Self-dual Harmonicity: the planar case

TL;DR

This paper introduces a geometrically self-dual formulation of the Fano harmonicity axiom specifically for the projective plane, highlighting symmetry in geometric axioms.

Contribution

It provides a new self-dual version of the Fano harmonicity axiom, emphasizing geometric symmetry in the projective plane.

Findings

A self-dual harmonicity axiom for the projective plane

Enhanced understanding of duality in projective geometry

Potential applications in geometric axiom systems

Abstract

We present a manifestly geometrically self-dual version of the Fano harmonicity axiom for the projective plane.

Equations

$$D_1=P_0{P}_1\cdot P_2{P}_3,D_2=P_0{P}_2\cdot P_3{P}_1,D_3=P_0{P}_3\cdot P_1{P}_2.$$

$$d_1=(p_0\cdot p_1)(p_2\cdot p_3),d_2=(p_0\cdot p_2)(p_3\cdot p_1),d_3=(p_0\cdot p_3)(p_1\cdot p_2).$$

$$P_0=P_0{P}_1\cdot P_0{P}_2=D=P_2{P}_3\cdot P_3{P}_1=P_3.$$

$$p_0\ni D_2{D}_3\cdot P_2{P}_3,D_3{D}_1\cdot P_3{P}_1,D_1{D}_2\cdot P_1{P}_2.$$

$$p_1\ni D_2{D}_3\cdot P_3{P}_2,D_3{D}_1\cdot P_2{P}_0,D_1{D}_2\cdot P_0{P}_3,$$

$$p_2\ni D_2{D}_3\cdot P_0{P}_1,D_3{D}_1\cdot P_1{P}_3,D_1{D}_2\cdot P_3{P}_0,$$

$$p_3\ni D_2{D}_3\cdot P_1{P}_0,D_3{D}_1\cdot P_0{P}_2,D_1{D}_2\cdot P_2{P}_1.$$

$$d_1=D_2{D}_3,d_2=D_3{D}_1,d_3=D_1{D}_2$$

$$D_1=d_2\cdot d_3,D_2=d_3\cdot d_1,D_3=d_1\cdot d_2.$$

$$d_1=D_2{D}_3,d_2=D_3{D}_1,d_3=D_1{D}_2$$

$$D_1=d_2\cdot d_3,D_2=d_3\cdot d_1,D_3=d_1\cdot d_2$$

$$p_i\cdot p_j=P_0{P}_k\cdot D_i{D}_j,p_0\cdot p_k=P_i{P}_j\cdot D_i{D}_j$$

$$P_i{P}_j=(p_0\cdot p_k)(d_i\cdot d_j),P_0{P}_k=(p_i\cdot p_j)(d_i\cdot d_j)$$

Full text

Introduction

In a recent sequence of papers, we have explored manifestly self-dual formulations of three-dimensional projective space: [2] begins the sequence, basing projective space on lines and their abstract incidence, from which points and planes are derived as secondary elements; [4] pursues the complementary approach, taking points and planes to be primary elements from which lines are derived. The most recent paper [6] addresses the axioms (or assumptions) of harmonicity and projectvity: traditionally, harmonicity asserts (with Fano) that ‘the diagonal points of a complete quadrangle are noncollinear’ while projectivity asserts that ‘if a projectivity leaves each of three distinct points of a line invariant, it leaves every point of the line invariant’; see [7] Section 18 and [7] Section 35 respectively.

In keeping with our aim to present projective space in a formulation that is manifestly self-dual, it is desirable to express projectivity and harmonicity in manifestly self-dual forms. The axiom of projectivity is so expressed in [3] and [6]: indeed, [7] Section 103 contains a self-dual expression of projectivity in its discussion of reguli. In [5] and [6] our versions of the axiom of harmonicity are self-dual only in being the conjunction of a pair of dual statements, each of which implies the other; we might say that these versions are self-dual in rather a logical sense. Our purpose in the present paper and its sequel is to offer expressions of harmonicity that are more geometrically self-dual: the present paper deals with harmonicity in the plane; its sequel deals with the more elaborate spatial harmonicity.

