### Properties Concerning the Hessian Complex in Relation to Quadrics, Conics, and a Twisted Cubic

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*Abstract*

*Abstract*

This paper is concerned mainly with the interpretation and properties of the hessian complex; how it arises from a quadric defined by three generators, its relationship to a conic as a plane section of the quadric, and to a twisted cubic. Geometrical interpretations of many invariants and covariants arise naturally, some of which appear to be new.

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If the three generators are A, B, C the equation of the quadric may be written^{*} as [x ABC x] = o or ×.Q.× = o where Q = A.B.C — C.B.A, that is, the symmetric part of A.B.C. The antisymmetric part leads naturally to the hessian complex H defined by H = Σ (A.B.C.+ C.B.A), where Σ denotes summation over cyclic permutations of A, B, C, which in turn reduces to H = (BC) A + (CA) B + (AB) C.

It is clear that all generators of the opposite system of the quadric to A, B, C belong to the complex. It is also clear that all generators of the quadric of the system A, B, C are given in terms of a parameter λ by

G = (1—λ) (BC) A + λ (CA) B + λ (λ—1) (AB) C,

and of these lines, two only belong to the hessian complex, namely those for which λ = — ω, and λ = — ω^{2}, (where ω^{3} = 1) These two lines

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h_{1} = (BC) A + ω (CA) B + ω^{2} (AB) C,

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h_{2} = (BC) A + ω^{2} (CA) B + ω (AB) C,

are called the hessian pair of lines. In terms of H, h_{1}, h_{2} the quadric takes a point equation of the form [x H h_{1} h_{2} x] = o and a line equation, (h_{1} h_{2}) (pH)^{2} = (HH) (ph_{1}) (ph_{2}), where p is the variable line co-ordinate.

The hessian complex also gives harmonic properties: For if A^{1} is the polar of A with regard to the complex H, then A, A^{1} separate B, C harmonically, in particular A^{1} is given by

A^{1} = — ½ (BC) A + (CA) B + (AB) C.

If B^{1}, C^{1} are formed in the same way, it is of importance that the hessian complex formed from A^{1}, B^{1}, C^{1} is the original hessian complex H. The hessian line pair are the double lines of the involution defined by the pairs AA^{1}, BB^{1}, CC^{1}. Alternatively^{†} the hessian complex may be defined as containing all the generators of the second system and two lines h_{1} h_{2} of the first system. The lines h_{1}, h_{2}, together with any pair of generators of the second system, form a skew quadrilateral whose diagonals are polar lines both with regard to the hessian complex and the quadric.

Now take a section of the quadric by an arbitrary plane π and, on the conic, denote the points where the generators A, B, C meet the plane by the same letters and similarly for the polars A^{1}, B^{1}, C^{1} and for h_{1}, h_{2} Let H represent the pole of the plane π with respect to the complex H. Points on the conic will be given para-

[Footnote] * [x ABC x] ≡ x_{mu;} A_{μ ν} B^{νσ} C_{σρ} x^{ρ} = (x a B) (x a′ C) − (x a′ B) (x a C) A.B.C. ≡ A_{μν} B^{νσ} C_{σρ} (BC) ≡ (b b′ c c′)

[Footnote] † It is realised that this complex is represented by a prime in 5 dimensional space, but it is of no special significance in the *properties* developed here.

metrically by the above λ and the parameters for the triad A, B, C will be the roots of a binary cubic. Similarly the triad A^{1}. B^{1}, C^{1} leads to the cubic covariant of this cubic, and h_{1}, h_{2} correspond to the roots of the hessian quadratic. The lines AA^{1}, BB^{1}, CC^{1} meet in the point H which is also the pole with respect to the conic of the hessian line h_{1} h_{2}. Since the same hessian complex arose from the lines A^{1}, B^{1}, C^{1}, the hessian pair of its triad on the conic, derived from the cubic covariant, will be the same as the hessian pair derived from the triad A, B, C. The configuration in the plane is completed by the introduction of the three complexes (ABCp), (BCAp), (CABp), where^{*} (CABp) = — (BC) (Ap) + (CA) (Bp) + (AB) (Cp). This complex contains the lines A, C and all the generators of the opposite system of the quadric. Taking the traces of these lines and complexes in the plane π, it is found that the tangents to the conic at B, C will belong to the complex (CABp) and hence these tangents will intersect in the pole O_{1} of the plane π with respect to this complex, and this point lies on the line AA_{1}. Similarly for the other two complexes (ABCp), (BCAp), giving poles O_{2}, O_{3}; thus the lines AHO_{1}, BHO_{2}, CHO_{3}, contain respectively the points A^{1}, B^{1}, C^{1}, a well-known configuration for the conic. This follows immediately, since the four complexes are linearly related.

