I've seen numerous different proofs that the tangent space to a Lie group is closed under $[\cdot,\cdot]$, i.e. that the Lie Bracket of two derivations is a derivation -- e.g. considering and differentiating the curve $e^{\sqrt{t}X}e^{\sqrt{t}Y}e^{-\sqrt{t}X}e^{-\sqrt{t}Y}$, or just showing that $[D_1,D_2]$ follows the product rule.

But one derivation I don't get comes from Timothy Goldberg's set of lecture notes The Lie Bracket and the Commutator of Flows. Here's the process:

  1. Define the curve $c(t)=\Phi_X^t\Phi_Y^t\Phi_X^{-t}\Phi_Y^{-t}(e)$.
  2. Show that $[X,Y]=\frac12c''(0)$.
  3. Define an operation $D:f(t)\mapsto (f\circ c)''(0)$.
  4. Show that $D$ is a derivation.

It's Step 3 I don't get. How do we know this operator $D$ is what "upgrades" $[X,Y]$ into a vector field? How can we show that $[X,Y]$ is the direction in which $D$ differentiates $f$?


Ah, never mind, it's obvious -- I just got confused because it's not true for all curves. Given $(f\circ c)''(t)$, it's clearly equal to

$$c''(t)\cdot\nabla f(t)+c'(t)\cdot\frac{d}{dt}\nabla f(t)$$

And since $c'(0)=0$ for the given curve, this is just equal to the first term.


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