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I want to find conditions under which one can pull-back vector fields (if it is at all possible).

Let $F:M \to N$ be a smooth surjective map between two $C^{\infty}$ manifolds of the same dimension. Let $Y$ be a vector field on $N$ (i.e. smooth section of the tangent bundle $TN$). Define: $T^\ast Y(p)(f)=Y(F(p))(f\circ F^{-1})$, where $f \in C^{\infty}(M,\mathbb R)$. We check that $T^\ast Y$ is a derivation, and this is true since:

$T^\ast Y(p)(fg)=Y(F(p)(fg \circ F^{-1})=Y(F(p)((f \circ F^{-1})(g \circ F^{-1}))$

My question: Is it enough for $F$ to be a local diffeomorphism for this to work?

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    How is $f \circ F^{-1}$ an element of $C^\infty(N,\mathbb{R})$, if you don't assume that $F^{-1}$ is a smooth map $N \to M$? In other words, it appears to me that you need to assume that $F$ is a diffeomorphism, and then, AFAICT, your definition reduces to the *pushforward* of $Y$ under $F^{-1}$.2012-04-27
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    @MartinWanvik: Actually, I just need $f \in C_p^{\infty}(M,\mathbb R)$, i.e, the germ at $p$. I am trying to avoid a full diffeomorphism.2012-04-27
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    Right, of course - it's been a long time since I've studied this stuff. It might make things clearer if you put some square brackets around the functions, to indicate that you're considering germs instead of actual functions, i.e. $[f \circ F^{-1}]_p$. As you probably know, there are several definitions of tangent spaces, and the one I looked at says that a tangent vector is a derivation on $C^\infty(M,\mathbb{R})$ - of course, it comes out to the same thing, but in that context your formula would require a word or two of explanation.2012-04-27
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    So, from what I can tell, there appears to be no question that question that you're able to associate a derivation to each point in $M$, so what remains is to prove smoothness. As an alternative, if you assume (for example), that both manifolds are Riemannian, you can pull back vector fields without any conditions on the map $F$ (only smoothness).2012-04-27
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    @MartinWanvik: Unfortunately that wont work for my purposes, one of the manifolds is $\mathbb R P^n$2012-04-27
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    That shouldn't matter - any smooth manifold admits a Riemannian metric, and $\mathbb{R}P^n$ apparently has a rather natural one coming from $\mathbb{S}^n$ (http://en.wikipedia.org/wiki/Real_projective_space). Or did I misunderstand?2012-04-27
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    @MartinWanvik: No, this is great. I don't need my question answered anymore. Should have just checked the wiki page. Thanks2012-04-27
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    I should add that I have some doubts as to whether the pullback obtained in that way behaves as you would expect it to in general - if $F$ actually happens to be a diffeomorphism, I'm not sure the pushforward of the resulting vector field is the same as the one you pulled back (unless the map is an isometry, I think). That may or may not matter to you, depending on your particular situation.2012-04-27

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