## New answers tagged vector-bundles

1

We can actually define this in terms of hom, but first one needs to extend $A$ and $B$ to bundles on $X\times Y$.
Let $\pi_X:X\times Y \to X$ and $\pi_Y:X\times Y \to Y$ be the projection maps. The pullback $\pi_X^*(A)$ will be, in your notation, $\{((x,y),A_x)|(x,y)\in X\times Y\}$, and similarly for $\pi_Y^*(B)$. The bundle you are looking for is then ...

1

Some references, e.g., Berline–Getzler–Vergne, Heat kernels and Dirac operators, will define, for convenience, a modified tensor product $A \boxtimes B \to X \times Y$ by $$A \boxtimes B := \operatorname{proj}_1^\ast A \otimes \operatorname{proj}_2^\ast B,$$ where $\operatorname{proj}_1 : X \times Y \to X$ and $\operatorname{proj}_2 : X \times Y \to Y$ are ...

0

For explicit descriptions of the relation between principal connections and induced connections on associated bundles including vector bundles, see section 19 of
this book.

5

This answer just extends the remark by Liviu Nicolaescu above. For a general connection on a vector bundle with structure group $G$, you cannot say anything about the connection coefficients. (As an extreme example look at the case of a trivial bundle which has structure group $\{e\}$ and still admits lots of non-trivial connections.) But the $G$-structure ...

2

I am not sure what you mean by equivariant. What you want is certainly true, provided the line bundle is ample (positive), over any projective variety. This follows e.g. from asymptotic Riemann-Roch (which gives control of the Euler characteristic), together with the fact that tensoring with an ample line bundle kills higher cohomology. I do not know the ...

4

Anton has already answered, but here's a (very) slightly different explanation: for any locally free sheaf $E$ of finite rank on a scheme $X$, the 'geometric' vector bundle $V=\underline{\mathrm{Spec}}_{X}(\mathrm{Sym}(E^\vee))$ has the following functorial description: for a scheme $f\colon T\to X$ over $X$, the set $\mathrm{Hom}_X(T,V)$ is naturally in ...

3

What you have is the fiber diagram.
$$
\begin{array}[c]{ccc}
X\times T & {\stackrel{\beta}{\rightarrow}} & X\times S \\
{\scriptstyle \delta} \downarrow & & \downarrow\scriptstyle{\pi}\\
T & {\stackrel{\alpha}{\rightarrow}} & S
\end{array}
$$
Now, there is a natural morphism
$\alpha^* V = \alpha^*\pi_*\mathcal{F}\to ...

1

The answer to your question is yes, if $F$ is a direct sum of line bundles. So, let us assume that and its rank is $m$. And as you observed, we will assume that we are working over $\mathbb{P}^2$. Let $0\to F_0\to F_1\to E\to 0$ be the minimal resolution of $E$, where $F_i$ are direct sum of line bundles with rank of $F_0=n$ so that rank of $F_1=n+m$. ...

8

In full generality, the answer is no. There are examples of Parimala of non-extended torsors for special orthogonal groups over $\mathbb{A}^2_{\mathbb{R}}$, see e.g. Amer J. Math. 100 (1978), 913-924, (admittedly the group is not simply-connected in this case). These examples are also discussed in Lam's book "Serre's problem on projective modules". Further ...

3

Let me give a more detailed answer:
A rank $n$ vector bundle on $X$ is precisely given by a homotopy class of maps $X\rightarrow BO(n)$. The Stiefel-Whitney classes of the vector bundle describe how that map acts on cohomology with $\mathbb{F}_2$ coefficients.
In general, not all maps on cohomology can be realized by an actual map of spaces. (For spectra ...

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