A well-known theorem in projective geometry states that the three Pascal lines of an arbitrary hexagon inscribed in a quadric intersect in one point. I found an algebraic reformulation, which states that for any six lines $l_i=a_ix+b_iy+c_iz$, if the quadrics $l_i^2$ are linearly dependent, then $l_1l_3$, $l_3l_5$, $l_1l_5$, $l_2l_4$, $l_2l_6$, $l_4l_6$ are linearly dependent. But I can not understand the algebraic meaning of this fact. Is it just an accident or a special case of some duality?
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Pascal's theorem is a special case of the purely algebraic Cayley-Bacharach theorem. For a brief discussion see e.g. http://terrytao.wordpress.com/2011/07/15/pappuss-theorem-and-elliptic-curves/
$\begingroup$
$\endgroup$
You might want to look at http://ssdut.dlut.edu.cn/lzxjj/uploadfile/200905/20090512022236594.pdf
(I am not certain that it answers your question...)
answered Sep 26, 2011 at 20:24
$\begingroup$
$\endgroup$
1
Thank you very much! This is interesting article- I have read it. But what I am concerned is interpretation of the result from the perspective of geometry of space of quadrics. First, I thought it follows from the polar duality in $P^5$, but it seems it does not.
-
4$\begingroup$ You should add this as a comment instead of an answer. $\endgroup$ Commented Dec 6, 2011 at 3:06