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For example, Fano's Geometry exemplifies a triangle with a circle within it. How is this possible? Are lines not defined to be straight? Is this geometry projective geometry(or is projective geometry a different set of axioms)?

cakey
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  • What is the definition you are using? – Braindead Feb 04 '14 at 06:08
  • Fano's geometry on slide 18 http://www-math.ucdenver.edu/~wcherowi/courses/m3210/lecture1.pdf with the usual axioms of projective geometry http://www.math.cornell.edu/~web4520/CG3-0.pdf – cakey Feb 04 '14 at 06:23
  • Okay, so if you look at pg. 6, it says that "points" and "lines" are undefined terms. So they don't really have to be "anything." They are just words. What's important is that these words satisfied the set of axioms that are given for the particular geometry. – Braindead Feb 04 '14 at 06:27

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The axioms which define a projective plane only speak about points, lines and incidences. They say nothing about straightness, they say nothing about an embedding into $\mathbb R^2$. The common representation as a triangle with an inscribed circle is just a combinatoric illustration. The circle in the center is just a device to denote that the three points it connects have a line in common. But the points are not really at the positions where that picture shows them. They cannot really be correctly shown in the plane at all. The right way to see them is algebraically, in $\mathbb Z_2\mathbb P^2$, and algebraically that “circular” line is no different from the other six lines.

MvG
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  • So, in most instances these figures just represent what we want to connotate about the incidence of points and lines? In addition, are all representations of projective geometries have the symmetry quality? ie, #of points=# of lines? or is this just in Fano? – cakey Feb 06 '14 at 05:49
  • @cakey: Right. You could even draw the plane as a bipartite graph, with nodes for both points and lines, and edges for incidence. That way you'd be less likely to mistake the edges for geometric lines at specific positions. All finite projective planes have $n^2+n+1$ planes, by arguments similar to (and in fact easier than) these. This $n$ is called the order of the plane. If the plane is over a finite field, then $n$ is its cardinality, but not every projective plane has an underlying field. – MvG Feb 06 '14 at 05:53
  • $n^2+n+1$ points and the same number of lines is what I meant to write above. – MvG Feb 06 '14 at 06:42