User marc - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:14:40Z http://mathoverflow.net/feeds/user/18013 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/124802/decomposition-theorem-for-principally-polarized-abelian-varieties-in-positive-cha Decomposition theorem for principally polarized abelian varieties in positive characteristic. Marc 2013-03-17T17:46:41Z 2013-03-23T03:13:34Z <p>In characteristic zero we have the following decomposition theorem for polarized abelian varieties: it gives an isomorphism between a PPAV and a product of PPAV's of lower dimension and is valid (as far as I have been able to find) in characteristic 0. I would like to know whether or not this holds in characteristic $p>0$. Below is the statement as it can be found in [Lange-Birkenhake, Theorem 4.3.1].</p> <p>Let $(X,L)$ be a principally polarized complex abelian variety and decompose the linear system $|L|$ as $$|L|=|M|+F_1+\cdots+F_r$$ where $M$ is the moving part and $F_1+\cdots+F_r$ is the decomposition of the fixed part into irreducible components. Denote $N_{\ell}=O_X(F_{\ell})$.</p> <p>Denote by <code>$p_M:M\rightarrow X_M=X/K(M)_0$</code> and $p_{N_{\ell}}:N \rightarrow X_{N_{\ell}}=N/K(N_{\ell})_0$ the canonical projections, where for a pair $(X,L\in Pic(X))$, if $\phi_L:X\rightarrow \hat{X}$ denotes the map $x\mapsto \phi_L(x)=t_x^{\ast}L\otimes L^{-1}$, we denote by $K(X)_0$ the connected component of $\ker\phi_L$ containing $0$.</p> <p>There are positive line bundles $\bar{M}\in Pic(X_M)$ and <code>$\bar{N}_{\ell}\in Pic(X_{N_{\ell}})$</code> such that $M=p_M^{\ast}\bar{M}$ and $N_{\ell}=p_{N_{\ell}}^{\ast}\bar{N}_{\ell}$ and the pairs $(X_M,\bar{M})$ and $(X_{N_{\ell}},\bar{N}_{\ell})$ are polarized abelian varieties. \medskip</p> <p>Consider the product $X_M\times X_{N_1} \times \cdots \times X_{N_r}$ and denote be $q_M$ and $q_{N_{\ell}}$ the corresponding projections. Then we have:</p> <p><strong>Decomposition Theorem</strong>. The map $$(p_M,p_{N_1},\ldots,p_{N_r}):X\longrightarrow X_M\times X_{N_1} \times \cdots \times X_{N_r}$$ is an isomorphism of polarized abelian varieties <code>$$(X,L)\simeq (X_M\times X_{N_1} \times \cdots \times X_{N_r}, q_M\bar{M}\otimes q_{N_1}^{\ast}\bar{N}_1 \otimes \cdots \otimes q_{N_r}^{\ast}\bar{N}_r)$$</code></p> <p>Thanks in advance for any insights.</p> http://mathoverflow.net/questions/76174/any-two-curves-over-k-homeomorphic Any two curves over k homeomorphic Marc 2011-09-23T01:32:23Z 2013-03-18T16:16:00Z <p>Why are two curves over a field k homeomorphic?</p> <p>I have been able to prove that any variety of positive dimension over a field k has the same cardinality as k.</p> http://mathoverflow.net/questions/120627/conjugation-of-dolbeault-cohomology-and-cup-product Conjugation of Dolbeault cohomology and cup product. Marc 2013-02-02T22:09:06Z 2013-02-02T22:09:06Z <p>Let $H^p(X,\Omega_X^q)$ denote the (p,q) Dolbeault cohomology group of a Kähler manifold X. Conjugation of forms induces an isomorphism $$H^p(X,\Omega_X^q) \simeq \overline{H^q(X,\Omega_X^p)}$$</p> <p>Let $v\in H^1(X,\mathcal{O}_X)$ and denote by $\bullet \cup v$ the cup product with this class, and consider the following diagram </p> <p><code>$\begin{array}{ccc} H^p(X,\Omega_X^q) &amp; \stackrel{\cup v}{\longrightarrow} &amp; H^{p+1}(X, \Omega_X^q) \\ \downarrow &amp; &amp; \downarrow \\ H^q(X,\Omega_X^p) &amp; \longrightarrow &amp;H^q(X,\Omega_X^{p+1}) \end{array}$</code></p> <p>where the vertical maps are isomorphisms induced by conjugation. The bottom horizontal map looks like wedge product with the conjugate class <code>$\bar{v}\in H^0(X,\Omega_X^1)$</code>. Is this true?</p> http://mathoverflow.net/questions/120499/extension-class-and-cup-product Extension class and cup product Marc 2013-02-01T07:31:40Z 2013-02-01T12:41:23Z <p>Recall that the group $Ext^1(F'',F')$ parametrizes extensions $$0 \rightarrow F' \rightarrow F \rightarrow F'' \rightarrow 0$$ as follos: given one such extension, consider the long exact cohomology sequence arising from the functor $Hom(F'',\bullet)$. If $\delta$ is the connecting coboundary map $$\delta:Hom(F'',F'') \rightarrow Ext^1(F'',F')$$ and we set $\theta\in Ext^1(F'',F')$ to be the image of the identity map on $F''$ under $\delta$, mthis process gives a 1-1 correspondence between isomorphism classes of extensions of $F''$ by $F'$, and elements of the group $Ext^1(F'',F')$.