User jvp - MathOverflow

Answer by jvp for Strictness of the inequality relating the Iitaka dimension and algebraic dimension

2013-02-14T23:30:20Z

Given a compact complex manifold $X$, there exists a meromorphic map $F: X\dashrightarrow Z$ to a projective manifold such that every meromorphic function on $X$ is the pull back under $F$ of a rational function on $Z$. The pull-back of an ample line-bundle on $Z$ realizes the equality between algebraic and Iitaka's dimensions.

Answer by jvp for which homogeneous polynomials split into linear factors?

2012-10-10T23:31:38Z

There are Brill's equations. Look for them at the book by Gelfand, Kapranov, and Zelevinski. In general Brill's equations do not generate the ideal of totally decomposable polynomials, see this paper.

Answer by jvp for Is SL(2,C)/SL(2,Z) a quasi-projective variety?

2012-08-30T21:55:46Z

No, the quotient is not quasi-projective. In the paper Invariant meromorphic functions on complex semisimple Lie groups by D. N. Ahiezer you can find the following result.

Theorem. Let $G$ be a connected semisimple linear algebraic group defined over the rationals and $\Gamma$ be a subgroup of $G(\mathbb Q)$ which is Zariski dense in $G$. Then there are no invariant analytic hypersurfaces in $G$ invariant by the action of $\Gamma$. In particular, if the quotient $G/ \Gamma$ exists as a complex variety then every meromorphic function on it is constant.

This implies that the quotient is not quasi-projective and even more: it cannot be holomorphically embedded in any algebraic variety.

Answer by jvp for Affine Cremona group

2011-09-14T13:28:57Z

There is one obvious representation which takes an automorphism of $\mathbb A^n$ to the determinant of its Jacobian. Let $G_n$ be its kernel. It carries a natural structure of infinite-dimensional algebraic group. Shafarevich proved that $G_n$ is simple as an algebraic group in this paper. Perhaps this suffices to answer your question but I am not sure.

Edit. According to Jérémy's comment below the proof of Shafarevich does not work. Apparently, the simplicity of $G_n$ as an algebraic group is an open problem for $n \ge 3$.

However it is known that $G_2$, as an abstract group, is not simple. This was proved by Danilov in this other paper. According to Furter and Lamy, Danilov shows that the normal subgroup generated by $(ea)^{13}$, where $a = (y,−x)$ and $e = (x,y+ 3x^5 − 5x^4)$, is a strict subgroup of $G_2$. For a more general statement see the paper By Furter and Lamy.

It is also known that planar Cremona group ${\mathrm{Bir}}(\mathbb P^2)$ is not simple. This was proved recently by Cantat and Lamy.

Answer by jvp for Embedding algebraic surfaces in projective space

2012-06-17T14:58:07Z

Perhaps this paper of Harris is what you are looking for: A bound on the geometric genus of projective varieties. It generalizes Castelnuovo's bound to smooth projective varieties of arbitrary dimension. In page 44, we find the following statement:

Theorem. Let $V$ be a non-degenerated smooth and irreducible variety of dimension $k$ and degree $d$ in $\mathbb P^n$. If we set $M=\left[ \frac{d-1}{n-k}\right]$ then $$ h^0(V,\Omega^k_V) \le \binom{M}{k+1} (n-k) + \binom{M}{k} (d-1-M(n-k)). $$ In particular, if $d \le k(n-k) + 1$ then $h^0(V, \Omega^k_V)=0.$

Answer by jvp for Non-bimeromorphic compactifications

2012-01-18T16:07:47Z

Another example, different but diffeomorphic to Polizzi's, is given by
$E \times \mathbb C$, where $E$ is a fixed elliptic curve say $ \mathbb C^* / \lbrace z \mapsto 2z \rbrace$. It can be compactified as the projective surface $E \times \mathbb P^1$ or as the (non-Kähler) Hopf surface $\mathbb C^2-{0} / \lbrace (z,w) \mapsto (2z,2w) \rbrace$.

Notice that the two compactifications have fields of meromorphic functions of different transcendence degree over $\mathbb C$, i.e., they have different algebraic dimensions.

Answer by jvp for Can a Lagrangian submanifold of ${\mathbb R}^{2n}$ be dense ($n>1$)?

2012-01-17T12:53:03Z

This is more a remark than an answer.

The typical solution of the typical polynomial ODE is uniformized by the Poincaré disc not by the complex line.

Indeed, after the work of McQuillan, it is known that the existence of a non-algebraic leaf uniformized by $\mathbb C$ imposes strong restrictions on the polynomial vector field. It turns out that there exits a projective surface birational to $\mathbb C^2$ where the foliation defined by the vector field has at worst canonical singularities and its cotangent sheaf has Kodaira dimension zero or one.

Answer by jvp for Example of rational projective variety of Picard number 1

2012-01-05T17:12:42Z

Hyperquadrics of dimension at least three.

Answer by jvp for Subtlety in the definition of the Kobayashi metric

2011-11-11T13:49:25Z

Let me give an example where $d \neq \delta$. I learned it from A Survey on Hyperbolicity of Projective Hypersurfaces, Example 1.2.1.

