## New answers tagged vector-bundles

2

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 ...

2

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 ...

4

According to lemma 4 of
M.F. Atiyah, Vector bundles over an elliptic curve, Proc. London Math. Soc. (3) 7 (1957), 414-452.
you could take $N(g,n) = - 2 g$.

2

The only invertible sheaf on $U := \text{Spec}A \setminus \langle x,y,z\rangle$ that extends to all of $\text{Spec} A$ is the trivial invertible sheaf. As you correctly surmise, the invertible sheaves on $W$ are precisely the $\mu_k$-linearized invertible sheaves on $V := \text{Spec}\mathbb{C}[u,v] \setminus \langle u,v \rangle$. Of course every invertible ...

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