Duality projective geometry

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.

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

$\displaystyle P_1 : (x_1, y_1),$
$\displaystyle P_2 : (x_2, y_2),$
$\displaystyle P_3 : (x_3, y_3);$

these three points are collinear iff their coordinates satisfy the equation

$\displaystyle {y_2 - y_1 \over x_2 - x_1} = {y_3 - y_2 \over x_3 - x_2} = {y_3 - y_1 \over x_3 - x_1} \qquad \qquad (1)$ .

Likewise, given three lines

$\displaystyle L_1 : (m_1, b_1),$
$\displaystyle L_2 : (m_2, b_2),$
$\displaystyle L_3 : (m_3, b_3);$

one can verify that these three lines are concurrent iff their parameters satisfy the equation

$\displaystyle {b_2 - b_1 \over m_2 - m_1} = {b_3 - b_2 \over m_3 - m_2} = {b_3 - b_1 \over m_3 - m_1}. \qquad \qquad (2)$

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 $\displaystyle (m,-b)^\star.$ Then it can be verified that the intersection L1.L2 of a pair of lines L1 and L2 is

$\displaystyle (m_1,b_1)^\star.(m_2,b_2)^\star = \left( {b_2 - b_1 \over m_2 - m_1}, {m_1 b_2 - m_2 b_1 \over m_2 - m_1} \right) \qquad \qquad (3)$

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

$\displaystyle (x_1,y_1).(x_2,y_2) = \left( {y_2 - y_1 \over x_2 - x_1}, {x_1 y_2 - x_2 y_1 \over x_2 - x_1} \right)^\star. \qquad \qquad (4)$

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:

$\displaystyle A^{\star \star} = A, \qquad \qquad (5)$
$\displaystyle A = B \rightarrow A^\star = B^\star, \qquad \qquad (6)$
$\displaystyle (A.B)^\star = A^\star.B^\star = B^\star.A^\star. \qquad \qquad (7)$

Indeed, if equation (3) is represented as

$\displaystyle A^\star.B^\star = C$

then applying rule (6) yields

$\displaystyle (A^\star.B^\star)^\star = C^\star.$

Applying rule (7) then yields

$\displaystyle A^{\star \star}.B^{\star \star} = C^\star$

and applying rule (5) finally yields

$\displaystyle A.B = C^\star,$

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.

• 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

$\displaystyle f:[0,\pi/2] \times [0,2 \pi] \rightarrow \mathbb{R}P^2$ .

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

$\displaystyle f:(\theta, \phi) \mapsto [\cos \phi : \sin \phi : \cot \theta],$
$\displaystyle f^{-1} : [x : y : z] \mapsto \left( \arctan \sqrt{\left({x \over z}\right)^2 + \left({y\over z}\right)^2}, \arctan_2 (y,x) \right)$

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.

File:ProjectiveDuality.PNG
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:

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

such that

$\displaystyle g : [m : b : 1]_L \mapsto [m : -1 : b]$

and

$\displaystyle g : [x : y : 1] \mapsto [x : 1 : -y]_L$

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 [m : −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 [0:0:1], then its dual is [0:1:0]L which is the line at infinity. If the point is [x:0:1], 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

$\displaystyle G = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}$

whose inverse is

$\displaystyle G^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix}.$

Matrix G has one real eigenvalue: one, whose eigenvector is [1:0:0]. The line [1:0:0]L is the y-axis, whose dual is the ideal point [1:0:0] which is the intersection of the ideal line with the x-axis.

Notice that [1:0:0]L is the y-axis, [0:1:0]L is the line at infinity, and [0:0:1]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:

$\displaystyle g^{-1} (L_1 \cap L_2) = gL_1 . gL_2$ .

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−1P1g−1P2 = P then

$\displaystyle g^{-1}P = g^{-1}(g^{-1}P_1 \cap g^{-1}P_2) = g(g^{-1}P_1).g(g^{-1}P_2)$
$\displaystyle = P_1.P_2$
$\displaystyle = L$

so that

$\displaystyle g(g^{-1}P) = gL$
$\displaystyle P = gL$
$\displaystyle g(P_1.P_2) = g^{-1}P_1 \cap g^{-1}P_2$

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

$\displaystyle [x_1:y_1:1].[x_2:y_2:1] = g^{-1} ([x_1:y_1:1] \times [x_2:y_2:1])$

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:

$\displaystyle g[m_1:b_1:1]_L . g[m_2:b_2:1]_L = g^{-1}(g[m_1:b_1:1]_L \times g[m_2:b_2:1]_L)$
$\displaystyle g(g[m_1:b_1:1]_L . g[m_2:b_2:1]_L) = g[m_1:b_1:1]_L \times g[m_2:b_2:1]_L$
$\displaystyle [m_1:b_1:1]_L \cap [m_2:b_2:1]_L = g([m_1:b_1:1]_L \times [m_2:b_2:1]_L)$

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 $\displaystyle (a:b:c) \times (d:e:f) = (b f - c e : c d - a f : a e - b d)$ but

$\displaystyle g(a:b:c) = (a:-c:b),$
$\displaystyle g(d:e:f) = (d:-f:e),$
$\displaystyle (a:-c:b) \times (d:-f:e) = (-c e + b f : b d - a e : -a f + c d)$
$\displaystyle = g(b f - c e : c d - a f : a e - b d)$ .

Therefore

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