User marc - MathOverflow

Decomposition theorem for principally polarized abelian varieties in positive characteristic.

2013-03-17T17:46:41Z

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].

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 $N_{\ell}=O_X(F_{\ell})$.

Denote by $p_M:M\rightarrow X_M=X/K(M)_0$ 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 $0$.

There are positive line bundles $\bar{M}\in Pic(X_M)$ and $\bar{N}_{\ell}\in Pic(X_{N_{\ell}})$ 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

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:

Decomposition Theorem. 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 $$(X,L)\simeq (X_M\times X_{N_1} \times \cdots \times X_{N_r}, q_M\bar{M}\otimes q_{N_1}^{\ast}\bar{N}_1 \otimes \cdots \otimes q_{N_r}^{\ast}\bar{N}_r)$$

Thanks in advance for any insights.

Any two curves over k homeomorphic

2011-09-23T01:32:23Z

Why are two curves over a field k homeomorphic?

I have been able to prove that any variety of positive dimension over a field k has the same cardinality as k.

Conjugation of Dolbeault cohomology and cup product.

2013-02-02T22:09:06Z

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 $$H^p(X,\Omega_X^q) \simeq \overline{H^q(X,\Omega_X^p)}$$

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

$\begin{array}{ccc} H^p(X,\Omega_X^q) & \stackrel{\cup v}{\longrightarrow} & H^{p+1}(X, \Omega_X^q) \\ \downarrow & & \downarrow \\ H^q(X,\Omega_X^p) & \longrightarrow &H^q(X,\Omega_X^{p+1}) \end{array}$

where the vertical maps are isomorphisms induced by conjugation. The bottom horizontal map looks like wedge product with the conjugate class $\bar{v}\in H^0(X,\Omega_X^1)$ . Is this true?

Extension class and cup product

2013-02-01T07:31:40Z

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 $Ext^1(F'',F')$.

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 $\theta$.

I was wondering whether someone could provide some insight on this last statement.

Projective bundle given by vanishing of a section

2012-11-16T06:43:48Z

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)

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 $$0\rightarrow N \rightarrow E \rightarrow K \rightarrow 0$$

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.

The projectibve bundle $\pi:\mathbb{P}(E^{\ast})\rightarrow Y$ comes with a tautological surjection $$\pi^{\ast}E^{\ast}\rightarrow \mathcal{O}_{\mathbb{P}(E^{\ast})}(1) \rightarrow 0$$

(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

$$s\in \Gamma(\mathbb{P}(E),\pi^{\ast}K \otimes \mathcal{O}_{\mathbb{P}(E^{\ast})}(1))$$

Then it is claimed that the zero-locus of this section gives precisely the subvariety $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$.

I am wondering whether this is obvious or not, but this correspondence is not apparent to me.

Thanks in advance for any insight.

Amplitude and bigness issues

2012-10-29T01:50:27Z

There are a couple of statements that I have read which are made as though they were trivial, but I am doubtful about them.

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?
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?

Thanks in advance for any insight.

Proper nonprojective surface

2012-05-07T02:34:06Z

I am trying to construct a proper non-projective surface following the indications in section III.5 in Hartshorne's 'Alebraic geometry'.

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 $\cdots \rightarrow \underbrace{H^1(X,\omega_X)}_0 \rightarrow \underbrace{H^1(X',\mathcal{O}_{X'}^{\ast})}_{Pic(X')} \rightarrow \underbrace{H^1(X,\mathcal{O}_X^{\ast})}_{Pic(X)} \stackrel{\delta}{\longrightarrow} \underbrace{H^2(X,\omega_X)}_k \rightarrow \cdots$

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

$\begin{array}{ccccccccc} 0 & \rightarrow & \check{C}^1(U,\omega) & \rightarrow & \check{C}^1(U,\mathcal{O}_{X'}^{\ast}) & \rightarrow & \check{C}^1(U,\mathcal{O}_X^{\ast}) & \rightarrow & 0 \\ && \downarrow && \downarrow && \downarrow && \\ 0 & \rightarrow & \check{C}^2(U,\omega) & \rightarrow & \check{C}^2(U,\mathcal{O}_{X'}^{\ast}) & \rightarrow & \check{C}^2(U,\mathcal{O}_X^{\ast}) & \rightarrow & 0 \end{array}$

where $U$ is the standard cover of $\mathbb{P}^2$.

