The Pappus Line
The text which is unclear to me:
Let the range of points $(A_i)$ be projectively related to the range of points $(B_i)$ under the correspondence $A_i \leftrightarrow B_i$. Fix $i$: then the pencil of lines through $A_i$ is projectively related to the pencil of lines through $B_i$ by the sequence: $$(A_iB_j){\overset{A_i}{\wedge}}B_j\wedge A_j {\overset{B_i}{\wedge}}(B_iA_j)$$ The line $(A_i B_i)$ is self-corresponding and so this projectivity is a perspectivity from a line $c_i$. If we interchange the role of $i$ and $j$ we see that $c_i$ is independent of $i$. This line is called the Pappus line of the projectivity. It has the following property: the intersection of $A_iB_j$ with $A_jB_i$ lies on the Pappus line. In fact once the Pappus line is known then it can be used as an alternative geometric construction of the point $B$ corresponding to $A$.
It's a paragraph from a book on geometry (Roger Fenn geometry), section "Projectivities and Perspectivities"
Here are my questions:
- a) How the line $(A_iB_i)$ can be self-correspondent?(self-correspondent line is not defined before. But self-correspondent point is: it's a point which corresponds to itself after composition of perspectives). b) What is a self-correspondent line and how to imagine/draw it?
- The notation $(A_iB_j){\overset{A_i}{\wedge}}B_j\wedge A_j {\overset{B_i}{\wedge}}(B_iA_j)$ itself. I even can't understand what it means as it is not defined earlier in this book. Notations $(A_1,A_2,A_3,A_4){\overset{O}{\wedge}}(B_1,B_2,B_3,B_4)$ and $(A){\overset{O}{\wedge}}(B)$ are defined, however and clear to me.
- What is a perspectivity from a line? Where it arises from?(it is not defined as well earlier). How to imagine draw it?
- Finally, what the points $A$ and $B$ are? Is it $(A_i) = (A_1, A_2, A_3, A)$ and $(B_i) = (B_1, B_2, B_3, B)$ in the very beginning, so $A$ and $B$ are 4th points, or, $A$ and $B$ are just some abstract points where given a Pappus line and some $A$ we can construct corresponding $B$?
Edit: what adds mess is it's not mentioned which plane is used for this proof - Projective or Euclidean. In case Projective is used, then, maybe, I have to consider duality, but I'm not sure...
P.S. I googled for it, no results. As everywhere Pappus theorem is being proved without introducing notation above.
P.P.S. if the notation is a complete mess, then just let me know please. So I'll give up with this very construction. I can't tell if it is, as i'm not too proficient in projective geometry.
Solution 1:
"Perspectivity from a line" is the dual of the "perspectivity from a point": The perspectivity of the set of lines through point $P$ to the set of lines through point $Q$ from the line $l$ maps any line $p$ through $P$ to the line of $p\cap l$ and $Q$.
A projectivity from a set of lines through a point to the set of lines through another point is a composition of countable perspectivities from lines - so it is just the dual of projectivities among lines.
We can also define a perspectivity from a set of lines through a point to a line. If $l$ is a line and $P$ is a point, we just map any line $p$ through $P$ to $p\cap l$.
So the construction is as follows:
- A projectivity $\pi$ is given from a line $e$ to a line $f$.
- We fix a point $A_i$ on the line $e$ and another point $B_i$ on line $f$.
- We construct the following projectivity from the set of lines through $A_i$ to the set of lines through $B_i$:
- we consider a line $a$ through $A_i$ and let its intersection with $f$ be $B_j$.
- $\pi^{-1}$ maps $B_j$ to $A_j$.
- Then the image of $a=A_iB_j$ is $B_jA_i$. This is a projectivity indeed, since it is a finite composition of projectivities.
- The above projectivity maps $A_iB_i$ to itself, so it is called a self-correspondent element.
- It is known that if a projectivity among the sets of lines through two point maps the line of the two points to itself (i.e. it is self-correspondent), then the projecitivity is just a perspectivity from a line. It is the dual of a well-known statement for projectivities among lines. Now this line is denoted by $c_i$ and called the Pappos-line.
- In the above construction if we interchange the roles of $A_i$ and $B_i$ with $A_j$ and $B_j$ we get the same line $c_i$.
- $A$ is just an arbitrary point of line $e$ and $B$ is the image of $A$ under $\pi$. So the last statement is that we can use the Pappos line to construct the image of any point on $e$ under the projectivity $\pi$ if the images of three points are given.