In the geometry of projective spaces, including the projective plane, duality concerns the interchangeability between points and lines which preserves incidence properties. It is a special case of a duality for projective space in general, that interchanges

dimension + 1

and

codimension.

That is, in an ambient projective space of dimension n, the points (dimension 0) are made to correspond with hyperplanes (codimension 1), the lines joining two points (dimension 1) are made to correspond with the intersection of two hyperplanes (codimension 2), and so on.

Duality in terms of vector space

---

The points of n-dimensional projective space over a field F, called CFⁿ, can be taken to be the nonzero vectors in the n+1 dimensional vector space over F, where we identity two vectors which differ by a scalar factor. Another way to put it is that the points of n-dimensional projective space are the lines through the origin in Fⁿ⁺¹. Also the n-dimensional subspaces of Fⁿ⁺¹ represent the (n-1)-dimensional hyperplanes of projective n-space over F.

A nonzero vector u in Fⁿ⁺¹ also determines an n-dimensional subspace, by means of the equation

$u\_0x\_0\; +\; \backslash cdots\; +\; u\_nx\_n\; =\; 0$

where u_i is the ith coordinate of u, starting from zero. In terms of the dot product we can write this as $u\; \backslash cdot\; x=0$, where u is the vector representing a hyperplane, and x the vector representing a point. The dot product is symmetrical, and the same vector u represents both a point and a hyperplane. Hence we have a duality between points and hyperplanes, which extends to a duality between the line generated by two points and the intersection of two hyperplanes, and so forth.

Points and lines in the plane

---

Notice that both points and lines can be represented (on a plane) by means of ordered pairs. A point is represented by the ordered pair (x,y), where x is the abscissa and y is the ordinate, which together are coordinates of the point. A line can likewise be represented by an ordered pair (m,b) where m is the slope and b is the y-intercept.

Given three points

$P\_1\; :\; (x\_1,\; y\_1),$

$P\_2\; :\; (x\_2,\; y\_2),$

$P\_3\; :\; (x\_3,\; y\_3);$

these three points are collinear if and only if their coordinates satisfy the equation

$\{y\_2\; -\; y\_1\; \backslash over\; x\_2\; -\; x\_1\}\; =\; \{y\_3\; -\; y\_2\; \backslash over\; x\_3\; -\; x\_2\}\; =\; \{y\_3\; -\; y\_1\; \backslash over\; x\_3\; -\; x\_1\}\; \backslash qquad\; \backslash qquad\; (1)$.

Likewise, given three lines

$L\_1\; :\; (m\_1,\; b\_1),$

$L\_2\; :\; (m\_2,\; b\_2),$

$L\_3\; :\; (m\_3,\; b\_3);$

one can verify that these three lines are concurrent (i.e., all share an intersection) if and only if their parameters satisfy the equation

$\{b\_2\; -\; b\_1\; \backslash over\; m\_2\; -\; m\_1\}\; =\; \{b\_3\; -\; b\_2\; \backslash over\; m\_3\; -\; m\_2\}\; =\; \{b\_3\; -\; b\_1\; \backslash over\; m\_3\; -\; m\_1\}.\; \backslash qquad\; \backslash qquad\; (2)$

Equations (1) and (2) are equivalent to each other up to an exchange of x with m and y with b. Therefore there exists is a way to exchange lines with points in such a way that concurrency is exchanged with collinearity.

It is possible to distinguish lines from points by conjugating ordered pairs. That is, let line (m,b) be represented instead by its conjugate $(m,-b)^\backslash star.$ Then it can be verified that the intersection L₁.L₂ of a pair of lines L₁ and L₂ is

$(m\_1,b\_1)^\backslash star.(m\_2,b\_2)^\backslash star\; =\; \backslash left(\; \{b\_2\; -\; b\_1\; \backslash over\; m\_2\; -\; m\_1\},\; \{m\_1\; b\_2\; -\; m\_2\; b\_1\; \backslash over\; m\_2\; -\; m\_1\}\; \backslash right)\; \backslash qquad\; \backslash qquad\; (3)$

where b₁ and b₂ are negative y-intercepts. Also, the common line P₁.P₂ passing through a pair of points P₁ and P₂ is

$(x\_1,y\_1).(x\_2,y\_2)\; =\; \backslash left(\; \{y\_2\; -\; y\_1\; \backslash over\; x\_2\; -\; x\_1\},\; \{x\_1\; y\_2\; -\; x\_2\; y\_1\; \backslash over\; x\_2\; -\; x\_1\}\; \backslash right)^\backslash star.\; \backslash qquad\; \backslash qquad\; (4)$

Equation (4) can be seen to be the same as equation (3), after exchanging m with x and b with y, and applying the following rules of conjugation:

$A^\{\backslash star\; \backslash star\}\; =\; A,\; \backslash qquad\; \backslash qquad\; (5)$

$A\; =\; B\; \backslash rightarrow\; A^\backslash star\; =\; B^\backslash star,\; \backslash qquad\; \backslash qquad\; (6)$

$(A.B)^\backslash star\; =\; A^\backslash star.B^\backslash star\; =\; B^\backslash star.A^\backslash star.\; \backslash qquad\; \backslash qquad\; (7)$

Indeed, if equation (3) is represented as

$A^\backslash star.B^\backslash star\; =\; C$

then applying rule (6) yields

$(A^\backslash star.B^\backslash star)^\backslash star\; =\; C^\backslash star.$

Applying rule (7) then yields

$A^\{\backslash star\; \backslash star\}.B^\{\backslash star\; \backslash star\}\; =\; C^\backslash star$

and applying rule (5) finally yields

$A.B\; =\; C^\backslash star,$

which is equation (4).

