inflection points and canonical forms of non-singular cubic curves
In this entry, we shall investigate the inflection points of non-singular complex cubic curves and show that the equations of such curves can always be put in various canonical forms. 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 where for some third- order homogenous polynomial . For the purpose at hand, the term “inflection point” may be taken to mean a point on the curve where the tangent intersects the curve with multiplicity 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.
Suppose is a non-singular complex cubic curve and a point on that curve. Then we may choose homogenous coordiantes in which the coordiantes of are and equation of the curve looks as follows:
The most general equation of a cubic is as follows:
Requiring that the curve pass through means that . Since our curve is not singular, it has a tangent at all points, in particular, a tangent through . We may ask that the equation of the tangent through be , which requires that . Since our curve is not singular at , both and cannot be zero so, without loss of generality, we may take .
We may then make the following linear transform:
Upon doing so, our equation assumes the form
where the coefficients are defined as follows:
Next, we turn our attention to the Hessian to begin our investigation of inflection points.
If is a non-singular complex cubic curve with homogenous equation , then the Hessian of does not equal a multiple of .
Let us examine the Hessian at the point :
If , then the Hessian cannot me a multiple of because . Assume then that . In that case, we must have because, if , then would factor as a product of and a quadratic polynomial, which is impossible because the curve is assumed not to be singular.11Were for some quadratic polynomial , then the gradient of would vanish at the intersection of the line and the conic , so the variety described by would be singular at the point(s) of intersection. Evaluating the Hessian on the line ,
Since cannnot be zero, this does not vanish identically and is clearly not a non-zero multiple of . ∎
This theorem tells us that is the equation of a cubic curve distinct from . 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 as an inflection point. With a little tidying up, this will give us Weierstrass’ canonical form.
Suppose is a non-singular complex cubic curve and a point of inflection of on that curve. Then we may choose homogenous coordinates in which the coordiantes of are 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 are . By the calculation made in the proof of the previous theorem, we know that . Since is a point of inflection, this implies that .
As noted in the previous proof, we cannot have because that would make our curve singular. Hence, we can divide by to make the following coordinate transformation:
After canceling a factor of , our equation becomes
Using this canonical form, it is easy to show that we have the maximum number of inflection points possible.
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 system in which that point has coordinates and the equation of the curve assumes the form where
Computing the Hessian, we have
Computing its gradient at , we find that
For comparison, we have
Thus, we see that the curves and are both smooth at and intersect transversely, hence they have intersection multiplicity 1 there. Because we could choose our coordinates to place any inflection point at , this means that every intersection of our curve with its Hessian has intersection multiplicity 1. Since both and 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.
Suppose is a non-singular complex cubic curve and and are points of inflection of on that curve. Then we may choose homogeneous coordinates in which the coordinates of are , the coordinates of are and the equation of the curve looks as follows:
Since is a point of its inflection, the tangent to at intersects with multiplicity 3 at . Since is a cubic, this means that this tangent cannot intersect at any other points; in particular, it cannot pass through . Likewise, the tangent through cannot pass through . Hence, the tangent at and the tangent at must intersect at some third point which is not collinear with and .
We may choose our coordinates so as to place at , at and at . The equation of our curve is , where
Since and , we must have and in order for and to lie on . The equation of the line is . Restricting to this line, we have
Since is an inflection point, the curve must have third order contact with this tangent line, the restriction of to this line should have a triple root at , i.e. should be a multiple of . Hence, and .
Likewise, the equation of the line is . Restricting to this line, we have
Again, since is an inflection point, we must have a triple root, so this quantity should be a multiple of , so we must have and .
Summarizing our progress so far, we have found that
Next, we note that cannot be zero because that would imply that had a singularity at . Likewise, we must not have be zero because that would mean a singularity at . Also, were , we would have which would have a singularity at . Thus, by rescaling , , and if necessarry, we can take , so our equation assumes the form
Upon examining the form of the equation just derived more closely, we learn an important geometric fact:
Suppose that is a non-singular complex cubic curve and that and are inflection points of . Then the line intersects in a third point , which is also an inflection point.
By the foregoing result, we know that we can write the equation for as with
with located at and located at . The line then has equation . Restricting to this line, , so , the third point of intersection of with , has coordinates .
Next, we compute the Hessian determinant at :
Since it equals zero, 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 collineation. 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 symmetry of the curve apparent.
Suppose 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 which intersects in exactly three points, all of which are inflection points of . Choose a homogenous coordinate system in which the equation of is . Since the three points of intersection of with are distinct, we may place them at any three locations; we shall choose , , and , where is a primitive cube root of unity.
With this choice, the equation of 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 and unchanged.
Next, we impose the requirement that the three points lying on be inflecton points of . Setting to zero, the Hessian becomes
In order for the three points intersection of and to be inflection points, we must have this be a multiple of . This requires that and both be zero.
Were zero as well, our cubic would have a singularity at . Since is assumed to not be singular, that means that , so we can make the further transform
to put the equation for in the form . ∎
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.
A equation describes a singular curve if and only if .
The curve described by the equation is singular if and only if there exists a point on the curve at which all three partial derivatives of go zero. Taking derivatives and doing a little bit of algebraic 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 , or . If , the only way for this to be satisfied is to have either or or . However, suppose that . Then, by the last two equations, we would also have and , so the solution would be trivial. Likewise, if , then it follows that and ; and, if , it follows that and . Hence, when , we only have the trivial solution, so the curve is non-singular.
We will finish by explicitly showing that the curve degenerates when , i.e. when or or . (As before, 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 , our curve degenerates into a triangle (whose vertices are the singularities). ∎
Having obtained this canonical form, it is quite easy to exhibit all nine inflection points.
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, denotes a primitive third root of unity.
For convenience, set and let be times the Hessian of . Computing, we have
Thus the Hessian is of the same form, but with replaced by . Next, we form two combinations of ande :
As we have just seen, the condition for the curve not to be singular is exactly that , so we may cancel to conclude that
The latter equation will be satisfied if either or or . When , the former equation reduces to , which gives us the solutions
Likewise, when , it reduces to with the solutions
and, when , it reduces to , with solutions
|Title||inflection points and canonical forms of non-singular cubic curves|
|Date of creation||2014-01-16 16:32:44|
|Last modified on||2014-01-16 16:32:44|
|Last modified by||rspuzio (6075)|