Like the title, with a graph if convenient, I was reading Mumford's book, a picture make me confused:
Could someone explain it to me, thanks.
Like the title, with a graph if convenient, I was reading Mumford's book, a picture make me confused:
Could someone explain it to me, thanks.
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 $[1:-i:0]$.
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.
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 $