Consider $D$ as the following open subset of $\mathbb C^2$
$$ D = \lbrace (z,w) \in \mathbb C^2 ; |z| < 1, |zw|< 1 \rbrace \setminus \lbrace (0,w) | |w| \ge 1 \rbrace . $$ The Kobayashi distance of $p=(0,0)$ and $q=(0,1/2)$ is zero. Indeed, if $p_n = ( 1/n,0)$ and $q_n = ( 1/n, 1/2)$ then $$\lim_{n \to \infty} \delta(p_n,q_n) = 0.$$ And we can verify that this implies $d(p,q)=0$.

If $f: \Delta \to D$ is such that $f(0)=p$ and $f(a)=q$ then applying Schwarz lemma to $f_2$, the second component of $f=(f_1,f_2)$, we see that $|a|\ge 1/2$. Therefore $\delta(p,q) = 1/2$.

Notice that the Kobayashi pseudo-distance is continuous while $\delta$ is not. It seems reasonable to expect that the continuity of $\delta$ implies $\delta=d$.

Answer by jvp for Obstructions to being a hyperplane section or a fibre of a Lefschetz pencil

2011-11-04T11:28:44Z

Your question is related to the problem of the existence of non-trivial extensions of subvarieties $X \subset \mathbb P^N$ to $\mathbb P^{N+1}$. An extension of $X$ is just a subvariety $Y$ of $\mathbb P^{N+1}$ such that $X= Y \cap \mathbb P^{N}$. It is called trivial if $Y$ is the join of $X$ and a point outside of $\mathbb P^N$.

This is a classical question that was studied by the Italian school of algebraic geometry. For instance, Scorza proved that the Veronese surface in $\mathbb P^5$ does not admit non-trivial extensions.

More recently, the problem has been studied by Zak, S. L'vovsky, L. Badescu, among many others. In Extensions of projective varieties and deformations by S. L'vovsky you will find the following result:

Theorem. Suppose $X$ is not $\mathbb P^N$ nor a quadric. If $\dim X \ge 2$ and $H^1(X,TX\otimes \mathcal O_{\mathbb P^N}(-1))=0$ then every extension of $X$ is trivial.

For a very nice introduction to this circle of ideas see the first chapter of the book Projective geometry and formal geometry by L. Badescu. Unfortunately, the relevant Chapter doesn't seem to be available from Google books.

Of course this does not answer your question as the embedding of $X$ into $\mathbb P^N$ is fixed.

Answer by jvp for Does the diffeomorphism group preserving a particular section act transitively?

2011-11-02T02:36:57Z

I am not sure I understand the meaning of the equation $\psi\circ g=\psi$ as I don't see how to compare a section at two different points without fixing an isomorphism between the line-bundle and its pull-back under $g$. Anyway, in any possible interpretation, I don't think the group preserving $\psi$ acts transitively as it should preserve the zero locus of $\psi$.

Answer by jvp for Reference Request: Deformations of a map bijective to global sections of the pullback of the tangent sheaf

2011-11-01T21:00:04Z

Eiji HORIKAWA, On deformations of holomorphic maps I, J. Math. Soc. Japan Volume 25, Number 3 (1973), 372-396.

Answer by jvp for Holomorphic vector fields on $\mathbb{P}^n$ that extend to the blow up

2011-10-31T16:31:08Z

1) No. There are many more vector fields. The vector fields you are looking for are precisely those which vanish at $p_0$. Since $h^0( \mathbb P^n, T \mathbb P^n) = (n+1)^2 -1$ and you are imposing $n$ linearly independent conditions you should get $n^2 + n - 1$ vector fields. In homegeneous coordinates they can be written as $$ l_0 \frac{\partial}{\partial z_0} + \sum_{i=1}^n \sum_{j=1}^n c_{ij} z_i \frac{\partial}{\partial z_j} . $$ Notice that the homogeneous vector field $\sum_{i=0}^n z_i \frac{\partial}{\partial z_i}$ does not contribute to the counting since it corresponds to the zero vector field on $\mathbb P^n$.

2) You should get $k^2 + (n+1-k) n -1$ vector fields. As pointed out in the comments the result remains unchanged if we swap $k$ and $n-k+1$. The point is that $PGL(n+1)$ acts on $\mathbb P^n$ as well as on the dual projective space $\check{\mathbb P}^n$. If we look at the subgroup preserving a linear subspace of codimension $k$ on $\mathbb P^n$, then the natural action on $\check{\mathbb P}^n$ will preserve its dual: a linear subspace of codimension $n-k+1$.

3) You will get a holomorphic vector field on the blow-up if and only if the curve is left invariant by the original vector field.

If your vector field vanishes on a linear subspace of dimension at least two then every curve contained in it will be invariant by the vector field. Thus on the blow-up along these curves you will still get holomorphic vector fields.

If instead you restrict your attention to vector fields with zero set of dimension at most one then while you can find curves of arbitrarily high degree invariant by this class of vector fields (think on orbits of $\mathbb C^*$-actions on $\mathbb P^n$), they all have (geometric) genus zero.

