Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mathbb R^3\subset \mathbb RP^3$) defining the 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.

share|improve this question
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

6 Answers 6

up vote 12 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.

share|improve this answer
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

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.

share|improve this answer
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

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: $x^6y + y^3(x^4+y^4+z^4+t^4) +x^3t^3z - y^4x^3-x^3z^3t$.

share|improve this answer
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
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

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 $2\deg(f) - 2 - \deg(Z)$.

share|improve this answer
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
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

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

share|improve this answer

Here is a code in Macaulay2 for checking that some plurisubharmonic function determines a contact structure

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


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 = degffdiff(y,f)-nR*diff(x,f)

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

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

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


re = sub(cxyz,t=>1)

factor re

So, $f2=x^2+y^2+z^2+t^2\ $ (in the above notation) 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 the induced structure is contact ($d\eta$ tames complex structure on $\mathbb R^4$ )

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.