User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:07:26Z http://mathoverflow.net/feeds/user/10849 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62300/morphism-of-artin-stacks/62316#62316 Answer by profilesdroxford54 for Morphism of Artin stacks profilesdroxford54 2011-04-19T18:50:59Z 2011-04-19T20:05:17Z <p>No. $B\mathbb{G}_m\to\mathrm{pt}$ is surjective with source of dimension -1.</p> <p>As commented below this is nonsense, since this is not representable. </p> <p>Let me try to make amends for my flip response to the question. Of course what I am about to write could be equally stupid. However, $\phi$ surjective means that the induced map $\mathcal{F}\times_{\mathcal{G}}G\to G$ is surjective, where $G\to \mathcal{G}$ is smooth surjective with $G$ a scheme. Suppose that the dimension of $G$ is $n$ and that of the scheme $\mathcal{F}\times_{\mathcal{G}}G$ is $n+p$ where $p\geq 0$. Then the dimension of $\mathcal{G}$ is $n-q$ where $q$ is the relative dimension of the smooth surjective morphism $G\times_{\mathcal{G}}G\to G$. Now $\mathcal{F}\times_{\mathcal{G}}G\to\mathcal{F}$ is smooth and surjective. Furthermore: $$(\mathcal{F}\times_{\mathcal{G}}G)\times_\mathcal{F}(\mathcal{F}\times_{\mathcal{G}}G)\simeq (\mathcal{F}\times_{\mathcal{G}}G)\times_G(G\times_\mathcal{G}G)$$ and hence is of relative dimension $q$ over $(\mathcal{F}\times_{\mathcal{G}}G)$. Therefore $$\mathrm{dim}(\mathcal{F})= n+p-q\geq n-q=\mathrm{dim}(\mathcal{G}).$$</p> http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve/62120#62120 Answer by profilesdroxford54 for A question on K_1 of an elliptic curve profilesdroxford54 2011-04-18T13:19:35Z 2011-04-18T20:11:36Z <p>This is not a full answer, more a lenghty comment, since I think the key part of your question is whether there are elements in $K_1(\mathcal{E})^{(2)}$ with non-trivial regulator.</p> <p>Conjecturally $K_1(\mathcal{E})^{(2)}$ is finitely generated -- this a particular case of what is known as Bass's conjecture which is that the K-theory of a regular f.g. $\mathbb Z$-algebra is finitely generated in each degree. For rings of integers in global fields this is a theorem of Quillen, but as I wrote in my comment above, I don't believe that this is known for any non-rational arithmetic surfaces. The map $K_1(\mathcal{E})^{(2)} \to K_1(E)^{(2)}$ is far from being bijective, since by the localization sequence its cokernel is the kernel of the map $$\bigoplus_p K'_0(\mathcal{E}_p)^{(1)}\to K_0(\mathcal{E})^{(2)}$$ where $\mathcal{E}_p$ is the curve mod $p$. By Bloch-Kato-Saito, the target of this map is finite, while the source is an infinite sum of non-trivial groups. This is analogous to the situation for number fields.</p> <p>There is a completely trivial reason that $r\circ \iota$ is not onto: the source is countable while the target is not. As I wrote above what is of more interest is to exhibit elements of $K_1(\mathcal{E})^{(2)}$ with non-trivial regulator. My impression is that the standard method of constructing elements that I described in the comment should give such elements, but I do know where this might have been done. </p> <p>Additional Comments. </p> <p>(1) When you write "surjective after tensoring with $R$" I guess you really mean does the image generate the real vector space $H_D^3(E_{/ \mathbb{R}} , \mathbb{R}(2) )$? </p> <p>(2) I think one can construct elements of $K_1^{(2)}$ with non trivial regulator as follows. Take a curve $E$ with rank at least one, so that there is a rational point $P$ which is not torsion. Take a conjugate pair of points $Q_1$, $Q_2$ in a real quadratic extension $F$ of $\mathbb Q$ such that $Q_1+Q_2+P=0$ in the elliptic curve (such points exist by taking a line with rational slope through $P$, when we embed $E$ in the projective plane). Now