Let $X'$ be the non-trivial extension given by the cocyle $\xi\in H^1(X,\Omega_X^1)$ given by $\xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_i}{x_j}\right)$.

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 $\check{C}^2(U,\omega_X)$.

In order to lift $\alpha$ to $\beta=(\beta_0,\beta_1,\beta_2)\in \check{C}^1(U,\mathcal{O}^{\ast}_{X'})$ we first need a description of $\check{C}^1(U,\mathcal{O}^{\ast}_{X'})=\bigoplus_{i<j} \Gamma(U_{ij},\mathcal{O}_{X'}^{\ast})$

I recall having read somewhere that

$\Gamma(U_{ij},\mathcal{O}_{X'})\cong \Gamma(U_{ij},\mathcal{O}_X)[\eta_{ij}]=k\left[\frac{x_i}{x_j},\frac{x_j}{x_i},\frac{x_k}{x_i}\right][\eta_{ij}]$

where $\eta_{ij}^2=0$. How does this isomorphism follow? What is the description of the lift $\beta$?

Thanks in advance for any insight.

Variational problem

2012-05-27T06:11:51Z

I would like to verify the following computation:

Consider the subspace $V=\left\{u\in H^1(\Omega): \quad \frac{1}{|\partial\Omega} \int_{\partial \Omega} u\,d\sigma=0 \right\}$ of $H^1(\Omega)$, which is a Hilbert space with inner product $<,>_{1,2}$ : I intend to show that the boundary value problem with weak formulation

$$\int_{\Omega} (\nabla u\cdot \nabla v+uv)\,dx=\int_{\Omega} fv \,dx, \quad \forall v\in V$$ has a unique solution

We verify that $(V,<,>_{1,2})$ 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 $(H^1(\Omega),<,>_{1,2})$ , which is complete, so there exists $v\in H^1(\Omega)$ such that $\lim_{k\to \infty} ||v_k-v||_{1,2}=0$ and it suffices to observe that actually $v\in V$, namely that $\frac{1}{|\partial \Omega|} \int_{\partial \Omega} v\,d\sigma=0$ .

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 $$\int_{\Omega} uv-v\Delta u\,dx+\int_{\partial \Omega} v\partial_{\nu}u\,d\sigma=\int_{\Omega} fv\,dx, \quad \forall v\in V$$

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

$$\left\{ \begin{array}{ll} u-\Delta u=f, & \Omega \\ \partial_{\nu}u=const, & \partial\Omega \end{array} \right.$$

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 $Lv=\int_{\Omega} fv \,dx$.

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 $V$.

With regards to coercivity, it follows from $$B(u,u)=\int_{\Omega} |\nabla u|^2\,dx+\int_{\Omega} u^2\,dx\geq ||\nabla u||_{1,2}^2$$

Continuity of $L$ follows from Cauchy-Schwarz's and Poincaré's inequalities, namely

$$|Lu|=\leq \int_{\Omega} |uv| \,dx\leq ||u||_{1,2}||v||_{1,2}\leq C_P||u||_{1,2}||\nabla v||_{1,2}$$

Hence $L\in H^{-1}(\Omega)$ and $||L||_{H^{-1}(\Omega)}\leq C_P||u||_{1,2}$ .

Thanks in advance for any insight.

Mistake in variational formulation of Neumann problem?

2012-05-14T19:17:20Z

Consider the problem

$$\left\{\begin{array}{ll} -\Delta u+c\cdot \nabla u=f, & \Omega \\ \partial_{\nu} u=0, & \partial \Omega \end{array} \right.$$

with $\Omega$ smooth, $c\in C^1(\bar{\Omega})$ and $f\in L^2(\Omega)$. Assume further that $div c=0$ and that $c\cdot \nu\geq c_0>0$.

