User ariyan javanpeykar - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:25:45Z http://mathoverflow.net/feeds/user/4333 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/124558/algebraicity-of-the-canonical-projection-x-gamma-to-x1-and-of-x-gamma/124569#124569 Answer by Ariyan Javanpeykar for Algebraicity of the canonical projection $X(\Gamma)\to X(1)$ and of $X(\Gamma)$ Ariyan Javanpeykar 2013-03-14T23:30:26Z 2013-03-14T23:30:26Z <p>Let <code>$X(\Gamma)\to \mathbf{P}^1_{\mathbf C}$</code> be the composition of the natural map $X(\Gamma)\to X(1)$ associated to the inclusion <code>$\Gamma\subset$</code> SL<code>$_2(\mathbf Z)$</code>, and the $j$-invariant <code>$j:X(1)\to \mathbf{P}^1_{\mathbf C}$</code>. This map is, as you said, ramified over at most three points. Since these three points are algebraic numbers, by a theorem of Grothendieck and Weil, there exists a unique pair <code>$(Y,Y\to \mathbf{P}^1_{\overline{\mathbf Q}})$</code> with $Y$ a smooth projective connected curve over <code>$\overline{\mathbf Q}$</code> and <code>$Y\to \mathbf{P}^1_{\overline{\mathbf Q}}$</code> a finite morphism which gives the natural map <code>$X(\Gamma)\to \mathbf{P}^1_{\mathbf C}$</code> after base-change. If you just want a model for your <code>$X(\Gamma)$</code> over <code>$\overline {\mathbf Q}$</code> as a curve (and not as a cover) you have more choice. </p> <p>The theorem of Grothendieck and Weil is more general. Let $k\subset K$ be an extension of algebraically closed fields of characteristic zero. Then, for any smooth quasi-projective connected variety $U$ over $k$, the base-change functor is an equivalence of categories from "finite etale covers of $U$" to "finite etale covers of $U_K$". We are applying this theorem to <code>$\mathbf P^1_{k} -\{0,1,\infty\}$</code> with <code>$k=\overline {\mathbf Q}$</code> and <code>$K=\mathbf C$</code>. </p> http://mathoverflow.net/questions/21370/where-does-the-splitting-principle-come-from-and-does-it-generalize Where does the splitting principle come from and does it generalize Ariyan Javanpeykar 2010-04-14T18:09:35Z 2013-02-27T14:08:21Z <p>Basically, I'm aware of "splitting principles" for the following three objects (which are all isomorphic modulo torsion). </p> <p><strong>1</strong>. The Chow group a la Fulton.</p> <p><strong>2</strong>. The classical Grothendieck group of vector bundles or coherent sheaves.</p> <p><strong>3</strong>. The $\gamma$-graded Grothendieck group.</p> <p>I was just wondering where the idea of "the splitting principle" comes from. I'm guessing somewhere in topology when one wanted to define Chern classes and show some properties. But I don't know.</p> <p>And above that, is there some more general way of looking at this? I know there is a theorem that connects higher K-groups with Chow groups in a sense. So I ask, is there a way of deducing the splitting principle for one of the above objects from the other? (It's easy if we want to do this modulo torsion, of course.)</p> http://mathoverflow.net/questions/119392/top-chern-class-under-finite-unramified-dominant-morphism/119469#119469 Answer by Ariyan Javanpeykar for Top chern class under finite, unramified, dominant morphism Ariyan Javanpeykar 2013-01-21T14:00:04Z 2013-01-29T11:58:44Z <p>Angelo's answer is complete, but I think you would be interested in the following (which is more about Euler characteristics than Chern classes).</p> <p>I will assume $k= \mathbf C$, but what I will write holds for $k$ algebraically closed of characteristic zero once you replace "cohomology with compact support and coefficients in $\mathbf Q$ on the category of para-compact Hausdorff topological spaces" by "etale cohomology with compact support and coefficients in $\mathbf Q_\ell$ for some prime $\ell$ on the category of finite type separated $k$-schemes". </p> <p>Let $H^\cdot_c(-,\mathbf Q)$ denote cohomology with compact support and coefficients in $\mathbf Q$ on the category of para-compact Hausdorff topological spaces. For a finite type separated $\mathbf C$-scheme, write $e_c(X)$ for the Euler characteristic of $X$, i.e., $e_c(X) = \sum_{i} (-1)^i \dim_{\mathbf Q} H^i_c(X,\mathbf Q)$. Since $X$ is separated and of finite type, this is a well-defined integer. (Of course, I'm implicitly utilizing the analytification of $X$ here.)</p> <p><strong>Theorem.</strong> Let $\pi:X\to Y$ be a finite etale morphism of finite type separated $\mathbf C$-schemes. Then $e_c(X) = \deg \pi e_c(Y)$.</p> <p><em>Proof.</em> We may and do assume $X$ and $Y$ are connected. Also, we may and do assume $\pi:X\to Y$ is Galois. (In fact, let $P\to Y$ be a Galois closure of $X\to Y$. Let $G$ be the Galois group of $P\to Y$. Let $H$ be the subgroup of $G$ such that $P/H = X$. Then <code>$$e_c(Y) = \frac{e_c(P)}{\# G} = \frac{\# H}{\# G} e_c(X) = \frac{1}{\deg \pi} e_c(X)$$</code> and so the result follows in the general case.)</p> <p>Thus, we have a finite group $G$ acting freely (without fixed points) on $Y$ such that $X=Y/G$. Note that $\deg \pi = \vert G\vert$. Apply the Lefschetz trace formula to see that $Tr(g,H^\ast_c(Y)) =0$ for any $g\neq e$ in $G$. By character theory for $\mathbf Q_\ell[G]$, we may conclude that the element $$[H^\ast_c(Y,\mathbf Q_\ell)] := \sum (-1)^i [ H^i_c(Y,\mathbf Q_\ell)]$$ in the Grothendieck group $K_0(\mathbf Q_\ell[G])$ of finitely generated $\mathbf Q_\ell[G]$-modules is given by an integer multiple of $[\mathbf Q_\ell[G]]$; the class of the regular representation. So we may write $$[H^\ast_c(Y,\mathbf Q_\ell)] = m [\mathbf Q_\ell[G]],$$ where $m\in \mathbf Z$. Now, note that $H^i_c(X,\mathbf Q_\ell) = \left(H^i_c(Y,\mathbf Q_\ell)\right)^G$ for any $i\in \mathbf Z$. Therefore, we have that $$[H^\ast_c(X,\mathbf Q_\ell)] = m$$ in $K_0(\mathbf Q_\ell[G])$. In particular, we see that $e_c(X) = \dim_{\mathbf Q_\ell} [H^\ast_c(X,\mathbf Q_\ell)] = m$. We conclude that <code>$$e_c(Y) = \dim_{\mathbf Q_\ell} [H^\ast_c(Y,\mathbf Q_{\ell})]= m \vert G\vert = e_c(X) \vert G \vert = \deg \pi e_c(X).$$</code> QED.</p> <p>For completeness, here is what you can do for "ramified covers". Not surprisingly, the same equality holds up to a "correction term" coming from the branch locus.</p> <p><strong>Lemma.</strong> Let $M$ be a finite type separated $\mathbf C$-scheme. Let $N$ be a closed subscheme of $M$. Then $e_c(M) = e_c(N) + e_c(M\backslash N)$.</p> <p><em>Proof.</em> Mayer-Vietoris. QED</p> <p><strong>Corollary.</strong> Let $\pi:X\to Y$ be a finite flat surjective morphism, and let $D$ be a closed subscheme of $Y$ such that $\pi$ is etale over $Y\backslash D$. Then $$e_c(X) = \deg \pi e_c(Y) + e_c(\pi^{-1}D) - \deg\pi e_c(D) .$$</p> <p><em>Proof.</em> Write $U=Y\backslash D$ and $V=\pi^{-1}(U)$. Then $$e_c(X) = e_c(V) + e_c(\pi^{-1}D) = \deg \pi e_c(U) + e_c(\pi^{-1}D) = \deg \pi(e_c(Y) - e_c(D)) + e_c(\pi^{-1}D).$$ The first equality follows from the Lemma, the second from the Theorem and the third from the Lemma. QED</p> <p>We can use this Corollary to obtain a more precise description of the "error term" under some mild hypotheses. Recall that a strict normal crossings divisor on a smooth projective variety is a divisor whose irreducible components are smooth and intersect transversally.</p> <p><strong>Theorem.</strong> Let $D$ be a strict normal crossings divisor on a smooth projective connected variety $X$ over $k$. Let $U$ be the complement of the support of $D$ in $X$ and let $V\to U$ be a finite etale morphism with $V$ connected. Let $\pi:Y\to X$ be the normalization of $X$ in the function field of $V$. Then</p> <ol> <li>The singularities of $Y$ are quotient singularities (and thus rational singularities);</li> <li>The singularities of $Y$ lie in $\pi^{-1}D^{sing}$, where $D^{sing}$ is the singular locus of $D$;</li> <li>The morphism obtained by restriction <code>$\pi^{-1}(D-D^{sing})\to D-D^{sing}$</code> is etale;</li> <li>We have $$e_c(Y) = \deg \pi e_c(X) + e_c(\pi^{-1}(D^{sing}))-\deg \pi e_c(D^{sing}) +$$ $$e_c(\pi^{-1}(D-D^{sing})) - \deg \pi e_c(D-D^{sing}).$$</li> </ol> <p><em>Proof.</em> This is a long but not difficult proof. I can include the details if you'd like. For now, let me say that if you prove $Y$ has quotient singularities, it follows that $Y$ has rational singularities by a theorem of Viehweg; see "Rational singularities of higher dimensional schemes". To prove (1), (2) and (3) you use results from SGA1 on the fundamental group. Note that (4) follows from the Corollary, the Lemma and (3). QED</p> <p><strong>Final Remark.</strong> In the last formula $$e_c(\pi^{-1}(D-D^{sing})) - \deg \pi e_c(D-D^{sing})= e_c(D-D^{sing})(\deg \pi - d^\prime),$$ where $d^\prime$ is the degree of the finite etale morphism <code>$\pi^{-1}(D-D^{sing})\to D-D^{sing}$</code>. (In a previous version I thought this was always zero, because I mistakingly assumed $d^\prime = \deg \pi$.)