# Missing gap in proof

1. Nov 3, 2005

### Diophantus

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. Aug 21, 2009

### Reb

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.

Last edited: Aug 21, 2009