Definition 3. An immersion ℐ : M⟶N is called “free” if ℐ ≡ ℐ∘ϕ for ϕ ∈ Diff(M) implies that ϕ is the identity.

Proposition 3 (see [5] Lemma 1.3.)If ℐ is immersed and ℐ(ϕ(t)) = ℐ(t) for all t and ϕ(t) = t for a t, then ϕ = Id.

Proof. Indeed, it is easily seen that

is closed; and it is also open, since an immersion is also a local diffeomorphism with its image.

Corollary 1 (see [5] Lemma 1.4.)If ℐ is an immersion and there is a x ∈ ℝⁿ s.t. ℐ(t) = x for one and only one t, then ℐ is a free immersion.

Definition 4. The isotropy group (a.k.a. “stabilizer subgroup” or “little group”) $$ is the set of all ϕ ∈ Diff(M) such that ℐ ≡ ℐ∘ϕ; it is a subgroup of Diff(M).

Remark 2. If we reparameterize $$, then $$ changes by conjugation:

Remark 3. If M is orientable, then Diff(M) has a subgroup Diff⁺(M) of orientation preserving diffeomorphisms; for the case of curves, then we obtain that Diff(S¹) has two connected components

where

• (i)

  Diff⁺(S¹) is the family of diffeomorphisms with ϕ^′ > 0, and is a normal subgroup

• (ii)

  Diff⁻(S¹) is the family of diffeomorphisms with ϕ^′ < 0

Definition 7. If the curve c is immersed, we can define the derivation with respect to the arc parameter

Definition 8. We define the tangent vector

Definition 9. The length of the curve c is

Definition 10. We define the integration by arc-parameter of a function g : S¹⟶ℝⁿ along the curve c by

Definition 11 (H). If c is C² regular and immersed, we can define the (mean) curvature H of c as

Definition 12 (N). When the curve c is planar, we can define a normal vector N to the curve, by requiring that |N| = 1, N⊥T and N is rotated π/2 degree anticlockwise with respect to T.

Definition 13. (κ) If c is in ℝ² and C², then we can define a signed scalar curvature κ = 〈H, N〉, so that

Remark 5. Note that T, κ, N, H are geometrical quantities. If ψ ∈ Diff⁺(S¹) and $$, then $$, $$, $$ and $$.

Lemma 1 (constant speed reparameterization). A curve c ∈ C¹ can be reparameterized to $$ using a φ ∈ Diff⁺(S¹) so that $$ where ℓ = len(c)/(2π) is constant.

Proof. For simplicity, we assume that S¹ = [0,2π]. Let L = len(c), let $$. Then ψ : [0,2π]⟶[0,2π] is a diffeomorphism, let φ = ψ⁻¹.

Proposition 4 (angle function and rotation index). If c ∈ C¹ is planar and is immersed, then T = c^′/|c^′| is continuous and |T| = 1, so there exists a continuous function α : ℝ⟶ℝ satisfying

and α(s) is unique, up to adding the constant k2π with k ∈ ℤ. α is called the angle function.

Moreover α(s + 2π) − α(s) = 2πI, where I is an integer, known as rotation index of c. This number is unaltered if c is deformed by a smooth homotopy (Figure 1).

Remark 6. We can use the angle function to compute the scalar curvature κ, that was defined in Definition 13 by ∂_sT = κN, indeed deriving (14) and combining this with

we obtain

Example 4. The following two curves have the same trace, are freely immersed, are smooth, but have rotation indexes 0 and 1.

• (1)

  This immersed closed curve c : [0,2π]⟶ℝ² with components

  $$ (22)

• (2)

  This immersed closed curve c : [0,2π]⟶ℝ² with components

See Figure 4 on the following page.

Definition 16. Suppose c : S¹⟶ℝⁿ is C¹. Let $$. When $$ there are two arcs in S¹ connecting σ to $$. By

we will mean the minimum of the lengths of c when restricted to one of the two arcs connecting σ to $$.

Remark 7. When c is not parameterized by constant velocity, the above may lead to some confusion. The interval $$ in the notation (28) implicitly refers to the choice of arc in S¹ that provides the above minimum. Note that this may not be the shortest arc connecting σ to $$ in S¹. This may happen if the parameterization of c has regions of fast and slow velocity, as in this example.

Example 5. Let $$ be the standard circle, and

(see plot in Figure 5) then smooth out the corners of ϕ so that it becomes a diffeomorphism of S¹; let let θ₀ = 0, θ₁ = π/3 in S¹ ~ ℝ/(2π), then $$, and is given by the arc moving counterclockwise from θ₀ to θ₁, while is given by the arc moving clockwise from θ₀ to θ₁.

