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

We say that a surface $f(x,y,z)=0$ is ruled if for each point $p$ in the surface there is a line that passes through $p$ and is contained in the surface. See http://en.wikipedia.org/wiki/Ruled_surface for more information.

Does anybody know if there is a partial differential equation whose solutions are all ruled surfaces and only them? And what is the equation? Some reference would be helpful, too (different from Salmon's old book about surfaces which I unfortunately find unreadable).

share|improve this question
5  
When you write "Does anybody know if there is a partial differential equation whose solutions are all ruled surfaces and only them?", do you mean to ask whether there is a partial differential equation that characterizes ruled surfaces? The natural PDE that does this is the one that sets equal to zero the product of the curvatures of the asymptotic curves. It's a 3rd order equation, and the general solution depends on 3 functions of 1 variable, as you'd expect. This equation is projectively invariant, so it is better to express it directly in terms of projective invariants. –  Robert Bryant Oct 12 '11 at 0:46
    
yes, i'm looking for a partial equation that characterizes the ruled surfaces, but preferably for surfaces given by $f(x,y,z)=0$ and not in the parametric form. so if we're given a surface $f(x,y,z)=0$ we just plug in $f$ in the partial equation and we know that it's ruled iff we get 0. –  filipm Oct 12 '11 at 8:33
    
here faculty.fairfield.edu/jmac/rs/halftw.htm they mention an equation $x^2z_{xx}+2xyz_{xy}+y^2z_{yy}=0$, but it's not the right one, it misses e.g. $z=xy$ (probably this equation gives only a subclass of the ruled surfaces). –  filipm Oct 12 '11 at 8:36
3  
@filipm: In that case, set $II = f_{xx} dx^2 + 2f_{xy}dxdy + f_{yy}dy^2$. If the discriminant of $II$ vanishes, then the surface is ruled. If the discriminant is positive, it is not ruled. If the discriminant is negative, compute $III = f_{xxx}dx^3+3f_{xxy}dx^2dy+3f_{xyy}dxdy^2+f_{yyy}dy^3$ and let $III_0$ be the $II$-trace-free part of $III$. Then the surface is ruled if and only if the discriminant of $III_0$ vanishes. –  Robert Bryant Oct 12 '11 at 13:56
2  
@Felipe: For $z=f(x,y)$, it's the equation ${f_{yy}}^3{f_{xxx}}^2+6{f_{yy}}{f_{xxx}}{f_{yyy}}{f_{xy}}{f_{xx}}$ $-6{f_{yy}}^2{f_{xxx}}{f_{xyy}}{f_{xx}}-6{f_{yyy}}{f_{xy}}{f_{xx}}^2{f_{xyy}}$ $+9{f_{yy}}{f_{xyy}}^2{f_{xx}}^2-6{f_{xy}}{f_{yy}}^2{f_{xxy}}{f_{xxx}}$ $+12{f_{xy}}^2{f_{xxy}}{f_{yyy}}{f_{xx}}-18{f_{xy}}{f_{yy}}{f_{xxy}}{f_{xyy}}{f_‌​{xx}}$ $+12{f_{yy}}{f_{xyy}}{f_{xy}}^2{f_{xxx}}-8{f_{yyy}}{f_{xy}}^3{f_{xxx}}$ $+9{f_{xx}}{f_{yy}}^2{f_{xxy}}^2-6{f_{yy}}{f_{xxy}}{f_{yyy}}{f_{xx}}^2$ $+{f_{yyy}}^2{f_{xx}}^3 = 0$. –  Robert Bryant Oct 13 '11 at 18:45

5 Answers 5

up vote 13 down vote accepted

Here is a test for when a surface of the form $z = f(x,y)$, where $f$ is a sufficiently smooth function of two variables, is ruled.

To begin, set $II = f_{xx} dx^2 + 2f_{xy}dxdy + f_{yy}dy^2$. If $II$ vanishes identically, then the surface is a plane, so it is ruled.

Suppose that $II$ is nonzero. The discriminant of $II$ is defined to be $$ \Delta(II) = f_{xx}f_{yy}- {f_{xy}}^2. $$

If $\Delta(II) >0$, then the surface is locally strictly convex and so cannot be ruled.

If $\Delta(II) = 0$, then the surface is ruled. In fact, it has vanishing Gauss curvature. Moreover, $II = \pm \alpha^2$ for some nonzero $1$-form $\alpha$ on the domain of $f$, and the curves in this domain defined by $\alpha = 0$ (which turn out to be straight lines) lift to the graph $z = f(x,y)$ to be straight lines.

If $\Delta(II) < 0$, set $III = f_{xxx}\ dx^3+3f_{xxy}\ dx^2dy+3f_{xyy}\ dxdy^2+f_{yyy}\ dy^3$ and let $III_0$ be the $II$-trace-free part of $III$. Then the surface $z = f(x,y)$ is ruled if and only if the discriminant of $III_0$ vanishes.

(Added later: This latter condition turns out to be equivalent to the condition that $II$ and $III$ have a common linear factor, say, $\alpha$, and hence is equivalent to the vanishing of the resultant of $II$ and $III$, i.e., $\textrm{Reslt}(II,III) = 0$. When such an $\alpha$ exists, the leaves of $\alpha=0$ are lines on the surface.)

