Throughout this post we shall always work in the smooth category, thus all manifolds, maps, coordinate charts, and functions are assumed to be smooth unless explicitly stated otherwise.

A (real) manifold ${M}$ can be defined in at least two ways. On one hand, one can define the manifold extrinsically, as a subset of some standard space such as a Euclidean space ${{\bf R}^d}$. On the other hand, one can define the manifold intrinsically, as a topological space equipped with an atlas of coordinate charts. The fundamental embedding theorems show that, under reasonable assumptions, the intrinsic and extrinsic approaches give the same classes of manifolds (up to isomorphism in various categories). For instance, we have the following (special case of) the Whitney embedding theorem:

Theorem 1 (Whitney embedding theorem) Let ${M}$ be a compact manifold. Then there exists an embedding ${u: M \rightarrow {\bf R}^d}$ from ${M}$ to a Euclidean space ${{\bf R}^d}$.

In fact, if ${M}$ is ${n}$-dimensional, one can take ${d}$ to equal ${2n}$, which is often best possible (easy examples include the circle ${{\bf R}/{\bf Z}}$ which embeds into ${{\bf R}^2}$ but not ${{\bf R}^1}$, or the Klein bottle that embeds into ${{\bf R}^4}$ but not ${{\bf R}^3}$). One can also relax the compactness hypothesis on ${M}$ to second countability, but we will not pursue this extension here. We give a “cheap” proof of this theorem below the fold which allows one to take ${d}$ equal to ${2n+1}$.

A significant strengthening of the Whitney embedding theorem is (a special case of) the Nash embedding theorem:

Theorem 2 (Nash embedding theorem) Let ${(M,g)}$ be a compact Riemannian manifold. Then there exists a isometric embedding ${u: M \rightarrow {\bf R}^d}$ from ${M}$ to a Euclidean space ${{\bf R}^d}$.

In order to obtain the isometric embedding, the dimension ${d}$ has to be a bit larger than what is needed for the Whitney embedding theorem; in this article of Gunther the bound

$\displaystyle d = \max( n(n+5)/2, n(n+3)/2 + 5) \ \ \ \ \ (1)$

is attained, which I believe is still the record for large ${n}$. (In the converse direction, one cannot do better than ${d = \frac{n(n+1)}{2}}$, basically because this is the number of degrees of freedom in the Riemannian metric ${g}$.) Nash’s original proof of theorem used what is now known as Nash-Moser inverse function theorem, but a subsequent simplification of Gunther allowed one to proceed using just the ordinary inverse function theorem (in Banach spaces).

I recently had the need to invoke the Nash embedding theorem to establish a blowup result for a nonlinear wave equation, which motivated me to go through the proof of the theorem more carefully. Below the fold I give a proof of the theorem that does not attempt to give an optimal value of ${d}$, but which hopefully isolates the main ideas of the argument (as simplified by Gunther). One advantage of not optimising in ${d}$ is that it allows one to freely exploit the very useful tool of pairing together two maps ${u_1: M \rightarrow {\bf R}^{d_1}}$, ${u_2: M \rightarrow {\bf R}^{d_2}}$ to form a combined map ${(u_1,u_2): M \rightarrow {\bf R}^{d_1+d_2}}$ that can be closer to an embedding or an isometric embedding than the original maps ${u_1,u_2}$. This lets one perform a “divide and conquer” strategy in which one first starts with the simpler problem of constructing some “partial” embeddings of ${M}$ and then pairs them together to form a “better” embedding.

In preparing these notes, I found the articles of Deane Yang and of Siyuan Lu to be helpful.

— 1. The Whitney embedding theorem —

To prove the Whitney embedding theorem, we first prove a weaker version in which the embedding is replaced by an immersion:

Theorem 3 (Weak Whitney embedding theorem) Let ${M}$ be a compact manifold. Then there exists an immersion ${u: M \rightarrow {\bf R}^d}$ from ${M}$ to a Euclidean space ${{\bf R}^d}$.

