User anon - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:25:34Z http://mathoverflow.net/feeds/user/30149 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122596/rigid-monoidal-abelian-category-without-an-exact-tensor-functor-to-vect/122602#122602 Answer by anon for Rigid monoidal abelian category without an exact tensor functor to Vect anon 2013-02-22T05:38:22Z 2013-02-22T05:38:22Z <p>There's a naturally occurring one, namely, the category of motives over a field is a rigid monoidal category without a fiber functor because the rank of the motive of an algebraic variety is its Euler characteristic, which may be negative. To make the category Tannakian, you have to assume that the Kunneth components of the diagonal are algebraic, and change the commutativity constraint.</p> http://mathoverflow.net/questions/117229/etale-site-is-useful-examples-of-using-the-small-fppf-site/117435#117435 Answer by anon for Etale site is useful - examples of using the small fppf site? anon 2012-12-28T18:24:35Z 2012-12-28T18:24:35Z <p>Briefly, to understand $p$-phenomena in characteristic $p$ you need to replace the etale site by the fppf site. For example, to understand the $p$-torsion in the Brauer group you need the $p$-Kummer sequence and the cohomology of $\mu_{p^n}$, and the study of the $p$-torsion in the Tate-Shafarevich group entails the study of the cohomology of finite group schemes of $p$-power order over curves. There are also many purely geometric applications, e.g., to the Picard functor. You should think of the fppf cohomology as being THE cohomology theory, but the etale topology is fine enough for computing the cohomology of smooth group schemes, and the Zariski topology is fine enough for computing the cohomology of coherent sheaves.</p> http://mathoverflow.net/questions/117033/center-of-the-algebraic-group-g-mathbbr-for-a-centerless-g/117040#117040 Answer by anon for Center of the algebraic group G_{\mathbb{R}} for a centerless G anon 2012-12-22T18:06:12Z 2012-12-22T19:15:49Z <p>The short answer is that the center Z(G) of a semisimple algebraic group is a well-defined (finite) algebraic subgroup which commutes with extension of the base field, so if it's trivial over $\mathbb{Q}$, then its trivial over every field. But presumably, that is not what you meant to ask. Perhaps you mean: does $Z(\mathbb{Q})$ trivial imply $Z(\mathbb{R})$ trivial? The answer is no. You have a finite group $Z(\bar{\mathbb{Q}})$ with an action of the absolute Galois group of $\mathbb{Q}$ and you are asking: if only 1 is fixed by the full Galois group is only 1 fixed by complex conjugation? Obviously, not necessarily. It's easy to makes lots of different Galois modules as centers of semisimple groups over $\mathbb{Q}$. Or perhaps you mean: if $Z(G)$ is trivial (as an algebraic subgroup), is the center of $G(\mathbb{R})$ trivial? The answer is yes, because $G(\mathbb{R})$ is dense in $G$ for the Zariski topology. </p> http://mathoverflow.net/questions/122776/property-of-lattices-in-lie-groups/122785#122785 Comment by anon anon 2013-02-24T18:17:44Z 2013-02-24T18:17:44Z The rank one case is exceptional in that there are lots of nonarithmetic irreducible lattices. No one was saying that it isn't important. http://mathoverflow.net/questions/121853/extension-of-unipotent-algebraic-groups Comment by anon anon 2013-02-15T01:24:52Z 2013-02-15T01:24:52Z @Jim In order to define the unipotent radical, you need to know enough about unipotent groups to answer the question. And connectedness is not a problem in any characteristic, at least, not if you are talking about algebraic group schemes. http://mathoverflow.net/questions/121853/extension-of-unipotent-algebraic-groups Comment by anon anon 2013-02-15T00:24:51Z 2013-02-15T00:24:51Z Yes, see for example Milne's notes AGS, XV, 2.5. http://mathoverflow.net/questions/121765/field-of-definition-of-isogenies-of-abelian-varieties Comment by anon anon 2013-02-14T04:30:17Z 2013-02-14T04:30:17Z This is in Mumford's book on Abelian Varieties (for a finite subgroup scheme). Mumford assumes