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I'm currently reading through some notes on Lie Theory online, and I've stumbled across the following question, which I'm totally stumped by.

"Let X,Y be two commuting complete vector fields on a manifold M, that is $[X,Y]=0$. Show that the vector field X+Y is complete and that the flow of X+Y is given by $\phi_{t,X+Y}(p)=\phi_{X,t}\circ \phi_{Y,t}(p)$, where $\phi_{t,X}$ stands for the flow of the vector field X,and so on."

I have no problems showing that the vector field is complete. However, it's the flow part that bugs me. So far I've tried the following: Look at $h(s,t,p) = \phi_{X,t}\circ \phi_{Y,s}(p)$, for some point p. Set $c(t,p) = h(t,t,p)$. We then have, after differentiating that $\frac{d}{dt}_{t=0}c(t,p) = D_1h(0,0,p)+D_2h(0,0,p)$.Since $h(t,0,p)=\phi_{t,x}(p)$ and $h(0,t,p) = \phi_{t,Y}(p)$ we get that $D_1h(0,0,p) = X(p)$, and $D_2h(0,0,p) = Y(p)$ and thus, the flow is $X(p)+Y(p)$.

Shaf_math
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1 Answers1

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It is easy to show that if $\phi_t$ is a one-parameter group of diffeomorphisms with the property that $\frac{d}{dt}\vert_{t=0} \phi_t(p) = X_p$ for all $p$, then $\phi_t$ is the flow of $X$. So based on what you've shown, you now need to show that $\phi_{t,X} \circ \phi_{t,Y}$ is a one-paraemter group of diffeomorphisms. To prove this you will need to show that the flows of $X$ and $Y$ commute, which follows from the commutativity of $X$ and $Y$ (I haven't worked out all the details but this is a known fact).

Eric O. Korman
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  • Dear Eric, if am not wrong, then your second paragraph should be ``So based on what you've shown, you now need to show that $\phi_{t,X}\circ\phi_{t,Y}$ is a one-parameter group of diffeomorphisms''. – agt Oct 16 '11 at 07:46
  • Eric: yes, I get that the flows commute. But how could I from this extract that it is an one parameter group of diffeo? – Shaf_math Oct 16 '11 at 09:17
  • Eric, I think I see it. We simply have now to show that it is a one-parameter group of diffeomorphisms, which is easy since the flows commute. And further, a composition of diffeomorphisms is a diffeomorphism, so we get our theorem. Thank you! – Shaf_math Oct 16 '11 at 11:02
  • @Giuseppe: ah, thanks I just edited it. – Eric O. Korman Oct 16 '11 at 14:03