take a non trivial unit $\alpha$ in the ring of integers $\mathcal{O}_F$ of $F$. The pair $(Q_1,Q_2)$ determines a point $q:Spec(\mathcal{O}_F)\to \mathcal{E}(\mathcal{O}_F)$. Push $\alpha$ forward by $q$. Then I think (but have not double checked) that the regulator of this class will be non-zero, and essentially equal to the regulator of $\alpha$. </p> http://mathoverflow.net/questions/58471/analytic-torsion-in-the-derived-category/58474#58474 Answer by profilesdroxford54 for Analytic Torsion in the Derived Category profilesdroxford54 2011-03-14T20:52:16Z 2011-03-14T20:52:16Z <p>What does depend on the "derived category" is the notion of determinant line. Thus take any local system $\mathcal E$ of $R$-modules on $M$ (though one should probably work with constructible sheaves of some kind), then compute cohomology any way you want -- de Rham, etc. So long as the the cohomology is a perfect complex of $R$ modules, it has a determinant -- a free rank one projective $R$-module.</p> <p>So far this has nothing to do with the Riemannanian structure. Now suppose there is a second structure -- say $M$ is Riemannian and $\mathcal E$ is a local system of inner product spaces, where $R$ is a subring of the reals. Then you can view the DeRham cx of $\mathcal E$ as a perfect complex of (infinite dimensional) inner-product spaces, and the zeta normalization gives a way of defining the determinant of this as a one dimensional inner product space.</p> <p>The torsion is the relationship between these two structures. EG if $R={\mathbf Z}$ then it is the real number given by the length of a generator of the line.</p> <p>To give this part of the construction a more "derived category" feel, one would need a notion of quasi-isomorphism "preserving inner products". I am not sure anyone has really thought hard about this.</p> http://mathoverflow.net/questions/56207/grothendieck-riemann-roch-involving-higher-k/56230#56230 Answer by profilesdroxford54 for Grothendieck Riemann Roch involving Higher K ? profilesdroxford54 2011-02-22T00:14:44Z 2011-02-22T00:14:44Z <p>MR0624666 (83m:14013) Gillet, Henri Riemann-Roch theorems for higher algebraic K-theory. Adv. in Math. 40 (1981), no. 3, 203–289.</p> http://mathoverflow.net/questions/54832/extending-holomorphic-connections/54833#54833 Answer by profilesdroxford54 for Extending holomorphic connections profilesdroxford54 2011-02-08T23:13:08Z 2011-02-08T23:13:08Z <p>Connections on the punctured disk extend (up to isomorphism) to connections with log poles, but not in general to holomorphic connections. </p> <p>A standard reference is: Pierre Deligne. "Equations differentielles a points singuliers reguliers. Springer-Verlag, Berlin, 1970. Lecture Notes in Mathematics, Vol. 163.</p> <p>I'd say the key point is that in one dimension the connection is integrable corresponds to a local system defined by its monodromy - an $r\times r$ matrix A, which will have a logarithm.</p> <p>Perhaps someone else can suggest a more recent exposition.</p> http://mathoverflow.net/questions/54028/are-affine-groups-over-rings-of-integers-finitely-generated/54044#54044 Answer by profilesdroxford54 for Are affine groups over rings of integers finitely generated? profilesdroxford54 2011-02-02T00:20:21Z 2011-02-02T00:20:21Z <p>It seems to me that if you are thinking of affine groups, then the appropriate result is that $S$-arithmetic subgroups of reductive linear algebraic groups over number fields are finitely generated. Over function fields there are exceptions (which I think are known explicitly). </p> <p>This includes examples such as $SL_2(\mathbb{Z})$, as well as $S$-units in number fields. </p> <p>For number fields the proof depends on the existence of compact (equivariant) retracts of fundamental domains for the action of the group on suitable spaces -- reduction theory, combined with the finite generation of the group of units.