As an illustration of the application of the hessian complex, we may interpret the vanishing of the invariant,

(BC) (AD) + (CA) (BD) + (AB) (CD)

that is, the line D belongs to the hessian complex H, and, incidentally, each of the four lines A, B, C, D belong to the hessian complex of the remaining three lines This implies that the two generators of the same system as A, B, C which meet D, say with parameters λ_{1}, λ_{2}, will separate harmonically, the two special lines h_{1}, h_{2} whose parameters are —ω, —ω^{2}. The two generators which meet D are given by the equation.

(1 − λ) (BC) (AD) + λ (CA) (BD) + λ (λ − 1) (CA) (BD) = o,

or.

λ^{2}(CA) (BD) − λ{(BC) (AD) − (CA) (BD) + (CA) (BD)} + (BC) (AD) = 0

or (λ − λ_{1}) (λ − λ_{2}) = o.

Thus the cross-ratio {λ_{1},λ_{2}; — ω, — ω^{2}} = — 1 provided 2λ_{1} λ_{2} — λ_{1} — λ_{2} + 2 = o, that is, provided (AB) (CD) + (BC) (AD) + (CA) (BD) = o.

Further, the vanishing of the ring product (ABCD) implies that these two generators, harmonically separate the lines A, C. Again these two generators coincide if

√(BC) (AD) + √(CA) (BD) + √(AB) (CD) = o,

which is the condition that D is a tangent or a generator of the quadric. Further properties have been derived by similar processes. We consider next the relationship between the twisted cubic through three points a, b, c whose tangents at these points are the lines A, B, C, the quadric which has the lines A, B, C as generators, and the corresponding hessian complex. An explicit equation for the quadric is obtained in terms of the plane [abc] and the three fundamental quadrics containing the twisted cubic. Interesting configurations and an interpretation of several invariants arise from a consideration of the two quadrics derived from the tangents at two triads of points on the twisted cubic, in particular, the cases when the two triads are apolar and when the triads form covariant sets. A particularly simple invariant arises when the two quadrics have a common generator, and some properties of the residual cubic curve are derived.

These properties lead to interpretations of all the fundamental invariants and covariants of the binary cubics defining the two triads; several of these are investiated.

[Footnote] *(CABp) is a ring product given by C^{λμ} A_{μν} B_{νρ} P_{ρλ}

Any point **k**, of the twisted cubic, with parameter *t* is given by **k** = **a** + **b***t* + c*t*^{2} + d*t*^{3} where the points **a, d** correspond to values o, ∞ of the parameter, the osculating plane at **k̈** is [**k k̇ k̈**] where [**k k k**] = [**a b c**] − 3 [**a d k**] + [b c d] *t*^{3}, (the upper dots denoting differentiation with respect to the parameter *t*); thus the osculating planes at **a** and **d** are [**a b c**] and [**b c d**] respectively, and clearly the point of intersection of the osculating planes at **a, d, k** is the point b + c*t* and this point also lies in the plane [**a d k**]. This point will be shown to be the pole (or trace) in the plane [**a d k**] of the hessian complex formed from the tangents to the cubic at **a, d, k** and also of the nul complex to which all the tangents to the cubic belong.

The tangent to the twisted cubic at a point **k** may be written as [**k k̇**] where [**k k**] = [**a b**] + 2*t* [**a c**] + 3*t*^{2} [**a d**] + *t*^{2} [**b c**] + 2*t*^{3} [**b d**] + *t*^{4} [**c d**], and hence all such tangents belong to a nul complex [**b c**] — 3 [**a d**]. Further, the tangents at **a, d, k** are [**a b**], [**c d**], [**k k̇**] respectively, and the hessian complex formed from these three lines is.

*H* = (**a b c d**) { [**a b**] + [**k k̇**] + [**c d**] *t*^{4}} (i)

and its pole in the plane [**a d k**], namely [*H* **a d k**] reduces to **b** + *t* c. Hence the hessian pole H is the intersection of the three osculating planes at **a, d, k** and lies in the plane [**a d k**]. Also each osculating plane is the polar of **a, d, k** with respect to the nul complex and hence their point H of intersection is the pole of the nul complex with respect to the plane [**a d k**].

