# The Gaussian curvatures of spheres

1. An example of the Gaussian curvature

Example 1 Compute the Gaussian curvature of sphere

$\displaystyle S = \{(x, y, z) \in \mathbb{R}^3 | x^2 + y^2 + z^2 = R^2\}.$

Parametrizing ${X:}$

$\displaystyle U \subset \mathbb{R}^2 \rightarrow S$

$\displaystyle (u, v)\mapsto (u, v, \sqrt{R^2 - u^2 - v^2}).$

we have

$\displaystyle X_u = (1, 0, -\frac{u}{\sqrt{R^2 - u^2 - v^2}}), X_v = (0, 1, -\frac{v}{\sqrt{R^2 - u^2 - v^2}})$

The coefficients of the second fundamental form:

$\displaystyle E = \langle X_u, X_u \rangle = 1 + \frac{u^2}{R^2 - u^2 - v^2}$

$\displaystyle G= \langle X_v, X_v \rangle = 1 + \frac{v^2}{R^2 - u^2 - v^2}$

$\displaystyle F = \langle X_u, X_v \rangle = \frac{uv}{R^2 - u^2 - v^2}$

$\displaystyle \Rightarrow EG - F^2 = (1 + \frac{u^2}{R^2 - u^2 - v^2})(1 + \frac{v^2}{R^2 - u^2 - v^2}) - \frac{u^2v^2}{(R^2 - u^2 - v^2)^2}$

$\displaystyle = 1 + \frac{u^2}{R^2 - u^2 - v^2}+ \frac{v^2}{R^2 - u^2 - v^2}$

$\displaystyle EG - F^2= \frac{R^2}{R^2 - u^2 - v^2}$

we compute coefficients of the first fundamental form:

$\displaystyle N= -\frac{X_u \wedge X_v}{\|X_u \wedge X_v\|}$

$\displaystyle X_u \wedge X_v = (\frac{u}{R^2 - u^2 - v^2}, \frac{v}{R^2 - u^2 - v^2}, 1)$

$\displaystyle \|X_u \wedge X_v\| = \sqrt{\frac{R^2}{R^2 - u^2 - v^2}}$

$\displaystyle = \frac{R}{\sqrt{R^2 - u^2 - v^2}}$

$\displaystyle \Rightarrow N = -\frac{X_u \wedge X_v}{\|X_u \wedge X_v\|} = -(\frac{u}{R}, \frac{v}{R}, \frac{\sqrt{R^2 - u^2 - v^2}}{R})$

$\displaystyle X_{uu} = (0, 0, - \frac{R^2 - v^2}{\sqrt{(R^2 - u^2 - v^2)^3}})$

$\displaystyle X_{vv} = (0, 0, -\frac{R^2-u^2}{\sqrt{(R^2-u^2-v^2)^3}})$

$\displaystyle e = \langle N, X_{uu} \rangle = \frac{v^2 - R^2}{R}\cdot \frac{1}{R^2 - u^2 - v^2}$

$\displaystyle g = \langle N, X_{vv} \rangle = \frac{u^2 - R^2}{R}\cdot \frac{1}{R^2 - u^2 - v^2}$

$\displaystyle f = - \langle N_u, X_v \rangle = \frac{uv}{R(R^2 - u^2 - v^2)}$

where

$\displaystyle N_u = (-\frac{1}{R}, 0, \frac{u}{R\sqrt{R^2 - u^2 - v^2}})$

$\displaystyle X_v = (0, 1, -\frac{v}{\sqrt{R^2 - u^2 - v^2}})$

These imply that

$\displaystyle eg - f^2 = \frac{1}{R^2(R^2 - u^2 - v^2)^2}[(u^2 - R^2)(v^2 - R^2) - u^2v^2]$

$\displaystyle = \frac{1}{R^2(R^2 - u^2 - v^2)^2}[- (u^2 + v^2)R^2 + R^4] = \frac{R^2[R^2 - u^2 - v^2]}{R^2(R^2 - u^2 - v^2)^2}$

$\displaystyle = \frac{1}{R^2 - u^2 - v^2}.$

By the above computation, the curvature of sphere is

$\displaystyle K = \frac{eg - f^2}{EG - F^2} = \frac{1}{R^2}$.

# MATH-F-420: Differential geometry of Verbitsky

## MATH-F-420: Differential geometry

##### Monday 16:00-18:00, P.OF.2058

Announcement for this course.

Slides:

Handouts:

Miscellanea: test problems, exam, etc.

# Math 216: Foundations of algebraic geometry 2007-08 of Prof. Ravi Vakil

Here is the materials of Prof. Vakil ‘s course on Foundations of Algebraic Geometry 07-08 in Stanford University:

http://math.stanford.edu/~vakil/0708-216/

# On the diffetential of a mapping 2

In the case ${\psi : \mathbb{R}^n \rightarrow \mathbb{R}^m}$ case, there is a linear map, which is “linear approximation” of ${\psi}$. In the manifold case, there is a similar linear map, but now it acts between tangent spaces. If ${M}$ and ${N}$ are smooth manifolds and ${\psi \colon M \rightarrow N}$ is a smooth map then for each ${m \in M}$, the map

$\displaystyle d\psi \colon T_mM \rightarrow T_{\psi(m)}N$

is defined by

$\displaystyle d\psi(v)(f) = v(f \circ \psi)$

is called the pushforward. Actually,

$\displaystyle d\psi \colon TM \rightarrow TN.$

Suppose that ${\dim{M} \ge \dim{N}}$ and ${f \colon M \rightarrow N}$ is a differentiable mapping. We have

Definition 1 The mapping ${f}$ is called a trivial fibration (differentiable) on ${N}$ if there exists a differential manifold ${F}$, is called fibre of ${f}$, and a diffeomorphism

$\displaystyle \phi \colon M \rightarrow N \times F$

such that the following diagram is commutative

# Representation of a linear functional

We review results of Haviland and Riesz on the reperesentation of a linear functional.