that the ground field is algebraically closed, but the proof doesn't need this. Alternatively, if you are willing to assume that the ground field is perfect, then the statement follows from the algebraically closed case + descent. http://mathoverflow.net/questions/121247/differential-geometry-study-materials/121248#121248 Comment by anon anon 2013-02-09T01:31:52Z 2013-02-09T01:31:52Z It's a good book, but it takes 327 pages to get to metrics. For someone wanting to learn differential &lt;i&gt;geometry&lt;/i&gt;, there are faster routes. http://mathoverflow.net/questions/120951/intersection-cohomology-and-etale-cohomology Comment by anon anon 2013-02-06T18:30:38Z 2013-02-06T18:30:38Z Intersection cohomology (in the context of the etale topology) generalizes the usual l-adic cohomology. With the appropriate choice of the perversity, intersection cohomology gives the usual l-adic cohomology, with or without compact support. http://mathoverflow.net/questions/120553/should-the-etale-cohomology-of-a-smooth-projective-variety-over-rationals-be-se Comment by anon anon 2013-02-02T05:47:20Z 2013-02-02T05:47:20Z According to the Tate conjecture, l-adic realization gives an equivalence of categories from motives tensor $Q_l$ to the category of l-adic Galois representations generated by the cohomology of smooth projective algebraic varieties over $Q$. The standard conjectures imply the first is semisimple, hence also the second. http://mathoverflow.net/questions/120260/a-question-on-the-picard-group Comment by anon anon 2013-01-30T07:59:22Z 2013-01-30T07:59:22Z I'd guess that, by using etale cohomology, you can give an algebraic proof valid over any algebraically closed field (use the same argument you use over $\mathbb{C}$). http://mathoverflow.net/questions/118708/primitive-cohomology-useful Comment by anon anon 2013-01-12T18:09:41Z 2013-01-12T18:09:41Z The question should be, why do we need to decompose the cohomology into its primitive parts? You answered your own question: in order to be able to state the Hodge index theorem, or, more generally, Grothendieck's standard conjectures. http://mathoverflow.net/questions/118141/does-every-polynomial-diophantine-equation-have-solutions-modulo-p/118149#118149 Comment by anon anon 2013-01-06T04:32:11Z 2013-01-06T04:32:11Z My recollection is that they worked with projective varieties, but removing a lower dimensional subvariety is not going to change much. I suggest you look at their article --- it is quite short and readable. http://mathoverflow.net/questions/117700/why-is-the-brauer-group-of-a-local-field-is-mathbb-q-z-is-it-an-accident Comment by anon anon 2013-01-01T04:45:47Z 2013-01-01T04:45:47Z &quot;When Tate was finding local duality, How could he know cup product make dual relation exactly?&quot; Actually, he didn't. He originally proved a duality theorem for abelian varieties over local fields, observed that it implied a local duality for modules occurring in the abelian varieties, and only later realized that the local duality held for all modules. Concerning your general question of why all this holds. Well, we can prove it. I'm sure there a vague philosophical reasons why it must hold, but they are probably not very helpful. http://mathoverflow.net/questions/117229/etale-site-is-useful-examples-of-using-the-small-fppf-site/117435#117435 Comment by anon anon 2012-12-31T04:41:47Z 2012-12-31T04:41:47Z Crystalline cohomology is a Weil cohomology, which explains its importance, but is not very useful for the things I mentioned in my answer. There are relations between flat cohomology and crystalline cohomology, but they are rather complicated to explain. As far as I know, only for the Zariski topology, the etale topology, and topologies in between, do we have a really explicit description of the &quot;points&quot;. http://mathoverflow.net/questions/117033/center-of-the-algebraic-group-g-mathbbr-for-a-centerless-g/117040#117040 Comment by anon anon 2012-12-22T22:17:20Z 2012-12-22T22:17:20Z Actually, having read Jack's comments, I still haven't a clue what he asking.