# Question about the $\operatorname{Exp}$ map on $\operatorname{End}(V)$ and the left invariant vector field

Let $$V$$ be a finite dimensional vector space over $$\mathbb{R}$$. We define for $$T \in \operatorname{End}(V)$$ with $$\| T \| < \infty$$ $$\operatorname{Exp}(T) = \sum_{k=0}^{\infty} \frac{T^k}{k!}.$$ Then it can be shown that $$\frac{d}{dt} \operatorname{Exp}(tT)= \operatorname{Exp}(tT)\cdot T.$$

Then the notes I am reading states : "Therefore, $$t \rightarrow \operatorname{Exp}(tT)$$ is tangential to the left-invariant vector field evaluated at $$e$$ is $$T$$."

I am struggling to understand this sentence here. I would appreciate if someone could explain me in more detail what this sentence means... In particular, which left-invariant vector field are they talking about? Thank you.

Recall that the commutator map $$(S, T) \mapsto [S, T] := S \circ T - T \circ S$$ defines a Lie algebra structure on the vector space $$\operatorname{End}(V)$$ of linear maps $$V \to V$$. We can identify this Lie algebra with the Lie algebra $$\mathfrak{gl}(V)$$ of the Lie group $$GL(V)$$ of invertible linear transformations of $$V$$. (If we choose a basis of $$V$$, we can identify both of these with the space $$M(n, \Bbb R)$$ of $$n \times n$$ real matrices, where $$n := \dim V$$, endowed with the commutator $$(A, B) \mapsto A B - B A$$ of matrices.)
In particular, we can regard any linear transformation $$T \in \operatorname{End}(V)$$ as an element of $$\mathfrak{gl}(V)$$, and so it determines a left-invariant vector field $$\tilde T$$ on $$GL(V)$$ characterized by $$\tilde T_1 = T .$$
Now, the map $$\gamma : \Bbb R \to GL(V)$$ defined by $$\gamma : t \mapsto \operatorname{Exp}(t T)$$ is a smooth curve in $$GL(V)$$, and unwinding definitions shows that it is an integral curve of $$\tilde T$$, that is, that $$\gamma'(t) = \tilde T_{\gamma(t)}$$ for all $$t$$.
• Thank you for your answer, I think I'm almost getting it... Could you possibly elaborate on how to see that $End(V)$ is identified with $gl(V)$? I know that $gl(V)$ is isomorphic to the tangent space to $GL(V)$ at identity, but how do you identify this with $End(V)$? Thank you – Takeshi Gouda Apr 2 '19 at 11:51
• There are a few ways to see this. One is that $GL(V)$ is an open subset of the vector space $\operatorname{End}(V)$, so we can identify $\mathfrak{gl}(V) \cong T_1 GL(V)$ with $T_1 \operatorname{End}(V)$, but for any vector space $W$ (in particular for $W = \operatorname{End}(V)$) and any $w \in W$ there is a canonical isomorphism $T_w W \cong W$. We can make this more concrete by fixing a basis of $V$, which determines isomorphisms $\operatorname{End}(V) \cong M(n, \Bbb R)$ and $GL(V) \cong GL(n, \Bbb R)$, where $n := \dim V$. – Travis Willse Apr 3 '19 at 0:48