inflection points and canonical forms of non-singular cubic curves

In this entry, we shall investigate the inflection pointsof non-singular complex cubic curves and show that theequations of such curves can always be put in variouscanonical forms. The reason for lumping these two subjectsin a single entry is that, being related, it is moreefficient to develop them both at the same time than it wouldbe to treat either one in isolation — on the one hand,putting the equation of the curve in a canonical formsimplifies calculations relating to inflection points whilst,on the other hand, information about inflection points allowsone to construct canonical forms.

We shall consider our curves as projective curves and describethem with homogenous equations; i.e. a curve is a locus in$ℂ{\mathbb{P}}^2$ where $F(x,y,z)=0$ for some third-order homogenous polynomial $F$. For the purpose at hand, theterm “inflection point” may be taken to mean a point on thecurve where the tangent intersects the curve with multiplicity3 — a point on the curve will have this property if and onlyif it is a zero of the Hessian.

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

Theorem 1.

Suppose $C$ is a non-singular complex cubic curve and $P$ a pointon that curve. Then we may choose homogenous coordiantes in whichthe coordiantes of $P$ are $\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{)}$ and equation of the curvelooks as follows:

$$y{z}^2+a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3+e{x}^{2}z=0$$

Proof.

The most general equation of a cubic is as follows:

$$A{X}^3+B{X}^{2}Y+C{X}^{2}Z+DX{Y}^2+EXYZ+FX{Z}^2+G{Y}^3+H{Y}^{2}Z+KY{Z}^2+L{Z}^3=0$$

Requiring that the curve pass through $(0,0,1)$ means that$L=0$. Since our curve is not singular, it has a tangentat all points, in particular, a tangent through $P$. Wemay ask that the equation of the tangent through $P$ be$Y=0$, which requires that $F=0$. Since our curve isnot singular at $(0,0,1)$, both $F$ and $K$ cannot be zeroso, 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

$$y{z}^2+a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3+e{x}^{2}z=0$$

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

$a$$=A-CE/2$

$b$$=B-CH/2-E^2/4$

$c$$=D-EH/2$

$d$$=G-H^2/4$

$e$$=C$

∎

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

Theorem 2.

If $C$ is a non-singular complex cubic curve with homogenousequation $F\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{,}z\mathrm{)}\mathrm{=}\mathrm{0}$, then the Hessian of $F$ does not equala multiple of $F$.

Proof.

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

$$F(x,y,z)=y{z}^2+a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3+e{x}^{2}z.$$

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

If $e\ne 0$, then the Hessian cannot me a multiple of $F$because $F(0,0,1)=0$. Assume then that $e=0$. In thatcase, we must have $a\ne 0$ because, if $a=0$, then $F$would factor as a product of $y$ and a quadratic polynomial,which is impossible because the curve $C$ is assumed notto be singular. Evaluating the Hessian on the line $y=0$,

Since $a$ cannnot be zero, this does not vanish identicallyand 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 ofa cubic curve distinct from $C$. Hence, we know that theremust be at least one and at most nine point of intersectionof these two curves, i.e. at least one and at most nineinflection points. Knowing that inflection points exist, wecan improve our canonical form by choosing $P$ as aninflection point. With a little tidying up, this will give usWeierstrass’ canonical form.

Theorem 3.

Suppose $C$ is a non-singular complex cubic curve and $P$ a pointof inflection of on that curve. Then we may choose homogenouscoordinates in which the coordiantes of $P$ are $\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{)}$ andequation of the curve looks as follows:

$$y{z}^2+4x^3+g_{2}x{y}^2+g_3{y}^3=0$$

Proof.

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

$$Y{Z}^2+a{X}^3+b{X}^{2}Y+cX{Y}^2+d{Y}^3+e{X}^{2}Z=0$$

and the coordinates of $P$ are $(0,0,1)$. By thecalculation made in the proof of the previous theorem,we know that $H(0,0,1)=8e$. Since $P$ is a pointof 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 $a^{1/3}$ to make the following coordinatetransformation:

$X$$=x-by/3$

$Y$$=ay/4$

$Z$$=z$

After canceling a factor of $a$, our equation becomes

$$y{z}^2+4x^3+g_{2}x{y}^2+g_3{y}^3=0,$$

where

$g_2$$={\displaystyle \frac{1}{4}}ac-{\displaystyle \frac{7}{12}}{b}^2$

$g_3$$={\displaystyle \frac{11}{432}}{b}^3+{\displaystyle \frac{1}{48}}abc+{\displaystyle \frac{1}{16}}{a}^{2}d$

∎

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

Theorem 4.

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

Proof.

