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 $t$.

  • $\begingroup$ 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 $\endgroup$ – Takeshi Gouda Apr 2 '19 at 11:51
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    $\begingroup$ 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 $n := \dim V$. $\endgroup$ – Travis Willse Apr 3 '19 at 0:48

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