When will a complex representation of a semi-simple lie group, have an invariant hermitian form? I know for compact groups, there is always one, and it can be made positive/negative definite. Does this remain true for non-compact groups, allowing for indefinite signature? I cannot think of an example where there isn't one, but I'm a physicist, not a mathematician.
 A: Right, for compact real Lie groups $K$, by averaging (with respect to Haar measure) a random hermitian inner product on a $K$-representation space $\mathbb C^n$, we get a $K$-invariant one. It is critical that the total measure of the group $K$ is finite. In fact, with a (continuous) infinite-dimensional Hilbert space representation, the same averaging trick (with some technicalities) succeeds.
For non-compact (even very nice) semi-simple real Lie groups, such as $G=SL_2(\mathbb R)$, the invariant measure is infinite, so an averaging trick cannot work so simply. Anyway, a group homomorphism of $G$ to the unitary group $U(n)$ of isometries of a hermitian inner product is a continuous homomorphism of $G$ to the compact group $U(n)$. This constrains $G$.
For that matter, for $SL_2(\mathbb R)$, the classification of finite-dimensional irreducibles shows that none (apart from the trivial, one-dimensional repn) preserves a hermitian form.
Nevertheless, there are infinite-dimensional Hilbert spaces with hermitian-inner-product-preserving representations of $SL_2(\mathbb R)$ and such groups on them. Certainly the non-finite-dimensional-ness creates complications... but these complications are interesting both in physics and in mathematics.
EDIT: if by "hermitian form" we allow indefinite signatures, but maybe not degenerate, then semi-simple real groups acting on themselves by Adjoint action preserve the Killing form $B(x,y)=\mathrm{tr}(\mathrm{ad}x\circ\mathrm{ad}y)$, which is non-degenerate for semi-simple algebras.
A: Given a representation $\rho:G\to GL(n,\mathbb{C})$ of a Lie group $G$, a $\rho$-invariant Hermitian product on $\mathbb{C}^n$ exists iff the image of $\rho$ is compact.
As an example of when this does not occur for a semisimple Lie group, you can complexify the defining representation of any noncompact semisimple matrix Lie group, such as $SL(2,\mathbb{R})$.