Thus it is possible to imagine a pair of planes S₁ and S₂, and a bijective relation between loci of points in the two planes, such that points in S₂ correspond to lines in S₁, and points in S₁ correspond to lines in S₂.

Great circles

---

One way to establish such bijection is to model the real projective plane, not as an extended affine plane, but as a "unit sphere modulo antipodes", i.e. a unit sphere in which antipodal points are equivalent. Then through points P₁ and P₂ in S₁ passes a geodesic line L₃ which is actually a great circle. But to these two original points correspond a pair of great circles L₁ and L₂ in S₂, such that if S₂ and S₁ are superposed, then L₁ is the unique great circle perpendicular to the line through the pair of points P₁ *¹, and L₂ is the unique great circle perpendicular to the line through the pair of points P₂. These great circles L₁ and L₂ intersect at a pair of points P₃ in S₂. The vector through P₃ is the cross product of the vectors through P₁ and P₂. Then the unique great circle perpendicular to the line passing through the pair of points P₃ is geodesic line L₃ in S₁.

Therefore to every great circle in S₁ corresponds a unique pair of points (which are actually the same point) in S₂, such that if S₁ and S₂ are superposed, then the (3-D) line passing through the pair of points is perpendicular to (the plane in 3-D of) the great circle. The above sentence remains true if S₁ and S₂ are exchanged. This establishes the bijective nature of the duality in the projective plane.

It must be noted that in the "unit sphere modulo antipodes", one "geodesic line", i.e. great circle, must be chosen to be the line at infinity if the surface is to be mapped to an extended affine plane. This line may be chosen to be the equator by convention.

Three dimensions

---

There is also a duality in projective 3-space, in which points correspond to planes, and lines correspond to lines. This is analogous to duality of polyhedra in solid geometry, where points are dual to faces, and sides are dual to sides, so that the icosahedron is dual to the dodecahedron, and the cube is dual to the octahedron.

• *¹ When we say a line through pair of points P, or simply a line through P, we refer to the three dimensional euclidian line that passes through the antipodal points represented by P. When we say that a geodesic line, or a great circle, L is perpendicular to the line passing through the pair of points P, we mean that L lies on the plane that is perpendicular to, and intersects at the midpoint of, the straight line segment in euclidian space that connects the antipodal points that is represented by P. In other words, L is the set of points equidistant in euclidian space to the antipodal points represented by P. L is unique for P.

Mapping the sphere onto the plane

---

The unit sphere modulo −1 model of the projective plane is isomorphic (w.r.t. incidence properties) to the planar model: the affine plane extended with a projective line at infinity.

To map a point on the sphere to a point on the plane, let the plane be tangent to the sphere at some point which shall be the origin of the plane's coordinate system (2-D origin). Then construct a line passing through the center of the sphere (3-D origin) and the point on the sphere. This line intersects the plane at a point which is the projection of the point on the sphere onto the plane (or vice versa).

This projection can be used to define a one-to-one onto mapping

$f:\backslash times\; [0,2\; \backslash pi\backslash rightarrow\; \backslash mathbb\{R\}P^2$.

If points in $\backslash mathbb\{R\}P^2$ are expressed in homogeneous coordinates, then

$f:(\backslash theta,\; \backslash phi)\; \backslash mapsto\backslash phi\; :\; \backslash sin\; \backslash phi\; :\; \backslash cot\; \backslash theta,$

$f^\{-1\}\; ::\; y\; :\; z\backslash mapsto\; \backslash left(\; \backslash arctan\; \backslash sqrt\{\backslash left(\{x\; \backslash over\; z\}\backslash right)^2\; +\; \backslash left(\{y\backslash over\; z\}\backslash right)^2\},\; \backslash arctan\_2\; (y,x)\; \backslash right)$

Also, lines in the planar model are projections of great circles of the sphere. This is so because through any line in the plane pass an infinitude of different planes: one of these planes passes through the 3-D origin, but a plane passing through the 3-D origin intersects the sphere along a great circle.

As we have seen, any great circle in the unit sphere has a projective point perpendicular to it, which can be defined as its dual. But this point is a pair of antipodal points on the unit sphere, through both of which passes a unique 3-D line, and this line extended past the unit sphere intersects the tangent plane at a point, which means that there is a geometric way to associate a unique point on the plane to every line on the plane, such that the point is the dual of the line.

