21 Curve[1NT]

Let \((X,d)\) be a metric space.

Definition 415

[1NV] Let \(I⊆ ℝ\) be an interval.

A continuous function \(𝛾:I→ X\) is called parametric curve, or more simply in the following curve.
If \(𝛾\) is injective, the curve is said to be simple.
If \(𝛾\) is a homeomorphism onto its image, the curve is said to be embedded.
If \(X=ℝ^ n\) and \(𝛾\) is of class \(C^ 1\) and \(𝛾'(t)≠ 0\) for every \(t∈ I\), then \(𝛾\) is called an immersed curve or regular curve.

We will call support or trace the image \(𝛾(I)\) of a curve.

The term arc is also used as a synonym for curve; ¹ this term is mainly used when the curve is not (necessarily) closed.

We postpone the study of closed curves to the next section.

Here are two notions of equivalence of curves. The first was taken from an earlier version of the the lecture notes [ 2 ] .

Definition 416

[1NW] Let \(I,J⊆ {\mathbb {R}}\) be intervals. Let \(𝛾:I→ X\) and \(𝛿:J→ X\) be two curves. We will write \(𝛾∼𝛿\) if there exists an increasing homeomorphism ² \(𝜑:I→ J\) such that \(𝛾=𝛿◦𝜑\).

The second is Definition 7.5.4 from chapter 7 section 6 in the notes [ 2 ] .

Definition 417

[1NX] Let \(I,J⊆ {\mathbb {R}}\) be intervals. Let \(𝛾:I→{\mathbb {R}}^ n\) and \(𝛿:J→{\mathbb {R}}^ n\) be two regular curves. We will write \(𝛾≈𝛿\) if there is a diffeomorphism ³ \(𝜑:I→ J\) monotonic increasing, such that \(𝛾=𝛿◦𝜑\).

E417

[1J8]Prerequisites:416,417.

Show that the relation \(𝛾∼𝛿\) is an equivalence relation.

Show that the relation \(𝛾≈𝛿\) is an equivalence relation.

E417

[1NY]Let \(A⊆ ℝ^ n\) be open and let \(f:A→ℝ\) be a function. Show that \(f\) is continuous if and only if, for each curve \(𝛾:[0,1]→ A\) we have that \(f◦ 𝛾\) is continuous. Hidden solution: [UNACCESSIBLE UUID ’1NZ’]

E417

[1P0] Suppose \(I\) is a closed and bounded interval; use the exercise 3 to show that a simple arc \(𝛾:I\to X\) is a homeomorphism with its image, so the curve is embedded.

Is the result still true if \(I\) is not closed? What if \(I\) is not bounded?

E417

[1P1] Prerequisites:3.Difficulty:*.

Fix a curve \(𝛾:I→ℝ^ n\). We define in the following \(\hat I=\{ t∈ℝ:-t∈ I\} \) and \(\hat𝛾:\hat I→ℝ^ n\) via \(\hat𝛾(t)=𝛾(-t)\).

We want to show that, in certain hypotheses, two curves have the same support if and only if they are equivalent.

Let \(𝛾,𝛿:[0,1]→ℝ^ n\) be simple curves, but not closed, and with the same support. Show that if \(𝛾(0)=𝛿(t)\) then \(t=0\) or \(t=1\). In case \(𝛾(0)=𝛿(0)\), show that \(𝛾∼𝛿\). If instead \(𝛾(0)=𝛿(1)\) then \(\hat𝛾∼𝛿\).
Let \(𝛾,𝛿:[0,1]→ℝ^ n\) be simple immersed curves, but not closed, and with the same support, and let \(𝛾(0)=𝛿(0)\): show that \(𝛾≈𝛿\). If instead \(𝛾(0)=𝛿(1)\) then \(\hat𝛾≈𝛿\).

(For the case of closed curves see 5)

Hidden solution: [UNACCESSIBLE UUID ’1P2’]

E417

[1P3] Show that \([0,1]\) and \([0,1]^ 2\) are not homeomorphic. Hidden solution: [UNACCESSIBLE UUID ’1P4’]

E417

[1P5]Prerequisites:3, 5.Show that you can’t find a curve \(c:[0,1]\to [0,1]^ 2\) continuous and bijective. Therefore a curve \(c:[0,1]\to [0,1]^ 2\) that is continuous and surjective cannot be injective; such as the Peano curve, the Hilbert curve.

Hidden solution: [UNACCESSIBLE UUID ’1P6’]

E417

[1P7] Note:Nice formula taken from [ 67 ] .

Let \(S=S(0,1)⊆ ℝ^ n\) be the unit sphere \(S=\{ x: |x|=1\} \). Let \(v,w∈ S\) with \(v≠ w\) and \(v≠ -w\); let \(T = \arccos ( v⋅ w )\) so that \(T∈(0,𝜋)\); then the geodesic (that is, the arc-parameterized minimal length curve) \(𝛾(t):[0,T]→ S\) connecting \(v\) to \(w\) inside \(S\) is

\[ 𝛾(t)=\frac{\sin \big(T-t\big) }{\sin (T)} v + \frac{\sin \big(t\big) }{\sin (T)} w\quad , \]

and its length is \(T\).
(You may assume that, when \(v⋅ w=0\) that is \(T=𝜋/2\), then the geodesic is \(𝛾(t) = v \cos (t) + w \sin (t)\)). Hidden solution: [UNACCESSIBLE UUID ’1P8’]

[UNACCESSIBLE UUID ’1P9’]