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Let \(F: U \subset R^{2} \rightarrow R^{3}\) be given by $$ \begin{aligned} F(u, v) &=(u \sin \alpha \cos v, u \sin \alpha \sin v, u \cos \alpha), \\ (u, v) \in U &=\left\\{(u, v) \in R^{2} ; u>0\right\\}, \quad \alpha=\text { const. } \end{aligned} $$ a. Prove that \(F\) is a local diffeomorphism of \(U\) onto a cone \(C\) with the vertex at the origin and \(2 \alpha\) as the angle of the vertex. b. Is \(F\) a local isometry?

Short Answer

Expert verified
F is a local diffeomorphism but not a local isometry.

Step by step solution

01

Definition of Local Diffeomorphism

A function is a local diffeomorphism if it is differentiable, one-to-one, onto, and its differential (Jacobian matrix) is invertible at every point in its domain.
02

Write the Jacobian Matrix

The Jacobian matrix of function \(F\) is defined as the matrix of all first-order partial derivatives. Compute the partial derivatives of \(F\) with respect to \(u\) and \(v\):\(F(u, v) = (u \sin \alpha \cos v, u \sin \alpha \sin v, u \cos \alpha)\)
03

Compute Partial Derivatives

Calculate the partial derivatives: \(\frac{\partial F}{\partial u} = (\sin \alpha \cos v, \sin \alpha \sin v, \cos \alpha)\) \(\frac{\partial F}{\partial v} = (-u \sin \alpha \sin v, u \sin \alpha \cos v, 0)\)
04

Form the Jacobian Matrix

After computing the partial derivatives, the Jacobian matrix \(J_F\) is:\[ J_F = \begin{pmatrix} \sin \alpha \cos v & -u \sin \alpha \sin v \ \sin \alpha \sin v & u \sin \alpha \cos v \ \cos \alpha & 0 \end{pmatrix} \]
05

Check the Invertibility of the Jacobian Matrix

A matrix is invertible if its determinant is non-zero. Compute the determinant of \(J_F\):\[ \det(J_F) = \sin \alpha \begin{vmatrix} \sin \alpha \sin v & u \cos v \cos \alpha \ \cos \alpha & 0 \end{vmatrix} = \sin \alpha (u \sin^2 \alpha \cos v) = u \sin^3 \alpha \cos v \]
06

Verify the Local Diffeomorphism Property

Check for non-zero determinant. Since \(u > 0\), \(\alpha\) is constant, \(\sin \alpha \) and \(\cos \alpha\) are non-zero. Hence, the determinant is non-zero and \(J_F\) is invertible, confirming \(F\) is a local diffeomorphism.
07

Identify the Cone

The cone vertex is at the origin and the parameter \(2\alpha\) defines the vertex angle. Therefore, \(F(u, v)\) maps \(U\) onto a cone with vertex origin and angle \(2\alpha\).
08

Determine if F is a Local Isometry

A local isometry preserves the metric. The lengths of vectors are invariants under \(F\). Calculating the first fundamental form gives \[ ds^2 = du^2 + u^2 dv^2 \] for domain \((u,v)\) but \[ dx^2 + dy^2 + dz^2 eq ds^2 \] for \(\mathbb{R^3}\). Therefore, \(F\) is not a local isometry.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Jacobian matrix
To understand if a function is a local diffeomorphism, we must examine its Jacobian matrix. The Jacobian matrix is crucial in differential calculus as it encapsulates all first-order partial derivatives of a function. For a function \(F:\, U \subset \mathbb{R}^{2} \to \mathbb{R}^{3}\), its Jacobian matrix, denoted as \(J_F\), is a 3x2 matrix. Each element of this matrix is a partial derivative of a component of \(F\) with respect to one of its variables. For example, given:

