# Isometric embedding as a graph

Question

Let $M$ be a (finite dimensional) smooth manifold and $g,\bar{g}$ be Riemannian metrics on $M$.

1. Under what conditions can we guarantee that there exists another finite dimensional Riemannian manifold $(N,h)$ and a smooth map $f:M\to N$ such that $(M,\bar{g})$ is realised as the graph of $f$ in the product manifold $(M\times N, g\oplus h)$?

To put it another way, when is it possible to write $\bar{g} = g + f^*h$?

2. Is there a way to bound the dimension of $N$ required?

1. Clearly by definition $\bar{g} - g$ must be positive semidefinite for this to work. But we can equally well ask the question in the context of pseudo-Riemannian manifolds where this requirement is unnecessary.

2. There is a trivial lower bound on the dimension of $N$ from the fact that the maximal rank of $f^*h$ (equivalently of $\mathrm{d}f$) is bounded above by the dimension of $N$. So if in local coordinates $\bar{g} - g$ is a rank $k$ matrix somewhere, we know that $N$ has to be at least dimension $k$.

3. The global question aside, what is the correct integrability condition for the local problem? This probably just requires a suitable rephrasing of the question, but I'm having a bit of problem seeing the right geometric picture.

4. The rank 1 case is not too hard (I think). Without loss of too much generality we can let $N$ be $\mathbb{R}$ with the standard metric. Using that the gradient vector field is orthogonal to the level sets, we have additionally an integrability condition (roughly speaking, let $v$ be the smooth vector field of unit eigenvectors of $\bar{g} - g$ relative to $g$ with non-zero eigenvalue $\lambda^2$, then we need the vector field $\lambda v$ to be hypersurface orthogonal (in the metric $g$); this gives necessity. For sufficiency take a hypersurface orthogonal to $\lambda v$ and set $f = 0$ on there, and integrate along $\lambda v$ to get the desired function).

• If I am not mistaken, point 4 also lead to a sufficient condition for the case where $\bar{g} - g$ has rank $k$ everywhere with distinct eigenvalues. If in addition the corresponding fields $\lambda_i v_i$ are holonomic then the same integration argument says that we can take $N = \mathbb{R}^k$. But this chain of conditions is clearly not necessary. – Willie Wong Aug 20 '13 at 20:47

Moving in the distribution of $\mathrm{Ker} (\bar g-g)$ does not change the image point in $N$. So, yes, this distribution has to be integrable.

Now let us do the local problem. Pass to a small open set where $k=\mathop{\rm rank }(\bar g-g)$ at any point. Take a $k$-submanifold $S$ which is transveral to the distribution. The same argument shows that locally we have $f(S)=f(M)$. So the generic $k$-submanifolds of $M$ equipped with metric $\bar g-g$ have to be locally isometric to $f(S)$. This gives some rigidity. If it holds for $\bar g-g$, take $f$ to be the projection map along the distribution to $N=(S,\bar g-g)$.

• P.S. If $k=\dim M$ then by Nash--Kuiper theorem you can do it for any $N$ if $\dim N>k$ and if $\dim N=k$ you have to take $N=(M,\bar g-g)$. – Anton Petrunin Aug 21 '13 at 2:02
• A couple questions: (1) $f^{-1}(q)$ is only a submanifold if $q$ is a regular point, no? Or are you only answering the question when $\bar{g} - g$ has constant rank? Do you have any insights when the rank is not constant? (2) I am sorry that I didn't make it clear, I intend for smooth $C^\infty$ case, and not the $C^1$ case that Nash-Kuiper applies. – Willie Wong Aug 21 '13 at 7:08
• @WillieWong Yes, I consider this case only. I may try to think about general case, but tell me little more; say do you want to construct such $N$ and $f$, or $N$ is given and you want to check if such $f$ exists, or do you want to show that there is no $f$ and $N$... – Anton Petrunin Aug 21 '13 at 14:50
• In the case $k=\dim M$, anyway, you have the standard embedding problem for $(M,\bar g -g)$ --- if you need $C^\infty$-smooth solution use Nash's theorem, it will only make the dimension of $N$ higher. – Anton Petrunin Aug 21 '13 at 14:56
• I had three main questions, and your previous comment partially solved the first. Let me list the questions. (1) Is the construction always possible if we allow sufficiently high dimensions? In the case $k = \dim M$ your previous comment indicates that it is indeed the case, and that is very helpful indeed! I had that part in the back of my mind already but I forgot to mention it in the question statement. I was wondering particular in the case whether non-constant $k$ will introduce additional obstructions. – Willie Wong Aug 21 '13 at 15:03

As Anton points out, in the case that $q = \bar g - g$ has constant rank $k>0$, it is necessary that $K = \ker(q)\subset TM$ be integrable in order that $q = f^*h$ for some smooth map $f:M\to N$, where $h$ is a Riemannian metric on $N$.

For local solvability, these conditions plus the (necessary) condition that $\mathcal{L}_X q = 0$ for all $X$ that are sections of $K$ is sufficient. (This latter integrability condition is the infinitesimal version of the invariance with respect local sections $S$ that Anton mentions.) This is because this latter condition ensures that locally one can write $q$ as a quadratic form in the variables that are constant on the leaves of $K$.

However, these necessary conditions are not sufficient for global solvability. For example, take $M$ to be the $2$-torus $\mathbb{R}^2/\mathbb{Z}^2$ with standard metric $g = dx^2 + dy^2$, let $q = (\cos\theta\ dx + \sin\theta\ dy)^2$ for some constant $\theta$ that is not a rational multiple of $\pi$, and let $\bar g = g + q$. This satisfies all of the local conditions but there is no rank $1$ mapping $f:M\to N$ for a Riemannian manifold $(N,h)$ such that $f^*h = q$ because each nonempty fiber of such an $f$ would have to be dense in $M$.

The case in which $q\ge0$ has variable rank is subtle because it is not at all obvious how to tell when such a $q$ can be written as a sum of squares of $1$-forms (or how many it would need).

• Robert: thanks for your answer! My thoughts (which I described) on the constant rank case is indeed muddled. Your answer cleared it up very nicely. – Willie Wong Aug 21 '13 at 15:22