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Manifolds with symmetries

  1. Apr 1, 2014 #1

    Matterwave

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    Hello,

    I'm reading the book Geometrical methods of mathematial physics by Brian Schutz. In chapter 3, on Lie groups, he states and proves that the vector fields on a manifold over which a particular tensor is invariant (i.e. has 0 Lie derivative over) form a Lie algebra. And associated with every Lie algebra is a Lie group.

    Does this have any implications on the manifold somehow "containing" a Lie group in it (as some sort of submanifold maybe?), since the exponentiation of the Lie algebra gives you the identity component of the Lie group it's associated with? This seems not right since, for example, the Lie group of symmetries of a manifold may be of higher dimension than the manifold itself (e.g. Minkowski spacetime has dimension 4, while its group of symmetries, the Poincare group is of dimension 10).

    Nevertheless, it seems that there has to be some sort of association going on? Or am I just taking crazy pills?
     
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  3. Apr 5, 2014 #2
    There's no implication that the manifold contains a Lie group in it. The manifold isn't necessary a group or anything, so it doesn't even make sense to talk about that, unless you mean a diffeomorphic copy of the group that has "forgotten" its group structure. But I think there should be a Lie group homomorphism from the Lie group associated to the Lie algebra of vector fields into the diffeomorphism group of the manifold, corresponding to flowing along the vector fields.

    When you say symmetry group of the manifold, you have to specify what kind of symmetry. By default (since it's only assumed to be a manifold), that will be the entire big fat diffeomorphism group, rather than something like the Poincare group, which is something that preserves the metric. Something like the diffeomorphism group will be very infinite-dimensional. Just think of the diffeomorphism group of R (there are different topologies you can put on this baby--I'm not sure exactly how that affects things here). That's all smooth functions with smooth inverses from R to itself. It should be easy to cook up infinitely many linearly independent functions from R to itself that are like that (maybe f(x) = x + localized smooth perturbations at different, non-overlapping spots).
     
  4. Apr 5, 2014 #3
    Oh, and generally speaking, you only get a local diffeomorphism when you flow along a vector field. But maybe in this case, with the invariant tensors, you could get a globally defined one. I'm not sure.
     
  5. Apr 7, 2014 #4

    George Jones

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    As homeomorphic has written, there is a global/local issue here. I am going to assume global.

    No, it has implications for a symmetry group (also a manifold) that has action on the given manifold.

    I am pressed for time today, but I hope to get back to the details of this, as I have found some interesting Killing vectors for a particular manifold, so I want to work through this.
     
  6. Apr 7, 2014 #5

    Matterwave

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    Hmmm, I think you guys must be right, since by dimensional analysis what I have stated must be false (as I suspected). But let me just rephrase my question because I don't think I phrased it very well. Say I have a manifold of dimension n on which is defined a tensor field [itex]\bf{T}[/itex]. This tensor is invariant over some sets of Lie dragging defined by a set of vector fields [itex]\left\{\vec{v}_a\right\}[/itex] (the a's just label which vector field). We know that the vector fields define a Lie algebra with the regular Lie bracket on our manifold defined as usual.

    Now look at the expression [itex]e^{\epsilon\vec{v}_a}[/itex] for [itex]\epsilon[/itex] a small number. Doesn't this expression both generate a neighborhood of the identity of the Lie group associated with the Lie algebra defined earlier, as well as some neighborhood of our manifold? The answer must be no as we figured from dimensional analysis alone, but I just don't know why the answer is no.

    As far as the group structure goes, obviously we cannot generate the global properties of the group from the Lie algebra alone, but we can generate the connected component of the identity, and the group structure would be imposed by the inducement from the Lie bracket operation. In other words, we could then use the Lie algebra to give our manifold a group structure, can we not?
     
    Last edited: Apr 7, 2014
  7. Apr 8, 2014 #6
    I haven't seen that notation before, so I'm not entirely sure what you're trying to say, but it seems like the idea of generating a "neighborhood of the manifold" is questionable. I think you want to say neighborhood of a point of the manifold. But those are vector fields, so they are globally defined things. So, I don't see how they are going to have anything to do with neighborhoods of a point.

    You must mean the exponential map for a Riemannian manifold, but that's a bit different from the exponential map for a Lie Group. I believe if you put an appropriate metric on the Lie group, the exponential map there is a special case of the Riemannian manifold exponential map at the tangent space to the identity of the Lie group. But it's a more general thing, and you shouldn't expect that it will work the same way or have the same sort of properties.
     
  8. Apr 8, 2014 #7

    Matterwave

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    Hmmm, I think I'm just grasping at straws here. I didn't really have much to begin with, and it seems I did not end up going anywhere haha. Thanks for the help anyways.
     
  9. Apr 9, 2014 #8

    George Jones

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    Instead of "Reimannian manifold", I think you mean differentiable manifold with connection, i.e., a metric tensor is not needed.

    The two notions of exp in play here are:

    1) ##\mathrm{exp}: \mathcal{G} \rightarrow G##, where ##\mathcal G## is Lie algebra and ##G## is Lie group, and where the mapping uses left translation;

    2) ##\mathrm{exp}: T_p M \rightarrow M##, where ##M## is a differentiable manifold with connection, and where the mapping uses geodesics.

    It turns out that left-transaltion for Lie groups can be used to define a notion of a parallel transport (equivalent to a connection that has zero curvature and non-zero torsion). When this is done, 1) and 2) are the same for Lie groups.
     
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