Theorem 2. Fix an immersed curve c : S¹⟶ℝⁿ; let

• (i)

  For any θ₀, θ₁ in S¹ such that

  $$ (37)

•  

  the shortest arc connecting them in S¹ is also the arc where

  $$ (38)

•  

  is computed

• (ii)

  For any θ₀, θ₁ in S¹ such that

  $$ (39)

•  

  the arc where

  $$ (40)

•  

  is computed is also the shortest arc connecting them in S¹, whose length is

  $$ (41)

• (iii)

  In any of the above cases,

Proposition 5. Let α(s) be the angle function, for s ∈ J = [0,2π]. The fact that the curve is closed imposes lower bounds on max_J|α^′|.

• (i)

  If the rotation index ℐ of the curve is not zero, then |α(2π) − α(0)| = 2π|I| so necessarily max_J|α^′| ≥ |I|

• (ii)

  If the rotation index of the curve is zero, then necessarily max_J|α^′| ≥ 1/2

Indeed, we can prove that

otherwise, let taking v = (cos  β,   sin  β), we would have for all θ, hence the curve would not be closed.

Definition 17. Given c : S¹⟶ℝ², a C² immersed closed curve, we recall that κ is the scalar curvature of c; we define

Note that since the curve is closed, κ cannot be identically zero.

Remark 8. Note that if we rescale the curve c by a factor λ, then δ_c, τ_c and $$ are multiplied by λ as well. If we rotate or translate c, then δ_c, τ_c and $$ are unaffected. If we reparameterize, then δ_c, τ_c are unchanged, whereas if ψ ∈ Diff(S¹) and $$, we have

In all (with the exception of relation (100) in Lemma 7) following definitions, propositions, and theorems, the formulas are built to be “geometrical”: this means that, if the curves are reparameterized, rescaled, translated or rotated, then the formulas change in predictable ways (as explained above).

This simplifies the proofs: in the proofs we can assume, with no loss of generality, that the curve is parameterized by arc parameter.

Remark 9. Note that δ_c ≤ len(c)/3 for curves of index zero and δ_c ≤ len(c)/(6|I|) for curves of index I ≠ 0.

Proof. We use Proposition 5. The formula in the thesis is invariant for reparameterizations and scaling; we rescale the curve so that len(c) = 2π and reparameterize by arc parameter so that |c^′| ≡ 1. For curves of index zero, the thesis π/(3max_J|κ|) ≤ len(c)/3, that is, π ≤ max_J|κ|len(c); since |c^′| ≡ 1 by (16), this last becomes 1/2 ≤ max_J|α^′| that was proved above. For curves of index I, the thesis π/(3max_J|κ|) ≤ len(c)/(6|I|), that is, 2π|I| ≤ max_J|κ|len(c), then becomes |I| ≤ max_J|α^′| that was proved above.

Proposition 6 (local embedding). Let c : S¹⟶ℝ² a C² immersed curve. Define δ_c as in Definition 17. For any a, b ∈ S¹, let $$; assume that L ≤ 2δ_c, then |c(b) − c(a)| ≥ L/2; so, c_{|[a, b]} is embedded.

Proof. For simplicity, we assume that c is periodically extended to c : ℝ⟶ℝ²; then, we identify the interval in ℝ that is associated to the arc of the curve where the length len c_{|[a, b]} is computed; for simplicity, we call this interval [a, b] again. (if the arc is short enough, then by Theorem 2, no ambiguity is possible).

Using Lemma 1 and Remark 8, assume that |c^′| ≡ ℓ = len(c)/(2π); then, ∂_s = (1/ℓ)∂_θ, so T = c^′/ℓ and

As noted in (16)

so δ_c = ℓπ/(3max|α^′|). Let m = (a + b)/2 be the middle point. Let α(t) be the angle function (14). Up to rotation, suppose c^′(m) = (ℓ, 0), T(m) = (1,0); so, we can assume α(m) = 0. Let L = len c_{|[a, b]} = ℓ(b − a) ≤ 2δ_c. For any θ ∈ [a, b], we have ℓ|θ − m| ≤ δ_c; hence, |θ − m| ≤ π/(3max|α^′|); hence, for all a ≤ θ ≤ b, we have and hence, cos(α(θ)) ≥ 1/2; hence, for a ≤ θ₁ ≤ θ₂ ≤ b, for the abscissa, we can write

Example 6. Let 0 < α < 1 and c(θ) = (θ, |θ|^1+α); then, for θ > 0,

and this meets the y axes for

So, by symmetry,

and at the same time,

Proposition 8. U_τ is also the set of points at distance at most τ from the trace c(S¹).