</p> http://mathoverflow.net/questions/118849/is-the-set-of-surfaces-over-spec-z-with-ample-canonical-sheaf-empty Is the set of surfaces over Spec Z with ample canonical sheaf empty Ariyan Javanpeykar 2013-01-13T22:14:46Z 2013-01-25T18:16:08Z <p><strong>Main question.</strong> Does there exist a smooth projective morphism $X\to$ Spec $\mathbf Z$ of relative dimension two such that the canonical sheaf $\omega_{X_{\mathbf Q}}$ of the generic fibre $X_{\mathbf Q}$ is ample?</p> <p>Replacing "relative dimension two" by "relative dimension one", the answer is negative by a theorem of Abrashkin-Fontaine. I highly suspect the answer to be negative in this case too. Unfortunately, it is not known yet though as confirmed by Sándor.</p> <p><strong>Question 2.</strong> Does there exist a number field $K$ such that there are infinitely many $K$-isomorphism classes of smooth projective geometrically connected surfaces over $K$ with ample canonical sheaf and a smooth projective model over $O_K$? </p> <p>The answer is positive if we replace "surfaces" by "curves". And as Will points out the answer is positive in the higher-dimensional case.</p> <p>My main question is part of the arithmetic Shafarevich conjecture. As the terminology suggests, this conjecture is the <strong>arithmetic analogue</strong> of a conjecture for geometric objects. The latter (geometric) conjecture has been resolved by Arakelov, Bedulev, Kovács, Lieblich, Möller, Parshin, Viehweg, Zuo, et al. (Edit: Please see the references in Sándor's answer.) Its arithmetic analogue remains widely open for relative dimension $\geq 2$ to my knowledge, and was resolved in 1983 by Faltings for relative dimension 1.</p> <p>With my second question I would like to assure myself of the non-triviality of a higher-dimensional arithmetic Shafarevich conjecture. It turns out to be trivial.</p> <p>Let me state the results (due to the before-mentioned) in algebraic geometry relevant to this question. The base field is an algebraically closed field $k$ of characteristic zero.</p> <p><strong>Theorem 1.</strong> <em>(Higher-dimensional geometric analogue of main question)</em> There are no smooth projective (strongly?) non-isotrivial morphisms $X\to \mathbf P^1_k$ such that the canonical sheaf of the generic fibre of $X\to \mathbf P^1_k$ is ample. </p> <p><strong>Theorem 2.</strong> <em>("Folklore?" Higher-dimensional geometric analogue of second question)</em> Fix $d\geq 0$. There exists a smooth projective connected curve $C$ such that there are infinitely many isomorphism classes of (strongly?) non-isotrivial smooth projective morphisms $X\to C$ of relative dimension $d$ whose generic fibre has ample canonical sheaf.</p> <p>Now, Theorem 2 is one of the reasons that the following grand finiteness theorem is difficult.</p> <p><strong>Theorem 3.</strong> Let $C$ be a smooth projective connected curve and let $h$ be a polynomial. Then, there are only finitely many isomorphism classes of smooth projective (strongly?) non-isotrivial morphisms $X\to C$ whose generic fibre is canonically polarized with Hilbert polynomial $h$.</p> <p>Let me note that I am considering function fields over a field of characteristic zero to be analogous to Spec $\mathbf O_K$. I know some of you prefer function fields over finite fields, but regarding these questions the analogy also "works" to a certain extent.</p> <p>I might have stated Theorems 1-3 slightly incorrectly. In this case I apologize. (Also, I didn't state Theorem 3 in its full generality. The base curve doesn't need to be compact for instance.) Maybe, I should have only considered <strong>deformation types of families over $C$</strong> in the statements. </p> <p>Finally, let me point out some related MO questions:</p> <p><a href="http://mathoverflow.net/questions/111708/what-can-be-the-dimension-of-a-pointless-smooth-proper-z-scheme" rel="nofollow">http://mathoverflow.net/questions/111708/what-can-be-the-dimension-of-a-pointless-smooth-proper-z-scheme</a></p> <p><a href="http://mathoverflow.net/questions/9576/smooth-proper-scheme-over-z" rel="nofollow">http://mathoverflow.net/questions/9576/smooth-proper-scheme-over-z</a></p> http://mathoverflow.net/questions/111556/does-the-moduli-space-of-genus-three-curves-contain-a-complete-genus-two-curve Does the moduli space of genus three curves contain a complete genus two curve Ariyan Javanpeykar 2012-11-05T13:41:55Z 2012-11-06T19:18:13Z <p>Inspired by the question </p> <p><a href="http://mathoverflow.net/questions/103120/does-the-moduli-space-of-smooth-curves-of-genus-g-contain-an-elliptic-curve" rel="nofollow">http://mathoverflow.net/questions/103120/does-the-moduli-space-of-smooth-curves-of-genus-g-contain-an-elliptic-curve</a></p> <p>and its amazing answers, I ask (pure out of curiosity) whether the moduli space $M_3$ of (smooth projective connected) curves of genus $3$ contains a (smooth projective connected) curve of genus $2$. </p> <p>The existence of such a genus two curve is (Edit: <strong>stronger</strong>) than the existence of a surface $S$, a genus two curve $C$ and a smooth projective non-isotrivial morphism $S\to C$ whose fibres are genus three curves. </p> <p>If the answer is positive, how explicit can our answer be made? I'm already aware of the fact that $M_g$ contains a complete curve for all $g\geq 3$. For instance, in the paper by Chris Zaal </p> <p><a href="http://dare.uva.nl/document/38546" rel="nofollow">http://dare.uva.nl/document/38546</a></p> <p>many curves of some genus (I think 513) are shown to embed into $M_3$.</p> <p>Of course, by Shafarevich' conjecture, if $K(C)$ denotes the function field of $C$, there are only finitely many $K(C)$-isomorphism classes of genus three curves over $K(C)$ with good reduction over $C$. I'm asking whether there exists some genus two curve $C$ such that there exists a genus three curve over $K(C)$ with good reduction over $C$. </p> <p><strong>Edit:</strong> the arithmetic analogue also has a negative answer.</p> <p>The latter (<strong>weaker</strong>) phrasing of my question allows us to formulate an arithmetic analogue of the above question. (I know that I'm considering function fields over $\mathbf{C}$ and that some of you might argue function fields over $\mathbf{F}_p$ are a better analogue of number fields.) This arithmetic analogue reads as follows. There exists a number field of "genus two" such that there exist a genus three curve over $K$ with good reduction over the ring of integers of $K$. Here a number field of "genus two" should be a number field of absolute discriminant $e^2$. I'll take this to mean discriminant at most $8$.</p> <p><strong>Arithmetic analogue. (Abrashkin-Fontaine)</strong> There do not exist non-zero smooth abelian schemes over the ring of integers of a number field of absolute discriminant at most 8.</p> <p>There are many related questions I'd also like to ask. For example, what is the minimal $g$ such that $M_g$ contains a genus two curve? Or, what is the minimal $g$ such that $M_3$ contains a genus $g$ curve? And, finally, is there an example of a complete curve in $M_g$ which is defined over $\overline{\mathbf{Q}}$? (Edit: The answer to the last question is positive. This is explained in the comments below.)</p> http://mathoverflow.net/questions/110810/dual-reduction-graph-of-a-curve/110973#110973 Answer by Ariyan Javanpeykar for Dual (/reduction?) graph of a curve Ariyan Javanpeykar 2012-10-29T08:08:28Z 2012-10-29T08:08:28Z <p>I'm not sure if this should be posted as an answer, but it became too long to post as a comment.</p> <p>Let $X$ be a smooth projective geometrically connected curve of positive genus over a number field $K$. (The condition on the genus will be used below.) Let me explain some elements of the theory of models for $X$ over $O_K$.</p> <p>Firstly, there exists a model of $X$ over $O_K$. In fact, there is a closed immersion $X\to\mathbf{P}^n_{K}$. The Zariski closure of $X$ in $\mathbf{P}^n_{O_K}$ via $X\to \mathbf{P}^n_K \subset \mathbf{P}^n_{O_K}$ gives a model for $X$ over $\mathcal{O}_K$. It is a projective model. It is irreducible and reduced as a scheme. Normalizing this scheme gives a normal (projective) model. Now, you can use "resolution of singularities", e.g., Lipman's theorem, to obtain a regular model for $X$ over $O_K$. Then, subsequently contracting all the $-1$-curves (also called exceptional curves) on this regular model you will obtain a regular projective model $\mathcal{X}_{min}$ for $X$ over $O_K$ which is "minimal". We call $\mathcal{X}_{min}$ the minimal regular model of $X$ over $O_K$. It is this minimality condition which is quite natural (for curves of positive genus). </p> <p>Before continuing, let me give some references for the above paragraph. You can find them all in Liu's book.