inflection points and canonical forms of non-singular cubic curves

In this entry, we shall investigate the inflection pointsMathworldPlanetmath of non-singularPlanetmathPlanetmath complex cubic curves and show that the equations of such curves can always be put in various canonical formsMathworldPlanetmath. The reason for lumping these two subjects in a single entry is that, being related, it is more efficient to develop them both at the same time than it would be to treat either one in isolation — on the one hand, putting the equation of the curve in a canonical form simplifies calculations relating to inflection points whilst, on the other hand, information about inflection points allows one to construct canonical forms.

We shall consider our curves as projective curves and describe them with homogenous equations; i.e. a curve is a locus in 2 where F(x,y,z)=0 for some third- order homogenous polynomialPlanetmathPlanetmath F. For the purpose at hand, the term “inflection point” may be taken to mean a point on the curve where the tangentMathworldPlanetmathPlanetmathPlanetmath intersects the curve with multiplicityMathworldPlanetmath 3 — a point on the curve will have this property if and only if it is a zero of the Hessian.

We begin by presenting a crude canonical form. With further refinement, this will become the Weierstrass canonical form.

Theorem 1.

Suppose C is a non-singular complex cubic curve and P a point on that curve. Then we may choose homogenous coordiantes in which the coordiantes of P are (0,0,1) and equation of the curve looks as follows:


The most general equation of a cubic is as follows:


Requiring that the curve pass through (0,0,1) means that L=0. Since our curve is not singular, it has a tangent at all points, in particular, a tangent through P. We may ask that the equation of the tangent through P be Y=0, which requires that F=0. Since our curve is not singular at (0,0,1), both F and K cannot be zero so, without loss of generality, we may take K=1.

We may then make the following linear transform:

X =x
Y =y
Z =z-Ex/2-Hy/2

Upon doing so, our equation assumes the form


where the coefficients a,b,c,d,e are defined as follows:

a =A-CE/2
b =B-CH/2-E2/4
c =D-EH/2
d =G-H2/4
e =C

Next, we turn our attention to the Hessian to begin our investigation of inflection points.

Theorem 2.

If C is a non-singular complex cubic curve with homogenous equation F(x,y,z)=0, then the Hessian of F does not equal a multipleMathworldPlanetmath of F.


By the foregoing theorem, we know that there exists a system of coordinates in which


Let us examine the Hessian at the point (0,0,1):


If e0, then the Hessian cannot me a multiple of F because F(0,0,1)=0. Assume then that e=0. In that case, we must have a0 because, if a=0, then F would factor as a productPlanetmathPlanetmath of y and a quadratic polynomial, which is impossible because the curve C is assumed not to be singular.11Were F(x,y,z)=yQ(x,y,z) for some quadratic polynomial Q, then the gradient of F would vanish at the intersection of the line y=0 and the conic Q(x,y,z)=0, so the varietyMathworldPlanetmathPlanetmath described by F(x,y,z)=0 would be singular at the point(s) of intersection. Evaluating the Hessian on the line y=0,


Since a cannnot be zero, this does not vanish identically and is clearly not a non-zero multiple of F(x,0,z). ∎

This theorem tells us that H(x,y,z)=0 is the equation of a cubic curve distinct from C. Hence, we know that there must be at least one and at most nine point of intersection of these two curves, i.e. at least one and at most nine inflection points. Knowing that inflection points exist, we can improve our canonical form by choosing P as an inflection point. With a little tidying up, this will give us Weierstrass’ canonical form.

Theorem 3.

Suppose C is a non-singular complex cubic curve and P a point of inflection of on that curve. Then we may choose homogenous coordinates in which the coordiantes of P are (0,0,1) and equation of the curve looks as follows:


By our earlier result, we know that there exist coordinates in which the equation of the curve is


and the coordinates of P are (0,0,1). By the calculation made in the proof of the previous theorem, we know that H(0,0,1)=8e. Since P is a point of inflection, this implies that e=0.