Proof: Our objective is to construct a map ${u: M \rightarrow {\bf R}^d}$ such that the derivatives ${\partial_\alpha u(x)}$ are linearly independent in ${{\bf R}^d}$ for each ${x \in M}$. For any given point ${x_0 \in M}$, we have a coordinate chart ${\phi_{x_0}: U_{x_0} \rightarrow {\bf R}^n}$ from some neighbourhood ${U_{x_0}}$ of ${x_0}$ to ${{\bf R}^n}$. If we set ${u_{x_0}: M \rightarrow {\bf R}^n}$ to be ${\phi_{x_0}}$ multiplied by a suitable cutoff function supported near ${x_0}$, we see that ${u_{x_0}}$ is an immersion in a neighbourhood of ${x_0}$. Pairing together finitely many of the ${u_{x_0}}$ and using compactness, we obtain the claim. $\Box$

Now we upgrade the immersion ${u}$ from the above theorem to an embedding by further use of pairing. First observe that as ${M}$ is smooth and compact, an embedding is nothing more than an immersion that is injective. Let ${u: M \rightarrow {\bf R}^d}$ be an immersion. Let ${\Sigma \subset M \times M}$ be the set of pairs ${(x_1,x_2)}$ of distinct points ${(x_1,x_2)\in M}$ such that ${u(x_1)=u(x_2)}$; note that this set is compact since ${u}$ is an immersion (and so there is no failure of injectivity when ${(x_1,x_2)}$ is near the diagonal). If ${\Sigma}$ is empty then ${u}$ is injective and we are done. If ${\Sigma}$ contains a point ${(x_1,x_2)}$, then by pairing ${u}$ with some scalar function ${\eta: M \rightarrow {\bf R}}$ that separates ${x_1}$ and ${x_2}$, we can replace ${u}$ by another immersion (in one higher dimension ${{\bf R}^{d+1}}$) such that a neighbourhood of ${x_1}$ and a neighbourhood of ${x_2}$ get mapped to disjoint sets, thus effectively removing an open neighbourhood of ${(x_1,x_2)}$ from ${\Sigma}$. Repeating these procedures finitely many times, using the compactness of ${\Sigma}$, we end up with an immersion which is injective, giving the Whitney embedding theorem.

At present, the embedding ${u: M \rightarrow {\bf R}^d}$ of an ${n}$-dimensional compact manifold ${M}$ could be extremely high dimensional. However, if ${d > 2n+1}$, then it is possible to project ${u}$ from ${{\bf R}^d}$ to ${{\bf R}^{d-1}}$ by the random projection trick (discussed in this previous post). Indeed, if one picks a random element ${\omega \in S^{d-1}}$ of the unit sphere, and then lets ${T: {\bf R}^d \rightarrow \omega^\perp}$ be the (random) orthogonal projection to the hyperplane ${\omega^\perp}$ orthogonal to ${\omega}$, then it is geometrically obvious that ${T \circ u: M \rightarrow \omega^\perp}$ will remain an embedding unless ${\omega}$ either is of the form ${\pm \frac{u(x)-u(y)}{\|u(x)-u(y)\|}}$ for some distinct ${x,y \in M}$, or lies in the tangent plane to ${u(M)}$ at ${u(x)}$ for some ${x \in M}$. But the set of all such excluded ${\omega}$ is of dimension at most ${2n}$ (using, for instance, the Hausdorff notion of dimension), and so for ${d > 2n+1}$ almost every ${\omega}$ in ${S^{d-1}}$ will avoid this set. Thus one can use these projections to cut the dimension ${d}$ down by one for ${d>2n+1}$; iterating this observation we can end up with the final value of ${d=2n+1}$ for the Whitney embedding theorem.

Remark 4 The Whitney embedding theorem for ${d=2n}$ is more difficult to prove. Using the random projection trick, one can arrive at an immersion ${u: M \rightarrow {\bf R}^{2n}}$ which is injective except at a finite number of “double points” where ${u(M)}$ meets itself transversally (think of projecting a knot in ${{\bf R}^3}$ randomly down to ${{\bf R}^2}$). One then needs to “push” the double points out of existence using a device known as the “Whitney trick”.