Proof. Let K = c(S¹) be the trace of the curve (it is a compact subset of ℝ²). We use the distance function d_K : ℝ²⟶ℝ defined as

(for an introduction to this object, see [14] and references therein).

Let x ∈ U_τ, there is a θ, t ∈ V_τ such that

So,

then let $$ be a minimum for

So, clearly,

Vice versa if d_K(x) ≤ τ let $$ be a minimum as above, then geometrical considerations tell that the segment from x to $$ is orthogonal to the tangent $$ at $$.

Lemma 5 (nearby projection). Fix a C^R immersed curve c, with R ≥ 2.

• (1)

  If x ∈ ℝ² and $$ and

  $$ (73)

•  

  then there is an a ∈ ℝ with |a| ≤ d and a σ ∈ S¹ with

  $$ (74)

•  

  such that

  $$ (75)

•  

  Note also that a is uniquely identified by σ.

• (2)

  They are unique in the family of σ, a such that |a| ≤ τ_c and

  $$ (76)

•  

  so we can see σ, a as functions of x, as follows.

• (3)

  Consider $$ and $$ for which

let ε > 0 small such that d + ε < δ_c/4 and let $$ for convenience. There is a choice of function a, φ : B⟶ℝ of class C^R−1 such that for all x ∈ B, and they are unique as specified above.

Note that τ_c/2 < δ_c/4 < τ_c.

Proof. Suppose c is by arc parameter (with no loss of generality as explained in Remark 8); so we write (recalling (32)) |a − b| instead of len c_{|[a, b]}.

• (i)

  Choose $$ as in the statement and let $$, then consider any minimum point $$ of

  $$ (79)

•  

  note that the minimum value has to be less than d; so,

  $$ (80)

•  

  but at the same time (since $$ and $$ are at arc distance at most δ_c), by the previous Proposition 6.

  $$ (81)

•  

  So, combining the two

  $$ (82)

•  

  but 4d < δ_c so $$ is not at extremes. Then, any $$ providing the minimum must be internal in the interval $$: by geometrical reasoning the segment from x to $$ is orthogonal to the curve so there is a a such that

  $$ (83)

• (ii)

  Recall that δ_c/4 ≤ τ_c; the map Φ is injective for $$ and |a| ≤ τ_c, so $$ are unique.

• (iii)

  For any x ∈ B, we have

  $$ (84)
  and since d + ε < δ_c/4, then there is an unique $$ and a with |a| ≤ τ_c such that
  $$ (85)
  and we denote them by σ = φ(x), a = a(x). Moreover, we can invert the function
  $$ (86)
  and write
  $$ (87)
  for x ∈ B. This proves that φ, a ∈ C¹(B).

Proposition 9 (global lifting). Suppose c : S¹⟶ℝ² is a C^R immersed curve and $$ is C^R−1, with R ≥ 2. Fix 0 < τ < δ_c/4. Suppose that we have $$ for all θ. There exists choice of a : S¹⟶ℝ and φ : S¹⟶S¹ such that

with |a(σ)| ≤ τ and holding for all σ. And, they are unique in the class of C^R−1 functions such that |a| ≤ τ_c and

Proof. We just substitute $$ in the previous Lemma. By the second point, we can define functions φ(θ) and a(θ) uniquely as prescribed. By the third point, they are C¹.

Remark 10. Suppose we are given two curves c₁, c₂ and we know that there exists a choice (a, φ) such that

as in (88). If we rotate or translate the two curves, then the above relation will hold, with the same (a, φ). If we rescale the two curves by λ > 0, then the relation will hold with (λa, φ).

If we choose ψ ∈ Diff⁺(S¹) and we reparameterize all curves at the same time by $$, then

holds for

This follows from direct computation and Remark 5.

Example 7. So far so good, but φ may fail to be a diffeomorphism, as in this simple example in Figure 7. where the curve c is blue and the curve $$ is red.

Lemma 6. Let 0 < α < 1. If we have w, v ∈ ℝⁿ such that

then the angle β between v and w satisfies |β| ≤   arcsin(α), and moreover, See Figure 8 on the next page.

Remark 11. Let α > 0. Suppose c₁, c₂, c₃ : S¹⟶ℝ² are C¹ maps and

if ψ ∈ Diff⁺(S¹) and we reparameterize all curves at the same time by $$, then

Similarly, if we rescale, rotate, or translate all curves at the same time.

