User daniel loughran - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:11:52Z http://mathoverflow.net/feeds/user/5101 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126889/what-does-the-tensor-product-of-two-central-simple-algebras-correspond-to-geometr What does the tensor product of two central simple algebras correspond to geometrically? Daniel Loughran 2013-04-08T18:19:15Z 2013-04-10T11:05:43Z <p>Let $k$ be a field, assumed to have characteristic $0$ for simplicity (though this probably isn't necessary). </p> <p>Let $A$ be a central simple algebra over $k$ of dimension $n^2$. Then the collection of left ideals of rank $n$ in $A$ may be given the structure of a variety $B(A)$, which turns out to be a Brauer-Severi variety (i.e. a form of projective space) of dimension $n-1$. Moreover a classical theorem states that every Brauer-Severi variety of dimension $n-1$ arises in this way.</p> <p>Next recall that given two central simple algebras $A_1$ and $A_2$ over $k$, their tensor product $A_1 \otimes_k A_2$ is also a central simple algebra over $k$. My question concerns how the corresponding Brauer-Severi varieties are related.</p> <blockquote> <blockquote> <p>How may one visualise $B(A_1 \otimes_k A_2)$ in terms of $B(A_1)$ and $B(A_2)$? In particular, is there a geometrical operation that one may perform on $B(A_1)$ and $B(A_2)$ to obtain $B(A_1 \otimes_k A_2)$?</p> </blockquote> </blockquote> <p>This question is slightly vague; so let me give you an example of the kind of thing I want, but which unfortunately does not work. Namely, the only natural construction which I can think of here is the fibre product. But $B(A_1) \times B(A_2)$ is not isomorphic to $B(A_1 \otimes_k A_2)$ for two obvious reasons:</p> <ul> <li><p>It has the wrong dimension.</p></li> <li><p>It is isomorphic to a product of projective spaces, and not projective space, over an algebraic closure $\overline{k}$ of $k$.</p></li> </ul> http://mathoverflow.net/questions/125010/producing-2-curves-on-a-smooth-surface/125015#125015 Answer by Daniel Loughran for Producing $(-2)$ curves on a smooth surface Daniel Loughran 2013-03-19T20:27:52Z 2013-03-19T20:27:52Z <p>Here is something that might help answer your question:</p> <p>Blowing-up rational double points on normal singular surfaces produces $(-2)$-curves of genus zero. Conversely Artin [1, Thm. 2.7] showed that (under suitable conditions) every such $(-2)$-curve of genus zero arises in this way.</p> <p>[1] Artin - Some numerical criterion for contractability of curves on surfaces.</p> http://mathoverflow.net/questions/122922/group-actions-on-blow-ups/122926#122926 Answer by Daniel Loughran for group actions on blow-ups Daniel Loughran 2013-02-25T20:46:40Z 2013-02-26T09:05:25Z <p>Yes you can extend the action. One can prove this using the universal property of blow-ups (see Hartshorne Corollary II.7.15). </p> <p>As $Z$ is invariant under the action of $G$, the inverse image of $Z$ with respect to the morphism $G \times X \to X$ is $G \times Z$. Therefore on applying the universal property of blow-ups to this morphism we obtain a morphism $G \times \widetilde{X} \to \widetilde{X}$. Now, by assumption this morphism satisifies the identities $$(gh)x = g(hx), \quad ex = x$$ for all $x$ in $\widetilde{X}\setminus E$, where $E$ denotes the exceptional divisor of the blow-up. However since any two morphisms which are equal on an open dense subset must be equal on the whole space, we see that these identities hold for all $x$ in $\widetilde{X}$, i.e. the morphism $G \times \widetilde{X} \to \widetilde{X}$ gives an action of $G$ on $X$.</p> <p>Note that in this argument we did not use the fact that $G$ was finite, it works for any algebraic group.</p> http://mathoverflow.net/questions/119140/what-are-the-general-techniques-for-proving-a-variety-is-not-toric/119192#119192 Answer by Daniel Loughran for What are the general techniques for proving a variety is not toric? Daniel Loughran 2013-01-17T17:09:53Z 2013-01-17T17:09:53Z <p>Here is another way to see why this is true (though as already noted you need $s >3$).</p> <p>Note that if $X$ is a toric variety with respect to an algebraic torus $T$, then by definition $T$ acts faithfully on $X$. In particular there is an injective homomorphism $T \to \mathrm{Aut}(X)$.</p> <p>Now if $X$ the blow-up of $\mathbb{P}^2$ in $s >3$ general points, then it is "well-known" that the automorphism group of $X$ is finite (see e.g. The main theorem of Koitabashi - Automorphism groups of generic rational surfaces). In particular such surfaces are not toric varieties. Moreover Koitabashi even shows that the automorphism group is trivial for $s \geq 9$.</p> <p>Note that this method also shows that such $X$ do not admit a faithful action of any algebraic group of positive dimension, so they are very far from being homogeneous. </p> http://mathoverflow.net/questions/118449/brauer-group-elements-associated-to-conic-bundles Brauer group elements associated to conic bundles Daniel Loughran 2013-01-09T13:49:34Z 2013-01-11T11:32:45Z <p>Let $X$ be a non-singular projective variety over a field $k$ (perhaps not of characteristic $2$), and let $\pi:Y\to X$ be a conic bundle over $X$ i.e. a proper morphism all of whose fibres are isomorphic to plane conics. Let $Z \subset X$ denote the discriminant locus, i.e. the closed subset of $X$ consisting of those point whose fibres are singular (this is a divisor on $X$). Finally let $U = X \setminus Z$.</p> <p>Then, $\pi$ restricts to give a smooth morphism over $U$, which locally for the étale topology is a trivial bundle of $\mathbb{P}^1$'s, hence we obtain an element $A \in H^1(U,PGL_2)$ associated to $X$. A standard argument using the exact sequence defining $PGL_2$ gives rise to an injective map $H^1(U,PGL_2) \to H^2(U,\mathbb{G}_m)=\mathrm{Br}~U$ by which we obtain an element of $\mathrm{Br}~U[2]$ (which by abuse of notation we also denote by $A$).