— 2. Reduction to a local isometric embedding theorem —

We now begin the proof of the Nash embedding theorem. In this section we make a series of reductions that reduce the “global” problem of isometric embedding a compact manifold to a “local” problem of turning a near-isometric embedding of a torus into a true isometric embedding.

We first make a convenient (though not absolutely necessary) reduction: in order to prove Theorem 2, it suffices to do so in the case when ${M}$ is a torus ${({\bf R}/{\bf Z})^n}$ (equipped with some metric ${g}$ which is not necessarily flat). Indeed, if ${M}$ is not a torus, we can use the Whitney embedding theorem to embed ${M}$ (non-isometrically) into some Euclidean space ${{\bf R}^m}$, which by rescaling and then quotienting out by ${{\bf Z}^m}$ lets one assume without loss of generality that ${M}$ is some submanifold of a torus ${({\bf R}/{\bf Z})^m}$ equipped with some metric ${g}$. One can then use a smooth version of the Tietze extension theorem to extend the metric ${g}$ smoothly from ${M}$ to all of ${({\bf R}/{\bf Z})^m}$; this extended metric ${\tilde g}$ will remain positive definite in some neighbourhood of ${M}$, so by using a suitable (smooth) partition of unity and taking a convex combination of ${\tilde g}$ with the flat metric on ${({\bf R}/{\bf Z})^m}$, one can find another extension ${g'}$ of ${g}$ to ${({\bf R}/{\bf Z})^m}$ that remains positive definite (and symmetric) on all of ${({\bf R}/{\bf Z})^m}$, giving rise to a Riemannian torus ${(({\bf R}/{\bf Z})^{m}, g')}$. Any isometric embedding of this torus into ${{\bf R}^d}$ will induce an isometric embedding of the original manifold ${M}$, completing the reduction.

The main advantage of this reduction to the torus case is that it gives us a global system of (periodic) coordinates on ${M}$, so that we no longer need to work with local coordinate charts. Also, one can easily use Fourier analysis on the torus to verify the ellipticity properties of the Laplacian that we will need later in the proof. These are however fairly minor conveniences, and it would not be difficult to continue the argument below without having first reduced to the torus case.

Henceforth our manifold ${M}$ is assumed to be the torus ${({\bf R}/{\bf Z})^n}$ equipped with a Riemannian metric ${g = g_{\alpha \beta}}$, where the indices ${\alpha,\beta}$ run from ${1}$ to ${n}$. Our task is to find an injective map ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ which is isometric in the sense that it obeys the system of partial differential equations

$\displaystyle \partial_\alpha u \cdot \partial_\beta u = g_{\alpha \beta}$

for ${\alpha,\beta=1,\dots,n}$, where ${\cdot}$ denotes the usual dot product on ${{\bf R}^d}$. Let us write this equation as

$\displaystyle Q(u) = g$

where ${Q(u)}$ is the symmetric tensor

$\displaystyle Q(u)_{\alpha \beta} := \partial_\alpha u \cdot \partial_\beta u.$

The operator ${Q}$ is a nonlinear differential operator, but it behaves very well with respect to pairing:

