Geometric Proof that Fubini-Study Metric is Round

2010-10-11T02:09:48Z

The Fubini-Study metric d(x,y) on $CP^1$ is defined as follows: for x and y in $CP^1$ let v and w be unit vectors in $C^2$ representing x and y. Then $d(x,y)=2arccos(\langle v,w\rangle)$. The round metric on $S^2$ is given by taking a pair of points to the angle between those two points. It is a fact that under stereographic projection the round metric corresponds to the Fubini-Study metric. This can be proven by some complicated and unenlightening algebraic manipulation. On the other hand, it seems like maybe there should be a purely (or almost purely) geometric proof of this fact for the following reason. Embed $S^2$ into $C^2$ by taking it to the unit sphere around (0,0,1,0) in v=0, where $C^2$ has coordinates z=x+iy and w=u+iv,so the north pole of $S^2$ is just the origin (0,0,0,0). Then the for two points a and b on the sphere, the angle metric is the angle aCb where C is the center of the sphere, (0,0,1,0). On the other hand, the Fubini-Study metric is just $2arccos(\langle x,y\rangle/\mid x\mid \mid y\mid)$ , which, if this were the ordinary Euclidean inner product, would just be twice the angle aOb where O is the origin (0,0,0,0). The statement that these two are equal is highly reminiscent of the high-school geometry fact that the angle measure between two points on a circle measured from the center is twice the angle measure between two points measured from another point on the circle. Of course, this is not a circle, and the inner product is not the Euclidean inner product. Is the similarity just a coincidence, or is there some geometric fact that explains what's going on?

Geometric Proof that Fubini-Study Metric is Round - MathOverflow

Geometric Proof that Fubini-Study Metric is Round