Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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 ${}^fD$?

share|cite|improve this question
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 0:37
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 1:02
@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
up vote 4 down vote accepted

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 $f^{-1}(U)\subset M$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.