# closed irreducible subspaces in L²(R)

Let R be the set of real numbers, then R acts on L²(R) by translation. This is a unitary representation, which is far away from being irreducible. So what are the closed irreducible subspaces?

This might be a too naive question and not appropriate here.

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Isn't this Fourier analysis? – Steve Huntsman Mar 13 '10 at 0:51
And a nice application too (IMHO). Motivates group C*-algebras and all that jazz. – Yemon Choi Mar 13 '10 at 0:55
Not only that but it's the fundamental example that shows that if you want to decompose a representation of a Lie group on a Hilbert space then you need the notion of a direct integral as well as a direct sum. @unknown: if you're happy to accept that an irreducible representation of an abelian group is 1-dimensional, then you're asking for a character chi:R-->C^* and a non-zero L^2 function f such that f(x+z)=chi(z)f(x) for all x. But this latter property forces f to be non-L^2. This at least proves there are no 1-d subspaces. And then you go on to invent the Fourier transform to fix this! – Kevin Buzzard Mar 13 '10 at 9:07

Taking Fourier transforms, you get the action by multiplication. (Really, $x$ acts by multiplying with $e^{ix}$.) The closed invariant subspaces will thus be of the form $\{f\in L^2(\mathbb{R})\colon\hat f|_E=0\}$ for measurable sets $E$.
I was typing a more cumbersome version, but HHO beat me to it :) I'd also suggest considering the discrete version, with $Z$ acting on $l^2(Z)$. The resulting characterisation is exactly analogous, but the Fourier analysis you'd want to use is a bit easier since we only need to think about Fourier series rather than Fourier transforms, etc. – Yemon Choi Mar 13 '10 at 0:55
@Harald: What about $L^2(\mathbb{R},\mathcal{B},\nu)$ for $\nu$ different than Lebesgue measure? – Steve Huntsman Mar 13 '10 at 2:18