Duality mapping defined

---

Given a line L in the projective plane, what is its dual point? Draw a line L′ passing through the 2-D origin and perpendicular to line L. Then pick a point P on line L′ on the other side of the origin from line L, such that the distance of point P to the origin is the reciprocal of the distance of line L to the origin.

Figure 1. Three pairs of dual points and lines: one red pair, one yellow pair,
and one blue pair. The duality is an isomorphism of incidence, so that, e.g.,
the line passing through the red and yellow points is dual to the intersection
of the red and yellow lines.

Expressed algebraically, let g be a one-to-one mapping from the projective plane onto itself:

$g\; :\; \backslash mathbb\{R\}P^2\; \backslash rightarrow\; \backslash mathbb\{R\}P^2$

such that

$g\; ::\; b\; :\; 1\_L\; \backslash mapsto:\; -1\; :\; b$

and

$g\; ::\; y\; :\; 1\backslash mapsto:\; 1\; :\; -y\_L$

where the L subscript is used to semantically distinguish line coordinates from point coordinates. In words, affine line (m, b) with slope m and y-intercept b is the dual of point (m/b, −1/b). If b=0 then the line passes through the 2-D origin and its dual is the ideal point : −1 : 0.

The affine point with Cartesian coordinates (x,y) has as its dual the line whose slope is −x/y and whose y-intercept is −1/y. If the point is the 2-D origin then its dual is x:0:1" target="_blank" >^*," target="_blank" >on the x-axis, then its dual is line [x:1:0_L which shall be interpreted as a line whose slope is vertical and whose x-intercept is −1/x.

If a point or a line's homogeneous coordinates are represented as a vector in 3x1 matrix form, then the duality mapping g can be represented as a trilinear transformation, a 3x3 matrix

whose inverse is

Matrix G has one real eigenvalue: one, whose eigenvector is The line ^* which is the intersection of the ideal line with the x-axis.

Notice that is the y-axis, ^*_L is the x-axis. In 3-space, matrix G is a 90° rotation about the x-axis which turns the y-axis into the z-axis. In projective 2-space, matrix G is a projective transformation which maps points to points, lines to lines, conic sections to conic sections: it exchanges the line at infinity with the x-axis and maps the y-axis onto itself through a Möbius transformation. As a duality, matrix G pairs up each projective line with its dual projective point.

Preservation of incidence

---

The duality mapping g is an isomorphism with respect to the incidence properties (such as collinearity and concurrency). The mapping g has this property: given a pair of lines L₁ and L₂ which intersect at a point P, then their dual points gL₁ and gL₂ define the unique line g^−1P:

$g^\{-1\}\; (L\_1\; \backslash cap\; L\_2)\; =\; gL\_1\; .\; gL\_2$.

Given points P₁ and P₂ through which passes line L, P₁.P₂ = L, then what is the intersection of lines g^−1P₁ and g^−1P₂? If g⁻¹P₁ ∩ g⁻¹P₂ = P then

$g^\{-1\}P\; =\; g^\{-1\}(g^\{-1\}P\_1\; \backslash cap\; g^\{-1\}P\_2)\; =\; g(g^\{-1\}P\_1).g(g^\{-1\}P\_2)$

$=\; P\_1.P\_2$

$=\; L$

so that

$g(g^\{-1\}P)\; =\; gL$

$P\; =\; gL$

∴$g(P\_1.P\_2)\; =\; g^\{-1\}P\_1\; \backslash cap\; g^\{-1\}P\_2$

Given a pair of affine points in homogeneous coordinates, the line passing through them is

where the cross product is computed just as it would for an ordinary pair vectors in 3-space.

From this last equation can be derived the intersection of lines, by using the mapping g to "plug in" the lines into the slots for points:

where mapping g is seen to distribute with respect to the cross product: i.e. g is an isomorphism of cross product.

Theorem. The duality mapping g is an isomorphism of cross product. I.e. g is distributive w.r.t. cross product.

Proof. Given points A=(a:b:c) and B=(d:e:f), their cross product is $(a:b:c)\; \backslash times\; (d:e:f)\; =\; (b\; f\; -\; c\; e\; :\; c\; d\; -\; a\; f\; :\; a\; e\; -\; b\; d)$ but

$g(a:b:c)\; =\; (a:-c:b),$

$g(d:e:f)\; =\; (d:-f:e),$

$(a:-c:b)\; \backslash times\; (d:-f:e)\; =\; (-c\; e\; +\; b\; f\; :\; b\; d\; -\; a\; e\; :\; -a\; f\; +\; c\; d)$

$=\; g(b\; f\; -\; c\; e\; :\; c\; d\; -\; a\; f\; :\; a\; e\; -\; b\; d)$.

Therefore

$g(A\; \backslash times\; B)\; =\; gA\; \backslash times\; gB$.

Q.E.D.

Projective geometry | Duality theories

dualité (géométrie projective)