User abd&#243; roig-maranges - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:00:29Z http://mathoverflow.net/feeds/user/322 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1931/bimodules-in-geometry/2056#2056 Answer by Abdó Roig-Maranges for Bimodules in geometry Abdó Roig-Maranges 2009-10-23T08:44:27Z 2009-10-23T08:44:27Z <p>The <a href="http://en.wikipedia.org/wiki/Mukai-Fourier_transform" rel="nofollow">Fourier-Mukai transform</a> 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<sub>A</sub>-O<sub>Â</sub>-bimodule to produce a functor between the derived categories of coherent sheaves on A and Â via a push-pull construction.</p> http://mathoverflow.net/questions/761/undergraduate-level-math-books/1463#1463 Answer by Abdó Roig-Maranges for Undergraduate Level Math Books Abdó Roig-Maranges 2009-10-20T18:02:51Z 2009-10-20T18:08:11Z <p>As an undergraduate I loved Shafarevich's book <a href="http://www.amazon.com/Basic-Notions-Algebra-I-R-Shafarevich/dp/3540612211" rel="nofollow">Basic notions of algebra</a>. This is not a textbook, but gives small beautiful tastes of a broad choice of topics in algebra, emphasizing connections with other fields.</p> <p>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.</p> http://mathoverflow.net/questions/1359/functorial-characterization-of-open-subschemes/1417#1417 Answer by Abdó Roig-Maranges for Functorial characterization of open subschemes? Abdó Roig-Maranges 2009-10-20T12:43:27Z 2009-10-20T12:43:27Z <p>The abelian category of quasicoherent sheaves on a schemes determine the scheme. This is an old result of Gabriel ("<a href="http://www.numdam.org/item?id=BSMF_1962__90__323_0" rel="nofollow">des categories abeliennes</a>" 1962), proved in full generality by <a href="http://www.mpim-bonn.mpg.de/preprints/send?bid=3948" rel="nofollow">Rosenberg</a>. 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...</p> <p>An abelian subcategory B of an abelian category A is said to be a <b>thick subcategory</b> if it is full and for any exact sequence in A</p> <p>0 ---> M'---> M ---> M'' ---> 0</p> <p>M belongs to B if, and only if M' and M'' do.</p> <p>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.</p> <p>Let T:A ---> A/B be the localization functor. Then B is said to be a <b>localizing subcategory</b> 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).</p> <p>Finally, if M is an object of A, we denote by &lt;M&gt; the smallest localizing subcategory containing M. </p> <p>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<sub>* </sub>:QCoh(Y) ---> QCoh(X) has a left adjoint i<sup>* </sup>:QCoh(X) ---> QCoh(Y) and a right adjoint i<sup>! </sup>:QCoh(X) ---> QCoh(Y). On the other hand, the functor j<sup>* </sup>:QCoh(X) ---> QCoh(U) has a left adjoint j<sub>! </sub>:QCoh(U)--->QCoh(X) and a right adjoint j<sub>* </sub>:QCoh(U) ---> QCoh(X). This is sometimes called a recollement.</p> <p>Let's assume that X is Noetherian and let A = QCoh(X). We have an exact sequence of abelian categories</p> <p>0 ---> QCoh(Y) ---> A ---> QCoh(U) ---> 0</p> <p>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<sub>* </sub> and the second j<sup>* </sup>. Moreover, I think that QCoh(Y) is the smallest localizing subcategory of QCoh(X) containing i<sub>* </sub>O<sub>Y</sub>. Gabriel proves that there are no more such localizing subcategories, that is closed subschemes of X correspond exactly to localizing subcategories &lt;M&gt; 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 &lt;I&gt; 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.</p> <p>In particular, an open subscheme U of X gives a complementary closed subscheme Y, which is in correspondence with a localizing subcategory &lt;M&gt; and, moreover, QCoh(U) = A/&lt;M&gt;. So, responding to the queston above, for any f:U ---> X, U is an open subscheme if, and only if the kernel of f<sup>* </sup>:QCoh(X) ---> QCoh(U) is a localizing subcategory of the form &lt;M&gt; for a coherent sheaf M.</p> <p>Regarding the structure sheaf O<sub>X</sub> there is an isomorphism O<sub>X</sub>(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.</p> <p>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 <a href="http://ncatlab.org/nlab/show/noncommutative+algebraic+geometry" rel="nofollow">this entry</a> in nlab.</p> http://mathoverflow.net/questions/984/algebraic-k-theory-and-tensor-products/1038#1038 Answer by Abdó Roig-Maranges for algebraic K-theory and tensor products Abdó Roig-Maranges 2009-10-18T14:45:16Z 2009-10-18T14:45:16Z <p>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) /\<sup>L</sup> K(Y) when X, Y are smooth schemes over k, and one of them (say Y) is a linear variety. Here /\<sup>L</sup> is the derived smash product over K(Spec k).