- #1

caffeinemachine

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We have an action of $\mathbb Z$ on $L^2(S^1, \mu)$ defined by $n\cdot \varphi = e^{2\pi i n}\varphi$.

The following is a standard theorem in functional analysis:

**Spectral Theorem.**Let $\mathbb Z$ act unitarily on a Hilbert space $H$. Let $f$ be any element in $H$ and let $H_f$ be the closure of the span of the orbit of $f$. Then there is a unique finite Borel measure $\mu_f$ on $S^1$ and a unitary isomorphism $L^2(S^1, \mu_f)\to H_f$ such that this isomorphism intertwines the action of $\mathbb Z$ on $L^2(S^1, \mu_f)$ with the action of $\mathbb Z$ on $H_f$.

So the above theorem allows us to view the action of $\mathbb Z$ restricted to the cyclic subspace generated by $f$ as a concrete action of $\mathbb Z$ on $L^2(S^1, \mu_f)$, where the sacrifice for this concreteness is that we now have an abstract Borel measure $\mu_f$ appearing.Consider the group $G$ formed by the semidirect product of $\mathbb Z$ and the multiplicative group $\{\pm 1\}$, where the semidirect product is with respect to the homomorphism $\{\pm 1\}\to \text{aut}(\mathbb Z)$ which sends $-1$ to the "reflection map" $\mathbb Z\to \mathbb Z: n \mapsto -n$.

In other words, $G$ is the group of reflection about the origin and translations.

**Question.**Suppose we are given a unitary action of $G$ on a Hilbert space $H$. Let $f\in H$ and $H_f$ be the closure of the span of the $G$-orbit of $f$. Can we identify concretely the action of $G$ restricted to $H_f$?

Any character $\chi:G\to S^1$ will factor through the abelianization of $G$.

But the commutator of $G$ is $\{(2n, 1):\ n\in \mathbb Z\}$, and thus the abelianization of G is finite.

So the characters of $G$ do not give us much information (Unlike the case of $\mathbb Z$, where the characters together form the group $S^1$ and this is why $S^1$ features in the theorem above).

However, $G$ has a finite index abelian group sitting in it, so I suspect one may be able to get a similar result even for $G$.