</p> <p>For any discrete valutation $v$ of the function field $k(U)$, we have a residue map $$\mathrm{res}_v:\mathrm{Br}~U \to H^1(k(v),\mathbb{Q}/\mathbb{Z}).$$ My first question is about how the geometric properties of $\pi$ are related to the algebraic properties of $A$.</p> <blockquote> <blockquote> <p>Question 1. Let $v$ be a valuation of $k(U)$ corresponding to an irreducible divisor $D$ supported on $Z$. What is the a relationship between $\mathrm{res}_v(A)$ and the fibre of $\pi$ over $D$?</p> </blockquote> </blockquote> <p>This is slightly vague so I will try to make more precise what I am after.</p> <p>Naively, one might expect that $\mathrm{res}_v(A)$ is non-zero since the fibre over $D$ is singular. However, this is not quite true as birational conic bundles should give rise to the same Brauer group element on sufficiently small open subsets of $X$. So there may be some another conic bundle also giving rise to $A$ for which the fibre over $D$ is non-singular. The following is a more precise version of the previous question which hopes to get around this issue (we use the same notation as Question 1).</p> <blockquote> <blockquote> <p>Question 2. Is $\mathbb{res}_v(A) =0$ if and only if the singular conic $\pi^{-1}(D)$ (considered over $k(v)=k(D)$) is split over $k(D)$?</p> </blockquote> </blockquote> <p>Here by split I mean that the two singular lines are defined over $k(D)$, rather than conjugate over some quadratic field extension. Also I am slightly abusing notation, as really it is the generic fibre of $\pi^{-1}(D)$ over $D$ which is a conic over $k(D)$.</p> <p>Finally, one might expect that conic bundles for which all singular conics $\pi^{-1}(D)$ are not split over $k(D)$ are in some sense "relatively minimal". Unfortunately I can't make this precise except in the case where $\mathrm{dim}~ Y = 2$, where relatively minimal means that one may not contract any of divisors lying in the fibres of $\pi$. I would like a higher dimensional analogue of this.</p> <blockquote> <blockquote> <p>Question 3. Suppose that for some divisor $D \subset Z$ the corresponding singular conic $\pi^{-1}(D)$ over $k(D)$ is split. Then is it possible to contract one of the components of $\pi^{-1}(D)$? More precisely, does there exist a conic bundle $\pi':Y'\to X$ and a morphism $f:Y \to Y'$ (respecting $\pi$ and $\pi'$) such that $Z' = Z \setminus D$? Here $Z'$ denotes the discriminant locus of $Y'$.</p> </blockquote> </blockquote> http://mathoverflow.net/questions/116145/cohomology-of-vector-bundles-via-intersection-theory/116438#116438 Answer by Daniel Loughran for Cohomology of vector bundles via Intersection Theory Daniel Loughran 2012-12-15T11:06:26Z 2012-12-15T11:06:26Z <p>The answer to Question $4$ is also no. Let $C$ be a smooth projective curve of genus $g$ over $\mathbb{C}$. Then the chow ring $A(C)$ of $C$ is isomorphic to $\mathbb{Z}[x]/(x^2)$, whereas of course the betti numbers of $C$ are $1,2g,1$. So the chow ring will not determine the homology in general.</p> <p>However, for smooth projective varieties $X$ over $\mathbb{C}$ there is a cycle class map $A(X) \to H^*(X,\mathbb{Z})$ which is a ring homomorphism (this is all explained in the appendices of Hartshorne, though one needs to use $\ell$-adic cohomology over other fields). In particular algebraic cycles will give you cohomology classes. Determining which cohomology classes come from algebraic cycles is part of the Hodge conjecture.</p> http://mathoverflow.net/questions/115452/average-orders-of-multiplicative-functions Average orders of multiplicative functions Daniel Loughran 2012-12-04T23:18:13Z 2012-12-07T01:50:17Z <p>For a multiplicative function $f$ and $x>0$ let $$S_f(x)= \sum_{n \leq x} f(n).$$ Studying sums of this type is a favourite pastime of analytic number theorists. I'm trying to understand what kind of behaviour can occur for such sums. In particular, my question is the following.</p> <blockquote> <blockquote> <p>Does there exist a multiplicative function $f$ and a constant $c_f>0$ such that $$S_f(x) \sim c_f\frac{x}{\log x},$$ as $x \to \infty$?</p> </blockquote> </blockquote> <p>Here is some motivation for how I came across this problem. Analytic number theorists often study sums of the above type by studying the analytic properties of associated Diriclet series $$L(f,s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}.$$</p> <p>Here, if $L(f,s)$ has a pole of order $r>0$ at $s=1$ and is well-behaved for $\text{Re}( s) >1$, then one can often show (using e.g. a Tauberian theorem such as Perron's formula) that we have an asymptotic formula $$S_f(x) \sim c_f x (\log x)^{r-1}.$$</p> <p>More generally there is the Selberg-Delange method, here one works with complex powers $\zeta^z(s)$ of the Riemann zeta function. This method, when it works, will give an asymptotic formula of the shape $$S_f(x) \sim c_f x (\log x)^{z-1}.$$ In particular, it does not seem that one can obtain an asymptotic formula like the one I am seeking using this approach. </p> <p>Note that one cannot use something like the prime number theorem to construct an example of the shape I am looking for, since $\pi(n)$ is not a multiplicative function!</p> http://mathoverflow.net/questions/111890/geometrically-connected-curve/111892#111892 Answer by Daniel Loughran for Geometrically connected curve Daniel Loughran 2012-11-09T12:09:48Z 2012-11-09T14:11:16Z <p>For a variety over a non-algebraically closed field, "geometrically connected" means connected over the algebraic closure.</p> <p>As an example where this fails, note that the curve $x^2+1=0$ in $\mathbb{A}^2$ is connected over $\mathbb{Q}$, but not over $\mathbb{Q}[i]$, where is becomes $(x+i)(x-i)=0$, which is a union of two lines. Hence this curve is connected but not geometrically connected.