\( F(u, v) = (u \, \sin \alpha \, \cos v, u \, \sin \alpha \, \sin v, u \, \cos \alpha) \),

the Jacobian matrix \(J_F\) would be:

\[ J_F = \begin{pmatrix} \frac{\partial F_1}{\partial u} & \frac{\partial F_1}{\partial v} \ \frac{\partial F_2}{\partial u} & \frac{\partial F_2}{\partial v} \ \frac{\partial F_3}{\partial u} & \frac{\partial F_3}{\partial v} \end{pmatrix} = \begin{pmatrix} \sin \alpha \cos v & -u \, \sin \alpha \, \sin v \ \sin \alpha \sin v & u \, \sin \alpha \, \cos v \ \cos \alpha & 0 \end{pmatrix} \]

This matrix allows us to investigate the invertibility, and thus determine if \(F\) provides a one-to-one smooth mapping suitably.
cone vertex
A cone is a three-dimensional geometric shape that narrows smoothly from a flat base to a point called the vertex. In problems concerning mappings to a cone, the key is to identify the properties and the position of the vertex.

Given the function \(F(u, v)\), the image is that of a cone with the vertex at the origin. The coordinates \((u, v)\) map into a 3D space as \((u \, \sin \alpha \, \cos v, u \, \sin \alpha \, \sin v, u \, \cos \alpha)\), forming a cone as \(u\) (radius) changes while keeping \(\alpha\) (angle) constant. Therefore, the point where all lines of the cone meet—the vertex—is at the origin (0,0,0).

Furthermore, the angle of the vertex can be determined from the parameter \(\alpha\). Since the angle of the cone’s vertex is given by \(2\alpha\), this determines the spread of the cone in 3D space. Understanding this geometric interpretation is crucial in connecting 2D parameter space with three-dimensional surfaces.
local isometry
A local isometry is a function that preserves distances locally. In mathematical terms, it retains the length of vectors under the transformation. To verify if \(F\) is a local isometry, we need to compare the metrics (distances) in the original domain and the image.

The first fundamental form describes the metric on the surface induced by the embedding function. For our example, the form in the parameter space \((u, v)\) is given by:

\[ ds^2 = (du)^2 + (u \, dv)^2 \]

However, when we map this into the 3D space, the length element becomes:

\[ dx^2 + dy^2 + dz^2 = (\sin \alpha \, du \, \cos v - u \, \sin \alpha \, dv \, \sin v )^2 + (\sin \alpha \, du \, \sin v - u \, \sin \alpha \, dv \, \cos v )^2 + (\cos \alpha \, du )^2 \]

These forms do not match precisely, indicating \(F\) does not preserve the local metric. Thus, \(F\) is not a local isometry as the intrinsic geometry scattered in the parameter domain doesn’t retain the same distances after mapping.
partial derivatives
Partial derivatives represent the rate of change of a function with respect to one variable while keeping the others constant. In our case, for a function \(F(u, v)\), partial derivatives help construct the Jacobian matrix.

For \(F(u, v) = (u \, \sin \alpha \, \cos v, u \, \sin \alpha \, \sin v, u \, \cos \alpha)\), we compute the partial derivatives \(\frac{\partial F}{\partial u}\) and \(\frac{\partial F}{\partial v}\). These are:

\( \frac{\partial F}{\partial u} = (\sin \alpha \, \cos v, \sin \alpha \, \sin v, \cos \alpha) \)

and

\( \frac{\partial F}{\partial v} = (-u \, \sin \alpha \, \sin v, u \, \sin \alpha \, \cos v, 0)\)

These derivatives indicate how the function \(F\) changes as we slightly alter \(u\) or \(v\), forming rows in our Jacobian matrix. Explaining this concept clearly helps in understanding the building blocks of differential calculus and how smooth transformations work among spaces.
first fundamental form
The first fundamental form is a key concept in differential geometry, describing the way curves and surfaces are embedded in space. It measures lengths and angles of a parametric surface.

For a function \(F(u, v)\) mapping a 2D plane into 3D space, the first fundamental form \((ds^2)\) gives an intrinsic measure on how lengths transform under \(F\).

