On the enumerative geometry of Pascal’s hexagram

Fix a nonsingular conic $\mathbb{K}$ in the complex projective plane $\mathbb{P}^{2}$. Suppose that we are given six distinct points $A,B,C,D,E,F$ on $\mathbb{K}$, arranged into an array $\left[\begin{array}[]{ccc}A&B&C\\ F&E&D\end{array}\right]$. Then Pascal’s theorem says that the three intersection points

(corresponding to the three minors of the array) are collinear (see Diagram 1). This is one of the oldest theorems in classical projective geometry, and the picture is so striking that a verbal explanation is almost unnecessary.

The line containing the intersections is called the Pascal line (or just the pascal) of the array; we will tentatively denote it as $\left\{\begin{array}[]{ccc}A&B&C\\ F&E&D\end{array}\right\}$. The pascal evidently remains the same if we permute the rows or the columns of the array; thus the same line could also be written as $\left\{\begin{array}[]{ccc}B&C&A\\ E&D&F\end{array}\right\}$ or $\left\{\begin{array}[]{ccc}F&D&E\\ A&C&B\end{array}\right\}$ etc.

But any essentially different arrangement of the same points, say $\left\{\begin{array}[]{ccc}B&F&D\\ C&A&E\end{array}\right\}$, corresponds a priori to a different pascal. Since there are $6!$ ways to allocate the points into an array, and $2\times 3!=12$ ways to write the same pascal, the number of possible pascals is $\frac{6!}{2\times 3!}=60$. It is a theorem due to Pedoe [15] that these lines are pairwise distinct, if the initial six points are chosen generally.

To recapitulate, six general points on a conic give rise to a collection of sixty pascals. The entirety of this configuration¹¹1These sixty pascals satisfy some incidence theorems, leading to the so-called Steiner and Kirkman points (see Section 5 below). These points in turn satisfy some further incidences, leading to Cayley lines, Plücker lines and Salmon points. Sometimes these auxiliary geometric elements are also included in the term ‘mystic hexagram’ (see [4]). However, they will play no substantial role in this paper. is usually called Pascal’s hexagrammum mysticum, or ‘mystic hexagram’.

We will now give an informal description of the enumerative problem to be considered in this paper. The necessary technical terminology will be introduced later in Section 2.

Once and for all, fix the conic $\mathbb{K}$ inside the complex projective plane. Recall that $\mathbb{K}$ has dimension one (as an algebraic variety), and hence the set of sextuples $(A,B,\dots,F)\in\mathbb{K}^{6}$ has dimension six. As the sextuple varies in $\mathbb{K}^{6}$, so does the hexagram.

Now let $\ell$ be a line in $\mathbb{P}^{2}$, and consider the subset of sextuples $(A,B,\dots,F)$ such that

Such a subset will later be called a Pascal variety. Since the lines in $\mathbb{P}^{2}$ form a two-dimensional family, we expect the Pascal variety to be of codimension two in the set of all sextuples. The same would be true if one substitutes any of the sixty possible arrays in (1).

Now fix three general lines $\ell_{1},\ell_{2},\ell_{3}$ in the plane, and (by way of example) consider the subset of sextuples $(A,B,\dots,F)$ such that

(The three arrays were chosen randomly.) This subset is of codimension $2\times 3=6$; that is to say, there is a finite number of sextuples which satisfy the conditions in (2). We should like to find this number for each such triple of arrays. It will turn out that there are $77$ essentially distinct triples to be considered. The main result of this paper is a determination of this number in all of these cases. We will use computational techniques in commutative algebra to do so.

The problem is shown schematically in Diagram 2. The three coloured pascals are pre-specified, and we are to find all positions of $(A,B,\dots,F)$ which make the equalities in (2) come true. There is no obvious geometric construction which will lead us from the pascals to the sextuples, and the solution will have to be essentially algebraic.

The paper is organised as follows. In the next section, we formulate the notion of a Pascal variety. The required number is the cardinality of the intersection of three such varieties. We then describe the computational procedure to find this number. In order to list the cases, it is convenient to introduce the so-called ‘dual notation’ for Pascals, which makes crucial use of the unique nontrivial outer automorphism of the symmetric group $\mathbb{S}_{6}$. This is explained in Section 3. Using this notation, we tabulate the results on page 4.1 of Section 4. Some further lines of investigation are sketched in Section 5.

All the computations were done using a conjunction of Maple and Macaulay2.