</p> <p>You can also use the same adjective for many other properties, so that you can talk about something being geoemtrically integral, geometrically rational, etc...</p> http://mathoverflow.net/questions/110441/brauer-group-elements-of-order-2 Brauer group elements of order $2$ Daniel Loughran 2012-10-23T16:11:22Z 2012-10-24T07:53:14Z <p>Let $K$ be a field and let $Q$ be a quaternion algebra over $K$. Then it is well-known that the class $[Q]$ of $Q$ in $Br(K)$ has order $2$. One can show this by constructing an explicit isomorphism $Q \otimes_K Q \cong M_2(K)$. My question is about the converse. </p> <blockquote> <p>Does there exist a field $K$ and a division algebra $D$ over $K$ such that the class $[D]$ of $D$ in $Br(K)$ has order $2$, but such that $D$ is not isomorphic to a quaternion algebra over $K$?</p> </blockquote> <p>If such a $K$ and $D$ exist, then it would also be nice to see an explicit example. </p> <p>As a non-example, I believe that it follows from local and global class field theory that if $K$ is a local or global field, then every element of order $2$ in $Br(K)$ may indeed be represented by a quaternion algebra.</p> http://mathoverflow.net/questions/108945/defining-isogenies-over-smaller-fields/108946#108946 Answer by Daniel Loughran for Defining isogenies over smaller fields Daniel Loughran 2012-10-05T19:09:45Z 2012-10-05T19:09:45Z <p>No: Consider the elliptic curve $E: y^2 = x^3 + x$ defined over $\mathbb{Q}$. Then the isogeny $y \mapsto iy$ and $x \mapsto -x$ is defined over $\mathbb{Q}(i)$ but obviously not over $\mathbb{Q}$.</p> <p>In general if $K \subset L$ is Galois, then a morphism defined over $L$ comes from one over $K$ if and only if it is invariant under the action of $Gal(L/K)$. This follows from the so-called "theory of descent". See for example Bjorn Poonen's notes <a href="http://www-math.mit.edu/~poonen/papers/Qpoints.pdf" rel="nofollow">Rational points on varieties.</a></p> http://mathoverflow.net/questions/107448/birch-and-swinnerton-dyer-conjecture-in-positive-characteristic/107449#107449 Answer by Daniel Loughran for Birch and Swinnerton-Dyer conjecture in positive characteristic Daniel Loughran 2012-09-18T09:11:24Z 2012-09-18T11:22:29Z <p>Edit: This answer addresses an earlier version of the question, where the OP asked whether or not BSD made sense for elliptic curves over finite fields. It also however answers the current question.</p> <p>The Birch and Swinnerton-dyer conjecture for an elliptic curve over a number field relates the rank of the Mordell-Weil group to the L-function of the curve. To have a "BSD" over other fields, one needs an analogue of these objects.</p> <p>The appropriate positive characteristic analogue of BSD is for elliptic curves over function fields of curves (the other "global fields"). This is known in some cases, but not in full generality. In fact it is known that BSD for an elliptic curve over a function field is equivalent to the finiteness of the Tate-Shafarevich group of the curve. </p> <p>However, if you have an elliptic curve $E$ over a finite field $k$ and a function field $K$ of a curve in positive characteristic, then by base change you may consider $E_K$ as a "constant" curve over $K$. In which case I do believe that BSD is known, but hopefully someone else can help me with a precise reference as I cannot remember it.</p> http://mathoverflow.net/questions/101952/determination-of-rationality-and-computing-a-rational-parametrization/101977#101977 Answer by Daniel Loughran for Determination of rationality and computing a rational parametrization Daniel Loughran 2012-07-11T18:15:28Z 2012-07-11T19:37:56Z <p>Here is my attempt at a heuristic as to why the problem should be undecidable. </p> <p>Suppose we have a hypersurface $X$ of dimension $n$ and we wish to decide whether or not it is rational. I will assume that $n\geq2$. Then giving a rational map $\mathbb{P}^n \dashrightarrow X$ is the same as giving a $\mathbb{C}(t_1,\ldots,t_n)$-vauled point on $X$. However, "Hilbert's 10th problem" for such function fields is undecidable (see <a href="http://www.math.psu.edu/eisentra/varieties.pdf" rel="nofollow">http://www.math.psu.edu/eisentra/varieties.pdf</a>). Hence the problem you have asked for is undecidable. </p> <p>Edit: As noted in the comments, this reasoning is not quite correct as for Hilbert's 10th problem we fix $m$ and a field $\mathbb{C}(t_1,\ldots,t_m)$, then allow the dimension $n$ to vary. Hence why it is only a heuristic!</p> <p>Note that for Hilbert's 10th problem, the case $\mathbb{C}(t)$ is still open.</p> <p>Edit: As remarked below, rationality for curves is decidable. One just needs to compute the genus of the normalisation of the projective closure of the curve.</p> http://mathoverflow.net/questions/98418/what-are-the-truly-global-methods-in-number-theory/98419#98419 Answer by Daniel Loughran for What are the truly 'global methods' in number theory? Daniel Loughran 2012-05-30T21:37:21Z 2012-05-31T07:40:55Z <p>This is quite a brief answer as I am not quite sure your question is appropriately phrased for this site, but one can use the "Brauer-Manin obstruction" to show the non-existence of rational points, even if the variety is everywhere locally soluble. Try searching google - in particular papers by Colliot-Thélène, Sansuc, Harari, Swinnerton-Dyer, Skorobogatov and many others. </p> <p>I'm not quite sure if this classifies as "truely global" according to your definition, as the method works by cutting out a certain subset of the adèles which contains the set of rational points, and the adèles are built out of all the completions of the number field.