</p> <p>However the Mordell-Weil theorem (finite generation of rational (or equivalently integral) points on an abelian variety, is a rather deeper result. </p> http://mathoverflow.net/questions/52702/adams-graded-parts-of-rational-k-theory-of-a-number-field/53112#53112 Answer by profilesdroxford54 for Adams graded parts of rational K-theory of a number field. profilesdroxford54 2011-01-24T19:39:12Z 2011-01-24T19:39:12Z <p>One way of viewing Borel's theorem is that if $F$ is a number field the $p$-th Chern class with values in Deligne cohomology is a map from $K_{2p-1}(F)\to H^{1}_{\mathrm{Deligne}}(\mathrm{Spec}(F),\mathbb{R}(p))$ which embeds the $K$-group (mod torsion) as a lattice in a real vector space of the appropriate rank. (The point being that these Chern classes "are" the Borel regulator -- there is a nice book by Burgos explaining this; note that one must also take account of the action of complex conjugation.) But the p-th Chern class has the property that it converts the n-th Adams operation into multiplication by $n^p$. So the fact that a single Chern class detects $K_m(F)$ mod torsion implies that $K_m$ has pure weight.</p> http://mathoverflow.net/questions/50700/deformation-theory-over-the-field-of-algebraic-numbers/51429#51429 Answer by profilesdroxford54 for Deformation theory over the field of algebraic numbers profilesdroxford54 2011-01-07T20:00:55Z 2011-01-07T20:00:55Z <p>The answer is yes, with the modification that the descent may not be from $B$ to $\overline{\mathbb Q}$ but from some etale cover of $B$. I.e. it really is a result about algebraically closed fields of definition. (The intuition being that Kodaira-Spencer trivial implies that a family is isotrivial).</p> <p>The following was first proved by Buium:</p> <p><strong>Theorem</strong> Let $X$ be a variety, proper over an algebraically closed ﬁeld $K$. Then $X$ is deﬁned over the ﬁxed ﬁeld of the set of all derivations of $K$ which lift to derivations of the structure sheaf of $X$.</p> <p>See: </p> <p>Buium, Alexandru; Diﬀerential function ﬁelds and moduli of algebraic varieties. Lecture Notes n Mathematics, 1226. Springer-Verlag, Berlin, 1986.</p> <p>Buium, Alexandru; Fields of deﬁnition of algebraic varieties in characteristic zero. Compositio Math. 61 (1987), no. 3, 339–352.</p> <p>and also:</p> <p>Gillet, Henri; "Differential algebra - A Scheme Theory Approach", in <em>Differential algebra and related topics: proceedings of the International Workshop, Newark Campus of Rutgers, The State University of New Jersey, 2-3 November 2000</em>, Editors Li Guo, William F. Keigher, World Scientific</p> <p>The converse statement is of course trivial</p> http://mathoverflow.net/questions/34801/dirichlets-regulator-vs-beilinsons-regulator/51357#51357 Answer by profilesdroxford54 for Dirichlet's regulator vs Beilinson's regulator profilesdroxford54 2011-01-06T23:48:43Z 2011-01-06T23:48:43Z <p>As you comment, the Beilinson regulator is defined using Chern classes for Deligne cohomology. In particular, for $K_1(\mathbb C)$, the first Chern class induces the identity from $K_1(\mathbb C)\simeq \mathbb C^*$ to $H^1_{Deligne}(pt, \mathbb Z(1))\simeq \mathbb C^*$. One passes to $H^1_{Deligne}(pt, \mathbb R(1))\simeq \mathbb R$ by taking the logarithm. So the Belinson regulator for $K_1$ is induced by the map $\log |\; |:K_1(\mathbb C)\to \mathbb R$.