The literature on Pascal’s theorem is very large. The standard classical reference is by George Salmon (see [17, Notes]). The dual notation, together with a host of results discovered by Cremona and Richmond are explained by H. F. Baker in his note ‘On the Hexagrammum Mysticum of Pascal’ in [2, Note II, pp. 219–236]. One of the best modern surveys of this material is by Conway and Ryba [4, 5], which also contains a bibliography of older literature on this subject. We refer the reader to [6, 14, 18] for foundational notions in projective geometry; in particular each of them contains a proof of Pascal’s theorem. The modest amount of commutative algebra and algebraic geometry that we need may be found in [8, 10, 19].

During the past three centuries, scores of papers on various aspects of Pascal’s hexagram have been published. However, to the best of my knowledge, this specific enumerative problem has never been considered before. One possible explanation is that the computations cannot be done by hand. As we will see below, they involve calculating the primary decomposition of a rather complicated ideal inside a polynomial ring in six variables. In my experience, it usually takes several minutes of computation in Macaulay2 to work through a single case. For this reason, several natural questions in this territory still appear to be inaccessible (see Section 5).

Let $[z_{0},z_{1},z_{2}]$ be the homogeneous coordinates in $\mathbb{P}^{2}$. For instance, we will denote the line $3\,z_{0}+z_{1}+2\,z_{2}=0$ as $\langle 3,1,2\rangle$. Identify $\mathbb{K}$ with the conic $z_{0}\,z_{2}=z_{1}^{2}$, and fix an isomorphism $\tau:\mathbb{P}^{1}\longrightarrow\mathbb{K}$ by the formula $\tau(r)=[1,r,r^{2}]$ for $r\in\mathbb{C}$, and $\tau(\infty)=[0,0,1]$.

Consider the set of letters $\textsc{ltr}=\{\mathbb{A},\mathbb{B},\mathbb{C},\mathbb{D},\mathbb{E},\mathbb{F}\}$. These should be seen formally at the moment, but we will soon relate them to points on the conic. Define a Pascal symbol to be an array such as $s=\left\{\begin{array}[]{ccc}\mathbb{E}&\mathbb{C}&\mathbb{A}\\ \mathbb{B}&\mathbb{F}&\mathbb{D}\end{array}\right\}$, determined up to row and column shuffles. There are sixty such symbols.

Let $\mathbb{K}^{\textsc{ltr}}$ denote the set of maps $\textsc{ltr}\longrightarrow\mathbb{K}$. This is a projective variety isomorphic to $(\mathbb{P}^{1})^{6}$. A hexad is an injective map $\textsc{ltr}\stackrel{{\scriptstyle h}}{{\longrightarrow}}\mathbb{K}$; it corresponds to six distinct points

on the conic. If $\mathcal{H}$ denotes the set of all hexads, then $\mathcal{H}\subseteq\mathbb{K}^{\textsc{ltr}}$ is a quasiprojective algebraic variety. Given a Pascal symbol $s$, we get a morphism

which sends a hexad $h$ to the corresponding pascal obtained by substituting the points $A,\dots,F$ into the symbol $s$. Given a line $\ell\in({\mathbb{P}}^{2})^{*}$, define the Pascal variety

Proof. Without loss of generality, we may assume

First, assume that $\ell$ is not tangent to $\mathbb{K}$. Let $\mathbb{K}\cap\ell=\{P_{1},P_{2}\}$, and write $\mathbb{K}^{\prime}=\mathbb{K}\setminus\{P_{1},P_{2}\}$. Choose arbitrary distinct points $A,B,C,F$ in $\mathbb{K}^{\prime}$, and let $Q_{1}=BF\cap\ell,Q_{2}=CF\cap\ell$. Now let $D$ (respectively $E$) be the other intersection of $\mathbb{K}$ with the line $AQ_{2}$ (respectively $AQ_{1}$). This gives a hexad in $\Pi(s,\ell)$ and all hexads in it are obtained this way.