</p> <p>However in my opinion I would certainly say that it is a global method and the general set-up of how it works relies on results from global class field theory. </p> http://mathoverflow.net/questions/97204/real-vs-complex-surfaces/97218#97218 Answer by Daniel Loughran for Real vs complex surfaces Daniel Loughran 2012-05-17T12:46:23Z 2012-05-17T12:51:26Z <p>I think a lot of your questions apply to arbitrary Galois field extensions, not just the field extension $\mathbb{R} \subset \mathbb{C}$. This might help clarify the problems you are having by putting them into a general context. </p> <p>Let $E \subset F$ be a finite Galois extension of fields and let $X$ be a smooth variety over $E$. Then, the canonical divisor $K_X$ is always defined over the base field $E$. Moreover, it is well-behaved with respect to smooth base change, so that yes indeed the base change of $K_X$ to $X_F$ is the canonical divisor $K_{X_F}$ of $X_F$.</p> <p>With regards to divisors, we have natural homomorphisms $\mathrm{Div}(X) \to \mathrm{Div}(X_F)$ and $\mathrm{Pic}(X) \to \mathrm{Pic}(X_F)$ given by base change. Moreover, in the case of surfaces this morphism respects the intersection pairing as your desire, as the intersection number of two divisors is defined geometrically. Also in the first case, it is true that the image consists of those divisors which are invariant under the action of $\mathrm{Gal}(F/E)$.</p> <p>However, in general this does not hold for Picard groups. By which I mean, there may exist divisor <em>classes</em> which are Galois invariant, but nonetheless there does not exists a divisor in that class defined over $E$.</p> <p>As an example, consider a conic $X$ defined over $E$ which has no rational points, such that $X_F$ has rational points. Then the natural map $\mathrm{Pic}(X) \to \mathrm{Pic}(X_F)$ corresponds to the inclusion $2\mathbb{Z} \to \mathbb{Z}$, as lowest degree of any divisor is $2$ (given by the anticanonical divisor). However, the action of $\mathrm{Gal}(F/E)$ on $\mathrm{Pic}(X_F)$ is trivial as it preserves the degree of a divisor.</p> <p>As for blow-ups, one can define blow-ups of closed points in a similar manner to how one defines the blow-up of a rational point. Closed points correspond to Galois invariant collections of rational points on $X_F$, therefore the map given by blowing up each of these rational points is Galois invariant and so descends to a morphism defined over $E$.</p> <p>A lot of these ideas can be found in Manin's book on cubic forms. </p> http://mathoverflow.net/questions/93859/an-arithmetic-analogue-of-the-discriminant-curve-of-a-conic-bundle-threefold An arithmetic analogue of the discriminant curve of a conic bundle threefold Daniel Loughran 2012-04-12T11:59:31Z 2012-04-13T12:01:02Z <p>I am looking for an "arithmetic" analogue of a well known result on threefolds with a conic bundle structure. The following result can be found in [Iskovskikh - On the rationality problem for conic bundles, Lemma 1]. Note that Iskovskikh has some extra condition of relative minimality which I am pretty sure I don't need for the result I want.</p> <hr> <p>Let $X$ be a smooth irreducible threefold over $\mathbb{C}$ with a morphism $\pi:X \to B$ to a smooth rational surface $B$ such that every fibre is a (possibly degenerate) conic.</p> <p>Then, then there exists a reduced normal crossings divisor (the "discriminant curve") $\Delta \subset B$ such that for any $b \in B$ we have:</p> <p>(a) $\pi^{-1}(b) \cong \mathbb{P}^1$, if $b \not \in \Delta$<br> (b) $\pi^{-1}(b)$ is two intersecting lines if $b \in \Delta \backslash Sing (\Delta)$<br> (c) $\pi^{-1}(b)$ is a non-reduced line if $b \in Sing(\Delta)$<br> (d) In particular, there are only finitely many non-reduced fibres.</p> <hr> <p>In my situation, I have a smooth conic bundle surface $p:S \to \mathbb{P}^1$ defined over $\mathbb{Q}$, and I have chosen a regular model $\pi: X \to \mathbb{P}^1_{\mathbb{Z}}$, i.e. the morphism $\pi$ restricted to the generic fibre is exactly the morphism $p$ and every fibre is a conic.</p> <blockquote> <p>Does an analogue of the above result hold in my case? If so, does anyone have a reference to where it has been worked out in the literature?</p> </blockquote> <p>I hope it is clear, but just to clarify that I want a reduced normal crossings divisor $\Delta \subset \mathbb{P}^1_{\mathbb{Z}}$ which satisfies the appropriate analogues of conditions (a), (b), (c) and (d).</p> http://mathoverflow.net/questions/89395/is-there-a-classification-of-surfacesmooth-and-projective-over-arbitrary-field/89588#89588 Answer by Daniel Loughran for Is there a classification of surface(smooth and projective) over arbitrary field? Daniel Loughran 2012-02-26T15:15:58Z 2012-02-26T15:15:58Z <p>One of the main subtleties in trying to classify surfaces over non-algebraically closed fields is that there are minimal surfaces which become non-minimal over the algebraic closure.</p> <p>As an example I will focus on the case that I know best, that of (geometrically) rational surfaces. Over an algebraically closed field, it is well-known that the only such minimal surfaces are $\mathbb{P}^2$ and the rational ruled surfaces $\mathbb{F}_n$ for $n \geq 0$.</p> <p>If the field is not algebraically closed, then things are a lot more complicated. It is a theorem of Iskovskikh that a minimal rational surface over a perfect field is one of the following types:</p> <ul> <li>$\mathbb{P}^2$.</li> <li>A smooth quadric $X \subset \mathbb{P}^3$ with $\mathrm{Pic}(X) = \mathbb{Z}$.</li> <li>A Del Pezzo surface $X$ with $\mathrm{Pic}(X) = \mathbb{Z}K_X$, here $K_X$ denotes the canonical divisor.</li> <li>A conic bundle $f : X \to C$ over a rational curve $C$, with $\mathrm{Pic}(X) = \mathbb{Z} \oplus \mathbb{Z}$.</li> </ul> <p>In particular conic bundles form a very large family and can have arbitrarily many (geometrically) degenerate fibres.