## New answers tagged vector-bundles

0

I think Statement (3.6) in the following paper of Becker and Schultz
Becker, J.C.; Schultz, R.E.
Equivariant function spaces and stable homotopy theory. I.
Comment. Math. Helv. 49, 1-34 (1974).
is in the line of what you are after.

4

The Thom space is the homotopy cofiber of the Unit Sphere Projection:
$$ \mathbb{S}_\xi X \to X \to T(\xi) $$
while the unit sphere in a sumspace $\mathbb{S}[U\oplus V]$ is a join
$$ \begin{array}{c} \mathbb{S}[U] \times \mathbb{S}[V] & \rightarrow & \mathbb{S}[U] \\ \downarrow & & \downarrow \\
\mathbb{S}[V] & \rightarrow & ...

5

Since I was rather surprised that those authors would make such a claim, I looked up the original reference [Koszul, Malgrange, Sur certaines structures fibrÃ©es complexes.] Here is a rough translation of the relevant theorem:
Theorem 2. Let $G$ be a complex Lie group, $V$ a complex manifold, $P$ a principal bundle with group $G$ over $V$. Let $\gamma$ ...

8

Put $u_\alpha=[\mathbb{C}]-[L_\alpha]$.
It is a standard fact, known as the projective bundle theorem, that
$$K(\mathbb{P}(E))=K(X)[t]/\prod_{\alpha}(t-u_\alpha)$$
One can express $Fl(E)$ as the top of a tower in which each level is the projective bundle associated to a vector bundle over the level below. Using this one can obtain
$$ K(Fl(E)) = ...

1

This is not a complete answer but only deals with the rank 2 case with trivial determinant:
Let $u=\Phi\neq0$ be a nilpotent Higgs field, then $\Phi^2=0,$ so the kernel
line bundle $L\subset E$ (assuming $\Phi$ is not identically zero) contains the image of $\Phi.$
Hence, $\Phi$ is uniquely
determined by a holomorphic section $\phi\in H^0(X;L^2K).$
This ...

6

No, this is false. Take $X=\mathbb{R}^3$ and $U=\mathbb{R}^3\smallsetminus\{0\} $,
with $\mathcal{O}_X$ the sheaf of complex $C^{\infty}$ functions.
Line bundles on $U$ are parametrized by $H^1(U, \mathcal{O}_U^*)$, which is isomorphic to $H^2(U,\mathbb{Z})=\mathbb{Z}$ by the exponential exact sequence. Similarly $H^1(X, \mathcal{O}_X^*)\cong ...

1

The stack $\mathfrak{M}$ was discussed in section 7 of On the motivic class of the stack of bundles by Behrend and Dhillon, where the same stratification as discussed by Alexander Braverman was described. Moreover, the various strata are identified with classifying stacks (for the automorphism groups of the corresponding bundles).

2

Yes, it is complete. The flow is given by the operator $e^{tV}$ acting on the coordinate functions $x^i$. This is defined globally because on Zariski open sets the operator is a unipotent linear transformation, so given by polynomials in $t$.

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