The variational formlation of this problem consists of finding $u\in H^1(\Omega)$ such that $$\int_{\Omega} \nabla u\cdot \nabla v+(c\cdot \nabla u)v dx=\int_{\Omega} fv dx, \qquad \forall v\in H^1(\Omega)$$

Let $V=H^1(\Omega)$ and consider the bilinear form $$B(u,v)=\int_{\Omega} \nabla u\cdot \nabla v+(c\cdot \nabla u)v dx$$

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

$$\int_{\Omega} (c\cdot \nabla u)u dx=\frac{1}{2}\int_{\Omega} c\cdot \nabla (u^2) dx=\int_{\partial \Omega} u^2c\cdot \nu d\sigma$$

Recall also that in $H^1(\Omega)$, the norm $$||u||_{1,\partial}=\int_{\partial \Omega} u^2\,d\sigma+\int_{\Omega} |\nabla u|^2\,dx$$ is equivalent to the standard norm $||u||_{1,2}$

Thus, if $c\cdot \nu\geq c_0>0$ then $$B(u,u)=\int_{\Omega} |\nabla u|^2+(c\cdot \nabla u)u \,dx \geq \int_{\Omega} |\nabla u|^2+c_0\int_{\partial \Omega} u^2\,d\sigma\geq C||u||_{1,2}$$

so that $B$ is $V$-coercive. Checking that both B and $Lv=\int_{\Omega} fv\,dx$ are continuous, one concludes that the problem has a unique solution $u\in H^1(\Omega)$.

Variational formulation for bilaplacian

2012-05-13T21:31:15Z

I am trying to derive a variational formulation for the following problem $$\left\{ \begin{array}{ll} \Delta^2u=f, & \Omega \\ \Delta u+\rho \partial_{\nu}u=0, & \partial \Omega \end{array}\right.$$

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.

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.

Multiplying the equation by $v\in C_0^2(\Omega)$ and integrating over $\Omega$ we obtain $$\int_{\Omega} \Delta^2u\cdot v\,dx=\int_{\Omega} fv\,dx$$

and now integrating by parts (Green's identity) twice on the left hand side we obtain

\begin{eqnarray*} \int_{\Omega} \Delta^2u\cdot v\,dx &=& \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 \\ &\stackrel{Green}{=}& \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 \\ &=& \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*}

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 $$\int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx=\int_{\Omega} fv\,dx$$

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?

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 $$|B(u,v)|\leq \int_{\partial \Omega} \rho |\partial_{\nu}u \cdot \partial_{\nu}v|\,d\sigma + \int_{\Omega} |\Delta u\cdot \Delta v|\,dx$$

For the second integral we have $$\int_{\Omega} |\Delta u\cdot \Delta v|\,dx\leq ||\Delta u||_0||\Delta v||_0\leq ||u||_{H^2(\Omega)} ||v||_{H^2(\Omega)}$$ but what about the first one?

Thanks in advance for any insight.

Answer by Marc for Proper nonprojective surface

2012-05-07T22:45:57Z

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.

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 $\check{C}^1(U,\mathcal{O}_{X'}^{\ast})=\bigoplus_{i<j} \Gamma(U_{ij},\mathcal{O}_{X'})$ in the first place, where $\Gamma(U_{ij},\mathcal{O}_{X'})=k\left[\frac{x_i}{x_j},\frac{x_j}{x_i},\frac{x_k}{x_i}\right][\eta_{ij}]$.

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: $$d\left(\frac{x_i}{x_j}\right)\wedge \xi_{ij}=\frac{x_i}{x_j}\xi_{ij}\wedge \xi_{ij}=0, \quad d\left(\frac{x_j}{x_i}\right)\wedge \xi_{ij}=-\frac{x_j}{x_i}\xi_{ji}\wedge \xi_{ji}=0$$ $$d\left(\frac{x_k}{x_j}\right)\wedge \xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_k}{x_j}\right) \wedge d\left(\frac{x_i}{x_j}\right)$$

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 $\alpha=\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$ 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 $\check{C}^2(U,\mathcal{O}_{X'}^{\ast})=\Gamma(U_{012},\mathcal{O}_{X'}^{\ast})$ 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 $\check{C}^2)U,\omega)$.