</p> <p>For basic facts on the fibres of a model for $X$ over $O_K$ see Chapter 8.3.1.</p> <p>Desingularization of a normal model is explained in Chapter 8.3.4. For example, a precise statement of Lipman's theorem can be found in Theorem 8.3.44. </p> <p>Exceptional divisors on a model are defined in Definition 9.3.1. </p> <p>There are two notions of minimality (Definition 9.3.12) which are shown to coincide in Corollary 9.3.24 (when the generic fibre has positive genus).</p> <p>Finally, the existence of the minimal regular model is obtained in Therem 9.3.21.</p> <p>To summarize, the minimal regular model exists and is unique. Thus, it is "natural" to look at the dual reduction graphs of this model.</p> <p>The (geometric) fibres of $\mathcal{X}_{min}$ are, in general, very complicated. Of course, by Proposition 8.3.11, almost all of them are smooth. But, the geometric fibre over a "bad" place will be a singular curve with "complicated" singularities.</p> <p>This brings us to semi-stable reduction. In fact, if your reduction isn't a smooth curve, you could hope for the next best thing: semi-stability.</p> <p>A curve over an algebraically closed field is semi-stable if it is connected, reduced and has only ordinary double singularities. The model $\mathcal{X}_{min}$ doesn't have semi-stable geometric fibres in general, but a deep theorem of Grothendieck and Mumford states that there exists a finite field extension $L/K$ such that the minimal regular model of $X_L$ over $O_L$ is semi-stable, i.e., its geometric fibres are semi-stable. Considering the reduction graph of this model is also very natural; see Definition 10.3.17 in Liu's book.</p> <p>(Caution: just because the singularities of the geometric fibres have "easy" singularities, doesn't mean the configuration of the irreducible fibres is easy. In fact, determining the configuration is a difficult problem in arithmetic geometry. For example, the semi-stable reduction of the modular curve $X_0(n)$ (for all $n$) has only been achieved recently by Jared Weinstein; see <a href="http://arxiv.org/abs/1010.4241" rel="nofollow">http://arxiv.org/abs/1010.4241</a> .)</p> <p>There is also the notion of stability. This is stronger than semi-stability. The stable reduction of a curve is also "natural" to consider.</p> <p>I'll finish with a quick note on elliptic curves.</p> <p>Let $E/K$ be an elliptic curve. Then you can consider the minimal regular model and, for some suitable $L/K$, the semi-stable reduction of $E_L$ over $O_L$.</p> <p>It is also natural to ask whether $E$ has a model over $O_K$ which extends the group structure and the smoothness property of $E/K$. Such a model doesn't exist in general if we demand properness. If we drop the properness condition, then such a model exists. (The finiteness conditions being as usual: the model is of finite type and separated over the base scheme Spec $O_K$.) You can then ask for a model which extends this group structure in the "best possible" way. This brings us to Neron models. Such a model for $E$ over $O_K$ always exists. In fact, you can show that the smooth locus <code>$\mathcal{E}_{min}^{sm}$</code> of the minimal regular model $\mathcal E_{min}$ of $E$ over $O_K$ is the Neron model of $E$ over $O_K$; see Theorem 10.2 for a discussion of this beautiful theory. </p> http://mathoverflow.net/questions/103308/overview-of-arakelov-intersection-theory-and-the-arakelov-chow-ring/103312#103312 Answer by Ariyan Javanpeykar for Overview of Arakelov intersection theory and the Arakelov Chow ring Ariyan Javanpeykar 2012-07-27T14:31:12Z 2012-07-27T14:31:12Z <p>A good reference in my humble opinion is Bost's paper in Bourbaki:</p> <p>Théorie de l'intersection et théorème de Riemann-Roch arithmétiques</p> <p>Séminaire BOURBAKI. Novembre 1990. 43ème année, 1990-91, n° 731</p> <p>Another reference would be Soule's book on Arakelov geometry "Lectures on Arakelov geometry" written with Abramovich, Burnol and Kramer. </p> <p>Finally, I know you didn't ask this, but in the case of arithmetic surfaces there are more references (besides Faltings' "Calculus on arithmetic surfaces" and Arakelov's original paper). For example, Deligne's paper "le determinant de la cohomologie" and R. de Jong's Ph.D. thesis: <a href="http://www.math.leidenuniv.nl/~rdejong/publications/thesis.pdf" rel="nofollow">http://www.math.leidenuniv.nl/~rdejong/publications/thesis.pdf</a> Also, Moret-Bailly's paper "Metriques permises" in Szpiro's 1985 Asterisque is wonderful.</p> http://mathoverflow.net/questions/102642/is-the-class-of-k-gonal-curves-dominant Is the class of $k$-gonal curves dominant Ariyan Javanpeykar 2012-07-19T11:05:36Z 2012-07-20T09:49:29Z <p>Before I start, let me make a note on terminology. Curves are always smooth projective connected curves over an algebraically closed field of characteristic zero.</p> <p>Let $\mathcal C$ be a class of curves. We say that $\mathcal C$ is dominant if, for all curves $X$, there exists a curve $Y$ in $\mathcal C$ and a finite morphism $Y\to X$.</p> <p>Bogomolov and Tschinkel proved that the class of hyperelliptic curves and their unramified covers is dominant. Manin proved that the class of modular curves $X(n)$ and their unramified covers is dominant. Both proofs rely on Abhyankar's Lemma.</p> <p>Let $k\geq 2$ be an integer. Let $\mathcal C_k$ be the class of $k$-gonal morphisms, i.e., the class of curves for which the gonality equals $k$.</p> <p>Q1. Is $\mathcal{C}_2$ dominant?</p> <p>Q2. Is $\mathcal{C}_k$ dominant?</p> <p>Q3. Is $\cup_{2 \leq j \leq k} \mathcal{C}_j$ dominant if $k>>0$?</p> <p>Let me repeat this in words. Let $X$ be a curve. Does there exist a $k$-gonal curve $Y$ and a finite morphism $Y\to X$?</p> <p>I'm mainly interested in the case $k=2$. In this case, it suffices to answer the following question.</p> <p>Q1b. Let $X$ be a curve. Does $\mathbf{P}^1$ admit a closed immersion into the symmetric product $X^{(2)}$?</p> http://mathoverflow.net/questions/68213/families-of-curves-for-which-the-belyi-degree-can-be-easily-bounded Families of curves for which the Belyi degree can be easily bounded Ariyan Javanpeykar 2011-06-19T14:54:00Z 2012-06-18T19:32:06Z <p>I know (edit: three) families of smooth projective connected curves over $\bar{\mathbf{Q}}$ for which the Belyi degree is not hard to bound from above. </p> <ol> <li><p>The modular curves $X(n)$. They are constructed by compactifying the quotient $Y(n) = \Gamma(n)\backslash \mathbf{H}$. The natural morphism $X(n) \longrightarrow X(1)$ is Belyi of degree $n^2$ (up to a constant factor). This also bounds the Belyi degree of a modular curve given by a congruence subgroup $\Gamma$. In general, Zograf proves that the Belyi degree of a (classical congruence) modular curve is bounded by $128(g+1)$.</p></li> <li><p>The Fermat curves $F(n)$. They are given by the equation $x^n+y^n+z^n =0$ in $\mathbf{P}^2$. The morphism $(x:y:z)\mapsto (x^n:z^n)$ is Belyi of degree $n^2$. It is known that $F(n)$ is <strong>not</strong> a modular curve for $n$ big enough. So this example is really different than the one above. (Also note that $n^2\leq 10g+10$ by the Plucker formula.)</p></li> <li><p>Wolfart curves are curves $X$ over $\overline{\mathbf{Q}}$ with a Galois Belyi morphism $X\to \mathbf{P}^1$; I took this terminology from a preprint by Pete L. Clark. Such curves are also called Galois Belyi covers or Galois three-point covers in the literature. The Belyi degree of a Wolfart curve is bounded by $84(g-1)$. (In particular, the latter implies that there are only finitely many Wolfart curves of given genus.)</p></li> </ol> <p>The following family of curves is not so easily dealt with.</p> <ol> <li>For an elliptic curve $E$ over the rational numbers, the Belyi degree can be bounded in the height of the $j$-invariant of $E$ following Belyi's proof of his theorem. This was written down explicitly by Khadjavi and Scharaschkin.</li> </ol> <p>I'm looking for families of curves for which the Belyi degree is easy to read off''. That is, a collection (finite or infinite) of smooth projective connected curves $X_i$ over $\bar{\mathbf{Q}}$ for which the Belyi degree can be bounded easily. </p> <p>Are there any other nice examples?</p> http://mathoverflow.net/questions/96621/shimura-taniyama-weil-vs-grothendiecks-dessins/96763#96763 Answer by Ariyan Javanpeykar for Shimura-Taniyama-Weil VS Grothendieck's dessins Ariyan Javanpeykar 2012-05-12T11:02:18Z 2012-05-12T11:08:09Z <p>This does not answer your question. But it was a bit too long to put as a comment.</p> <p>Firstly, it seems that the following old question is of some relevance.