Usually by Duality we mean a bijection between a projective plane P2 and its dual, which is defined as the set of p-lines (projective lines) of P2. This set has again the structure of a projective plane. To illustrate the ideas consider the projective plane P2, identified with the set of e-lines (euclidean lines) through the origin of R3. Such an e-line is represented by [x], where x=(x1,x2,x3) is a non-zero point of R3. The projective lines of P2 (call them p-lines) consist of all e-lines of R3 lying in an e-plane (euclidean plane) through the origin. Hence the points [x] of this p-line satisfy an equation of the form a1*x1+a2*x2+a3*x3=0. Denote the set of non-zero multiples of a=(a1,a2,a3) by [a]. Each [a] represents a p-line of P2 and the set of all these [a]'s builds a copy of P2 called the dual of P2 and denoted by P2*.
The basic relation between points [x] of P2 and points [a] of P2* is that of coincidence, expressed by the equality a1*x1+a2*x2+a3*x3=0. The equation expressing that p-point [x] is on the p-line [a], or equivalently, that the p-line [a] passes through the p-point [x].
Relation a1*x1+a2*x2+a3*x3=0, can be read from two sides.
First, fixing [x] and considering all [a]'s satisfying this relation.
Second, fixing [a] and considering all [x]'s satisfying this relation.
In the first case all [a]'s define a p-line in P2* consisting of the bundle of p-lines passing through [x]. We denote this bundle by [x]*. In terms of our euclidean eyes this idendifies line [x] with all the planes containing this line. The map [x]-->[x]* establishes a natural bijection between the projective spaces P2 and (P2*)* (though there is no intrisec bijection between P2 and P2*).
In the second case all [x]'s define a p-line of P2 consisting of all p-points of P2 contained in the line [a]. We could denote it by [a]*, but this is identified with [a]. Abstractly we have the relation [x]** = [x].
The important thing is that the set of lines of P2 has the same structure as P2 itself. Further the lines of P2* are identified with [x]* i.e. the bundle of lines through point [x]. Thus the two conditions become equivalent:
(1) line [a] = join([x],[y]), for points [x], [y] of P2 and [a] a line of P2 (equivalently a point of P2*).
(2) point [a] = inter([x]*,[y]*) , [x]*, [y]* being lines of P2* and [a] being a point of P2*.
Thus, in theorems of projective geometry, intechanging the words point<--->line and verbs join<--->intersect we get dual new theorems. The new theorems being valid in P2*, which is isomorphic to P2. We say then proof by duality and prove only one version of the two dual theorems.
A typical case is the theorem of Desargues (see Desargues.html ).
Theorem
I and II below are equivalent.
(I) For triangles [x][y][z], [x'][y'][z'], lines [a]=join([x],[x']), [b]=join([y][y']), [c]=join([z][z']) are concurrent, i.e. there is a point [w] lying on all three lines [a], [b], [c].
(II) Intersection points of sides [p]=inter([xy],[x'y']), [q]=inter([yz],[y'z']), [r]=inter([zx],[z'x']) lie on the same line [e].
Here [xy] denotes the line [f]=join([x],[y]), [yz] denotes [g]=join([y],[z]), etc..
Assume we proved that I => II. Then to prove II => I take the dual of II:
[p]([xy],[x'y']) reads: line [p]* joining points [f] and [f'] (in the dual space P2*).
[q]([yz],[y'z']) reads: line [q]* joining points [g] and [g'] (in the dual space P2*).
[r]([zx],[z'x']) reads: line [r]* joining points [h] and [h'] (in the dual space P2*).
lie on the same line [e] reads: [p]*,[q]*,[r]* intersect at [e] (again in P2*).
Thus, II translates:
For triangles [f][g][h], [f'][g'][h'] of P2* lines join([f],[f']), join([g],[g']), join([h],[h']) are concurrent. But this is just I stated for the dual space P2*. By the proven part of the theorem follows that:
[u]([fg],[f'g']) etc. are on the same line [o]* (of P2*). But [fg] is [x], [f'g'] is [x'], [u] is line [xx'] etc.. Hence [u], [v], [w] being on the same line [o]* means that line [xx'] passes through [o] and similar does [yy'] and [zz'].
Thus, the dual (II') of (II) is (I) reformulated in the projective space P2*. Also the dual (I') of (I) is (II) reformulated in P2*. Discarding the proper nature of the projective spaces (i.e. considering P2 and P2* to be abstractly the same thing) we see that {I=>II} and {II=>I} expresses the same proposition in two different projective spaces.
A similar situation arises in the theorem of Pappus, which is self-dual. i.e. the dual expresses the same figure as the original (see PappusSelfDual.html ). Another instance of duality is that between the theorems of Brianchon and Pascal.