If $\ell$ is tangent to $\mathbb{K}$, then the same argument goes through with $\mathbb{K}^{\prime}=\mathbb{K}\setminus\{P\}$ where $P$ is the point of tangency. This proves the result. ∎

Now consider a triple

where $s_{1},s_{2},s_{3}$ are pairwise distinct Pascal symbols and $\ell_{1},\ell_{2},\ell_{3}$ are general lines in $\mathbb{P}^{2}$. Our basic enumerative problem is to find the number of points in the variety

Fix the symbol $s$ as in (4). Let $a,b,c,d,e,f$ be indeterminates, and consider the polynomial ring $R=\mathbb{C}[a,b,c,d,e,f]$. Write

Now it is easy to calculate the coordinates of the lines $AE,BF$ etc, and eventually those of the Pascal $\left\{\begin{array}[]{ccc}A&B&C\\ F&E&D\end{array}\right\}$. They turn out to be $\langle u_{0},u_{1},u_{2}\rangle$, where

(This computation was programmed in Maple.) Given the line $\ell=\langle\alpha_{0},\alpha_{1},\alpha_{2}\rangle$, we have

exactly when the $2\times 2$ minors of the matrix $\left[\begin{array}[]{ccc}u_{0}&u_{1}&u_{2}\\ \alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right]$ are zero. These minors generate the ideal $I(s,\ell)\subseteq R$ which locally defines the Zariski closure of the Pascal variety $\Pi(s,\ell)$. A similar construction works for any symbol $s$.

Let $\Delta=\mathbb{K}^{\textsc{ltr}}\setminus\mathcal{H}$, which is usually called the big diagonal. It corresponds to sextuples $(A,\dots,F)$ where some of the points coincide. By construction, $\Delta$ is disjoint from $\Pi(s,\ell)$.

Given a triple $T$ as above, define the ideal sum

and let $J_{T}=\sqrt{I_{T}}$. It is not quite the case that $J_{T}$ defines $\Pi_{T}$, since some irreducible components of the variety defined by it lie inside $\Delta$. In order to remove them, consider the minimal prime decomposition

and define the restricted intersection $\Lambda_{T}=\bigcap^{\prime}\;\mathfrak{p}_{i}$, where we exclude those prime ideals which contain any expression of the form $u-v$, for $u,v\in\{a,b,c,d,e,f\}$. If the $\ell_{i}$ are chosen generally, then $\Lambda_{T}\subseteq R$ is an ideal of height $6$. Now we have²²2The only points of $\mathbb{K}^{\textsc{ltr}}$ which lie outside the open subset $\text{Spec}\,R$ are those where one of the coordinates is $\infty$. If the $\ell_{i}$ are chosen generally, none of them will be included in $\Pi_{T}$.

We denote this number by $\mathcal{N}_{T}$, since it is independent of the $\ell_{i}$ as long as the lines are chosen generally. With this understood, we will simply write $T=\{s_{1},s_{2},s_{3}\}$ instead of (5).

I have used Maple to program the ideal sum in (8), and then programmed the rest of the computation in Macaulay2. There are several techniques for calculating the radical and the primary decomposition of an ideal inside a polynomial ring. By and large, they depend crucially upon the concept of a Gröbner basis. The reader will find an explanation and comparison of such techniques in [7].

There are practical difficulties arising from the fact that $I_{T}$ is generated by $9$ large polynomials in the variables $a,\dots,f$. This leads to a corresponding difficulty in calculating the prime decomposition of its radical. Hence, this is the procedure I have followed: instead of using ${\mathbb{Q}}$ as the base field (which slows down the computation) I have worked through each case thrice by using large prime fields $\mathbb{F}_{p}$ (such as $p=32003,43051,48619$) and varying the choice of $\ell_{i}$ each time. Since the results are consistent, I have a high degree of confidence in their correctness.

In order to tabulate the results in a compact form, we will use the dual notation for pascals. It is important in its own right, since it is used throughout all later development of the subject. Be that as it may, the mathematics underlying it has a charm of its own.

Recall that if $G$ is a group and $g\in G$, then

is called an inner automorphism of $G$. The subgroup of inner automorphisms (denoted $\text{Inn}(G)$) is a normal subgroup of $\text{Aut}(G)$ (the group of all automorphisms), and the quotient $\text{Out}(G)=\frac{\text{Aut}(G)}{\text{Inn}(G)}$ is the group of outer automorphisms. The automorphisms of the symmetric groups $\mathbb{S}_{n}$ are characterised by the following theorem due to Hölder (see [16, Ch. 7]):

Thus, $\mathbb{S}_{6}$ has an essentially unique nontrivial outer automorphism. We will not reproduce its construction here, since it may be found in several places (e.g., see [12, 13]). For our purposes, it will be more convenient to see it as an isomorphism between two different copies of $\mathbb{S}_{6}$.