</p> <p>The class of linear varieties is the smallest class of quasi-projective varieties such that</p> <ol> <li>Affine spaces are linear,</li> <li>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.</li> </ol> <p>For example, any toric variety is linear.</p> <p>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) /\<sup>L</sup> K(Y).</p> <p>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<sub>S</sub>Y may be singular, this might lead to trouble.</p> <p>This is a very special case though (Y is very special). For a general Y this result will not be true.</p> http://mathoverflow.net/questions/1006/motivation-interpretation-for-quillens-q-construction/1032#1032 Answer by Abdó Roig-Maranges for Motivation/interpretation for Quillen's Q-construction? Abdó Roig-Maranges 2009-10-18T13:19:35Z 2009-10-18T13:19:35Z <p>Expanding on Tyler Lawson's comment on the Q-construction, I would say the following. The K<sub>0</sub> 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</p> <p>A >---> B --->> C</p> <p>Now, the higher K-theory is a sort of categorification, making both processes above remember higher homotopical data. Then we make the definition</p> <p>K<sub>i</sub>(E) = pi<sub>i</sub> Omega BQE</p> <p>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.</p> http://mathoverflow.net/questions/827/embedding-abelian-categories-to-have-enough-projectives/892#892 Answer by Abdó Roig-Maranges for Embedding abelian categories to have enough projectives Abdó Roig-Maranges 2009-10-17T16:47:36Z 2009-10-17T16:47:36Z <p>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.</p> <p>I don't know of any universal way of adding enough projectives.</p> http://mathoverflow.net/questions/730/an-existence-contra-partition-of-unity-statement-for-integer-matrices/865#865 Answer by Abdó Roig-Maranges for An "existence contra partition of unity" statement for integer matrices? Abdó Roig-Maranges 2009-10-17T12:34:51Z 2009-10-17T12:34:51Z <p>I believe that what you say is true. I'll sketch an argument.</p> <p>Let f:Z<sup>n</sup> ---> Z<sup>2n</sup> be the map of free Z-modules given by the matrices B<sub>1</sub>, B<sub>2</sub> put in column (i.e. the direct sum of the morphisms given by B<sub>1</sub> and B<sub>2</sub>). Now we rephrase conditions (1) and (2) in a slightly more abstract way:</p> <ul> <li><p>(1) fails to hold if, and only if there exists p:Z<sup>2n</sup> ---> Z<sup>n</sup> such that, together with f, fit in a short exact sequence</p> <p>0 ---> Z<sup>n</sup> ---> Z<sup>2n</sup> ---> Z<sup>n</sup> ---> 0 (*)</p></li> </ul> <p>Indeed, the failure of (1) means that any v in Q<sup>n</sup> such f(v) in Z<sup>2n</sup> must be integral (i.e. v in Z<sup>n</sup>). In particular, this implies that f is injective. Moreover, take w in Z<sup>2n</sup> 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 Z<sup>n</sup> 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). </p> <ul> <li>(2) holds if, and only if there exists a morphism of Z-modules r:Z<sup>2n</sup> ----> Z<sup>n</sup> such that rf = id.</li> </ul> <p>Let r be represented by a matrix (A<sub>1</sub>,A<sub>2</sub>). Then gf has matrix A<sub>1</sub>B<sub>1</sub> + A<sub>2</sub>B<sub>2</sub>, and gf = id if, and only if (2) holds.</p> <p>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:Z<sup>2n</sup> ----> Z<sup>n</sup>, which is exactly condition (2).</p> http://mathoverflow.net/questions/274/non-zero-sheaf-cohomology/770#770 Answer by Abdó Roig-Maranges for Non-zero sheaf cohomology Abdó Roig-Maranges 2009-10-16T17:48:06Z 2009-10-16T17:48:06Z <p>The sheaf cohomology H<sup>i</sup>(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 "<a href="http://books.google.com/books?id=qfWcUSQRsX4C&amp;lpg=PA475&amp;ots=Dk-Z-JgqhP&amp;dq=kashiwara%20schapira%20sheaves%20on%20manifolds&amp;pg=PP1#v=onepage&amp;q=&amp;f=false" rel="nofollow">Sheaves on manifolds</a>" proposition III.3.2.2.</p> http://mathoverflow.net/questions/20184/flatness-in-algebraic-geometry-vs-fibration-in-topology/20187#20187 Comment by Abdó Roig-Maranges Abdó Roig-Maranges 2010-04-02T20:49:26Z 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. http://mathoverflow.net/questions/794/examples-of-rational-families-of-abelian-varieties/1230#1230 Comment by Abdó Roig-Maranges Abdó Roig-Maranges 2009-10-19T16:23:44Z 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.