User abdó roig-maranges - MathOverflow

Answer by Abdó Roig-Maranges for Bimodules in geometry

2009-10-23T08:44:27Z

The Fourier-Mukai transform comes from a bimodule: the Poincaré bundle. Let A be an abelian variety, the Poincaré bundle P is a vector bundle on AxÂ coming from the fact that the points in the dual abelian variety Â parametrize line bundles on A (P is the universal family). In the Fourier-Mukai construction, P is used as a O_A-O_Â-bimodule to produce a functor between the derived categories of coherent sheaves on A and Â via a push-pull construction.

Answer by Abdó Roig-Maranges for Undergraduate Level Math Books

2009-10-20T18:02:51Z

As an undergraduate I loved Shafarevich's book Basic notions of algebra. This is not a textbook, but gives small beautiful tastes of a broad choice of topics in algebra, emphasizing connections with other fields.

I found it very stimulating, in the sense that every example or overview of some topic in this book made me want to learn more details about it. In fact I became interested in algebraic geometry because of this book.

Answer by Abdó Roig-Maranges for Functorial characterization of open subschemes?

2009-10-20T12:43:27Z

The abelian category of quasicoherent sheaves on a schemes determine the scheme. This is an old result of Gabriel ("des categories abeliennes" 1962), proved in full generality by Rosenberg. This means that, QCoh(X) does not only tell you the open subschemes of X but also gives you the structure sheaf ! I've known this result for some time but I had never looked at it in detail until today. I'll sketch what I have just learned hoping not to make big mistakes...

An abelian subcategory B of an abelian category A is said to be a thick subcategory if it is full and for any exact sequence in A

0 ---> M'---> M ---> M'' ---> 0

M belongs to B if, and only if M' and M'' do.

If B is a thick subcategory of A there is a well defined localization A/B, which is again an abelian category. A/B has the same objects as A and a morphism f:M--->N in A/B is an isomorphism if, and only if ker f and coker f belong to B.

Let T:A ---> A/B be the localization functor. Then B is said to be a localizing subcategory if B is thick and T has a right adjoint. The condition of being localizing can be rephrased only in terms of A and B. see Gabriel's thesis above (proposition 4 in chapter III).

Finally, if M is an object of A, we denote by <M> the smallest localizing subcategory containing M.

Now let X be a scheme, j:U ---> X an open embedding and i:Y ---> X its closed complement. Then there is a bunch of adjunctions between the categories of quasicoherent sheaves of U,X,Y: i_*:QCoh(Y) ---> QCoh(X) has a left adjoint i^*:QCoh(X) ---> QCoh(Y) and a right adjoint i^!:QCoh(X) ---> QCoh(Y). On the other hand, the functor j^*:QCoh(X) ---> QCoh(U) has a left adjoint j_!:QCoh(U)--->QCoh(X) and a right adjoint j_*:QCoh(U) ---> QCoh(X). This is sometimes called a recollement.

Let's assume that X is Noetherian and let A = QCoh(X). We have an exact sequence of abelian categories

0 ---> QCoh(Y) ---> A ---> QCoh(U) ---> 0

in the sense that the category QCoh(Y) happens to be a localizing subcategory of A and its quotient is identified with QCoh(U). The first map in the exact sequence is i_* and the second j^*. Moreover, I think that QCoh(Y) is the smallest localizing subcategory of QCoh(X) containing i_*O_Y. Gabriel proves that there are no more such localizing subcategories, that is closed subschemes of X correspond exactly to localizing subcategories <M> generated by a single coherent sheaf (i.e. Noetherian object in A). Moreover, irreducible closed subsets (the points in the underlying topological space of X) are given by localizing subcategories <I> for I an indecomposable injective. We have described the points of X and its closed sets in terms of only the category A, so we can recover the underlying topological space of X from A.

In particular, an open subscheme U of X gives a complementary closed subscheme Y, which is in correspondence with a localizing subcategory <M> and, moreover, QCoh(U) = A/<M>. So, responding to the queston above, for any f:U ---> X, U is an open subscheme if, and only if the kernel of f^*:QCoh(X) ---> QCoh(U) is a localizing subcategory of the form <M> for a coherent sheaf M.

Regarding the structure sheaf O_X there is an isomorphism O_X(U) and the ring of endomorphism of the identity functor on QCoh(U) (which happens to be A/QCoh(Y)), so the structure sheaf can be recovered only in terms of the category A.

Finally, just say that there are other results in the spirit of reconstructing a scheme from some category of sheaves on it. This is the starting point for using such categories of sheaves as a definition of noncommutative scheme. There is more information on this entry in nlab.

Answer by Abdó Roig-Maranges for algebraic K-theory and tensor products

2009-10-18T14:45:16Z

I don't have a complete answer to this. However, there is an argument (which I have not checked carefully, but I believe it works) to proves that K(XxY) = K(X) /\^L K(Y) when X, Y are smooth schemes over k, and one of them (say Y) is a linear variety. Here /\^L is the derived smash product over K(Spec k).

The class of linear varieties is the smallest class of quasi-projective varieties such that

Affine spaces are linear,
Let X be a variety, U an open subvariety and Y its closed complement. If Y and either U or X is linear, so is the other.

