# Tag Info

2

No, such an hermitian metric does not necessarily split, even if $n=2$. As a matter of fact, vector bundles are a quite rigid sort of objects, while metrics are pretty smooth and can be tweaked locally to have any prescribed look and a given small enough open subset. By partitions of unity, it is easy to define metrics that have a prescribed form on some ...

-1

You can see from the special case when $X$ is a point that what you proposed cannot work.

1

You get two copies of $H^*(X)$. In general ($k$ pairwise disjoint sections), by excision it's just like disjoint union of $k$ copies of the original bundle, hence $k$ copies of $H^*(X)$. (An extra observation is that, for one section, the result does not depend on its choice, as any section is homotopic to $0$.) There's no generalized Thom class: the usual ...

4

Swan's paper "On Seminormality" shows that whenever the ring $R$ fails to be seminormal, there is an algebraic counterexample with $X=Spec(R)$, $E$ trivial and rank one over $X$, and $V$ rank one over $E$. Seminormality means that whenever $b^2=c^3\in R$, it follows that $b=a^3,c=a^2$ for some $a\in R$. In particular, the coordinate ring of the cusp ...

10

This is definitely false in the algebraic or holomorphic setting, even in dimension 1. There is a well-known example (see this post) of a rank 2 vector bundle $E$ on $\mathbb{C}\times \mathbb{P}^1$ such that $E_{|\{t\}\times \mathbb{P}^1 }=\mathcal{O}_{\mathbb{P}^1}^2$ for $t\neq 0$, but $E_{|\{0\}\times \mathbb{P}^1}=\mathcal{O}_{\mathbb{P}^1}(-1)\oplus ... 4 It is not trivial. Its characteristic classes were worked out by Dmitrii Fuchs. In particular, the Maslov class is one of its nontrivial characteristic classes, and was the subject of a famous paper of Vladimir Arnol'd. 2 No, the bundle is not trivial in general. If we consider the case$n=1$, then the associated Lagrangian Grassmannian is actually the ordinary Grassmannian, namely$\mathbb{R}\mathbb{P}^1$. Also, the tautological bundle over$\mathbb{R}\mathbb{P}^1$is non-trivial. 2 As in my answer to your previous question,$F=E\oplus E$. This should answer the rest (as much as it is possible to answer). A simple obstruction is that the total Stiefel--Whitney class should be a square. The same about the total rational Pontrjagin class. The class should be divisible by two in the$K$-group. I'm not sure that there are simple sufficient ... 1 It seems to me that$F_1=E\oplus TM$and$F_2=E\oplus E$(Whitney sums). These decompositions are not quite canonical: there are short exact sequences, but in this category they always split. Anyway, it follows that$F_1$and$F_2$are stably equivalent iff so are$E$and$TM$. 1 Here is an answer for this specific case. Let$Q$be the tautological bundle on$X$, the Grassmannian. If it can be extended to say$E$on 5-spcae, using the fact that$H^1_*(Q)=0$, one gets from the exact sequence,$0\to E(-2)\to E\to Q\to 0$, that$H^1_*(E)=0$and thus$H^0(E)\to H^0(Q)$is onto. In particular, you can lift the 4 sections of$Q$to$E$and ... 5 I would say it doesn't. The tautological Chern classes generate$H^*$of the Grassmanian, but the inclusion to$\mathbb{P}^4$does not induce an epimorphism. 4 View$\mathbb{P}^1\times \mathbb{P}^1$as a quadric$Q$in$\mathbb{P}^3$; let$\pi :Q\rightarrow \mathbb{P}^2$be a general projection. Let$D$be a line contained in$Q$. Then$\pi _*\mathcal{O}_Q(D)\cong \mathcal{O}_{\mathbb{P}^2}^2$, so$\mathcal{O}_Q(D)\$ is a Ulrich bundle.

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