3 Proof of the Theorem

Talk II: The Nash-Kuiper process for curves

Vincent Borrelli July 2, 2013

The Nash and Kuiper process [1, 2] converts a short embedding of a Rie- mannian manifold into an Euclidean space into a C¹-isometric embedding.

A short embedding is a mapf0 : (M^m, g)−→E^q = (R^q,h., .i) that shortens the length of curves, that is:

Length(f ◦γ) =Length(γ)

for every rectifiable curve γ : [a, b]−→M^m.Since these maps are plentiful, the Nash and Kuiper process must be thought of as a machinery to produce C¹-isometric maps in great profusion.

Here we present the Nash and Kuiper process in the simplest situation where the source manifold is a Riemannian segment or a circle and the target manifold is the Euclidean 2-planeE².

1 One dimensional Isometric Problem

Let us begin with the following elementary problem.

One dimensional Isometric Problem, first version.–Letf0 : [0,1] ^C

∞

−→

E² 'C be a given regular curve and let r : [0,1] −→ R^∗+ be any C^∞ map such that

∀x∈[0,1], r(x)>kf₀⁰(x)k.

Buildf : [0,1] ^C

∞

−→E² such that:

i) ∀x∈[0,1], kf⁰(x)k=r(x) ii) kf−f0k_C0 < δ

where δ >0 is given.

We would like to solve this problem by using the One-dimensional Convex Integration Theory. To do so, we need to find aC^∞ family of loops

h: [0,1] −→ C^∞(R/Z,C) x 7−→ h(x, .) such that

∀x∈[0,1], Z 1

0

h(x, s)ds=f₀⁰(x) and

∀x∈[0,1],∀x∈E/Z, kh(x, s)k=r(x).

We will then simply have to definef by

∀x∈[0,1], f(x) :=f₀(0) + Z x

0

h(s,{N s})ds

whereN ∈N^∗.By the very definition off we will have

∀x∈[0,1], kf⁰(x)k=kh(x,{N x})k=r(x), and by theC⁰-density property, we will also have

kf −f₀k_C0 =O 1

N

.

Therefore, ifN is large enough, f := IC(f0, h, N) will be a solution of our problem.

The choice of h.– In that construction, we have a huge freedom in the choice of the family of loopsh and we must be careful to make a choice as natural as possible. It sounds clear to choose hin the form

h(x, s) :=r(x)e^iψ(x,s)t₀(x) wheret₀(x) := _kf^f000^(x)

0(x)k andψ: [0,1]×E/Z−→R is any map such that Z 1

0

e^iψ(x,s)ds= kf₀⁰(x)k r(x) . But there is no obvious choice forψ.

A choice in the form h(x, s) :=r(x)e^iψ(x,s)t0(x)

For some reasons that will become clear latter (in Talk III), we decide to take

ψ(x, s) :=α(x) cos 2πs whereα(x) must be chosen such that

Z 1 0

eiα(x) cos 2πsds= kf₀⁰(x)k r(x) . Note that

Z 1 0

sin(α(x) cos 2πs)ds= 0 thus

Z 1 0

eiα(x) cos 2πsds= Z 1

0

cos(α(x) cos 2πs)ds

The looph(x, .)is round-trip Definition.– The map

J0: R+ −→ R

α 7−→ 1

π Z π

0

cos(αsins)ds

is called theBessel function of the first kind and of order 0.

Graph of the Bessel functionJ0. Lemma 1.– We have:

J0(α) = Z 1

0

cos(αcos 2πs)ds.

Proof.– Indeed J₀(α) = 1

π Z π

0

cos(αsins)ds

= 2

π Z π/2

0

cos(αsins)ds (since sin(π−s) = sins)

= 2

π Z π/2

0

cos(αsinu)du (u= ^π₂ −s)

= 1

2π Z 2π

0

cos(αsinu)du

= Z ₁

0

cos(αcos 2πs)ds (u= 2πs).

Letκ= 2.404...be the first positive zero ofJ₀.The restricted Bessel function

J₀: [0, κ]−→[0,1]

is 1:1. We denote by J₀⁻¹ its inverse. We must choose α(x) :=J₀⁻¹

kf₀⁰(x)k r(x)

to ensure

Z 1 0

eiα(x) cos 2πsds= kf₀⁰(x)k r(x) .