Given an inflection point, the foregoing theorem tells usthat we can choose a coordinate system in which that pointhas coordinates $(0,0,1)$ and the equation of the curveassumes the form $F(x,y,z)=0$ where

$$F(x,y,z)=y{z}^2+4x^3+g_{2}x{y}^2+g_3{y}^3.$$

Computing the Hessian, we have

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

$$(\frac{\partial H}{\partial x},\frac{\partial H}{\partial y},\frac{\partial H}{\partial z})(0,0,1)=(-96,0,0).$$

For comparison, we have

$$(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z})(0,0,1)=(0,1,0).$$

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

Knowing that there exist multiple inflection points,we will now cast the equation in a form which placestwo 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 wemay choose homogeneous coordinates in which the coordinatesof $P$ are $\mathrm{(}\mathrm{1}\mathrm{,}\mathrm{0}\mathrm{,}\mathrm{0}\mathrm{)}$, the coordinates of $Q$ are $\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{,}\mathrm{0}\mathrm{)}$and the equation of the curve looks as follows:

$$x^{2}y+y^{2}x+z^3+axyz=0$$

Proof.

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 cannotintersect $C$ at any other points; in particular, itcannot 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 collinear 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 ourcurve is $F(x,y,z)=0$, where

$$F(x,y,z)=axyz+b{z}^3+cy{z}^2+d{y}^{2}z+e{y}^3+fx{z}^2+g{x}^{2}z+hx{y}^2+j{x}^{2}y+k{x}^3.$$

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

$$F(x,0,z)=b{z}^3+f{z}^2+g{x}^{2}z.$$

Since $P$ is an inflection point, the curve must havethird order contact with this tangent line, the restrictionof $F$ to this line should have a triple root at $P$,i.e. should be a multiple of $z^3$. Hence, $f=0$ and$g=0$.

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

$$F(0,y,z)=b{z}^3+cy{z}^2+d{y}^{2}z.$$

Again, since $Q$ is an inflection point, we must havea triple root, so this quantity should be a multipleof $z^3$, so we must have $c=0$ and $d=0$.

Summarizing our progress so far, we have found that

$$F(x,y,z)=axyz+b{z}^3+hx{y}^2+j{x}^{2}y.$$

Next, we note that $h$ cannot be zero because thatwould imply that $C$ had a singularity at $Q$.Likewise, we must not have $j$ be zero because thatwould mean a singularity at $P$. Also, were $b=0$,we would have $F(x,y,z)=xy(z+hy+jx)$ which wouldhave 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

$$F(x,y,z)=axyz+z^3+x{y}^2+x^{2}y.$$

∎

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

Theorem 6.

Suppose that $C$ is a non-singular complex cubic curveand that $P$ and $Q$ are inflection points of $C$. Thenthe line $P\mathit{}Q$ intersects $C$ in a third point $R$, whichis also an inflection point.

Proof.

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

$$F(x,y,z)=x{y}^2+x^{2}y+z^3+axyz$$

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, $F(x,y,0)=x{y}^2+x^{2}y=x(x+y)y$, so$R$, the third point of intersection of $C$ with $PQ$, hascoordinates $(1,-1,0)$.

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 cannotintersect more than three points. Hence, we see thata line passing through two inflection points of anon-singular complex cubic plane curve must pass throughexactly three inflection points. As it turns out, thisobservation, together with the fact that there are exactlynine inflection points suffices to determine the locationsof the inflection points up to collineation. However, ratherthan pursuing this line of reasoning here, we will insteadcast the equation of the curve in a form which makes it easyto locate all the inflection points and makes the symmetryof the curve apparent.

Theorem 7.

Suppose $C$ is a non-singular complex cubic curve. Then wemay choose homogeneous coordinates the equation of the curveassumes the form

$$x^3+y^3+z^3+6mxyz=0.$$

Proof.

We know that there exists a straight line $L$ which intersects$C$ in exactly three points, all of which are inflection pointsof $C$. Choose a homogenous coordinate system in which theequation of $L$ is $Z=0$. Since the three points of intersectionof $C$ with $L$ are distinct, we may place them at any threelocations; we shall choose$(1,-1,0)$, $(1,-\rho ,0)$, and $(1,-{\rho }^2,0)$, where$\rho =(-1+\sqrt{-3})/2$ is a primitive cube root of unity.

With this choice, the equation of $C$ becomes

$$X^3+Y^3+D{X}^{2}Z+EXYZ+F{Y}^{2}Z+BX{Z}^2+CY{Z}^2+A{Z}^3=0.$$

By making the change of variables

$$x=X-FZ/3$$

$$y^{\prime }=Y-DZ/3$$

we simplify the equation of the curve to

$$x^3+y^3+ExyZ+bx{Z}^2+cy{Z}^2+a{Z}^3=0.$$

where

$$a=A-\frac{BD}{3}+\frac{2D^3}{27}-\frac{CF}{3}+\frac{DEF}{9}+\frac{2F^3}{27}$$

$$b=B-\frac{D^2}{3}-\frac{EF}{3}$$

$$c=C-\frac{DE}{3}-\frac{F^2}{3}$$

Note that this transformation leaves the coordinates of the threepoints of intersection of $L$ and $C$ unchanged.