</p> <p>If you want to learn more about this result, I heartily recommend the notes "Rational surfaces over nonclosed fields" by Brendan Hassett, which can be found on his webpage. </p> http://mathoverflow.net/questions/22040/is-the-complement-of-an-affine-variety-always-a-divisor Is the complement of an affine variety always a divisor? Daniel Loughran 2010-04-21T11:19:25Z 2012-01-28T23:26:13Z <p>Let $X$ be a connected affine variety over an algebraically closed field $k$, and let $X \subset Y$ be a compactification, by which I mean $Y$ is a proper variety (or projective if you prefer), and $X$ is embedded as an open dense subset. </p> <p>I am guessing that it is not always the case that $Y\setminus X$ is a divisor, one could imagine it being a single point with a horrible singularity. But if $Y$ is smooth or even normal, is it the case that $Y\setminus X$ is always a divisor? Does anybody know a proof of such a result?</p> <p>Thanks, Dan</p> http://mathoverflow.net/questions/79536/converse-to-a-theorem-of-landau-on-dirichlet-series/79540#79540 Answer by Daniel Loughran for Converse to a theorem of Landau on Dirichlet series Daniel Loughran 2011-10-30T19:33:27Z 2011-10-30T20:25:08Z <p>I'm not sure you can hope for much. For example consider the case $c_n=1$ if $n$ is not a square, and $c_n=-1$ otherwise. The associated Dirichlet series has a pole at $s=1$, but of course the terms are not of fixed sign for sufficiently large $n$.</p> <p>In general, knowing analytic properties of a Dirichlet series (such as convergence) cannot tell you much about any of the individual terms $c_n$, since you can always change infinitely many of the $c_n$ for $n$ in a "sparse" set.</p> http://mathoverflow.net/questions/75393/does-isomorphic-generic-fibre-imply-isomorphic-special-fibre-for-smooth-morphisms Does isomorphic generic fibre imply isomorphic special fibre for smooth morphisms? Daniel Loughran 2011-09-14T12:49:53Z 2011-09-15T18:06:20Z <p>Let $X$ and $Y$ be regular integral Noetherian schemes. Assume that $X$ and $Y$ are smooth and proper over a base scheme $S=Spec R$, where $R$ is a discrete valuation ring.</p> <p>If $X$ and $Y$ have isomorphic generic fibres, is it also the case that their special fibres are isomorphic?</p> <p>Remarks:</p> <ol> <li><p>The answer is yes when $X$ and $Y$ are abelian schemes (this follows from the theory of the Néron model). In general though it is not the case that morphisms between the generic fibres extend to the special fibre.</p></li> <li><p>I am also particularly interested in case where $R$ is the localisation of $\mathbb{Z}$ at some prime, and hence the generic fibres are smooth proper varieties over $\mathbb{Q}$.</p></li> </ol> http://mathoverflow.net/questions/74679/euler-summation-and-its-transformation/74683#74683 Answer by Daniel Loughran for Euler summation and its transformation Daniel Loughran 2011-09-06T20:46:09Z 2011-09-07T07:31:15Z <p>This is not a very detailed answer, but I can give you an idea how to solve these problems.</p> <p>For the first equality you can deal with the coprimality condition using Mobius inversion. Then it is simply applying Euler-Maclaurin summation and playing around with identities of arithmetic functions. For the second equality, use partial summation. </p> <p>These techniques can be found in Apostol's book on analytic number theory. Also try Iwaniec and Kowalski's book on analytic number theory.</p> <p>Edit: You can find the statement of the partial summation trick (sometimes attributed to Abel) in Apostol's book, Theorem 4.2.</p> http://mathoverflow.net/questions/69493/integral-roots-to-degree-d-forms-in-four-variables-inside-a-box/69529#69529 Answer by Daniel Loughran for Integral roots to degree $d$-forms in four variables inside a box Daniel Loughran 2011-07-05T11:09:19Z 2011-07-05T17:16:05Z <p>Im not sure if this is exactly what you are looking for, but try searching google for the "dimension growth conjecture". This is essentially the conjecture that $N(X,B) \ll_{d,\varepsilon} B^{dim X + \varepsilon}$ for any projective variety $X$ of degree $d \geq 2$.</p> <p>I think this is now known for any variety whose degree is not three. This follows from work of Heath-Brown, Browning, Salberger, Marmon and others. Also, Salberger has recently announced a proof for the case of degree three, however the implied constant is not uniform with respect to $X$.</p> <p>These results are generally proved using a higher dimensional analogue of the determinant method of Bombieri and Pila, in particular Heath-Brown has developed a $p$-adic version of the determinant method that has proved fruitful. You might be able to use this to get the kind of result that you are looking for.</p> <p>For an overview of these results I would recommend: <a href="http://www.maths.bris.ac.uk/~matdb/preprints/pila.pdf" rel="nofollow">http://www.maths.bris.ac.uk/~matdb/preprints/pila.pdf</a></p> http://mathoverflow.net/questions/68322/a-heuristic-for-the-density-of-solutions-to-diophantine-equations/68362#68362 Answer by Daniel Loughran for A heuristic for the density of solutions to Diophantine equations Daniel Loughran 2011-06-21T10:32:42Z 2011-06-22T10:46:05Z <p>In 1989, Manin and his collaborators formed a series of conjectures on the asymptotic behaviour of the number of solutions to diophantine equations. Let $X\subset \mathbb{P}^n$ be a fano variety (that is $-K_X$ is ample) under its anticanonical embedding, and let $H$ be the associated height function. Then it is expected that there exists a zariski open subset $U \subset X$ such that the number of rational points of height less than $B$ (e.g. the number of solutions in an expanding ball or box) is asymptotic to $c_X B(\log B)^{r_X},$ as $B \to \infty$ for some constants $c_X$ and $r_X$.</p> <p>This result is true in some cases, for example complete intersections with many variables and small degree via the circle method, quadratic forms, toric varieties, flag varieties and also for some del Pezzo surfaces. But there are counter-examples showing that it is not true for all fano varieties (namely one expects $r_X=\textrm{rank } \textrm{Pic}(X)-1$, but this is not true in general).