Past exam of the short options course of the University of Oxford / 2017 / 3

The real projective plane is the set of one-dimensional linear subspaces of ${\mathbb{R}}^3$. Equivalently,A projective point is written in homogeneous coordinates $[x:y:z]$. A projective line is the set of projective points whose representatives belong to one fixed two-dimensional linear subspace of ${\mathbb{R}}^3$. Such a linear subspace is the kernel of a linear map of a nonzero linear functional, so its equation has the form $\alpha x+\beta y+\gamma z=0$ with $(\alpha ,\beta ,\gamma )\ne 0$.

Two distinct projective points have linearly independent representatives $p,q$. Their linear span has dimension two and hence defines a projective line containing both. Conversely, any two-dimensional linear subspace containing both representatives contains their entire linear span and therefore equals it. Thus exactly one projective line passes through two distinct projective points.

For representatives $p=(a,b,c)$ and $p^{\prime }=(a^{\prime },b^{\prime },c^{\prime })$, a point $[x:y:z]$ belongs to their projective line exactly when these three vectors are linearly dependent. The determinant criterion givesExpanding along the first row givesThe coefficient vector is the cross product $p\times p^{\prime }$. It is nonzero because the projective points are distinct. Rescaling either representative merely rescales all three coefficients, leaving the projective line unchanged. This is the join formula in projective joins and meets by cross products.

Two projective triangles are in perspectivity from a point $O$ when their three corresponding-vertex projective lines $P{P}^{\prime }$, $Q{Q}^{\prime }$, and $R{R}^{\prime }$ all pass through $O$.

Desargues's theorem: perspectivity from a point implies perspectivity from a line. More explicitly, if $P{P}^{\prime },Q{Q}^{\prime },R{R}^{\prime }$ are concurrent, then the three corresponding-side intersectionsare collinear. The usual statement assumes nondegenerate projective triangles and well-defined, distinct corresponding-side intersections. The projective line containing $X,Y,Z$ is the axis of perspectivity from a line. All intersections are projective: parallel lines in an affine chart meet on its projective line at infinity.

The dual projective space is $\mathbb{P}(({\mathbb{R}}^3{)}^{\ast })$. A projective line with coefficients $[\alpha :\beta :\gamma ]$ becomes a projective point in this dual plane; a projective point $[a:b:c]$ becomes the dual projective line $a\alpha +b\beta +c\gamma =0$.

The point-line duality reverses incidence:Thus the dual statement is that the dual point $l^{\ast }$ lies on the dual line $P^{\ast }$. A join of projective points becomes an intersection of dual projective lines, and the converse holds too. In particular, collinear projective points correspond to concurrent dual projective lines.

The dual of Desargues's theorem is its converse: if the intersections of the three pairs of corresponding sides are collinear, then the three lines joining corresponding vertices are concurrent. In the same nondegenerate setting, perspectivity from a line therefore implies perspectivity from a point.

Apply projective joins and meets by cross products to each corresponding pair. The cross products of their representatives are proportional to $(0,1,0)$, $(0,0,1)$, and $(0,1,-1)$, respectively. HenceTheir common point isIt is a valid projective point, because its representative is nonzero. The three equations force $y=z=0$, so this common projective point is unique. Thus the two projective triangles are in perspectivity from a point $O$.

The corresponding side equations, computed using projective joins and meets by cross products, areSubtracting the two equations in each row and substituting back gives the three projective pointsFor example, the first pair forces $y=z$ and $x=3z$; the second pair forces $y=0$ and $x=-z$; the last pair forces $x=z=2y$. All three satisfysince $3-4+1=0$, $1-0-1=0$, and $2-4+2=0$. The coefficient vector is nonzero, so this is a projective line. Thus the corresponding-side intersections are collinear, explicitly confirming Desargues's theorem.

The diagram uses the affine chart $x+y+z\ne 0$ with coordinates $u=y/(x+y+z)$ and $v=z/(x+y+z)$. Here the center $O$ is finite, the axis is $u=1/5$, and $Y$ lies on the projective line at infinity. The apparent parallelism of $QR$ and $Q^{\prime }{R}^{\prime }$ records their common vertical point at infinity, which also belongs to the axis.