More generally, the hessian complex *H* of the tangents at the points **p, q, r** of the cubic, with parameters λ, μ, ν is

*H* = (μ − ν)^{4} [**p p**] + (ν − λ)^{4} [**q q**] + (λ − μ)^{4} [**r r**] (ii)

and this contains the tangent at the point s with parameter σ provided Σ (λ — μ)^{4} (ν — σ)^{4} = o, and in virtue of the identity Σ (λ − μ) (ν − σ) = o this implies that if A, B, C, D are the line co-ordinates of the four tangents at the points λ, μ, ν, σ, then the invariant.

{ (AB) (CD) }^{¼} + { (BC) (AD) }^{¼} + { (CA (BD) }^{¾}

vanishes. However, as will be shown later, by fixing λ, μ, ν in (ii), this quartic in σ degenerates and gives the two tangents to the twisted cubic at the two hessian points of the cubic whose roots are λ, μ, ν.

Now consider the quadric Q whose generators are the tangents [**p p**], [**q q**], [**r r**] at the points **p, q, r** of the twisted cubic. A generator G of this quadric, with the parameter *l* is given by

G = *l* (μ-ν)^{4} [**pp**] + *l* (*l* − 1) (ν-λ)^{4} [**qq**] + (*l*-1) (λ-μ)^{1} [**rr**](iii)

and since [**p p. p. q.r**] = − (λ − μ)^{2} (λ − ν)^{2} (μ − ν) **p** with similar expressions for [**q q . p q r**], [**r r . p q r**]; the hessian pole H of the plane [**p q r**] is H = Σ (μ − ν)^{4} [**pp . pqr**] ≡ (μ − ν)^{3} **p** + (ν − λ)^{3} **q** + (λ − μ)^{3} **r** and G will meet the plane [**p q r**] in the point **g** where

g = *l* (μ − ν)^{3} p. + *l* (*l* − 1) (ν − λ)^{3} **q** + (*l* − 1) (λ − μ)^{3} **r** ≡ × **p** + y **q** + z **r**

where x^{-1} (μ − ν)^{3} + y^{-1} (ν − λ)^{3} + z^{-1} (λ − μ)^{3} = o(iv)

This gives the equation of the conic in which the quadric Q meets the plane [**p q r**] referred to the points **p, q, r** as base points.

Further, we may obtain an explicit expression for the quadric whose generators are the tangents to the twisted cubic at the points **p, q, r** in terms of the plane a_{x}^{3} ≡ a_{3} (x p) (x q) (x r) = o which meets the cubic in the points with parameters **p, q, r**. We have, in terms of the base points **a, b, c, d**,

**p = a** + pb + p^{2} c + p^{3} d

x = x_{0} a + x_{1} b + **x _{2} c + x_{3} d ≡ a + xb** + x

^{2}c + x

^{3}d

then (**p p. q x**) = − (p-q)^{2} (p-x)^{2} (q-x),

(**p p. q̇ x**) = 2 (p-q) (p-x)^{2} (q-x) − (p-q)^{2} (p-x)^{2}

Now Q = [**x ṗ. p . q q . r ṙ x**] = (**x p. ṗ q̇**) (**q r ṙ x**) − (**x p. ṗ q̇**) (**q̇ r ṙ x**), whence substituting in the values of the bracket factors and after some reduction, the quadric Q takes the simple symbolic form

Q = 4 a_{x}^{3} b_{x}^{3} − 9 (x y)^{2} (a b)^{2} a_{x} b_{y} (v) in which a, b and x, y are equivalent symbols respectively. The second term involves only the coefficients in the hessian quadratic ht^{2} = (a b)^{2} a̖ b̖ Written in full Q = (a_{x}^{3})^{2} − 9 (h_{0} Q_{0} + h_{1} Q_{1} + h_{2} Q_{2}) where a̖_{3} = a_{3}*t* + 3 a_{2} z + 3 a_{1} y + a_{0} x. h_{0} = a_{0} a_{2} − a_{1}^{2}, 2 h_{1} = a_{0} a_{3} − a_{1} a_{2}, h_{2} = a_{1} a_{1}a_{3} − a_{2}^{2}, Q_{0} = × z − y^{2}, Q_{1} = × *t* − y z, Q_{2} = y *t* − z^{2}, and it is noticed that Q_{0} = 0, Q_{1} = 0, Q_{2} = 0 are three independent quadrics each containing the twisted cubic.

