1

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.

Craig
  • 974

3 Answers3

1

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.

paul garrett
  • 55,317
  • What about split-real forms, such as SU(1,1)? It is non-compact but preserves a hermitian form, no? Or is there a more apt name for the form it preserves? – Craig Mar 06 '22 at 23:00
  • Ah, $SU(1,1)$ preserves an indefinite hermitian form, indeed. I had supposed you meant a positive-definite (or negative-definite) one, so that the ambient space would be a Hilbert space, etc. Do you really intend to allow indefinite signatures? Please clarify... :) – paul garrett Mar 06 '22 at 23:04
  • Yes I do mean of a general signature! Apologies. Should have said "not necessarily positive definite" – Craig Mar 07 '22 at 02:20
0

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 $SL(2,\mathbb{R})$.

Kajelad
  • 15,356
0

The simplest answer to this question to this question, is that a representation of a group will have an invariant (Semi-)Hermitian form when There is a $G$-invariant anti-linear map from $V \rightarrow V^*$. This means you have some map $h$ for which all Lie algebra elements satisfy $h\overline{\lambda}h^{-1} = -\lambda^\intercal$

Craig
  • 974