Some times ago I posted this question here. There I carelessly assumed that if you have a set of sections of a vector bundle which span every fiber pointwise, they also generate the module of smooth sections over the smooth functions on the base manifold, I think that is true if your set of sections is finite. But in my situation for the horizontal bundle this is not the case. So I want to ask the following question:

Let $(P,\pi,B,G)$ be a principal bundle with total space $P$, base $B$, projection $\pi$ and structure group $G$. Let $\gamma \colon TP \to \mathrm{Lie}(G)$ a connection one-form, which induces the horizontal space $HP$. For a vector field $X \in \Gamma^\infty(TB)$ I write $X^h \in \Gamma^\infty(HP)$ for the horizontal lift of $X$ with respect to $\gamma$.

Is it true that $\Gamma^\infty(HP) = C^\infty(P)-\mathrm{Span}(X^h \mid X \in \Gamma^\infty(TB))$, i.e. that the horizontal lifts generate the $C^\infty(P)$-module of horizontal vector fields.

If so, can you give me a proof?

(For a vector bundle $E \to M$ I denote the smooth sections on $E$ by $\Gamma^\infty(E)$. $C^\infty(M)$ are the smooth functions on the manifold $M$)

Note that every manifold is assumed to be real, finite dimensional, $T_2$, paracompact and smooth.

  • $\begingroup$ Is there a more direct and elementary proof than Micheal ones, which uses Serre-Swan? $\endgroup$
    – student
    Commented Oct 2, 2010 at 14:52

2 Answers 2


There is a result (for which I don't recall a reference, but you might prove it maybe with the smooth Serre-Swan theorem. See for example Nestruev, Smooth manifolds and observables), which says that for the sections of the pullback of a vector bundle $E\to M$ along a map $\phi:N\to M$ one has $$\Gamma(\phi^*(E))=C^\infty(N)\otimes_{C^\infty(M)}\Gamma (E)$$ This is a canonical isomorphism obtained from the natural map $\Gamma(E)\to\Gamma(\phi^*(E))$. Combining this with the fact that the horizontal bundle in your case is canonically isomorphic with the pullback of $TB$ to $P$ would give you a proof.

  • $\begingroup$ Thanks very much! After reading your solution I managed it to find a reference for your lemma. See Greub, Halperin, Vanstone: "Connections Curvature and Cohomology Volume I", Ch. II. 2.26 Prop. XVI. $\endgroup$
    – student
    Commented Oct 2, 2010 at 13:52

For those who are interested I give here a slightly more detailed version of Michael's answer.

First note that $T\pi|_{HP} \colon HP \to TB$ is a vector bundle map covering $\pi \colon P \to B$ and a vector space isomorphism on each fiber. This is clear because $T_v\pi$ is obviously onto for every $v \in HP$ and because of dimensional reasons also injective.

Then we can apply Ch. II. 2.26 Prop. XVI from Greub, Halperin, Vanstone: "Connections Curvature and Cohomology Volume I" and get immediately that the map

$$ \alpha \colon C^\infty(P) \otimes_{C^\infty(B)} \Gamma^\infty(TB) \to \Gamma^\infty(HP), f \otimes X \mapsto f \cdot (T\pi|_{HP})^* X $$

is an isomorphism of $C^\infty(P)$-modules.

Now by the very definition of $(T\pi|_{HP})^*$ and of the horizontal lift, the equation

$$ ((T\pi|_{HP})^*X)(p) = T\pi|_{H_vP}^{-1}(X(\pi(p)) = X^h(p) $$

holds for every $p \in P$ and $X \in \Gamma^\infty(TB)$, i.e. $$(T\pi|_{HP})^* X = X^h$$ for every $X \in \Gamma^\infty(TB)$. Consequently given a $\sigma \in \Gamma^\infty(HP)$ there is a $k \in \mathbb{N}$ and for every $i \in \{1,\dots,k\}$ $f^i \in C^\infty(P)$ and $X_i \in \Gamma^\infty(TB)$ with

$$\sigma = \alpha\left(\sum_{i = 1}^k f^i \otimes X_i\right) = \sum_{i = 1}^k f^i X_i^h.$$

This proves the theorem.

So the only nontrivial part is the proposition from Greub et. al. Just for convenience let my cite it here:

Let $\hat \pi \colon \hat E \to \hat M$ and $\pi \colon E \to M$ be two vector bundles and let $\Phi \colon \hat E \to E$ a vector bundle map over some smooth map $\phi \colon \hat M \to M$ restricting to linear isomorphisms in the fibers. Make $C^\infty(\hat M)$ into an $C^\infty(M)$-module by setting

$$ f \cdot \hat f := \phi^*f \cdot \hat f $$

for all $f \in C^\infty(M)$ and $\hat f \in C^\infty(\hat M)$. Then the map

$$ \alpha_{\Phi} \colon C^\infty(\hat M) \otimes_{C^\infty(M)} \Gamma^\infty(E) \to \Gamma^\infty(\hat E), \hat f \otimes \sigma \mapsto \hat f \cdot \Phi^* \sigma $$

is an isomorphism of $C^\infty(\hat M)$-modules. Therby for $\sigma \in \Gamma^\infty(E)$ the smooth section $\Phi^* \sigma \in \Gamma^\infty(\hat E)$ is the pullback of $\sigma$, defined by $(\Phi^*\sigma)(\hat p) := \Phi|_{{\hat E}{\hat p}}^{-1}(\sigma(\hat p))$ for all $\hat p \in \hat M$.

That's essentially the same theorem as Michael stated. Greub et. al. proof it by using that the modules of sections of a vector bundle is finitely generated (Serre-Swan) and that every vector bundle can enlarged via whitney sum to a trivial vector bundle. Again to prove this they also use Serre-Swan.


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