$\displaystyle Q( (u_1,u_2) ) = Q(u_1) + Q(u_2). \ \ \ \ \ (2)$

We can use (2) to obtain a number of very useful reductions (at the cost of worsening the eventual value of ${d}$, which as stated in the introduction we will not be attempting to optimise). First we claim that we can drop the injectivity requirement on ${u}$, that is to say it suffices to show that every Riemannian metric ${g}$ on ${({\bf R}/{\bf Z})^n}$ is of the form ${g = Q(u)}$ for some map ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ into some Euclidean space ${{\bf R}^d}$. Indeed, suppose that this were the case. Let ${u_1: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ be any (not necessarily isometric) embedding (the existence of which is guaranteed by the Whitney embedding theorem; alternatively, one can use the usual exponential map ${\theta \mapsto (\cos \theta, \sin \theta)}$ to embed ${{\bf R}/{\bf Z}}$ into ${{\bf R}^2}$). For ${\varepsilon>0}$ small enough, the map ${\varepsilon u_1}$ is short in the sense that ${Q(\varepsilon u_1) < g}$ pointwise in the sense of symmetric tensors (or equivalently, the map ${\varepsilon u_1}$ is a contraction from ${(M,g)}$ to ${{\bf R}^d}$). For such an ${\varepsilon}$, we can write ${g = Q(\varepsilon u_1) + g'}$ for some Riemannian metric ${g'}$. If we then write ${g' = Q(u')}$ for some (not necessarily injective) map ${u': ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^{d'}}$, then from (2) we see that ${g = Q( (\varepsilon u_1, u') )}$; since ${(\varepsilon u_1, u')}$ inherits its injectivity from the component map ${u_1}$, this gives the desired isometric embedding.

Call a metric ${g}$ on ${({\bf R}/{\bf Z})^n}$ good if it is of the form ${Q(u)}$ for some map ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ into a Euclidean space ${{\bf R}^d}$. Our task is now to show that every metric is good; the relation (2) tells us that the sum of any two good metrics is good.

In order to make the local theory work later, it will be convenient to introduce the following notion: a map ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ is said to be free if, for every point ${x \in ({\bf R}/{\bf Z})^n}$, the ${n}$ vectors ${\partial_\alpha u(x)}$, ${\alpha=1,\dots,n}$ and the ${\frac{n(n+1)}{2}}$ vectors ${\partial_\alpha \partial_\beta u(x)}$, ${1 \leq \alpha \leq \beta \leq n}$ are all linearly independent; equivalently, given a further map ${v: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$, there are no dependencies whatsoever between the ${n + \frac{n(n+1)}{2}}$ scalar functions ${\partial_\alpha u \cdot v}$, ${\alpha=1,\dots,n}$ and ${\partial_{\alpha \beta} u \cdot v}$, ${1 \leq \alpha \leq \beta \leq n}$. Clearly, a free map into ${{\bf R}^d}$ is only possible for ${d \geq n + \frac{n(n+1)}{2}}$, and this explains the bulk of the formula (1) of the best known value of ${d}$.

For any natural number ${m}$, the “Veronese embedding” ${\iota: {\bf R}^m \rightarrow {\bf R}^{m + \frac{m(m+1)}{2}}}$ defined by

$\displaystyle \iota(x_1,\dots,x_m) := ( (x_\alpha)_{1 \leq \alpha \leq m}, (x_\alpha x_\beta)_{1 \leq \alpha \leq \beta \leq m} )$

can easily be verified to be free. From this, one can construct a free map ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^{m + \frac{m(m+1)}{2}}}$ by starting with an arbitrary immersion ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^m}$ and composing it with the Veronese embedding (the fact that the composition is free will follow after several applications of the chain rule).

Given a Riemannian metric ${g}$, one can find a free map ${u}$ which is short in the sense that ${Q(u) < g}$, by taking an arbitrary free map and scaling it down by some small scaling factor ${\varepsilon>0}$. This gives us a decomposition

$\displaystyle g = Q(u) + g'$

for some Riemannian metric ${g'}$.

The metric ${Q(u)}$ is clearly good, so by (2) it would suffice to show that ${g'}$ is good. What is easy to show is that ${g'}$ is approximately good:

Proposition 5 Let ${g'}$ be a Riemannian metric on ${({\bf R}/{\bf Z})^n}$. Then there exists a smooth symmetric tensor ${h}$ on ${({\bf R}/{\bf Z})^n}$ with the property that ${g' + \varepsilon^2 h}$ is good for every ${\varepsilon>0}$.

Proof: Roughly speaking, the idea here is to use “tightly wound spirals” to capture various “rank one” components of the metric ${g'}$, the point being that if a map ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ “oscillates” at some high frequency ${\xi \in {\bf R}^n}$ with some “amplitude” ${A}$, then ${Q(u)}$ is approximately equal to the rank one tensor ${A \xi_\alpha \xi_\beta}$. The argument here is related to the technique of convex integration, which among other things leads to one way to establish the ${h}$-principle of Gromov.

