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Wikipedia gives two definitions of the Lie bracket $[X, Y]$ of vector fields $X$ and $Y$ on a manifold $M$, which it claims are equivalent:

Let $\Phi_t^Z$ be the flow associated with the vector field $Z$. At a point $x \in M$,

$[X, Y]_x := \left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} (\mathrm{d}\Phi^X_{-t}) Y_{\Phi^X_t(x)}$.

This is the definition that appears in the texts that I have seen. However, the article claims that the Lie bracket can also be defined as follows:

$[X, Y]_x := \left.\frac12\frac{\mathrm{d}^2}{\mathrm{dt}^2}\right|_{t=0} (\Phi^Y_{-t} \circ \Phi^X_{-t} \circ \Phi^Y_{t} \circ \Phi^X_{t})(x) = \left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} (\Phi^Y_{-\sqrt{t}} \circ \Phi^X_{-\sqrt{t}} \circ \Phi^Y_{\sqrt{t}} \circ \Phi^X_{\sqrt{t}})(x)$.

I find the latter definition quite intuitive in geometric terms. But I don't see how to show that the two are equivalent. Can someone point me in the right direction, or provide a reference?

Thanks

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This is an exercise in Warner Foundations of Differentiable Manifolds and Lie Groups. Let $M$ be a smooth manifold and let $X,Y \in \mathfrak{X}(M)$. Suppose $\phi_t$ and $\varphi_t$ are flows of $X$ and $Y$ respectively (for $t \in (−e, e)$ sufficiently small). Let: $$\beta(t)=\varphi_{-\sqrt[]{t}}\circ \phi_{-\sqrt[]{t}}\circ \varphi_{\sqrt[]{t}}\circ \phi_{\sqrt[]{t}}(m)$$ Show that for every $f$ we have: $$([X,Y]f)(m)=\lim_{t \to 0}\frac{f(\beta(t))-f(\beta(0))}{t}$$

The function is only defined for $t \ge 0$ so we have to compute the limit from the right. This is basically computing the limit interchanging $t$ with $t^2$ in the formula: $$([X,Y]f)(m)=\lim_{t \to 0^+}\frac{f(\beta(t^2))-f(\beta(0))}{t^2} $$ Define $F(t_1,t_2,t_3,t_4)= \varphi_{-t_1}\circ \phi_{-t_2}\circ \varphi_{t_3}\circ \phi_{t_4}(m)$ so we have: $$([X,Y]f)(m)=\lim_{t \to 0^+}\frac{f(F(t,t,t,t))-f(\beta(0))}{t^2} $$ We will use the Taylor rule to $g(t_1,t_2,t_3,t_4)=f \circ F$ as: $$g (t_1,t_2,t_3,t_4) =g(0,0,0,0)+\sum_i \frac{\partial g}{\partial t_i}|_{0}(t_i)+\frac{1}{2}\sum_{i,j}t_i \cdot t_j\int_0^1(1-s)\frac{\partial^2 g}{\partial t_i t_j}(s (t_1,t_2,t_3,t_4))ds $$ This means that we can write the limit as $$\lim_{t \to 0^+}\frac{\sum_i \frac{\partial g}{\partial t_i}|_{0}(t)+\frac{1}{2}\sum_{i,j}t ^2 \int_0^1(1-s)\frac{\partial^2 g}{\partial t_i t_j}(s(t,t,t,t))ds }{t^2} $$ We now observe using the fact that $\varphi_0=Id=\phi_0$ that $$\sum_i \frac{\partial}{\partial t_i}=-Y(f)(m)-X(f)(m)+Y(f)(m)+X(f)(m)=0$$ So we can write the limit as $$ \lim_{t \to 0^+}\frac{1}{2}\sum_{ij}\int_0^1(1-s)\frac{\partial^2 g}{\partial t_i t_j}(s(t,t,t,t))ds= \frac{1}{2}\sum_{ij}\int_0^1 \lim_{t \to 0^+}(1-s)\frac{\partial^2 g}{\partial t_i t_j}(s(t,t,t,t))ds=\frac{1}{2}\sum_{ij}\int_0^1 (1-s)\frac{\partial^2 g}{\partial t_i t_j}(0)ds$$ So we get $$\frac{1}{2}\sum_{ij}\frac{-1}{2}(1-s)^2\frac{\partial^2 g}{\partial t_i t_j}(0)\big|^{s=1}_{s=0}$$ And this will be equal to $$ \frac{1}{2}\sum_{ij}\frac{1}{2}\frac{\partial^2 g}{\partial t_i t_j}(0)$$ After calculating the derivatives will get the desired result $$ \frac{1}{2}\sum_{ij}\frac{1}{2}\frac{\partial^2 g}{\partial t_i t_j}(0)=XY(f)(m)-YX(f)(m) $$

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  • $\begingroup$ Great, thanks for taking the time to give such a detailed answer $\endgroup$ – Simon Jul 30 '18 at 18:19
  • $\begingroup$ I am happy you found it useful $\endgroup$ – Elad Jul 31 '18 at 7:32

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