# Tag Info

1

On the sphere, any two "lines" intersect twice; in the projective plane, just once. The sphere is oriented; the projective plane is not. The letter R has more asymmetry than other letters of the alphabet; move and "R" around the projective plane and keep it going in the same direction until it returns to where it started. It will be a backwards, ...

0

$\Sigma_{2,0} \cong \Bbb P^2$, and $\Sigma_{2,1} \cong \Bbb P^1\subset\Bbb P^2$, so $\tilde\Sigma_{2,0} \cong \Bbb A^2\subset\Bbb P^2$.

0

If $a^2 + b^2 + c^2 = 1$ and $u^2 + v^2 + w^2 = 1$, the requested distance is $$\tfrac{\pi}{2} - \arccos|au + bv + cw| = \arcsin|au + bv + cw|.$$ The "usual" metric on the projective plane is the Riemannian quotient of the round unit sphere by the antipodal map. To find the distance from a projective line $\ell$ to a point $p$, interpret the homogeneous ...

0

No, the hyperbolic distance is not the only function satisfying $d(z_1,z_3)=d(z_1,z_2)+d(z_2,z_3)$. You could assign arbitrary (not even neccessarily continuous) values to all the points, interpret them as “distance from some origin”, and use differences between these values as distances. Your equation would be satisfied in all these cases, even if $z_2$ ...

1

The result still holds. Yay to projective geom. In fact, there is an interesting 'degenerate case': $ABCD$ is a cyclic quad (or quadrilateral on a conic), then $AB \cap CD$, $BC \cap DA$, the point intersection of tangents at $A$ and $C$, and the point of intersection of tangents at $B$ and $D$ are collinear.

0

No less a geometer than David Hilbert wrote an accessible book on the subject "Geometry and the Imagination". After finishing Hartshorne's book "Geometry: Euclid and Beyond", I wanted to learn more about configurations and incidence structures. You can find a pdf of Hilbert's book by googling the title, or you can buy a dead tree copy for around thirty ...

1

Funny, I did the proof which excludes $1$ in a lecture today. The basic idea is that you can show that if the cross ratio is $1$, then two points must coincide. Without loss of generality you may assume  v_1=\begin{pmatrix}a\\1\end{pmatrix} \qquad v_2=\begin{pmatrix}b\\1\end{pmatrix} \qquad v_3=\begin{pmatrix}c\\1\end{pmatrix} \qquad ...

0

Start with your second attempt, with a given bijection $f$ and a matching projective transformation $f'$ defined using $P_1,P_2,P_3$ and their images. Now consider the composition $f^{-1}\circ f'$. By construction of $f'$, it has $P_1,P_2,P_3$ as fixed points. And by the already shown implication b$\Rightarrow$a you know that it preserves cross ratios. In ...

Top 50 recent answers are included