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The Nash theorem

Khang Manh Huynh

March 13, 2018

Abstract This is an attempt to present an elementary exposition of the for the graduate student who at least knows what a vector field is. We mainly rely on [Tao16] and [How99].

1 Preliminary definitions

Definition 1. A Riemannian (M, g) is a smooth manifold M equipped with a smooth Rieman- nian g. In other words, for any vector fields X, Y (i.e. X, Y ∈ XM), g(X, Y) is a smooth function on M and for any p in M, there is a positive-definite inner product gp : Tp M × Tp M → R such that ∞ g(X, Y)(p) = gp(Xp, Yp). As a consequence, for any f ∈ C (M) : g( f X, Y) = f g(X, Y). An isometric embedding φ from (M1, g1) to (M2, g2) (both Riemannian ) is a smooth embed- ding φ : M1 → M2 that preserves the metric, i.e.

∀X, Y ∈ XM1, ∀p ∈ M1 : (g1)p (Xp, Yp) = (g2)φ(p) (dφ · Xp, dφ · Yp)

We write e = h·, ·i for the Euclidean metric on Rn, formed by the usual Euclidean . Unless indicated otherwise, Rn is always equipped with e and we write Rn for (Rn, e).

Throughout this note, everything we work with is assumed to be smooth unless indicated otherwise. Every metric is a Riemannian metric unless indicated otherwise. Now we can state the main theorem:

Theorem 2 (Nash embedding). Any compact (M, g) without boundary can be isometrically embedded into Rn for some n.

2 Sketch of proof

We will first give a sketch of the proof, leaving the technical details as lemmas to be proven later. The first tool we require is Whitney embedding. For any u ∈ C∞(M, Rn) and v ∈ C∞(M, Rk), we can define u ⊕ v : p 7→ (u(p), v(p)). Glueing chart functions (properly cut off) leads to Whitney.

Lemma 3 (Baby Whitney). Any compact smooth manifold can be smoothly embedded into Rn for some n.

Because of this, and because M is compact, we can embed M into a , which will simplify our calculations greatly.

Lemma 4 (Torus embedding). WLOG, we can assume M = Tm = (R/Z)m, i.e. the m-dimensional torus, equipped with an arbitrary Riemannian metric g.

1 Then, we introduce some definitions:

Definition 5. Let Sym denote the set of symmetric tensors on Tm. In other words, h ∈ Sym when h = m (hαβ)1≤α,β≤m is a smooth function from T into Symm(R) (the set of symmetric m × m matrices). Any m metric g of T can be considered a symmetric tensor and gαβ = g(∂α, ∂β) where (∂α) are the standard coordinate vector fields on Tm. Conversely, if h ∈ Sym and h > 0 (i.e. h(p) is a positive matrix ∀p), h is a metric. ∞ m n m n Let Map = ∪n∈NC (T , R ) be the set of all maps from T into some Euclidean R . We define the function Q : Map → Sym such that for any u ∈ Map:

Q(u)αβ = ∂αu, ∂βu

where h·, ·i is the usual Euclidean dot product. We note that Q(u) ≥ 0. A metric g on Tm is called good when we can find u ∈ Map such that Q(u) = g (and such u would have to be immersions, though not necessarily ). We write Good for the set of good metrics. Because Q(u ⊕ v) = Q(u) + Q(v), Good is closed under addition. Let Emb ⊂ Map be the set of maps that are embeddings. Nash’s theorem says every metric is in Q(Emb). However, if every metric is good, Nash’s theorem is proven. Indeed, let g be a metric and W ∈ Emb be a Whitney embedding. By rescaling W, WLOG Q(W) < g. Then g = Q(W) + g0 where g0 is a metric, therefore good and g0 = Q(v) for some v ∈ Map. Then W ⊕ v ∈ Emb and Q(W ⊕ v) = g. We’re done.

Before we can prove every metric is good, we can prove it is approximately good by finding a very clever symmetric tensor.

Lemma 6 (Approximation). For any metric g, there is h ∈ Sym such that g + ε2h ∈ Good ∀ε > 0.