For example, any toric variety is linear.

Now using the localization exact triangle for the variety Y, the homotopy invariance of K-theory of smooth schemes (i.e. K(XxA^k) = K(X)) and the fact that derived-smashing with K(X) preserves exact triangles, I believe one can use an inductive five-lema to show that K(XxY) = K(X) /\^L K(Y).

Maybe this argument can be extended to deal with more general fibre products over a general base S. But as it uses homotopy invariance of K-theory, which does not hold for singular schemes, and as Xx_SY may be singular, this might lead to trouble.

This is a very special case though (Y is very special). For a general Y this result will not be true.

Answer by Abdó Roig-Maranges for Motivation/interpretation for Quillen's Q-construction?

2009-10-18T13:19:35Z

Expanding on Tyler Lawson's comment on the Q-construction, I would say the following. The K₀ functor involves two processes, a group completion (of the monoid structure given by direct sum) and an identification of all the extensions of any two objects. That is, K_0 of an exact category E is the free abelian group generated by the objects of E (group completion) together with the relations [B] = [A] + [C] for every exact sequence

A >---> B --->> C

Now, the higher K-theory is a sort of categorification, making both processes above remember higher homotopical data. Then we make the definition

K_i(E) = pi_i Omega BQE

Here Omega B corresponds to a homotopical group completion as explained by Tyler. Quillen's Q-construction changes the morphisms of the category E in a way that when group-completed, "the extensions become split" (giving the relation [B] = [A] + [C] above). Strictly speaking, the Q-construction does not make exact sequences split in QE, as QE has the same isomorphisms as E, only the non-isos change: a morphism A ---> B in QE correspond to an identification of A with a subquotient of B.

Answer by Abdó Roig-Maranges for Embedding abelian categories to have enough projectives

2009-10-17T16:47:36Z

It seems that Pro(A) does not have enough projectives in general. In Kashiwara-Schapira's book "Categories and Sheaves" they prove (corollary 15.1.3) that Ind(k-Mod) does not have enough injectives. This means, taking opposite categories, that Pro(k-Mod^{op}) does not have enough projectives.

I don't know of any universal way of adding enough projectives.

Answer by Abdó Roig-Maranges for An "existence contra partition of unity" statement for integer matrices?

2009-10-17T12:34:51Z

I believe that what you say is true. I'll sketch an argument.

Let f:Zⁿ ---> Z²ⁿ be the map of free Z-modules given by the matrices B₁, B₂ put in column (i.e. the direct sum of the morphisms given by B₁ and B₂). Now we rephrase conditions (1) and (2) in a slightly more abstract way:

(1) fails to hold if, and only if there exists p:Z²ⁿ ---> Zⁿ such that, together with f, fit in a short exact sequence

0 ---> Zⁿ ---> Z²ⁿ ---> Zⁿ ---> 0 (*)

Indeed, the failure of (1) means that any v in Qⁿ such f(v) in Z²ⁿ must be integral (i.e. v in Zⁿ). In particular, this implies that f is injective. Moreover, take w in Z²ⁿ representing a nonzero torsion element in the cokernel of f. As w represents a torsion element, Nw belongs to the image of f for some big enough positive integer N, so there is v in Zⁿ such that f(v) = Nw. But now f(1/N v) = w, and this means, by the failure of (1), that 1/N v is integral, so w is in the image of f and the cokernel of f has no torsion. As a finitely generated torsion-free Z-module is free, we get an exact sequence like (*) above. This argument can easily be reversed, to show the equivalence between the existence of this exact sequence and the failure of (1).

(2) holds if, and only if there exists a morphism of Z-modules r:Z²ⁿ ----> Zⁿ such that rf = id.

Let r be represented by a matrix (A₁,A₂). Then gf has matrix A₁B₁ + A₂B₂, and gf = id if, and only if (2) holds.

Now, the proof of what you asked for is easy. (1) fails if, and only if we can form the exact sequence (*), but such an exact sequence is always split because Z^n is projective, so we can form such exact sequence if, and only if there exists a splitting r:Z²ⁿ ----> Zⁿ, which is exactly condition (2).

Answer by Abdó Roig-Maranges for Non-zero sheaf cohomology

2009-10-16T17:48:06Z

The sheaf cohomology Hⁱ(X,F) of a (topological) manifold X of dimension n vanishes for i > n. This is a topological version of Grothendieck's vanishing theorem above. You can find this result in Kashiwara-Schapira's "Sheaves on manifolds" proposition III.3.2.2.

Comment by Abdó Roig-Maranges

2010-04-02T20:49:26Z

The inclusion of an open affine, as well as the map $\mathbb{A}^1\sqcup \mathbb{A}^1 \to \mathbb{P}^1$ are flat. The thing is that the theorem of Ehresmann applies only when the map is proper.

Comment by Abdó Roig-Maranges

2009-10-19T16:23:44Z

There are different ways to make intermediate Jacobians. But as far as I know, either they produce abelian varieties, that is they come with a polarization (the Weil ones) or deform holomorphically in families (the Griffiths ones), but not both at the same time.