By the spectral theorem, every positive definite tensor ${g_{\alpha \beta}}$ can be written as a positive linear combination of symmetric rank one tensors ${v_\alpha v_\beta}$ for some vector ${v \in {\bf R}^n}$. By adding some additional rank one tensors if necessary, one can make this decomposition stable, in the sense that any nearby tensor ${g_{\alpha \beta}}$ is also a positive linear combination of the ${v_\alpha v_\beta}$. One can think of ${v_\alpha}$ as the gradient ${\partial_\alpha \psi}$ of some linear function ${\psi: {\bf R}^n \rightarrow {\bf R}^n}$. Using compactness and a smooth partition of unity, one can then arrive at a decomposition

$\displaystyle g'_{\alpha \beta} = \sum_{i=1}^m (\eta^{(i)})^2 \partial_\alpha \psi^{(i)} \partial_\beta \psi^{(i)}$

for some finite ${m}$, some smooth scalar functions ${\eta^{(i)}, \psi^{(i)}: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}}$ (one can take ${\psi^{(i)}}$ to be linear functions on small coordinate charts, and ${\eta^{(i)}}$ to basically be cutoffs to these charts).

For any ${\varepsilon>0}$ and ${i=1,\dots,m}$, consider the “spiral” map ${u_\varepsilon^{(i)}: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^2}$ defined by

$\displaystyle u_\varepsilon^{(i)} := \varepsilon \eta^{(i)} ( \cos( \frac{\psi^{(i)}}{\varepsilon} ), \sin( \frac{\psi^{(i)}}{\varepsilon} ) )$

Direct computation shows that

$\displaystyle Q(u_\varepsilon^{(i)})_{\alpha \beta} = (\eta^{(i)})^2 \partial_\alpha \psi^{(i)} \partial_\beta \psi^{(i)}$

$\displaystyle + \varepsilon^2 \eta^{(i)}_\alpha \eta^{(i)}_\beta$

and the claim follows by summing in ${i}$ (using (2)) and taking ${h := \sum_{i=1}^m \eta^{(i)}_\alpha \eta^{(i)}_\beta}$. $\Box$

The claim then reduces to the following local (perturbative) statement, that shows that the property of being good is stable around a free map:

Theorem 6 (Local embedding) Let ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ be a free map. Then ${Q(u)+h}$ is good for all symmetric tensors ${h}$ sufficiently close to zero in the ${C^\infty}$ topology.

Indeed, assuming Theorem 6, and with ${h}$ as in Proposition 5, we have ${Q(u) - \varepsilon^2 h}$ good for ${\varepsilon}$ small enough. By (2) and Proposition 5, we then have ${g = (Q(u) - \varepsilon^2 h) + (g' + \varepsilon^2 h)}$ good, as required.

The remaining task is to prove Theorem 6. This is a problem in perturbative PDE, to which we now turn.

— 3. Proof of local embedding —

We are given a free map ${u: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ and a small tensor ${h}$. It will suffice to find a perturbation ${u+v: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ of ${u}$ that solves the PDE

$\displaystyle Q(u+v) = Q(u) + h.$

We can expand the left-hand side and cancel off ${Q(u)}$ to write this as

$\displaystyle L(v)= h - Q(v) \ \ \ \ \ (3)$

where the symmetric tensor-valued first-order linear operator ${L}$ is defined (in terms of the fixed free map ${u}$) as

$\displaystyle L(v)_{\alpha \beta} := \partial_\alpha u \partial_\beta v + \partial_\beta u \partial_\alpha v.$

To exploit the free nature of ${u}$, we would like to write the operator ${L}$ in terms of the inner products ${\partial_\alpha u \cdot v}$ and ${\partial_{\alpha \beta} u \cdot v}$. After some rearranging using the product rule, we arrive at the representation

$\displaystyle L(v) = \partial_\beta ( \partial_\alpha u \cdot v) + \partial_\alpha( \partial_\beta u \cdot v ) - 2 \partial_{\alpha \beta} u \cdot v.$