</p> <p>At any rate, Peyre formed a conjecture on the leading constant $c_X$ which occurs in the asymptotic formula. It is very close to what you describe. One defines a measure on the set of adelic points on $X$, and then the leading constant is essentially the volume of the closure of $X(\mathbb{Q})$ inside the adeles. This is really an adelic integral and not a real integral, but for suitable varieties (namely those which satisfy weak approximation), the local factors at the primes come out as the $c_p$ in the way you describe. In general though one needs to introduce convergence factors to insure that the product over the $c_p$ converges. These come from an Artin L-function associated to the Picard group.</p> <p>There are however some extra factors $\alpha$ and $\beta$ present in the constant, related to the position of the anticanonical divisor in the effective cone and the Brauer group of $X$. For conics with a rational point, we have $\alpha=1/2$ and $\beta =1$. This might explain your missing factor of two.</p> <p>Papers:</p> <p>J. Franke, Y. I. Manin and Y. Tschinkel, Rational Points of Bounded Height on Fano Varieties. Invent. Math. 95, 421--435 (1989).</p> <p>E. Peyre, Hauteurs et measures de Tamagawa sur les variétiés de Fano. Duke Math. J., 79(1), 101--218 (1995).</p> http://mathoverflow.net/questions/65868/picard-number-and-torsion-of-neron-severi-group-of-abelian-varieties-over-a-numbe/65880#65880 Answer by Daniel Loughran for Picard number and torsion of Neron-Severi group of abelian varieties over a number field Daniel Loughran 2011-05-24T17:26:10Z 2011-05-24T19:06:06Z <p>There were quite a few different questions, so forgive me if my answer is somewhat fragmented.</p> <p>The Néron-Severi group $NS(X)$ (divisors modulo algebraic equivalence) is finitely generated over any field for any non-singular projective variety $X$, this is Severi's theorem of the base (at least for the case of characteristic zero). In what follows however I am assume that the variety is defined over $\mathbb{C}$ for simplicity. Some answers to your other questions:</p> <ul> <li><p>In the case of curves, $NS(X)\cong\mathbb{Z}$, with the isomorphism given by the degree map. </p></li> <li><p>More generally, there is a natural injective morphism $NS(X) \to H^2(X,\mathbb{Z})$. This can then be used to get an upper bound for the Picard number.</p></li> <li><p>Torsion in $NS(X)$ also naturally lives inside $H_1(X,\mathbb{Z})$, so if this is torsion free then so is $NS(X)$. This is the case free for abelian varieties (over $\mathbb{C}$ say), since $H_1(X,\mathbb{Z})$ can identified with a lattice $\Lambda \subset \mathbb{C}^g$ such that $X\cong\mathbb{C}^g/\Lambda$.</p></li> <li><p>Finally, torsion divisors can never be ample because they are numerically trivial, and ampleness is preserved under numerical equivalence. </p></li> </ul> http://mathoverflow.net/questions/65676/the-closure-of-the-set-of-rational-points-in-the-adeles The closure of the set of rational points in the Adeles Daniel Loughran 2011-05-21T20:42:31Z 2011-05-22T22:37:50Z <p>Let $X$ be a smooth geometrically integral projective variety over $\mathbb{Q}$. Then we may consider the closure $\overline{X(\mathbb{Q})}$ of $X(\mathbb{Q})$ inside the adelic points $X(\mathbb{A})=\prod_v X(\mathbb{Q}_v)$ of $X$. However, we may also take the closure $\overline{X(\mathbb{Q})}^v$ of $X(\mathbb{Q})$ inside $X(\mathbb{Q}_v)$ for any place $v$ of $\mathbb{Q}$. Obviously we have $$\overline{X(\mathbb{Q})} \subset \prod_v \overline{X(\mathbb{Q})}^v \subset X(\mathbb{A}).$$</p> <p>My question whether this first inequality is actually an equality?</p> <p>My motivation is that I am trying to understand better $\overline{X(\mathbb{Q})}$ and what it looks like. I will simply note that the answer to my question is yes in the easy cases where $X$ satisfies weak approximation and when $X(\mathbb{Q})$ is empty.</p> <p>Edit: To make sure there are not simple counter-examples like the one David pointed out below, I am assuming that $X(\mathbb{Q})$ is Zariski dense. I should also note that I am particularly interested in the case where $X$ is a fano variety.</p> http://mathoverflow.net/questions/64266/triviality-of-line-bundle-over-complex-manifold/64276#64276 Answer by Daniel Loughran for Triviality of line bundle over complex manifold Daniel Loughran 2011-05-08T09:37:36Z 2011-05-08T09:37:36Z <p>There is one useful fact that might by relevant to you. A line bundle $L$ and a complex projective variety $X$ is trivial if and only if $H^0(X,L) \neq 0$ and $H^0(X,L^{-1})\neq0$. </p> <p>Indeed one implication is clear, and for the other choose two non-zero elements $s \in H^0(X,L)$ and $s' \in H^0(X,L^{-1})$. Then $ss' \in H^0(X,\mathcal{O})$ is non-zero. Since $H^0(X,\mathcal{O}) = \mathbb{C}$, we see that in fact $ss'$ is nowhere vanishing. Thus $L$ admits a nowhere vanishing section, and so is trivial.</p> <p>Naturally a similar result holds for more general fields.</p> http://mathoverflow.net/questions/61382/rationality-of-flag-varieties Rationality of flag varieties Daniel Loughran 2011-04-12T10:47:47Z 2011-04-12T17:35:03Z <p>Let $X$ be a (generalised) flag variety over an algebraically closed field $k$ of characteristic zero, that is to say, $X$ is a projective variety which is a homogeneous space for some algebraic group $G$. Any such $X$ can be realised as a quotient $X=G/P$, where $P$ is a parabolic subgroup (please correct me if this is wrong!)</p> <p>Some basic properties of $X$ are:</p> <ul> <li>$X$ is a fano variety</li> <li>$X$ is unirational</li> </ul> <p>The fact that it is fano is not perhaps immediately obvious, but it is clear that it is unirational since it is dominated by $G$, which is a rational variety. There is no reason to expect your average fano variety to be rational, however are flag varieties rational? I am particularly interested in the case where $dim X = 3$.