</p> <p><a href="http://mathoverflow.net/questions/68213/families-of-curves-for-which-the-belyi-degree-can-be-easily-bounded" rel="nofollow">http://mathoverflow.net/questions/68213/families-of-curves-for-which-the-belyi-degree-can-be-easily-bounded</a></p> <p>In fact, dessins $X\to \mathbf{P}^1$ are also called Belyi maps/morphisms/functions on $X$. I wanted to know of curves for which one has explicit bounds on the Belyi degree, i.e., the minimal degree of a dessin $X\to \mathbf{P}^1$. Here are the examples </p> <ol> <li>Fermat curves</li> <li>Modular curves (congruence or non-congruence)</li> <li>Hurwitz spaces (see JSE's answer to the above question)</li> <li>Galois Belyi curves = Wolfart-curves = Galois three-point covers</li> <li>Elkies' curves (see his answer to the above question).</li> </ol> <p>Let me elaborate on 2. If <code>$\Gamma\subset \mathrm{SL}_2(\mathbf{Z})$</code> is a finite index subgroup, you can consider the quotient <code>$Y_\Gamma = \Gamma\backslash \mathbf{H}$</code>, where $\mathbf{H}$ is the complex upper half-plane and <code>$\mathrm{SL}_2(\mathbf{Z})$</code> acts on $\mathbf{H}$ by Mobius transformations. The curve <code>$Y_\Gamma$</code> naturally inherits the structure of a connected Riemann surface from $\mathbf{H}$. We compactify $Y_\Gamma$ by adding "cusps". The compactification of $Y_\Gamma$ is usually denoted by $X_\Gamma$. Note that there is a natural map <code>$Y_\Gamma \to Y_{\mathrm{SL}_2(\mathbf{Z})} = Y(1)$</code> induced by the inclusion <code>$\Gamma\subset \mathrm{SL}_2(\mathbf{Z})$</code>. This morphism extends to the compactifications $X_\Gamma \to X(1)$ and induces a dessin $X_\Gamma \to \mathbf{P}^1(\mathbf{C})$ after you compose with the isomorphism given by the $j$-invariant <code>$j:X(1)\to\mathbf{P}^1(\mathbf{C}$</code>. (The branch points are the elliptic points $0$, $1728$ and the cusp $\infty$ of $X(1)$.)</p> <p>Let me adress your third question. The above is about your second question. I don't have much to say about your first question, unfortunately. What do you mean by a dessin which "captures" all elliptic curves over $\mathbf{Q}$? </p> <p>Firstly, assume that <code>$X\to \mathbf{P}^1$</code> is a dessin of prime degree. It's clear that this morphism will not factor.</p> <p>I get the feeling (but I might be wrong) that you are interested in modular parametrizations of elliptic curves in the following sense. You want to know whether the above explicit dessins on <code>$X_0(n)$</code> can be shown to factor through some elliptic curve. If this is the case, the answer is likely to be no for $n$ big.</p> <p>Now, you can bound the number of dessins on a curve $X$ of given degree $d$ by the number of dessins of degree $d$, i.e., the number of topological covers of <code>$\mathbf{P}^1-\{0,1,\infty\}$</code>.</p> <p>But your $H_{X,Y}$ will be zero or infinite. </p> <p>In fact, if it not zero then there exists a dessin $X\to \mathbf{P}^1$ which factors through a dessin $Y\to \mathbf{P}^1$. But Belyi proved that for any finite set $B\subset \mathbf{P}^1(\overline{\mathbf{Q}})$ there exists a dessin $R:\mathbf{P}^1_{\mathbf{Q}}\to\mathbf{P}^1_{\mathbf{Q}}$ (defined over $\mathbf{Q}$ even!) such that $R$ sends $B$ to the set <code>$\{0,1,\infty\}$</code>. So from a given factorization $X\to Y\to \mathbf{P}^1$ you can construct an infinite number of really different dessins (and associated factorizations).</p> <p>The former paragraph is just applying the fact that given a dessin $f:X\to \mathbf{P}^1$ you can construct an infinite number of dessins $g :X\to \mathbf{P}^1$ by composing $f$ with an arbitrary dessin on $\mathbf{P}^1$. (Belyi actually gave an algorithm to compute a dessin $R$ on $\mathbf{P}^1$ associated to $B$ as above.)</p> <p>So to make sense of your last "crazy" question, you might want to fix a dessin $X\to \mathbf{P}^1$ on $X$ and try to look at possible factorizations, where $Y\to \mathbf{P}^1$ is a dessin and $Y$ is not fixed. Thus, let $H_{\pi}$ be the number of pairs $(Y,f)$ up to isomorphism, where $f:Y\to \mathbf{P}^1$ is a dessin and there exists a factorization $g:X\to Y$ such that $\pi = fg$.</p> <p>I don't think it is possible to give a precise formula for $H_\pi$ easily, but it is certainly possible to bound this number in terms of the degree of your dessin.</p> http://mathoverflow.net/questions/95729/what-are-supersingular-varieties/95784#95784 Answer by Ariyan Javanpeykar for What are supersingular varieties? Ariyan Javanpeykar 2012-05-02T16:26:07Z 2012-05-02T17:48:45Z <p>This doesn't really answer your question. But I think you might find these comments interesting (even though I might not be saying anything you don't know).</p> <p>By a theorem of Mazur-Ogus (Katz' conjecture) the $m$-dimensional Newton polygon of a variety lies above or is equal to its $m$-dimensional Hodge polygon.</p> <p>A variety is ordinary if these polygons are equal (for all $m$).</p> <p>For abelian varieties the $m=1$ case suffices and you see that an abelian variety is ordinary iff it is ordinary in the usual sense.</p> <p>By a theorem of Grothendieck-Katz most varieties are ordinary. This is stated more precisely also in Illusie's paper you mention. </p> <p>You should take a look at Mazur's beautiful paper on Katz' conjecture.</p> <p><a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183533965" rel="nofollow">http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183533965</a></p> <p>Let me also talk about another nice subject which has to do with Newton polygons and supersingular varieties. Namely, "constructing" varieties with given Newton polygon.</p> <p>If you stick to the case of curves there are many open questions (to my knowledge). For example, Mazur asks in <em>loc. cit.</em> (page 659) if all five different possible Newton polygons arising from a smooth projective curve of genus $3$ allowed by the restraint of Poincaré duality really arise from some curve or not. I don't know if this question has been answered by now (and let me add that it might actually be answered by now). </p> <p>I can't really tell you anything else on supersingular varieties. It does seem to be a nice sport to look at "strata" in the moduli spaces of abelian varieties (=Shimura varieties). For example, every "symmetric" Newton polygon arises from an abelian variety (and the Newton polygon of an abelian variety is symmetric). See <a href="http://arxiv.org/abs/math/0007201" rel="nofollow">http://arxiv.org/abs/math/0007201</a> for even more beautiful statements.</p> <p>For "strata" of Shimura varieties see </p> <p><a href="http://arxiv.org/abs/1011.3230" rel="nofollow">http://arxiv.org/abs/1011.3230</a> (Wedhorn-Viehmann) <a href="http://arxiv.org/abs/1111.6830" rel="nofollow">http://arxiv.org/abs/1111.6830</a> (Kret)</p> <p>These have to do with showing existence of abelian varieties with certain Newton polygons.</p> <p>So the moral of the story is that it's already pretty difficult to prove that certain polygons arise from geometric objects, e.g., the case of the genus $3$ curves and the Shimura variety business.</p> http://mathoverflow.net/questions/88351/zografs-bound-on-the-index-of-a-modular-curve-for-shimura-curves Zograf's bound on the index of a modular curve for Shimura curves Ariyan Javanpeykar 2012-02-13T15:18:27Z 2012-02-13T21:44:56Z <p>I've been reading Voight's paper on Shimura curves and it prompted the following question; see <a href="http://www.cems.uvm.edu/~voight/articles/shimura-clay-proceedings-071707.pdf" rel="nofollow">http://www.cems.uvm.edu/~voight/articles/shimura-clay-proceedings-071707.pdf</a> for which notes I'm talking about</p> <p>Let $F$ be a totally real number field, $B$ a quaternion algebra which splits at exactly one real place, and $\mathcal{O}\subset B$ a maximal order. Let $\Gamma^B(1)$ be the associated discrete arithmetic subgroup of $\mathrm{PSL}_2(\mathbf{R})$. This group is the analogue of $\Gamma(1)=\mathrm{SL}_2(\mathbf{Z})$. Let $X^B(1) = \Gamma^B(1)\backslash \mathbf{H}$. </p> <p><strong>Question 1.</strong> Let $\Gamma \subset \Gamma^B(1)$ be a finite index subgroup. Do I understand correctly that $X^B(\Gamma) = \Gamma \backslash \mathbf{H}$ is a compact Riemann surface, and that the inclusion $\Gamma \subset \Gamma^B(1)$ induces a finite morphism $X^B(\Gamma) \to X^B(1)$? What are the branch points of this morphism?</p> <p>Zograf showed that, if $\Gamma \subset \mathrm{SL}_2(\mathbf{Z})$ is a congruence subgroup, the index of $\Gamma$ in $\mathrm{SL}_2(\mathbf{Z})$ is bounded by $128(g+1)$, where $g$ is the genus of the compactification of the Riemann surface $\Gamma \backslash \mathbf{H}$. (This is how he proved that there are only finitely many congruence subgroups of $\mathrm{SL}_2(\mathbf{Z})$ of given genus.)</p> <p><strong>Question 2.