Answer by jvp for Thom polynomial for contact algebraic structures

2011-10-23T01:41:33Z

Let $i:C \to \mathbb P^3$ be the normalization of an irreducible curve $C_0\subset \mathbb P^3$ of degree $d$ and geometric genus $g$.

If $\mathcal D$ is a distribution on $\mathbb P^3$ of degree $p$ then it is defined by a section $\omega$ of $\Omega^1_{\mathbb P^3} \otimes \mathcal O_{\mathbb P^3}(p+2)$. To compute the tangencies between $C_0$ and $\mathcal D$ we pull-back $\omega$ to $C$ using $i$. What we get is a section of $\Omega^1_C \otimes \mathcal O_C(p+2)$. Notice that $\mathcal O_C(p+2) = i^* \mathcal O_{\mathbb P^3}(p+2)$ is a line-bundle of degree $d(p+2)$ over $C$.

If $i^* \omega$ vanishes identically then $C_0$ is everywhere tangent to $\mathcal D$. Otherwise, it is a section of a line-bundle of degree equal to $\deg(\Omega^1_C) + \deg(\mathcal O_C(p+2)$. As such it has exactly $2g -2 + d(p+2)$ zeros counted with multiplicities. Therefore $$ f(d,g,p) = 2g -2 + d(p+2) +1
$$ is a polynomial. If one recalls that the genus of degree $d$ irreducible curve is bounded by $(d-1)(d-2)/2$ then one sees that $$ f(d,g,p) \le d( p + d -1) + 1 . $$ This is in accordance with Serge R. claim that the bound can be taken independent of $g$.

The argument above works equally well for integrable and non-integrable distributions, and also works for codimension one distributions on $\mathbb P^n$, $n\ge 2$.

Answer by jvp for Polynomial contact structures on $RP^3$

2011-03-23T14:21:40Z

Polynomial distributions on $\mathbb P^n$. The following works for any field $k$. The polynomial $1$-forms defined on $\mathbb A^{n+1}$ which induce distributions on $\mathbb P^n$ are those invariant by homotheties and annihilated by the Euler vector field $R = \sum_{i=0}^n x_i \partial_i$. Explictly these can be written as $$ \omega = \sum_{i=0}^n A_i dx_i $$ with $A_0, \ldots, A_n$ being homogeneous polynomials of degree $d+1$ satisfying the relation $$ \sum_{i=0}^n x_i A_i =0 . $$

In more intrinsic terms $\omega$ is section of $\Omega^1_{\mathbb P^n}(d+2)$. The integer $d$ appearing above has a nice geometric interpretation when $k=\overline k$ is an algebraically closed field. If we consider a linear inclusion $i: \mathbb P^1 \to \mathbb P^n$ then $i^* \omega$ is a section of $\Omega^1_{\mathbb P^1}(d+2) \simeq \mathcal O_{\mathbb P^1}(d)$ and therefore $d$ counts the number of tangencies between the distribution defined by $\omega$ with a generic line. We say that $d$ is the degree of the distribution.

Be careful: the degree of a distribution on $\mathbb P^n$ as defined above does not coincide with the degree of the coefficients of a polynomial $1$-form defining the same distribution in affine coordinates. Indeed the (maximal) degree of the affine polynomials defining the distribution on $\mathbb A^{n}$ is equal to $d+1$.

Examples of polynomial contact structures on $\mathbb R\mathbb P^3$ of even degree. The contact structures on $\mathbb R^3$ defined by $$ (qy−rz+a)dx+(pz−qx+b)dy+(rx−py+c)dz , $$ with $ ap+br+cq \neq 0 $, all have degree zero as they can be written in homogenous coordinates $(x:y:z:w) \in \mathbb P^3$ as $$ (qy−rz+aw)dx+(pz−qx+bw)dy+(rx−py+cw)dz + (-ax -by - cz ) dw . $$ It can also be checked that the induced distributions are all on the $PGL(4,\mathbb R)$-orbit of the one defined by $$ \omega_0 = xdy- ydx + zdw- w dz . $$ Indeed, the action of $\mathrm{PGL}(4,\mathbb C)$ on $\mathbb P H^0 ( \mathbb P^3, \Omega_{\mathbb P^3}(2))$ has only two orbits. The closed one corresponds to the integrable $1$-forms ( foliations singular along a line ) while the open one corresponds to contact structures.

Clarification. The space $\mathbb PH^0(\mathbb P^3, \Omega^1(2))$ can be naturally identified with $\mathbb P ( \bigwedge^2 \mathbb C^4)$. Indeed, the exterior differential is an injective map from linear homogeneous $1$-forms annihilated by Euler's vector field to constant $2$-forms; and the interior product with Euler's vector field sends constant $2$-forms to linear homogeneous $1$-forms annihilated by Euler's vector field. Under these maps the integrable $1$-forms correspond to decomposable $2$-forms. In other words, the foliations in $\mathbb P H^0(\mathbb P^3, \Omega^1(2))$ correspond to the Plucker embedding of the Grasmannian of lines in $\mathbb P^3$ into $\mathbb P (\bigwedge^2 \mathbb C^4)$.