Definition 1 Let ${X}$ be a subset of ${\mathbb{R}^n}$ and ${C(X)}$ be algebra of continuous functions on ${X}$. A positive linear functional on ${C(X)}$ is a linear functional ${L}$ with ${L(f) \ge 0}$ for all ${f \in C(X)}$ such that ${f(a) \ge 0, \forall a \in X}$.

We recall Haviland’s result in \cite{Marshall2} (also see \cite{Ha1, Ha2}), with ${\mathbb{R}[x_1, \dots, x_n]}$ denotes the ring of real multivariable polynomials:

Theorem 2 (Haviland) For a linear functional ${L: \mathbb{R}[x_1, \dots, x_n]}$ and closed set ${K}$ in ${\mathbb{R}^n}$, the following are equivalent:

1. ${L}$ comes from a Borel measure on ${K}$, i.e., ${\exists}$ a Borel measure ${\mu}$ on ${K}$ such that, ${\forall f \in \mathbb{R}[x_1, \dots, x_n], L(f) = \int f d\mu.}$
2. ${L(f) \ge 0}$ holds for all ${f \in \mathbb{R}[x_1, \dots, x_n]}$ such that ${f \ge 0}$ on ${K}$.

In Haviland’s theorem, a positive linear functional extended from ring of real multivariable polynomials to larger subalgebra and this theorem can be derived as a consequence of the following Riesz Representation Theorem (see \cite[p. 77]{KS}):

Theorem 3 (Riesz Representation Theorem) Let ${X}$ be a locally compact Hausdorff space and let ${L: C_c(X) \rightarrow \mathbb{R}}$ be a positive linear functional. Then there exists a unique Borel measure ${\mu}$ on ${X}$ such that

$\displaystyle L(f) = \int f d\mu, \forall f \in C_c(X).$

${C_c(X)}$ is the algebra of continuous functions with compact support.

# On the differential of a mapping

From on Warner’s book.

Smooth curve on manifold ${M}$:

A ${C^\infty}$ mapping ${\alpha : (a, b) \rightarrow M}$. Let ${t \in (a, b)}$, we define the tangent vector of the curve ${\alpha}$ at ${t}$ is the vector

$\displaystyle d\alpha\bigg(\dfrac{d}{dt}\bigg|_{t = 0}\bigg) \in T_{\alpha(0)}M.$

we apply the formula

$\displaystyle d\psi(v)(g) = v(g \circ \psi),$

where ${g}$ is an any function on ${M}$.

Put ${\psi = \alpha(t)}$ and ${v = \dfrac{d}{dt}\alpha(t)\big|_{t = 0}}$, the above formula implies

$\displaystyle d\alpha(\frac{d}{dt}\big|_{t=0})(f) = (\frac{d}{dt}\big|_{t=0})(f \circ \alpha) = \frac{d}{dt}(f \circ \alpha)\big|_{t=0}.$

This is directional derivative.

# Riesz Representation Theorem – 1

This paragraph we follow the book of Rudin [1]. On linear functionals, there is special relationship between integration and linear functionals. In ${L^1(\mu)}$, a vector space, for any positive measure ${\mu}$, the mapping

$\displaystyle f \mapsto \int_X f d\mu.$

example, let ${C([0,1])}$ be the set of all continuous functions on the unit interval ${I = [0,1]}$. Then

$\displaystyle \Lambda f = \int_0^1 f(x)dx \quad \ (f \in C([0,1])),$

has two properties:

• ${\Lambda(f + g) = \int_0^1[f(x) + g(x)]dx = \int_0^1f(x)dx + \int_0^1g(x)dx = \Lambda(f) + \Lambda(g)}$.
• ${\Lambda(c.f) =\int_0^1cf(x)dx = c\int_0^1f(x)dx= c.\Lambda(f)}$.

so it is a linear functional on ${C([0,1])}$. Moreover, this is a positive linear functional: if ${f \ge 0}$ then ${\Lambda(f) \ge 0}$.

Consider a segment ${(a,b) \subset I}$ and the class of all ${f \in C(I)}$ with ${0 \le f(x) \le 1, \forall x \in I}$ and ${f(x) = 0, \forall x \notin (a,b)}$ or support of ${f}$ is ${(a,b) \subset I}$. So we get ${\Lambda(f) = \int_a^b f(x)dx \le \int_a^bdx = b-a}$. It is important that the length of ${(a,b)}$ related to the values of the functional ${\Lambda}$.

There is an important theorem of F. Riesz, this illustrates to above event

Theorem 1 (F. Riesz) Let $X$ be a closed subset of $\mathbb{R}$. Then every positive linear functional ${\Lambda}$ on ${C(X)}$ there corresponds a finite positive Borel measure ${\mu}$ on ${X}$ such that

$\displaystyle \Lambda(f) = \int_X fd\mu \quad \ (f \in C(X)).$

References

[1] W. Rudin, Real and Complex analysis, Mc.Graw-Hill, 1970.