This reduces Nash’s theorem into a local perturbation problem, where Nash made his fundamental contribution. We need another definition:

m n Definition 7. (Gromov-Rohklin) An injective smooth map φ : T → R is called free when {∂αφ(p)}α ∪ m {∂α∂βφ(p)}α≤β is linearly independent for any p ∈ T . Note that injective immersions on a compact manifold are automatically embeddings. Let Free ⊂ Emb be the set of embeddings that are free. We can also define free maps on open of Tm and Rm.

C∞ Ck Now we can state the perturbation lemma. Recall that h −→ 0 means h −→ 0 ∀k.

Lemma 8 (Perturbation). Let u ∈ Free. Then for any h ∈ Sym small enough (in the C∞ topology), Q(u) + h ∈ Good.

We use the flexibility of free maps make the symmetric tensor good. Now let us prove Nash’s theorem.

Proof. Let g be any metric on Tm. We first find a free map. m m m ∞ Because T is just like R locally, we can find a finite open cover (Ui) of T with cutoffs ψi ∈ Cc (Ui), 0 ≤ m ψi ≤ 1 such that Vi = int{ψi = 1} also form an open cover of T , and there are φi : Ui → BRm (0, 2) such that φi(Vi) = BRm (0, 1) and dφi · ∂α = ∂α ∀α. m L Obviously ψi and ψiφi can be defined on T by zero extension. Then we define Φ = i (ψi ⊕ ψiφi). So Φ is an injective , therefore embedding from Tm into Rk for some k. k(k+1) k k+ 2 α k α α β Then define the Veronese embedding ιk : R → R , (x )α=1 7→ (x , x x )1≤α≤β≤k and let u = ◦ | ιk Φ. By looking at u Vi , it’s easy to see u is a free embedding. m By rescaling u, and by the compactness of T , we can WLOG assume Q(u)(p) < gp ∀p (as positive matrices). Then g = g0 + Q(u) where g0 is a metric.

2 By approximation, there is h ∈ Sym such that g0 + ε2h ∈ Good ∀ε > 0. By perturbation, for ε small enough, Q(u) − ε2h ∈ Good. So g = (g0 + ε2h) + (Q(u) − ε2h) ∈ Good. So every metric is good and Nash’s theorem is proven.

So we now only need to prove our lemmas.

3 Embedding into the torus

Proof of baby Whitney. Let M be any compact manifold. We can find a finite open cover (Ui) of M with ∞ cutoffs ψi ∈ Cc (Ui), 0 ≤ ψi ≤ 1 such that Vi = int{ψi = 1} also form an open cover of M, and there are diffeomorphisms φi : Ui → BRm (0, 2) such that φi(Vi) = BRm (0, 1). Obviously ψi and ψiφi can be defined L on M by zero extension. Then we define W = i (ψi ⊕ ψiφi). So W is an injective immersion, therefore embedding from M into Rk for some k.

Proof of torus embedding. Let (M, g) be any compact Riemannian manifold. By Whitney, there is an em- bedding W : M → Rm for some m. Because M is compact, by translation and rescaling, WLOG assume W(M) ⊂ (0, 1)m. So WLOG, M ⊂ Tm. We want to extend g from M to Tm. We first do this locally. Let p ∈ M. As M is a regular of Tm, there is a neighborhood U m containing p, open in T with a Φ : U → BRm (0, 1) such that for U = U ∩ M, Φ(U) = m l m−l BRl (0, 1) × {0} where l = dim M. Parametrize R = {(y, z) : y ∈ R , z ∈ R }, then WLOG, via the diffeomorphism, U = {(y, z) : |y|2 + |z|2 < 1} and U = {(y, 0) : |y|2 < 1} where U is equipped with a

metric g. Then simply define g on U: g(y,z)((a1, b1), (a2, b2)) = gy(a1, b1) + hb1, b2i where h·, ·i is the usual Euclidean dot product. Then going back via the diffeomorphism we get g on the original U. Finally we use partition of unity to extend g globally. We can find a finite collection of Riemannian manifolds (U , g )N such that U are open in Tm and cover M, while g are extensions of g| . Let i i i=1 i i M∩Ui m U0 = T \M. Then U0 is open and (U0, g0) is a Riemannian manifold where g0 is the usual Euclidean ( )N N = N metric. Then we can find a partition of unity ψi i=0 subordinate to Ui i=0 and we define g ∑i=0 ψi gi. So we can find an isometric embedding (M, g) → (Tm, g). So we just need to prove Nash’s theorem for the torus.