In addition to $\textsc{ltr}=\{\mathbb{A},\mathbb{B},\mathbb{C},\mathbb{D},\mathbb{E},\mathbb{F}\}$, consider the set $\textsc{six}=\{1,2,3,4,5,6\}$. For any set $X$, let $\mathbb{S}(X)$ denote the group of bijections $X\rightarrow X$. Then the following table gives an isomorphism

which is one realisation of the outer automorphism. For instance, it is to be understood as saying that $\zeta$ takes the transposition $(\mathbb{A}\,\mathbb{B})$ to the cycle $(1\,4)\,(2\,5)\,(3\,6)$ of type $2+2+2$.

Similarly, $\zeta^{-1}$ is given by the following table:

For instance, it takes $(1\,2)$ to $(\mathbb{A}\,\mathbb{E})\,(\mathbb{B}\,\mathbb{D})\,(\mathbb{C}\,\mathbb{F})$.

Our labelling schema uses the fact that the $15$ elements of type $2+2+2$ in $\mathbb{S}(\textsc{ltr})$ are bijectively mapped by $\zeta$ onto the $15$ transpositions in $\mathbb{S}(\textsc{six})$.

By way of example, consider the Pascal symbol $\left\{\begin{array}[]{ccc}\mathbb{A}&\mathbb{D}&\mathbb{E}\\ \mathbb{C}&\mathbb{B}&\mathbb{F}\end{array}\right\}$. There are six lines going between the rows; separate them into two diagrams as follows:

In either of the diagrams, each letter has a unique edge joining it to another letter lying in a different column. Read both diagrams as cycles of type $2+2+2$, and observe that

Single out the $2$ common to both cycles, leaving $3$ and $5$. Hence we label this pascal as $k(2,\{3,5\})$. It is customary to flatten this out for readability, and write it indifferently as either $k(2,35)$ or $k(2,53)$. A moment’s reflection will show that a row or column shuffle in a symbol does not change its label.

Thus any pascal is labelled as $k(w,xy)$ for some $w\in\textsc{six}$, and $x,y\in\textsc{six}\setminus\{w\}$, where the order of $x,y$ is immaterial. There are $6\times\binom{5}{2}=60$ such labels, bijectively corresponding to the Pascal symbols. It is easy to reconstruct the array from the label; for instance, starting from $k(3,15)=k(3,51)$,

which must correspond to $\left\{\begin{array}[]{ccc}\mathbb{A}&\mathbb{B}&\mathbb{E}\\ \mathbb{F}&\mathbb{C}&\mathbb{D}\end{array}\right\}$. Of course, the array is determined only up to row and column shuffles. As an exercise, the reader may wish to check that $\left\{\begin{array}[]{ccc}\mathbb{A}&\mathbb{B}&\mathbb{C}\\ \mathbb{F}&\mathbb{E}&\mathbb{D}\end{array}\right\}$ corresponds to $k(1,23)$.

The group $\mathbb{S}(\textsc{ltr})$ acts on $\mathbb{K}^{\textsc{ltr}}$ and hence on $\mathcal{H}$ in an obvious way. Let $\Omega$ denote the set of unordered triples of Pascal symbols. This set has $\binom{60}{3}=34,220$ elements. For instance,

is an element of $\Omega$. It is clear that $\mathbb{S}(\textsc{ltr})$ acts on $\Omega$ as well. If $\sigma\in\mathbb{S}(\textsc{ltr})$ takes the triple $T_{1}$ to $T_{2}$, then we have a bijection

for a fixed choice of $\ell_{1},\ell_{2},\ell_{3}$. In other words, every hexad in $\Pi_{T_{2}}$ is obtained by relabelling a unique hexad in $\Pi_{T_{1}}$ as dictated by the permutation $\sigma$. It follows that $\mathcal{N}_{T_{1}}=\mathcal{N}_{T_{2}}$. Hence it is only necessary to decompose $\Omega$ into its $\mathbb{S}(\textsc{ltr})$-orbits, and find $\mathcal{N}_{T}$ for any one triple $T$ in each orbit.

Using the labelling scheme, we might as well regard $\Omega$ as the set of unordered triples of Pascal labels with a natural action of $\mathbb{S}(\textsc{six})$. For instance, the triple written above now appears in a more compact form as

It is routine to program this group action in Maple. It turns out that there are altogether $77$ orbits. In each case, I have carried out the computational procedure from Section 2.4 on lexicographically³³3We describe this order in brief, although its details are of minor importance. Within each triple, elements of the form $k(1,xy)$ are listed first, followed by those of the form $k(2,xy)$ and so on. Ties are broken by writing $x,y$ in increasing order. The order between two triples is decided by comparing the first element in each, following by the second in each, and so on. the smallest element in each orbit.