</p> <p>Note that if $E''$ is locally free, we have an isomorphism. $Ext^1(F'',F')=H^1(F''^{\ast}\otimes F')$. I have read that the coboundary map $\delta$ is actually obtained by taking cup-product with the extension class $\theta$.</p> <p>I was wondering whether someone could provide some insight on this last statement.</p> http://mathoverflow.net/questions/112556/projective-bundle-given-by-vanishing-of-a-section Projective bundle given by vanishing of a section Marc 2012-11-16T06:43:48Z 2012-11-16T06:55:10Z <p>This question might be tautological. It comes from a statement in the proof of the non-emptiness of the degeneracy loci of a vector bundle homomorphism that Prof. Lazarsfeld gives in his book "Positivity in AG II" (Theorem 7.2.1)</p> <p>Take a homomorphism of vector bundles $v:E\rightarrow F$ with kernel $F=\ker v$ and image $K=Im v$ and consider the short exact sequence $$0\rightarrow N \rightarrow E \rightarrow K \rightarrow 0$$</p> <p>The surjection $E^{\ast}\to N^{\ast}$ gives an embedding $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$ and we seek to realize $\mathbb{P}(N^{\ast})$ as the zero locus of the section of some vector bundle.</p> <p>The projectibve bundle $\pi:\mathbb{P}(E^{\ast})\rightarrow Y$ comes with a tautological surjection $$\pi^{\ast}E^{\ast}\rightarrow \mathcal{O}_{\mathbb{P}(E^{\ast})}(1) \rightarrow 0$$</p> <p>(given by the identity $E^{\ast}\rightarrow E^{\ast}$) and the composition of its dual $\mathcal{O}_{\mathbb{P}(E^{\ast})}(-1) \rightarrow \pi^{\ast}E$ with the pullback homomorphism $\pi^{\ast}v:\pi^{\ast}E \rightarrow \pi^{\ast}K$ gives a section </p> <p>$$s\in \Gamma(\mathbb{P}(E),\pi^{\ast}K \otimes \mathcal{O}_{\mathbb{P}(E^{\ast})}(1))$$</p> <p>Then it is claimed that the zero-locus of this section gives precisely the subvariety $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$.</p> <p>I am wondering whether this is obvious or not, but this correspondence is not apparent to me.</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/110955/amplitude-and-bigness-issues Amplitude and bigness issues Marc 2012-10-29T01:50:27Z 2012-10-29T16:22:36Z <p>There are a couple of statements that I have read which are made as though they were trivial, but I am doubtful about them.</p> <ol> <li><p>One is related to an example showing that the s-invariant of an ample line bundle on a projective variety X is an algebraic integer of degree $\leq dim X$. Recall that, given an ideal sheaf $\mathcal{I}\subset \mathcal{O}_X$ on a projective irreducible variety X, the s-invariant of $\mathcal{I}$ with respect to an ample line bundle L is $s_L(\mathcal{I})$ is the minimum $s\in \mathbb{R}$ such that $\mu^{\ast}(sL)-E$ is nef on $X'$, where $$\mu:X'=Bl_{\mathcal{I}}X\rightarrow X$$ is the blow-up along the ideal $\mathcal{I}$, with exceptional divisor E. The author claims that the class $s_L(\mathcal{I})L-E$ is nef (by definition) but not ample and then uses the Campana-Peternell theorem to conclude the result. How does the non-ampleness follow?</p></li> <li><p>Now let L be a big and nef divisor on a smooth projective variety X of dimension n and assume that the Seshadri constant of L at some point $x\in X$ is $\epsilon(L;x)>2n$ (we hence have the same inequality at a very general point). Take two general points $x,y\in X$ and consider the blow-up $\mu:X'=Bl_{{x,y}}X\rightarrow X$, with corresponding exceptional divisors $E_x$ and $E_y$. The divisor $\mu^{\ast}(\frac{1}{2}L)-nE_x$ is then nef (by definition) and big. How does bigness follow?</p></li> </ol> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/96177/proper-nonprojective-surface Proper nonprojective surface Marc 2012-05-07T02:34:06Z 2012-06-04T23:22:00Z <p>I am trying to construct a proper non-projective surface following the indications in section III.5 in Hartshorne's 'Alebraic geometry'.