Given a vector field $X$ over a smooth manifold, under what conditions over $f$ is $df(X)$ a smooth field?
Conditions for $df(X)$ to be a smooth field
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0As 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. – 2011-06-29
2 Answers
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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3@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, – 2011-06-29
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).