The results are summarised in the following theorem, which refers to the table on page 4.1.

The triples are in lexicographic order when read from left to right and top to bottom. The first label is always $k(1,23)$. The colour-coding of the entries has the following meaning:

• $\triangleright$

  The three red entries have intersection number zero. The geometry of these cases is discussed in Section 5 below.

• $\triangleright$

  If $T=\{s_{1},s_{2},s_{3}\}$ is a triple, then define its stabiliser $G_{T}$ to be the set of elements $\sigma\in\mathbb{S}(\textsc{six})$ such that $\sigma(s_{i})=s_{i}$ for $i=1,2,3$. It is easy to determine this group for each triple; then it turns out that the only possible nontrivial stabilisers are $\mathbb{Z}_{2}$ and $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. The blue entries are the ones with $G_{T}\simeq{\mathbb{Z}}_{2}$. For instance, if

  $$T=\{(1,23),(1,45),(2,45)\}$$

  (the entry in row 10, column 2), then $G_{T}=\{e,(4\,5)\}$. A complete example for this case is given in Section 4.2 below. Notice that

  $$(4\,5)\stackrel{{\scriptstyle\zeta^{-1}}}{{\longrightarrow}}(\mathbb{A}\,\mathbb{D})(\mathbb{B}\,\mathbb{E})(\mathbb{C}\,\mathbb{F}).$$

  Hence, starting from a hexad in $\Pi_{T}$, we get another one by the simultaneous interchange

  $$a\leftrightarrow d,\quad b\leftrightarrow e,\quad c\leftrightarrow f.$$

  This implies that the $8$-element set $\Pi_{T}$ can be partitioned into four subsets of two elements each, where the hexads in each subset are obtained from each other by this interchange. Of course, a similar remark applies to each blue entry.

• $\triangleright$

  There is a unique triple $Q$ with stabiliser $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. It appears in row 11, column 2 and is coloured brown. It is easy to see that

  $$G_{Q}=\{e,(2\,3),(4\,5),(2\,3)(4\,5)\}.$$

  As above, this implies that the $8$-element set $\Pi_{Q}$ can be partitioned into two subsets of four elements each, where the hexads in each subset are permuted by the elements of the stabiliser.

The entry in row 25, column 3 is especially simple and elegant. It corresponds to the smallest nonzero intersection number. This case is treated at length in [1]. Its discovery was something of an inspiration for this paper.

Given a triple $T=\{(s_{1},\ell_{1}),(s_{2},\ell_{2}),(s_{3},\ell_{3})\}$, one should like to write down the set $\Pi_{T}$ explicitly at least as an illustration. This is awkward to do over the complex numbers, since the generators of the ideal $\Lambda_{T}$ are usually unwieldy. As a compromise, I have written down such an example where the base field is $\overline{\mathbb{F}}_{101}$, namely the algebraic closure of the finite field with $101$ elements. The prime is large enough to give an interesting example, and small enough to keep the coefficients tidy.

Let

and

As explained on page $\triangleright$ ‣ 4.1, the set $\Pi_{T}$ breaks up into four $G_{T}$-orbits of two elements each. Within each orbit, we get one solution from another by simultaneously interchanging $a$ with $d$, $b$ with $e$, and $c$ with $f$. I have written down the elements in $\Pi_{T}$ by following the procedure in Section 2.4.

The first orbit is given by the solution

together with its interchange.

The second orbit is given by

where $a$ is either of the roots of the equation $x^{2}-4\,x+63=0$. Observe that the two roots add up to $4$. If we substitute $4-a$ for $a$ throughout, then $b,e$ get exchanged and so do $c,f$.

The third orbit is given by

where $a$ is either of the roots of the equation $x^{2}-51\,x+4=0$. The exchange property is exactly as above when $a$ is replaced by $51-a$.

The fourth orbit is given by

where $a$ is either of the roots of the equation $x^{2}-56\,x+4=0$. The exchange property is exactly as above when $a$ is replaced by $56-a$.

This pattern of solutions is not an intrinsic feature of the triple, but depends on the choice of the prime and the coefficients of the $\ell_{i}$. For instance, the polynomial $x^{2}-4\,x+63$ in the second orbit does not split over $\mathbb{F}_{101}$ because its discriminant $-236$ is a non-residue modulo $101$. If it did, the second orbit would have looked like the first. One expects to see different patterns in different examples; it is only the cardinality of $\Pi_{T}$ which is intrinsic to $T$.