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

In order for the three points intersection of $L$ and $C$ to beinflection points, we must have this be a multiple of $x^3+y^3$.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 $a\ne 0$, sowe can make the further transform

$$z=a^{1/3}Z$$

$$m=a^{-1/3}E/6$$

to put the equation for $C$ in the form $x^3+y^3+z^3+6mxyz=0$.∎

While we have shown that we can transform the equation of anynon-singular curve into our canonical form, there remains thepossibility that a curve whose equation is expressible inour canonical form may be singular or degenerate. We willnow examine this possibility.

Theorem 8.

A equation $x^{\mathrm{3}}\mathrm{+}{y}^{\mathrm{3}}\mathrm{+}{z}^{\mathrm{3}}\mathrm{+}\mathrm{6}\mathit{}m\mathit{}x\mathit{}y\mathit{}z\mathrm{=}\mathrm{0}$ describes a singularcurve if and only if $\mathrm{8}\mathit{}{m}^{\mathrm{3}}\mathrm{+}\mathrm{1}\mathrm{=}\mathrm{0}$.

Proof.

The curve described by the equation $f(x,y,z)=0$ is singularif and only if there exists a point on the curve at which allthree partial derivatives of $f$ go zero. Taking derivativesand doing a little bit of algebraic manipulation, this meansthat, for the curve to be singular, there must be a non-trivialsolution to the system of equations

$$x^2=-2myz$$

$$y^2=-2mxz$$

$$z^2=-2mxy.$$

Multiplying the last three equations together, we obtain$x^2{y}^2{z}^2=-8m^3{x}^2{y}^2{z}^2=0$, or$(8m^3+1)x^2{y}^2{z}^2=0$. If $8m^3+1\ne 0$, the onlyway 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 twoequations, we would also have $y=0$ and $z=0$, so the solutionwould 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 $8m^3+1\ne 0$, we only have the trivial solution,so the curve is non-singular.

We will finish by explicitly showing that the curve degenerateswhen $8m^3+1=0$, i.e. when $m=-1/2$ or $m=-\rho /2$ or$m=-{\rho }^2/2$. (As before, $\rho$ is a primitive cuberoot of unity.) In each of these cases, we can factor ourequation into a product of three linear terms:

$$x^3+y^3+z^3-3xyz=(x+y+z)(x+\rho y+{\rho }^{2}z)(x+{\rho }^{2}y+\rho z)$$

$$x^3+y^3+z^3-3\rho xyz={\rho }^2(x+y+\rho z)(x+\rho y+z)(\rho x+y+z)$$

$$x^3+y^3+z^3-3{\rho }^{2}xyz=\rho (x+y+{\rho }^{2}z)(x+{\rho }^{2}y+z)({\rho }^{2}x+y+z)$$

Thus, when $8m^3+1=0$, our curve degenerates into a triangle(whose vertices are the singularities).∎

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

Theorem 9.

Given a non-singular complex cubic plane curve, thereexists a coordinate system in which the inflectionpoints of that curve have the following coordinates:

As previously, $\rho$ denotes a primitive third root of unity.

Proof.

For convenience, set $f=x^3+y^3+z^3+6mxyz$ andlet $h$ be $1/216$ times the Hessian of $f$.Computing, we have

Thus the Hessian is of the same form, but with $m$ replaced by$-(2m^3+1)/6m^2$. Next, we form two combinations of $f$ande $h$:

$$m^{2}f+h=(8m^3+1)xyz$$

$$(2m^3+1)f-6mh=(8m^3+1)(x^3+y^3+z^3)$$

As we have just seen, the condition for the curve not to besingular is exactly that $8m^3+1\ne 0$, so we may cancelto conclude that

$$x^3+y^3+z^3=0$$

and

$$xyz=0.$$

The latter equation will be satisfied if either $x=0$ or$y=0$ or $z=0$. When $x=0$, the former equation reducesto $y^3+z^3=0$, which gives us the solutions

$$(0,1,-\rho ),(0,1,-1),(0,1,-{\rho }^2)$$

Likewise, when $y=0$, it reduces to $x^3+z^3=0$ withthe solutions

$$(-\rho ,0,1),(-1,0,1),(-{\rho }^2,0,1)$$

and, when $z=0$, it reduces to $x^3+y^3=0$, withsolutions

$$(1,-\rho ,0),(1,-1,0),(1,-{\rho }^2,0).$$

∎