April 7, 2013

http://coursework.tylerlogic.com/courses/math501/homework07

Taken off the homework.

Let X be a parametrization of a surface with normal N. Define Y to be

| (3.1) |

for some positive a.

We know the following to hold

and

resulting in

Let F be the homeomorphism from S to the parallel surface defined by F(p) = p + aN(p). Thus we have

which results in

| (3.2) |

Now because of part (a), we know that the normal field for the parallel surface, call it M, at F(p) is the same as the normal field for S at p, i.e.

| (3.3) |

We will make use of this to determine the Gaussian and mean curvatures.

Gaussian Curvature Using the chain rule with subsection 3.3 gives us that

| (3.4) |

The combination of subsection 3.2 and paragraph 3.4 leads to the parallel surface having a Gaussian curvature, K, of

Mean Curvature Again using the chain rule with subsection 3.3 and employing Lemma A.1 and paragraph 3.4 we find that

Let S

(a) Inverses of isometries are isometries

The proof in Problem section 8 part (a) holds for this when λ is the constant function of 1.

(b) Composition of isometries is an isometry

The proof in Problem section 8 part (b) holds for this when λ

Let φ : S → S be a rotation about the axis of a surface of revolution, S. Because it is simply a rotation, φ is the restriction of some linear map of rotation, R : ℝ

but because A is not dependent on t

resulting in dφ_{p}(v) = Av. Thus since A is orthogonal then for any v,w ∈ T_{p}S

thereby giving us that φ is an isometry.

The drawing in Figure 1 has two points p and q in a planar surface for which there is no curve between them with length equal to the intrinsic distance between the two points. A curve that would potentially have a length of the intrinsic distance would need to go through the hole in the middle of the surface, but it obviously cannot while remaining a curve of the surface.

From our first homework assignment, we know that for any curve α in S with α(a) = p and α(b) = q, L(α)

(a) Coefficients of the first fundamental form.

Since is the map from the domain of X to the domain of Y , then we have

| (7.5) |

for (ϕ,θ) in the domain of X.

(c) Coefficients of the first fundamental form of X

By subsection 7.5 we have

which results in

In order to have M be a conformal map, we must satisfy

From the result of the previous part of the problem we know that

assuming z(0) = 0 and z(θ) > 0 for θ ∈ (0,π∕2).

Let S

(a) Inverses of conformal maps are conformal

Since S

which in turn implies

however, φ ∘ φ^{-1} is the identity map implying that the above equation simplifies to

which gives us what we’re looking for

(b) Composition of conformal maps is conformal

Let p ∈ S

giving us the fact that φ ∘ ψ is conformal since it is a diffeomorphism.

Proof. Let A be the matrix denoted by then

which in turn leads to

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