Note If a conic in $\mathbb{R}P^{2}$ contains a line, then it is degenerate.
Theorem. Through five points $P_{1},P_{2},P_{3},P_{4},P_{5} \in \mathbb{R}P^{2}$ there exists a conic in $\mathbb{R}P^{2}$.
$i)$ If no four points lie on a line, then the conic is unique.
$ii)$ If no three lie on one line (five points in general position), then the conic is non-degenerate.
Lemma. If a conic contains three collinear points, then it contains the line through these points. Continue reading
Definition.
Let $V$ be a vector space over $\mathbb{R}$ (or $\mathbb{C}$). A map
$b\colon V \times V \to \mathbb{R}$ is a symmetric bilinear form, if
$b$ is non-degenerate, if
\[b(v,w)=0 \quad \forall w \in V \Rightarrow v=0\,.\]
The corresponding quadratic form is defined by
\[q(v)=b(v,v) \quad \forall v \in V\,.\]
Cross-ratio of lines through a point (again).
Claim. The map which maps a point $g^{\ast} \in p^{\ast}$ to the intersection point of $g \cap \ell$ is a projective map. Continue reading
Fundamental Theorem of real projective geometry: Let $f\colon \RP^n \to \RP^n$, $n \ge 2$, be a bijective map that maps lines to lines. Then $f$ is a projective transformation.
Remark: The theorem does not hold for arbitrary fields. For example $f\colon \CP^n \to \CP^n$ ($n \ge 2$) with
\[f\left(\sqvector{z_1\\ \vdots\\ z_{n+1}}\right) = \sqvector{\bar z_1\\ \vdots\\ \bar z_{n+1}}\]
is a bijective map, mapping complex projective lines to complex projective lines but it is not a projective transformation of $\CP^n$.
Theorem (complete quadrilateral). Consider four lines with intersection points $A$, $B$, $C$, $D$, $P$, and $Q$ as shown in the picture. Then the cross-ratio of the intersection points $X$ and $Y$ of the diagonals with the line $PQ$ and $P$ and $Q$ is:
\[ \cr(P,X,Q,Y)=−1.\]
Proof of theorem on complete quadrilateral (multi-ratio).
Consider the multi-ratio with
$P_{1} = Q_{1} = P$, $P_{3} = Q_{3} = Q$, $Q_{2} = X$, $P_{2} = Y$,
where $p$ is the affine coordinates of $P$, $q$ the ones of $Q$ etc.
\begin{align}
-1 &= \mathrm{m}(P, X, Q, P, Y, Q) \\
&= \frac{p-x}{x-q} \frac{q-p}{p-y} \frac{y-q}{q-p} \\
&= \frac{p-x}{x-q} \frac{q-y}{y-p} \\
&= \mathrm{cr}(P, X, Q, Y). \\
\end{align}
$\square$
Definition (complete quadrilateral). A configuration consisting of four lines in the projective plane – no three through one point – and the six intersection points, one for each pair of lines, form a complete quadrilateral.
Cross-ratio (Doppelverhältnis)
We are looking for invariants with respect to projective transformations.
Definition: Let $ P_{i} =[v_i]= \ssqvector{x_{i} \\ y_{i}}$, $i=1,\ldots,4$, be four distinct points on a projective line $ \RP^{1} $. Then the cross-ratio of these points is
\begin{align*}
cr(P_{1},P_{2},P_{3},P_{4})
&= \frac{\det(v_1v_2)}{\det(v_2v_3)}\frac{\det(v_3v_4)}{\det(v_4v_1)}\\
&= \dfrac{x_{1}y_{2}-x_{2}y_{1}}{x_{2}y_{3}-x_{3}y_{2}} \dfrac{x_{3}y_{4}-x_{4}y_{3}}{x_{4}y_{1}-x_{1}y_{4}}.\\
\end{align*}
If $y_{i} \neq 0$, we may introduce affine coordinates $ u_{i} = \frac{x_{i}}{y_{i}}$. This yields
\begin{align*}
cr(P_{1},P_{2},P_{3},P_{4}) =
&= \dfrac{y_{1}y_{2}( \frac{x_{1}}{y_{1}}-\frac{x_{2}}{y_{2}} ) }{y_{2}y_{3}( \frac{x_{2}}{y_{2}}-\frac{x_{3}}{y_{3}} )} \dfrac{y_{3}y_{4}( \frac{x_{3}}{y_{3}}-\frac{x_{4}}{y_{4}} ) }{y_{4}y_{1}( \frac{x_{4}}{y_{4}}-\frac{x_{1}}{y_{1}} )} \\
&= \dfrac{u_{1}-u_{2}}{u_{2}-u_{3}} \dfrac{u_{3}-u_{4}}{u_{4}-u_{1}}\,.
\end{align*}
Let $V$, $W$ be two vectorspaces over the same field and of the same dimension and $F\colon V \rightarrow W$ a linear isomorphism. In particular $ker(F) = \{0\}$, so F maps 1-dimensional subspaces to 1-dimensional subspaces.
Hence $F$ induces a map from $P(V)$ to $P(W)$.
Definition: A projective transformation $f$ from $P(V)$ to $P(W)$ is a map defined by a linear isomorphism $F\colon V \rightarrow W$ such that
\begin{equation*}
f([v]) = [F(v)] \quad \forall [v] \in P(V)\,.
\end{equation*} Continue reading
Definition: Two triangles $\triangle_1 = \triangle(A_1, B_1, C_1)$ and $\triangle_2 = \triangle(A_2, B_2, C_2)$ are in perspective w.r.t. a point $S$ if
\[
S = (A_1 A_2) \cap (B_1 B_2) \cap (C_1 C_2)
\]
The triangles are in perspective w.r.t. a line $\ell$ if
\begin{align*}
A’ = (B_1 C_1) \cap (B_2 C_2)\\
B’ = (A_1 C_1) \cap (A_2 C_2)\\
C’ = (A_1 B_1) \cap (A_2 B_2)\\
\end{align*}
lie on $\ell$.