4 Approximating the metric

Firstly, by the Gram-Schmidt process, for any p, we can find an orthonormal basis for the inner product m  m m α m space TpT , gp . Identify TpT with R . Let v = (v ) ∈ R be any vector, then define the rank-1 tensor T α β m v ⊗ v = vv = (v v )αβ ∈ Symm(R). So there are vectors (vi)i=1 such that gp = ∑i vi ⊗ vi. We want to do this in a stable way around p.

Lemma 9 (Stable decomposition). Let g be a metric on Tm and p ∈ Tm. Then on a neighborhood U of p, we can m find a finite collection of vectors vi in R and smooth functions ai such that ai > 0 and g = ∑i ai (vi ⊗ vi) Proof. The principle is simple, and best illustrated by a detailed example in the case when m = 3. Because  1 0 0  m we are working locally, WLOG p = 0 ∈ U ⊂ R . By a change of basis, WLOG gp =  0 1 0 . 0 0 1

3  a x y  ε By shrinking U if needed, WLOG g|U =  x b z  where |x|, |y|, |z|, |a − 1|, |b − 1|, |c − 1| < 2 and y z c 1 ε ∈ (0, 100 ) is a small positive constant we can choose. 3 +   − Let e1, e2, e3 be the basis vectors of R . Define wi = ei ⊗ ei and wij = ei + ej ⊗ ei + ej , wij =   ei − ej ⊗ ei − ej . These are all the rank-1 tensors we need.  0 0 0   1 0 −1  − ε ε For example, v2 =  0 1 0  and w13 =  0 0 0  . Then because 2 > x, y, z > − 2 on U 0 0 0 −1 0 1 and 1  ε, there are unique smooth functions a1, a2, a3, a12, a13, a23 positive on U such that

− + g = ε ∑ wij + ∑ aijwij + ∑ aiwi on U i

Now we can prove the approximation lemma.

Proof of approximation. For any metric g, there is h ∈ Sym such that g + ε2h ∈ Good ∀ε > 0. 2 Consider the previous lemma. As ai > 0 on U, we can rewrite ai = bi where bi is smooth on U. As Tm is compact, we can find a finite open cover Uj such that the above lemma applies for all Uj and 2  j  j j j g = ∑i bi vi ⊗ vi on U . j j For any u ∈ Map, write Du for (∂αu)α. Because vi are constant vectors, and by shrinking U if needed, 2 j ∞ j j j j  j j we can find functions ui ∈ C U such that Du = vi on U . Then g|Uj = ∑i bi Q(ui ). There exists j j j j ψ j2 a partition of unity ψ subordinate to U . Define ψe = q . Then ( ψe ) is another partition l 2 ∑l (ψ ) 2 j j2  j j j m of unity subordinate to U and by zero extension, ψe g = ∑i ψe bi Q(ui ) on T . Therefore g = 2  j j j m ∑i,j ψe bi Q(ui ) on T . 2 ∞ m To simplify notation, WLOG, we can reindex and rewrite this as g = ∑j ηj Q(uj) where ηj ∈ Cc (T ). This is called the Kuratowski-Weyl-Nash decomposition. Then we introduce the very clever spiral map      uj uj uε = εη cos , sin j j ε ε

ε 2 2  ε 2  L ε 2 L  Then Q(uj ) = ηj Q(uj) + ε Q(ηj) and g = ∑j Q(uj ) − ε Q(ηj) = Q j uj − ε Q j ηj . L  2 Let h = Q j ηj ≥ 0. Then g + ε h ∈ Good ∀ε > 0.

5 Perturbation

Nash’s original method to prove the perturbation lemma is now called the Nash-Moser iteration scheme, which has played an important role in nonlinear PDE. Here, we present Gunther’s¨ approach which uses elliptic operators to reduce the lemma into a usual contraction mapping problem. Although we’ve tried to make the note as self-contained as possible, here we will need to borrow one important tool from the theory of elliptic operators. See, for instance, chapter 6 of [GT01].

4 Theorem 10 (Schauder estimates). For any k ≥ 0 and a ∈ (0, 1), the linear operator ∆ − 1 : Ck+2,a(Tm) → − Ck,a(Tm) is a . In other words, we can define the continuous linear operator (∆ − 1) 1 (also called − − the Bessel potential) from Ck,a(Tm) to Ck+2,a(Tm). We say (∆ − 1) 1 is of order -2. Note that because (∆ − 1) 1 commutes with , the constants for the bounds do not depend on k.

Proof of perturbation. Let u ∈ Free. Then for any h ∈ Sym small enough (in the C∞ topology), Q(u) + h ∈ Good. For the whole problem, u ∈ C∞(Tm, Rn) is fixed. Then we want to find v ∈ C∞(Tm, Rn) such that Q(u + v) = Q(u) + h. To do more algebra with Q, we define its bilinear symmetric form. For any k ≥ 1, define B : C∞(Tm, Rk) × ∞ m k ∞ m k C (T , R ) → Sym such that for any v1, v2 ∈ C (T , R ):

B(v1, v2)αβ = ∂αv1, ∂βv2 + ∂βv1, ∂αv2 = ∂β h∂αv1, v2i + ∂α ∂βv1, v2 − 2 ∂αβv1, v2

Then B(v, v) = 2Q(v, v) and Q(u + v) − Q(u) = Q(v) + B(u, v). So we define L : C∞(Tm, Rn) → Sym, v 7→ B(u, v). We want Q(v) + L(v) = h. It turns out that we can get rid of L by exploring its algebraic properties. The crucial fact is that u is free, so we can freely control quantities like h∂αu, vi and ∂αβu, v by using the duals of ∂αu and ∂αβu. More precisely: Lemma 11. There are smooth functions {w } w Tm → Rn such that when ≤ 0 ≤ 0: α α , αβ α≤β : α β, α β  hwα, ∂α0 ui = δαα0 D E wαβ, ∂α0 β0 u = δαα0 δββ0 D E  wα, ∂α0 β0 u = wαβ, ∂α0 u = 0

Proof. Consider the linearly independent set {∂αu(p)}α ∪ {∂αβu(p)}α≤β. We can give it an order so that, say, ∂1u(p) comes last. Then apply the Gram-Schmidt process to the ordered sequence, and we will have an orthogonal sequence of vectors, in which the last vector is, say, we1(p). Then hwe1(p), ∂1u(p)i 6= 0,

hwe1(p), ∂αu(p)i = 0 ∀α 6= 1 and we1(p), ∂αβu(p) = 0 ∀α, β. As the process is stable, we1 is smooth. we1  Define w1 = . Then hw1, ∂α0 ui = δ1α0 . Do similar things to get all of {wα} , wαβ . hwe1,∂1ui α α≤β Note that these functions depend on u, which is fixed for this problem. Now we can explore L. ∞ m n Lemma 12. There is an operator M : Sym → C (T , R ) such that LM = IdSym. Moreover, M is of order zero, i.e. ∀k ≥ 1, ∀ f ∈ Sym : ||M f ||Ck,a ≤ C|| f ||Ck,a for some constant C = C(u, m, a, k). Proof. Let f ∈ Sym. We want f = LM f = B(u, M f ). This means

fαβ = B(u, M f )αβ = ∂β h∂αu, M f i + ∂α ∂βu, M f − 2 ∂αβu, M f

−1 Then define M f = ∑α≤β fαβwαβ. Then h∂αu, M f i = 0 and for any α ≤ β, −2 ∂αβu, M f = 2 ∂αβu, fαβwαβ = fαβ. Obviously M is of order zero. The next step is Gunther’s¨ contribution. The use of smoothing elliptic operators helps make the operator zero-order (preventing the loss of derivatives phenomenon). ∞ m n ∞ m n Lemma 13 (Gunther’s¨ lemma). There is a zero-order operator Q0 : C (T , R ) → C (T , R ) such that ∞ m n Q = LQ0 on C (T , R ). Moreover, it has a bilinear symmetric form B0 such that B0(v, v) = 2Q0(v) and

