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If $X, Y$ are two smooth vector fields on a smooth manifold $M$, we have that $[X,Y]=0$ implies that the flows $\phi^X_s \circ \phi^Y_t = \phi^Y_t \circ \phi^X_s$ commute, wherever both sides are defined.

However, I'm troubled by a "counterexample" that I thought up. parking garage counterexample

This diagram evokes a two floor parking garage. If you start in the bottom left corner, you can either end up on the upper level or lower level, depending on which flow you take first. We can make sure both flows are global by smoothly decreasing the vector fields $X, Y$ to zero at the boundaries. Then $\phi^X_s \circ \phi^Y_t$ and $\phi^Y_t \circ \phi^X_s$ are defined everywhere, yet they are not the same.

Why is this counterexample wrong?

Edit: User @QiaochuYuan pointed out that the globalization argument does not work because the bracket may be modified. Instead, we can fix the counterexample by appropriately choosing the domains of the flows.

fixed diagram

Mark
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    Are you sure that if you smoothly decrease the vector fields to zero at the boundaries that they'll continue to bracket to zero? – Qiaochu Yuan Oct 08 '23 at 04:28
  • @QiaochuYuan I hadn't thought of that. Indeed, if $g$ is some smooth modifier function that dies on the boundary, we have $[gX, gY] = gX(gY) - gY(gX) = g X(g) Y + g^2 [X,Y] - g Y(g) X - g^2 [Y,X] = g X(g) Y - g Y(g) X$ which is not automatically zero. I guess this is why the counterexample breaks. – Mark Oct 08 '23 at 14:12
  • I updated the question with a fixed counterexample. – Mark Oct 08 '23 at 14:34
  • The differential forms formulation of the Frobenius Theorem (and its proof) may provide some intuition here. I would need to review it myself first before I could be precise, but you can see how this has the flavor of differential forms. You're trying to show that that flow rectangle is in fact closed, so that you have to end up where you started. Dually, you're trying to show that some one-form is closed. Somehow you will use the identity $d\omega(X,Y) = X\omega(Y) - Y\omega(X) - \omega([X, Y])$. – Charles Hudgins Oct 08 '23 at 14:46

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I discussed this with my professor and she thinks it actually is a valid counterxample. She provided an additional hypothesis to make the theorem correct.

It suffices to additionally assume that there are open intervals $I,J$ containing zero, such that $\phi^X_s \circ\phi^Y_t (p) = \phi^Y_t \circ \phi^X_s(p)$ are defined for all $(t,s) \in I \times J$.

The incorrect version of the theorem actually appears in Jeffery Lee's Manifolds and Differential Geometry as Theorem 2.109 (iii)

Mark
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  • @MarianoSuárez-Álvarez what about Kobayashi-Nomizu's corollary 1.11? is that also wrong? – Jackozee Hakkiuz Oct 09 '23 at 02:36
  • @JackozeeHakkiuz I deleted my previous reply about your question. The reference you cite actually uses the stronger condition that $\phi^X_s$ and $\phi^X_t$ are one parameter groups. So if my answer here is correct, then Kobayashi's version follows. – Mark Oct 09 '23 at 03:27
  • @Mark it is interesting, in proposition 1.5 Kobayashi and Nomizu also say that every vector field induces a local $1$-parameter group of diffeomorphisms around any point. – Jackozee Hakkiuz Oct 09 '23 at 04:14
  • Looking again, I think the Kobayahsi version may be faulty, by the counterexample here, since they only require local one parameter groups. – Mark Oct 09 '23 at 16:54