7
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.