It is a theorem due to Steiner that the pascals

are concurrent (see [4]). Of course, this is understood to apply to any three pascals following the same pattern; that is to say,

are concurrent for any $x,y,z\in\textsc{six}$. This explains why the entry in row 12, column 1 is zero, since three general lines are not concurrent.

However, this suggests the following modification of the original set-up: choose a general point $P$ in the plane and three lines $\ell_{1},\ell_{2},\ell_{3}$ passing through $P$. Now consider the variety

A simple count shows that $X$ is of codimension five in $\mathcal{H}$; that is to say, it is a curve. (This is so because $P$ has two degrees of freedom and each $\ell_{i}$ has one, leading to a total of $5$.) Since $X\subseteq\mathbb{K}^{\textsc{ltr}}$, there is a natural projection map $X\stackrel{{\scriptstyle\pi_{L}}}{{\longrightarrow}}\mathbb{K}$ for each of the letters $L\in\textsc{ltr}$.

I have calculated the equations of $X$ in a few cases. They show that $X\stackrel{{\scriptstyle\pi_{L}}}{{\longrightarrow}}\mathbb{K}$ is a $7$ to $1$ map for any $L$. In other words, if any one letter (say $a$) is given a fixed value, then there are exactly $7$ such points in $X$.

It is a theorem due to Kirkman that the pascals

are concurrent (see [4]). As before, this is understood to apply to any three pascals following the same pattern; that is to say,

are concurrent for any $x,y,z,w\in\textsc{six}$. This explains why the entry in row 1, column 2 is zero.

We can construct a curve

as before. Calculations show that $Y\stackrel{{\scriptstyle\pi_{L}}}{{\longrightarrow}}\mathbb{K}$ is a $4$ to $1$ map for any letter $L$.

One should like to know more about the geometry of the curves $X$ and $Y$, and specifically whether they are rational. However, their equations are far too complicated to make this feasible at the moment.

The triple in the $14$th row and $3$rd column corresponds to the arrays

These pascals are concurrent, since they all pass through $AE\cap BF$. Hence the corresponding entry is zero. It is possible to construct a curve as in the previous two sections; we leave these details to the reader.

In the end, we will outline some further lines of investigation.

The common point which lies on

is called the Steiner point corresponding the set $\{x,y,z\}\subseteq\textsc{six}$. It is customary to write it simply as $G[xyz]$, where the order of $x,y,z$ is immaterial. There are $\binom{6}{3}=20$ such points.

Given $\{x,y,z\}$ and a general point $P$, it is natural to consider the Steiner variety consisting of sextuples $(A,B,\dots,F)$ such that $G[xyz]=P$. This is expected to be of codimension two, and as before we have the enumerative question of finding the intersection degree of three such varieties.

Similarly, the point common to the pascals

is called the Kirkman point $K[x,yzw]$. There are $60$ such points, and exactly the same question can be formulated for them.

There is another (somewhat speculative) approach to the enumerative problem considered in this paper. The relevant geometric terminology can be found in [9, 11].

The pascal $\left\{\begin{array}[]{ccc}A&B&C\\ F&E&D\end{array}\right\}$ still remains well-defined if the sextuple $(A,B,\dots,F)$ has a double point, i.e., if two of the points come together on the conic. However, it may no longer be well-defined if we have a triple point or two double points. (A precise statement of possibilities is given in [3, §2].) It follows that

from Section 2.2 is only a rational map, and its indeterminacy locus consists of a union of polydiagonals. It may be possible to use successive blow-ups (cf. [11, Ch. II.7]) in order to construct a smooth projective variety $\Gamma$ such that,

is a morphism, and $\Gamma$ contains $\mathcal{H}$ as a dense open subset on which $p_{s}$ agrees with the earlier geometric definition.⁵⁵5In the paper [3], Sergio Da Silva and I have partly succeeded in resolving the indeterminacies of the rational map above. However, for reasons outlined there, this falls short of a full solution. Once this is accomplished, $p_{s}^{-1}(\{\ell\})$ would be a codimension two algebraic cycle on $\Gamma$. Then, assuming that the Chow ring of $\Gamma$ can be effectively computed, one can interpret $\mathcal{N}_{T}$ as the degree of the intersection of three such cycles. This approach, if carried out in full, would prove to be interesting and informative.

As mentioned earlier, this is very likely the first paper on this enumerative problem. Much remains to be done.