</p> http://mathoverflow.net/questions/60359/parametrization-of-unit-varieties/60362#60362 Answer by Daniel Loughran for Parametrization of unit varieties Daniel Loughran 2011-04-02T15:03:47Z 2011-04-02T15:16:51Z <p>Thanks to Dror for pointing out a nice geometrical way to think of such varieties. This paper shows that such varieties are not rational in general:</p> <p><a href="http://www.math.jussieu.fr/~florence/norm_one.pdf" rel="nofollow">http://www.math.jussieu.fr/~florence/norm_one.pdf</a></p> <p>However, any algebraic torus over a number field $K$ of dimension one or two is rational over $K$. Indeed, the case of dimension one is easy, and the case of dimension two is a theorem of VoskresenskiÄ­, but algebraic tori of dimension three and above do not have to be rational over the ground field in general.</p> http://mathoverflow.net/questions/58601/contour-integration-of-zetas-zeta2s/58629#58629 Answer by Daniel Loughran for Contour integration of $\zeta(s)\zeta(2s)$ Daniel Loughran 2011-03-16T10:37:08Z 2011-03-17T03:59:36Z <p>If you havn't done so already, you might find it useful to look at the proof of theorem 12.2 on the divisor problem in Titchmarsh - The theory of the Riemann zeta function. Here, he goes through a detailed application of Perron's formula for the function $\zeta^k(s)$, which I believe to be very similar to your case.</p> <p>Indeed, for $s=\sigma + it$ and $\sigma>1/2$, $\zeta(2s)$ is absolutley convergent and hence uniformly bounded with respect to $t$. So this will not contribute to the contours that you choose (as long as $\sigma>1/2$!). What you need then is good upper bounds for the order of the zeta function in the critical strip. </p> <p>To get these, one normally finds the order of the function at two points, and then uses the <a href="http://en.wikipedia.org/wiki/Lindel%C3%B6f%27s_theorem" rel="nofollow">Phragménâ€“Lindelöf principle</a> for strips to get estimates for the function between these two points. For example, it is known that $\zeta(1/2 + it) = O(t^{1/4})$ (see <a href="http://en.wikipedia.org/wiki/Lindel%C3%B6f_hypothesis" rel="nofollow">The Lindelöf hypothesis</a>), although there are much better bounds available than that. This is all done in Titchmarsh's book.</p> <p>I hope this helps!</p> http://mathoverflow.net/questions/33774/existence-of-zero-cycles-of-degree-one-vs-existence-of-rational-points Existence of zero cycles of degree one vs existence of rational points Daniel Loughran 2010-07-29T11:47:12Z 2011-01-24T19:33:45Z <p>Let $k$ be a field (I'm mainly interested in the case where $k$ is a number field, however results for other fields would be interesting), and $X$ a smooth projective variety over $k$. </p> <p>By a zero cycle on $X$ over $k$ I mean a formal sum of finitely many (geometric) points on $X$, which is fixed under the action of the absolute Galois group of $k$. We can define the degree of a zero cycle to be the sum of the multiplicities of the points. </p> <p>Now, if $X$ contains a $k$-rational point then it is clear that $X$ contains a zero cycle of degree one over $k$. </p> <p>What is known in general about the converse? That is, which classes of varieties are known to satisfy the property that the existence of a zero cycle of degree one over $k$ implies the existence of a $k$-rational point? For example what about rational varieties and abelian varieties?</p> <p>As motivation I shall briefly mention that the case of curves is easy. Since here zero cycles are the same as divisors we can use Riemann-Roch to show that the converse result holds if the genus of the curve is zero or one, and there are plently of counter-examples for curves of higher genus. However in higher dimensions this kind of cohomological argument seems to fail as we don't (to my knowledge) have such tools available to us.</p> http://mathoverflow.net/questions/23011/graphs-as-cycles-and-intersection-theory Graphs as cycles and intersection theory Daniel Loughran 2010-04-29T17:27:38Z 2011-01-07T14:14:21Z <p>I'm guessing that the answer to this question is well-known, but I'm struggling to find anything to help me.</p> <p>Let $X,Y$ be compact manifolds of dimension $n,m$ respectively. Let $f:X \to Y$ be a smooth map. Then one can consider the graph $\Delta_f$ of $f$ as a cycle in $X \times Y$. </p> <p>Firstly what is "known" about $\Delta_f$ considered as a homology class? (I appreciate that this is a little vague). There might need to be some extra conditions placed of $f$, as clearly if for example $f$ maps everything to a point then there is nothing to be said.</p> <p>Secondly (and related to the first question), suppose that $n=m$, and $f$ is an immersion. Then the self intersection $\Delta_f^2$of $\Delta_f$ with itself is well-defined. Is there a simple expression for this in terms of the basic properties of $f$?</p> <p>Thanks!