The Planar Axiom

In this section, we present a manifestly self-dual version of the harmonicity axiom in the projective plane. Recall that harmonicity is traditionally formulated as follows:

‘the diagonal points of a complete quadrangle are noncollinear’

and that the planar dual of this reads as follows:

‘the diagonal lines of a complete quadrilateral are noncurrent’.

These two dual statements are equivalent within the Veblen-Young axiomatization [7]: granted the Veblen-Young assumptions of alignment and extension, each statement implies the other. Of course, we may take the conjunction of this dual pair of equivalent statements as a ‘self-dual’ formulation of the planar harmonicity axiom. It might be objected that the self-dual nature of this version is as much logical as it is geometrical. Our purpose here is to offer a version in which the self-duality is more purely geometrical. In order to present this alternative version, we recall some elementary facts pertaining to quadrangles and quadrilaterals.

Explicitly, let $P_{0}P_{1}P_{2}P_{3}$ be a complete quadrangle in the plane: thus, its four vertices $P_{0},P_{1},P_{2},P_{3}$ are in general position, in the sense that no three of them are collinear; by definition, the three diagonal points of this quadrangle are

[TABLE]

Dually, let $p_{0}p_{1}p_{2}p_{3}$ be a complete quadrilateral in the plane: its four sides $p_{0},p_{1},p_{2},p_{3}$ are in general position, in the sense that no three of them are concurrent; by definition, the three diagonal lines of this quadrilateral are

[TABLE]

We note that the diagonal points of a complete quadrangle are distinct: indeed, were we to suppose that $D_{1}=D_{2}=D$ say, then $D$ would lie on $P_{0}P_{1},P_{0}P_{2},P_{2}P_{3},P_{3}P_{1}$ and hence support the absurd conclusion

[TABLE]

Dually, the diagonal lines of a complete quadrilateral are distinct.

Now, let let us start with a complete quadrangle $P_{0}P_{1}P_{2}P_{3}$. We shall assume that its diagonal points $D_{1},D_{2},D_{3}$ are noncollinear; they are therefore the vertices of its diagonal triangle $D_{1}D_{2}D_{3}$. By the very definition of the diagonal points, the triangles $D_{1}D_{2}D_{3}$ and $P_{1}P_{2}P_{3}$ are perspective from $P_{0}$: by the Desargues theorem, they are therefore perspective from a line $p_{0}$; thus,

[TABLE]

The diagonal triangle $D_{1}D_{2}D_{3}$ is similarly perspective to $P_{0}P_{3}P_{2}$ (from $P_{1}$), $P_{3}P_{0}P_{1}$ (from $P_{2}$), $P_{2}P_{1}P_{0}$ (from $P_{3}$); accordingly, the Desargues theorem likewise furnishes respective lines

[TABLE]

[TABLE]

[TABLE]

The four lines $p_{0},p_{1},p_{2},p_{3}$ so constructed constitute a complete quadrilateral. For example, suppose $p_{1},p_{2},p_{3}$ were to concur at $Q$: then $Q=p_{2}\cdot p_{3}=D_{2}D_{3}\cdot P_{0}P_{1}$ would lie on $D_{2}D_{3}$ and $Q=p_{3}\cdot p_{1}=D_{3}D_{1}\cdot P_{0}P_{2}$ would lie on $D_{3}D_{1}$ so that $Q=D_{3}$ by noncollinearity of the diagonal points; but then $D_{3}=Q=p_{1}\cdot p_{2}=D_{1}D_{2}\cdot P_{0}P_{3}$ would lie on $D_{1}D_{2}$ and noncollinearity of the diagonal points would be contradicted. From above, $p_{0}\cdot p_{3}=D_{1}D_{2}\cdot P_{1}P_{2}$ and $p_{1}\cdot p_{2}=D_{1}D_{2}\cdot P_{0}P_{3}$: thus $D_{1}D_{2}$ contains both $p_{0}\cdot p_{3}$ and $p_{1}\cdot p_{2}$ and so coincides with the diagonal line $d_{3}=(p_{0}\cdot p_{3})(p_{1}\cdot p_{2})$ of the quadrilateral; likewise, $D_{2}D_{3}=d_{1}$ and $D_{3}D_{1}=d_{2}$.

