20
$\begingroup$

It is easy to check that the immersion $\mathbb{T}^2=\mathbb{S}^1\times \mathbb{S}^1\longrightarrow\mathbb{R}^4$, $(\alpha,\beta)\longmapsto(\cos\alpha,\sin\alpha,\cos\beta,\sin\beta)$ induces the flat metric in the torus, and then, from the mathematical point of view, this immersion is highly preferred to the standard 'doughnut' one into $\mathbb{R}^3$.

I have always wondered whether there exists something similar for higher genus surfaces, for instance a genus 2 surface – double torus. I imagine that, since the hyperbolic plane is the space used as universal cover for the double torus with constant curvature $-1$, one should search some smooth functions from the hyperbolic plane (say, the Poincaré disc in $\mathbb{C}$) to a higher Euclidean or hyperbolic space which are periodic respect to the action of some Fuchsian group. Having these functions, the point is that the immersion induces the $-1$ constant curvature.

Of course, there must be some such embedding, according to the theorems of isometric embeddability. But the aim of this question is to ask whether there exist some nice and natural functions which do the job as in the case of the torus.

Any suggestion is welcome.

$\endgroup$
5
  • 3
    $\begingroup$ I wonder if first embedding a lemniscate and then "fattening" it might lead to a clean embedding? $\endgroup$ Commented Jan 21, 2015 at 12:04
  • 4
    $\begingroup$ Each harmonic form gives you a map to $\mathbb S^1$; you can take a basis for harmonic forms and map your surface in an $n$-torus, the later can be embedded into $\mathbb R^{2{\cdot}n}$ if you want. This embedding is not isometric, but it is kind of canonical; so maybe something can be build on it. $\endgroup$ Commented Jan 21, 2015 at 18:59
  • 1
    $\begingroup$ @AntonPetrunin: this is the Jacobian mapping from algebraic geometry, I think. $\endgroup$
    – Ben McKay
    Commented Dec 20, 2016 at 13:00
  • $\begingroup$ A related question is: is there any explicity (maybe algebraic) isometric embedding of a genus 2 surface endowed with a metric of curvature K = -1? $\endgroup$ Commented Jul 30, 2020 at 14:42
  • $\begingroup$ The sphere and the torus are highly symmetric in a way that higher-genus surfaces are not. Your 4-dimensional embedding of the torus (and the standard embedding of the sphere) are special in the sense that they realize all the symmetries of these surfaces. Since these surfaces have many symmetries, such embeddings are "rare". But, for example, the double torus has not many symmetries and so there are many somehow good embeddings, but none of these stand out as especially symmetric or favorable. $\endgroup$
    – M. Winter
    Commented Jul 30, 2020 at 22:37

1 Answer 1

6
$\begingroup$

This is not really an answer, but rather a longish comment and a suggestion about how one might focus the question a bit better.

First, when one asks for a 'canonical' isometric embedding into some Euclidean space of a genus $2$ surface endowed with a metric of curvature $K=-1$, one might want to consider in what sense even the Clifford torus is 'canonical'.

In fact, there are many flat metrics on the 1-holed torus: If $\Lambda\subset\mathbb{C}$ is a lattice (i.e., a discrete subgroup of rank $2$, say generated by $1$ and $\tau\in\mathbb{C}$ with positive imaginary part), then the standard metric $\mathrm{d}z\circ\mathrm{d}\bar z$ induces a flat metric $g_\Lambda$ on $T = \mathbb{C}/\Lambda$, and these flat metrics are not globally isometric as $\tau$ varies. Nevertheless, as Montiel and Ross showed in this paper, there is a $g_\Lambda$-isometric immersion $f_\Lambda:T\to\mathbb{E}^6$ by first eigenfunctions (that actually maps into a round $5$-sphere) that, moreover, is equivariant with respect to the identity component of the isometry group $G_\Lambda$ of $g_\Lambda$ in the sense that there is an embedding $G_\Lambda\to\mathrm{SO}(6)$ so that $f_\Lambda(g\cdot p) = g\cdot f_\Lambda(p)$ for all $p\in T$. In particular, this isometric embedding is as 'homogeneous' as possible. When $\tau=i$ (so that $\Lambda$ is the square lattice), this reduces to the Clifford torus example that the OP listed.

Of course, these examples are simply quotients of global equivariant isometric immersions of $\mathbb{E}^2$ into $\mathbb{E}^n$ for $n\ge 2$.

Meanwhile, it is well-known that the isometry group of the Poincaré disk is isomorphic to $\mathrm{PSL}(2,\mathbb{R})$ and that this group does not have any nontrivial homomorphism to the isometry group of Euclidean space in any (finite) dimension. Thus, there is no hope to construct an equivariant isometric immersion of the Poincaré disk into any (finite dimensional) Euclidean space, much less to find one that 'closes up' under some Fuchsian subgroup to give a 'canonical' isometric embedding of the $2$-holed torus into Euclidean space.

Thus, one must look elsewhere for a notion of 'canonical isometric embedding'. One possibility one might try to generalize the above flat case would be to ask for an isometric embedding by the eigenfunctions of the Laplace operator associated to a particular eigenvalue. A related (non-existence result) is the following: If $(M^2,ds^2)$ is a surface of constant negative Gauss curvature then there is no nontrivial map $f:M\to S^n\subset\mathbb{E}^{n+1}$ that satisfies $\Delta f = -\lambda f$ for any $\lambda\not=0$. This follows by the same techniques used to prove Theorem 2.3 in this paper of mine. Thus, some other kind of extra system of equations would be needed to determine what you mean by 'canonical'.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .