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The Polynomial Method Lecture 23

In this lecture, we discuss Cayley’s theorem on flecnodes and ruled surfaces
Subject

Diploma in civil engineering

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Academic year: 2023/2024
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Winnacunnet High School

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In this lecture, we discuss Cayley’s theorem on flecnodes and ruled surfaces.

Theorem 0. If P is a polynomial in C[z 1 , z 2 , z 3 ], and if F P vanishes on Z(P ), then Z(P ) is ruled.

We know from last lecture that F P (z) = 0 if and only if z is flecnodal. So for each z ∈ Z(P ), we know that there is a non-zero vector V so that P vanishes in the direction V to fourth order. Informally, this means that Z(P ) locally looks ruled. We want to put the local information together and prove that there are actual global lines contained in Z(P ). Here is the basic difficulty with the proof. Suppose that V (z) is a smooth non- vanishing vector field on Z(P ) which obeys the flecnodal equation at each point of Z(P ). How can we use V to find lines? A natural method is to look at the integral curves of V. But consider the following example. The surface Z(P ) may be a plane. At each point z in the plane Z(P ), every tangent vector obeys the flecnodal equation. So let V be any smooth (tangent) vector field in Z(P ). It obeys the flecnodal equation at every point, but the integral curves of V are basically arbitrary curves in the plane. If Z(P ) is irreducible and not a plane, then this method actually works, but we can see the proof needs to be a little subtle because we need to use the fact that Z(P ) is not a plane. There are also unfortunately a couple of cases in the proof. We won’t give a complete proof. Instead we will carefully do one case, which I think of as the main case. Moreover, this one case is enough to give the full proof of the regulus detection lemma. In our model case, we will work over the real numbers, which is technically easier (and all we need in the regulus detection lemma). The argument works over the complex numbers with minor modifications, but we think it’s easier to see the main ideas over R. Let’s recall/clarify our notation for derivatives and higher derivatives, because we will need to be clear-headed about it. If F : R 3 → R is a function, we write ∂iF to abbreviate the standard partial derivative ∂∂x i F. If V is a vector, we write ∇V F (x) for

∑

i Vi∂iF (x). The most important role in our story is played by second derivatives. If V, W are two vectors, then we write 1

∇ 2 V,W F (x) =

∑

ViWj ∂i∂j F (x). i,j We abbreviate ∇ 2 V = ∇ 2 V,V. Higher derivatives are similar. Now we can state our special case.

Proposition 0. Suppose that P ∈ R[x 1 , x 2 , x 3 ]. Let O ⊂ Z(P ) be an open subset of Z(P ). Suppose that V is a smooth, non-zero vector field on O, obeying the flecnodal equation:

0 = ∇sV P (x), for all x ∈ O, s = 1, 2 , 3. We add a technical assumption. Suppose that at each point x ∈ O, ∇P (x) = 0 and ∇ 2 P (x) : T Z × T Z → R is non-degenerate. Then the integral curves of V are straight line segments. Therefore, every point in O lies in a line in Z(P ).

A word about the technical assumption. We defined above ∇ 2 V,W P (x) for any vectors V, W. Therefore, ∇ 2 P (x) is a map from R 3 × R 3 → R. We restrict it to a map T Z × T Z → R. Being non-degenerate means that for each non-zero V ∈ T Z, there is some W ∈ T Z so that ∇ 2 V,W P (x) = 0. For most surfaces Z(P ), ∇ 2 P is non-degenenerate on a dense open set. In this case, our propisition allows us to find a line of Z(P ) thru almost every point. And since the non-degenerate points are dense, we can find a line of Z(P ) thru the other points by taking limits. There are, however, some surfaces where ∇ 2 P is degenerate at every point of Z(P ). These surfaces require a different argument - so we begin to see that the general theorem requires cases.

Proof. It suffices to show that at each point x ∈ O, ∇V V is a multiple of V. If we let V 1 be a unit length renormalization of V , then if follows that ∇V 1 V 1 = 0 on O. This equation implies that the integral curves of V 1 (or V ) are straight lines. (Suppose that γ : R → O is an integral curve of V 1. In other words, γ′(t) = V 1 (γ(t)). If we differentiate, we get γ′′(t) = ∇V 1 (γ(t))V 1 (γ(t)) = 0. ) To explain the argument, we need a different derivative - the Lie derivative. If V∑ is a vector field, we let LV denote the Lie derivative, defined by LV F (x) =

i Vi(x)∂iF (x). Actually, LV F (x) = ∇V F (x), but we come to second derivatives, there is an important difference:

LV (LV F ) = ∇ 2 V F! (1) Let’s clarify what the left-hand side means. LV F is a function. Then LV (LV F ) is the Lie derivative of that function. The reason that the two sides are different is that on the left-hand side, the outer differentiation hits the vector field V appearing

&

&

&

also many examples of Gauss flat algebraic surfaces Z(P ), but I’m not positive. If Z(P ) is Gauss flat and F P = 0 on Z(P ), prove that Z(P ) is still ruled.

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The Polynomial Method Lecture 23

Subject: Diploma in civil engineering

32 Documents
Students shared 32 documents in this course

School: Winnacunnet High School - Hampton

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FROM LOCAL TO GLOBAL
In this lecture, we discuss Cayley’s theorem on flecnodes and ruled surfaces.
Theorem 0.1. If Pis a polynomial in C[z1,z
2,z
3],andifFP vanishes on Z(P),
then Z(P)is ruled.
We know from last lecture that FP(z)=0ifandonlyifzis flecnodal. So for
each z∈Z(P), we know that there is a non-zero vector Vso that Pvanishes in the
direction Vto fourth order. Informally, this means that Z(P)locallylooksruled.
We want to put the lo cal information together and prove that there are actual global
lines contained in Z(P).
Here is the basic difficulty with the proof. Suppose that V(z)isasmoothnon-
vanishing vector field on Z(P)whichobeystheflecnodalequationateachpoint
of Z(P). How can we use Vto find lines? A natural method is to look at the
integral curves of V.Butconsiderthefollowingexample.ThesurfaceZ(P)maybe
aplane. Ateachpointzin the plane Z(P), every tangent vector obeys the flecnodal
equation. So let Vbe any smooth (tangent) vector field in Z(P). It obeys the
flecnodal equation at every point, but the integral curves of Vare basically arbitrary
curves in the plane. If Z(P)isirreducibleandnotaplane,thenthismethodactually
works, but we can see the proof needs to be a little subtle because we need to use
the fact that Z(P)isnotaplane.
There are also unfortunately a couple of cases in the proof. We won’t give a
complete proof. Instead we will carefully do one case, which I think of as the main
case. Moreover, this one case is enough to give the full proof of the regulus detection
lemma.
In our model case, we will work over the real numbers, which is technically easier
(and all we need in the regulus detection lemma). The argument works over the
complex numbers with minor modifications, but we think it’s easier to see the main
ideas over R.
Let’s recall/clarify our notation for derivatives and higher derivatives, because we
will need to be clear-headed about it.
If F:R3→Ris a function, we write ∂iFto abbreviate the standard partial
derivative ∂F
∂xi.IfVis a vector, we write ∇VF(x)for!iVi∂iF(x). The most
important role in our story is played by second derivatives. If V, W are two vectors,
then we write
1

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