Incidentally, we feel free here to use the Desargues theorem in the plane, because our ultimate concern is with projective space (in which all planes are automatically Desarguesian); of course, we might instead have adopted a framework for the projective plane that axiomatically incorporates the Desargues theorem, as in [1].

Thus, each complete quadrangle $P_{0}P_{1}P_{2}P_{3}$ with noncollinear diagonal points engenders a complete quadrilateral $p_{0}p_{1}p_{2}p_{3}$ sharing the very same diagonal triangle. Dually, each complete quadrilateral with nonconcurrent diagonal lines engenders a complete quadrangle sharing the very same diagonal triangle. In fact, these dual constructions are mutually inverse; application of the second construction to the quadrilateral that results from the first construction reproduces the original quadrangle $P_{0}P_{1}P_{2}P_{3}$. To see this, by symmetry we need only show that if $\{i,j,k\}=\{1,2,3\}$ then the line joining $d_{i}\cdot d_{j}$ (a vertex of the triangle $d_{1}d_{2}d_{3}$) and $p_{j}\cdot p_{j}$ (the corresponding vertex of $p_{1}p_{2}p_{3}$) passes through $P_{0}$; but this is clear, for $p_{j}\cdot p_{j}=D_{i}D_{j}\cdot P_{k}P_{0}$ lies on $P_{k}P_{0}$ and $d_{i}\cdot d_{j}=D_{j}D_{k}\cdot D_{k}D_{i}=D_{k}\notin D_{i}D_{j}$ lies on $P_{k}P_{0}$ so that $(d_{i}\cdot d_{j})(p_{j}\cdot p_{j})=P_{k}P_{0}$.

The figure comprising a complete quadrangle and a complete quadrilateral that share the same diagonal triangle is called a quadrangle-quadrilateral configuration. We may present this configuration in a way that does not make quite explicit the fact that the respective diagonals form a triangle. Thus, the configuration comprises a complete quadrangle $P_{0}P_{1}P_{2}P_{3}$ (with distinct diagonal points $D_{1},D_{2},D_{3}$) and a complete quadrilateral $p_{0}p_{1}p_{2}p_{3}$ (with distinct diagonal lines $d_{1},d_{2},d_{3}$) so related that

[TABLE]

and

[TABLE]

Notice that these two sets of relations are equivalent: if $\{i,j,k\}=\{1,2,3\}$ and the former set holds, then the distinct lines $d_{i}$ and $d_{j}$ have $D_{k}$ in common, so $D_{k}=d_{i}\cdot d_{j}$ and the latter set holds. As we note in the following paragraph, these relations imply that $D_{1},D_{2},D_{3}$ are noncollinear and that $d_{1},d_{2},d_{3}$ are nonconcurrent, so that $P_{0}P_{1}P_{2}P_{3}$ and $p_{0}p_{1}p_{2}p_{3}$ indeed share the same diagonal triangle. For more on the quadrangle-quadrilateral configuration, see the account in [7] Section 18; for a slightly different perspective on the construction that leads from a complete quadrangle to the complete quadrilateral sharing its diagonal triangle, see also [1] Section 3.1 Exercise 2.

To summarize the foregoing discussion, if the diagonal points of a complete quadrangle $P_{0}P_{1}P_{2}P_{3}$ are not collinear, then there exists a corresponding complete quadrilateral $p_{0}p_{1}p_{2}p_{3}$ making up a quadrangle-quadrilateral configuration; dually, if the diagonal lines of a complete quadrilateral $p_{0}p_{1}p_{2}p_{3}$ are not concurrent, then there exists a corresponding complete quadrangle $P_{0}P_{1}P_{2}P_{3}$ making up a quadrangle-quadrilateral configuration. In the opposite direction, if a complete quadrangle $P_{0}P_{1}P_{2}P_{3}$ (with distinct diagonal points $D_{1},D_{2},D_{3}$) and a complete quadrilateral $p_{0}p_{1}p_{2}p_{3}$ (with distinct diagonal lines $d_{1},d_{2},d_{3}$) are so related as to satisfy the relations