</strong> Is there a similar theorem for Shimura curves? That is, assume that $\Gamma \subset \Gamma^B(1)$ is a <em>congruence subgroup</em> (does this term make sense? what should it mean?). Is the index of $\Gamma$ in $\Gamma^B(1)$ bounded by the genus of $X^B(\Gamma)$? If yes, can we make this bound explicit?</p> <p>I'm not sure what I mean by a congruence subgroup of $\Gamma^B(1)$ yet, but I want the curve $X_\Gamma(B)$ to be a <strong><em>Shimura curve</em></strong>. </p> http://mathoverflow.net/questions/83440/do-divisors-of-degree-g-with-this-property-exist-in-general Do divisors of degree g with this property exist in general Ariyan Javanpeykar 2011-12-14T17:15:43Z 2011-12-14T21:29:55Z <p>I have the following question. It's a long shot, but worth the try.</p> <p>Let X be a compact connected Riemann surface of genus $g\geq 2$. Does there exist an effective divisor $D$ on $X$ of degree $g$ such that, if $D=\sum_{i=1}^g D_i$ is the prime decomposition of $D$, the image of the line bundle $\mathcal{O}_X(D_i-D_j)$ is torsion in the Picard group of $X$ for all $i,j = 1,\ldots,g$?</p> <p>(Note that the $D_i$ are not necessarily distinct.) </p> <p>The answer is yes for modular curves by Manin-Drinfeld. Namely, take $D$ to be a degree $g$ divisor supported on the cusps. The answer is also yes for Fermat curves.</p> <p>The answer would be no if there would exist a Jacobian variety without torsion. Fortunately, Jacobians have a lot of torsion. </p> <p>If it helps assume that $X$ can be defined over some number field (as an algebraic curve).</p> <p>If the answer to my question is not completely trivial, there should be some general theory about such divisors. If yes, does there exist a good reference?</p> http://mathoverflow.net/questions/83380/manin-drinfeld-and-constructing-a-finite-morphism-with-two-given-ramification-poi Manin-Drinfeld and constructing a finite morphism with two given ramification points Ariyan Javanpeykar 2011-12-14T00:07:02Z 2011-12-14T03:07:07Z <p>Fix a smooth projective connected curve $X$ over <code>$\overline{\mathbf{Q}}$</code> of genus $g\geq 1$ and distinct points <code>$x,y \in X$</code> such that $x-y$ has infinite order in the Jacobian. </p> <p>Can we always find a finite morphism $X\to \mathbf{P}^1$ which ramifies at $x$ and $y$?</p> <p>If not, can we always find $x,y\in X$ such that $x-y$ has infinite order in $\mathrm{Jac}(X)$ and a finite morphism $X\to \mathbf{P}^1$ which ramifies at $x$ and $y$?</p> <p><strong>Application:</strong> If one of the above questions has a positive answer, Belyi's theorem gives the existence of a finite morphism $X\to \mathbf{P}^1$ which ramifies over exactly three points and such that $x$ and $y$ ramify. From this it is easy to see that there exists a subgroup $\Gamma\subset \Gamma(2)$ of finite index such that $\Gamma$ gives a Belyi uniformization of $X$ and such that the Manin-Drinfeld theorem doesn't hold for $\Gamma$. </p> http://mathoverflow.net/questions/82232/does-each-finite-morphism-of-curves-have-a-model-whose-minimal-resolution-is-semi Does each finite morphism of curves have a model whose minimal resolution is semi-stable Ariyan Javanpeykar 2011-11-29T23:43:26Z 2011-11-30T23:29:30Z <p>Let $\pi:Y\to X$ be a finite morphism of smooth projective geometrically connected curves over a number field $K$. </p> <p><strong>Question.</strong> Does there exist a finite field extension $L/K$ and a regular model $\mathcal{X}/O_L$ for $X_L/L$ such that the minimal resolution of singularities of the normalization of $\mathcal{X}$ in $Y_L$ is semi-stable over $O_L$?</p> <p>The answer to this question is yes if $X=\mathbf{P}^1_K$. (Take $\mathcal{X} = \mathbf{P}^1_{O_L}$, where $L/K$ can be chosen using Corollary 2.8 in Liu's <em>Stable reduction of finite covers of curves</em>. ) More generally, if $X$ has potentially good reduction over $K$, the same argument works.</p> <p>Unfortunately, I can't seem to make this work for higher genus curves. What am I missing?</p> http://mathoverflow.net/questions/81040/do-etale-neighhbourhoods-of-a-subvariety-descend-along-base-field-extensions-doe/81158#81158 Answer by Ariyan Javanpeykar for Do etale neighhbourhoods of a subvariety descend along base field extensions; does normalization commute with etale base change? Ariyan Javanpeykar 2011-11-17T09:07:54Z 2011-11-17T09:07:54Z <p>The answers above completely answer your questions. Let me just point out the following useful reference (to me at least).</p> <p>For question 2, you could look at Chapter 4.1.2. of Liu's Algebraic geometry and arithmetic curves. See Definition 4.1.24.</p> <p>Using a standard argument with the trace form one can show that, for any integral normal noetherian scheme $X$ with function field $K(X)$, the normalization $X^\prime \to X$ of $X$ in a finite separable extension of $K(X)$ is a finite morphism (Proposition 4.1.25).</p> http://mathoverflow.net/questions/67571/can-we-bound-the-minimal-degree-of-a-field-extension-required-to-obtain-semi-stab Can we bound the minimal degree of a field extension required to obtain semi-stable reduction Ariyan Javanpeykar 2011-06-12T11:18:17Z 2011-11-14T19:55:39Z <p>Let $K$ be a number field and let $X$ be a smooth projective geometrically connected curve over $K$.</p> <p>There exists a finite field extension $L/K$ such that <code>$X_L=X\otimes_K L$</code> has semi-stable reduction, i.e., there exists a semi-stable arithmetic surface $\mathcal{X}$ over the ring of integers $O_L$ with generic fibre $L$-isomorphic to $X_L$. Let $L_m$ be such an extension of minimal degree over $K$.</p> <p><strong>Question 1.</strong> Can we bound $[L_m:K]$ in terms of data depending only on $X$? </p> http://mathoverflow.net/questions/80698/is-every-weil-divisor-on-an-arithmetic-surface-q-cartier Is every Weil divisor on an arithmetic surface Q-Cartier Ariyan Javanpeykar 2011-11-11T17:34:59Z 2011-11-11T18:54:49Z <p>This question is about a technical issue I ran into.</p> <p>Let $S$ be a connected 1-dimensional Dedekind scheme, and let $X\to S$ be a flat projective integral normal 2-dimensional scheme. (For simplicity, we can also assume that the generic fibre of $X\to S$ is smooth.)</p> <p>Is every Weil divisor on $X$ a $\mathbf{Q}$-Cartier divisor? That is, does every Weil divisor on $X$ have the property that a certain integer multiple is a Cartier divisor on $X$?</p> <p>The answer to this question is positive by Lemma 3.3 in Moret-Bailly's <em>Groupes de Picard et problemes de Skolem,I</em> if $S$ is affine, excellent and satisfies Condition (T) on page 162 of Moret-Bailly's article. In particular, if $S$ is the spectrum of the ring of integers of a number field the answer to this question is positive. </p> <p>I suspect that the answer to the above question is positive for any excellent Dedekind scheme. Is this known? </p> http://mathoverflow.net/questions/79281/dimension-of-affine-variety/79291#79291 Answer by Ariyan Javanpeykar for Dimension of affine variety Ariyan Javanpeykar 2011-10-27T17:58:42Z 2011-10-27T19:56:43Z <p>The codimension of $X=Z(f_1,\ldots,f_k)$ in $\mathbf{A}^n$ equals $k$, or equivalently, the dimension of $X$ is $n-k$, if $(f_1,\ldots,f_k)$ is a regular sequence. </p> <p>Let me explain what a regular sequence is. Sorry if I'm writing things you already know.</p> <p>Let $A$ be a noetherian ring. An element $x\in A$ is called <em>regular</em> if the multiplication by $x$ is injective. A sequence $(x_1,\ldots,x_n)$ of elements $x_1,\ldots,x_n\in A$ is said to be a <em>regular sequence</em> if $x_1$ is regular and the image of $x_i$ in $A/(x_1A+\ldots+ x_{i-1}A)$ is regular for all $i=2,\ldots,n$. </p> <p>You can use Krull's principal ideal theorem to show that any ideal $I$ of $A$ which can be generated by a regular sequence $(x_1,\ldots,x_r)$ satisfies $\textrm{ht}( I) = r$.</p> <p>So one way to find out if the dimension of $X$ is $n-k$ is to check the above condition. </p> <p>If $k=1$ and $f_1\neq 0$ we're good.</p> <p>Let's see how it goes for $k=2$. Let's suppose that $f_1\neq 0$ and that $f_2$ is not contained in the ideal $(f_1)$. Now, you have to check that the image of $f_2$ in $k[x_1,\ldots,x_n]/(f_1)$ is regular. So you compute the quotient and check if it's an integral domain. If it's an integral domain, we're good. If not, it might be a bit more difficult to check if $f_2$ is regular in $k[x_1,\ldots,x_n]/(f_1)$. I wouldn't know a fast way of checking if this element is a non-zero divisor at the moment. </p> <p>This is not a complete answer but I hope it at least helped a bit.</p> http://mathoverflow.net/questions/79003/is-there-an-easier-argument-to-prove-that-almost-all-of-these-curves-have-no-semi Is there an easier argument to prove that almost all of these curves have no semi-stable reduction Ariyan Javanpeykar 2011-10-24T17:51:55Z 2011-10-24T20:29:12Z <p>Fix a number field $K$ and a polynomial $F(x)\in K[x]$ of degree at least $4$. For a squarefree integer $d$, define the curve $X_d$ over $K$ by the equation $dy^2 = F(x)$. Note that the curves $X_d$ are isomorphic over $\overline{\mathbf{Q}}$. Therefore, they have the same (stable) Faltings height and the same genus.