</p> http://mathoverflow.net/questions/51065/explain-the-relation-between-k-0-and-morphisms-of-chow-motives/51075#51075 Answer by profilesdroxford54 for Explain the relation between $K_0$ and morphisms of Chow motives profilesdroxford54 2011-01-03T23:49:03Z 2011-01-03T23:49:03Z <p>If we consider just the question of constructing the map from <code>$K_0(X)$</code> to <code>$Hom_{Chow\otimes \mathbb Q}(M(X),M(\mathbb P^n))$</code>, and leave the question of proving that it is an isomorphism to later, I would say that this follows directly from the expression for the Chern classes of a vector bundle via the splitting principle. I.e., given a vector bundle $E$ of rank $d$ on $X$ look at the $d$-th power of the tautological bundle <code>$\xi$</code> on <code>$\mathbb P(E)$</code> as usual. This gives a cycle on <code>$\mathbb P(E)$</code> which can be written as a linear combination of chern classes times lower powers of <code>$\xi$</code>. We can identify $CH^d(\mathbb P(E))$ with the appropriate $Hom$ of motives by, for example, embedding <code>$\mathbb P(E)$</code> into <code>$X\times\mathbb P^n$</code> for some large <code>$n$</code>. (Assuming we don't want to simply appeal to the expression for the motive of a projective bundle.) So it seems to me that if the map is given by sending $E$ to $\xi^d$, viewed as a morphism of motives. I imagine that one can then get properties like Whitney sum in terms of this homomorphism of motives by the classical geometric arguments. (So that the usual expression in terms of individual Chern classes would only appear after the fact.) </p> <p>I don't see a way to prove that the map is an isomorphism without going through an analog of one of the usual proofs.</p> <p>Not sure if this is the kind of thing you were looking for.</p> http://mathoverflow.net/questions/43507/impact-of-poincare-duality-on-functional-equation/46160#46160 Answer by profilesdroxford54 for impact of Poincaré duality on functional equation profilesdroxford54 2010-11-15T22:54:34Z 2010-11-15T23:23:16Z <p>Perhaps I am missing something here. But my quick reaction is that surely they are the same. Assuming that $D$ is the full Grothendieck-Verdier duality functor, then <code>$H^i(\overline{X},\mathcal{F})^*\simeq H^{2n-i}(\overline{X},D(\mathcal{F}))$</code> (Here base change from <code>$\mathbb{F}_q$</code> t<code>$D(\mathcal{F})$</code>o its algebraic closure commutes with duality).</p> <p>Then since the characteristic polynomial of an endomorphism of a vector space is the same as that of its transpose, and n is even, nothing changes.</p> <p>One (possibly esoteric) way of thinking about the (cohomological) L-function, is as action induced by <code>$1-t\sigma$</code> (where $\sigma$ is Frobenius) on the determinant line (tensored with <code>$\mathbb{Q}_l(t)$</code>, strictly speaking) of the perfect complex $R\pi_!(\mathcal{F})$ of <code>$\mathbb{Q}_l$</code> vector spaces. Then the determinant of the dual is the dual of the determinant -- but these are both scalars acting on dual lines and so are the same.</p> <p>PS Perhaps what you really want is the proof of the functional equation along these lines. The key point is to not only replace <code>$\mathcal{F}$</code> by <code>$D(\mathcal{F})$</code> but also to replace $t$ by $1/(q^nt)$.</p> http://mathoverflow.net/questions/46116/a-valued-points-of-projective-space/46129#46129 Answer by profilesdroxford54 for A-valued points of projective space profilesdroxford54 2010-11-15T18:03:37Z 2010-11-15T18:22:54Z <p>Indeed, the functor from commutative rings to sets: $A\mapsto {(n+1)\mbox{-tuples of elements of }A\mbox{ that generate the unit ideal }}$ is the scheme <code>$\mathbb{A}_\mathbb{Z}^{n+1}\backslash \{0\}$</code>.</p> <p>The exercise is not even true over field -- after all you have to impose an equivalence relation on the <code>$(n+1)$</code>-tuples; for general rings we have:</p> <p>The <em>quotient</em> of <code>$\mathbb{A}_\mathbb{Z}^{n+1}\backslash \{0\}$</code> by the obvious action of $\mathbb{G}_m$ is <code>$\mathbb{P}^n_\mathbb{Z}$</code></p> <p>This is what the exercise should have asked.