[TABLE]

and

[TABLE]

then the diagonal points $D_{1},D_{2},D_{3}$ of $P_{0}P_{1}P_{2}P_{3}$ are not collinear and the diagonal lines $d_{1},d_{2},d_{3}$ of $p_{0}p_{1}p_{2}p_{3}$ are not concurrent: for instance, $D_{1}\neq D_{2}$ implies that $d_{2}\cdot d_{3}\neq d_{3}\cdot d_{1}$ so that $d_{1},d_{2},d_{3}$ do not concur.

Recalling that if one complete quadrangle has noncollinear diagonal points then so do all, we have now established the following alternative self-dual version of the planar harmonicity axiom:

[H] There exists a quadrangle-quadrilateral configuration.

Notice that the self-dual nature of this version is manifestly geometrical. Of course, this version is equivalent both to the assertion that each quadrangle/quadrilateral is part of a quadrangle-quadrilateral configuration and to the original version according to which the diagonals of each quadrangle/quadrilateral form a triangle.

It is of some interest to observe that we may interrupt the present discussion at an intermediate point. Again let $P_{0}P_{1}P_{2}P_{3}$ be a complete quadrangle with noncollinear diagonal points $D_{1},D_{2},D_{3}$: our construction furnishes a complete quadrilateral $p_{0}p_{1}p_{2}p_{3}$ such that if $\{i,j,k\}=\{1,2,3\}$ then $p_{i}\cdot p_{j}=P_{0}P_{k}\cdot D_{i}D_{j}$ and $p_{0}\cdot p_{k}=P_{i}P_{j}\cdot D_{i}D_{j}$; of course, the sides $p_{0},p_{1},p_{2},p_{3}$ themselves can be recovered from these points. Our dual construction leads from a complete quadrilateral $p_{0}p_{1}p_{2}p_{3}$ with nonconcurrent diagonal lines $d_{1},d_{2},d_{3}$ to a complete quadrangle $P_{0}P_{1}P_{2}P_{3}$ whose vertices can be recovered from the lines $P_{i}P_{j}=(p_{0}\cdot p_{k})(d_{i}\cdot d_{j})$ and $P_{0}P_{k}=(p_{i}\cdot p_{j})(d_{i}\cdot d_{j})$. Temporarily suspending the harmonicity axiom, we may say that the complete quadrangle $P_{0}P_{1}P_{2}P_{3}$ (with diagonal points $D_{1},D_{2},D_{3}$) and the complete quadrilateral $p_{0}p_{1}p_{2}p_{3}$ (with diagonal lines $d_{1},d_{2},d_{3}$) are mated iff they satisfy the relations

[TABLE]

and

[TABLE]

whenever $\{i,j,k\}=\{1,2,3\}$. It may be readily verified that these two sets of conditions are both equivalent to each other and equivalent to the condition that $P_{0}P_{1}P_{2}P_{3}$ and $p_{0}p_{1}p_{2}p_{3}$ make up a quadrangle-quadrilateral configuration.

Accordingly, the following is another manifestly self-dual version of planar harmonicity:

[${\bf H}^{\prime}$] There exists a mated quadrangle-quadrilateral pair.

Equivalently, each quadrangle/quadrilateral has a mate.

REFERENCES

[1] H.S.M. Coxeter, Projective Geometry, Second Edition, Springer-Verlag (1987).

[2] P.L. Robinson, Projective Space: Lines and Duality, arXiv 1506.06051 (2015).

[3] P.L. Robinson, Projective Space: Reguli and Projectivity, arXiv 1506.08217 (2015).

[4] P.L. Robinson, Projective Space: Points and Planes, arXiv 1611.06852 (2016).

[5] P.L. Robinson, Projective Space: Tetrads and Harmonicity, arXiv 1612.01913 (2016).

[6] P.L. Robinson, Projective Space: Harmonicity and Projectivity, arXiv 1612.08422 (2016).

[7] O. Veblen and J.W. Young, Projective Geometry, Volume I, Ginn and Company (1910).