</p> <p>By a theorem of Faltings, the set of curves over $K$ of fixed genus which have semi-stable reduction over $K$ and bounded Faltings height is finite. </p> <p>Therefore, the curve $X_d$ has no semi-stable reduction over $K$ for all but finitely many $d$.</p> <p>It seems to me that using Faltings' theorem isa bit of overkill. We should be able to arrive at the above conclusion more directly. In fact, we should be able to say for which $d$ the curve has or has no semi-stable reduction. </p> <p>Can we prove that the curve $X_d$ has no semi-stable reduction over $K$ for all but finitely many $d$ without appealing to Faltings' theorem? Can we make our conclusion more explicit?</p> http://mathoverflow.net/questions/78460/learning-arakelov-geometry/78463#78463 Answer by Ariyan Javanpeykar for Learning Arakelov geometry Ariyan Javanpeykar 2011-10-18T15:20:23Z 2011-10-18T21:54:33Z <p>Dear Vamsi,</p> <p>A while ago I wrote my point of view on what "you should and shouldn't read" before studying Arakelov geometry. See</p> <p><a href="http://mathoverflow.net/questions/54603/what-should-i-read-before-reading-about-arakelov-theory/54615#54615" rel="nofollow">http://mathoverflow.net/questions/54603/what-should-i-read-before-reading-about-arakelov-theory/54615#54615</a></p> <p>Taking another look at that answer, it seems that my answer is written for people with a more algebraic background. I think the "road to Arakelov geometry" for someone from analysis is a bit different, but I'm convinced that the following is a good way to start for everyone.</p> <p>If you're more comfortable with analysis than algebraic geometry, I think a good idea would be to start with the analytic part of Arakelov geometry. This is explained very well in Chapter 1.1 of R. de Jong's thesis</p> <p><a href="http://www.math.leidenuniv.nl/~rdejong/publications/" rel="nofollow">http://www.math.leidenuniv.nl/~rdejong/publications/</a></p> <p>and P. Bruin's master's thesis (written under the supervision of R. de Jong and B. Edixhoven)</p> <p><a href="http://www.math.leidenuniv.nl/~pbruin/" rel="nofollow">http://www.math.leidenuniv.nl/~pbruin/</a></p> <p>These two explain very well what Faltings and Arakelov did in their articles. </p> <p>Since you don't want to apply the analysis to do intersection theory on an arithmetic surface, you don't have to go into this, I believe. (This is where schemes and number theory come into play.)</p> <p>Now, I think after reading the relevant parts in the above references, you could start reading papers about analytic torsion (assuming you're already familiar with what this is). There's many of these, but I'm not the person to tell you which one is the best to start with.</p> <p>Good luck!</p> http://mathoverflow.net/questions/72829/which-curves-have-stable-faltings-height-greater-or-equal-to-1 Which curves have stable Faltings height greater or equal to 1 Ariyan Javanpeykar 2011-08-13T10:56:18Z 2011-10-14T07:20:18Z <p>Let $Y$ be a smooth projective connected curve of genus $g>0$ over $\overline{\mathbf{Q}}$. Let $h_{\textrm{Fal}}(Y)$ be the Faltings height of $Y$. </p> <p><strong>Question 1.</strong> Can one classify or describe the curves $Y$ such that $h_{\textrm{Fal}}(Y) \geq 1$?</p> <p><strong>Question 2.</strong> For any $g>0$, does there exist a curve $Y$ of genus $g$ such that $h_{\textrm{Fal}}(Y) &lt;1$? </p> <p>Essentially, I would like to know which curves one is excluding by looking at curves $Y$ such that $h_{\textrm{Fal}}(Y) \geq 1$. </p> <p>A result of Bost says that the stable Faltings height of an abelian variety $A$ over $\overline{\mathbf{Q}}$ of dimension $g$ is bounded from below by $-\frac{1}{2}\log(2\pi)g$.</p> <p>By the Northcott property of the Faltings height, the set of curves of genus $g$ with $h_{\textrm{Fal}}(Y) &lt;1$ is finite. This means that I'm looking at the finite set of curves of genus $g$ with Faltings height not in the interval $$[-\frac{1}{2}\log(2\pi)g,1)\subset[-2/5g, 1).$$</p> <p><strong>Added:</strong> To answer Junkie's question, I'm aware of only one definition of the Faltings height of a curve over <code>$\overline{\mathbf{Q}}$</code>. There are several equivalent definitions, though. </p> <p>Let $X$ be a smooth projective curve of genus $g>0$ over $\overline{\mathbf{Q}}$. Let $K$ be a number field such that $X$ has a semi-stable regular model $p:\mathcal{X}\to \mathrm{Spec} O_K$ over the ring of integers $O_K$ of $K$. Then, the Faltings height $h_{\mathrm{Fal}}(X)$ of $X$ is the arithmetic degree $$h_{\mathrm{Fal}}(X):=\frac{\widehat{\mathrm{deg}} Rp_\ast \mathcal{O}_{\mathcal{X}}}{[K:\mathbf{Q}]},$$ where we endow the determinant of cohomology with the Arakelov-Faltings metric. This is well-defined, i.e., independent of the field $K$. By Serre duality, it coincides with <code>$$h_{\mathrm{Fal}}(X)=\frac{\widehat{\mathrm{deg}} p_\ast \mathcal{\omega}_{\mathcal{X}/O_K}}{[K:\mathbf{Q}]}.$$</code> It also coincides with the Faltings height of the Jacobian. All of this is explained in Section 4.4 of </p> <p><a href="http://www.math.univ-toulouse.fr/~couveig/book.htm" rel="nofollow">http://www.math.univ-toulouse.fr/~couveig/book.htm</a> </p> <p>For a curve over a number field, there is also the important <em>relative Faltings height</em>. This invariant depends on the number field $K$, though. </p> http://mathoverflow.net/questions/75943/the-locus-of-cyclic-covers-in-the-moduli-space-of-curves The locus of cyclic covers in the moduli space of curves Ariyan Javanpeykar 2011-09-20T13:10:45Z 2011-09-20T16:21:43Z <p>Let $\mathcal{M}_g$ be the moduli space of smooth curves of genus $g$. Let $Z$ be the closure in $\mathcal{M}_g$ of the set of smooth curves of genus $g$ which are a cyclic cover of the projective line. </p> <p><strong>Question.</strong> Is $Z$ irreducible?</p> <p><strong>Question.</strong> What is the dimension of $Z$? Do we have non-trivial bounds?</p> <p><strong>Question.</strong> Is $Z$ affine?</p> <p><em>Remark.</em> Let $W$ be the closure of the set of smooth curves which are a cyclic cover of the projective line of prime degree. Then it is known that $W$ is affine. Note that $W\subset Z$.</p> http://mathoverflow.net/questions/75678/the-smallest-positive-eigenvalue-and-the-length-of-the-shortest-geodesic The smallest positive eigenvalue and the length of the shortest geodesic Ariyan Javanpeykar 2011-09-17T16:01:48Z 2011-09-17T17:46:44Z <p>I'm confused about some things concerning lengths of geodesics on Riemann surfaces and positive eigenvalues of the Laplacian. Moreover, I'm also interested in the relation between these two.</p> <p>Let $X$ be a compact (connected) Riemann surface of genus $g\geq 2$.</p> <p>Since the complex upper half plane $\mathfrak{h}$ is the universal cover of $X$, we have that $X$ inherits the structure of a Riemannian manifold from $\mathfrak{h}$. The length of the shortest geodesic with respect to the smooth volume form on $X$ induced by the hyperbolic metric, denoted by $\ell_X$, is well-defined in this case. Let $\lambda_X$ be the smallest positive eigenvalue of the Laplace operator on $L^2(X)$.</p> <p><strong>Question.</strong> What's the relation between $\ell_X$ and $\lambda_X$? Is there some kind of correspondance? </p> <p>Now, let's suppose that $b_1,\ldots,b_n$ are points in $X$. Then $X$ is the compactification of a quotient <code>$G\backslash \mathfrak{h} = X\backslash \{b_1,\ldots,b_n\}$</code> by adding the cusps'' $b_1,\ldots,b_n$. (Note that $G \backslash \mathfrak{h}$ inherits the structure of Riemannian manifold from $\mathfrak{h}$.) In this case there is no shortest geodesic on $X$ (due to the existence of cusps). Let $\lambda_G$ be the smallest positive eigenvalue of the Laplace operator on $L^2(G\backslash \mathfrak{h}$.</p> <p><strong>Question.</strong> Does $\lambda_X$ equal $\lambda_G$?</p> http://mathoverflow.net/questions/75521/reference-request-parametrizing-covers-of-the-projective-line Reference request: parametrizing covers of the projective line Ariyan Javanpeykar 2011-09-15T14:17:09Z 2011-09-15T15:45:51Z <p>Hurwitz spaces (or Hurwitz schemes) parametrize covers of the projective line. One can do this in many ways. </p> <p>For example, one could fix the number $r$ of branch points, the degree $n$ of the cover and look only at simple covers of $\mathbf{P}^1$. This is usually denoted by $H_{r,n}$. Fulton defined this space as a scheme over $\textrm{Spec} \mathbf{Z}$ and showed that $H_{r,n} \otimes \mathbf{F}_p$ is irreducible for $p$ big enough.