Specialising these results, consider the points **a, k, d** on the twisted cubic, with parameters o, *t*, ∞ and the binary cubic τ (τ-*t*) = 0 (vi) with these parameters as roots. The hessian pair of the binary cubic is given by the roots − ω*t*, − ω^{2}*t* and will give the hessian points H_{1}, H_{2} on the twisted cubic. The join of the hessian pair H_{1} H_{2} will have the equation

[H_{1} H_{2}] ≡ [a b] + [a c] *t* + [b c] *t*^{2} + [b d] *t*^{3} + [c d] *t*^{4} (vii)

The generators h_{1}, h_{2} of the quadric (v) formed from the tangents to the cubic at the points **a, k, d** and belonging to this system of generators and to the hessian complex, are found by assigning the values —ω, —ω^{2} to the parameter *l* in the equation (iii), namely

G = *l* (**c d k k**) [**a b**] + *l* (*l*-1) (**a b c d**) [**k k**] + (1-*l*) (**a b k k**) [c d] ≡ *l* [**a b**] + *l* (*l*-1) [**k k**] + (1-*l*) [**c d**] *t*^{4}

which meets the plane [**a k d**] in the point

g = [**a k d . G**] ≡ *l* **a** + *l*-1) (-**k**) + (1-*l*) (*t*^{3} **d**);

hence substituting −ω, −ω^{2} for *l* it is found that the line joining the points in which the hessian pair h_{1}, h_{2} meets the conic of intersection of the quadric with the plane [**a k d**] is.

[ω **a − k** + ω^{2} *t*^{3} **d** . ω^{2} **a-k** + ω *t*^{3} **d**] ≡ [**k** + *t*^{3} **d** -**a** + *t*^{3} **d**] thus [h_{1} h_{2}]_{conte} = *t* [**a b**] + *t*^{2} [**a c**] + 3 *t*^{3} [**a d**] + *t*^{4} [**b d**] + *t*^{5} [**c d**] (viii) and by comparison with [H_{1} H_{2}]_{cubic} it follows that,

[H_{1} H_{2}]_{cuble} − [h_{1} h_{2}] _{conte} = *t*^{2} { [**b c**] − 3 [**a d**] } (ix)

Hence these two hessian lines in space are pole and polar with respect to the nul complex of the tangents; in the same manner it may be shown that [H_{1} H_{2}] + [h_{1} h_{2}] = *H* and hence these two lines are also conjugate with respect to the hessian complex. Further, these two complexes are apolar and the two lines are therefore the directrices of the congruence common to the two complexes.

It is well known that, in general, four tangents of a twisted cubic belong to a general linear complex, but in the case of the hessian complex formed from the tangents at **a, k d** only two tangents of the cubic belong to this complex, and these are the tangents at the two hessian points H_{1}, H_{2} on the cubic. Since these two tangents belong both to the nul complex and the hessian complex, they will intersect the directrices of their linear congruence, and consequently will also intersect the line joining h_{1} h_{2} on the conic section of the quadric by the plane [**a k d**]. (It is shown later that these two tangents pass through the points h_{1} h_{2}.)

Consider now the cubic covariant of the cubic (vi) where for convenience, the parameter of k is taken as unity. This cubic covariant will be

2 τ^{3} − 3 τ^{2} − 3 τ + 2 = 0 (x)

It is well known that the three osculating planes at the points on the cubic curve whose parameters satisfy (x) will intersect in a point *K* such that the cross-ratio {H_{1} H_{2}, H K} = -1. Denote by Q_{1}, Q_{2} the quadrics whose generators

are the tangents to the twisted cubic at the three points (i) **a, k, d** whose parameters satisfy (vi) and (ii) **l, m, n** whose parameters satisfy (x). It is found that the hessian complex corresponding to the tangents at **a k d** is the same as the hessian complex corresponding to the tangents at **l m n**, and that the planes [**a k d**], [**l m n**] meet in the line h_{1} h_{2} and further that K is the pole of the hessian complex in the plane [**l m n**].

