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Sobolev class of loops to a compact lie Group

  1. Oct 31, 2012 #1
    I am currently reading a paper discussing the convexity of the image of moment maps for loop groups. In particular, if G is a compact Lie group and [itex] S^1 [/itex] is the circle, the paper defines the loops group to be the set of function [itex] f: S^1 \to G [/itex] of "Sobolev class [itex] H^1 [/itex]." Now in the traditional sense, being of Sobolev class [itex] H^1 [/itex] means that the function and its first derivative are both square integrable. The question is, what does this mean for functions with codomain a compact Lie group?

    So for real-valued functions whose domain is a compact Lie group, we could use the Haar integral to talk about square integrability. In this case, [itex] S^1 [/itex] conveniently just happens to be a compact, connected, abelian Lie group, though I'm sure we could even avoid the Haar measure by identifying [itex] S^1 [/itex] with the appropriate circles in R or C. But if the image of the function lies in a group, what do we do then? I suppose that we could use Whitney embedding to embed [itex] G \subseteq \mathbb R^n [/itex], or Peter-Weyl to embed [itex] G \subseteq U(n) [/itex], but then I would need to choose a norm on these spaces.

    Even if that is the correct procedure, the better question is "what functions then are not in this space?" Since G is compact, composition with any norm would ensure that continuous functions are bounded. But we also have that the derivative exists, implying that all such functions would be continuous by necessity (or are we imposing weak-derivative conditions on this space)? That is, would it not then be sufficient to restrict ourselves to continuous functions?

    Any insight would be appreciated.
  2. jcsd
  3. Nov 9, 2012 #2
    Yes, presumably you have an inner product on G. Of course f(t) is automatically bounded, but f'(t) need not be (its image is in the Lie algebra), so having an L2 derivative is a restriction on the function.

    Yes, the derivative is a weak derivative, but according to the Sobolev embedding theorem, an H1 function on R is Holder continuous with constant 1/2. So you still get something better than continuity. In that sense, you could not restrict yourself to continuous functions because they are a larger class of functions.
  4. Nov 9, 2012 #3
    Hey Vargo, thanks for the reply.

    I was actually able to deduce what was meant by Sobolev in the first part of my post. It simply means that in any coordinate chart [itex] \Phi: G \to \mathbb R^n [/itex] the function [itex] f : S^1 \to G [/itex] is Sobolev if [itex] \Phi \circ f [/itex] is Sobolev H1 under the Euclidean norm in [itex] \mathbb R^n [/itex].

    On a related note, I was thinking about trying to put a norm on G as follows: I can fix a G-invariant metric on G (G acting on itself) and hence get an induced complete metric on G by Hopf-Rinow. Since I believe one of the typical obstructions for that metric to be a norm is the choice of a suitable base point, we are in luck since G has an obvious natural basepoint, the identity. However, I never got around to actually testing this out since I discovered the "true" definition of Sobolev class on the way.

    I believe that f'(t) should also be automatically bounded, unless I'm just crazy. Perhaps I mis-stated the question? Anyway, here is my rationale. For our purposes, we are identifying [itex] S^1 [/itex] with either [itex] \mathbb R/2\pi\mathbb Z [/itex] or as the unit circle in [itex] \mathbb C[/itex]. In either case, the derivative of a function [itex] f: S^1 \to G [/itex] is the map [itex] f': S^1 \to TG [/itex] (we are thinking of this as a path, not a formal map of manifolds, so this is not the pushforward!). The fact the image lies in (what may be identified with) [itex] \mathfrak g [/itex] should not matter, as the domain is still compact. This would imply that derivative is also bounded, no?
  5. Nov 9, 2012 #4
    Right, the local coordinates determine the sobolev functions. I forgot that part.

    You are saying that f' is like a loop in R^n and therefore closed. But if f' is not necessarily continuous how would you deduce that? Aren't you using the fact that the image of a compact set by a continuous map is compact? f' need not be continuous.

    [itex]g(x)=\int_0^x (2/3)1/t^{1/3}\, dt = x^{2/3},\,\,\, 0\leq x\leq 1 [/itex].

    Extend g periodically.

    [itex] f(x)=e^{2\pi ig(x)} [/itex].

    f maps S1 into S1. f'(x) is an L2 function which is not continuous and not bounded for x near 0.
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