</p> <p>One could also fix a subset $B\subset \mathbf{P}^1$, the degree $n$ of the cover and look at covers of degree $n$ unramified outside $B\cup {\lambda }$, where $\lambda \in \mathbf{P}^1-B$ is allowed to vary. One can show that most curves arise as an irreducible component of such a space (Diaz, Donagi, Harbater).</p> <p>One could also look at Galois covers with a fixed Galois group, etc.</p> <p>In the end, there are many ways to parametrize covers of the projective line.</p> <p>Are there any standard references that contain the basics of Hurwitz spaces?</p> <p>At the moment I have at my disposal</p> <p>Work of M. Romagny, J. Bertin and S. Wewers (available on Romagny's website). These are very <em>stacky</em>.</p> <p>The article of Fulton <em>Hurwitz Schemes and Irreducibility of Moduli of Algebraic Curves</em>.</p> <p>Notes by Brian Osserman available on his website <em>(The representation theory, geometry, and combinatorics of branched covers.)</em>.</p> <p>The article of Diaz, Donagi and Harbater: <em>Every curve is a Hurwitz space</em>.</p> <p><strong>Question.</strong> What are the standard references for the basics of Hurwitz spaces/schemes?</p> http://mathoverflow.net/questions/22120/what-is-the-difference-between-grothendieck-groups-k-0x-vs-k0x-on-schemes/22158#22158 Answer by Ariyan Javanpeykar for What is the difference between Grothendieck groups K_0(X) vs K^0(X) on schemes? Ariyan Javanpeykar 2010-04-22T07:31:17Z 2011-09-14T20:47:59Z <p>Asking $K^0(X)$ to be isomorphic to $K_0(X)$ is not always "good enough". Of course, it will allow you to carry over constructions for $K_0(X)$ to $K^0(X)$, but not canonically. And it can happen that $K^0(X)\cong K_0(X)$ without $X$ being regular. For example, take $X= \textrm{Spec} A$, $A=k[x]/(x^n)$ with $n\geq 2$. Then you have an infinite resolution as given in David's answer for $k$. Computing $Tor^A_i(k,k)$ shows that $k$ has no finite resolution. (In fact, $Tor_i^A(k,k) = k^2$ for all $i>0$.) Now, although the above "existence of finite resolution" fails, it is not hard to see that $K^0(X)\cong \mathbf{Z}\cong K_0(X)$ in this case. (Use that $A$ is a local ring and the length map on $A$.) Of course, the natural map $K^0(X) \longrightarrow K_0(X)$ is not an isomorphism. (It is given by $1\mapsto n$.) </p> <p>[Edit: I added another example]</p> <p>[Edit 2: There was something wrong with the example below as noted by Michael. I fixed the problem]</p> <p>Let me also add to my answer the following "snake in the grass". If you work with general schemes, even if regular, one requires the extra assumption of "finite-dimensionality". For example, take the scheme $X=\textrm{Spec} (k \times k[t_1]\times k[t_1,t_2] \times \ldots)$. Now, even though $A = k\times k[t_1]\times\ldots$ is regular, there is an infinite resolution for $k$ of the form $$\ldots \longrightarrow A\longrightarrow A\longrightarrow A \longrightarrow k \longrightarrow 0$$ which corresponds geometrically to taking a point, then adding a line, then adding a plane, etc. Again, take the Tor's to see that $k$ has no finite resolution. Do note that $X$ is not noetherian.</p> <p>[Edit 3: I added the following for completeness]</p> <p>Let $X$ be a regular finite-dimensional scheme. Assume that $X$ has enough locally frees. (This notion also arose in <a href="http://mathoverflow.net/questions/25122/are-schemes-that-have-enough-locally-frees-necessarily-separated" rel="nofollow">http://mathoverflow.net/questions/25122/are-schemes-that-have-enough-locally-frees-necessarily-separated</a> ). Then the canonical morphism $K^0(X) \longrightarrow K_0(X)$ is an isomorphism. In the second example, $X=\textrm{Spec} \ A$ is regular, but not finite-dimensional. Does $X$ have enough locally frees?</p> http://mathoverflow.net/questions/73269/given-a-curve-under-which-condition-is-the-set-of-gonal-morphisms-finite Given a curve, under which condition is the set of gonal morphisms finite Ariyan Javanpeykar 2011-08-20T10:38:22Z 2011-08-20T13:02:22Z <p>Recently, in my research I bumped onto gonal morphisms. At the moment, my knowledge is based upon some things I read on the internet. Before stating my questions, I added some definitions/facts that might motivate the questions below.</p> <p>By a curve, I mean a smooth projective connected curve over $\mathbf{C}$. A non-constant morphism $\pi:X\longrightarrow \mathbf{P}^1$ is <em>gonal</em> if $\deg \pi$ is minimal. The <em>gonality</em> of a curve $X$, denoted by $\gamma_X$, is the degree of a gonal morphism $\pi:X\longrightarrow \mathbf{P}^1$. Thus, for example, a curve of genus $g\geq 2$ is hyperelliptic iff it is $2$-gonal. </p> <p>The hyperelliptic map of a hyperelliptic curve is unique. (Of course, here by unique we mean <em>unique up to composition with an isomorphism of the projective line</em>.) </p> <p>Edit: In the questions below, we consider the set of gonal morphisms of a curve modulo the action of Aut$(\mathbf{P}^1)$.</p> <p><strong>Fact 1.</strong> For any curve $X$ of genus $g\geq 2$, we have that $\gamma_X \leq [\frac{g+3}{2}]$.</p> <p><strong>Fact 2.</strong> For any integer $\gamma \geq 2$, the closure of the locus of $\gamma$-gonal curves in the moduli space $\mathcal{M}_g$ of smooth curves of genus $g\geq 2$ is irreducible of dimension $2g-5+2\gamma$.</p> <p><strong>Fact 3.</strong> For any prime number $p$ and integer $g\geq 2$ such that $g\geq (p-1)^2$, Accola showed that any $p$-gonal curve of genus $g$ has a unique gonal morphism.</p> <p>I can't prove these facts, but I do remember where I got them from. So if necessary I could give the references.</p> <p><strong>Question 1.</strong> Let $X$ be a $\gamma$-gonal curve of genus $g\geq 2$. Is the set of gonal morphisms for $X$ <em>modulo the action of Aut$(\mathbf{P}^1)$</em> finite?</p> <p>I expect the answer to this question to be negative if $g-\gamma$ is small. In view of Fact 3, I would like to propose the following question.</p> <p><strong>Question 2a.</strong> Fix $\gamma\geq 3$. Does there exist an integer $g_\gamma$ such that for any $g\geq g_\gamma$ and any $\gamma$-gonal curve $X$ of genus $g$, the gonal morphism for $X$ is unique?</p> <p><strong>Question 2b.</strong> Fix $\gamma\geq 3$. Does there exist an integer $g_\gamma$ such that for any $g\geq g_\gamma$ and any $\gamma$-gonal curve X of genus $g$, the set of gonal morphisms for $X$ is finite?</p> <p><strong>Question 3.</strong> Does there exist a positive integer $g_0$ with the following property? For any $g\geq g_0$ and curve $X$ of genus $g$, the set of gonal morphisms of $X$ is finite?</p> <p><strong>Question 4.</strong> Do there exist curves with infinitely many gonal morphisms? (Edit: In hindsight, this question is the same as Question 1.)</p> <p>I think it's clear that these questions aren't unrelated. They are all related to the set of gonal morphisms associated to a curve. It would be wonderful to know when this set is finite.</p> http://mathoverflow.net/questions/73162/dualizing-sheaf-on-varieties/73206#73206 Answer by Ariyan Javanpeykar for Dualizing sheaf on varieties Ariyan Javanpeykar 2011-08-19T08:15:23Z 2011-08-19T08:15:23Z <p>The answer to your question is positive and follows from Theorem 6.4.32 in Qing Liu's book <em>Algebraic geometry and arithmetic curves</em>. </p> <p>Note that Liu uses Corollary 6.4.13 in the statement of his Theorem. Moreover, the base scheme is a locally Noetherian scheme, e.g., the spectrum of a field.</p> http://mathoverflow.net/questions/71735/moving-a-weil-divisor-on-a-normal-surface-away-from-a-finite-set-of-closed-points Moving a Weil divisor on a normal surface away from a finite set of closed points Ariyan Javanpeykar 2011-07-31T16:38:28Z 2011-07-31T22:07:09Z <p>Let $Y$ be a normal surface and let $X$ be a closed subscheme of codimension 2, i.e., $X$ is a finite set of closed points.</p> <p>Let $D$ be a Weil divisor on $Y$. </p> <p><strong>Question.</strong> Does there exist a Weil divisor $E$ on $Y$ which is linearly equivalent to $D$ and does not go through $X$? (Edit: I do not assume $E$ to be effective.)</p> <p>Of course, when $D$ does not go through $X$ the answer is yes.</p> <p>When I say that $E$ does not go through $X$, I mean that the support of $E$ and $X$ are disjoint.</p> <p>I'm interested in this question in the most general set-up known. For example, $Y$ is an integral noetherian separated excellent normal 2-dimensional scheme. If you prefer sticking to algebraic surfaces, I would be glad to hear about what's possible in that case too.</p> <p>The motivation for this question comes from an article by Mumford in which he defines an intersection pairing on a normal surface. In this case $X$ is the singular locus of $Y$.