2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $t$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.