</p> http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve/62120#62120 Comment by 2011-04-21T03:07:42Z 2011-04-21T03:07:42Z @Fran&#231;ois As for whether the element in $K_1(E)$ is &quot;integral&quot; -- if you mean that it comes from $K_1$ of a regular model, the answer is yes. http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve/62120#62120 Comment by 2011-04-21T03:06:54Z 2011-04-21T03:06:54Z @Fran&#231;ois Since all the objects in question are $mathbf Z$ modules, I rather think we are tensoring over $math bf Z$. $mathbf R$ is arational vector space of uncountable dimension -- so $\mathbf{R}\otimes\mathbf{R}$ is rather large. For example, if one takes the regulator for a real quadratic field, it maps the group of units of the ring of integers to a free rank one subgroup of the reals, which therefore &quot;spans&quot; the reals - BUT the map is NOT surjective after tensoring with $\mathbf R$. http://mathoverflow.net/questions/62300/morphism-of-artin-stacks/62316#62316 Comment by 2011-04-19T22:24:35Z 2011-04-19T22:24:35Z The argument does not assert that the \emph{automorphisms} over $\mathcal F$ and $\mathcal G$ have the same dimension. You should think of the fiber of $G\times_{\mathcal{G}}G\to G$ over a point $g$ of $G$ as the morphisms with source (or target) $g$. That of course combines that dimension of the automorphisms with the dimension of the objects -- and is better behaved, since the dimension of the space of automorphisms can (and does) jump. http://mathoverflow.net/questions/62300/morphism-of-artin-stacks/62316#62316 Comment by 2011-04-19T19:36:26Z 2011-04-19T19:36:26Z Good point, I wasn't paying attention. http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve/62120#62120 Comment by 2011-04-19T17:44:43Z 2011-04-19T17:44:43Z I meant size OF order $p$ in the comment above. http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve/62120#62120 Comment by 2011-04-19T17:44:04Z 2011-04-19T17:44:04Z $K'_0(\mathcal{E}_p)^{(1)}$ can be more clearly be written as the Chow homology group of 0-dimensional cycles. This will always be $\mathbf Z$ (for the degree) plus a torsion group, which by Hasse (i.e. the Weil conjectures) has size or order $p$. So the cokernel will contain a copy of $\mathbf{Q}^*$ (which you take to be sitting at the zero-section of the elliptic curve, plus, in all likelihood, a large torsion group. As for tensoring with $\mathbf R$ -- if it isn't surjective before doing so, it will not be afterwards. http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve Comment by 2011-04-15T22:49:29Z 2011-04-15T22:49:29Z I believe that one does not know finite generation for these groups. As for explicit elements, $K_1(E)^{(2)}$ is generated by $K_1(P)$ as $P$ runs over the closed points of $E$. One has a homomorphism from $Pic(E\otimes F)\otimes F^*$ to $K_1(E)^{(2)}$ for any finite extensions $F$ of $\mathbb Q$ by taking the cup product followed by the norm. The union of the images is the whole group. http://mathoverflow.net/questions/58471/analytic-torsion-in-the-derived-category/58474#58474 Comment by 2011-03-15T03:00:26Z 2011-03-15T03:00:26Z @John Klein. As for your second comment -- starting with &quot;the answer depends ...&quot; Any quasi-iso of bounded cxs of fg projectives induces a canonical iso of determinants. Look at Deligne, Le d&#233;terminant de la cohomologie. Current trends in arithmetical algebraic geometry (Arcata, Calif., 1985), 93--177, Contemp. Math., 67. http://mathoverflow.net/questions/58471/analytic-torsion-in-the-derived-category/58474#58474 Comment by 2011-03-15T02:55:53Z 2011-03-15T02:55:53Z @John_Klein This is a good example. Lets start with an acyclic complex of Real Vector Spaces, $0\to V\to W\to 0$. Then we have an iso $V\to W$ and so det($V$) to det($W$), and so the