</p> <p>In $X=\mathbb{P}_k^2$ consider the sheaf of differential 2-forms $\omega_X$ and let $(X',\mathcal{I})$ be a nontrivial extension of $X$ by $\omega_X$,namely a scheme $X'$ together with a sheaf of ideals $\mathcal{I}$ with $\mathcal{I}^2=0$ such that $(X',\mathcal{O}_{X'}/\mathcal{I})\cong (X,\mathcal{O}_X)$. This gives a short exact sequence of sheaves $0\rightarrow \omega_X \rightarrow \mathcal{O}_{X'}^{\ast} \rightarrow \mathcal{O}_X^{\ast} \rightarrow 0$ inducing a long exact cohomology sequence <code>$\cdots \rightarrow \underbrace{H^1(X,\omega_X)}_0 \rightarrow \underbrace{H^1(X',\mathcal{O}_{X'}^{\ast})}_{Pic(X')} \rightarrow \underbrace{H^1(X,\mathcal{O}_X^{\ast})}_{Pic(X)} \stackrel{\delta}{\longrightarrow} \underbrace{H^2(X,\omega_X)}_k \rightarrow \cdots$</code> </p> <p>Non-projectivity of $X'$ shall follow from the fact that $Pic(X')=0$ and in order to see this it suffices to prove that $\delta$ is injective and nonzero. Since $Pic X\cong \mathbb{Z}$, any invertible sheaf is of the form $\mathcal{L}=\mathcal{O}_X(d)\cong \mathcal{O}_X(1)^{\otimes d}$ and it suffices to see that $\delta(\mathcal{O}_X(1))\neq 0$. I am confused as to how to carry out this computation since I guess I still do not understand very well the correspondence between infinitessimal extensions and the cohomology group. What I intend to do is to compute $\delta$ explicitly in the standard way, namely via the diagram</p> <p><code>$\begin{array}{ccccccccc} 0 &amp; \rightarrow &amp; \check{C}^1(U,\omega) &amp; \rightarrow &amp; \check{C}^1(U,\mathcal{O}_{X'}^{\ast}) &amp; \rightarrow &amp; \check{C}^1(U,\mathcal{O}_X^{\ast}) &amp; \rightarrow &amp; 0 \\ &amp;&amp; \downarrow &amp;&amp; \downarrow &amp;&amp; \downarrow &amp;&amp; \\ 0 &amp; \rightarrow &amp; \check{C}^2(U,\omega) &amp; \rightarrow &amp; \check{C}^2(U,\mathcal{O}_{X'}^{\ast}) &amp; \rightarrow &amp; \check{C}^2(U,\mathcal{O}_X^{\ast}) &amp; \rightarrow &amp; 0 \end{array}$</code></p> <p>where $U$ is the standard cover of $\mathbb{P}^2$. </p> <p>Let $X'$ be the non-trivial extension given by the cocyle $\xi\in H^1(X,\Omega_X^1)$ given by $\xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_i}{x_j}\right)$.</p> <p>The 1-cocycle $\alpha$ corresponding to $\mathcal{O}_X(1)$ in $\check{C}^1(U,\mathcal{O}_X^{\ast})$ is $\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$ and we only have to prove that it maps to some nonzero element in $\check{C}^2(U,\omega_X)$.</p> <p>In order to lift $\alpha$ to <code>$\beta=(\beta_0,\beta_1,\beta_2)\in \check{C}^1(U,\mathcal{O}^{\ast}_{X'})$</code> we first need a description of <code>$\check{C}^1(U,\mathcal{O}^{\ast}_{X'})=\bigoplus_{i&lt;j} \Gamma(U_{ij},\mathcal{O}_{X'}^{\ast})$</code> </p> <p>I recall having read somewhere that </p> <p><code>$\Gamma(U_{ij},\mathcal{O}_{X'})\cong \Gamma(U_{ij},\mathcal{O}_X)[\eta_{ij}]=k\left[\frac{x_i}{x_j},\frac{x_j}{x_i},\frac{x_k}{x_i}\right][\eta_{ij}]$</code></p> <p>where $\eta_{ij}^2=0$. How does this isomorphism follow? What is the description of the lift $\beta$?</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/98092/variational-problem Variational problem Marc 2012-05-27T06:11:51Z 2012-05-27T14:01:53Z <p>I would like to verify the following computation: </p> <p>Consider the subspace <code>$V=\left\{u\in H^1(\Omega): \quad \frac{1}{|\partial\Omega} \int_{\partial \Omega} u\,d\sigma=0 \right\}$</code> of $H^1(\Omega)$, which is a Hilbert space with inner product <code>$&lt;,&gt;_{1,2}$</code>: I intend to show that the boundary value problem with weak formulation</p> <p><code>$$\int_{\Omega} (\nabla u\cdot \nabla v+uv)\,dx=\int_{\Omega} fv \,dx, \quad \forall v\in V$$</code> has a unique solution</p> <p>We verify that $(V,&lt;,>_{1,2})$ is a Hilbert space in the first place, for which it suffices to prove that it is complete. Let ${v_k}\subset V$ be a Cauchy sequence: then it is also Cauchy in <code>$(H^1(\Omega),&lt;,&gt;_{1,2})$</code>, which is complete, so there exists $v\in H^1(\Omega)$ such that <code>$\lim_{k\to \infty} ||v_k-v||_{1,2}=0$</code> and it suffices to observe that actually $v\in V$, namely that <code>$\frac{1}{|\partial \Omega|} \int_{\partial \Omega} v\,d\sigma=0$</code>.