</p> http://mathoverflow.net/questions/131089/zeroes-of-a-homogeneous-function Comment by Daniel Loughran Daniel Loughran 2013-05-19T09:18:22Z 2013-05-19T09:18:22Z I'm not sure if this completely relevant to your question, but Pila and Wilkie have studied rational points on zero sets of analytic functions (or more generally, sets definable in certain O-minimal structures). Here they conjectured that &quot;most&quot; rational points lie on the algebraic part of these subsets (algebraic part being defined in a suitable sense). They have proved this conjecture in certain cases (see also the work of Butler). http://mathoverflow.net/questions/128100/non-singular-cubics-are-not-rational/128105#128105 Comment by Daniel Loughran Daniel Loughran 2013-04-21T09:34:01Z 2013-04-21T09:34:01Z @ J&#233;r&#233;my: I think Ashwath is pointing out a minor typo, and suggesting that you replace &quot;surface&quot; with &quot;threefold&quot; in 4). http://mathoverflow.net/questions/126889/what-does-the-tensor-product-of-two-central-simple-algebras-correspond-to-geometr Comment by Daniel Loughran Daniel Loughran 2013-04-09T07:33:21Z 2013-04-09T07:33:21Z @Michael and Will: Thanks for your comments. This shows that there is always a morphism $B(A_1) \times B(A_2) \to B(A_1 \otimes_k A_2)$. It seems quite possible that $B(A_1 \otimes_k A_2)$ is in some respects determined by this morphism, i.e. this morphism should satisfy some kind of universal property. Perhaps $B(A_1 \otimes_k A_2)$ is the Brauer-Severi variety of smallest dimension that $B(A_1) \times B(A_2)$ embeds into? http://mathoverflow.net/questions/126886/relation-between-riemann-zeta-function-and-hilbert-polynomial Comment by Daniel Loughran Daniel Loughran 2013-04-08T18:36:33Z 2013-04-08T18:36:33Z Also the term &quot;Riemann zeta function&quot; normally refers to exactly that, the Riemann zeta function. The zeta functions which you are talking about are usually just referred to as the zeta function of the variety. http://mathoverflow.net/questions/126886/relation-between-riemann-zeta-function-and-hilbert-polynomial Comment by Daniel Loughran Daniel Loughran 2013-04-08T18:32:38Z 2013-04-08T18:32:38Z I don't think you can hope for much in general. For example any two curves in $\mathbb{P}^2$ of the same degree will have the same Hilbert polynomial, whereas their zeta functions can be quite different. I don't think this question has been properly thought out. http://mathoverflow.net/questions/126151/meromorphic-continuation-of-a-dirichlet-series-associated-to-an-irrational-number Comment by Daniel Loughran Daniel Loughran 2013-04-02T07:45:09Z 2013-04-02T07:45:09Z @Pieter: Tschinkel has done lots of work on height zeta functions. In particular I think that his recent work with Chambert-Loir on integral points of bounded height on toric varieties applies to your setting (see his webpage). The analytic behaviour of these zeta functions is closely related to corresponding number of rational/integral points of bounded height. In their paper they use smooth norms like you, but I seem to remember some trick which allows you to pass to non-smooth norms as I mention above, however the precise details of this trick currently elude me... http://mathoverflow.net/questions/126151/meromorphic-continuation-of-a-dirichlet-series-associated-to-an-irrational-number Comment by Daniel Loughran Daniel Loughran 2013-04-01T08:55:24Z 2013-04-01T08:55:24Z Also, do you know about Height zeta functions and Epstein zeta functions? The zeta functions $\zeta_d$ which you are studying are very closely related to these. For example your zeta function with $d(x,y)=\mathrm{max}\{|y|,|x|/\theta\}$ is a height zeta function for integral points in the affine plane. Here meromorphic continuation is known in some region past $s=2$ (though I am not sure if meromorphic continuation is know to the whole complex plane). http://mathoverflow.net/questions/126151/meromorphic-continuation-of-a-dirichlet-series-associated-to-an-irrational-number Comment by Daniel Loughran Daniel Loughran 2013-04-01T08:03:08Z 2013-04-01T08:03:08Z Here is one idea for studying $\zeta_d$ with $d$ not smooth: Find a sequence of smooth norms $(d_t)$ with $t \in \mathbb{R}$ which converge to the norm $d$ you are interested in. If you have enough control over the convergence you might be able to deduce the meromorphic continuation of $\zeta_d$ from the meromorphic continuation of each $\zeta_{d_t}$. http://mathoverflow.net/questions/125528/the-closed-form-of-sum-k1-infty-left-frac23-rightk2-frac3k2 Comment by Daniel Loughran Daniel Loughran 2013-03-26T08:53:43Z 2013-03-26T08:53:43Z Have you tried using Poisson summation? http://mathoverflow.net/questions/125281/coutour-integral-of-gamma-functions Comment by Daniel Loughran Daniel Loughran 2013-03-22T17:00:52Z 2013-03-22T17:00:52Z What does $ii$ mean? And is $j^2=-1$? http://mathoverflow.net/questions/124937/how-to-refer-to-these-multiplicative-characters/124941#124941 Comment by Daniel Loughran Daniel Loughran 2013-03-19T11:25:50Z 2013-03-19T11:25:50Z You are quite right, I misread the question... Time to delete! http://mathoverflow.net/questions/120818/birational-automorphisms-and-infinite-divisibility/120896#120896 Comment by Daniel Loughran Daniel Loughran 2013-02-05T20:35:13Z 2013-02-05T20:35:13Z Right I see. I got confused with the notation and thought that $\mathbb{Z}[1/2]$ meant $(1/2)\mathbb{Z}$. Thanks for clearing it up. http://mathoverflow.net/questions/120818/birational-automorphisms-and-infinite-divisibility/120896#120896 Comment by Daniel Loughran Daniel Loughran 2013-02-05T20:03:05Z 2013-02-05T20:03:05Z Why is it clear that $\mathbb{Z}[1/2]$ belongs to $ker(\phi)$? I'm thinking about something like an elliptic surface which could have a copy of $\mathbb{Z} \cong \mathbb{Z}[1/2]$ in its automorphism group, given by translation by a non-torsion section. It is not clear to me that this copy of $\mathbb{Z}$ acts trivially on the cohomology of the surface. http://mathoverflow.net/questions/120434/galois-cohomology-over-finite-field Comment by Daniel Loughran Daniel Loughran 2013-01-31T18:14:19Z 2013-01-31T18:14:19Z I think you might be able to deal with such a cohomology group using global tate duality. Try for example the book &quot;Cohomology of number fields&quot; by Neukirch, Schmidt and Winberg. Note that despite the title, they do indeed study all global fields http://mathoverflow.net/questions/119073/functorial-properties-of-blow-up Comment by Daniel Loughran Daniel Loughran 2013-01-16T22:58:35Z 2013-01-16T22:58:35Z Perhaps you know this already, but the universal property of blow-ups should give you what you need in the case that $\phi$ is algebraic. See e.g. Hartshorne Corollary II.7.15.