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Let $\mathcal{V}\stackrel{\pi_N}{\longrightarrow}N$ be a $V$-vector bundle on a smooth manifold $N$, let $D$ a connection on this bundle and let $f:M\rightarrow N$ be a smooth function. Then we can define the pullback bundle $f^*\mathcal{V}\stackrel{\pi_M}{\longrightarrow}M$ as the bundle given by all the elements of type $\{(p,v)|p\in M,\ \pi_N(v)=f(p)\}$ (with the obvious projection $\pi_M$). Now I have been told that $D$ induces a connection ${}^fD$ on the pullback bundle which is completely determined by the fact that if $\eta$ is a section of $\mathcal{V}\stackrel{\pi_N}{\longrightarrow}N$ and $v\in T_pM$, then

$${}^fD_vf^*\eta=D_{f^*v}\eta$$

I have a feeling that it is not true, because there could be sections of $f^*\mathcal{V}\stackrel{\pi_M}{\longrightarrow}M$ that are not the pullback of any section of $\mathcal{V}\stackrel{\pi_N}{\longrightarrow}N$. Am I wrong or there is really something missing to describe completely ${}^fD$?

Daniel Robert-Nicoud
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  • There are certainly sections of $\mathcal{f*V}$ that aren't pullbacks of sections of $\mathcal{V}$ in general; you need to somehow express the the former in terms of the latter and extend the definition of the connection by linearity/product rule. – Anthony Carapetis Aug 21 '13 at 00:37
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    You have a typo in the formula? All you need to determine the connection is what happens to local bases when you trivialize the bundle. If you can trivialize $\mathcal V$ over $U\subset N$, then you can trivialize $f^*\mathcal V$ over $f^{-1}(U)\subset M$. Right? – Ted Shifrin Aug 21 '13 at 01:02
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    @TedShifrin Oh, so I just pull-back a local basis of $\mathcal{V}$ and use it as basis of the pullback bundle and work with that? If that's correct, please write your comment as an answer and I will accept it. – Daniel Robert-Nicoud Aug 21 '13 at 11:19

1 Answers1

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The correct formula should be $\ {}^fD_v(f^*\eta) = D_{f_*v}\eta$. The connection is determined by what it does to a basis of sections on open sets over which the bundle is trivial. Of course, having a trivialization of $\mathcal V$ over $U\subset N$ gives a trivialization of $f^*\mathcal V$ over $f^{-1}(U)\subset M$.

Ted Shifrin
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    the idea is that although there are sections that are not pullbacks, the pullback of a basis will give a basis and this is well defined for those, and extended by linearity in the general case? Did I get this right or my understanding is wrong? – Gold Sep 23 '17 at 13:39
  • @Gold You will also need the Leibniz rule to extend your connection. Think of the case $f:M \rightarrow { \text{point} }$. Then all pulled back sections are parallel and you will need the Leibniz rule to know what your pullback connection does on a function times such a parallel section. – user505117 Mar 24 '21 at 11:38