$S^1$ action on vector space implies $\mathbb{Z}$-grading?

Question

Suppose I have an action of $S^1$ on $V$, where $V$ is a complex vector space. This means I have a homomorphism $$ S^1\to\operatorname{GL}(V,\mathbb{C})\\ \lambda\mapsto D_\lambda, $$ where $$ D_{\lambda\mu}=D_\lambda D_\mu. $$ Does this entail that I get a $\mathbb{Z}$-grading on $V$? If so, how? I think that if we show that $D_\lambda$ is diagonalizable for each $\lambda$ then we are done since these operators commute.

Answer 1

The answer is yes if $V$ is equipped with an inner product making it a Hilbert space and the action of $S^1$ is continuous. Then we can average the inner product over the $S^1$-action to get an invariant inner product making this a unitary representation, which is then completely reducible by the Peter-Weyl theorem. $V$ is then graded by isotypic components: for $n \in \mathbb{Z}$ the $n^{th}$ graded component $V_n$ is the subspace of $V$ on which $z \in S^1$ acts by $z^n$.

The most famous special case of this is the regular representation on $L^2(S^1)$, where the decomposition above recovers the theory of Fourier series.

This takes care of the case that $V$ is finite-dimensional, since finite-dimensional inner product spaces are always Hilbert spaces. It's less clear to me what happens in general. If we really want to work with bare vector spaces without an inner product or anything like that, then we can replace actions of $S^1$ with actions of the multiplicative group scheme $\mathbb{G}_m$, which correspond to $\mathbb{Z}$-gradings for vector spaces over any field.