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I am doing some selfstudy on Linear Lie groups and trying do an exercise.

Let G be a linear Lie group and $\phi$ a differentiable morphism from GL(n,$\mathbb{R}$) into G. Define $\Phi = (D_\phi)_I$.

a) show that for every $X \in M(n,\mathbb{R})$ that $\phi(expX) = exp(\Phi X)$
b) Deduce $det(expX) = e^{trX}$

So in a) i am a stuck, because if i insert the defintion on the right hand side i get $$ exp(\Phi X) = exp((D_\phi)_I X) $$ How would i move forward from here? Thanks.

user622405
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1 Answers1

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First: for $g\in GL(n,\mathbb R)$ and $X\in Mat(n,\mathbb R)$ we have $$ (D_\phi)_g(X) = \lim_{\epsilon\to 0}\frac{\phi(g+\epsilon X)}{\epsilon} =\lim_{\epsilon\to 0}\phi(g)\frac{\phi(I+\epsilon g^{-1}X)}{\epsilon} =\phi(g)(D_\phi)_I(g^{-1}X)=\phi(g)\Phi(g^{-1}X). $$

Second: set $\gamma(t):=\phi(\exp (tX))$. Differentiating in $t$ by the chain rule and the first formula $$ \frac d{dt}\gamma(t) = \frac d{dt}\phi(\exp(t X)) = (D_\phi)_{\exp(t X)}(X\exp(tX))=\phi(\exp (tX))\Phi(X)=\gamma(t)\Phi(X) $$ hence $\gamma(t)$ is the solution to the Cauchy problem $$ \frac d{dt}\gamma(t)=\gamma(t)\Phi(X),\qquad \gamma(0)=I. $$ But $\exp(t\Phi (X))$ solves the same Cauchy problem, hence $\gamma(t) = \exp(t\Phi (X))$, and taking $t=1$ gives the statement.

Giulio R
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  • Thanks! I am just wondering why its not done after the formula?? Because if you then calculated $(D_\phi)_I(X) = \phi(I) \Phi(I^{-1}X) = \Phi(X)$ right?? – user622405 Sep 29 '21 at 14:26
  • Isn't $(D_\phi)_I = \Phi$ the definition of $\Phi$? I think that to conclude something about the exponentials one has to look at the flows of the corresponding vector fields... – Giulio R Sep 29 '21 at 14:34
  • Ah yea of course, that makes sense! – user622405 Sep 29 '21 at 14:36