# Debug proof of lie derivative equals lie bracket

I am following Nakahara's "Geometry, Topology and Physics", and I am stuck at trying to understand the proof of the equivalence of lie derivatives and lie brackets.

I'm going to lay down the definitions, since it looks like there are more definitions for objects in differential geometry than textbooks ;).

We define a curve $$c$$ as a differential map $$c: (-1, 1) \rightarrow M$$.

We define the action of a curve on a function $$f: M \rightarrow \mathbb R$$ as the directional derivative of the curve $$C$$ with respect to $$f$$, that is $$f[c] \equiv \frac{d f(c(t))}{dt}|_{t = 0}$$.

Now, the set of tangent vectors is the set of all curves quotiented by an equivalence relation: $$C \equiv \{ c: (-1, 1) \rightarrow \mathbb R \} / \sim$$ where $$c_1 \sim c_2 \equiv c_1(0) = c_2(0) \land \left(\forall f: M \rightarrow \mathbb R, f[c_1] = f[c_2] \right)$$

A tangent space at point $$x$$ is the vector space of all curves: $$T_x(M) \equiv \{ c \in C | c(0) = x\}$$.

A vector field $$V: (p: M) \rightarrow T_p(M)$$ is a function that maps each point $$p$$ to a vector in the tangent space $$T_p(M)$$.

The flow of a vector field $$V$$ around a point $$p$$ is a function $$\sigma: M \times \mathbb R \rightarrow M$$ which is a solution to the system of differential equations such that $$\sigma(x, 0) = x$$, and $$\frac{\sigma(x, t)}{dt}|_{x = x_0} = V(x_0)$$ (NOTE: this needs to be written in terms of local coordinates to make sense, but that is a composition of the correct chart map everywhere which I have elided with abuse of notation)

Let $$X$$ be a vector field on a manifold $$M$$ with flow $$\sigma_t: M \rightarrow M$$ at point $$x_0$$, and let $$Y$$ be another vector field on the manifold $$M$$.

Now, we define: $$L_XY(p) \equiv \lim_{\epsilon \rightarrow 0} \frac{\sigma_{-\epsilon}^*(Y(\sigma_{\epsilon}(p)) - Y(p)}{\epsilon}$$

We want to show that this equal to the quantity $$[X, Y](p) = X(Y(p)) - Y(X(p))$$

There are quite a few things that immediately don't make sense:

First of all, how does $$X(Y(p))$$ make sense? $$Y(p)$$ lives in $$T_pM$$, while $$X$$ takes arguments in $$M$$. Similarly, I don't understand how $$Y(X(p))$$ is well-typed either.

So, I googled, and I found that for this to work, you need to use the defintion of vectors using derivations. So, let's define those:

Let the space of smooth functions over $$M$$ be $$C^\infty(M)$$. A derivation at a point $$p \in M$$ is a linear map $$D_p: C^\infty(M) \rightarrow \mathbb R$$, which obeys the Leibniz rule $$D_p(f * g) = f(p) D_p(g) + g(p) D_p(f)$$ where $$*$$ is pointwise function multiplication. The set of derivations at a point form a vector space, and this is our new definition of tangent space at a point.

Now, we need to massage the definition of a vector field to be able to compose them together. First, we will look at them as: \begin{align*} &X: M \rightarrow (C^\infty(M) \rightarrow \mathbb R) \\ &X': C^\infty(M) \rightarrow C^\infty(M) \quad X'(f)(x) = X(x)(f) \end{align*}

This allows us to cleanly define $$X \circ Y - Y \circ X$$ since the types line up.

However, we are now left with the proof that $$L_xY = X \circ Y - Y \circ X$$, with different definitions of the vector field on the left hand side and the right hand side, so I'm totally lost at this point. I don't even know how to attack the problem.

• Where did you get the expression $X(Y(p))$? This, as you point out, makes no sense. What you should be doing is proving that $(\mathscr L_XY)(f)(p) = X(Y(f))(p) - Y(X(f))(p)$ for every smooth function $f$. – Ted Shifrin Jun 27 at 17:08

First of all i will use the definition of tangent vectors as derivations. Then one can show that for each $$p\in M$$ a linear map $$D_p: C^\infty(M) \rightarrow \mathbb R$$ is a derivation iff there exists a curve $$\gamma:(-\varepsilon,\varepsilon)\to M$$ with $$\gamma(0)=p$$ such that $$D_p(f)=\frac{d}{dt}_{|t=0}(f\circ\gamma)(t)$$ for all $$f\in C^\infty(M)$$.

If $$X$$ is a vectorfield and $$f\in C^\infty(M)$$ we define $$Xf\in C^\infty(M)$$ by $$(Xf)(p)=X(p)(f)$$. Then $$[X,Y](p)$$ is the tangent vector (!) at $$p$$ defined by $$[X,Y](p)(f)=X(p)(Yf)-Y(p)(Xf)$$.

Now what you want to show is

$$\lim_{\epsilon \rightarrow 0} \frac{\sigma_{-\epsilon}^*Y(\sigma_{\epsilon}(p)) - Y(p)}{\epsilon}=[X,Y](p)$$

This is an equality of tangent vectors in $$T_pM$$ so for the limit to make sense $$T_pM$$ should have a topololgy. As in all finite dimensional vectorspaces each norm on $$T_pM$$ induces the same topology and it is this one. Then one can show that a sequence $$X_n$$ converges to $$X$$ in $$T_pM$$ iff for all $$f\in C^\infty(M)$$ $$X_n(f)$$ converges to $$X(f)$$. Thus if you want to avoid choosing coordinates you should show

$$\lim_{\epsilon \rightarrow 0} \frac{\sigma_{-\epsilon}^*Y(\sigma_{\epsilon}(p))f- Y(p)f}{\epsilon}=[X,Y](p)f$$

For startes, if the left hand side exists it is equal to

$$\frac d{dt}_{|t=0} {\sigma_{-t}^*Y(\sigma_{t}(p))f}=\frac d{dt}_{|t=0} {Y(\sigma_{t}(p))(f\circ\sigma_{-t})}$$ Maybe some answers here will also be helpfull: Lie derivative of a vector field equals the lie bracket Good luck!

• Thanks for the explanation, I think this helped some pieces click! I'll try the derivation again :D – Siddharth Bhat Jun 30 at 11:02