Naturality of Flows

This is something I always forget exists and has a name, so I end up reproving it. Since this sequence of posts is a hodge-podge of things to help me take a differential geometry test, hopefully this will lodge the result in my brain and save me time if it comes up.

I’m not sure whether to call it a lemma or not, but the setup is you have a smooth map {F:M\rightarrow N} and a vector field on {M}, say {X} and a vector field on {N} say {Y} such that {X} and {Y} are {F}-related. Define {M_t} and {N_t} to be the image of flowing for time {t} and let {\theta} and {\eta} be the flows of {X} and {Y} respectively. Then the lemma says for all {t} we have {F(M_t)\subset N_t} and {\eta_t\circ F=F\circ \theta_t} on {M_t}.

This is a “naturality” condition because all it really says is that the following diagram commutes:

{\begin{matrix} M_t & \stackrel{F}{\longrightarrow} & N_t \\ \theta_t \downarrow & & \downarrow \eta_t \\ M_{-t} & \stackrel{\longrightarrow}{F} & N_{-t} \end{matrix}}

Proof: Let {p\in M}, then {F\circ \theta^p: \mathbb{R}\rightarrow N} is a curve that satisfies the property \displaystyle {\frac{d}{dt}\Big|_{t=t_0}(F\circ \theta^p)(t)=DF_{\theta^p(t_0)}(\frac{d}{dt}\theta^p (t)\Big|_{t=t_0})=DF_{\theta^p(t_0)}(X_{\theta^p(t_0)})=Y_{F\circ \theta^p(t_0)}}. Since {F\circ \theta^p(0)=F(p)}, and integral curves are unique, we get that {F\circ\theta^p(t)=\eta^{F(p)}(t)} at least on the domain of {\theta^p}.

Thus if {p\in M_t} then {F(p)\in N_t}, or equivalently {F(M_t)\subset N_t}. But we just wrote that {F(\theta^p(t))=\eta^{F(p)}(t)} where defined, which is just a different form of the equation {\eta_t\circ F=F\circ \theta_t(p)}.

We get a nice corollary out of this. If our function {F:M\rightarrow N} was actually a diffeo, then take {Y=F_*X} the pushforward, and we get that the flow of the pushforward is {\eta_t=F\circ \theta_t\circ F^{-1}} and the flow domain is actually equal {N_t=F(M_t)}.

In algebraic geometry we care a lot about families of things. In the differentiable world, the nicest case of this would be when you have a smooth submersion: {F: M\rightarrow N}, where {M} is compact and both are connected. Then since all values are regular, {F^{-1}(n_0)} is smooth embedded submanifold. If {N} were say {\mathbb{R}} (of course, {M} couldn’t be compact in this case), then we would have a nice 1-dimensional family of manifolds that are parametrized in a nice way.

It turns out to be quite easy to prove that in the above circumstance all fibers are diffeomorphic. In AG we often call this an “iso-trivial” family, although I’m not sure that is the best analogy. The proof basically comes down to the naturality of flows. Given any vector field {Y} on {N}, we can lift it to a vector field {X} on {M} that is {F}-related. I won’t do the details, but it can be done clearly in nice choice of coordinates {(x^1, \ldots, x^n)\mapsto (x^1, \ldots, x^{n-k})} and then just patch together with a partition of unity.

Let {M_x} be the notation for {F^{-1}(x)}. Fix an {x\in N}, then by the above naturality lemma {\theta_t\Big|_{M_x} : M_x\rightarrow M_{\eta_t(x)}} is well-defined and hence a diffeomorphism since it has smooth inverse {\theta_{-t}}. Let {y\in N}. Then as long as there is a vector field on {N} which flows {x} to {y}, then we’ve shown that {M_x\simeq M_y}, so since {x}, {y} were arbitrary, all fibers are diffeomorphic. But there is such a vector field, since {N} is connected.


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