ISEGORIA / MATH ENCYCLOPEDIA
Projective geometry: where parallel lines meet
Draw a road to the horizon and the parallel edges meet. Projective geometry takes that seriously: add one point at infinity to every direction and the plane becomes a space in which any two lines cross, exactly once. Points are lines through the origin of three-space, lines are planes through it, and projecting from one plane onto another is just a change of the chart. What survives every projection is not length or angle but a single ratio of ratios, the cross-ratio, and under projection the ellipse, the parabola and the hyperbola turn out to be one curve seen from different angles.
Before you begin: Vectors, the cross product and determinants, and the idea of a line through the origin as a one-dimensional subspace
Predict, manipulate, then test yourself: every experiment ends with practice questions that are generated fresh at three levels. Graphs illustrate the mathematics; they do not replace a proof.
1. Points are lines through the origin
Put the plane z = 1 above an origin O. A line through O pierces the plane in one point, so points of the plane and lines through O correspond. A line through O that lies flat (z = 0) pierces the plane nowhere: these are the points at infinity, one for each direction. A line of the plane z = 1 spans, together with O, a plane through O, so lines of the projective plane are planes through O. Two such planes always meet in a line through O, which is why two lines always meet. Drag A, B and C, or tilt the second line towards parallel, and watch the common line through O swing down to the horizon. The violet traveller walks along the first line: it leaves the chart in one direction and comes back from the other, because a line is a closed loop through its single point at infinity.
Worked example. Take \(\ell_1: y=x\) and \(\ell_2: y=x+2\), parallel. As homogeneous lines, \(\ell_1=(1,-1,0)\) and \(\ell_2=(1,-1,2)\) (the coefficients of \(aX+bY+cZ=0\)). Their cross product is \((-2,-2,0)\propto[1:1:0]\): the point at infinity in the direction \((1,1)\). Change \(\ell_2\) to \(y=2x+2\), that is \((2,-1,2)\), and the cross product becomes \((-2,-2,1)\), which has \(Z\neq0\): the point \((-2,-2)\) of the chart, as solving \(x=2x+2\) confirms.
Watch out. The 3D picture shows both halves of each line through O, because [X : Y : Z] and [−X : −Y : −Z] are the same point; the readout always flips the sign so that Z is not negative. The plane z = 1 is only a chart: it misses the horizon, and a different chart (say y = 1) would put some of the former points at infinity in view and send others away.
Why do two parallel lines meet at one point, and not at two, one at each end?
Because the point at infinity is a line through O, and a line through O has two directions that are the same object. Going to infinity to the right and going to infinity to the left arrive at the same point, so a projective line is a circle, not a segment with two ends.
2. The cross-ratio survives projection
Light from a point O throws four points A, B, C, D of a line onto a screen. The spacing is distorted, the ratio of two lengths changes, even the order of the gaps can change as the screen tilts. One number survives: the cross-ratio (A,B;C,D) = AC·BD / (BC·AD), the ratio of two ratios. It can be read off the four rays at O alone, from the sines of the angles between them, which is why it cannot depend on the screen. The graph on the right is the projection itself as a function from position on ℓ₁ to position on ℓ₂: a hyperbola, with a pole where the ray is parallel to the screen and an asymptote at the image of infinity. Tilt the screen through that parallel position and one image goes to infinity and comes back on the other side, while the cross-ratio does not flinch. Set D to the harmonic position and (A,B;C,D) = −1 stays −1 on every screen.
Worked example. Take A, B, C, D at positions 0, 1, 3, 6 on a line. Then \(AC=3\), \(BD=5\), \(BC=2\), \(AD=6\), so \((A,B;C,D)=15/12=5/4\). For a harmonic fourth point to A=0, B=1, C=3 solve \(\frac{3\,(d-1)}{2\,d}=-1\): \(d=3/5\). Then \(AC=3\), \(BD=-\tfrac25\), \(BC=2\), \(AD=\tfrac35\), and \(3\cdot(-\tfrac25)/(2\cdot\tfrac35)=-1\). Every screen reads \(-1\) again.
Watch out. Lengths are signed, measured along the line, so the cross-ratio can be negative: it is negative exactly when the pairs A,B and C,D separate each other. When one image is at infinity the lengths involving it cancel in the ratio and the cross-ratio is computed from the other three; the readout then reports the value from the rays instead. Permuting the four points changes the value among six numbers, λ, 1/λ, 1−λ, 1/(1−λ), (λ−1)/λ, λ/(λ−1), and no other.
Why is the cross-ratio of four points on a line the same as that of four rays through O?
Because the length of the gap between two points on a line is, up to a factor that depends only on the line, the determinant of their two rays from O. Writing each gap that way, the factors for the line cancel between numerator and denominator, leaving the sines of the angles at O.
3. Conics: one curve, and Pascal’s hexagon
Light from the apex through a circle makes a cone, and every plane section is the circle seen through a tilted screen. Horizontal: a circle. Tilt it and the circle stretches into an ellipse, until the plane is parallel to a generator of the cone and one point of the curve escapes to infinity: a parabola. Tilt further and the plane cuts both nappes, two points escape, and the curve is a hyperbola. They are one curve in the projective plane, differing only in how many points land at infinity (0, 1 or 2). Because projection sends lines to lines, any theorem about the circle that is about points, lines and incidence carries over to every conic. Pascal’s theorem is the classic: six points on a conic, join them in any order to make a hexagon, and the three pairs of opposite sides cross in three points on one line. The six teal points slide on the circle; the same six are drawn on the section on the right, where you can also drag them.
Worked example. Tilt the plane by \(\tau=30^\circ\). Then \(e=\sqrt2\sin30^\circ=0.707\), the section is the ellipse \(\tfrac12u^2+v^2+\sqrt3\,u-\tfrac12=0\), with semi-axes \(a=1/\cos60^\circ=2\) and \(b=\sqrt2\) about the centre \(u=-\tan60^\circ=-1.732\), and indeed \(1-b^2/a^2=0.5=e^2\). At \(\tau=45^\circ\) the \(u^2\) term vanishes, leaving the parabola \(v^2+2u=0\). At \(\tau=60^\circ\), \(e=1.225\) and the hyperbola has two points at infinity.
Watch out. The cone has half-angle 45° and the plane is always at distance 1 from the apex, which fixes the semi-latus rectum at 1 whatever the tilt, so an ellipse close to the parabolic case is very long and the right-hand view shows only part of it. The cone and the plane are drawn out to a finite size, but they extend without limit. When two opposite sides of the hexagon are parallel, their crossing is a point at infinity and is not drawn, but the Pascal line still passes through it.
Pascal’s theorem is stated for a circle in school geometry. Why is it true for a hyperbola?
Because a hyperbola is a projection of a circle, and projection sends points to points, lines to lines and intersections to intersections. The three crossings of opposite sides on the circle lie on a line; their images are the three crossings on the hyperbola and lie on the image of that line.