Reciprocation (pole and polar)

Like the Fano plane, the real projective plane is self-dual. How can we exchange the points and lines in such a way that collinearity becomes concurrency and vice versa? The answer lies in the pole of a line and the polar of a point.

Polar: Fix a circle $\omega$ in the Euclidean plane with center $O$. Let $A$ be any point in the projective plane, and let $A’$ be its inverse about $\omega$ (where all the points at infinity invert to $O$), and let $a$ be the line through $A’$ that is perpendicular to line $OA’$ (or $OA$, if $A’=O$). Then $a$ is the polar of $A$.

Pole: of a line $a$ is the reverse construction. First, the pole of the line at infinity is $O$. Now, given a line $a$ other than the line at infinity, let $A’$ be the foot of the altitude from $O$ to $a$ and let $A$ be the inverse of $A’$ about $\omega$. Then $A$ is the pole of $a$ (and $a$ is the polar of $A$.)

Exercise. Show that $A$ lies on the polar of $B$ if and only if the polar of $A$ passes through $B$.

We now know that reciprocation, the transformation about a circle that replaces each point with its polar and each line with its pole, interchanges lines and points in a way that preserves indicence. This enables us to “dualize” many theorems about concurrency to obtain a theorem about collinearity, or vice versa.

Reciprocation has the added benefit that it preserves conics. One way to define a conic is as the reciprocal of some circle. Alternatively, we can define a conic as the locus of solutions $(a:b:c)$ to some homogeneous equation of degree $2$ in $x,y,z$, such as $x^2+2y^2-z^2=0$ or $x^2-yz=0$.

Exercise. What kind of conics do the two quadratic equations above describe when restricted to the Euclidean plane ($c\neq 0$)?

Projective transformations

In homogeneous coordinates, a \textit{projective transformation} is a map of the form $$(x:y:z)\mapsto (ax+by+cz:dx+ey+fz:gx+hy+iz)$$ for some real $a,b,c,d,e,f,g,h,i$. The main facts we need to know about projective transformations are:

• Projective transformations send lines to lines and conics to conics.
• Projective transformations preserve incidence (collinearity and concurrency).
• Projective transformations preserve cross ratios of collinear points. (The cross ratio of four collinear points $A,B,C,D$ is $$(A,B;C,D)=\frac{AC\cdot BD}{AD\cdot BC}.$$ Each factor is a directed length according to a fixed orientation of the line.)
• Given four points $A,B,C,D$ in the projective plane, no three collinear, and another such choice of points $X,Y,Z,W$, there is a unique projective transformation sending $A$ to $X$, $B$ to $Y$, $C$ to $Z$, and $D$ to $W$.

It is often useful to choose an important line in a diagram and apply a projective transformation sending that line to the line at infinity.

Some theorems and their duals

We finally have the tools to prove some powerful about Euclidean geometry that are more natural in the context of projective geometry. Can you see why reciprocation sends each theorem to its dual?

Let’s illustrate the power of projective transformations by proving Desargues’ Theorem. Let $ABC$ and $DEF$ be triangles and assume $X=AB\cap DE$, $Y=BC\cap EF$, and $Z=CA\cap FD$ are collinear. Consider a projective transformation that sends $X$ and $Y$ to two points $X’$ and $Y’$ on the line at infinity. Since projective transformations preserve incidence, we only need to show that the image of $Z$ under this transformation, $Z’$, is on the line at infinity if and only if the images of the triangles, $A’B’C’$ and $D’E’F’$, have the property that $A’D’$, $B’E’$, and $C’F’$ are concurrent. (In this case we say the triangles are perspective from a point.)

Since $X’=A’B’\cap D’E’$ and $Y’=B’C’\cap E’F’$ and these are both on the line at infinity, we have $A’B’||D’E’$ and $B’C’||E’F’$. So we’ve created parallel lines to work with. Now, if $Z’$ is also on the line at infinity, then the third pair of sides of the triangles, $A’C’$ and $D’F’$, are also parallel, and so the triangles are homothetic (that is, one can be mapped to another via a dilation of the plane centered at some chosen point). The center of this homothety must lie on each of $A’D’$, $B’E’$, and $C’F’$, and so the triangles $A’B’C’$ and $D’E’F’$ are perspective from this point.

Conversely, if $A’D’$, $B’E’$, and $C’F’$ intersect at $P$, then the homothety centered at $P$ sending $B’$ to $E’$ must send $A’$ to $D’$ and $C’$ to $F’$ because of the two pairs of parallel lines. So, the third pair of lines must also be parallel, and so $Z’$ is on the line at infinity.