</p> <p>Recall that by Green's identity we have that $$\int_{\Omega} v\Delta u\,dx=\int_{\partial \Omega} v\partial_{\nu}u\,d\sigma-\int_{\Omega} \nabla u\cdot \nabla v\,dx$$ so that our variational problem is equivalent to finding $u\in V$ such that $$\int_{\Omega} uv-v\Delta u\,dx+\int_{\partial \Omega} v\partial_{\nu}u\,d\sigma=\int_{\Omega} fv\,dx, \quad \forall v\in V$$</p> <p>In particular this must hold for functions $v\in V_0$ (with zero trace) so that the integral equation $$\int_{\Omega} uv-v\Delta u\,dx=\int_{\Omega} fv\,dx, \quad \forall v\in V_0$$ must hold. This in turn implies that $$\int_{\partial \Omega} v\partial_{\nu}u\,d\sigma=0, \quad \forall v\in V$$ Since these functions $v\in V$ also satisfy $\frac{1}{|\partial\Omega} \int_{\partial \Omega} u\,d\sigma=0$, we conclude that $\partial_{\nu}u$ must be constant, whence the boundary value problem associated with the variational formulation above is given by </p> <p><code>$$\left\{ \begin{array}{ll} u-\Delta u=f, &amp; \Omega \\ \partial_{\nu}u=const, &amp; \partial\Omega \end{array} \right.$$</code></p> <p>In order to establish the well posedness of the problem, we verify the conditions of the Lax-Milgram theorem for the bilinear form $B(u,v)=\int_{\Omega} (\nabla u\cdot \nabla v+uv)\,dx$ and the linear functional $Lv=\int_{\Omega} fv \,dx$.</p> <p>Continuity of $B$ follows from $$|B(u,v)|=\left|\int_{\Omega} (\nabla u\cdot \nabla v+uv)\,dx\right|\stackrel{[1]}{\leq} ||\nabla u||_0||\nabla v||_0+||u||_0||v||_0 \stackrel{[2]}{\leq} ||\nabla u||_0||\nabla v||_0+C_P^2||\nabla u||_0||\nabla v||_0$$ where in [1] we use the Cauchy-Schwarz inequality and in [2] we use Poincare's inequality, which can be easily proved to hold in $V$.</p> <p>With regards to coercivity, it follows from $$B(u,u)=\int_{\Omega} |\nabla u|^2\,dx+\int_{\Omega} u^2\,dx\geq ||\nabla u||_{1,2}^2$$</p> <p>Continuity of $L$ follows from Cauchy-Schwarz's and Poincaré's inequalities, namely </p> <p><code>$$|Lu|=\leq \int_{\Omega} |uv| \,dx\leq ||u||_{1,2}||v||_{1,2}\leq C_P||u||_{1,2}||\nabla v||_{1,2}$$</code></p> <p>Hence <code>$L\in H^{-1}(\Omega)$</code> and <code>$||L||_{H^{-1}(\Omega)}\leq C_P||u||_{1,2}$</code>.</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/96938/mistake-in-variational-formulation-of-neumann-problem Mistake in variational formulation of Neumann problem? Marc 2012-05-14T19:17:20Z 2012-05-14T19:17:20Z <p>Consider the problem</p> <p><code>$$\left\{\begin{array}{ll} -\Delta u+c\cdot \nabla u=f, &amp; \Omega \\ \partial_{\nu} u=0, &amp; \partial \Omega \end{array} \right.$$</code></p> <p>with $\Omega$ smooth, $c\in C^1(\bar{\Omega})$ and $f\in L^2(\Omega)$. Assume further that $div c=0$ and that $c\cdot \nu\geq c_0>0$. </p> <p>The variational formlation of this problem consists of finding $u\in H^1(\Omega)$ such that $$\int_{\Omega} \nabla u\cdot \nabla v+(c\cdot \nabla u)v dx=\int_{\Omega} fv dx, \qquad \forall v\in H^1(\Omega)$$</p> <p>Let $V=H^1(\Omega)$ and consider the bilinear form $$B(u,v)=\int_{\Omega} \nabla u\cdot \nabla v+(c\cdot \nabla u)v dx$$ </p> <p>I am wondering whether there is something wrong with the following argument showing existence and uniqueness of a solution to the previous problem using the Lax-Milgram theorem. By direct computation, from the above assumptions we easily conclude that</p> <p>$$\int_{\Omega} (c\cdot \nabla u)u dx=\frac{1}{2}\int_{\Omega} c\cdot \nabla (u^2) dx=\int_{\partial \Omega} u^2c\cdot \nu d\sigma$$ </p> <p>Recall also that in $H^1(\Omega)$, the norm <code>$$||u||_{1,\partial}=\int_{\partial \Omega} u^2\,d\sigma+\int_{\Omega} |\nabla u|^2\,dx$$</code> is equivalent to the standard norm $||u||_{1,2}$</p> <p>Thus, if $c\cdot \nu\geq c_0>0$ then <code>$$B(u,u)=\int_{\Omega} |\nabla u|^2+(c\cdot \nabla u)u \,dx \geq \int_{\Omega} |\nabla u|^2+c_0\int_{\partial \Omega} u^2\,d\sigma\geq C||u||_{1,2}$$</code></p> <p>so that $B$ is $V$-coercive. Checking that both B and <code>$Lv=\int_{\Omega} fv\,dx$</code> are continuous, one concludes that the problem has a unique solution $u\in H^1(\Omega)$.</p> http://mathoverflow.net/questions/96854/variational-formulation-for-bilaplacian Variational formulation for bilaplacian Marc 2012-05-13T21:31:15Z 2012-05-14T05:45:56Z <p>I am trying to derive a variational formulation for the following problem <code>$$\left\{ \begin{array}{ll} \Delta^2u=f, &amp; \Omega \\ \Delta u+\rho \partial_{\nu}u=0, &amp; \partial \Omega \end{array}\right.