</p> http://mathoverflow.net/questions/71741/moving-a-canonical-divisor-on-a-normal-surface-away-from-the-singular-locus Moving a canonical divisor on a normal surface away from the singular locus Ariyan Javanpeykar 2011-07-31T17:33:09Z 2011-07-31T18:28:26Z <p>In a previous question <a href="http://mathoverflow.net/questions/71735/moving-a-weil-divisor-on-a-normal-surface-away-from-a-finite-set-of-closed-points" rel="nofollow">http://mathoverflow.net/questions/71735/moving-a-weil-divisor-on-a-normal-surface-away-from-a-finite-set-of-closed-points</a> I probably asked for too much. As J.C. Ottem pointed out, it is not always possible to move a Weil divisor on a normal surface away from a given closed set of points. Fortunately in my set-up this generality is not required. Therefore, I propose the following more mild set-up and hope that the answer will be positive.</p> <p>Let $Y$ be a normal surface and let $K_Y$ be the Weil divisor obtained by taking the closure of a canonical divisor on the nonsingular locus of $Y$. </p> <p><strong>Question.</strong> Is $K_Y$ linearly equivalent to a divisor which does not go through the singular locus of $Y$?</p> <p>Again, by a normal surface I mean an integral normal excellent separated noetherian 2-dimensional scheme. I will also assume $Y$ to be (locally?) $\mathbf{Q}$-factorial in the motivation below.</p> <p><strong>Motivation.</strong> Given a resolution of singularities $\rho:Y^\prime\longrightarrow Y$, I would like to show that the intersection number $(\psi^\ast K_{Y},E) =0$, where $E$ is an exceptional component of $\psi$ and $\psi^\ast K_Y$ is the pull-back of the $\mathbf{Q}$-Cartier divisor $K_Y$. This will hold if I can move $K_Y$ away from the singular locus.</p> http://mathoverflow.net/questions/127500/localization-sequence-for-k0x Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-04-13T23:04:57Z 2013-04-13T23:04:57Z Hi john, have you seen these questions: <a href="http://mathoverflow.net/questions/35788/extending-vector-bundles-on-a-given-open-subscheme-reprise" rel="nofollow" title="extending vector bundles on a given open subscheme reprise">mathoverflow.net/questions/35788/&hellip;</a> &amp; <a href="http://mathoverflow.net/questions/22111/extending-vector-bundles-on-a-given-open-subscheme" rel="nofollow" title="extending vector bundles on a given open subscheme">mathoverflow.net/questions/22111/&hellip;</a> . You seem to be asking a related question. http://mathoverflow.net/questions/126076/simple-field-extension-and-rational-points/126130#126130 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-04-01T17:21:11Z 2013-04-01T17:21:11Z This is very nice, I must say! If I understand correctly, you show that for the polynomial $f$ we must add at least $two$ elements to $F$ to get a root of $f$ inside $F^\prime$. I think, but I might be wrong, that one can slightly alter your construction (by adding more variables, and writing down $F$, $f$ and $F^\prime$ analogous to your construction) to obtain examples where one must add at least $n$ (with $n$ fixed, but arbitrarily large) variables to $F$ to obtain a root of $f$ within $F^\prime$. http://mathoverflow.net/questions/126076/simple-field-extension-and-rational-points/126130#126130 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-03-31T23:35:49Z 2013-03-31T23:35:49Z May I ask how one sees that $x^p-sz^p$ equals $A(u^p, v^p,\sigma,\tau)$ for some $A$ over $\mathbf F_p$? http://mathoverflow.net/questions/124558/algebraicity-of-the-canonical-projection-x-gamma-to-x1-and-of-x-gamma/124569#124569 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-03-17T15:03:03Z 2013-03-17T15:03:03Z @pranavk You are absolutely right. The point wccanard makes is certainly important. The above point of view doesn't take any moduli interpretation into account which Guillaume probably wants. I recommend Guillaume to keep my answer in mind (also when dealing with certain higher-dimensional Shimura varieties), but to always try and use the moduli-interpretation as much as possible as wccanard does. It's more natural when working with modular curves. http://mathoverflow.net/questions/124558/algebraicity-of-the-canonical-projection-x-gamma-to-x1-and-of-x-gamma/124569#124569 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-03-16T20:40:47Z 2013-03-16T20:40:47Z 1. Yes. 2. V is the inverse image of U as you already suspected. Let me know if anything else seems obscure. http://mathoverflow.net/questions/121880/sections-of-the-cotangent-bundle-of-elliptic-surfaces Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-17T13:38:53Z 2013-02-17T13:38:53Z @rvarma. You're right I was thinking about the smooth case. http://mathoverflow.net/questions/121880/sections-of-the-cotangent-bundle-of-elliptic-surfaces Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-17T13:38:35Z 2013-02-17T13:38:35Z @Damian My apologies. I was implicitly thinking about the smooth case. http://mathoverflow.net/questions/121880/sections-of-the-cotangent-bundle-of-elliptic-surfaces Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-15T18:17:29Z 2013-02-15T18:17:29Z You could write down what the entire exact sequence should be. If I'm not mistaken, it should be $0\to f^\ast K\to \Omega\to \omega_f\to 0$, where $\omega_f$ is the relative dualizing sheaf. This is a line bundle on $X$. It is zero because $\omega_f$ reduces to the canonical sheaf on the generic fibre of $X\to C$ , and the generic fibre is an elliptic curve, right? If you don't like &quot;relative dualizing sheaves&quot;, then just note that the quotient of $\Omega$ by $f^\ast K$ reduces to the canonical sheaf on the generic fibre. So your &quot;surjective map&quot; should be an isomorphism. http://mathoverflow.net/questions/121295/are-the-chern-numbers-of-a-hyperbolic-type-compact-complex-manifold-bounded-in-te Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-10T10:37:35Z 2013-02-10T10:37:35Z Ow I misread your question. Thanks for the clarification. http://mathoverflow.net/questions/121310/points-of-minimum-arakelov-height-and-harmonic-arithmetical-varieties/121335#121335 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-10T00:28:13Z 2013-02-10T00:28:13Z Yes, I think it is a good idea to include the important parts of this discussion in your question. http://mathoverflow.net/questions/121310/points-of-minimum-arakelov-height-and-harmonic-arithmetical-varieties/121335#121335 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-09T22:35:47Z 2013-02-09T22:35:47Z Ok so now we have a positive answer conditional on a conjecture nobody has any idea about. :) I still need to check the Lemmata you cited from Szpiro's festschrift for myself, but that should be fine. For now, maybe you could try to disprove the Neron-Tate conjecture by proving these arithmetic surfaces are actually harmonic. Ha! xD http://mathoverflow.net/questions/121310/points-of-minimum-arakelov-height-and-harmonic-arithmetical-varieties/121335#121335 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-09T22:22:13Z 2013-02-09T22:22:13Z In the function field case we are just working with integers. I think it would be interesting to write down what the analogue of the above formulas involving the Neron-Tate height are, and then conclude in a similar fashion (maybe even unconditionally because there are no &quot;log&quot;'s appearing) http://mathoverflow.net/questions/121310/points-of-minimum-arakelov-height-and-harmonic-arithmetical-varieties/121335#121335 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-09T22:15:25Z 2013-02-09T22:15:25Z When I say &quot;has little chance&quot; I actually mean &quot;I have no idea&quot;. But translating all of this to the function field setting should give us satisfying answers. The problem here is really $\delta_{Fal}(X)$. http://mathoverflow.net/questions/121310/points-of-minimum-arakelov-height-and-harmonic-arithmetical-varieties/121335#121335 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-09T22:13:22Z 2013-02-09T22:13:22Z Yeah this will most probably never happen. The Noether formula says that $h_{Fal}(X) = e(X) + \Delta(X) +\delta_{Fal}(X) -4g\log(2\pi)$. Of course, $\Delta(X)$ is of the right form by definition: <code>$\Delta(X) = \log(\prod&#95;{\mathfrak p \subset O&#95;K} \# k(\mathfrak p)^{\delta&#95;{\mathfrak p}})/[K:\mathbf Q]$</code>. Unfortunately, the real number $h_{Fal}(X) - \delta_{Fal}(X) +4g\log(2\pi)$ has little chance of being of the form $\log(n)$ with $n\geq 2$ an integer. Maybe something weaker suffices? http://mathoverflow.net/questions/121310/points-of-minimum-arakelov-height-and-harmonic-arithmetical-varieties/121335#121335 Comment by Ariyan Javanpeykar Ariyan Javanpeykar 2013-02-09T22:01:52Z 2013-02-09T22:01:52Z Ok, that sounds good. Do I understand correctly that for genus $g\geq 2$ curves such that $e(X) = \log n$ for some positive integer $n\geq 2$ we can conclude that the pair $(\mathcal X,\omega_{\mathcal X/O_K})$ is also <b>not</b> strict harmonic?