To produce polynomial contact structures of any even degree $2d$ we have just to multiply $\omega_0$ by an even homogenous polynomial $P_{2d} \in \mathbb R[x,y,z,w]$ without non-trivial real solutions and perturb the result in $H^0(\mathbb R \mathbb P^3, \Omega^1(2d+2))$. Since $$ (P_{2d} \omega_0) \wedge d (P_{2d} \omega_0) = P_{2d}^2 \omega_0 \wedge d \omega_0 $$ does not vanish at any point of $\mathbb R \mathbb P^3$, we obtain that any section of $ \Omega^1(2d+2) $ in a ~~Zariski~~ sufficiently small (analytic) neighborhood of $P_{2d}\omega_0$ also defines a contact structure.

There are no polynomial contact structures of odd degree on $\mathbb R \mathbb P^3$. If we have a nowhere zero section of real vector bundle $E$ on a compact manifold $X$ then the top Stiefel-Whitney class of $E$ vanishes. From Euler's sequence $$ 0 \to \Omega^1_{\mathbb R \mathbb P^n} \to \mathcal O_{\mathbb R \mathbb P^n}(-1)^{\oplus n+1} \to \mathcal O_{\mathbb R \mathbb P^n} \to 0 $$ we can deduce that $$ w_n( \Omega^1_{\mathbb R \mathbb P^n}(d+2) ) = \sum_{i=0}^n (-1)^i (d+1)^{n-i} \mod 2 . $$ Notice that the same formula (without the $\mod 2$) counts the number of singularities of a polynomial distribution over an algebraically closed field if the singularities are isolated.

Specializing to $\mathbb R \mathbb P^3$ we get $$ w_3 ( \Omega^1_{\mathbb R \mathbb P^3}(d+2) ) = \left\lbrace \begin{array} 00 &\text{ if } d \text{ is even} \newline 1 &\text{ if } d \text{ is odd} \end{array}\right. $$ and we see that there are no contact distributions of odd degree on $\mathbb R \mathbb P ^3$.

Historical remark. The inexistence result above can be traced back to Habicht (1948). He dealt with a somewhat different problem which admits an equivalent algebraic formulation. His motivation came from Poincaré-Brower Theorem about the inexistence of continuous vector fields on the sphere $S^2$. If one looks for homogeneous polynomial vector fields on $\mathbb R^{n+1}$ tangent to the unitary sphere $S^n$ one ends up with $n+1$ homogeneous polynomials $(f_0, \ldots, f_n)$ satisfying $\sum x_i f_i=0$. Of course, this is the same as homogeneous polynomial $1$-forms annihilated by Euler's vector field.

Answer by jvp for References for holomorphic foliations

2011-06-17T20:04:02Z

As far as I can remember right now, the great general introduction to the theory of holomorphic foliations is yet to be written. Anyway let me mention some of the books that I know and which you may find useful. Let me warn you that none of them address your specific question.

Brunella - Birational geometry of foliations
Suwa - Indices of vector fields and residues of holomorphic foliations
Gomez-Mont, Bobadilla - Sistemas Dinamicos Holomorfos en Superficies ( in Spanish )
Loray - Pseudo-groupe d'une singularité de feuilletage holomorphe en dimension deux (in French )
Camacho, Sad - Pontos singulares de equações diferenciais analiticas ( in Portuguese )
Lins Neto, Scárdua - Folheações algébricas complexas ( in Portuguese )
Lins Neto - Componentes irredutíveis dos espaços de folheações ( in Portuguese )

Let me also mention that F. Touzet recently studied foliations admitting a transversal Kähler metric in this paper.

Answer by jvp for Kahler manifolds with special submanifolds

2011-09-26T17:28:17Z

If your $X$ is projective and $E$ is an ample vector bundle over $M$ then the field of meromorphic functions of $X$ is a finite extension of the field of meromorphic functions of $\mathbb P(E \oplus \mathcal O_M )$. This follows from Corollary 6.8 of Hartshorne's Cohomological dimension of algebraic varieties.

There is also an analytic version by Andreotti, which precedes Hartshorne's paper, implying the same result.

Answer by jvp for Wanted: example of a non-algebraic singularity

2011-09-24T02:59:48Z

This is not answer, but a complement to the comments above about isolated singularities of hypersurfaces.

Every isolated hypersurface singularity, $R=\mathbb C[[x_1, \ldots, x_n]]/f(x_1,\ldots,x_n)$, is not just algebraic but also $k$-determined, for some $k \in \mathbb N$.

If $\mathfrak m \subset \mathbb C[[x_1,\ldots, x_n]]$ then we say that $f$ is $k$-determined if for every $g\in \mathbb C[[x_1, \ldots, x_n]]$ satisfying $f-g \in \mathfrak m^k$ there exists an automorphism $\varphi :\mathbb C[[x_1,\ldots, x_n]] \to \mathbb C[[x_1, \ldots, x_n]]$ such that $\varphi(g)= f$.

Moreover, always assuming we have isolated singularities, the natural number $k$ can be easily determined from $f$. For instance, in this book you will find the following result:

If $\mathfrak m^{k+1} \subset \mathfrak m^2 J(f)$, where $J(f)$ is the Jacobian ideal of $f$, then $f$ is $k$-determined.