Now denote the conics of intersection of the quadrics Q_{1} Q_{2} with the plane [**a k d**] by C, C′ respectively; the hessian pole h of this plane will be the same for both conics. If the generators at **l, m, n** to Q_{2} meet the conic C′ in l′, m′, n′ then the symmetry between the cubic and the cubic covariant persists, namely, that the points given on the first conic C by **a, k, d** and their cubic covariant points **a′ k′ d′** are such that the lines [**a a**′], [**k k′**],[**d d″**] are concurrent in h, but also l′ a′ h a l” are collinear, where **l″** lies on the conic C′; similarly for **m′, m″, n′, n”** and **l′ m′ n′** would correspond parametrically to a binary cubic on the conic C′ whilst **l″ m” n″** would represent its cubic covariant.

Exactly the same property holds if the quadrics Q_{1}, Q_{2} are intersected by the plane [**l m n**], a very interesting reciprocity between the binary cubic on the twisted cubic and the two pairs of conics; incidentally this leads to a configuration similar to that in the above section.

In particular from (v) the quadrics Q_{1}, Q_{2} reduce to the simple forms.

Q_{1} = Q_{x} + 2 P_{1}^{2}, Q_{2} = 27/4 Q̖ + ½ P_{2}^{2},

where P_{1} P_{2} are the planes [**a k d**], [**l m n**] respectively, so that.

P_{1} = y-z, P_{2} = 2 × − 3 y − 3 z + 2 *t*

and Q̖ = 2 (*t* y − z^{2}) − (*t* x-y z) + 2 (xz − y^{2})

Thus the quadrics Q_{1}, Q_{2} have double contact with the quadric Q_{x} along the two conics C_{1}, C_{2} in which Q_{x} is cut by the planes P_{1}, P_{2}.

By considering the characteristic equations the following properties arise, (a) Cones of the pencil Q_{1} + λ Q_{2} have vertices (i) on the line [h, h_{2}], (ii) at the points H, K. For the pencil 8 Q_{2} + 2 λ Q_{1}, the characteristic roots are (1) λ = − - 27 (twice) giving the line h_{1} h_{2}, (ii) λ = − 3 giving H, and (iii) λ = − 234 giving K.

(b) The pencil 2 Q_{x} + λ P_{1} P_{2} contains the conics C_{1}, C_{2} and specializes, for λ = ± √3 into two cones. Let O_{1}, O_{2} be the vertices of these two cones, then by suitable weighting,

O_{1} + O_{2} = K where K = 2**a + b** − c − 2d,

O_{1} − O_{2} = H. H.= √3 (**b + c**)

and the hessian pair H_{1} H_{2} on the cubic are given by.

H_{1} = a − ω b + ω^{2} c − d, H_{2} = a − ω^{2} b + ω c − d

whence H_{1} = H_{2} = K and H_{1} − H_{2} = 1 H,

verifying that H_{1} H_{2} H K are collinear and form a harmonic range.

(c) The planes P_{1}, P_{2} are conjugate with respect to the quadric Q_{x} since the coefficient of λ in |2 Q̖ − λ P_{1} P_{2} | = 0 gives the invariant relation {Q_{x}^{3}, P_{1} P_{2}} 0 or “θ” = 0. The whole configuration gives a very compact closed system.

Various Invariants and Covariants

If the point × (x_{0} x_{1} x_{2} x_{3}) is written symbolically as (l, x, x^{2}, x^{3}) then by equating the coefficients of the various powers of *t* the identity a*t*^{3} ≡ (*t*x)^{3} defines a point x ≡ (a_{3}, − a_{2}, a_{1}, − a_{0}) if a*t*^{3} = 0 is the plane of the three points **p, q, r** of the twisted cubic, then the new point so defined, gives the point of intersection of the osculating planes at **p, q, r**, it is also the pole of the hessian complex formed from the three tangents to the cubic at **p, q, r** in the plane [**p q r**] and also the pole of the nul complex in this plane.

If, further, the two cubics a_{t}^{3}, b_{0}^{t} are apolar so that (a b)^{3} = 0 or.

a_{3} b_{0} − 3 a_{2} b_{1} + 3 a_{1} b_{2} − a_{0} b_{3} = 0

which immediately shows that the hessian pole for the plane a̖^{3} = 0 lies in the

plane b_{x}^{3} = 0 and since the relation is symmetrical, the hessian pole for the plane b_{x}^{3} = 0 lies in the plane a_{x}^{3} = 0. In general there is one cubic which is apolar to three given cubics, and this cubic is clearly given by the plane through the hessian poles of the three cubics.

