the bracket of a family of automorphisms associated with derivations

Question

I was reading through my class notes when I came across something I didn't know. Why is it true that if $D_1$, $D_2$ are derivations associated to the automorphisms $\varphi_t$, $\psi_t$ respectively, then $\varphi_t\psi_t\varphi_t^{-1}\psi_t^{-1}$ has zero first derivative and second derivative given by $[D_1,D_2]$?

Answer 1

There are more and less straightforward ways to see it but at the end, it boils down to a long calculation using the chain rule that is best done in the privacy of one's home.

Fix a smooth function $f$ and a point $p \in M$ and define $F \colon (-\varepsilon, \varepsilon)^4 \rightarrow \mathbb{R}$ by

$$ F(x_1,x_2,x_3,x_4) = (f \circ \varphi_{x_1} \circ \psi_{x_2} \circ \varphi_{-x_3} \circ \psi_{-x_4})(p) $$

and $G \colon (-\varepsilon, \varepsilon) \rightarrow \mathbb{R}$ by $G(t) := F(t,t,t,t)$. Then calculate all partial derivatives $\frac{\partial F}{\partial x_i}, \frac{\partial^2 F}{\partial x_i \partial x_j}$ at $(0,0,0,0)$ and finally calculate $G'(0)$ and $G''(0)$ using

$$ G'(0) = \sum_{i=1}^4 \frac{\partial F}{\partial x_i}|_{(0,0,0,0)}, \\ G''(0) = \sum_{i,j=1}^4 \frac{\partial^2 F}{\partial x_i \partial x_j}|_{(0,0,0,0)}. $$

The calculation can be made less painful by exploiting the symmetries of $F$ so that you don't really have to calculate 20 quantities (all first and second order partial derivatives) but only two. For example,

$$ (f \circ \varphi_t)(p) = F(t,0,0,0) = F(0,0,-t,0), \\ (f \circ \psi_t)(p) = F(0,t,0,0) = F(0,0,0,-t) $$

and so we must have

$$ \frac{\partial f}{\partial x_1}|_{(0,0,0,0)} = - \frac{\partial f}{\partial x_3}|_{(0,0,0,0)}, \\ \frac{\partial f}{\partial x_2}|_{(0,0,0,0)} = - \frac{\partial f}{\partial x_4}|_{(0,0,0,0)} $$

and we don't even have to calculate the derivatives to deduce that $G'(0) = 0$.

For a full proof, you can see page 160 of Spivak's first volume. The proof is very unilluminating and basically boils down to what I described above.