Answer by jvp for Extension of integrable distribution over a subset

2011-08-24T03:34:13Z

The answer in general is no.

If $K$ is a submanifold of $M$ then tangent bundle of $K$ defines an integrable distribution on $K$. To wit we are talking about the foliation with just one leaf: $K$.

If we can extend this foliation to a neighborhood of $K$ then the restriction of Bott's connection to $K$ induces a flat connection on the normal bundle of $K$. This imposes restrictions on $K$. For instance $K$ cannot be a projective curve in $\mathbb P^2(\mathbb C)$.

Recall that Bott's connection is the partial connection on the normal bundle of $\mathcal F $ defined as follows. If $T\mathcal F$ is the tangent bundle of a foliation and $N\mathcal F$ is its normal bundle then they fit into the exact sequence: $$ 0 \to T \mathcal F \to TM \stackrel{\pi}{\to} N\mathcal F \to 0 . $$ Now, if $v$ is a local section of $T\mathcal F$ and $w$ is a local section of $N\mathcal F$ then Bott's partial connection $\nabla : T\mathcal F \times N\mathcal F \to N \mathcal F$ is defined by the formula
$$ \nabla_{v}(w) = \pi ([v, \pi^{-1}(w)]) $$ where $\pi^{-1}(w)$ is an arbitrary lifting of $w$ to $TM$. The involutiviness of $T\mathcal F$ implies that $\nabla$ is well-defined. It is not hard to check that $\nabla^2=0$.

Answer by jvp for Rosenlicht theorem about uniruledeness and zeroes of holomorphic vector field on complex projective manifold

2011-07-07T02:28:47Z

You can look at Lieberman's paper Holomorphic Vector Fields on Projective Manifolds.

His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component of the identity of the automorphism group of $X$, is an algebraic group which acts algebraically on $X$.

Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian.

If $p \in X$ is a zero of your vector field then $p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O_{X,p}}{\mathfrak{m}_p^k}, $$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull. Thus $G$ is isomorphic to a linear algebraic group and yet another result of Rosenlicht says that a Zariski-closed abelian subgroup of a linear algebraic group is of the form $(\mathbb Cˆ*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves.

Added later: For an alternative proof see Theorem 6.4 of this paper. There it is proved that the existence of a non-zero section of $\bigwedge^q TX$ vanishing at a point suffices to ensure that $X$ is uniruled.

Rational maps with all critical points fixed

2009-10-18T02:07:51Z

What can be said about rational self-maps of $\mathbb P^1$ for which all critical points are also fixed points ?

If all but one of the fixed points are critical, there is a characterization in http://arxiv.org/abs/math/0411604v1 ( see Corollary 1 and the discussion just after the statement ).

Still assuming that all critical points are fixed: Is it possible to bound the degree of the rational map if all but two of the fixed points are critical ?

I think that the answer is probably no, but I would really love to hear the contrary.

Motivation. The question is motivated by a rather specific problem I like to think about from time to time. It concerns the classification of some special arrangements of lines on the projective plane. More specifically, I would like to classify arrangements of $3d$ lines(or rather hyperplanes through the origin of $\mathbb C^3$) invariant by degree $d$ homogeneous polynomial vector fields on $\mathbb C^3$. Given one arrangement like that one can produce a degree $d$ rational map having all its critical points fixed.

Answer by jvp for Geometry of complex elliptic curves

2011-06-27T20:01:07Z

According to this paper by Linda Ness the Gaussian curvature of a curve $C\subset \mathbb P^2$ defined by the zeros of a degree $d>1$ homogeneous polynomial $F \in \mathbb C[x,y,z]$ at a smooth point $p$ is given by $$ K(p) = 2- \frac{\|p\|^6 \cdot | \rm{Hessian}(F)(p)|^2}{ (d-1)^4 \cdot \| \nabla F(p) \|^6} , $$ where $\| \cdot \|$ stands for the usual norm in $\mathbb C^3$, and $\nabla F$ is the gradient of $F$.

In particular, if $p$ is a smooth inflection point of $C$ then $K(p) = 2$. Thus, there are no smooth cubics in $\mathbb P^2$ which are Euclidean flat, since these have $9$ inflection points.

N.B. : Ness normalizes the Fubiny-Study metric to have sectional curvature $2$.

After googling a bit I've found the paper The Riemannian geometry of holomorphic curves by Blaine Lawson which is strictly related to the subject. There he says that Eugenio Calabi proved, in Isometric imbedding of complex manifolds, that

($\ldots$) modulo holomorphic congruences, there is only one curve $C_n$ of constant Gauss curvature in $\mathbb C P^n$ which does not lie in any linear subspace. This curve has curvature $1/n$ and is given by the following embedding of $\mathbb C P^1\to \mathbb C P^n$: $$(z_0,z_1) \mapsto \left(z_0^n, \sqrt{n} z_0^{n-1} z_1, \ldots, \sqrt{\binom{n}{k}}z_0^{n-k}z_1^k, \ldots, z_1^n \right). $$

I could not find this statement in Calabi's paper, but this does not exclude the possibility that it is indeed there. The paper is the published version of Calabi's Phd thesis, so another possibility is that the statement is in the thesis but did not make its way into the paper.

