The Nash embedding theorem
Khang Manh Huynh
March 13, 2018
Abstract This is an attempt to present an elementary exposition of the Nash embedding theorem 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 manifold (M, g) is a smooth manifold M equipped with a smooth Rieman- nian metric 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 manifolds) 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 dot product. 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 Riemannian manifold (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 torus, 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 space 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 embeddings). 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 submanifolds 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 diffeomorphisms 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 immersion, 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 submanifold of Tm, there is a neighborhood U m containing p, open in T with a diffeomorphism 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 homeomorphism. 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 derivatives, 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 metric space. 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. 7