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4

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.

edited Mar 13 2010 at 0:54

Harald Hanche-Olsen
5,750●1●13●34

asked Mar 13 2010 at 0:41

unknown (google)
1,357●1●4●10

Isn't this Fourier analysis? – Steve Huntsman Mar 13 2010 at 0:51

And a nice application too (IMHO). Motivates group C*-algebras and all that jazz. – Yemon Choi Mar 13 2010 at 0:55

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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 2010 at 9:07

Harald Hanche-Olsen · Answer 1 · 2010-03-13 00:52:35Z UTC

9

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 $E$.

But you were asking for closed irreducible subspaces: There aren't any, because the Lebesgue measure has no atoms.

answered Mar 13 2010 at 0:52

Harald Hanche-Olsen
5,750●1●13●34

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 2010 at 0:55

@Yemon: I added the second paragraph while you were commenting. (I had forgotten about the irreducible part of the question.) To get irreducible subspaces, go for the circle group T acting on L²(T), since the dual group Z certainly has atoms. – Harald Hanche-Olsen Mar 13 2010 at 1:04

@Harald: What about $L^2(\mathbb{R},\mathcal{B},\nu)$ for $\nu$ different than Lebesgue measure? – Steve Huntsman Mar 13 2010 at 2:18

3

@Steve: but you still want R to act on L^2(R) in a norm-preserving way, right? and this limits the choices... – Yemon Choi Mar 13 2010 at 4:27

@Yemon--Good point. – Steve Huntsman Mar 13 2010 at 14:55

closed irreducible subspaces in L²(R)

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