## New answers tagged vector-bundles

2

I think the Reeb foliation can be generated by two independent global vector fields so the Frobenius rank of $S^{3}$ is "2".
The reason is that every real vector bundle on $S^{3}$ is a trivial bundle. so the two dimensional subvector bundle of $TS^{3}$ tangent to the Reeb foliation, is a trivial bundle. So there are two global continuous ...

5

Here is one possible way of answering the question, using the simplicial model for $EG$. In this context, we can show very explicitly that the map $\Sigma X\to BG$ built (via adjointness) from the clutching function for $E$ classifies $E$ (and then if $X$ is a CW complex, any other classifying map for $E$ is homotopic to this one, by a standard cell-by-cell ...

4

If $E$ has rank $0$ or $1$, 'yes', otherwise, 'no'. Just do a dimension count. You'll find that you have (many) more unknowns than equations and, for general $h$ and $h''$, there will be no solution $x$.
The answer for your modified question is still 'no, in general when $r$ and $s$ are both greater than $1$.' This is purely a pointwise linear algebra ...

4

As I mentioned earlier, Peskine (and possibly Kollar too) asked whether given a family of smooth curves in 3-space with general member a complete intersection, is the special member also a complete intersection. To the best of my knowledge, the answer is not known (over complex numbers). Under the above hypothesis, it is immediate that ...

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