det of the complex is $\mathrm{det}(V)\otimes \mathrm{det}(W)^{-1}\simeq \mathbb{R}$. BUT, if you in addition choose bases (or even just orientations of $V$ and $W$, then you get additional data -- the sign that you mention. The key point here is that torsion is secondary data, which compares two ways of computing determinants -- one from the acyclicity, one from choosing bases. http://mathoverflow.net/questions/58471/analytic-torsion-in-the-derived-category/58474#58474 Comment by 2011-03-14T23:44:57Z 2011-03-14T23:44:57Z In response to both comments. To my knowledge you always need the complex to be perfect i.e. locally quasi-iso to a bounded complex of projectives. This is the construction explained by Deligne. To every object in the derived &quot;category of perfect complexes&quot; it associates a rank one projective, turning triangles into tensor products. The method is to choose a bounded complex of f.g. projectives quasi-iso to the given perfect complex, takes its determinant, and then check the answer does not depend on choices. The key point is that an acyclic cx of f.g. projectives has trivial determinant. http://mathoverflow.net/questions/56431/hurewicz-theorem-related-to-galois-group-or-tannakian-categories Comment by 2011-02-23T20:02:51Z 2011-02-23T20:02:51Z Sorry -- forgot a dollar sign: If one defines $H^1(X,\mathbb{Z})$ as the group of $\mathrm Z$-torsors over $X$, then one can see that $H^1(X,\mathbb{Z})$ is the group of homomorphisms from $\pi_1(X)^{\mathrm{ab}}$ to $\mathbb Z$ by thinking about how deck transformations act on torsors. But that is $H^1$ not $H_1$. http://mathoverflow.net/questions/56431/hurewicz-theorem-related-to-galois-group-or-tannakian-categories Comment by 2011-02-23T19:57:10Z 2011-02-23T19:57:10Z If one defines $H^1(X,{\mathbb Z})$ as the group of ${\mathbb Z})$ torsors over $X$, then one can see that $H^1(X,{\mathbb Z})$ is the group of homomorphisms from $\pi_1(X)^{mathrm ab}$ to ${\mathbb Z}) by thinking about how deck transformations act on torsors. But that is$H^1$not$H_1$. http://mathoverflow.net/questions/33198/on-gamma-graded-pieces-of-the-localization-sequence-for-g-theory-i-e-for-k/33203#33203 Comment by 2011-01-25T18:02:14Z 2011-01-25T18:02:14Z For X smooth over a field, see Gillet &amp; Soul&#233;, &quot;Filtrations on higher algebraic K-theory&quot; AMS Proc. Sympos. Pure Math. vol 67 (MR1743238) for the result on higher K-theory http://mathoverflow.net/questions/53078/gysin-maps-between-smooth-schemes-generically-surjective/53092#53092 Comment by 2011-01-24T18:53:58Z 2011-01-24T18:53:58Z Never mind about finite generation. As you say, the restriction map on Chow groups from X to U is surjective, and so the same would have be true of the restriction from X to Z. So for any projective variety one would have to have all the Chow groups cyclic, which fails even for many elliptic curves over finite fields, and indeed just about any randomly chosen projective variety. Perhaps the intuition driving this question is that for smooth manifolds, there is a tubular nbd of Z which is diffeomorphic to the normal bundle, which however is false in the algebraic category. http://mathoverflow.net/questions/11958/transcendental-galois-theory Comment by 2011-01-06T21:36:31Z 2011-01-06T21:36:31Z I would argue that once one moves beyond etale extensions that one should not look at discrete Galois groups. E.g. Jacobson's theory for exponent purely inseparable extensions uses Lie algebras. For a Galois theory for extensions$E/F$I would be more inclined to look at subalgebras of the derivations (or of the differential operators) of$E$over$F\$. See e.g. &quot;On Galois correspondence between intermediate fields and closed derivation subalgebras&quot; Teppei Kikuchi J. Math. Kyoto Univ. 23, (1983), 281-287.