As best as I understand, the Frobenius Schur indicator tells us for a finite group $G$ and an irreducible representation $\rho: G \rightarrow \operatorname{Aut}(V)$ (where $V$ is a vector is a vector space over the complex numbers):
what $G$ invariant bilinear forms does $V$ possess?
I am extremely confused; Why doesn't the standard inner product:
$$ \langle\cdot,\cdot\rangle: V \otimes V \rightarrow \mathbb R; \langle v, w \rangle \equiv \sum_{g \in G} [\rho(g)(v), \rho(g)(w)] $$ for some inner product $[\cdot,\cdot]: V \otimes V \rightarrow \mathbb R$ qualify?
I also know that separately, the space of $G$ invariant bilinear forms must be 1-dimensional. This is because a $G$ invariant bilinear form carries the same data as a G-invariant mapping $f: V \mapsto V^\star$. However, since in the finite dimensional case, $V \simeq V^*$, we have by Schur's lemma that $f = \lambda I$ for some $I$.
Combining the two "facts", we have that
- $[\cdot, \cdot]$ is a $G$ invariant symmetric bilinear form on $V$
- The space of $G$-invariant symmetric bilinear forms is one dimensional
- Thus, all $G-$invariant bilinear forms are 1-dimensional and symmetric
This is clearly absurd, since the entire point of the Frobenius Schur indicator is to tell us if the space of bilinear forms is symmetric or antisymmetric.
What am I missing?
You said yourself the indicator applies to complex vector spaces, then proceed to talk about invariant inner products on real vector spaces. See what the issue is?
For complex inner product spaces, the inner product is sesquilinear, not symmetric bilinear.