Among other things, this allows for a way to right-invert the underdetermined linear operator ${L}$. As ${u}$ is free, we can use Cramer’s rule to find smooth maps ${w_{\alpha \beta}: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ for ${\alpha,\beta=1,\dots,n}$ (with ${w_{\alpha \beta} = w_{\beta \alpha}}$) that is dual to ${\partial_{\alpha \beta} u}$ in the sense that

$\displaystyle \partial_\alpha u \cdot w_{\alpha' \beta'} = 0$

$\displaystyle \partial_{\alpha \beta} u \cdot w_{\alpha' \beta'} = \delta_{\alpha \alpha'} \delta_{\beta \beta'} + \delta_{\alpha \beta'} \delta_{\beta \alpha'}$

where ${\delta}$ denotes the Kronecker delta. If one then defines the linear zeroth-order operator ${M}$ from symmetric tensors ${f = f_{\alpha \beta}}$ to maps ${Mf: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ by the formula

$\displaystyle Mf := -\frac{1}{4} \sum_{1 \leq \alpha,\beta \leq n} f_{\alpha \beta} w_{\alpha \beta}$

then direct computation shows that ${LMf = f}$ for any sufficiently regular ${F}$. As a consequence of this, one could try to use the ansatz ${v = Mf}$ and transform the equation (3) to the fixed point equation

$\displaystyle f = h - Q(Mf). \ \ \ \ \ (4)$

One can hope to solve this equation by standard perturbative techniques, such as the inverse function theorem or the contraction mapping theorem, hopefully exploiting the smallness of ${h}$ to obtain the required contraction. Unfortunately we run into a fundamental loss of derivatives problem, in that the quadratic differential operator ${Q}$ loses a degree of regularity, and this loss is not recovered by the operator ${M}$ (which has no smoothing properties).

We know of two ways around this difficulty. The original argument of Nash used what is now known as the Nash-Moser iteration scheme to overcome the loss of derivatives by replacing the simple iterative scheme used in the contraction mapping theorem with a much more rapidly convergent scheme that generalises Newton’s method; see this previous blog post for a similar idea. The other way out, due to Gunther, is to observe that ${Q(v)}$ can be factored as

$\displaystyle Q(v) = L Q_0(v) \ \ \ \ \ (5)$

where ${Q_0}$ is a zeroth order quadratic operator ${Q_0}$, so that (3) can be written instead as

$\displaystyle L( v + Q_0(v) ) = h,$

and using the right-inverse ${M}$, it now suffices to solve the equation

$\displaystyle v = Mh - Q_0(v) \ \ \ \ \ (6)$

(compare with (4)), which can be done perturbatively if ${Q_0}$ is indeed zeroth order (e.g. if it is bounded on Hölder spaces such as ${C^{2,\alpha}}$).

It remains to achieve the desired factoring (5). We can bilinearise ${Q(v)}$ as ${\frac{1}{2} B(v,v)}$, where

$\displaystyle B(v,w)_{\alpha \beta} := \partial_\alpha v \cdot \partial_\beta w + \partial_\beta v \cdot \partial_\alpha w.$

The basic point is that when ${v}$ is much higher frequency than ${w}$, then

$\displaystyle B(v,w)_{\alpha \beta} \approx \partial_\alpha ( v \cdot \partial_\beta w ) + \partial_\beta ( v \cdot \partial_\alpha w ) \ \ \ \ \ (7)$

which can be approximated by ${L}$ applied to some quantity relating to the vector field ${v \cdot \partial_\beta w}$; similarly if ${w}$ is much higher frequency than ${v}$. One can formalise these notions of “much higher frequency” using the machinery of paraproducts, but one can proceed in a slightly more elementary fashion by using the Laplacian operator ${\Delta = \sum_{\alpha=1}^n \partial_{\alpha \alpha}}$ and its (modified) inverse operator ${(1-\Delta)^{-1}}$ (which is easily defined on the torus using the Fourier transform, and has good smoothing properties) as a substitute for the paraproduct calculus. We begin by writing

