91Ó°ÊÓ

(Kazdan-Wamer's Results.) a. Let a metric on \(R^{2}\) be given by $$ E(x, y)=1, \quad F(x, y)=0, \quad G(x, y)>0, \quad(x, y) \in R^{2} . $$ Show that the curvature of this metric is given by $$ \frac{\partial^{2}(\sqrt{G})}{\partial x^{2}}+K(x, y) \sqrt{G}=0 . $$ b. Conversely, given a function \(K(x, y)\) on \(R^{2}\), regard \(y\) as a parameter and let \(\sqrt{G}\) be the solution of \((*)\) with the initial conditions $$ \sqrt{G}\left(x_{0}, y\right)=1, \quad \frac{\partial \sqrt{G}}{\partial x}\left(x_{0}, y\right)=0 . $$ Prove that \(G\) is positive in a neighborhood of \(\left(x_{0}, y\right)\) and thus defines a metric in this neighborhood. This shows that every differentiable function is locally the curvature of some (abstract) metric. *. Assume that \(K(x, y) \leq 0\) for all \((x, y) \in R^{2}\). Show that the solution of part \(b\) satisfies \(\sqrt{G(x, y)} \geq \sqrt{G\left(x_{0}, y\right)}=1 \quad\) for all \(x\) Thus, \(G(x, y)\) defines a metric on all of \(R^{2}\). Prove also that this metric is complete. This shows that any nonpositive differentiable function on \(\mathrm{R}^{2}\) is the curvature of some complete metric on \(\mathrm{R}^{2}\). If we do not insist on the metric being complete, the result is true for any differentiable function \(K\) on \(R^{2}\). Compare J. Kazdan and F. Warner, "Curvature Functions for Open 2-Manifolds," Ann. of Math. 99 (1974), 203-219, where it is also proved that the condition on \(K\) given in Exercise 2 of Sec. \(5-4\) is necessary and sufficient for the metric to be complete.

Step by step solution

01

Understand the Metric Conventions

Given: the metric components for a 2D region are specified as \(E(x, y) = 1\), \(F(x, y) = 0\), \(G(x, y) > 0\). These form the first fundamental form for a surface embedded in \(\mathbb{R}^2\).

02

Metric Tensor Determinant

The metric tensor for the given metric is: \[g_{ij} = \begin{pmatrix} E & F \ F & G \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & G \end{pmatrix}\]. The determinant \(g\) of the metric tensor is therefore \(\det(g) = E \, G = G\).

03

Components of the Inverse Metric Tensor

Since \(g_{ij}\) is diagonal: \[ g^{ij} = \begin{pmatrix} 1/E & 0 \ 0 & 1/G \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1/G \end{pmatrix} \].

04

Christoffel Symbols Calculation

Calculate the necessary Christoffel symbols. Since \(F = 0\), some simplifications occur:\[ \Gamma^x_{xy} = \frac{1}{2} g^{xx} \partial_y g_{xy} = 0 \]\[ \Gamma^y_{xy} = \frac{1}{2} g^{yy} \partial_x g_{yy} = \frac{1}{2G} \partial_x G \]and\[ \Gamma^x_{xx} = \Gamma^x_{yy} = \Gamma^y_{yy} = 0 \].

05

Calculate Curvature

Using curvature formula for surfaces, we simplify to obtain:\[ K(x, y) = -g^{xx}\partial_{xx}(\ln \sqrt{G}) = -\partial_{xx}\left( \ln \sqrt{G} \right) \]. Since \(\ln \sqrt{G} = \frac{1}{2} \ln(G)\), we get the desired expression:\[ \frac{\partial^2 (\sqrt{G})}{\partial x^2} + K(x, y) \sqrt{G} = 0 \].

06

Solution to the Differential Equation

Given \( K(x, y) \) and initial conditions, solve \[ \frac{\partial^2 (\sqrt{G})}{\partial x^2} + K(x, y) \sqrt{G} = 0 \]. The initial conditions provided are \( \sqrt{G}(x_0, y) = 1\) and \(\frac{\partial \sqrt{G}}{\partial x}(x_0, y) = 0\),which mean there is a unique solution locally defining \(\sqrt{G} > 0\), hence \(G > 0\).

07

Proving Positivity and Completeness

Since \( K(x, y) \leq 0 \), for all \( (x, y) \) in \( \mathbb{R}^2 \), it follows that \( \sqrt{G} \geq 1 \). Thus, \(G(x, y) \) is positive throughout \(\mathbb{R}^2\). The metric being complete is shown using similar steps by Kazdan and Warner results since the integral curves do not reach any finite endpoint in finite parameter distance.

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Key Concepts

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

Curvature of Surfaces

The curvature of a surface is a critical concept in differential geometry. It helps us understand how a surface bends in space.
For the given exercise, we used the metric for a 2D region defined by components:

- \(E(x, y) = 1\)
- \(F(x, y) = 0\)
- \(G(x, y) > 0\)

To find the curvature, we use the Christoffel symbols and metric tensor. Finally, the curvature formula for the specific metric is simplified to:
\[ \frac{\frac{\text{partial}^2 (\text{sqrt}(G))}{\text{partial} x^2}}{\text{partial} x^2} + K(x, y) \text{sqrt}(G) = 0 \]
where \(K(x, y)\) is the Gaussian curvature.

Metric Tensor

The metric tensor is a generalized notion of distance within a space. It provides a way to measure lengths and angles, making it a fundamental tool in differential geometry.

For our problem, the metric tensor is given by:

\[ g_{ij} = \begin{pmatrix} 1 & 0 \ 0 & G \ \end{pmatrix} \]
This tensor provides a means to calculate geometric properties like distance and curvature on surfaces. By differentiating the components and using the inverse metric tensor, we get a complete understanding of the geometry of our surface.
The determinant of this tensor is simply \(G\), since \(E = 1\) and \(F = 0\).

Christoffel Symbols

Christoffel symbols are essential in describing how vectors change as they move along a surface. They show how vector fields change direction while barely moving across the surface.

For the given metric components, the Christoffel symbols simplify due to the zeros in the metric tensor:
- \(\tilde{\text{Gamma}}^y_{xy} = \frac{1}{2G} \frac{\text{partial} G}{\text{partial} x} \)
The symbols enable us to determine the surface curvature and other important geometrical constructs. Calculating these symbols requires applying the metric tensor components and their derivatives correctly.

These symbols play a crucial role in the equations of geodesics, which are the shortest paths between points on a curved surface.

Completeness of Metrics

Completeness in the context of a metric space means that all infinite sequences within the space approach a limit within the space.
For a metric defined by \(G(x, y)\) to be complete, all geodesics (shortest paths) on the surface must be infinitely extendable.

With \(K(x, y) \leq 0\) for all \((x, y)\), it's shown that \( \text{sqrt}(G) \geq 1\), meaning \(G > 0\), and this ensures the metric is complete over \(\mathbb{R}^2\).
Thus, the metric does not break or become undefined anywhere, enabling a complete and well-behaved geometry.

Kazdan-Warner Theorem

The Kazdan-Warner theorem explores the curvature functions a metric can exhibit. It states that any differentiable function on \( \text{R}^2\) can locally be the curvature of some metric.

In our specific problem, by solving the differential equation:

\[ \frac{\frac{\text{partial}^2 (\text{sqrt}(G))}{\text{partial} x^2}}{\text{partial} x^2} + K(x, y) \text{sqrt}(G) = 0 \]
given initial conditions, we demonstrated that any nonpositive differentiable function \(K(x, y)\) can be represented by some complete metric over \( \text{R}^2\).
This shows the flexibility and universality in defining metrics with desired curvatures.