## Math 501: Differential Geometry

Homework 7 <me@tylerlogic.com>

April 7, 2013
http://coursework.tylerlogic.com/courses/math501/homework07

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Taken off the homework.

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3 do Carmo pg 212 problem 11

Let X be a parametrization of a surface with normal N. Define Y to be
 (3.1)

for some positive a.

(a)

We know the following to hold
where (aij) is the matrix representation of the differential of N. section 3.1 yields
and thus combining the two sets of equations above we are left with
With this, we can take the cross product of Y u and Y v revealing that
Now since K = det([dN]) and H = -12tr([dN]) for Gaussian and mean curvatures of X, then

and

resulting in

(b)

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

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

Hence the mean curvature, H, is

(c)

4 do Carmo pg 229 problem 9

Let S1 and S2 be regular surfaces with a conformal maps φ : S1 S2 and ψ : S2 S3.

(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 λφ and λψ are both the constant functions of 1.

5 do Carmo pg 229 problem 10

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 : 3 3, to S. Hence for v S, φ(p) = Ap for some matrix of rotation A. Note that rotational matrices such as A are orthogonal. Thus we have the following for p S and v TpS with some curve α such that α(0) = p and α(0) = v

but because A is not dependent on t

resulting in p(v) = Av. Thus since A is orthogonal then for any v,w TpS

thereby giving us that φ is an isometry.

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(a)

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.

(b)

From our first homework assignment, we know that for any curve α in S with α(a) = p and α(b) = q, L(α)ab ≥|p-q|. Thus because d(p,q) is the infimum a set of the lengths (from a to b) of curves in S which pass through p and q at a and b, respectively, then d(p,q) ≥|p - q|.

(c)

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(a) Coefficients of the first fundamental form.

(b) Relation of M, , X, and Y

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

(d)

In order to have M be a conformal map, we must satisfy
and
indicating that z(θ) must be secθ.

(e)

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).

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Let S1 and S2 be regular surfaces with a conformal maps φ : S1 S2 and ψ : S2 S3.

(a) Inverses of conformal maps are conformal

Since S1 and S2 are diffeomorphic, then TpS1 = Tφ(p)S2 and p = φ(p)-1 for p S2. Thus for vectors v1,v2 TpS2 = Tφ-1(p)S1, the vectors p-1(v1) and p-1(v2) are vectors in Tp(S1). So we have

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 S1 and v1,v2 TpS. Then we have the following
so for λ(p) = λψ(φ(p))λφ(p) we have

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

(c)