Solution of the One dimensional Isometric Problem.– Obviously, the map f defined by

∀x∈[0,1], f(x) :=f0(0) + Z x

0

r(x)eiα(x) cos 2πN sds where

e^iθ:= cosθt₀+ sinθn₀, n₀ =it₀ is a solution of problem 1.

The two following pictures shows the image off =IC(f₀, h, N) with x7→f₀(x) :=

x,1

π cos 3

2πx²

and r≡4.

The numberN of oscillation is 9

Beware that f(0) =f0(0)butf(1)6=f0(1) even if it seems that the endpoints coincide on the picture. HereN is 20

2 Isometric Problem for E / Z

We now adress the question of the isometric problem for the circleS¹ =E/Z. Isometric Problem forE/Z.–Letf0 :E/Z ^C

∞

−→E² 'Cbe a given regular closed curve and letr :E/Z−→R^∗+ be any C^∞ map such that

∀x∈[0,1], r(x)>kf₀⁰(x)k.

Buildf :E/Z ^C

∞

−→E² such that:

i) ∀x∈[0,1], kf⁰(x)k=r(x) ii) kf−f0k_C0 < δ

where δ >0 is given.

Not that, even iff₀ is defined over E/Z, the curve x7−→f0(0) +

Z x

0

r(x)eiα(x) cos 2πN sds

is not closed in general. This defect can be easily corrected by the following modification of the convex integration formula:

f(x) :=f₀(0) + Z x

0

r(x)eiα(x) cos 2πN sds−x Z 1

0

r(x)eiα(x) cos 2πN sds.

For short, we write f =IC(ff 0, h, N) with

h(x, s) =r(x)eiα(x) cos 2πs.

LetF =IC(f₀, h, N), from the C⁰-density property we havekF −f₀k_C0 = O _N¹

and since f(x) =F(x)−x(F(1)−F(0)) we also have kf−f₀k_C0 ≤2kF−f₀k_C0 =O

1 N

.

Modified integration convex formula withN = 3, 5,10and20 The speedkf⁰kis only approximately equal to r(x),precisely

∀x∈E/Z,

kf⁰(x)k −r(x)

≤ kF−f0k_C0 =O 1

N

. We shall obtain a mapf∞:E/Z−→E² such that

∀x∈E/Z, kf_∞⁰ (x)k=r(x)

by iteratively applying the modifying convex integration formula. The idea behind the construction off∞ is due to Nash.

The Nash and Kuiper Process.– Let (δ_k)k∈N^∗ such that δ_k ↑ 1 and δk>0.For allk∈N^∗ and for all x∈E/Z we set

r_k²(x) :=r²₀(x) +δ_k r²(x)−r₀²(x) withr0(x) :=kf₀⁰(x)k and then define

f_k :=ICf(fk−1, h_k, N_k)

with

h_k(x, s) :=r_k(x)e^iαk(x) cos 2πs

where

α_k(x) =J₀⁻¹

kf_k−1⁰ (x)k r_k(x)

and e^iθ := cosθtk−1+ sinθnk−1.

Eachf_k has a speed which is approximatelyr_k: kf_k⁰(x)k −r_k(x)

=O 1

N_k

.

Since the sequencer_k(x) is strictly increasing for everyx∈E/Z,the number N_k can be chosen large enough such that

∀x∈E/Z, r_k+1(x)>kf_k(x)k.

From now on, we implicitely assume that every Nk is iteratively chosen such that to fulfill the above inequality. This is crucial to define f_k+1 as IC(ff _k, h_k+1, N_k+1).

Theorem (solution of the Isometric Problem for E/Z).– If the se- quence (δ_k)k∈N^∗ is chosen so that

X pδk−δk−1<+∞

then there exists a sequence of integers(N_k)k∈N^∗ such thatf_k:=IC(ff k−1, h_k, N_k) isC¹ converging toward a C¹ limit f∞ with speed kf_∞⁰ k=r.

The Nash-Kuiper process,f0,f1,f2 andf_∞.

3 Proof of the Theorem

Lemma 2.– The inequality

1 +J₀²(α)−2J0(α) cos(α)≤7(1−J₀²(α))

holds for every α∈[0, κ](recall that κ is the first positive root of J0).

Proof of lemma 2.– Subtracting the right hand side from the left hand side, we rewrite this inequality as

4J₀²(α)−J₀(α) cos(α)−3≤0.