N.B. : Lawson normalizes the Fubiny-Study metric to have sectional curvature $1$.

Answer by jvp for what are the subgroups of an algebraic group with codimension one

2011-06-27T20:17:37Z

Perhaps it is better to phrase the question in terms of Lie algebras. For instance, if you want to know which are the possible codimension one Lie subalgebras of a given finite dimensional Lie algebra then there is a result of Tits which address exactly this.

Let $\mathfrak g$ be a finite dimensional Lie algebra over a field of characteristic zero. If $\mathfrak h$ is a codimension one subalgebra then there exists a morphism $\phi : \mathfrak g \to \mathfrak{sl}(2)$ with kernel contained in $\mathfrak h$.

This result has been explored by Hoffman to provide a classification of codimension one subalgebras of Lie algebras in this paper.

Answer by jvp for Inequality on Chern classes of surfaces

2011-06-08T18:57:32Z

The integer $c_1^2(S) - c_2(S)$ is the second Segre class of the surface $S$. For surfaces of general type a Riemann-Roch computation shows that its positivity implies that $$ \lim \sup \frac{\log h^0\big(S,\mathrm{Sym}^i(\Omega^1_S) \otimes \mathcal L\big) }{\log i} =3 $$ for any line-bundle $\mathcal L$.

This fact has been explored by Bogomolov back in the seventies, to prove the finiteness of rational and elliptic curves on surfaces of general type with positive second Segre class, see for instance this Bourbaki seminar. The point of Bogomolov's argument is that a symmetric $1$-form defines a multi-foliation (also called web) on $S$. If $i: \mathbb P^1 \to S$ is a non-trivial morphism and $\omega \in H^0(S,\mathrm{Sym}^i \Omega^1_S)$ then $$i^* \omega \in H^0(\mathbb P^1, \mathrm{Sym}^i \Omega^1_{\mathbb P^1}) = H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(-2i)) . $$ We deduce that $i^* \omega$ vanishes identically, i.e., the image of $i$ is a leaf of the multi-foliation defined by $\omega$. If there are infinitely many of them, a theorem by Jouanolou implies that we have a $1$-parameter family of rational curves on $S$, thus $S$ is uniruled and cannot be of general type. If we start with a section of $\mathrm{Sym}^i \Omega^1_S \otimes \mathcal L$ with $\mathcal L^*$ ample, the very same argument shows the finiteness of elliptic curves on $S$. A more involved argument, but following the same lines, shows the boundeness of curves of bounded genus.

More recently, McQuillan proved that surfaces of general type with positive sencond Segre class do not admit Zariski dense entire curves in Diophantine approximations and foliations. This work lead to a birational classification of foliations on projective surfaces (by McQuillan, Brunella, and Mendes) very much in the spirit of Enriques-Kodaira classification, see this paper and references therein.

Similar results are not known for surfaces satisfying the inequality you impose: $c_2 -c_1^2\ge 0$. Already the starting point, the existence of symmetric differential forms, is rather non-trivial. Very recently Demailly proved the existence higher order differential equations with coefficients in duals of ample line-bundles.

Concerning surfaces with non-positive second Segre class let me mention that any smooth surface in $\mathbb P^3$ of degree at least $5$ is of general type and satisfies the inequality $c_1^2(S) - c_2(S) \le 0$. As far as I know, even the finiteness of rational curves on a generic quintic surface is unknown.

If, by any chance, one knows that they are minimal then the graph below borrowed from this wikipedia page might tell something. Notice that the line containing the ruled surfaces is $2c_2 = c_1^2$.

Otherwise the situation is even less encouraging. After a blow-up $c_2$ increases by one while $c_1^2$ decreases by one. After sufficiently many blow-ups we always end up with a surface with negative second Segre class.

Answer by jvp for Curves with negative self intersection in the product of two curves

2011-06-06T02:17:24Z

Disclaimer. The answer below is a variation of Bogomolov's argument and it would not come to be without Dmitri's answer. If you feel like upvoting this, please upvote his answer.

Curves on products of isogeneous elliptic curves. As already suggested in the body of the question, if we start with a pair of elliptic curves, say $E_1$ and $E_2$, admitting a non-constant morphism $f : E_1 \to E_2$ then given any point $p \in X=E_1 \times E_2$ we have infinitely many elliptic curves with self-intersection zero on $X$ passing through $p$. It suffices to consider translates of the graphs of endomorphisms of $E_2$ (there are at least $\mathbb Z$ of them) composed with $f$.

If we blow-up $p$ then we get a surface $S$ containing infinitely many (elliptic) curves with negative self-intersection.

Jacobians of genus $2$ curves. As pointed out in Dmitri's answer the natural morphism $$ \mathrm{Sym}^2 C \to \rm{Pic}^2(C) \cong \rm{Jac}(C) $$ identifies $\mathrm{Sym}^2 C$ with the blow-up of $\rm{Jac}(C)$ at a point. Thus if we have a genus $2$ curve with Jacobian isogenous to the square of an elliptic curve then the discussion in the previous paragraph shows that $C^2$ has infinitely many curves of negative self-intersection since we can pull-back the negative curves on $\mathrm{Sym}^2 C$ through the natural morphism $C^2 \to \rm{Sym}^2 C$. Notice also that the negative curves have unbounded intersection with the diagonal $\Delta \subset C^2$. It is not hard to verify that the pull-backs of the negative elliptic curves to $C^2$ will have unbounded genus.

