Let $G$ be a linear Lie group, it is known that the exponential map on $G$ is $\exp(X)=\sum_{k=0}^\infty \frac{X^k}{k!}$.
If $G$ is not necessarily a linear Lie group, then the exponential map defined (in a modern way) by the one-parameter subgroup.
Why is the exponential map with the one-parameter subgroup coincide with $\exp(X)=\sum_{k=0}^\infty \frac{X^k}{k!}$ where $G$ is a linear Lie group? More specifically, what is the one-parameter subgroup (corresponding to $x\in T_eG$) where $G=Gl_n(\mathbb{R})$ or $G=Gl_n(\mathbb{C})$?
Best Answer
The one-parameter subgroup corresponding to $X\in\mathfrak{gl}_n(\Bbb R)(=T_eGL_n(\Bbb R))$ is the map$$\begin{array}{rccc}\varphi_X\colon&\Bbb R&\longrightarrow&GL_n(\Bbb R)\\&t&\mapsto&\exp(tX).\end{array}$$It's clearly a group homomorphism and $\varphi_X'(0)=X$. The same thing works with $\Bbb C$.