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Chapter 4: Problem 3

Verify that the surfaces $$ \begin{aligned} &\mathbf{x}(u, v)=(u \cos v, u \sin v, \log u), \quad u>0 \\ &\overline{\mathbf{x}}(u, v)=(u \cos v, u \sin v, v) \end{aligned} $$ have equal Gaussian curvature at the points \(\mathbf{x}(u, v)\) and \(\overline{\mathbf{x}}(u, v)\) but that the mapping \(\overline{\mathbf{x}} \circ \mathbf{x}^{-1}\) is not an isometry. This shows that the "converse" of the Gauss theorem is not true.

Short Answer

Expert verified
Both surfaces have Gaussian curvature K = -1, but the mapping is not an isometry. Thus, the converse of Gauss's Theorem is not true.

Step by step solution

01

Parametric Equations and Surface Definitions

Given surfaces are defined by: \(\mathbf{x}(u, v)=(u \cos v, u \sin v, \log u), \quad u> 0\) and \(\overline{\mathbf{x}}(u, v)=(u \cos v, u \sin v, v).\) Identify and note down these parametric equations for the respective surfaces.
02

Compute the First Fundamental Form

Compute the coefficients of the first fundamental form (E, F, and G) for both surfaces.For \(\mathbf{x}(u, v)\): \(E = 1 + \frac{1}{u^2}, \quad F = 0, \quad G = u^2\).For \(\overline{\mathbf{x}}(u, v)\): \(E = 1, \quad F = 0, \quad G = u^2\).These coefficients are derived from the dot products of the partial derivatives of the parametric equations.
03

Calculate the Second Fundamental Form

Compute the second fundamental form (L, M, N) for both surfaces:For \(\mathbf{x}(u, v)\): \(L = \frac{-1}{u}, \quad M = 0, \quad N = u\).For \(\overline{\mathbf{x}}(u, v)\): \(L = \frac{-1}{u}, \quad M = 0, \quad N = u\).These are computed using the normal vector to the surface which can be found by the cross product of the partial derivatives.
04

Gaussian Curvature Calculation

Calculate the Gaussian curvature (K) using the formula: \(K = \frac{LN - M^2}{EG - F^2}\).Substitute the previously found values for both surfaces.For \(\mathbf{x}(u, v)\), K: \(K = \frac{\left(\frac{-1}{u}\right)u - 0^2 }{\left(1 + \frac{1}{u^2}\right)u^2 - 0^2 } = \frac{-1}{u(1 + \frac{1}{u^2})} = -1\).For \(\overline{\mathbf{x}}(u, v)\), K: \(K = \frac{\left(\frac{-1}{u}\right)u - 0^2}{1*u^2 - 0^2} = \frac{\left(\frac{-1}{u}\right)u}{u^2} = -1\).Both surfaces have the same Gaussian curvature, K = -1.
05

Checking for Isometry

Check whether the mapping \(\overline{\mathbf{x}} \circ \mathbf{x}^{-1}\) preserves the first fundamental form. Since the first fundamental forms are different: For \(\mathbf{x}(u, v)\): \(E = 1 + \frac{1}{u^2}, \quad G = u^2\).For \(\overline{\mathbf{x}}(u, v)\): \(E = 1, \quad G = u^2\).The forms do not match, thus, \(\overline{\mathbf{x}} \circ \mathbf{x}^{-1}\) is not an isometry.
06

Conclusion

This shows that although both surfaces have the same Gaussian curvature, they are not isometric. Therefore, the converse of Gauss鈥檚 Theorem is not true.

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Key Concepts

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

First Fundamental Form
The first fundamental form is crucial in understanding the geometry of surfaces. It provides a way to measure lengths and angles on the surface. For a surface parameterized by \(\mathbf{x}(u, v)\), the first fundamental form is given by:
\[ E = \mathbf{x}_u \cdot \mathbf{x}_u, \quad F = \mathbf{x}_u \cdot \mathbf{x}_v, \quad G = \mathbf{x}_v \cdot \mathbf{x}_v \]

Here, \(\mathbf{x}_u\) and \(\mathbf{x}_v\) are the partial derivatives of \(\mathbf{x}\) with respect to \(u\) and \(v\) respectively. These dot products yield the coefficients of the first fundamental form, which reveal important geometrical properties like distances and angles on the surface.