By considering the alternating Taylor series ofJ₀ and cos, we get J₀(α)≤1−α²

4 +α⁴

64 and cos(α)≥1−α² 2 .

Whence

0≤4J0(α)−cos(α)≤3−α² 2 +α⁴

16 ≤3 +α² 2 ,

where the last inequality follows from−^α₂² +^α₁₆⁴ ≤ ^α₂² for all α∈[0, κ]. We can now write

4J₀²(α)−J0(α) cos(α)−3 = J0(α)(4J0(α)−cos(α))−3

≤ (1− α² 4 +α⁴

64)(3 + α² 2 )−3.

Puttingx=α²/4, this last polynomial can be rewritten (1−x+x²

4 )(3 + 2x)−3 = x

2(x−x₁)(x−x₂),

where x₁ < 0 < κ²/4 < x₂. It ensues that this polynomial is negative for

α∈[0, κ].

Lemma 3.– For all x∈E/Z,we have kf_k⁰(x)−f_k−1⁰ (x)k ≤ √

7r(x)p

δk−δk−1

+p

14r(x)kF_k−1−fk−2k^1/2_C0 +kF_k−fk−1k_C0

Proof of lemma 3.– SinceJ0(α_k(x)) =kf_k−1⁰ (x)k/r_k(x), we have

kF_k⁰(x)−f_k−1⁰ (x)k²=r_k²(x) +kf_k−1⁰ (x)k²−2r_k(x)kf_k−1⁰ (x)kcos(α_k(x) cos(2πN x))

=r_k²(x) 1 +J₀(α_k(x))²−2J₀(α_k(x)) cos(α_k(x) cos(2πN x)) . We also have cos(αcos(2πN x)) ≥ cos(α) for every α ∈ [0, κ] ⊂ [0, π]. By using the previous lemma, we get

kF_k⁰(x)−f_k−1⁰ (x)k² ≤ r²_k(x) 1 +J₀(α_k(x))²−2J₀(α_k(x)) cos(α_k(x))

≤ 7r²_k(x)(1−J0(α_k(x))²)

≤ 7(r_k²(x)− kf_k−1⁰ (x)k²).

Since

kf_k−1⁰ (x)k −rk−1(x)

≤ kF_k−1−fk−2k_C0

and

kf_k−1⁰ (x)k+rk−1(x)≤2r(x)

we deduce

kf_k−1⁰ (x)k²−r_k−1² (x)

≤2r(x)kF_k−1−fk−2k_C0. Therefore

kF_k⁰(x)−f_k−1⁰ (x)k² ≤ 7(r²_k(x)−r_k−1² (x) + 2r(x)kF_k−1−fk−2k_C0)

≤ 7(δ_k−δk−1)(r²(x)−r²₀(x)) + 14r(x)kF_k−1−fk−2k_C0

and

kF_k⁰(x)−f_k−1⁰ (x)k ≤√

7r(x)p

δk−δk−1+p

14r(x)kF_k−1−fk−2k^1/2_C0. Finally,

kf_k⁰(x)−f_k−1⁰ (x)k ≤ √

7r(x)p

δk−δk−1

+p

14r(x)kF_k−1−f_k−2k^1/2_C0 +kF_k−f_k−1k_C0

Proof of the theorem (solution of the Isometric Problem forE/Z).–

SincekF_k−fk−1k_C0 =O 1

N_k

, we can choose the sequence (Nk)k∈N^∗ such that

XkF_k−f_k−1k_C0 <+∞

and thus the maps f_ksare C⁰-converging toward aC⁰-mapf∞.If X pδ_k−δk−1<+∞

and if the sequence (N_k)k∈N^∗ is also chosen such that XkF_k−fk−1k^1/2_C0 <+∞

then, by Lemma 3,

Xkf_k⁰ −f_k−1⁰ k_C0 <+∞

and the maps fksare C¹-converging. In particular

∀x∈E/Z, lim

k→+∞kf_k⁰(x)k=kf_∞⁰ (x)k and since

∀x∈E/Z, rk−1(x)<kf_k⁰(x)k< r_k+1(x) we obtain

∀x∈E/Z, kf_∞⁰ (x)k=r(x).

References

[1] N. Kuiper,OnC¹-isometric imbeddings, Indag. Math., 17 (1955), 545- 556.

[2] J. Nash,C¹-isometric imbeddings, Ann. of Math. (2), 60 (1954), 383- 396.