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Given a vector field $X$ over a smooth manifold, under what conditions over $f$ is $df(X)$ a smooth field?

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$df(X)$ is a function, not a vector field (though I suppose one could interpret a function as a "scalar field"...) –  Jesse Madnick Jun 29 '11 at 22:34
    
Are we assuming that $X$ is a smooth vector field? –  Jesse Madnick Jun 29 '11 at 22:36
    
I'm sorry I should have been more clear: $f: M \rightarrow N$ where $M$ and $N$ are smooth manifolds and $X$ is a smooth vector field on $M$ –  Chu Jun 29 '11 at 22:39
    
As Willie Wong and I both note in comments to Jesse Madnick's answer below, there is not generally a well-defined vector field $df(X)$ on $N$ unless $f$ is 1-1. –  Matt E Jun 29 '11 at 23:08

2 Answers 2

up vote 5 down vote accepted

The condition you're looking for is called $f$-relatedness:

  • A smooth vector field $X$ on $M$ is $f$-related to a smooth vector field $Y$ on $N$ iff $df(X_p) = Y_{f(p)}$ for every $p \in M$.

If we regard $X\colon C^\infty(M) \to C^\infty(M)$ and $Y\colon C^\infty(N) \to C^\infty(N)$ as derivations, then it is a fact that $X$ and $Y$ are $f$-related iff for every smooth function $g\colon U \to \mathbb{R}$ (where $U\subset N$ is open), we have $X(g\circ f) = (Yg)\circ f$. ("Introduction to Smooth Manifolds" by John Lee, Lemma 4.8)

It is also a fact that if $f\colon M \to N$ is a diffeomorphism, then $df(X)$ will in fact be a smooth vector field on $N$. (As above, Proposition 4.10).

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@Chu: You should seriously consider getting a copy of John Lee's "Introduction to Smooth Manifolds." A couple of the questions you've asked could be easily answered by referencing Lee's text. (Also, it's my hands-down favorite textbook.) –  Jesse Madnick Jun 29 '11 at 22:53
    
I think I'll grab it off Amazon. Thanks! By the way, in the Answer above, would the conclusion still be true if we only ask that f is a submersion? –  Chu Jun 29 '11 at 22:59
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@Chu: no. If $f$ is say, a covering map, then $df(X)$ may assign two different vectors to the same point $p\in N$. –  Willie Wong Jun 29 '11 at 23:05
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@Chu: Dear Chu, In regard to the case when $f$ is a submersion: the expression $df(X)$ is not well-defined in general if $f$ is not 1-1. If $p$ and $p'$ are two distinct points of $M$ such that $f(p) = f(p')$, there is no reason in general for $df(X_p)$ and $df(X_{p'})$ to be equal, and hence there is not generally a well-defined field $df(X)$ on $N$. Regards, –  Matt E Jun 29 '11 at 23:06

If $f$ is assumed to be an injective immersion, then $df(X)$ is a smooth vector field on $f(M)$ (it isn't necessarily true that it can be extended to all of $N$ though). This ties in with Jesse's answer because $df(X)$ is in fact the unique smooth vector field on $f(M)$ that is $f$-related to $X$ (see Proposition 8.27 in Lee's book).

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