# Lie bracket and flows on manifold

Suppose that $X$ and $Y$ are smooth vector fields with flows $\phi^X$ and $\phi^Y$ starting at some $p \in M$ ($M$ is a smooth manifold). Suppose we flow with $X$ for some time $\sqrt{t}$ and then flow with $Y$ for this same time. Then we flow backwards along $X$ for the same time, and then flow backwards along $Y$. All in all, we can define a curve dependent on $t$ as follows $$\alpha(t):= \phi_{-\sqrt(t)}^Y \circ \phi_{-\sqrt(t)}^X \circ \phi_{\sqrt(t)}^Y \circ \phi_{\sqrt (t)}^X$$

It is an exercise to show that $\frac{d}{dt}|_{t=0} \alpha(t) = [X,Y](p)$. In theory, this should be workable with just the chain rule, (assuming I know how to do these derivatives properly, which is something I'd like some clarification on) but this process is going to be excruciatingly long and painful, and it's really just something I want to avoid if I can help it. Is there another way to do this computation that will be less painful and more illustrative of why exactly this works out?

• If you draw the right picture I think you can see it. Where to find that picture... well, this math.stackexchange.com/q/168293/36530 is related... yep, math.stackexchange.com/questions/163262/… but, still something more before your question is seen... – James S. Cook Jul 3 '15 at 1:34
• Right I have a picture like this in my book. It is intuitively clear to me that you don't always end up back where you start, and that this is depends smoothly on the starting point and on the length of time $t$. I just don't want to perform this nasty calculation. – Alfred Yerger Jul 3 '15 at 1:50
• I see, well, I suppose Fredrick's answer explains why the square root is there. Although, I think he has a chain rule implicitly as he makes substitutions and exchanges a limit in one variable for another. – James S. Cook Jul 3 '15 at 14:42

I'm not sure if this is the answer you're looking for, since this is by computation. But it does not involve the chain rule, at least.

So... both sides of the equation are elements of the tangent space $T_pM$. To see that they are equal, we compute their action on a function $f:M \to \mathbb R$.

Now, by definition $$[X,Y](f)(p) = \lim_{h \to 0} \frac 1h \left[ (Yf) \circ \phi_h^X(p)-Y(f)\right] - \lim_{h \to 0} \frac 1h \left[ (Xf) \circ \phi_h^Y(p)-X(f)\right]$$ The first term is equal to: $$\lim_{h \to 0} \frac 1h \left[ (\lim_{k \to 0}\frac 1k [f \circ \phi_k^Y(p)-f(p)]) \circ \phi_h^X(p)-[\lim_{k \to 0}\frac 1k f \circ \phi_k^Y(p)-f(p) ]\right]$$ Setting $k=h$ (the functions are differentiable, so this shouldn't change the answer), we get $$= \lim_{h \to 0} \frac{1}{h^2} \left[ f \circ \phi_h^Y \circ \phi_h^X (p) - f \circ \phi_h^X(p) -f \circ \phi_h^Y(p) + f(p) \right]$$ Doing the same for the other term, we get: $$\lim_{h \to 0} \frac{1}{h^2} \left[ f \circ \phi_h^X \circ \phi_h^Y (p) - f \circ \phi_h^Y(p) -f \circ \phi_h^X(p) + f(p) \right]$$ Subtracting them, most terms cancel and we get $$\lim_{h \to 0} \frac {1}{h^2}\left[ f \circ \phi_h^Y \circ \phi_h^X(p) - f \circ \phi_h^X \circ \phi_h^Y(p) \right]$$ Now we use that $(\phi_h^X)^{-1}=\phi_{-h}^X$ to get $$\lim_{h \to 0} \frac {1}{h^2}\left[ f- f \circ \phi_h^X \circ \phi_h^Y \circ \phi_{-h}^X \circ \phi_{-h}^Y(p) \right]$$ But this is just the derivative of $\alpha$ (we traverse in the opposite direction, but that's okay)! Putting $t=h^2$ we get the result.

So all in all, this computation was not painfree, but it is clear why we need the square root signs.

• Nice answer. Just one small point, consider, $\lim_{h \rightarrow 0} \frac{e^{3h}-1}{h} = 3\lim_{h \rightarrow 0} \frac{e^{3h}-1}{3h} = \lim_{k \rightarrow 0} \frac{e^{k}-1}{k} = 3$. I traded limit in $h \rightarrow 0$ for $3h=k \rightarrow 0$ to pull-out a $3$. Recognize the equivalence to $f(x) = e^{3x}$ having $f'(x)=3e^x$ hence $f'(0)=3$. My point is just that when we make substitutions and change limits in a difference quotient that may be the nuts and bolts of a chain-rule. So, perhaps your answer does use a chain-rule (implicitly). – James S. Cook Jul 3 '15 at 14:51
• @James Thank you for the comment. So there might perhaps be another way to show this that does use the chain rule? – Fredrik Meyer Jul 3 '15 at 15:09
• @JamesS.Cook When I said I wasn't looking to use the chain rule, I suppose I meant I did not wan to calculate with it directly. This question was posed once before a few years back on this site, and instead of giving an answer, another user explains how to solve it by applying the chain rule on the left and then simplifying. This process looked absolutely horrifying, and since I'm already kind of okay with the idea from the picture, I was looking for other ways to see the calculation, or perhaps even a geometric interpretation that makes it clear. – Alfred Yerger Jul 3 '15 at 15:46
• How did you get $\lim_{h \to 0} \frac {1}{h^2}\left[ f- f \circ \phi_h^X \circ \phi_h^Y \circ \phi_{-h}^X \circ \phi_{-h}^Y(p) \right]$ from $\lim_{h \to 0} \frac {1}{h^2}\left[ f \circ \phi_h^Y \circ \phi_h^X(p) - f \circ \phi_h^X \circ \phi_h^Y(p) \right]$? I think what you got is $\lim_{h \to 0} \frac {1}{h^2}\left[ f\circ \phi_h^Y \circ \phi_h^X(p)- f \circ \phi_h^X \circ \phi_h^Y \circ \phi_{-h}^X \circ \phi_{-h}^Y \circ \phi_h^Y \circ \phi_h^X(p) \right]$ I don't know why we could take $\phi_h^Y \circ \phi_h^X(p)$ as $p$. – No One Apr 19 '16 at 21:56
• Intuitively $\phi_h^Y \circ \phi_h^X(p)$ somehow "converges" to $p$. But in the formal proof, I am afraid we can't take the limit of this part first. – No One Apr 19 '16 at 21:59