Lemma 7. Assume all hypotheses in Proposition 9. Assume moreover that τ ≤ τ_c/4 and assume that

then $$ is immersed and φ is a diffeomorphism. Moreover, when c is parameterized at constant speed, we can state that

Proof. We rescale and reparameterize c to arc parameter using a reparameterization ψ, and at the same time, we rescale and reparameterize $$ using the same rescaling and ψ (note that $$ is not necessarily by arc parameter); with no loss in generality, as explained in Remarks 10 and 11.

Let β be the angle between $$ and c^′(σ): by Lemma 6, β = arcsin(α), so it is at most π/6. Let θ = φ(σ); we know that |σ − θ| ≤ 4τ ≤ τ_c; the angle γ between c^′(σ) and c^′(θ) is at most |σ − θ|max|k| so γ ≤ 1/2: so the angle β + γ between $$ and $$ is at most 1/2 + π/6, and this is less than π/2.

Deriving in σ and assuming that c is by arc parameter

where T, N, κ are evaluated at (φ(σ)); then, now if |a| ≤ τ_c then (1 − κa) ≥ 1/2; moreover, by the above reasoning $$, so φ^′(σ) > 0. Moreover, we note that $$, 1/2 ≤ (1 − κa) ≤ 3/2 and cos(1/2 + π/6) ≤ 6/10 to prove (100). Relation (93) tells then that φ will always be a diffeomorphism, for any curve satisfying the hypotheses.

Theorem 3 (representation theorem). Suppose c : S¹⟶ℝ² is a C^R immersed curve and $$ is C^R−1, with R ≥ 2. Define δ_c, τ_c as in Definition 17. Fix 0 < τ ≤ τ_c/4. Suppose that we have $$ and

for all θ.

Then, $$ is immersed, there are φ ∈ Diff⁺(S¹) and a : S¹⟶[−τ, τ] of class C^R−1 such that

with |a(σ)| ≤ τ and holding for all σ.

They are unique in the class of C¹ functions such that |a| ≤ τ_c and

Proof. We can rescale and reparameterize c to arc parameter, and we rescale and reparameterize $$ at the same time ($$ will not be by arc parameter in general); as discussed in Remarks 8, 10, and 11, the hypotheses and theses are unaffected by this action. Then we apply all previous results. Just note that

for any diffeomorphism.

Remark 12. Actually, rerunning on the above proofs with some patience, we can improve the above thesis a bit. We add to the previous theorem these hypotheses: fix 0 < α ≤ 1/2 and then 0 < τ ≤ ατ_c/2 and suppose that we have $$ and

for all θ.

Then, all above thesis hold, moreover, there are two continuous functions f, g : [0, ∞]⟶ℝ (independent on α, τ) with f(0) = g(0) = 1, such that

Proposition 10 (derepresentation). Suppose c : S¹⟶ℝ² is a C² immersed curve and c₁, c₂ : S¹⟶ℝ² are given by tubular coordinates

where i = 1,2 and a₁, a₂ : S¹⟶ℝ are of class C¹; define (where κ is the curvature of c). Then, and (obviously) $$, for all θ.

Proof. We rescale and reparameterize c by arc parameter, and we rescale and reparameterize c₁, c₂ along with c, as explained in Remark 11; this operation is justified by Remarks 8 and 10; in particular, note that α, β are scale invariant; then,

so

Hence,

Example 8. Suppose that, for t near t = 0, we have

such a curve is C² but not C³; then, for t ≥ 0, (that can be easily computed using a standard formula for curvature of planar curves, see Sec. 1.7.1 in [21]).

Choose then a ≡ 1, so

but then and for t > 0,

So,

but $$ for t < 0.

Lemma 8 (local injectivity). Let c be a C² freely immersed planar curve. There exists a r = r_c > 0 such that, if

with where (∂/∂c) is the arc derivation: then a ≡ b and φ is the identity.

Proof. If rescale the curves $$ and the functions a, b by a constant λ > 0, and we rescale the constant r_c by the same constant λ, then all hypotheses and theses are unaffected. So, we can assume with no loss of generality that c has length 2π.

If we reparameterize $$ then (cf the relation (93) in Remark 10) the functions a, b are reparameterized as well; having $$, then $$; and similarly for b, again hypotheses and theses are unaffected.

So, we can assume that c is parameterized by arc parameter with no loss of generality.

Suppose that

with |a| ≤ τ_c and |a^′| ≤ 1/2, then

So,

Suppose moreover,

with |b^′| ≤ 1/2, then and then

By contradiction, we may write

with where φ_n is not the identity: then, when 1/n < τ_c/4, the uniqueness condition (90) is contradicted, so there is a σ_n such that

So, using Theorem 2,

Up to a subsequence, we can assume that $$ and

We know that

up to a subsequence φ_n⟶φ uniformly and $$ uniformly, where φ is a bi-Lip homeomorphism. Then, $$ uniformly, and passing to the limit so φ is a diffeomorphism. Moreover, so φ cannot be the identity; hence, c is not freely immersed.

