Math 501: Differential Geometry

Homework 3

Lawrence Tyler Rush

<me@tylerlogic.com>

February 10, 2013

http://coursework.tylerlogic.com/courses/math501/homework03

2 do Carmo pg 65 exercise 1

The set S is the cylinder {(x,y,z) ∈ ℝ³|x² + y² = 1}. Define f : ℝ³ → ℝ by the mapping

(x,y,z) ↦→ x2 + y2.

With this definition, f^-1(1) is exactly S since S is the set of all points of ℝ which have x² + y² = 1. Since the differential df_{(x₀,y₀,z₀)} is the left multiplication by the matrix (2x₀,2y₀,0), then the only point critical point of the differential is (0,0,0). Putting the previous two ideas together then gives us that 1 is a regular point of the mapping f since f^-1(1) = S doesn’t contain the origin of ℝ³. Hence, because f is smooth, then do Carmo’s Proposition 2 of Section 2-2 informs us that S is regular.

3 do Carmo pg 65 exercise 2

Denote the open disc, {(x,y,z) ∈ ℝ³ | z = 0 and x² + y² < 1}, by D and the closed disc {(x,y,z) ∈ ℝ³ | z = 0 and x² + y² ≤ 1}, by D.

Closed Disc The closed disc D is not a regular surface because if it were, then by the definition of a regular surface there would exist a neighborhood V ⊂ ℝ³ of (0,1) ∈D such that D∩V is homeomorphic to some open set of ℝ², say U. However, the point (0,1) is in D ∩ V but is not an interior point, which contradicts the fact that U, being open in ℝ², contains only interior points.

Open Disc Setting U ⊂ ℝ² to the open set {(x,y) ∈ ℝ² | x² + y² < 1} and defining f : U → ℝ by f(u,v) = 0, then D is simply the graph {(u,v,f(u,v)) | (u,v) ∈ U} of ℝ³. So D is a regular surface by do Carmo’s first proposition of chapter two.

4 do Carmo pg 66 exercise 11

Let S be the set {(x,y,z) ∈ ℝ³ | z = x² - y²}.

Regular Surface Define f : ℝ³ → ℝ by f(x,y,z) = x² -y² -z. With this definition, f is smooth and df_p for p = (x₀,y₀,z₀) is the linear map corresponding to the matrix (2x₀,-2y₀,-1). Since the third parameter of the matrix is -1, then df_p is surjective. Since p was arbitrary, we know the differential to be surjective at all points of ℝ³, and thus all points of ℝ are regular values of the map f. Hence by do Carmo’s second proposition of section 2-2, f^-1(0) is a regular surface, but f^-1(0) is all points that satify 0 = x² - y² - z, which is exactly the set S. So S is regular.