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I am going through "Perspectives on Projective Geometry" and have reached a section concerning projective involutions and conics.

I am specifically looking for an explanation of this concept (pg. 143): "If $\tau$ is an involution associated to $(a, a'; b, b'; c, c')$, then the fixed points of $\tau$ are exactly those two points that are simultaneously harmonic to all three point pairs of the quadrilateral set. In fact, any two of the point pairs of the quadrilateral set determine the position of $p$ and $q$ uniquely."

This fact is used on p.205 and p.206 to construct a conic passing through 4 points that is tangent to a line. If anyone could explain this in more detail it would be beneficial.

Thank you.

brainjam
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Frazer
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1 Answers1

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To paraphrase how I understand your question: knowing $a$, $\tau(a)$, $b$, $\tau(b)$ for some involution $\tau$ of a projective line, you want to find the fixed points $p$ and $q$ of said involution.

As theorem 3.4 of that book states (and my own answer to a different question on this site also details), a projective transformation of the plane is defined by the images of four points, no three of them collinear. On a line, a projective transformation is given in a similar way by the images of three distinct points. The calculation behind that is pretty much the same as for the planar case, but one dimension lower.

So in that sense, $\tau$ may be defined using $a\mapsto a'$, $a'\mapsto a$ and $b\mapsto b'$. If you have homogeneous coordinates established for the points on your line, then this will give you a $2\times2$ matrix representation for $\tau$. The eigenvectors of that matrix will be the fixed points of the involution, because an eigenvector will stay within the same equivalence class when it gets transformed.

MvG
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