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One-parameter subgroups and exponentials

  1. Jan 14, 2009 #1

    Fredrik

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    I'm trying to understand this proof of the claim that if A is a one-parameter subgroup of [itex]M_n(\mathbb C)[/itex], there exists a unique matrix X such that [itex]A(t)=e^{tX}[/itex]. (Theorem 2.13, page 37. Proof on page 38).

    To prove uniqueness, just note that A'(0)=X.

    To prove existence, the author defines

    [tex]X=\frac{1}{t_0} \log A(t_0)[/tex]

    and then shows that [itex]A(t)=e^{tX}[/itex] for all t in a dense subset of the real numbers. Continuity then implies that this identity holds for all real t.

    My problem with this is that he doesn't clearly state what t0 is, so did he really define what X is? I mean, if this t0 does the job, it looks like we could have used t0/2 instead, or any other positive real number that's smaller than t0. I'm probably missing something obvious, and I'm hoping someone can tell me what that is.
     
    Last edited: Jan 14, 2009
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  3. Jan 14, 2009 #2

    tiny-tim

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    Hi Fredrik! :smile:

    Yes, any value for t0 will do …

    just choose whichever value of t is most convenient! :wink:
     
  4. Jan 14, 2009 #3

    Fredrik

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    That contradicts the uniqueness unless we can prove that [itex]\frac{1}{t_0}\log A(t_0)[/itex] is independent of t0. I don't see how to do that.
     
  5. Jan 15, 2009 #4

    Fredrik

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    I think I got it. The proof shows that regardless of what t0 is (assuming that it's positive and small enough), we have [tex]A(t)=e^{t(\frac{1}{t_0}\log A(t_0))}[/tex] for all t. That, together with the result A'(0)=X, which guarantees uniqueness, implies that [itex]\frac{1}{t_0}\log A(t_0)[/itex] is independent of t0. This seemed circular to me at first, but I don't think it is.
     
  6. Jan 15, 2009 #5

    tiny-tim

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    Hi Fredrik! :smile:

    Yes, it's ok …

    it's just one of those things everyone has to work out for themselves! :redface:
     
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