find it's representations using something like it's lie algebra?

Answer 1

Indeed, the literal negative answer to the first question includes the standard counter-example as given by @JoseCarlosSantos. Still, modifying the question/example somewhat in various ways gives a more positive answer. First, if we insist on "algebraic" representations (in some sense...) this could be arranged to exclude the counter-example.

Or, for example, instead of $G=GL_2$, trying $G=SL_2$ avoids the counter-example.

In either style of avoiding that (and related) counter-examples, H. Weyl's "unitarian trick" tries to arrange that a finite-dimensional repn (under some hypotheses...) of a reductive (linear...) real Lie group can be complexified, and then restricted to a repn of the compact real form, thus reducing many questions to the repn theory of compact Lie groups.

Thus, under mild (but non-trivial) hypotheses such repns are completely reducible, and so on. But you are right, that they don't have invariant inner products for the (original) non-compact group, although they do have such for the compact real form, which enables Weyl's proofs (under various suitable hypotheses to make the argument possible).

And, yes, then a classification of ("algebraic"?) irreducibles by highest weights succeeds.

Answer 2

No. Consider the action $\rho$ of $GL_2(\mathbb{C})$ on $\mathbb{C}^2$ such that, for each $g\in GL_2(\mathbb{C})$, the matrix of $\rho(g)$ with respect to the standard basis is$$\begin{pmatrix}1&\log|\det g|\\0&1\end{pmatrix}.$$ It is not completely reducible, because $\{(z,0)\,|\,z\in\mathbb{C}\}$ is invariant, but it is the only one-dimensional invariant subspace.