5 ∞ m n ∀v1, v2 ∈ C (T , R ), ∀k ≥ 2, ∀a ∈ (0, 1) :

||B0(v1, v2)||Ck,a ≤ C1 (||v1||Ck,a ||v2||C2,a + ||v2||Ck,a ||v1||C2,a ) + C2 (||v1||Ck−1,a ||v2||Ck−1,a ) for some constants C1 = C1(u, m, a), C2 = C2(u, m, a, k). The fact that C1 does not depend on k is crucial.

∞ m n Proof. We want Q(v) = LQ0(v) = B(u, Q0(v)) ∀v ∈ C (T , R ). Then observe that

∆ ∂αv, ∂βv = ∂α∆v, ∂βv + ∂αv, ∂β∆v + 2 ∂αDv, ∂βDv = B(∆v, v)αβ + 2B(Dv, Dv)αβ

Therefore 1 ((∆ − 1) Q(v)) = B(∆v, v) + 2B(Dv, Dv) − B(v, v) αβ αβ αβ 2 αβ 1 = ∂ ∆v, ∂ v + ∂ h∆v, ∂ vi − 2 ∆v, ∂ v + 2B(Dv, Dw) − B(v, v) α β β α αβ αβ 2 αβ

0 −1 0 −1  1  By defining Qα(v) = (∆ − 1) h∆v, ∂αvi and Qαβ(v) = (∆ − 1) 2 ∆v, ∂αβv − 2B(Dv, Dv)αβ + 2 B(v, v)αβ , we have 0 0 0 Q(v)αβ = ∂αQβ(v) + ∂βQα(v) − Qαβ(v)

We want Q(v)αβ = B(u, Q0(v))αβ = ∂α ∂βu, Q0(v) + ∂β h∂αu, Q0(v)i − 2 ∂αβu, Q0(v) . So the obvious definition for Q0 is 0 1 0 Q0(v) = ∑ Qα(v)wα + ∑ Qαβ(v)wαβ α 2 α≤β

0 −1 Now we define the bilinear symmetric forms Bα(v1, v2) = (∆ − 1) (h∆v1, ∂αv2i + h∆v2, ∂αv1i) and   − 1 B0 (v , v ) = (∆ − 1) 1 2 ∆v , ∂ v − 2B(Dv , Dv ) + B(v , v ) αβ 1 2 ∑ i αβ j i j αβ 2 i j αβ {i,j}={1,2}

Then we can define 0 1 0 B0(v1, v2) = ∑ Bα(v1, v2)wα + ∑ Bαβ(v1, v2)wαβ α 2 α≤β

k,a m n To estimate B0, we note the following product estimate on C (T , R ) (by Leibniz’s product rule):

|| hv1, v2i ||Ck,a ≤ C(m, a) (||v1||Ck,a ||v2||C0,a + ||v2||Ck,a ||v1||C0,a ) + C(m, a, k) (||v1||Ck−1,a ||v2||Ck−1,a )

Then due to Schauder estimates, we have the estimate:

0 ||Bα(v1, v2)||Ck,a ≤ C(u, m, a) ||h∆v1, ∂αv2i + h∆v2, ∂αv1i||Ck−2,a

≤ C1 (||v1||Ck,a ||v2||C2,a + ||v2||Ck,a ||v1||C2,a ) + C2 (||v1||Ck−1,a ||v2||Ck−1,a )

0 where C1 = C1(u, m, a), C2 = C2(u, m, a, k). Same story for Bαβ and B0.

To sum up, we have Q = LQ0 and LM = 1. Then the equation h = Q(v) + L(v) becomes

LMh = LQ0v + Lv = L(Q0v + v)

∞ m n So we can forget about L and only need to find v ∈ C (T , R ) such that Mh = Q0v + v.

