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It is not hard to find the spectral theory of a single unitary operator $U$. This is the spectral theory for a $\mathbb{Z}$-action because we consider $U^n$ for $n\in\mathbb{Z}$. This is clear with eigenvalues because given an eigenvalue $\lambda$ with eigenvector $v$, we have $U^nv=\lambda^nv$. Given a Banach space $B$ and a group $G$ and a homomorphism $\pi:G\to L(B,B)$ where $\pi(g)$ unitary $\forall g \in G$. We can define an eigenvalue $\lambda$ of this $G$-action to be: $\exists v\ne0$ such that $\pi(g)v=\lambda(g)v$ where we see $\lambda$ is a group homomorphism from $G$ to $\mathbb{C}$. So we know eigenvalues, but what about generalizations of other parts of the spectrum? Is there a developed theory of this somewhere?

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Maybe not terrifically importnat, but did you really mean to say $\mathbb{Z}$-action, i.e., do you intend that $U$ be invertible? – Todd Trimble Jul 27 '11 at 14:52
Also, Is $U$ assumed unitary or something like that? If it's just a bounded linear operator (invertible or otherwise) I don't think there's much we can say about $U$ in the sense of spectral theory. The same applies to the case of a general group action. – Mark Jul 27 '11 at 16:33
What Mark just said, raised to a large power -- I find myself in what the British call "mild disagreement" with the first sentence. Also, there are well-behaved unitary group representations of $\mathbb Z$ where there are no eigenvalues -- just take the bilateral shift on $\ell^2({\mathbb Z})$ – Yemon Choi Jul 27 '11 at 16:56
I am trying to give the example of how $\mathbb{Z}$ pertains to the spectrum of a single operator. This is easiest seen though eigenvalues. I don't know the appropriate generalizations of other parts of spectrum, like the residual spectrum. Yes, I was not clear about the properties $U$ has. I have edited the above to make it a unitary group action. – Autoleech Jul 27 '11 at 22:09
For more general group actions, it is often the case that one is interested not only in "joint eigenvalues" (what you define as an eigenvalue of the action) but also in eigenvalues (and possibly other parts of the spectrum) of every operator $\pi (g)$ on its own. – Mark Jul 27 '11 at 23:17

As posed, the question turns into a question about repns about abelian groups $G$, since the $\lambda$ necessarily factors through the derived group $G/[G,G]$. This commutativity assures that eigenspaces (such as there are) are respected, for example. We might hope that the operators attached to $G$ are all normal, so that the usual spectral "calculus" applies, ... in which case there is a simultaneous "diagonalization" of all the operators coming from $G$. (Edit: Gelfand's theory of commutative $C^*$-algebras is a nice package for this.)

To lift the effectively-abelian constraint, one must more-or-less sacrifice the notion of "eigenspace/eigenvalue" and re-tool... to ask about irreducible sub-representations. There is some cost to this, but it's certainly worth it in the long term. Many natural situations require this, e.g., the action of the rotation group on $L^2$ functions on the 2-sphere (in Euclidean 3-space): there are no non-trivial group homs of the rotation group to $\mathbb C^\times$, but $L^2$ is the direct sum of the irreducible subspaces consisting of homogeneous harmonic polynomials (restricted to the sphere) of various degrees.

In the latter example, as in many others, but not universally, the irreducibles are also characterizable as eigenspaces for a single operator. In the case of spheres, this is the (non-Euclidean) Laplacian, which (by luck) distinguishes the irreducibles occuring in $L^2(S^2)$ by its eigenvalues.

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As a note to the first paragraph: bounded, SOT continuous representations of abelian groups on Hilbert space are similar/conjugate to unitary representations. This follows from results of Dixmier and Day (independently) in the 1950s but in this particular case was probably known earlier. – Yemon Choi Jul 28 '11 at 0:20
@Yemon: How come? Even an invertible linear operator on a finite-dimensional complex vector space isn't necessarily conjugate to a unitary (for instance, because conjugacy preserves eigenvalues) – Mark Jul 28 '11 at 12:37
Mark: if there is an eigenvalue of modulus $\neq 1$ then the representation isn't bounded (i.e. the original operator is not doubly power bounded). The same goes if the original operator is not diagonalizable (consider e.g. the Jordan normal form). – Yemon Choi Jul 30 '11 at 20:26

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