Explicit example. If $C$ is a genus $2$ curve admitting a morphism $\pi : C \to E$ to an elliptic curve $E$ then $\rm{Jac}(C)$ is isogeneous to the product of $E$ with another elliptic curve $E'$ ( the connected component of the kernel of $\pi$ through zero). Automorphisms of $C$ act naturally on $\rm{Jac}(C)$. If there is an element
$\varphi \in \mathrm{Aut}(C)$ with induced action on $\rm{Jac}(C)$ not preserving $E'$ then $E$ is isogeneous to $E'$ since $$\pi_* \circ \varphi_* : \rm{Jac}(C) \to \rm{Jac}(E)\cong E$$ restricted to $E'$ is an isogeny. Therefore $\rm{Jac}(C)$ is isogeneous to the square of $E$.

To have a concrete example we can take $C = \lbrace y^2 = x^6 - 1\rbrace$ which maps to $E =\lbrace y^2 = x^3 -1\rbrace$ and has automorphism group isomorphic to $\mathbb Z_3 \rtimes D_8$ (which is not the automorphism group of any elliptic curve). From the discussion above it follows that $C^2$ has infinitely many curves of negative self-intersection and unbounded genus.

Question. Suppose $C$ is genus $2$ curve such that $C^2$ contains infinitely many curves of negative self-intersection. Is the Jacobian of $C$ isogeneous to the square of an elliptic curve ?

Answer by jvp for when is a section of vector bundle determined by its zero locus?

2011-06-02T22:29:02Z

Take a look at On sections with isolated singularities of twisted bundles and applications to foliations by curves by Campillo and Olivares.

There you will find the following result.

Theorem. Let $X$ be a projective manifold of dimension $n$, $\mathcal L$ be an ample >line-bundle on $X$, and $E$ a rank $n$ vector bundle over $X$. If $k\gg 0$ and $s, s'$ >are sections of $E \otimes \mathcal L^{\otimes k}$ such that

the zero scheme $Z(s)$ of $s$ has dimension zero, and

the zero scheme $Z(s')$ of $s'$ contains $Z(s)$

then there exists $\varphi \in H^0(X,\mathrm{End}(E))$ such that $s' = \varphi(s)$. In particular, if $E$ is simple ( $H^0(X,\mathrm{End}(E))=\mathbb C$ ) then $s = \lambda s'$ for a suitable complex number $\lambda$.

In the particular case $E=T\mathbb P^n$, it suffices to take $\mathcal L= \mathcal O_{\mathbb P^n}(1)$ and $k\ge 1$.

Holomorphic Poisson brackets on Fano manifolds

2011-04-26T01:45:15Z

I am looking for the preprint

A. Bondal, Noncommutative deformations and Poisson brackets on projective spaces. Preprint MPI/93-67

which I could not find online. Does anyone have an eletronic version of it ?

I am interested in the conjecture made there which predicts that the locus where the rank of a holomorphic Poisson bracket on a Fano manifold is smaller than $k$ has dimension strictly greater than $k$, see for instance Section 2 of Beauville's problem list in holomorphic symplectic geometry. Some progress has been made by Polishchuck in Algebraic Geometry of Poisson Brackets (Journal of Mathematical Sciences 84 no. 5, 1997); and by Druel in Structures de Poisson sur les variétés algébriques de dimension 3 (Bull. Soc. Mat. France 127, 1999).

Any further references on this conjecture will be welcome.

Answer by jvp for When does the conormal bundle sequence split?

2011-04-13T23:37:28Z

You are right. This is a result due to Van de Ven.

[A. Van de Ven, A property of algebraic varieties in complex projective spaces. In: Colloque Géom. Diff. Globale (Bruxelles, 1958), 151–152, Centre Belge Rech. Math., Louvain 1959. MR0116361 (22 #7149) Zbl 0092.14004]

Even more is true. Recently Ionescu and Repetto proved the following generalization of Van de Ven's Theorem.

Let $X \subset \mathbb P^n$ be a smooth subvariety. If there exists a curve $C \subset X$ such that the restriction to $C$ of the conormal sequence of $X$ splits then $X$ is linear.

Let me sketch a short elementary proof (of Van de Ven's result not its generalization) in the case of hypersurfaces. I will phrase it in the analytic category but once it is translated to the algebraic category, working with infinitesimal neighborhoods, I believe that what will emerge is one of the proofs in the literature.

