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Missing gap in proof

  1. Nov 3, 2005 #1
    I am trying to costruct a proof relating to the tangent space of SL(n). I'm almost there but all I need to show that if B is an arbitary constant traceless nxn matrix then there exists an A(t) (nxn) with the following properties:
    i) detA = 1
    ii) A'(0) = B.
    Just a tip in the right direction would be great thanks.
  2. jcsd
  3. Aug 21, 2009 #2


    User Avatar

    Actually, B is a generator for the Lie Algebra. But since this the Calculus topic, let's drop all this algebraic formalism.

    Consider the ODE [tex]A'(t)=A(t)B, \ t\in \mathbb{R}[/tex]. (This should be no big surprise. Lie himself devised those famous groups by studying similarities in the solutions of differential equations, and his work is pedantically reproduced whenever a math software package attempts to solve a differential equation).

    So, back to the ODE. By specifying the initial conditions [tex]A(0)=I,A'(0)=B[/tex]
    standard arguments of linear differential equations theory, guarantee the existence of a unique smooth solution [tex]A(t),t\in \mathbb{R}[/tex].

    Let us now show that this solution also satisfies [tex]{\rm det}A(t)=1[/tex].

    Since [tex]{\rm det}A(0)=1[/tex], continuity implies that A is invertible at least at a neighbourhood [tex](-\delta, \delta)[/tex] of zero.
    By differentiating for those t, [tex]({\rm det}A(t))'={\rm det}A(t){\rm tr}(A^{-1}(t)A'(t))[/tex], and by remembering the ODE, [tex]({\rm det}A(t))'={\rm det}A(t){\rm tr}(A^{-1}(t)A(t)B)={\rm det}A(t){\rm tr}(B)=0[/tex]. This means [tex]{\rm det}A(t)[/tex] is constant throughout [tex](-\delta, \delta)[/tex], and so linearity again grants us that [tex]{\rm det}A(t)[/tex] is constant for all [tex]t\in \mathbb{R}[/tex]. Finally, the initial condition [tex]A(0)=I[/tex] requires that [tex]{\rm det}A(t)=1[/tex], qed.

    Ps. The solution to the ODE is -not unexpectedly- [tex]A(t)={\rm e}^{Bt}[/tex]. This is a semigroup, so Algebra achieves a small moral victory in this finishing act.
    Last edited: Aug 21, 2009
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