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Let $v_1$ and $v_2$ be two vector fields on a manifold $M$, and let $g(t)=\varphi_{v_1}^t\circ \varphi_{v_2}^t-\varphi_{v_2}^t\circ \varphi_{v_1}^t$. I want to prove that $$g''(0)=-2[v_1,v_2](p).$$ To do this, I tried writing $$v_1(p)=\sum_{j=1}^n a_j(p)\frac{\partial}{\partial x_j}\Big\vert_p \quad \text{and} \quad v_2(p)=\sum_{j=1}^n b_j(p)\frac{\partial}{\partial x_j}$$ Using that $\frac{\partial}{\partial t}\varphi^t_{v_i}=v_i(\varphi^t_{v_i})$ for $i=1,2$ I get that the first derivative is $$g'(t)=v_1(\varphi_{v_1}^t(\varphi_{v_2}^t(p)))\cdot v_2(\varphi_{v_2}^t(p)))-v_2(\varphi_{v_2}^t(\varphi_{v_1}^t(p)))\cdot v_1(\varphi_{v_1}^t(p)))$$ This is my attempt for the second derivative: $$g''(t)= \frac{\partial}{\partial p}v_1(p)\Big|_{\varphi_{v_1}^t(\varphi_{v_2}^t(p))}\cdot v_2(\varphi_{v_2}^t(p)))+v_1(\varphi_{v_1}^t(\varphi_{v_2}^t(p)))\cdot \frac{\partial}{\partial p}v_2(p)\Big|_{\varphi_{v_2}^t(p))}- \frac{\partial}{\partial p}v_2(p)\Big|_{\varphi_{v_2}^t(\varphi_{v_1}^t(p))}\cdot v_1(\varphi_{v_1}^t(p)))-v_2(\varphi_{v_2}^t(\varphi_{v_1}^t(p)))\cdot \frac{\partial}{\partial p}v_1(p)\Big|_{\varphi_{v_1}^t(p))}$$ However, as $\varphi^0_{v_i}(p)=p$ for $i=1,2$, I get that $g''(0)=0$. What am I doing wrong?

kubo
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    This is the third time I've seen this question posted in the past week. What does the negative sign even mean in your definition of $g$? – Ted Shifrin Apr 16 '22 at 18:12
  • Related to Ted Shifrin's comment: instead of a linear combination of flows, which does not make much sense on arbitrary manifolds, you probably want to take their commutator: $\phi^t_1\circ\phi^t_2\circ\phi^{-t}_1\circ\phi^{-t}_2$. – Johnny Lemmon Apr 16 '22 at 19:51
  • @JohnnyLemmon nope, this is what my exercise says – kubo Apr 17 '22 at 10:06
  • @kubo is it possible that there is a mistake in the statement? Anyway, see this post for a very much related question: https://math.stackexchange.com/questions/1209452/derivatives-of-the-commutator-of-flows-or-what-are-those-higher-derivatives-do – Johnny Lemmon Apr 17 '22 at 18:16
  • Nevermind, there is a way to interpret this so that it makes sense. See this question, that almost has the answer you look for https://math.stackexchange.com/q/3142486/261022 – Johnny Lemmon Apr 17 '22 at 20:47

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