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Like the title, with a graph if convenient, I was reading Mumford's book, a picture make me confused: enter image description here

Could someone explain it to me, thanks.

3 Answers 3

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The equation $x^2+y^2 = 1$ defines a surface in $\mathbb{C}P^2$ which is homeomorphic to a sphere. The unit circle $x^2 + y^2 = 1$ in $\mathbb{R}^2$ lies on this sphere, and can be thought of as the equator. Complex conjugation acts as a reflection of this sphere, with the real circle being the mirror of the reflection. Most of the points in the surface lie in $\mathbb{C}^2$, but the surface also contains two points at infinity, namely the points with homogeneous coordinates $[1:i:0]$ and $[1:-i:0]$.

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The picture is not meant to be taken too literally - it's really a 2-dimensional surface in a 4 (real) dimensional space. The points $(X,Y)$ where $X^2 + Y^2 = 1$ and $X$ and $Y$ are real form the "equator" - that really is a circle. All the other points are complex. Complex conjugation ($X \to \bar{X},\ Y \to \bar{Y}$) maps the surface to itself, leaving the real points fixed. The surface is not compact, but can be compactified by adding points at infinity.

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This surface is diffeomorphic to the cotangent bundle of the circle $T^*S^1$ as will be shown in the following explicit construction:

The cotangent bundle of the circle can be parameterized as a surface in $\mathbb{R}^4$ as follows:

$ u_1^2+u_2^2 = 1 $

$ u_1v_1+u_2v_2 = 0$

Of course, the first equation is the equation of the circle, while the second one expresses the orthogonality of the radius and the tangent vectors.

The first equation can be parameterized by: $u_1 = cos(\theta)$, $ u_2 = sin(\theta)$, we may define $\rho = v_2 u_1 - u_2 v_1$, then the transformation:

$X = cos(\theta + i \rho)$

$Y = sin(\theta + i \rho)$

lead to the required result: $X^2 + Y^2$ = 1. In order to see that this transformation is invertible, (thus constitutes a diffeomorphism), note that:

$XY = \frac{sinh(2\rho)}{2}+i cosh(2\rho)u_1 u_2$

The definition of $\rho$ is chosen such that the coordinates $\rho$ and $\theta$ considered as functions on $\mathbb{R}^4 \equiv T^*\mathbb{R}^2$ are canonically conjugate.

Here, the equator corresponds to the zero section $\rho = 0$. The points at infinity correspond to:

$\rho = \pm\infty $