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I don't think the modified question works either. Let $E$ be a general elliptic curve. Take $X$ to be the quotient of $E\times \mathbb{P}^{1}$ by an involution which is a translation by a point of or …

The statement is easily checked to be true for very stable bundles, so when you look for a counter example you will need to look for a wobbly (i.e. stable but not very stable) bundle. Suppose that $ …

There are many excellent references. It really depends on what you are after. One of the most comprehensive references is the thesis of Olivier Penacchio: http://arxiv.org/abs/math/0307156.

There is a natural pair $(X,\alpha)$ that you can construct. Only in this pair $X$ is not a smooth projective variety but is a smooth orbifold surface. If you choose a genus one pencil $Y \to \mathbb{ …

A slightly better variant of this question is to ask: is the Hurewitz image of $\pi_{2}(X)$ in $H_{2}(X)$ a sub Hodge structure? This is in fact an old question of Philippe Eyssidieux. In section 4.3 …

Yes, this has been achieved in some sense. There is a (unpublished and possibly not yet written) work of Gaitsgory and Lurie where they propose an answer to this question. Given a split reductive grou …

The branch locus in $Y$ need not be a normal crossings divisor even when $Y$ is a projective space. Suppose $X$ is a smooth complex projective surface with non-abelian fundamental group. By Noether no …

I am not completely sure what is meant by this statement since the topology of the torus fibration certainly changes if the fibration doesn't have a section. If on the other hand we follow your prescr …

Here is a very brief explanation of what is going on. You can find more details in the papers that Ana lists and in arxiv.org/abs/math/0604617 as you say. If $G$ is a reductive group, the Hitchin ba …

Quasi-unipotency is a well defined notion at any point of the discriminant. If we have a proper family $f : X \to S$ of varieties with a smooth total space and a smooth base, and if $p \in D \subset S …

I am not sure if you will count this but you have the examples from the other side of geometric Langlands. On any smooth variety the skyscraper sheaves of points are eigensheaves for the tensorization …

You will get such a pencil of genus two curves provided that the base point of the pencil of lines is not one of the nine points that you blew up to get your $E(1)$. But if the goal was to construct …

Do you want your K3 to be smooth? In that case the answer is no. For the double cover to have a trivial canonical class you will have to chose a branch divisor which is a section in half of the anti-c …

It appears that you are assuming that your varieties $C_{i}$ are smooth (you seem to assume that since you are talking about the tangent bundle). In this case each $C_{i}$ is an elliptic curve (I gues …

If $c$ is the codimension of $Z$ in $Y_{2}$, then $$\mathcal{E}xt^{i}_{X}(\mathcal{O}_{Y_{1}},\mathcal{O}_{Y_{2}}) = \wedge^{c} N_{Z/Y_{2}}\otimes \wedge^{i -c} \left( N_{Z/X}/N_{Z/Y_{2}}\right).$$ …