Z-stable complement

Discussion in 'Math Research' started by Anton Deitmar, Aug 1, 2006.

  1. Let Z denote the additive group of integers.
    Let H be the Hilbert space of all square summable complex valued functions on Z.
    Then Z acts unitarily on H by translations.
    The space C of all compactly supported functions is stable under this action.

    My question is: does C have a Z-stable complementary space W in H?
    In other words, is there a Z-stable linear subspace W of H such that H is the direct sum of C and W?

    Anton
     
    Anton Deitmar, Aug 1, 2006
    #1
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  2. I don't think it does. I'm no analyst so let's refomulate what I believe
    to be the question in terms of pure algebra. A K-vector space with
    an action of the group Z is just a K[Z]-module (the group ring). Let
    K denote the complexes, because you used C for something else, and let
    R denote the ring K[Z]; then R is isomorphic to the ring K[X,X^{-1}],
    with X an indeterminate.

    You have a big R-module H, a submodule C, and my understanding is
    that you're asking where
    there's a sub-R-module M of H such that H=M+C (direct sum) (in particular
    you are ignoring any topological questions). Say such
    an M exists. Then certainly M is isomorphic to H/C. A contradiction
    follows if we can prove that (a) H has no R-torsion (and hence M can have
    no R-torsion), but (b) H/C does have R-torsion.

    (a) is not so hard. R is a UFD and the primes are just (X-t) for t a
    non-zero complex number, so all we have to do is to check that if
    h is in H, and X.h=t*h, where X denotes the shift operator, then h=0.
    This is clear because there are problems either as you go to +infinity
    or -infinity, depending on whether |t|<=1 or |t|>=1.

    (b) is also not so hard if I've understood your definition of C
    correctly. Consider the function f on Z defined by f(n)=0 for n<0,
    and f(n)=2^{-n} for n>0. Then in H/C, X.f is either 2f or f/2
    depending on which way you're shifting, so X-2 or X-1/2
    has a non-trivial kernel on H/C. So done.

    You could now untranslate this back into a purely elementary proof
    which doesn't involve the ring R, if R is not to your taste.

    Kevin Buzzard
     
    Kevin Buzzard, Aug 1, 2006
    #2
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