As noted in the previous proof, we cannot have a=0 because that would make our curve singular. Hence, we can divide by a1/3 to make the following coordinate transformationMathworldPlanetmath:

X =x-by/3
Y =ay/4
Z =z

After canceling a factor of a, our equation becomes



g2 =14ac-712b2
g3 =11432b3+148abc+116a2d

Using this canonical form, it is easy to show that we have the maximum number of inflection points possible.

Theorem 4.

A non-singular complex cubic curve has nine inflection points.


Given an inflection point, the foregoing theorem tells us that we can choose a coordinate systemMathworldPlanetmath in which that point has coordinates (0,0,1) and the equation of the curve assumes the form F(x,y,z)=0 where


Computing the Hessian, we have


Computing its gradient at (0,0,1), we find that


For comparison, we have


Thus, we see that the curves H(x,y,z)=0 and F(x,y,z)=0 are both smooth at (0,0,1) and intersect transversely, hence they have intersection multiplicity 1 there. Because we could choose our coordinates to place any inflection point at (0,0,1), this means that every intersection of our curve with its Hessian has intersection multiplicity 1. Since both F and H are of third order, Bezout’s theorem implies that there are nine distinct intersection points, i.e. nine distinct points of inflection. ∎

Knowing that there exist multiple inflection points, we will now cast the equation in a form which places two inflection points at priveleged locations.

Theorem 5.

Suppose C is a non-singular complex cubic curve and P and Q are points of inflection of on that curve. Then we may choose homogeneous coordinatesMathworldPlanetmath in which the coordinates of P are (1,0,0), the coordinates of Q are (0,1,0) and the equation of the curve looks as follows:


Since P is a point of its inflection, the tangent to C at P intersects C with multiplicity 3 at P. Since C is a cubic, this means that this tangent cannot intersect C at any other points; in particular, it cannot pass through Q. Likewise, the tangent through Q cannot pass through P. Hence, the tangent at P and the tangent at Q must intersect at some third point R which is not collinearMathworldPlanetmath with P and Q.

We may choose our coordinates so as to place P at (1,0,0), Q at (0,1,0) and R at (0,0,1). The equation of our curve is F(x,y,z)=0, where


Since F(1,0,0)=k and F(0,1,0)=e, we must have k=0 and e=0 in order for P and Q to lie on C. The equation of the line PR is y=0. Restricting to this line, we have


Since P is an inflection point, the curve must have third order contact with this tangent line, the restrictionPlanetmathPlanetmathPlanetmath of F to this line should have a triple root at P, i.e. should be a multiple of z3. Hence, f=0 and g=0.

Likewise, the equation of the line QR is x=0. Restricting to this line, we have


Again, since Q is an inflection point, we must have a triple root, so this quantity should be a multiple of z3, so we must have c=0 and d=0.

Summarizing our progress so far, we have found that


Next, we note that h cannot be zero because that would imply that C had a singularity at Q. Likewise, we must not have j be zero because that would mean a singularity at P. Also, were b=0, we would have F(x,y,z)=xy(z+hy+jx) which would have a singularity at R. Thus, by rescaling x, y, and z if necessarry, we can take h=j=b=1, so our equation assumes the form


Upon examining the form of the equation just derived more closely, we learn an important geometric fact:

Theorem 6.

Suppose that C is a non-singular complex cubic curve and that P and Q are inflection points of C. Then the line PQ intersects C in a third point R, which is also an inflection point.


By the foregoing result, we know that we can write the equation for C as F(x,y,z)=0 with


with P located at (1,0,0) and Q located at (0,1,0). The line PQ then has equation z=0. Restricting F to this line, F(x,y,0)=xy2+x2y=x(x+y)y, so R, the third point of intersection of C with PQ, has coordinates (1,-1,0).

Next, we compute the Hessian determinant at R:


Since it equals zero, R is an inflection point. ∎

Since our curve is of the third order, a line cannot intersect more than three points. Hence, we see that a line passing through two inflection points of a non-singular complex cubic plane curve must pass through exactly three inflection points. As it turns out, this observation, together with the fact that there are exactly nine inflection points suffices to determine the locations of the inflection points up to collineationMathworldPlanetmath. However, rather than pursuing this line of reasoning here, we will instead cast the equation of the curve in a form which makes it easy to locate all the inflection points and makes the symmetryMathworldPlanetmathPlanetmath of the curve apparent.

