Are every two vector bundles weakly isomorphic?

Question

Let $E,F$ be two smooth vector bundles of the same rank over a smooth compact manifold $M$.

Does there exists a weak isomorphism $\Phi \in W^{1,2}(M,E^* \otimes F)$ between $E$ and $F$?

i.e I am asking if there exists a bundle morphism $\Phi:E \to F$ which lies in a suitable Sobolev space, and is an isomorphism (of the fibers) almost everywhere.

For a concrete example, we can start with $E=T\mathbb{S}^2, F=\mathbb{S}^2 \times \mathbb{R}^2$. Is there a measurable global frame on $T\mathbb{S}^2$?

Of course, in the smooth category what I am asking is nonsense- the point is that I am interested to know, if, when weakening the regularity requirements every two bundles of the same rank are indistinguishable.

Answer 1

You can get something even better. Ignoring sets of measure zero, all vector bundles are smoothly isomorphic.

Let's consider first your specific example of $TS^2$. Think of $S^2$ as the unit sphere in $\mathbb{R}^3$ and let $X_1(p)$ be the orthogonal projection of the vector $e_1 = (1,0,0)$ onto the tangent plane of $S^2$ at $p$. This is a global vector field on $S^2$ which vanishes at two points $(\pm 1, 0, 0)$. Take $X_2(p)$ to be the orthogonal projection of the vector $e_2 = (0,1,0)$ onto the tangent plane at $p$. This is again a global vector field on $S^2$ which vanishes at two points $(0, \pm 1, 0)$. The locus of points at which $X_1,X_2$ are linearly dependent is precisely the equator $\{ (x,y,0) \, | \, x^2 + y^2 = 1 \}$ so the map $S^2 \times \mathbb{R}^2 \rightarrow TS^2$ given by $(p,(a,b)) \mapsto aX_1(p) + bX_2(p)$ is a smooth bundle map which is an isomorphism of the fibers away from the equator.

This situation is actually quite general and generic. Given a rank $k$ bundle $E$ over $M$, one can define the submanifolds $\Sigma_r \subseteq \operatorname{Hom}(\mathbb{R}^k, E)$ which consists of bundle morphisms of rank $r$. Those are smooth submanifolds of codimension $(k - r)^2$. It turns out that a generic section $s$ of $\operatorname{Hom}(\mathbb{R}^k,E)$ intersects the submanifolds $\Sigma_r$ transversally so the degeneracy locus

$$ \mathcal{D}_r(s) := \{ p \in M \, | \operatorname{rank} s(p) = r \} $$

is a smooth manifold of codimension $(k - r)^2$. In particular, the singular locus $$ \Sigma(s) := \{ p \in M \, | \, \operatorname{rank} s(p) < k \}$$ is contained in the union of submanifolds of positive codimension and so has measure zero. Such a generic section will give you a smooth bundle map which is an isomorphism away from a set of measure zero between $E$ and the trivial vector bundle.

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To see that $\Sigma_r$ is a manifold, let's argue first why the space $\Delta_r$ of matrices of rank $r$ is a smooth submanifold of $M_k(\mathbb{R}) \cong \mathbb{R}^{k^2}$ of codimension $(k - r)^2$. Set

$$ U = \{ X \in M_k(\mathbb{R}) \, | \, \text{the top } r \times r \text{ minor of } X \text{ is non-zero} \}. $$

Then $U$ is an open subset of $M_k(\mathbb{R})$. Let's describe both a chart for $U \cap \Delta_r$ and a locally defining function. Take a matrix $X \in U \cap \Delta_r$ and write it as a block matrix

$$ X = \begin{pmatrix} A & B \\C & D \end{pmatrix} $$

with $A \in M_r(\mathbb{R}), \det(A) \neq 0$.

Performing "block Gauss elimination" to eliminate $B$, we get

$$ \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} I & -A^{-1}B \\ 0 & I \end{pmatrix} = \begin{pmatrix} A & 0 \\ C & D - CA^{-1}B \end{pmatrix} $$

but since the rank of the right hand side is $r$, we must have $D = CA^{-1}B$. This gives a local chart $\operatorname{GL}_r(\mathbb{R}) \times M_{r \times (k - r)}(\mathbb{R}) \times M_{(k - r) \times r}(\mathbb{R}) \rightarrow U \cap \Delta_r$ of the form

$$ (A,B,C) \mapsto \begin{pmatrix} A & B \\ C & CA^{-1}B \end{pmatrix} $$

and a locally defining function $f \colon U \rightarrow M_{k-r}(\mathbb{R})$

$$ f \begin{pmatrix} A & B \\ C & D \end{pmatrix} = D - CA^{-1}B $$

so that $f^{-1}(0_{(k - r) \times (k - r)}) = U \cap \Delta_r$. This shows that $U \cap \Delta_r$ is a submanifold of codimension $(k - r)^2$. Since a rank $r$ matrix must have some non-zero $r \times r$ minor, the argument above can be applied after permuting the coordinates around any rank $k$ matrix showing that $\Delta_r$ is a submanifold.

Finally, the argument above works as well for families and allows you to show that $\Sigma_r$ is a submanifold of codimension $(k - r)^2$.

Answer 2

This is an interesting question and I don't know the answer in general. For the sphere 2-sphere $S^2$, the tangent bundle is indeed trivializable when you weaken the regularity.

To prove that, take a local smooth section $g$ of the frame bundle of $p:{\cal F}(TS^2)\to S^2$ defined on the open subset $U=S^2-\{\ast\}$, where $\ast\in S^2$ is any point. Define the function $f:S^2\to {\cal F}(TS^2)$ as $$f(x)=\left\{\begin{array}{cc} g(x) & x\neq \ast \\ L\in p^{-1}(\ast) & x=\ast \end{array} \right. $$ where $L$ is any element in the fiber $p^{-1}(\ast)$. Since $\{\ast\}$ has measure zero in $S^2$, this defines a global section of the frame bundle with $W^{k,2}$ regularity for any $k$.

Edit: Since measure theory is involved here, I would also expect the number of equivalence classes of $W^{1,2}$-isomorphic bundles over a smooth manifold to change as you deform by a smooth homotopy equivalence: something that does not happen in the smooth case.

Answer 3

As explained here, we can always choose a coordinate chart which covers the entire manifold except a measure zero set. Over this chart we can of course smoothly trivialize any bundle.