Taking three points **a, k, d** on the twisted cubic to have parameters o, k, ∞ with respect to both the twisted cubic and also the conic of intersection of the quadric, with the tangents at **a, k, d** as generators, with the plane [**a k d**] there is a (l, l) correspondence between the points of the cubic (l, λ, λ^{2}, λ^{3}) and the points of the conic given by k (k − λ) a + λ k (-**k**) + λ (λ − k) (k^{3} d) and it follows easily that the tangent plane to the quadric at the point λ of the come passes through the point λ on the cubic; in fact the equation of the tangent plane is (x λ) a_{x}^{2} aλ = 0 and this plane cuts the twisted cubic in two further points whose parameters are given by the roots of the quadratic a_{t}^{2} a_{λ} = 0 which provides an interpretation of the first polar a_{t}^{2} a_{λ}. It should also be noticed that the chord of the cubic through these two points meets the plane [**a k d**] in the point λ of the conic. It is of interest that the quadratic in *t*, a_{t}^{2} a^{λ} becomes a perfect square provided its hessian quadratic h_{λ}^{2} = (a a′) a_{λ} a′_{λ} vanishes. Hence it follows that the tangent plane to the quadric at one of the hessian points h_{1}, on the conic, will pass through the hessian point H_{1} on the cubic, and this tangent plane will also meet the cubic again in two points which are coincident at the second hessian point H_{2}; hence the tangents at H_{1}, H_{2} to the twisted cubic will meet the plane a_{x}^{3} = 0 in the hessian points h_{1}, h_{2} respectively of the conic.

Again, by expansion

(*t* λ) a_{t}^{2} a_{λ} = (λ a_{3} + a_{2}) *t*^{3} + (λ a_{2} + 2 a_{1} − λ^{2} a_{3}) *t*^{2} + (a_{0} − 2 λ^{2} a_{2} − λ a_{1}) *t* − λ^{2} a_{1} − λ a_{0}

whence the point of intersection of the tangent planes to the quadric at the points λ, μ, ν of the conic is given by the three column minors
(a_{3} a_{2} − a_{1} − a_{0}
a_{2} 2a_{1} a_{0} 0
0 a_{3} 2 a_{2} a_{1})
and these co-ordinates are in fact those derivable from the covaniant cubic: for writing (*t* x)^{3} ≡ (a b)^{2} (a c) b_{t} c_{t}^{2}, (a, b, c being equivalent symbols) then (x_{0} x_{1} x_{2} x_{3}) takes these required values. This point is also the pole of the plane a_{x}^{3} = 0 with respect to the quadric whose generators are the tangents to the cubic at the points in which the cubic is met by the plane a_{x}^{3} = 0; hence given a triad of points a_{t}^{3} = 0, the plane of the covariant triad passes through the pole of the plane a_{x}^{3} = 0 with respect to the quadric, and further this pole is the pole of the covariant plane C_{x}^{3} = 0 with respect to the hessian complex, and with respect to the nul complex. If any plane d_{x}^{3} = 0 is apolat to C_{x}^{3} = 0, it passes through the pole of this plane with respect to the quadric, so that the plane d_{x}^{3} = 0 and the covariant plane of C_{t}^{3} = 0 are then conjugate with respect to the quadric. If two triads are apolar, the plane of one triad and the plane of the covariant triad of the second triad are conjugate with respect to the quadric whose generators are the tangents to the cubic at the points of the second triad.

Further, the tangent planes to the quadric at the points λ, μ of the conic will meet the twisted cubic in points with parameters λ, μ and in a further four points, it follows easily that if the cubics giving the two tangent planes are apolar, then these four points form an harmonic range.

Again, the three tangents to the twisted cubic at the points **a, k, d** are the generators of a quadric: consider the three generators of the quadric of the opposite system and passing through **a, k, d.** They are easily found to be [**a, c** + 2 k d], [**k, a** + k **b**], [2 **a** + k **b, d**]

respectively, and it is of interest that their hessian complex is the nul complex itself.

Further these three new generators define an associated twisted cubic through three points **a k d** which has these three generators as the tangents at **a, k, d** and the parameterization on this cubic can be chosen so that **a, k, d** have parameters o, k, ∞. Hence a correspondence is set up between points T of the original cubic and points T* of the associated cubic such that T, T* correspond to the same parameter. It is found that the join of TT* passes through a fixed point σ, the pole of the plane [**a k d**] with respect to the quadric (and also the hessian pole of the plane of the covariant points) and that if TT* meets the plane [**a k d**] in **u** then the range {TT*, u σ} is harmonic. An involutory collineation connects T and T* and is given by (π σ) T* = T − 2 σ (π T) where π is the plane [**a k d**].