$$</code></p> <p>where $\rho>0$ is constant. I intend to show that the right functional setting is $H^2(\Omega)\cap H^1_0(\Omega)$ and to prove that the resulting problem is well posed.</p> <p>I am confused as to how to establish the right functional setting, so for a start I choose $C_0^2(\Omega)$ as a space of test functions (functions in $C^2(\Omega)$ compactly supported in $\Omega$) so that the boundary condition makes sense.</p> <p>Multiplying the equation by $v\in C_0^2(\Omega)$ and integrating over $\Omega$ we obtain $$\int_{\Omega} \Delta^2u\cdot v\,dx=\int_{\Omega} fv\,dx$$</p> <p>and now integrating by parts (Green's identity) twice on the left hand side we obtain </p> <p><code>\begin{eqnarray*} \int_{\Omega} \Delta^2u\cdot v\,dx &amp;=&amp; \int_{\Omega} div \nabla \Delta u)\,dx \stackrel{Green}{=} \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma- \int_{\Omega} \nabla \Delta u\cdot \nabla v \,dx \\ &amp;\stackrel{Green}{=}&amp; \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma - \int_{\partial \Omega} \Delta u\cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx \\ &amp;=&amp; \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma + \int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx \end{eqnarray*}</code></p> <p>where in the last equality I use the boundary condition. Now, enlarging the space of test functions by taking the closure of $C_0^2(\Omega)$ in $H^2(\Omega)$, namely $H_0^2(\Omega)\subset H_0^1(\Omega)\cap H^2(\Omega)$ the first integral vanishes (v has zero trace) so our variational formulation is $$\int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx=\int_{\Omega} fv\,dx$$</p> <p>How can I rigorously conclude that I need to take the whole $H_0^1(\Omega)\cap H^2(\Omega)$ as my space of test functions?</p> <p>In order to prove that the problem is well posed I intend to use Lax-Milgram theorem as usual, but I am confused as to how to tackle the integral over $\partial \Omega$. I define a bilinear form $$B(u,v)=\int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx$$ and I want to check continuity and coercitivity. For the first one I have $$|B(u,v)|\leq \int_{\partial \Omega} \rho |\partial_{\nu}u \cdot \partial_{\nu}v|\,d\sigma + \int_{\Omega} |\Delta u\cdot \Delta v|\,dx$$</p> <p>For the second integral we have <code>$$\int_{\Omega} |\Delta u\cdot \Delta v|\,dx\leq ||\Delta u||_0||\Delta v||_0\leq ||u||_{H^2(\Omega)} ||v||_{H^2(\Omega)}$$</code> but what about the first one?</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/96177/proper-nonprojective-surface/96274#96274 Answer by Marc for Proper nonprojective surface Marc 2012-05-07T22:45:57Z 2012-05-07T22:45:57Z <p>Thank you for your answer: regarding the second question, I just found some notes by prof. Mckernan in which he addresses this issue although there are some details that escape me.</p> <p>We have $\mathcal{O}_X(1)$ represented by the 1-cocyle $\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$ on $U_{01}$, $U_{12}$ and $U_{20}$ respectively, where $U_{ij}=U_i\cap U_j$, which we want to lift to <code>$\check{C}^1(U,\mathcal{O}_{X'}^{\ast})=\bigoplus_{i&lt;j} \Gamma(U_{ij},\mathcal{O}_{X'})$</code> in the first place, where $\Gamma(U_{ij},\mathcal{O}_{X'})=k\left[\frac{x_i}{x_j},\frac{x_j}{x_i},\frac{x_k}{x_i}\right][\eta_{ij}]$.</p> <p>The 1-cocycle $\xi\in H^1(X,\omega_X^1)$ defining $X'$ determines a derivation $D\in Hom(\Omega_X^1,\omega_X)$ on $U_{ij}$ as follows: we have an identification $\Omega_X \rightarrow Hom(\Omega_X,\omega_X)$ given by $v \mapsto (w\mapsto w\wedge v)$. The 1-cocycle $\xi=(\xi_{ij})$ hence determines the derivation: $$d\left(\frac{x_i}{x_j}\right)\wedge \xi_{ij}=\frac{x_i}{x_j}\xi_{ij}\wedge \xi_{ij}=0, \quad d\left(\frac{x_j}{x_i}\right)\wedge \xi_{ij}=-\frac{x_j}{x_i}\xi_{ji}\wedge \xi_{ji}=0$$ $$d\left(\frac{x_k}{x_j}\right)\wedge \xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_k}{x_j}\right) \wedge d\left(\frac{x_i}{x_j}\right)$$</p> <p>Then, computing in terms of the isomorphism $\phi:\Gamma(U_{12},\mathcal{O}_{X'}^{\ast})\rightarrow k\left[\frac{x_1}{x_2},\frac{x_2}{x_1},\frac{x_0}{x_1}\right]$, the lift $\beta=(\beta_2,\beta_0,\beta_1)\in \check{C}(U,\mathcal{O}_{X'}^{\ast})$ of <code>$\alpha=\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$</code> satisfies $$\phi(\beta_2)=\frac{x_1}{x_0}, \quad \phi(\beta_0)=\frac{x_2}{x_1}+\eta_{12}\frac{x_0}{x_1}d\left(\frac{x_2}{x_0}\right) \wedge d\left(\frac{x_1}{x_0}\right), \quad \phi(\beta_1)=\frac{x_0}{x_2}$$ I am not sure about why this holds yet, but then he concludes that the image of $\beta$ in <code>$\check{C}^2(U,\mathcal{O}_{X'}^{\ast})=\Gamma(U_{012},\mathcal{O}_{X'}^{\ast})$</code> is represented by $$\frac{x_1}{x_0}\left(\frac{x_2}{x_1}+\eta_{12}\frac{x_0}{x_1}d\left(\frac{x_2}{x_0}\right)\wedge d\left(\frac{x_1}{x_0}\right) \right) \frac{x_0}{x_2}=1+ \eta_{12}\frac{x_0}{x_2}d\left(\frac{x_2}{x_0}\right)\wedge d\left(\frac{x_1}{x_0}\right)$$ which lifts to the element $\frac{x_0}{x_2}d\left(\frac{x_2}{x_0}\right)\wedge d\left(\frac{x_1}{x_0}\right)$ of $\check{C}^2)U,\omega)$.</p> http://mathoverflow.net/questions/3624/nonprojective-surface/94702#94702 Answer by Marc for Nonprojective Surface Marc 2012-04-20T23:52:25Z 2012-04-21T00:56:39Z <p>There is also an example in an Exercise from Hartshorne's <em>Algebraic geometry</em> involving infinitessimal extensions which I am trying to understand.</p> <p>Let me recall some definitions and properties in the first place:</p> <ul> <li><p>Infinitessimal lifting property: given an algebraically closed field $k$, a finitely generated $k$-algebra $A$ with $X=\mbox{Spec } A$ non-singular, and an exact sequence $0\rightarrow \mathcal{I} \rightarrow B' \rightarrow B \rightarrow 0$, where $B,B'$ are k-agebras and $I\subset B'$ is an ideal such that $\mathcal{I}^2=0$, any k-algebra homomorphism $A\rightarrow B$ lifts to a h-algebra homomorphism $A\rightarrow B'$, and two such homomorphism differ by a k-derivation of A into $\mathcal{I}$, namely an element in $Hom_A(\Omega_{A/k},\mathcal{I})$.</p></li> <li><p>An infinitessimal extension of a k-scheme $X$ by a coherent sheaf $\mathcal{F}$ is a pair $(X',\mathcal{I})$ where $X'$ is a k-scheme and $\mathcal{I}$ is a sheaf of ideals on $X'$ with $\mathcal{I}^2=0$ and such that we have isomorphisms $(X',\mathcal{O}_{X'}/\mathcal{I})\cong (X,\mathcal{O}_X)$ (as k-schemes) and $\mathcal{I}\cong \mathcal{F}$ (as $\mathcal{O}_X$-modules). For instance, the trivial extension of $X$ by $\mathcal{F}$ is given by the pair $(X,\mathcal{F})$, where the $X$ has structure sheaf $\mathcal{O}_X'=\mathcal{O}_X\oplus \mathcal{F}$ with product $(a\oplus f)\cdot (a'\oplus f')=aa'\oplus (af'+a'f)$, so that $\mathcal{F}$ becomes an ideal sheaf in $X$.</p></li> <li><p>If $X=\mbox{Spec }A$ is affine and $\mathcal{F}$ is a coherent sheaf, then any extension is isomorphic to the trivial one: we just use the previous lifting property to construct a splitting of an appropriate short exact sequence.</p></li> <li><p>There is a correspondence between isomorphism classes of infinitessimal extensions of a k-scheme $X$ by a coherent sheaf $\mathcal{F}$ and the cohomology group $H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$ where $\mathcal{F}_X$ is the tangent sheaf of $X$. If $(X',\mathcal{I})$ is an infinitessimal extension of $X$ by $\mathcal{F}$ and and ${U_i}$ is an affine open cover of $X$ (so that sheaf cohomology is isomorphic to Cech cohomology) then on every open affine set the extension is trivial, namely of the form $(U_i,\mathcal{I}_{|U_i}=\mathcal{O}_{U_i}\oplus \mathcal{F}_{|U_i})$. It is easy to see from the construction of the trivialisation (choosing a lift, and noting that the difference of two lifts gives an element of $Hom_A(\Omega_{A/k},\mathcal{I})$) that this gives a cocyle in <code>$H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$</code>. Conversely, given a cocyle $\xi=(\xi_{ij})\in \check{H}^1(X,\mathcal{F}\otimes \mathcal{T}_X)$ and an open affine cover ${U_i}$, on each $U_i$ we have a trvial extension $(U_i,\mathcal{F}_{|U_i|}$ with $\mathcal{O}_{|U_i}'\cong\mathcal{O}_{U_i}\oplus \mathcal{F}_{|U_i}$ and we can glue them all via $\xi=(\xi_{ij})$ to give an extension of $X$ by $\mathcal{F}$.