Specifically, we have:

\[ ds^2 = Edu^2 + 2Fdudv + Gdv^2 \]

where \(E, F, G\) are elements derived from dot products of partial derivatives. For our example, the first fundamental form is:

\[ ds^2 = (\sum \partial_{u}F \, \cdot \, \partial_{u}F)du^2 + (\sum \partial_{u}F \, \cdot \, \partial_{v}F)dudv + (\sum \partial_{v}F \, \cdot \, \partial_{v}F)dv^2 \]

Here, \( E = 1, \; F = 0, \; G = u^2 \).

Thus:

\[ ds^2 = du^2 + u^2dv^2 \]

This form reflects how distance computations in the parameter space \((u, v)\) compare to Euclidean distances in \(\mathbb{R}^{3}\), integral in deciding properties like local isometries.

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Most popular questions from this chapter

Consider the torus of revolution generated by rotating the circle $$ (x-a)^{2}+z^{2}=r^{2}, y=0, $$ about the \(z\) axis \((a>r>0)\). The parallels generated by the points \((a+r, 0),(a-r, 0),(a, r)\) are called the maximum parallel, the minimum parallel, and the upper parallel, respectively. Check which of these parallels is a. A geodesic. b. An asymptotic curve. c. A line of curvature.

Justify why the surfaces below are not pairwise locally isometric: a. Sphere. b. Cylinder. c. Saddle \(z=x^{2}-y^{2}\).

Let \(S\) be a surface of constant Gaussian curvature. Choose points \(p_{1}, p_{1}^{\prime} \in S\) and let \(V, V^{\prime}\) be convex neighborhoods of \(p_{1}, p_{1}^{\prime}\), respectively. Choose geodesic triangles \(p_{1}, p_{2}, p_{3}\) in \(V\) (geodesic means that the sides \(\widetilde{p_{1} p_{2}, \widetilde{p_{2} p_{3}}, \widetilde{p_{3}} p_{1} \text { are }}\) geodesic arcs) in \(v\) and \(p_{1}^{\prime}, p_{2}^{\prime}, p_{3}^{\prime}\) in \(V^{\prime}\) in such a way that $$ \begin{aligned} &l\left(p_{1}, p_{2}\right)=l\left(p_{1}^{\prime}, p_{2}^{\prime}\right) \\ &l\left(p_{2}, p_{3}\right)=l\left(p_{2}^{\prime}, p_{3}^{\prime}\right) \\ &l\left(p_{3}, p_{1}\right)=l\left(p_{3}^{\prime}, p_{1}^{\prime}\right) \end{aligned} $$ (here \(l\) denotes the length of a geodesic arc). Show that there exists an isometry \(\theta: V \rightarrow V^{\prime}\) which maps the first triangle onto the second. (This is the local version, for surfaces of constant curvature, of the theorem of high school geometry that any two triangles in the plane with equal corresponding sides are congruent.)

Let \(V\) and \(W\) be ( \(n\)-dimensional) vector spaces with inner products \(\langle,\),\(rangle .\) Let \(G: V \rightarrow W\) be a linear map. Prove that the following conditions are equivalent: a. There exists a real constant \(\lambda \neq 0\) such that $$ \left\langle G\left(v_{1}\right), G\left(v_{2}\right)\right\rangle=\lambda^{2}\left\langle v_{1}, v_{2}\right\rangle \quad \text { for all } v_{1}, v_{2} \in V . $$ b. There exists a real constant \(\lambda>0\) such that $$ |G(v)|=\lambda|v| \quad \text { for all } v \in V $$ c. There exists an orthonormal basis \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) of \(V\) such that \(\left\\{G\left(v_{1}\right), \ldots, G\left(v_{n}\right)\right\\}\) is an orthogonal basis of \(W\) and, also, the vectors \(G\left(v_{i}\right), i=1, \ldots, n\), have the same (nonzero) length. If any of these conditions is satisfied, \(G\) is called a linear conformal map (or a similitude).

Let \(S \subset R^{3}\) be a regular, compact, connected, orientable surface which is not homeomorphic to a sphere. Prove that there are points on \(S\) where the Gaussian curvature is positive, negative, and zero.

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