Theorem 7.

Suppose C is a non-singular complex cubic curve. Then we may choose homogeneous coordinates the equation of the curve assumes the form


We know that there exists a straight line L which intersects C in exactly three points, all of which are inflection points of C. Choose a homogenous coordinate system in which the equation of L is Z=0. Since the three points of intersection of C with L are distinct, we may place them at any three locations; we shall choose (1,-1,0), (1,-ρ,0), and (1,-ρ2,0), where ρ=(-1+-3)/2 is a primitivePlanetmathPlanetmath cube root of unity.

With this choice, the equation of C becomes


By making the change of variables


we simplify the equation of the curve to




Note that this transformation leaves the coordinates of the three points of intersection of L and C unchanged.

Next, we impose the requirement that the three points lying on L be inflecton points of C. Setting Z to zero, the Hessian becomes


In order for the three points intersection of L and C to be inflection points, we must have this be a multiple of x3+y3. This requires that b and c both be zero.

Were a zero as well, our cubic would have a singularity at (0,0,1). Since C is assumed to not be singular, that means that a0, so we can make the further transform


to put the equation for C in the form x3+y3+z3+6mxyz=0. ∎

While we have shown that we can transform the equation of any non-singular curve into our canonical form, there remains the possibility that a curve whose equation is expressible in our canonical form may be singular or degenerate. We will now examine this possibility.

Theorem 8.

A equation x3+y3+z3+6mxyz=0 describes a singular curve if and only if 8m3+1=0.


The curve described by the equation f(x,y,z)=0 is singular if and only if there exists a point on the curve at which all three partial derivativesMathworldPlanetmath of f go zero. Taking derivativesPlanetmathPlanetmath and doing a little bit of algebraicMathworldPlanetmath manipulation, this means that, for the curve to be singular, there must be a non-trivial solution to the system of equations


Multiplying the last three equations together, we obtain x2y2z2=-8m3x2y2z2=0, or (8m3+1)x2y2z2=0. If 8m3+10, the only way for this to be satisfied is to have either x=0 or y=0 or z=0. However, suppose that x=0. Then, by the last two equations, we would also have y=0 and z=0, so the solution would be trivial. Likewise, if y=0, then it follows that x=0 and z=0; and, if z=0, it follows that x=0 and y=0. Hence, when 8m3+10, we only have the trivial solution, so the curve is non-singular.

We will finish by explicitly showing that the curve degenerates when 8m3+1=0, i.e. when m=-1/2 or m=-ρ/2 or m=-ρ2/2. (As before, ρ is a primitive cube root of unity.) In each of these cases, we can factor our equation into a product of three linear terms:


Thus, when 8m3+1=0, our curve degenerates into a triangleMathworldPlanetmath (whose vertices are the singularities). ∎

Having obtained this canonical form, it is quite easy to exhibit all nine inflection points.

Theorem 9.

Given a non-singular complex cubic plane curve, there exists a coordinate system in which the inflection points of that curve have the following coordinates:


As previously, ρ denotes a primitive third root of unityMathworldPlanetmath.


For convenience, set f=x3+y3+z3+6mxyz and let h be 1/216 times the Hessian of f. Computing, we have


Thus the Hessian is of the same form, but with m replaced by -(2m3+1)/6m2. Next, we form two combinationsMathworldPlanetmathPlanetmath of f ande h:


As we have just seen, the condition for the curve not to be singular is exactly that 8m3+10, so we may cancel to conclude that




The latter equation will be satisfied if either x=0 or y=0 or z=0. When x=0, the former equation reduces to y3+z3=0, which gives us the solutions


Likewise, when y=0, it reduces to x3+z3=0 with the solutions


and, when z=0, it reduces to x3+y3=0, with solutions


Title inflection points and canonical forms of non-singular cubic curves
Canonical name InflectionPointsAndCanonicalFormsOfNonsingularCubicCurves
Date of creation 2014-01-16 16:32:44
Last modified on 2014-01-16 16:32:44
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 42
Author rspuzio (6075)
Entry type Feature
Classification msc 14N10
Classification msc 14H52