How to understand $\frac{d}{dt}\{(\exp(tX))_*(Y)\}|_{t=0}=[X,Y]$?

Question

Let $G$ be a Lie group on which $X$ and $Y$ are two vector fields. Let $G\xrightarrow{\exp(tX)} G$ be the (Lie theory) exponential map corresponding to $X$. Then of fundamental importance is \begin{equation} \frac{d}{dt}\{(\exp(tX))_*(Y)\}|_{t=0}=[X,Y] \end{equation} where $[\cdot,\cdot]$ is the Lie bracket, and $\exp(tX)_*$ is the differential map induced from $\exp(tX)$.

Staring at this equation for a while, the only way I can prove it is by relating this exponential map to the (Remannian) exponential map. This proof involves a lot differential geometry, which seems rather unnecessary, and does not seem very instructive.

I am wondering whether someone can share a more elegant proof (more direct, less differential geometry) or just some intuitive idea about how one should understand this equation.

Thanks very much!

Answer 1

This is a very general fact, and does not need fancy extra structure like that of a Lie group (or even a Riemannian structure):

If you have two smooth vector fields $X,Y$ on a smooth manifold $M$ (for simplicity, assume complete vector fields), and if $\phi_t: M \to M$ denotes the flow of $X$, then $$\frac{d}{dt}\Big|_{t=0} (\phi_t)_* Y = [X,Y].$$ (Important: Note that the $(\phi_t)_*$ on the left hand side denotes the push-forward of a diffeomorphism acting on vector fields, and not mereley the fiber-wise map!)

Add.: There are two maps, which are sometimes denoted with a $*$: Suppose you have a smooth map $\psi: M \to N$ between manifolds (we could have $M=N$, but the general version might be clearer).

1. Given a tangent vector $v\in T_pM$ at a point, there is the fiber-wise map $\psi_*:T_pM \to T_{\psi(p)}N$ that maps $ v\mapsto \psi_* v$. It takes a vector at $p$ and maps it to a vector at the image point $\psi(p)$. This works for any smooth map $\psi:M \to N$.
2. If $\psi$ is a diffeomorphism, then we can also do the following. Given a vector field $X\in \Gamma(TM)$, we can map it via $\psi$ to vector field $\psi_*X\in \Gamma(TN)$. This is the push-forward map. We obtain $\psi_*X$ as follows: Given $q\in N$, let $p = \psi^{-1}(q)\in M$ be its preimage. Note that $X_p\in T_pM$ is a tangent vector. We can map $X_p$ to a tangent vector $\psi_* (X_p) \in T_{\psi(p)}N$ with the map from (1). The value of $\psi_*X$ at $q = \psi(p)$ is obtained by setting $(\psi_*X)_q := \psi_* (X_p)$. Doing this for every $q\in N$ gives us a vector field $[q\mapsto (\psi_*X)_q]\in \Gamma(TN)$. By this process, we obtain a map $\psi_*: \Gamma(TM)\to \Gamma(TN)$ which is explicitly given by $$\psi_*: \Gamma(TM)\to \Gamma(TN), \quad [p\mapsto X_p] \mapsto [q\mapsto \psi_*(X_{\psi^{-1}(q)})].$$ The push-forward acting on vector fields. It only makes sense for diffeomorphisms in general and is a way of transporting vector fields from one place to another.

Now let $f\in C^\infty(M)$ be an arbitrary smooth function. To prove the above identity, we need to show that $\left(\frac{d}{dt}|_{t=0} (\phi_t)_\ast Y \right) f = [X,Y]f$. Remember that the push-forward $(\phi_t)_\ast X$ is defined by $$((\phi_t)_* Y )_pf = (\phi_{t})_\ast (Y_{\phi_{-t}(p)})f = Y_{\phi_{-t}(p)}(f\circ \phi_{t})$$ for all $p\in M$. Thus \begin{align} \left(\frac{d}{dt}|_{t=0} (\phi_t)_\ast Y \right) {}_p f &= \lim_{t\to 0} \frac{((\phi_t)_* Y )_pf - Y_pf}{t} \\ &= \lim_{t\to 0} \frac{Y_{\phi_{-t}(p)}(f\circ \phi_{t}) - Y_pf}{t} \\ &= \lim_{t\to 0} \frac{Y_{\phi_{-t}(p)}(f\circ \phi_{t}) - Y_{\phi_{-t}}f}t +\lim_{t\to 0} \frac{Y_{\phi_{-t}(p)}f -Y_pf}{t} \\ &= \lim_{t\to 0} \frac{Y_{\phi_{-t}(p)}(f\circ \phi_{t}-f)}t +\lim_{t\to 0} \frac{(Yf)(\phi_{-t}(p)) -(Yf)(p)}{t} \\ &= Y_p\left(\lim_{t\to 0}\frac{f\circ \phi_{t} - f}{t}\right) +\lim_{t\to 0} \frac{(Yf)(\phi_{-t}(p)) -(Yf)(p)}{t} \\ &= Y_p(Xf) - X_p(Yf) \\ &= [X,Y]_p f. \end{align}

(the sign convention for the Lie brackets is $[X,Y]= YX-XY$ or $[X,Y] = XY-YX$, depending on preference.)

In your case $\phi_t = \exp(tX)$ is the flow of $X$.