</p></li> </ul> <p>Then Hartshorne suggests that we perform the following computation: let $X=P_k^2$ and consider the sheaf of differential 2-forms $\omega_X$; then $H^1(X,\Omega_X^1)\cong H^1(X,\omega_X\otimes \mathcal{T}_X)$ and a nontrivial extension $X'$ of $X$ by $\omega_X$ is given by the cocylce $\xi \in H^1(X,\omega_X^1)$ given over $U_{ij}=U_i\cap U_j$ (where the ${U_i}$ are the standard open subsets covering $P_k^2$) by $\xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_i}{x_j}\right)$. This is our target proper non-projective surface and in order to see that it is indeed non-projective we shall prove that it has no ample invertible sheafs (in fact no invertible sheaf at all, namely $Pic X'=0$). We have a short exact sequence</p> <p>$0\rightarrow \omega_X \rightarrow \mathcal{O}_{X'}^{\ast} \rightarrow \mathcal{O}_X^{\ast} \rightarrow 0$ inducing a long exact cohomology sequence <code>$\cdots \rightarrow \underbrace{H^1(X,\omega_X)}_0 \rightarrow \underbrace{H^1(X',\mathcal{O}_{X'}^{\ast})}_{Pic(X')} \rightarrow \underbrace{H^1(X,\mathcal{O}_X^{\ast})}_{Pic(X)} \stackrel{\delta}{\longrightarrow} \underbrace{H^2(X,\omega_X)}_k \rightarrow \cdots$</code> </p> <p>and in order to see that $Pic X'==$ it suffices to prove that $\delta$ is injective and nonzero. Since $Pic X\cong \mathbb{Z}$, any invertible sheaf is of the form $\mathcal{L}=\mathcal{O}_X(d)\cong \mathcal{O}_X(1)^{\otimes d}$ and it suffices to see that $\delta(\mathcal{O}_X(1))\neq 0$. I am confused as to how to carry out this computation since I guess I still do not understand very well the correspondence between infinitessimal extensions and the cohomology group. What I intend to do is to compute $\delta$ explicitly in the standard way, namely via the diagram</p> <p><code>$\begin{array}{ccccccccc} 0 &amp; \rightarrow &amp; \check{C}^1(U,\omega) &amp; \rightarrow &amp; \check{C}^1(U,\mathcal{O}_{X'}^{\ast}) &amp; \rightarrow &amp; \check{C}^1(U,\mathcal{O}_X^{\ast}) &amp; \rightarrow &amp; 0 \\ &amp;&amp; \downarrow &amp;&amp; \downarrow &amp;&amp; \downarrow &amp;&amp; \\ 0 &amp; \rightarrow &amp; \check{C}^2(U,\omega) &amp; \rightarrow &amp; \check{C}^2(U,\mathcal{O}_{X'}^{\ast}) &amp; \rightarrow &amp; \check{C}^2(U,\mathcal{O}_X^{\ast}) &amp; \rightarrow &amp; 0 \end{array}$</code></p> <p>The cycle corresponding to $\mathcal{O}_X(1)$ in $\check{C}^1(U,\mathcal{O}_X^{\ast})$ is $\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$. How does it map down to $\check{C}^2(U,\omega)$?</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/94365/isoperimetric-inequality Isoperimetric inequality Marc 2012-04-18T06:09:27Z 2012-04-19T19:13:40Z <p>Let $M$ be a complete, non-compact, simply connected Riemannian manifold of dimension $n$ whose sectional curvatures are bounded above by $\kappa&lt;0$. I want to prove that for any open subset $\Omega\subset M$ whose closure in $M$ is compact, the following inequality holds: $$\frac{Vol(\Omega)}{Vol(\partial \Omega)}\leq \frac{1}{(n-1)\sqrt{-\kappa}}$$</p> <p>The constant on the right gives a lower bound for the first Dirichlet eigenvalue of the Laplace operator. If the metric on $M$ is given by $ds^2=g_{ij}dx^idx^j$, then $$\Delta=\frac{1}{\sqrt{\det g}}\sum_{i,j} \frac{\partial}{\partial x^i}\left(\sqrt{\det g} g_{ij} \frac{\partial}{\partial x^j}\right)$$ If $0&lt;\lambda_1&lt;\lambda2&lt;\cdots$ are the Dirichlet eigenvalues of $-\Delta$, by a theorem of Mckean we have an inequality $$\lambda_1(M)\geq \frac{1}{4}(n-1)^2k$$ for a Riemannian manifold satisfying the conditions above.</p> <p>Is there a way to relate the first eigenvalue to the ratio of volumes so as to prove the isoperimetric inequality above or is all this the wrong strategy?</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/91434/inequality-for-the-solution-to-the-heat-equation Inequality for the solution to the heat equation Marc 2012-03-16T23:52:09Z 2012-03-18T20:00:57Z <p>I am trying to prove that the solution $u(x,t)$ to the heat equation $u_t=u_{xx}$ on an interval (a,b) which satisfies homogeneous Dirichlet boundary conditions $u(a,t)=u(b,t)=0$ and initial consition $u(x,0)=u_0(x)$ satisfies the inequality</p> <p>$$\int_a^b u^2(x,t)dx\leq e^{-\frac{2\pi^2}{(b-a)^2}t} \int_a^b u_0^2(x)dx$$</p> <p>So far, for any function $u\in \mathcal{C}^1[a,b]$ such that $u(a)=u(b)=0$, I have proved that the following holds </p> <p>$$\int_a^b u^2(x)dx\leq (b-a)^2\int_a^b (u')^2(x)dx$$ </p> <p>using the fundamental theorem of calculus and Schwarz's inequality, but I am unsure as to how to proceed next.