First of all, I must clarify at the outset that I am simply asking if there is an alternative way to solve an already known problem. It is known that the answer to my question is yes. The problem is as follows:

Consider $(\mathbb{R}^2, g_{H})$ with the hyperbolic metric, i.e. $$ ds^2 = e^{2y} dx^2 + dy^2 .$$ Does there exist a smooth isometric immersion into $(\mathbb{R}^5, g_{flat})$? In other words we are looking for a smooth map $ u : \mathbb{R}^2 \rightarrow \mathbb{R}^5$ such that $$ du_1(x,y)^2 + du_2(x,y)^2 + \ldots du_5(x,y)^2 = e^{2y} dx^2 + dy^2.$$ My question is the following: Is it conceivable that one can solve this question using the continuity method? More precisely, consider the family of metrics $g_t$ on $\mathbb{R}^2$ given by $$ ds^2 = e^{2ty} dx^2 + dy^2 .$$ Clearly one can solve his pde when $t=0$. It would seem to me naively that the set of $t$ for which the pde is solvable is open (probably one can show this from the Implicit Function theorem). Is it conceivable one can show this set is closed? Typically that is the difficult part of using the continuity method.

Remark: 1) There is an explicit solution to this immersion question. This can be found in the book "Isometric Embedding of Riemannian Manifolds in Euclidean Spaces" by Qing Han and Jia Xing Hong. The desired function $u$ is given explicitly.

2) Just in case the answer to my question is yes, i.e. one can actually use the continuity method to solve the immersion problem in $\mathbb{R}^5$, what are the obstructions to applying that same idea in $\mathbb{R}^4$? I believe it is an open question whether the hyperbolic plane can be immersed in $\mathbb{R}^4$.


I think you are asking the wrong question.

By scaling you have that if you can solve for

$$ \mathrm{d}s^2 = e^{2ty} \mathrm{d}x^2 + \mathrm{d}y^2 $$

for any non-zero $t$ you can also solve for

$$ \mathrm{d}s^2 = e^{2\tilde{t}y} \mathrm{d}x^2 + \mathrm{d}y^2 $$

for any non-zero $\tilde{t}$.

Proof: Let $u'(x,y) = u(\lambda x,\lambda y)$, you have $|\mathrm{d}u'|^2 = e^{2\lambda y} \lambda^2(\mathrm{d} x^2 + \mathrm{d}y^2)$. So you just need to define $\tilde{u} = \lambda^{-1} u'$ to get the desired embedding.

This means that if the solution contains any point $t_0 \in \mathbb{R}\setminus\{0\}$ it will also contain the entirety of $\mathbb{R}\setminus\{0\}$.

So the continuity method doesn't do anything for you: regardless of whether the solution set is closed, what you really need to prove is that there exists some $t_0$ for which the differential equations are solvable. Or, in other words, the hard part lies in showing

an open neighborhood of $\{0\}$ lies in the solution set,

the part which you implicitly considered to be obvious.

Another way to say the above is: the method of continuity is helpful if you want to get to "1" or "$\infty$" knowing that you have "0". If you can already reduce the problem to getting to "$\epsilon$", the method of continuity is not going to tell you much.

Also, the question of immersion of hyperbolic plane in $E^4$ is partially known, at least if you allow piecewise smooth solutions. See

where immersions are constructed which are globally $C^{0,1}$ and real analytic away from some singular lines.

  • $\begingroup$ I see. So in this case, proving openness is the difficult part. Proving the set is closed is obvious, as you have pointed out. Is there a reason why proving openness is hard here? Typically one uses the implicit function theorem on some appropriate Banach space to get openness. Is there some obvious reason that approach can't work? $\endgroup$
    – Ritwik
    Aug 16 '13 at 15:07
  • 1
    $\begingroup$ @Ritwik The devil is in the details! What Banach space are you going to use? Even if you mod out symmetries, because of the exponential weight $e^{2y}$, the class of $C^k$ functions relative to the hyperbolic metrics is quite different from the class relative to the Euclidean metric. (Similarly for Lebesgue type spaces.) Because of the exponential weight you will probably need a function space that can accommodate fast growth at infinity. Now, I've never tried your argument, since I don't even know where to start with the choice of functions spaces. So I cannot tell you why it is hard. $\endgroup$ Aug 16 '13 at 15:33

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