Notes:

  1. The discriminant of a cubic form $C = p\ dx^3 + 3q\ dx^2dy + 3r\ dxdy^2 + s\ dy^3$ is, by definition, $$ \Delta(C) = s^2p^2 + 4r^3p + 4 q^3s - 3 r^2q^2 - 6 sqrp. $$ It is, up to a multiple, the unique polynomial of degree $4$ in the coefficients that vanishes if and only if $C$ has a multiple factor.

  2. Given a quadratic form $Q = a\ dx^2 + 2b\ dxdy + c\ dy^2$ with nonvanishing discriminant $D$, the $Q$-trace of a form $C$ of degree $3$ is the linear form $$ tr_Q(C ) = \frac{(ar-2bq+cp)\ dx + (as-2br+cq)\ dy}{D}. $$ Any cubic form $C$ can be uniquely written in the form $$ C = C_0 + L\cdot Q $$ where $L$ is a linear form and $tr_Q(C_0) = 0$. (In fact, $L = \tfrac34 tr_Q(C)$.) The term $C_0$ is called the $Q$-trace-free part of $C$.

  3. The resultant $\textrm{Reslt}(Q,C)$ of a quadratic form $Q$ and a cubic form $C$ is the (unique up to nonzero multiples) polynomial that is cubic in the coefficients of $Q$, quadratic in the coefficients of $C$, and vanishes exactly when they have a common linear divisor.

share|improve this answer
    
Beautiful. Thanks, Robert. –  Deane Yang Oct 13 '11 at 0:49

From the paper to which 'lowerbound' linked, "Symmetry groups and Lagrangians associated with Tzitzeica surfaces," by Nicoleta Bila (also arXiv:math/9910138v1), here is Theorem 1 and its preamble: Consider $D \subset \mathbb{R}^2$ and let
   Tz1
   alt text


I take no credit (or blame!) for this; just posting as a community service.

share|improve this answer
    
Joseph, how do you insert a pdf document in an MO answer box ? You have even managed to insert material from two different pages without showing the page break. I would like to use the trick... –  Chandan Singh Dalawat Oct 14 '11 at 2:56
1  
@Chandan: I converted the PDF to JPEG (actually, to two JPEGs) and then included each as images with the HTML img-tag. –  Joseph O'Rourke Oct 14 '11 at 9:58

Does anybody know if there is a partial differential equation whose solutions are all ruled surfaces and only them?

An example is theorem 1 of http://www.kurims.kyoto-u.ac.jp/EMIS/journals/BJGA/10.1/bt-bil.pdf

Edit: I have made this a community wiki answer so that if anyone would care to type in the theorem then they are free to do so.

share|improve this answer
    
Would you care to type in the theorem? There is no guarantee that this pdf file will stay at that location for the lifetime of MO. –  David Roberts Oct 11 '11 at 21:43
    
@david: last time I did this for you the question was closed so I'm not going to spend the effort this time. see mathoverflow.net/questions/77643/pythagorean-analogue/77645 –  psd Oct 11 '11 at 21:53
1  
@David: If you look at the paper, you will see that typing in all the requisite notation is a bit painful. Perhaps including the math review pointer is a more practical solution. –  Igor Rivin Oct 11 '11 at 21:54
    
@david: I see that in my 'pythagorean analogue' answer you asked me to summarize the important points while at the same time you voted to close the question. –  psd Oct 11 '11 at 22:26

Isn't it just $K = 0$, where $K$ is the Gauss curvature?

Wrong. Although surface with zero Gauss curvature is necessarily ruled, the converse is not true. The catenoid is the best known counterexample. See comments below.

share|improve this answer
3  
A surface can still be ruled without $K = 0$. For example, one principal curvature could be positive and the other negative, so the second fundamental form still has null directions; if you're lucky enough, they line up in space. The catenoid in particular is ruled but curvy. –  Matt Noonan Oct 11 '11 at 22:19
    
The Wikipedia article en.wikipedia.org/wiki/Differential_geometry_of_surfaces shows a formula for the Gaussian curvature of a ruled surface, and gives conditions on when it vanishes. –  Joseph O'Rourke Oct 11 '11 at 22:21
    
Oops. That's right. You just need a null direction that propagates along a straight line. Gotta downvote my own answer. –  Deane Yang Oct 11 '11 at 22:49
1  
I think it's useful to leave it. I will edit it. –  Deane Yang Oct 12 '11 at 0:39
2  
An obvious example of a ruled surface with $K<0$: a one-sheet hyperboloid. –  Denis Serre Oct 12 '11 at 5:27

The only chracterization of ruled surfaces I know is the following: A surface S in the 3-dimensional Euclidean space of negative Gaussian curvature is a ruled surface iff its (equi)affine Pick-invariant vanishes identically on S.

This theorem can be found in the books of W. Blaschke on Affine Differential Geometry.

For surfaces of positive Gaussian curvature the following theorem is valid: A surface S in the n-dimensional Euclidean space of positive Gaussian curvature is a quadric iff its (equi)affine Pick-invariant vanishes identically on S.

share|improve this answer
    
How is this related to the question? –  Alex Degtyarev Feb 19 at 19:24

Your Answer

 
discard

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.