Consider next the possibility that any two quadrics, one (say) with the tangents at the points **p, q, r** of the cubic as generators, and the other with the tangents at **l, m, n** of the cubic as generators, should possess a common generator. Let the parameters corresponding to these points on the cubic be p, q, r; l, m, n respectively. Since the tangent at **p** is given by.

[**p p**] = [**a b**] + 2 p. [**a c**] + p^{2} {**3** [**a d**] + [**b c**]} + 2 p^{3} [**b d**] + p^{4} [**e d**]

and all the tangents belong to the nul complex 3 [**a d**] − [**b c**], it follows that a linear relation exists between any six tangents, which can therefore be written, [P Q L M N] R + [Q R L M N] P + [R P L M N] Q = [L M P Q R]N + [M N P Q R]L + [N L P Q R]M (xi)

where [P Q L M N] is the determinant |1, q, l^{2}, m^{3}, n^{4}|.

If the left side of this equation is special, then so also is the right side, and either side then gives the common generator of the two quadrics formed from the lines P, Q, R and the lines L, M, N. Writing φ (q) ≡ (q-I) (q-m) (q-n) and Δ = (l-m) (m-n) (n-l) then [Q R L M N] = Δ (q-r) φ (q) φ (r) so that we have to make special the complex.

q-r/φ (p)) + r-p/φ (q) Q + p-q/φ (r) R, (xii)

but a complex of the form α (QR) P + *β* (R P) Q + γ (P Q) R where (P Q) = (p-q)^{4} is special provided 1/α + 1/*β* + 1/γ = 0 or Σ (q-r)^{3} φ (p) =); Writing a_{3} 0 (p) = a_{p}^{3} so that a_{x}^{3} = 0 is the plane of **l, m, n** the condition becomes Σ (q-r)^{3} a_{p}^{3} = 0, and reduces immediately to.

(q-r) (r-p) (p-q) a_{p} a_{r} a_{i} = 0

Finally, writing b_{1}^{3} = b_{3} (*t*-p) (*t*-q) (*t*-r) so that b_{1}^{3} = 0 is the plane of the points **p q, r** then
a_{p} a_{q} a_{r} = a_{3} p. q r + a_{2} Σ q r + a_{1} Σ p. + a_{0} ≡ a_{3} b_{0} − 3 a_{2} b_{1} + 3 a_{1} b_{2} − a_{0} b_{3}
= (a b)^{3}

Hence the two quadrics will have a common generator provided the two triads a_{t}^{3} = 0, b_{t}^{3} = 0 are apolar.

Further, if the two triads a_{t}^{3} = 0, b_{t}^{3} = 0 are apolar, then any pair of members of the pencil a_{t}^{3} + k b_{t}^{3} = 0 will be apolar and the pair of corresponding quadrics (belonging to a single-parameter quadratic family) will have a common generator, which can be shown, by a determination of this generator, to be common to the whole family.

Any generator of one system of the quadric which has the tangents [p p], [q q], [r r] as generators, can, in terms of a parameter μ be written as.

1/(q-r) (μ-p) (**q q r r**) [**p p**] + 1/(r-p) (μ-q) (**r r p. p**) [q q] + 1/(p-q) (μ-r) (**p p. q q**) [**r r**]

and by identifying this with the right side of equation (xi) and after some calculation, it is found that the parameter μ is given by the transvectant, b_{μ} (b h)^{2} = 0 where h_{t}^{2} is the hessian quadratic (a a′)^{2} a_{t} a′_{t} formed from the cubic a_{t}^{3}.

This provides a geometrical determination of the parameter μ; for if *u*, ν are the roots of the hessian quadratic h_{t}^{2} = 0 then b_{μ} (b h)^{2} = b_{μ} b_{u} b_{v} = 0 so that the plane of the points on the twisted cubic, is apolar to the plane b_{x}^{3} = 0. Hence the plane through the hessian points H_{1}, H_{2} on the cubic and apolar to b_{x}^{3} = 0, that is, also passing through the hessian pole in the plane b_{x}^{3} = 0, will meet the twisted cubic again in the point whose parameter is μ. The corresponding point on the conic in the plane b_{x}^{3} = 0 gives a point on the common generator.

