Diophantus said:
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.
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 A'(t)=A(t)B, \ t\in \mathbb{R}. (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 A(0)=I,A'(0)=B
standard arguments of linear differential equations theory, guarantee the existence of a unique smooth solution A(t),t\in \mathbb{R}. Let us now show that this solution also satisfies {\rm det}A(t)=1.Since {\rm det}A(0)=1, continuity implies that A is invertible at least at a neighbourhood (-\delta, \delta) of zero.
By differentiating for those t, ({\rm det}A(t))'={\rm det}A(t){\rm tr}(A^{-1}(t)A'(t)), and by remembering the ODE, ({\rm det}A(t))'={\rm det}A(t){\rm tr}(A^{-1}(t)A(t)B)={\rm det}A(t){\rm tr}(B)=0. This means {\rm det}A(t) is constant throughout (-\delta, \delta), and so linearity again grants us that {\rm det}A(t) is constant for all t\in \mathbb{R}. Finally, the initial condition A(0)=I requires that {\rm det}A(t)=1, qed.
Ps. The solution to the ODE is -not unexpectedly- A(t)={\rm e}^{Bt}. This is a semigroup, so Algebra achieves a small moral victory in this finishing act.
