# Affine bundle of sections of a surjective morphism of vector bundles?

Everything that follows is over the real numbers, in the $$C^p$$ category for $$1\leq p\leq \infty$$.

Fix an affine bundle of class and consider a fiber-subbundle whose fibers are convex. A partition of unity can be used to average local sections of the subbundle (which exist because it's assumed to be locally trivial) into a global section.

This proves for instance the existence of vector bundle metrics.

Exercise 5.7 of this document suggests using this same result to prove that every surjective morphism of vector bundles has a section (without vector bundle metrics). The outline is to construct an affine bundle whose sections are sections of the vector bundle surjection, and apply the above.

I don't understand how to define this affine bundle of sections/splittings. I also don't understand how I might show it is a fiber bundle: for one, I have no idea why sections/splittings should exist even locally! In the $$C^p$$ category for $$p\geq 1$$ this is a consequence of the constant rank theorem, which I consider to be heavier artillery than existence of vector bundle metrics.

Question. What is the affine bundle whose sections are splittings of a given surjective morphism of vector bundles? Why is it a fiber bundle?

• If I understood you correctly, you want to show that if $X$ is a topological space and $u_x$, $x \in X$, is a continuous (or further regular) family of surjective linear applications $\mathbb{R}^n \rightarrow \mathbb{R}^p$, then there exists a continuous family $v_x$ (defined on a neighborhood of an arbitrary point in $X$) where $v_x: \mathbb{R}^p \rightarrow \mathbb{R}^n$ is such that $u_x \circ v_x=id$? – Mindlack Jan 18 at 23:39
• @Mindlack that's definitely the main part! – Arrow Jan 18 at 23:49

This answer is probably incomplete, it only addresses the question raised in my comment.

So let $$X$$ be a $$C^p$$ manifold, let $$p_0 \in X$$. Let us consider a $$C^p$$ map $$u: (p,v) \in X \times \mathbb{R}^n \longmapsto u_p(v) \in \mathbb{R}^p$$ such that each $$u_p$$ is linear and surjective.

In particular, $$u_{p_0}: \mathbb{R}^n \rightarrow \mathbb{R}^p$$ is linear surjective.

Let $$v_{p+1},\ldots,v_n$$ be a basis of its kernel. This is a free family of $$n-p$$ vectors, so there are indices $$1 \leq i_{p+1} < \ldots < i_n \leq n$$ such that the matrix $$(x_{i_j}(v_k))_{p < j,k \leq n}$$ is invertible.

Let, for every $$p \in X$$, $$v_p:\mathbb{R}^n \rightarrow \mathbb{R}^n$$ be such that $$v_p(y)=(u_p(y),x_{i_{p+1}}(y),\ldots,x_{i_n}(y))$$, so that $$(p,y) \longmapsto v_p(y)$$ is $$C^p$$ and $$v_{p_0}$$ is invertible.

In particular, $$\det{v_{p_0}} \neq 0$$. Let $$U=\{p \in X,\, \det{v_p} \neq 0$$: it is a neighborhood of $$p_0$$ in $$X$$. Let $$i:\mathbb{R}^p \rightarrow \mathbb{R}^n$$ be such that $$i(x)=(x,0,\ldots,0)$$.

Then $$(p,y) \in U \times \mathbb{R}^p \longmapsto v_p^{-1}(i(y))$$ is a $$C^p$$ local section of $$u$$ around $$p_0$$.

• Dear Mindlack, thank you very the answer! How to actually prove the local section you suggest is $C^p$? There's some currying of functions involved in passing from a map from $U\times \mathbb R^p$ to a map from $U$ into linear maps... – Arrow Jan 20 at 21:27
• I’m not sure if there is something more elegant, but you could write the coordinates of the matrix and use the explicit formula for the inverse. – Mindlack Jan 20 at 21:58
• If there's no other choice :) – Arrow Jan 20 at 22:03
• You don’t need the exact formulas: you just need to know that coordinates of the inverse of a matrix are given by rational functions of the coordinates, with the determinant as a denominator. – Mindlack Jan 20 at 22:12