6 ∞ m n ∞ m n Define the (nonlinear) operator Φh : C (T , R ) → C (T , R ), v 7→ Mh − Q0v. We want to find ∞ a fixed point of Φh by contraction mapping. Unfortunately there isn’t a convenient Banach norm for C . However, if we fix a ∈ (0, 1), by Gunther’s¨ lemma we have the estimate

1 ||Φ (v ) − Φ (v )|| 2,a = ||B (v − v , v ) + B (v , v − v )|| 2.a ≤ R||v − v || 2,a (||v || 2,a + ||v || 2,a ) h 1 h 2 C 2 0 1 2 1 0 2 1 2 C 1 2 C 1 C 2 C

∞ m n where R = R(u, m, a) is a constant, for any v1, v2 ∈ C (T , R ). We also note that ||Φh(0)||C2,a = ||Mh||C2,a ≤ r||h||C2,a where r = r(u, m, a) is a constant. So there is ε = ε(u, m, a) > 0 small enough ε that when ||h||C2,a < 2r and ||v1||C2,a , ||v2||C2,a ≤ ε: 1 ||Φ (v ) − Φ (v )|| 2,a ≤ 2εR||v − v || 2,a ≤ ||v − v || 2,a h 1 h 2 C 1 2 C 2 1 2 C ε 2 ε ||Φ (v )|| 2,a ≤ ||Φ (v ) − Φ (0)|| 2,a + ||Φ (0)|| 2,a ≤ R||v || 2,a ||v || 2,a + ≤ Rε + < ε h 1 C h 1 h C h C 1 C 1 C 2 2

ε 2,a So for any h ∈ Sym such that ||h||C2,a < 2r , there is a unique continuous (in the C topology) extension of Φh that makes it a contraction mapping on cl (BC2,a (0, ε)) (closed ball), which is a complete .  i  Then by the Banach fixed point theorem, the sequence of smooth functions (Φh) (0) eventually con- i∈N 2,a m verges in the C norm to some vh ∈ cl (BC2,a (0, ε)). We wish to show that vh is actually smooth. Since T is compact, by Arzela–Ascoli,` it suffices to show that the sequence is in BCk,a (0, εk) for some εk > 0 for any k ≥ 2. Note that the case k = 2 is already done. So we use induction on k. Assume k > 2 and εk−1 is already found. Then once again we rely on Gunther’s¨ lemma:

|| ( )i+1 ( )|| = || − ( )i ( )|| ≤ || || + || ( )i ( )|| || ( )i ( )|| + || ( )i ( )||2 Φh 0 Ck,a Mh Q0 Φh 0 Ck,a Mh Ck,a C1 Φh 0 Ck,a Φh 0 C2,a C2 Φh 0 Ck−1,a i 2 ≤ ||Mh||Ck,a + εC1|| (Φh) (0)||Ck,a + C2εk−1

where C1 = C1(u, m, a) and C2 = C2(u, m, a, k). Note that C1 does not depend on k, and neither does ε. As 1 we can make ε as small as we want, WLOG εC1 < 2 . Then we note that ||Mh||Ck,a is constant in i, so if we i+1 2 define Ai = || (Φh) (0)||Ck,a , and b = ||Mh||Ck,a + C2εk−1, we have the recurrence inequality:

Ai A + ≤ + b i 1 2

The final step is to show that the nonnegative sequence (Ai) has an upper bound as i → ∞ by the standard bootstrap argument: if Ai ≤ 2b then Ai+1 ≤ 2b, while if Ai > 2b then Ai+1 ≤ Ai. So Ai ≤ max{A0, 2b} ∀i. We are finally done.

References

[GT01] D. Gilbarg and N.S. Trudinger. Elliptic Partial Differential Equations of Second Order. Classics in Mathematics. U.S. Government Printing Office, 2001. ISBN: 9783540411604. URL: https://books. google.com/books?id=eoiGTf4cmhwC. [How99] Ralph Howard. Notes on Gunther’s Method and the Local Version of the Nash Isometric Embedding Theorem. 1999. URL: http://people.math.sc.edu/howard/Notes/nash.pdf.

[Tao16] Terence Tao. Notes on the Nash embedding theorem. 2016. URL: https://terrytao.wordpress. com/2016/05/11/notes-on-the-nash-embedding-theorem.

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