91Ó°ÊÓ

The concept of a local diffeomorphism is crucial in understanding how a surface of revolution can be related to a plane. A map \( \varphi : U \rightarrow \mathbb{R}^{2} \) is said to be a local diffeomorphism if it is differentiable and its Jacobian determinant is non-zero at every point. This means the map is locally invertible and the inverse is also differentiable. In simpler terms, a local diffeomorphism allows us to 'zoom in' on a small area of the surface and see that it behaves just like a flat plane.
For example, in the given exercise, we showed that \( \varphi(u, v) = \left(u, \int \frac{\sqrt{(f'(v))^2+(g'(v))^2}}{f(v)}dv\right) \) is a local diffeomorphism by computing its Jacobian matrix:
\[ J_\varphi = \begin{bmatrix} \frac{\partial \varphi_1}{\partial u} & \frac{\partial \varphi_1}{\partial v} \ \frac{\partial \varphi_2}{\partial u} & \frac{\partial \varphi_2}{\partial v} \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & \frac{\sqrt{(f'(v))^2 + (g'(v))^2}}{f(v)} \end{bmatrix} \]
Since this determinant is always positive, it proves that \(\varphi\) is a local diffeomorphism.

Parametrization is a method of describing a surface using a set of equations that express the coordinates of the points on the surface as functions of two variables. For a surface of revolution, we use specific functions to describe its points in 3D space.
In our exercise, the surface of revolution is parametrized by \( \mathbf{x}(u, v) = (f(v) \cos u, f(v) \sin u, g(v)) \), where \( f(v) > 0 \). This means that for any given angle \( u \) and height parameter \( v \), we can find a point on the surface.
The range of \( u \) is from \ 0 \ to \ 2\pi \, which describes a full circle, while the range of \ v \ is from \ a \ to \ b \ in the vertical direction. By combining these two parameters, we can describe every point on the surface.

A conformal map preserves angles and the shape of infinitesimally small figures, but not necessarily their size. This is useful for understanding how surfaces relate to each other in a local sense. For a surface of revolution, when we show that it is locally conformal to a plane, we are saying that small sections of the surface have the same angles as corresponding sections on a flat plane.
In Step 3 of the solution, using the local diffeomorphism \( \varphi \), we proved that a surface of revolution \( S \) is locally conformal to a plane. The map \( \theta: V \subset S \rightarrow \mathbb{R}^{2} \) takes parallels (constant \( v \)) to vertical lines and meridians (constant \( u \)) to horizontal lines. This means that in the \( \theta(V) \subset \mathbb{R}^{2} \), we have an orthogonal system of straight lines, which shows the local conformality.

An area-preserving map, also known as a symplectomorphism, is a type of transformation that maintains the area of any region it maps. For example, if you apply an area-preserving map to a shape, the area of the shape before and after the transformation remains the same.
In part d of the exercise, using the map \( \psi(u, v) = (u, \int f(v) \sqrt{(f'(v))^2 + (g'(v))^2} dv) \), we showed that \psi \) is also a local diffeomorphism. Its Jacobian matrix is:
\[ J_\psi = \begin{bmatrix} 1 & 0 \ 0 & f(v) \sqrt{(f'(v))^2 + (g'(v))^2} \end{bmatrix} \]
Because the determinant is positive, \( \psi \) is a local diffeomorphism. This lets us conclude that for each point \( p \) on surface \( S \), there exists a neighborhood \( V \subset S \, and a map \( \bar{\theta}: V \rightarrow \mathbb{R}^{2} \), that is area-preserving, thus maintaining the area when transforming \( V \) onto a plane.