## New answers tagged vector-bundles

10

For $n\geq 3$, there exists such non-trivial bundles. For example, take the bundle on $\mathbb{C}^n-\{0\}$ given by the quotient of $\mathcal{O}^n$ by the subbundle generated by the vector $(x_1,\ldots,x_n)$, where the $x_i$s are the coordinate functions. Then, for $n\geq 3$, this bundle is not trivial. The non-triviality will essentially follow from an ...

3

(i) is true and (ii) is false. In fact, (i) is true for all $m\ge1$. More precisely, dropping a coordinate yields a map from $\mathbb RP^{m+1}\setminus\{*\}\to\mathbb RP^m$ which is a locally trivial bundle with fiber $\mathbb R$. This bundle is not trivial for $m\ge1$ because, e.g., $\mathbb R P^m$ is orientable if and only if $m=0$ or $m$ is odd.

10

For $i = 2^k$ you can get an example with $X = \Bbb{RP}^{m}$ for any $m \geq 2^k$. If $L$ is the canonical line bundle on $\Bbb{RP}^{m}$ then let $E = \bigoplus_{i=1}^{2^k} L$ be the a sum of several copies of it. Then it has total Stiefel-Whitney class
$$
w(E) = \prod_{i=1}^{2^k} w(L) = (1+x)^{2^k} = 1 + x^{2^k},
$$
where $x$ is the generator of ...

4

After the comment of Jez, here is the corrected answer:
There is the well-defined vertical bundle $VTM\subset TTM\to TM$ given as the kernel of the (differential of the) projection.
Moreover, for any $v\in TM,$ there is a natural identification $\Phi$ of $V_vTM$ with $T_\pi(v)M.$
Using the Levi-Civita connection (or ans other connection), you can ...

3

As Hassan mentions, in the setting of singular metrics on vector bundles, the notion of curvature appears problematic, which is discussed in the paper of Raufi.
Still, as is also discussed in that paper, in the case that the singular metric is positively or negatively curved in the sense of Griffiths, one can give a reasonable meaning to $c_1(E,h)$ as a ...

4

Given a complex hermitian vector bundle $E$ of complex rank n over a smooth manifold $M$, a representative of each Chern class $ck(E)$ of $E$ are given as the coefficients of the characteristic polynomial of the curvature form $Ω$ of $E$.
$$\det \left(\frac {it\Omega}{2\pi} +I\right) = \sum_k c_k(E) t^k $$
Each Chern class
$c_k$
is a real cohomology class
...

2

There is nothing special about the tangent bundle: for any vector bundle $E \rightarrow X$ with projectivisation $\pi : \mathbb{P}E \rightarrow X$, and tautological bundle $\mathcal{L} \rightarrow \mathbb{P}E$, we have $T\mathbb{P}E \cong (\pi^*E / \mathcal{L}) \otimes \mathcal{L}^*$. In fact the base $X$ is not really involved, so you may as well work with ...

1

Motives and Algebraic cycles: A selection of conjectures and open questions, Joseph Ayoub.

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