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Let us consider polynomial contact structures on $\mathbb RP^3$, i.e. contact structures on $\mathbb R^3$ defined by a form $w=Pdx+Qdy+Rdz,\ P,Q,R\in \mathbb R[x,y,z]\ $ in an affine part and then extended to $\mathbb RP^3$, and $ w \wedge dw \ne 0$ everywhere.

One can find all such forms $w$ that $deg P, deg Q, deg R \leq 1$ by direct calculation:

$w=(qy-rz+a)dx+ (pz-qx+b)dy + (rx-py+c)dz,\ a,b,c,p,q,r\in \mathbb R;$ $ap+br+cq \ne 0$.

But I can't do anything for greater degrees. Do you know any criteria for coefficients of $P,Q,R$?

Does anybody know any contact polynomial form with $deg P, deg Q, deg R \geq 2$?

Added: What is the form (I mean form coefficients in $\mathbb R^3\subset \mathbb RP^3$) defines polynomial contact structure constructed by plurisubharmonic function $f=x^4+y^4+z^4+t^4$?

Answer: $f=x^4+y^4+z^4+t^4$ is not strictly plurisubharmonic (see on the plane $x=y=0$ on subspace generated by $dx,dy$). So it does not produce a contact structure.

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to jvp: Thanks! It seems to be very interesting. So, can you clarify situation with comment of Misha Verbitsky here? Is it true that we can produce polynomial contact structures on $\mathbb R^{2n-1}$ via convex polynomial function on $\mathbb R^{2n}$? –  Nikita Kalinin Mar 26 '11 at 8:34
    
I withdraw my previous comment. Indeed, Misha and Max are correct. The distribution is polynomial. I have added another answer trying to explain this fact. –  jvp Mar 29 '11 at 3:56
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6 Answers 6

up vote 10 down vote accepted

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.

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Thank you very much! Where can I find proves of all these theorems? Could you give me a references? "without non-trivial real solutions and perturb" - how should I perturb coefficient after multiplication? What do you mean? So, as I see, moving to affine chart $x_0=1$ is just equating $dx_0=0, x_0 = 1$. Right? –  Nikita Kalinin Mar 23 '11 at 19:59
    
You can take a look at Equations de Pfaff Algebriques, a LNM by J.-P. Jouanolou, for the basics about polynomial distributions. To perturb is the same as add a homogeneous polynomial 1-form of the right degree, annihilated by Euler vector field and with sufficiently small coefficients. You are right on how to write equations on the affine chart x0=1. –  jvp Mar 23 '11 at 22:12
    
jvp: do you know any other books about polynomial distributions in algebraic geometry? –  Nikita Kalinin Oct 5 '11 at 15:18
    
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/… –  jvp Oct 5 '11 at 18:55
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A plurisubharmonic function $\phi$ on ${\Bbb C}^2$ defines a contact structure on its level set. If $\phi$ is homogeneous, this level set has a real algebraic projection to ${\Bbb R} P^3$, which is a double covering and compatible with the contact structure. To find such $\phi$, take any homogeneous convex polynomial function on ${\Bbb R}^4$, it would be automatically plurisubharmonic.

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As I understand, you said that projection of form has polynomial coefficient. Why? –  Nikita Kalinin Mar 10 '11 at 19:42
    
a polynomial homogeneous function on ${\Bbb R}^4$ is the same as an algebraic section of $O(i)$ on ${\Bbb R} P^3$, where $i$ is its degree. –  Misha Verbitsky Mar 10 '11 at 22:50
    
