Projection is a projective map

Question

Let $l,m$ be two distinct lines in the projective plane and let $P$ be a point that is not on either of the lines. Prove that the projection $P$ of $l$ onto $m$ is a projective map.

My idea was to find a linear injective map in the underlying vector spaces that induces that projection. In the underlying vector space we thus must have two planes intersecting in a line though the origin and a line that is not in either of the planes that intersects the planes only in the origin. I thought of somehow mapping the basis vectors of the planes onto each other, which is of course injective but this does not involve the line $P$ in any way so there must be something I'm missing. Can anyone give me a hint? (I'm quite new to projective geometry)

Answer 1

The way I usually explain this to my students is the following: Suppose the vectors $A$ and $B$ form a basis on $l$, i.e. every point on $l$ can be described as $\lambda A + \mu B$. Project these points through $P$ to $m$, resulting in two vectors $A'$ and $B'$ such that every point on $m$ can be described as $\lambda'A'+\mu'B'$. Note that $\lambda,\mu$ as well as $\lambda',\mu'$ depend on the actual representatives you chose for your basis, but that need not trouble you at this point.

Now consider a point $Q=\lambda A+\mu B$ on $l$ and its image $Q'=\lambda'A'+\mu'B'$ on $m$. If these two points are preimage and image, then they must lie on a line with $P$. Three points in the projective plane are collinear if and only if their determinant is zero. So you have $\det(P,Q,Q')=0$. Now simply plug in the linear combination for $Q$, and you find a linear condition for $Q'$.

\begin{align*} 0 &= \det(P,\lambda A+\mu B,\lambda'A'+\mu'B') \\ &= \underbrace{\det(P,\lambda A,\lambda'A')}_{=0} + \det(P,\lambda A,\mu'B') + \det(P,\mu B,\lambda'A') + \underbrace{\det(P,\mu B,\mu'B')}_{=0} \\ &= \lambda\mu'\det(P,A,B')+\mu\lambda'\det(P,B,A') \end{align*}

One way to satisfy this equation is by writing

\begin{align*} \lambda'&=\lambda\det(P,A,B') & \mu'&=-\mu\det(P,B,A') \end{align*}

so you have the linear transformation

$$ \begin{pmatrix}\lambda'\\\mu'\end{pmatrix} = \begin{pmatrix}\det(P,A,B') & 0 \\ 0 & -\det(P,B,A')\end{pmatrix} \begin{pmatrix}\lambda\\\mu\end{pmatrix} $$

which clearly is a projective transformation on the equivalence classes.