Lemma 9. Let c be a C² immersed planar curve. There are only finitely many ways in which the curve can represent itself geometrically, that is, finitely many choices a, ϕ

with ϕ ∈ Diff(S¹) and a : S¹⟶ℝ continuous with ‖a‖_∞ ≤ τ_c/2. In particular, there is a ρ_c > 0 such that, if ‖a‖_∞ ≤ ρ_c then a ≡ 0.

Proof. Define δ_c, τ_c as in Definition 17 (see also Proposition 7). Let Φ(s, t) = c(s) + tN(s). Consider the family P of all the pairs a, φ with ‖a‖_∞ ≤ τ_c/2 and a≢0 and

we will prove that there are only finitely many such pairs.

Hence, we will let Pc be smaller than the minimum of ‖a‖_∞ for all such pairs:

In the example, in Figure 10, there are 3 pairs in P.

We have some very strong properties.

• (i)

  If (a₁, φ₁), (a₂, φ₂) ∈ P and there is a $$ s.t. $$, then φ₁ ≡ φ₂ and a₁ ≡ a₂. Indeed, there is a small interval J containing $$, where we can invert the map Φ and

  $$ (148)

•  

  that is, the first component of Φ⁻¹∘c is $$, so φ₁ ≡ φ₂ in J; so, the set {s ∈ S¹ : φ₁(s) = φ₂(s)} is both open and closed. The previous argument also proves that a₁ ≡ a₂.

• (ii)

  Figure 11 on the following page can be used as a visual guide in the following proof.

•  

  Let I₀ ⊂ S¹ be an open interval such that the length of $$ is less than δ_c and more than δ_c/2. As noted in Remark 9, δ_c ≤ len(c)/2. Let U₀ be the image of Φ for s ∈ I₀ and |t| < τ_c. Choose (a_i, φ_i) ∈ P with i ∈ 1,2, let

  $$ (149)

•  

  we will prove that either I₁∩I₂ = ∅, or I₁ = I₂, φ₁ ≡ φ₂, and a₁ ≡ a₂. Assume that $$, let $$, then s₁, s₂ ∈ I₀; so,

  $$ (150)

•  

  using the relation (146),

  $$ (151)

•  

  and the fact that Φ is a diffeomorphism for s ∈ I₀, |t| < τ_c and $$, we obtain that $$ so $$; hence, by the previous point φ₁ ≡ φ₂, a₁ ≡ a₂.

• (iii)

  The differential of Φ (computed using arc-length derivative) is

  $$ (152)
  so the smallest principal value is at least 1/2. We can estimate the length of c for s ∈ I_i to be at least δ_c/4. Hence, there can be only finitely many such intervals.

Example 9. The constants ρ_c in Lemma 9 and r_c in Lemma 8 though cannot be estimated apriori by using differential quantities such as max|κ|. These constants may arbitrarily smaller than the quantity τ_c (defined in Definition 17) that provides the width of the tubular neighborhood (Proposition 7). They really depend on how the curve is drawn.

This is seen in simple examples as follows:

• (i)

  c(z) = εz + z² for z ∈ S¹ ⊂ ℂ, or equivalently c(θ) = (ε  cos(θ) + cos(2θ), ε  sin(θ) + sin(2θ)) for θ ∈ ℝ/(2π) ~ S¹

that are small C^∞ perturbations of the doubly traversed circle c₂ seen in Example 1; see Figure 12. This curve is freely immersed, but it is quite near to the doubly traversed circle that is not freely immersed. For ε small, the curvature κ_c of c is approximatively 1; so, τ_c ~ 1/2, and while

Hence,

so we can use Theorem 3 to express using tubular coordinates when ε < 1/8, with a ~ ε: so $$; similarly, we can represent c(θ + π) using c, so we have $$.

Proposition 11. Consider the set

and this set is open in C^∞(S¹⟶ℝ²).

Proof. This is a simple case of the arguments of Section 3.7.6. Let $$, let $$ and

we know that α < τ, β < 1/3; if c₃ is a smooth curve and satisfies then by the results in Section 3$$.

Proposition 12. Let

be the set of all such a. This set is open.

Proof. The map (φ, a) ↦ (c + aN)∘φ⁻¹ is smooth, and $$ is the projection on the second component of the counterimage of $$ that is open.