Answer by Marc for Nonprojective Surface

2012-04-20T23:52:25Z

There is also an example in an Exercise from Hartshorne's Algebraic geometry involving infinitessimal extensions which I am trying to understand.

Let me recall some definitions and properties in the first place:

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 $Hom_A(\Omega_{A/k},\mathcal{I})$.
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 $X$.
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.
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 $H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$ . 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 $\mathcal{F}$.

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

$0\rightarrow \omega_X \rightarrow \mathcal{O}_{X'}^{\ast} \rightarrow \mathcal{O}_X^{\ast} \rightarrow 0$ inducing a long exact cohomology sequence $\cdots \rightarrow \underbrace{H^1(X,\omega_X)}_0 \rightarrow \underbrace{H^1(X',\mathcal{O}_{X'}^{\ast})}_{Pic(X')} \rightarrow \underbrace{H^1(X,\mathcal{O}_X^{\ast})}_{Pic(X)} \stackrel{\delta}{\longrightarrow} \underbrace{H^2(X,\omega_X)}_k \rightarrow \cdots$

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

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 $\check{C}^2(U,\omega)$?

Thanks in advance for any insight.

Isoperimetric inequality

2012-04-18T06:09:27Z

Let $M$ be a complete, non-compact, simply connected Riemannian manifold of dimension $n$ whose sectional curvatures are bounded above by $\kappa<0$. I want to prove that for any open subset $\Omega\subset M$ whose closure in $M$ is compact, the following inequality holds: $$\frac{Vol(\Omega)}{Vol(\partial \Omega)}\leq \frac{1}{(n-1)\sqrt{-\kappa}}$$

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 $0<\lambda_1<\lambda2<\cdots$ 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.

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?

Thanks in advance for any insight.

Inequality for the solution to the heat equation

2012-03-16T23:52:09Z

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

$$\int_a^b u^2(x,t)dx\leq e^{-\frac{2\pi^2}{(b-a)^2}t} \int_a^b u_0^2(x)dx$$

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

$$\int_a^b u^2(x)dx\leq (b-a)^2\int_a^b (u')^2(x)dx$$

using the fundamental theorem of calculus and Schwarz's inequality, but I am unsure as to how to proceed next.

The Fourier series expansion for $u(x,t)$ is given by

$$u(x,t)=\sum_{n\geq 1} b_n e^{-\lambda_n^2t} \sin\left(\frac{n\pi}{b-a}\right)x$$

where

$$b_n=\frac{2}{b-a}\int_a^b u_0(x) \sin\left(\frac{n\pi}{b-a}\right)x dx, \qquad \lambda_n=\frac{n\pi}{b-a}$$

so differentiating the solution and squaring doesn't seem to give the desired inequality and there is also the issue of the time dependence.

Hartshorne's proof of the birational invariance of the geometric genus

2012-03-15T06:56:30Z

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).

I shall sketch the proof and highlight my doubts.

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 $V\subset X$.

Along taxonomic lines, the proof goes like this:

We first prove that $f$ induces an injective map $f^{\ast}:\Gamma(X',\omega_{X'})\rightarrow \Gamma(V,\omega_V)$
Then we prove that the restriction map $\Gamma(X,\omega_X)\rightarrow \Gamma(V,\omega_V)$ is bijective, using the valuative criterion of properness.

From this it follows that $\rho_g(X')\leq \rho_g(X)$, and the reverse inequality follows by simmetry.

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?

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.

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.

Thanks in advance for any insight.

Local rings of non-closed points

2012-03-13T19:33:38Z

Non-singularity of an algebraic variety can be characterised in intrinsic terms by the fact that all local rings are regular local rings.

By a theorem of Serre, any localization of a regular local ring at a prime ideal is again a regular local ring.

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.

I am confused as to how to prove the former statement.

Comment by Marc

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.

Comment by Marc

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.