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}.

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MATH-F-420: Differential geometry of Verbitsky

Misha Verbitsky

Université Libre de Bruxelles

MATH-F-420: Differential geometry

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

Announcement for this course.

Slides:

Handouts:

Miscellanea: test problems, exam, etc.

 

Source: http://verbit.ru/ULB/GEOM-2015/

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

sodo

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.

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.