In our exercise, for \(\mathbf{x}(u, v) = (u \cos v, u \sin v, \log u)\), the coefficients are calculated as follows:
  • \(E = 1 + \frac{1}{u^2}\)
  • \(F = 0\)
  • \(G = u^2\)
For the surface \(\overline{\mathbf{x}}(u, v) = (u \cos v, u \sin v, v)\), they are determined to be:
  • \(E = 1\)
  • \(F = 0\)
  • \

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Most popular questions from this chapter

Show that if \(\mathbf{x}\) is an isothermal parametrization, that is, \(E=G=\lambda(u, v)\) and \(F=0\), then $$ K=-\frac{1}{2 \lambda} \Delta(\log \lambda) $$ where \(\Delta \varphi\) denotes the Laplacian \(\left(\partial^{2} \varphi / \partial u^{2}\right)+\left(\partial^{2} \varphi / \partial v^{2}\right)\) of the function \(\varphi\). Conclude that when \(E=G=\left(u^{2}+v^{2}+c\right)^{-2}\) and \(F=0\), then \(K=\) const. \(=4 c\).

Let \(S\) be an oriented regular surface and let \(\alpha: I \rightarrow S\) be a curve parametrized by arc length. At the point \(p=\alpha(s)\) consider the three unit vectors (the Darboux trihedron) \(T(s)=\alpha^{\prime}(s), N(s)=\) the normal vector to \(S\) at \(p, V(s)=N(s) \wedge T(s)\). Show tbat $$ \begin{aligned} &\frac{d T}{d s}=0+a V+b N \\ &\frac{d V}{d s}=-a T+0+c N \\ &\frac{d N}{d s}=-b T-c V+0 \end{aligned} $$ where \(a=a(s), b=b(s), c=c(s), s \in I\). The above formulas are the analogues of Frenet's formulas for the trihedron \(T, V, N\). To establish the geometrical meaning of the coefficients, prove that a. \(c=-\langle d N / d s, V\rangle\); conclude from this that \(\alpha(I) \subset S\) is a line of curvature if and only if \(c \equiv 0(-c\) is called the geodesic torsion of \(\alpha ; \mathrm{cf} .\) Exercise 19, Sec. 3-2). b. \(b\) is the normal curvature of \(\alpha(I) \subset S\) at \(p\). c. \(a\) is the geodesic curvature of \(\alpha(I) \subset S\) at \(p\).

Show that there exists no surface \(\mathbf{x}(u, v)\) such that \(E=G=1, F=0\) and \(e=1, g=-1, f=0\).

Let \(V\) and \(W\) be ( \(n\)-dimensional) vector spaces with inner products denoted by \(\langle,\),\(rangle and let F: V \rightarrow W\) be a linear map. Prove that the following conditions are equivalent: a. \(\left\langle F\left(v_{1}\right), F\left(v_{2}\right)\right\rangle=\left\langle v_{1}, v_{2}\right\rangle\) for all \(v_{1}, v_{2} \in V\). b. \(|F(v)|=|v|\) for all \(v \in V\). c. If \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) is an orthonormal basis in \(V\), then \(\left\\{F\left(v_{1}\right), \ldots, F\left(v_{n}\right)\right\\}\) is an orthonormal basis in \(W\). d. There exists an orthonormal basis \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) in \(V\) such that \(\left\\{F\left(v_{1}\right), \ldots, F\left(v_{n}\right)\right\\}\) is an orthonormal basis in \(W\). If any of these conditions is satisfied, \(F\) is called a linear isometry of \(V\) into \(W\). (When \(W=V\), a linear isometry is often called an orthogonal transformation.)

A diffeomorphism \(\varphi: S_{1} \rightarrow S_{2}\) is said to be a geodesic mapping if for every geodesic \(C \subset S_{1}\) of \(S_{1}\), the regular curve \(\varphi(C) \subset S_{2}\) is a geodesic of \(S_{2}\). If \(U\) is a neighborhood of \(p \in S_{1}\), then \(\varphi: U \rightarrow S_{2}\) is said to be a local geodesic mapping in \(p\) if there exists a neighborhood \(V\) of \(\varphi(p)\) in \(S_{2}\) such that \(\varphi: U \rightarrow V\) is a geodesic mapping. a. Show that if \(\varphi: S_{1} \rightarrow S_{2}\) is both a geodesic and a conformal mapping, then \(\varphi\) is a similarity; that is, $$ \langle v, w\rangle_{p}=\lambda\left\langle d \varphi_{p}(v), d \varphi_{p}(w)\right\rangle_{\varphi(p)}, \quad p \in S_{1}, v, w \in T_{p}\left(S_{1}\right), $$ where \(\lambda\) is constant. b. Let \(S^{2}=\left\\{(x, y, z) \in R^{3} ; x^{2}+y^{2}+z^{2}=1\right\\}\) be the unit sphere, \(S^{-}=\) \(\left\\{(x, y, z) \in S^{2} ; z<0\right\\}\) be its lower hemisphere, and \(P\) be the plane \(z=-1\). Prove that the map (central projection) \(\varphi: S^{-} \rightarrow P\) which takes a point \(p \in S^{-}\)to the intersection of \(P\) with the line that connects \(p\) to the center of \(S^{2}\) is a geodesic mapping. c. Show that a surface of constant curvature admits a local geodesic mapping into the plane for every \(p \in S\).
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