Also, since a_{μ} (a h)^{2} ≡ 0, it follows that if a_{t}^{3} and b_{t}^{3} are apolar, a_{t}^{3} + k b_{t}^{3} is apolar to each of a_{t}^{3} and b_{t}^{3} and that the quadric formed from the triad a_{t}^{3} + k b_{t}^{3} = 0 has a generator in common with the quadric formed from a_{t}^{3} = 0 given by b_{μ} (b h)^{2} = 0 and this generator is clearly independent of the value of k, and hence all the quadrics of the family have a generator in common.

In particular, taking the planes a_{x}^{3} = 0, b_{x}^{3} = 0 as the planes [**a k d**], [**p q r**] respectively, the value of μ for the common generator reduces to − b_{0}/ (k^{2} b_{3}) and, with points **a, k, d** as points of reference, this generator meets the plane [**a k d**] in the point (x′, y′, z′) where.

(x′)^{-1}: (y′)^{-1}: (z′)^{-1} = b_{0}: − (b_{0} + k^{3} b_{3}): k^{3} b_{3}

and in these co-ordinates the equation of the line of intersection of the two planes becomes x/x′ + y/y′ + z/z′ = 0, a second polar of (x′, y′, z′) with respect to the cubic × y z = 0.

As a further development, the equation of the residual cubic of intersection of the two quadrics may be deduced. Returning to the original base points **a, b, c, d** the quadric derived from the tangents at **a, k, d** has the equation 2 k × z − k × *t* − 3 k y z + 2 y *t* = 0 and any point on this quadric is given in terms of parameters λ, μ by
x/2λ-k = y/λk = z/μk^{2} = *t*/μk^{2} (2k−λ)
where λ = μ = 0 gives a, λ = μ = k gives **k**, λ = μ = ∞ gives **d**, and μ = constant corresponds to generators of the system [**a b**], [**k k**], [**c d**]. Inserting these values in the quadric derived from b_{t}^{3} = 0, namely.

(b_{0} × + 3 b_{1} y + 3 b_{2} z + b_{3} *t*)^{2} − 9 [h_{2} (y *t*-z^{2}) + h_{1} (x *t*-yz) + h_{0} (xz-y^{2})] = 0

and using the apolar condition (a b)^{3} = 0 it is found (after some reduction) that there is a factor b_{0} + μ k^{2} b_{3} whose vanishing gives of course the common generator, and leaving a (2, 1) relation between λ, μ which therefore gives the residual twisted cubic, namely.

b_{0} (2μ-k)^{2} + 3 b_{2} λ k^{2} (2k−λ) + μ k^{2} [b_{3} (2k−λ)^{2} − 3 b_{2} (2λ-k)] = 0

By setting λ = μ the residual cubic meets the plane [**a k d**] in three points whose parameters on the conic are given by (c a)^{3} bλ^{3} − 4 (b c)^{3} a_{λ}^{3} = 0 where c_{t}^{3} is the cubic covariant of a_{t}^{3} and since b_{t}^{3} are apolar, this triad of points is therefore apolar to both a_{t}^{3} = 0, b_{t}^{3} = 0 and the plane (c a)^{3} b_{x}^{3} − 4 (b c)^{3} a_{x}^{3} contains the line of intersection of the planes.

It may also be noticed that if the common generator meets the residual cubic in two points given by λ = λ_{1}, λ = λ_{2} the generators with parameters μ_{1}, μ_{2} of the system opposite to [**a b**], [**k k**], [**c d**] will meet the conic in the plane [**a k d**] in two points which are easily shown to be collinear with the hessian pole in the plane.

The condition that the two hessian completes H, K formed from the tangents to the twisted cubic at the points of the two triads given by a_{t}^{3} = 0, b_{t}^{3} = 0 b_{t}^{3} = 0

may be apolar involves the invariant of the hessian quadratics h_{t}^{2} = (a a′)^{2} a_{t} a_{t}′, k_{t}^{2} = (b b′)^{2} b_{t} b_{t}′ and may be written 4 (h h′)^{2} (k k′)^{2} = 3 { (h k)^{2} }^{2}, h h′, k k′ being equivalent symbols.

#### References

Babbage. Journal London Math. Soc. (25). 1950.

Forder. Calculus of Extension. p. 132.

Grace and Young. Algebra of Invariants, p. 196, pp. 240–2.

Meyer. Apolarital und rationale Curven, p. 463.

Turnbull. Theory of Determmants, Matrices and Invariants, chapter XII