</p> <p>The Fourier series expansion for $u(x,t)$ is given by</p> <p>$$u(x,t)=\sum_{n\geq 1} b_n e^{-\lambda_n^2t} \sin\left(\frac{n\pi}{b-a}\right)x$$</p> <p>where</p> <p>$$b_n=\frac{2}{b-a}\int_a^b u_0(x) \sin\left(\frac{n\pi}{b-a}\right)x dx, \qquad \lambda_n=\frac{n\pi}{b-a}$$</p> <p>so differentiating the solution and squaring doesn't seem to give the desired inequality and there is also the issue of the time dependence.</p> http://mathoverflow.net/questions/91256/hartshornes-proof-of-the-birational-invariance-of-the-geometric-genus Hartshorne's proof of the birational invariance of the geometric genus Marc 2012-03-15T06:56:30Z 2012-03-15T06:56:30Z <p>I am confused about a couple of steps in the proof of the birational invariance of the geometric genus (Theorem II.8.19 in Hartshorne's Algebraic Geometry).</p> <p>I shall sketch the proof and highlight my doubts.</p> <p>Let $X,X'$ be two birationally equivalent nonsingular projective varieties over a field k. Hence there is a birational map $X-->X'$ represented by a morphism $f:V\rightarrow X'$ for some largest open subset $V\subset X$.</p> <p>Along taxonomic lines, the proof goes like this:</p> <ol> <li>We first prove that $f$ induces an injective map $f^{\ast}:\Gamma(X',\omega_{X'})\rightarrow \Gamma(V,\omega_V)$</li> <li>Then we prove that the restriction map $\Gamma(X,\omega_X)\rightarrow \Gamma(V,\omega_V)$ is bijective, using the valuative criterion of properness.</li> </ol> <p>From this it follows that $\rho_g(X')\leq \rho_g(X)$, and the reverse inequality follows by simmetry.</p> <p>In the proof of step 1: the map $f$ induces an isomorphism $U\cong f(U)$ for some open subset $U\subset V\subset X$ and then Hartshorne claims that this implies that f induces an isomorphism $\omega_{V|U}\cong \omega_{X'|f(U)}$. Why is that?</p> <p>In the proof of step 2: from the valuative criterion of properness it follows that $\textrm{codim }(X\setminus V,X)\geq 2$. In order to prove that $\Gamma(X,\omega_X)\rightarrow \Gamma(V,\omega_V)$ is bijective it suffices to prove it on open sets $U\subset X$ trivializing the canonical sheaf $\omega_{X|U}\cong \mathcal{O}$, namely that $\Gamma(U,\mathcal{O}_U)\rightarrow \Gamma(U\cap V,\mathcal{O}_U\cap V)$ is bijective.</p> <p>Since $X$ is nonsingular, from the first remark in the previous paragraph we have that $\textrm{codim }(U\setminus U\cap V,U)\geq 2$ and then Hartsorne claims that the result (bijectivity) follows immediately from the fact that for an integrally closed Noetherian domain $A$, we have $A=\bigcap_{\textrm{ht } \mathfrak{p}=1} A_{\mathfrak{p}}$. I do not see this either.</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/91106/local-rings-of-non-closed-points Local rings of non-closed points Marc 2012-03-13T19:33:38Z 2012-03-13T20:21:03Z <p>Non-singularity of an algebraic variety can be characterised in intrinsic terms by the fact that all local rings are regular local rings.</p> <p>By a theorem of Serre, any localization of a regular local ring at a prime ideal is again a regular local ring.</p> <p>If ones proves that the local ring at any non-closed point is a localization of a local ring at a closed point, by the previous theorem it suffices to check non-singularity at closed points.</p> <p>I am confused as to how to prove the former statement.</p> http://mathoverflow.net/questions/124802/decomposition-theorem-for-principally-polarized-abelian-varieties-in-positive-cha Comment by Marc Marc 2013-03-18T19:26:49Z 2013-03-18T19:26:49Z Thanks for your reply. That is actually my doubt. I think the proof does extend to characteristic p, and I was asking for confirmation, since I have only seen it proven for complex abelian varieties. The proof is an easy inductive argument which relies on a Riemann-Roch computation and some intersection number computations. But I was wondering whether I am missing something, since I have been unable to find the theorem stated in positive characteristic. Regards. http://mathoverflow.net/questions/3624/nonprojective-surface/94702#94702 Comment by Marc Marc 2012-04-20T23:55:48Z 2012-04-20T23:55:48Z I don't undertand why it reads so bad. I have compiled this with a tex editor and there are no errors.