$\displaystyle Q(v) = \frac{1}{2} B(v,v) = \frac{1}{2} (1 - \Delta)^{-1} ( B(v,v) - \Delta B(v,v) ).$

The dangerous term here is ${\Delta B(v,v)}$. Using the product rule and symmetry, we can write

$\displaystyle \Delta B(v,v) = 2 B(\Delta v, v) + 2 \sum_{\gamma=1}^n B(\partial_\gamma v, \partial_\gamma v).$

The second term will be “lower order” in that it only involves second derivatives of ${v}$, rather than third derivatives. As for the higher order term ${B(\Delta v, v)}$, the main contribution will come from the terms where ${\Delta v}$ is higher frequency than ${v}$ (since the Laplacian accentuates high frequencies and dampens low frequencies, as can be seen by inspecting the Fourier symbol of the Laplacian). As such, we can profitably use the approximation (7) here. Indeed, from the product rule we have

$\displaystyle B(\Delta v, v)_{\alpha \beta} = \partial_\alpha ( \Delta v \cdot \partial_\beta v) + \partial_\beta (\Delta v \cdot \partial_\alpha v) - 2 \Delta v \cdot \partial_{\alpha \beta} v.$

Putting all this together, we obtain the decomposition

$\displaystyle Q(v)_{\alpha \beta} = \partial_\alpha Q_\beta(v) + \partial_\beta Q_\alpha(v) + Q'_{\alpha \beta}(v)$

where

$\displaystyle Q_\alpha(v) := - (1-\Delta)^{-1} ( \Delta v \cdot \partial_\alpha v )$

and

$\displaystyle Q'_{\alpha \beta}(v) := (1-\Delta)^{-1} ( \frac{1}{2} B(v,v)_{\alpha \beta} - \sum_{\gamma=1}^n B(\partial_\gamma v, \partial_\gamma v)_{\alpha \beta} + 2 \Delta v \cdot \partial_{\alpha \beta} v ).$

If we then use Cramer’s rule to create smooth functions ${w_\alpha}$ dual to the ${\partial_\alpha u}$ in the sense that

$\displaystyle \partial_\alpha u \cdot w_{\alpha'} = \delta_{\alpha \alpha'}$

$\displaystyle \partial_{\alpha \beta} u \cdot w_{\alpha' \beta'} = 0$

then we obtain the desired factorisation (5) with

$\displaystyle Q_0(v) := \sum_{\alpha=1}^n Q_\alpha(v) w_\alpha - \frac{1}{4} \sum_{1 \leq \alpha,\beta \leq n} Q'_{\alpha \beta} w_{\alpha \beta}.$

Note that ${Q_0(v)}$ is the smoothing operator ${(1-\Delta)^{-1}}$ applied to quadratic expressions of up to two derivatives of ${v}$. As such, one can show using elliptic (Schauder) estimates to show that ${Q_0}$ is Lipschitz continuous in the Holder spaces ${C^{2,\alpha}(({\bf R}/{\bf Z})^n)}$ for ${0 < \alpha < 1}$ (with the Lipschitz constant being small when ${v}$ has small norm); this together with the contraction mapping theorem in the Banach space ${C^{2,\alpha}}$ is already enough to solve the equation (6) in this space if ${h}$ is small enough. This is not quite enough because we also need ${v}$ to be smooth; but it is possible (using Schauder estimates and product Hölder estimates) to establish bounds of the form

$\displaystyle \| Q_0(v) \|_{C^{k,\alpha}} \lesssim \| v \|_{C^{k,\alpha}} \|v\|_{C^{2,\alpha}} + O_k( \|v\|_{C^{k-1,\alpha}}^2 )$

for any ${k \geq 2}$ (with implied constants depending on ${\alpha,u}$ but independent of ${k}$), which can be used (for ${k}$ small enough) to show that the solution ${v}$ constructed by the contraction mapping principle lies in ${C^{k,\alpha}}$ for any ${k \geq 2}$ (by showing that the iterates used in the construction remain bounded in these norms), and is thus smooth.