# Connection on pullback bundle.

Let $$E\to N$$ be a vector bundle and $$f:M\to N$$ be a smooth map. The pullback $$f^*:\Omega^k(N,E)\to \Omega^k(M,f^*E)$$ is then defined by $$(f^*\omega)_x(v_1,...,v_k)=\omega_{f(x)}((df)_x(v_1),...,(df)_x(v_k))$$ with $$x\in M$$ and $$v_1,...,v_k\in T_xM$$.

First question: how can I rewrite this pullback as a map $$f^*\omega:\mathfrak{X}(M)\times...\times \mathfrak{X}(M)\to \Gamma(M,f^*E)$$? My problem is that $$df:\mathfrak{X}(M)\to \mathfrak{X}(N)$$ is well defined as long as $$f$$ is a diffeomorphism. However, $$df:\mathfrak{X}(M)\to \Gamma(M,f^*TN)$$ is actually well defined.

Then, I know that the connection $$f^*\nabla$$ on $$f^*E\to M$$ is uniquely determined by $$(f^*\nabla)(f^*s):=f^*(\nabla(s))\in \Omega^1(M,f^*E)$$

Second question: In some books, I saw the notation $$(f^*\nabla)_X(f^*s)=f^*(\nabla_{df(X)}(s))=\nabla_{df(X)}(s)\circ f$$ but as long as $$f$$ is not a diffeomorphism $$df(X)$$ is not a vector field on $$N$$ so it doesn't make sense writing $$\nabla_{df(X)}$$. How can I solve this problem?

• Why are you trying to give a map $\mathfrak X(M)\to\mathfrak X(N)$? Don't you just need a vector field along the image of $M$? I don't understand what $f^*(\nabla s)$ means. A section of $f^*E$ is not the same as the pullback of a section of $E\to N$, is it? Oct 18, 2020 at 22:51
• If I want to write the pullback as $f^*\omega(X_1,...,X_k)=(\omega\circ f)(df(X_1),...,df(X_k))$, I need $df:\mathfrak{X}(M)\to \mathfrak{X}(N)$ to be well defined, right? $\nabla:\Gamma(N,E)\to \Omega^1(N,E)$, so $f^*(\nabla(s))$ is the pullback of the 1-form $\nabla(s)$ Oct 19, 2020 at 9:43
• There is also another induced map from the sections of $E\to N$ to the sections of $f^*E\to M$, $f^*:\Gamma(N,E)\to \Gamma(M,f^*E)$ given by precomposition. Oct 19, 2020 at 9:53

To write the pullback globally as a map of modules, you need to split the construction into two parts. To make everything less confusing, let me denote by $$f^{\star}(E)$$ the pullback bundle and denote by $$\mathfrak{X}_f(M) = \Gamma(M,f^{\star}(TN))$$ vector fields along $$f$$. Thus, the maps and constructions involving $$\star$$ have nothing to do with the derivative while the maps involving $$*$$ include the derivative. Now,
1. Given a form $$\omega \in \Omega^k(N,E)$$, we get an element $$f^{\star}(\omega) \in \Gamma \left( \operatorname{Alt}^k \left( f^{\star}(TN), f^{\star}(E)\right) \right)$$ given by $$f^{\star}(\omega)|_p \left( \xi_1, \dots, \xi_k \right) = \omega_{f(p)} \left( \xi_1, \dots, \xi_k \right)$$ where $$\xi_i \in \mathfrak{X}_f(M)$$ are vector fields along $$f$$ so that $$\xi_i(p) \in T_{f(p)}N$$ and the formula makes sense. Globally, this corresponds to an alternating map of $$C^{\infty}(M)$$-modules $$f^{\star}(\omega) \colon \mathfrak{X}_f(M) \times \dots \mathfrak{X}_f(M) \rightarrow \Gamma(f^{\star}(E)).$$
2. In addition, you have the map $$df \colon \mathfrak{X}(M) \rightarrow \mathfrak{X}_f(M)$$. Then $$f^{*}(\omega)(X_1, \dots, X_k) = f^{\star}(\omega)(df(X_1), \dots, df(X_k)).$$
Regarding the second question, the formula $$f^{*}(\nabla)_{X}(f^{\star}(s)) = \nabla_{df(X)}(s) \circ f$$ is meant to be understood locally as $$\left( f^{*}(\nabla)_{X}(f^{\star}(s)) \right)|_{p} = \left( \nabla_{df|_{p}(X)}(s) \right) \textrm{ (at }f(p)\textrm{)}.$$ It makes sense even when $$f$$ is not a diffeomorphism as a connection $$\nabla_X(s)$$ is tensorial in the $$X$$ variable. You can also understand the formula globally by splitting the construction of the pullback connection into two parts:
1. There is a map $$f^{\star}(\nabla) \colon \mathfrak{X}_f(M) \times \Gamma(E) \rightarrow \Gamma(f^{\star}(E))$$ given by $$\left( f^{\star}(\nabla)_{\xi} \right)(s)|_{p} = \left( \nabla_{\xi(p)} s \right)|_{f(p)}.$$
2. The pullback connection $$f^{*}(\nabla)$$ is the unique connection $$f^{*}(\nabla) \colon \mathfrak{X}(M) \times \Gamma(f^{\star}(E)) \rightarrow \Gamma(f^{\star}(E))$$ which satisfies $$f^{*}(\nabla)_{X}(f^{\star}(s)) = f^{\star}(\nabla)_{df(X)}(s).$$