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Still wondering exp(so(N)) and SO(N)

  1. Oct 28, 2014 #1
    I still don't know whether

    [tex]
    e^{\mathfrak{so}(N)}=\textrm{SO}(N)
    [/tex]

    is true or not with certainty. Somebody, please, prove this.

    Obviously I know

    [tex]
    e^{\mathfrak{so}(N)}\subset\textrm{SO}(N)
    [/tex]

    so that's not the problem. I have managed to prove the equality in cases N=2,3, but the cases N=4,5,6,... have remained a mystery to me.

    In my attempts to prove the equality, I think I have managed to prove a relation

    [tex]
    \textrm{SO}(N)\subset e^{\mathfrak{su}(N)}
    [/tex]

    through diagonalization of the rotation matrix, and this looks interesting, but hasn't resolved the original problem.
     
  2. jcsd
  3. Oct 29, 2014 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    It should be true, because ##SO(n)## is connected, and ##e^{\mathfrak{so}(n)}## should give the component connected to the identity.

    There must be some easy simplification that will give you the answer you want. However, it is difficult to figure out what you mean by "prove". In particular, from what assumptions, definitions, etc., are you starting?
     
  4. Oct 29, 2014 #3
    I have investigated this direction. A Corollary 0.20 in Anthony W. Knapp's book Lie Groups Beyond an Introduction states that If [itex]G[/itex] is a closed linear group and [itex]\mathfrak{g}[/itex] is its Lie algebra, then [itex]\exp\mathfrak{g}[/itex] generates the identity component [itex]G_0[/itex]

    Therefore, if we know that [itex]\textrm{SO}(N)[/itex] is connected, we also know that [itex]e^{\mathfrak{so}(N)}[/itex] is a set that generates the [itex]\textrm{SO}(N)[/itex]. This implies that [itex]e^{\mathfrak{so}(N)}[/itex] will be equal to [itex]\textrm{SO}(N)[/itex] precisely if it is a group. So how do you prove that [itex]e^{\mathfrak{so}(N)}[/itex] is a group? You must prove that for all [itex]X,Y\in\mathfrak{so}(N)[/itex] there exists a [itex]Z\in\mathfrak{so}(N)[/itex] such that [itex]e^Xe^Y=e^Z[/itex]. This looks like precisely as difficult as [itex]\textrm{SO}(N)\subset e^{\mathfrak{so}(N)}[/itex], so this result about generating the identity component isn't resolving the issue.

    I mean proving like proving in mathematics.

    The definitions are

    [tex]
    \textrm{SO}(N) = \big\{A\in\mathbb{R}^{N\times N}\;\big|\; \det(A)=1,\;\;A^{-1}=A^T\big\}
    [/tex]

    [tex]
    \mathfrak{so}(N) = \big\{X\in\mathbb{R}^{N\times N}\;\big|\; X+X^T=0\big\}
    [/tex]

    [tex]
    e^X = \sum_{n=0}^{\infty}\frac{1}{n!}X^n
    [/tex]
     
  5. Oct 29, 2014 #4
    It is certainly true that the exponential is surjective here but the previous post is missing one of the reasons why. The fact that [itex] SO(n) [/itex] is connected only implies that [itex] exp(\mathfrak{so(n)}) [/itex] is contained in the identity component of [itex] SO(n) [/itex]. To further conclude that it is all of the identity component requires the additional fact that [itex] SO(n) [/itex] is also compact.

    If you want an elementary proof of surjectivity using just the matrix exponential and some standard forms of matrices, try theorem 14.2.2 in the following notes:
    http://www.cis.upenn.edu/~cis610/geombchap14.pdf
    .
     
  6. Oct 29, 2014 #5
    Do you know what book that extract is from?
     
  7. Oct 29, 2014 #6
    I believe it is "Geometric Methods and Applications: For Computer Science and Engineering" by Jean Gallier. If you want to backtrack to the proofs of the various normal forms he uses to get surjectivity, they are also posted on his website:
    http://www.cis.upenn.edu/~cis610/geombchap11.pdf
    .

    Also the abstract proof that exp is surjective for any any connected compact Lie group, at least the way I've seen it done, is by appealing to some theorems from Riemannian geometry so I kind of doubt there will be any simple way to show that [itex] exp (\mathfrak{g}) [/itex] is a group as you were suggesting in a previous post.

    A brief sketch of the usual proof starts out by observing that every compact Lie group can be given a bi-invariant Riemannian metric (just take left/right translates of a nondegenerate, positive definite, Ad-invariant bilinear form on the Lie algebra .) By the Hopf-Rinow theorem [itex] G [/itex] is geodesically complete so you can connect any [itex] g\in G [/itex] to the identity using a geodesic. Then show that the geodesics of [itex] G [/itex] are precisely left/right translates of one parameter subgoups of [itex] G [/itex]. Hence the geodesic connecting the identity to [itex] g\in G [/itex] is of the form [itex] \mathrm{exp(tX)} [/itex] for some [itex] X\in \mathfrak{g} [/itex] and therefore [itex] g =\mathrm{exp}(X) [/itex].
     
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