If the normal sequence splits then we can define a foliation $\mathcal L$ by (germs of) lines everywhere transverse to $X$ at a neighborhood $U$ of $X$. Since the complement of $X$ is Stein we can extend $\mathcal L$ to the whole $\mathbb P^n$. Therefore $\mathcal L$ is defined by a global section of $T \mathbb P^n(d-1)$ for some $d \ge 0$. With the help of Euler's sequence, this section can be presented as a homogeneous vector field $v$ on $\mathbb C^{n+1}$ with coefficients of degree $d$. To compute the tangencies between $\mathcal L$ and $X$ we have just to contract the differential $dF$ of a defining equation $F$ of $X$ with $v$. If $F$ is not linear then the divisor on $X$ defined by the tangencies between $\mathcal L$ and $X$ (defined by $F=dF(v)=0$) will be non-empty contradicting the transversality between $X$ and $\mathcal L$.

Answer by jvp for Polynomial contact structures on $RP^3$

2011-03-29T03:57:23Z

I will try to clarify below the answers of Misha Verbitsky and Max Karev by reformulating then in the language of my other answer. Contrary to what I have wrote before in the comments of the main post, the distribution considered by Verbitsky is indeed polynomial.

Let $f(x,y,z,t)$ be a homogeneous polynomial and $H= f^{-1}(1)$. Consider the restriction of the $1$-form $$ \eta = \frac{\partial f}{\partial y} dx - \frac{\partial f}{\partial x} dy + \frac{\partial f}{\partial t} dz - \frac{\partial f}{\partial z} dt $$ at $H$. We want to extend the distribution defined on $H$ by it to the whole $\mathbb R^4$ in such a way that the result is invariant by homotheties.

Since $f$ does not vanish on $H$ we can multiply $\eta$ by $f$ and we will still get the same distribution on $H$. Similarly, since $df$ vanishes identically when restricted to $H$ we can sum multiples of $df$ to $\eta$ without changing the distribution on $H$. Therefore the sought homogeneous $1$-form is $$ \omega = \deg(f) f \eta - (\eta(R)) df , $$ where $R$ is Euler's vector field. Notice that $\omega$ defines a section of $\Omega_{\mathbb P^3}(2d)$. Its restriction to $H$ defines the very same distribution as $\eta$, and if $Z$ stands for the divisorial components of its zero set then $\omega$ defines a polynomial distribution on $\mathbb P^3_{\mathbb R}$ of degree $2\deg(f) - 2 - \deg(Z)$.

Hypersurfaces without real points

2011-03-26T04:42:05Z

Let $n, d$ be positive integers. I am interested in the open subset $\mathcal U_{n,d} \subset \mathbb P H^0 ( \mathbb P^n_{\mathbb R}, \mathcal O_{\mathbb P^n_{\mathbb R}}(d))$ corresponding to sections without real zeros. When $d$ is odd and $n$ arbitrary, the set $\mathcal U_{n,d}$ is empty since (homeogeneous) polynomials of odd degree always have (non-trivial) real roots.

Question. Suppose $d$ is even. How many connected components $\mathcal U_{n,d}$ has ? Do we know anything about the Betti numbers of $\mathcal U_{n,d}$ ?

Indeed, motivated by this other question, I am trying to figure out if it makes sense to ask for the number of connected components of the space of polynomial contact distributions of even degree on $\mathbb P^{2n+1}_{\mathbb R}$.

Comment by jvp

2012-12-03T10:58:20Z

Another related question: mathoverflow.net/questions/86000/…

Comment by jvp

2012-11-26T00:13:35Z

This is result can be traced back to the fifties, see mathoverflow.net/questions/61596/…

Comment by jvp

2012-08-31T09:17:24Z

Thank you. Corrected.

Comment by jvp

2012-06-19T21:07:04Z

Yes. Look at page 65 of Harris' paper.

Comment by jvp

2011-11-14T16:34:46Z

The paper by Lempert mentioned above is available at numdam.org/item?id=BSMF_1981__109__427_0

Comment by jvp

2011-10-31T18:54:35Z

@Will: I have edited my answer to reflect your comment. Thanks.

Comment by jvp

2011-10-31T18:24:30Z

@Jason: Thanks for catching that. You are of course right.

Comment by jvp

2011-10-24T20:04:56Z

Indeed we can avoid the dependance on g, as the genus is bounded by the degree.

Comment by jvp

2011-10-23T11:53:38Z

You look at the normalization (=desingularization) of your curve and use it to pull-back the $1$-form defining the distribution. In the case of a nodal curve you will not have necessarily tangencies at the nodes, only those branches which are indeed tangent to the distribution will contribute. If instead you have a cusp singularity, it will contribute since the normalization will have a critical point over it.

Comment by jvp

2011-10-05T18:55:49Z

There are some but most of them are in portuguese. For a reference list about holomorphic foliations you can look at my answer to this other question: mathoverflow.net/questions/68056/…

Comment by jvp

2011-09-26T20:02:56Z

You are welcome Dmitri.

Comment by jvp

2011-09-26T17:28:29Z

Comment by jvp

2011-09-25T00:51:08Z

Dmitri, you are probably thinking on Van de Ven Theorem. See mathoverflow.net/questions/61596/…

Comment by jvp

2011-09-04T05:11:47Z

Related question: mathoverflow.net/questions/2179/…

Comment by jvp

2011-07-17T11:21:46Z

Knuth's paper is on arxiv: arxiv.org/abs/math/9205211