## New answers tagged vector-bundles

2

The rank of $E, F$ essentially doesn't enter the discussion, since given a basis of $e_i$ and a basis $f_j$ of $E$ and $F$, you have a basis $e_i\otimes f_j$ of $E\otimes F$ and representing your connection against this basis you are down to dealing with scalar functions.
Suppose first that you can choose bases which are locally parallel. Then you see that ...

1

Any torsion free sheaf on a smooth surface can be written in the form $N \otimes I_\xi$, in particular this holds for $E/E_1$. Explicitly, one can write $N = \det(E)\otimes \det(E_1)^{-1}$, and also one can figure out the length of $\xi$ from Chern classes of $E_1$ and $E$. The actual subscheme $\xi$ is more difficult to understand.

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I suppose $V\to M$ is a $G$-equivariant vector bundle. Then $\bar V$ is defined as follows: the fiber $\bar V|_b$ over a point $b\in B$ is (canonically) equal to the space of $G$-invariant sections of $V|_{\pi^{-1}(b)}\to \pi^{-1}(b)$. The latter space has dimension equal to the rank of $V$.

1

I think that the answer is no in general. For instance, it seems that a locally free sheaf on $U$ does not even systematically extend to a locally free sheaf on $X$.
You can find a counterexample in http://www.mathunion.org/ICM/ICM1970.2/Main/icm1970.2.0619.0624.ocr.pdf for $X=\mathbb{C}^3$ and $p=0$. (at the bottom of page 620).

2

Let me expand my comments into an answer.
Before thinking about vector bundles, it makes sense to see what happens with vector spaces. If $(a,b,c)$ is a basis for a $3$-dimensional vector space $V$, then $\bigwedge^{2}V$ has basis $(a \wedge b,b \wedge c,c \wedge a)$. Then the determinant $\bigwedge^{3}(\bigwedge^{2}V)$ has basis $(a \wedge b) \wedge (b ...

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