# Flow on manifolds and Lie bracket.

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 $X(p)+Y(p)$.

-
It would be good if you showed the reasoning you used to prove that $X+Y$ is complete. –  Mariano Suárez-Alvarez Oct 16 '11 at 3:16

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).

-
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''. –  Giuseppe Tortorella Oct 16 '11 at 7:46
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 9:17
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