Sorry, I'm so stupid and can't do it even for $f=x^4+y^4+z^4+t^4$. $TM\cap JTM$ is specified by $x^3dx+y^3dy+z^3dz+t^3dt=y^3dx-x^3dy+t^3dz-z^3dt=0$. I want to project this intersection into affine cart $t=1$. It corresponds equating $dt=0$. And the sole condition is $\frac{x^3dx+y^3dy+z^3dz}{t^3}+\frac{y^3dx-x^3dy+t^3dz}{z^3}=0$ So, what should we do, to discard $t$. But it brings us radicals, such as $\frac{x^3dx+y^3dy+z^3dz}{(1-x^4-y^4-z^4)^{3/4}}+\frac{y^3dx-x^3dy+(1-x^4-y^4-z^‌​4)^{3/4}dz}{z^3}=0$ –  Nikita Kalinin Mar 20 '11 at 19:03
    
What is the form defines polynomial contact structure for plurisubharmonic $f=x^4+y^4+z^4+t^4$? –  Nikita Kalinin Mar 20 '11 at 19:03
    
I suppose we consider level set $f=1$. –  Nikita Kalinin Mar 20 '11 at 19:17
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Concerning your question about $f=x^4+y^4+z^4+t^4$. It indeed induces a polynomial distribution on $\mathbb RP^3$. You should consider a map from the unit sphere to $f^{-1}(1)$ and compute a pull-back of the form given by $df\circ J|_{x^4+y^4+z^4+t^4=1}$. It will become polynomial if you multiply it by an appropriate degree of $f$. I have computed the $dx$ coefficient of the resulting form, but it doesn't look very enlightening to me: $x^6y + y^3(x^4+y^4+z^4+t^4) +x^3t^3z - y^4x^3-x^3z^3t$.

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You can't discard $t$, I think. In my definition, contact structure is polynomial iff it is polynomial on any affine chart. –  Nikita Kalinin Mar 28 '11 at 7:23
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Nope, you can. The only problem is that this function is not strictly plurisubharmonic, so the induced distribution would be integrable along two lines in $\mathbb RP^3$. So the problem is to construct homogeneous strict plurisubharmonic function. –  Max Karev Mar 28 '11 at 20:21
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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 $2\deg(f) - 2 - \deg(Z)$.

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Thanks for the clarification. But I don't think if I understood it properly. Let's consider a plurisubharmonic function given by $f=x^2+y^2+z^2+t^2$. You wrote that the induced distribution should have degree 2, but it has degree 0! Where is the problem? –  Nikita Kalinin Mar 29 '11 at 17:17
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In this case $\eta(R)=0$, so $\eta$ already defines a distribution on $\mathbb P^3$. There is no need to multiply be $f$ and add a multiple of $df$. I will edit to correct this problem. –  jvp Mar 29 '11 at 19:04
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Here is a code in Macaulay for checking that some plurisubharmonic function determines a contact structure

R = QQ[x,y,z,t]

degf=4

f=x^degf+y^degf+z^degf+t^degf + 3*x^2*z^2 + x^2*t^2

--f2 = x^degf+y^degf+z^degf+t^degf

nR = diff(y,f)*x-diff(x,f)*y+diff(t,f)*z-diff(z,f)*t

cdx = degf*f*diff(y,f)-nR*diff(x,f)

cdy = -degf*f*diff(x,f)-nR*diff(y,f)

cdz = degf*f*diff(t,f)-nR*diff(z,f)

cdt = -degf*f*diff(z,f)-nR*diff(t,f)

cxyz=cdx*(diff(y,cdz)-diff(z,cdy))+cdy*(diff(z,cdx)-diff(x,cdz))+cdz*(diff(x,cdy)-diff(y,cdx))

re = sub(cxyz,t=>1)

factor re


So, $f2=x^2+y^2+z^2+t^2\ $ DOES not produce contact structure. Is convex but not strictly(!) plurisubharmonic (on the plane $x=y=0$). Here (http://www.math.ethz.ch/~evansj/lecture9.pdf) there is a good explanation why induced structure is contact ($d\eta$ tames complex structure on $\mathbb R^4$ )

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There is a very good article "Complex contact threefolds and their contact curves" of Yun-Gang Ye where on can find a classification of complex contact structures on threefolds

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