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O(V) and SO(V) algebras

  1. Feb 2, 2009 #1
    Hi,
    i have a consideration to post. Let V be a vector space over Reals with a symmetric inner product < , > : V x V -> R. We can introduce the concept of formal transposition of linear operator X on V, [tex]\forall x,y \in V[/tex] by the equation

    [tex]
    \left\langle x , X y \right\rangle = \left\langle X^t x , y \right\rangle
    [/tex]

    Now recalling that the orthogonal group and the special orthogonal group are given by

    [tex]
    O(V) = \left\{ X | X^t = X^{-1} \right\} = \left\{ X | det X = \pm 1 \right\}
    [/tex]

    [tex]
    SO(V) = \left\{ X | X^t = X^{-1}, det X = 1 \right\}
    [/tex]

    It can be proven that these groups are closed Lie Groups, in the convergence sense with the distance induced by usual operators norm, and further that the associated Lie Algebras, namely o(V) and so(V), are the same.

    How this can be possible? O(V) should contain SO(V) as a subgroup and the difference between them are "rotations" which do not preserve orientation (i.e. parity). And more, if this is true how can i build up an exponential map from an element of the algebra to recover a "rotation" belonging to O(V) but not to SO(V)?

    The technical aspects are clear to me, im looking for some clarification from the concrete point of view...thank you.
     
  2. jcsd
  3. Feb 2, 2009 #2

    Hurkyl

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    You can't. The other component of O(V) isn't path connected to the identity component.
     
  4. Feb 2, 2009 #3
    so when we look to O(V) as a Lie Group we are not able to describe reflections?
     
  5. Feb 2, 2009 #4

    Hurkyl

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    There are lots of ways to describe reflections. It's just that "product of exponentials of elements of o(V)" isn't one of them.
     
  6. Feb 2, 2009 #5
    so this an example of the fact that O(V) is "bigger" than exp(t X) with X in o(V). I guess this is true in general, for every Lie Group, since the assumption we make on a group to make it a Lie one are only local. A question arise: are there lie groups that coincides with the exponential map of the associated Lie algebra? And if this is the case, how to interpret such groups?

    I can try with an analogy. As manifolds locally has R^n geometry and the only ones for which this correspondance is global are flat ones. So we can think the Lie groups that coincides with the exponential of their Lie algebras in analogy to flat manifolds....can this be made?
     
  7. Feb 2, 2009 #6

    Hurkyl

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    Wikipedia claims that every element of a compact, connected Lie group is the exponential of a Lie algebra element. (I do not know if those conditions are necessary) I'm not really sure what you mean by "interpret"....


    Incidentally:

    Manifolds don't have geometry. But locally, they have the topology of R^n.

    Differentiable manifolds don't have geometry. But locally, they have the topology and differentiable structure of R^n.

    Riemann manifolds have geometry. Locally, they have the topology and differentiable structure of R^n, but usually do not have the same geometry.

    There are flat Riemann manifolds that are not globally diffeomorphic to R^n. (e.g. a circle is always flat. A Euclidean cylinder is flat. A torus can be given a flat geometry)

    There are lots of geometries one can put on the differentiable manifold R^n, very few are actually flat.
     
  8. Feb 3, 2009 #7
    I cannot find a definition of compactness for a group. Do you have one in mind?

    I mean that I usually try to establish a connection between a new mathematical concept or idea and the concepts I already have. The underlying motivation, in my opinion, is to find some path, some mechanics common to different concepts of mathematics in order to unifying them under the same logical contest.

    I try an example of this.

    Think to heat equation and schroedinger equation, they can be cast one in other sending t -> -it so they underlying the same mathematical structure. Then when I think to heat problems I can, carefully, use some idea or concepts from quantum mechanics by the fact that the two systems are governed by the same law. And this is quite useful.

    In some sense I'm trying to represent the Lie groups in my mind using an analogy with some other aspects of mathematics.
     
  9. Feb 3, 2009 #8

    Hurkyl

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    But you do have a definition of compactness for a topological space; a Lie group is both a group and a differentiable manifold....
     
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