3
$\begingroup$

I am looking for a reference for the following folklore theorem, which is the Lorentzian analogue of the Bonnet fundamental theorem of surface in Euclidean space, hyperbolic space or $3$d sphere.

If a Coddazzi tensor $b$ on a Riemannian surface $(S,g)$ of curvature $K$ satisfies the Gauss-equation $K=-det b+ k$ then there exits a unique (up to global isometries) isometric embedding of $(S,g)$ in a Lorentzian space $M$ of constant curvature $k$, such that $b$ is the pull-back by the immersion of the shape operator.

In the tome 4 of Spivak, the case when $M$ is Minkowski space is more or less given as an exercise, although it is not stated explicitly.

$\endgroup$

1 Answer 1

4
$\begingroup$

See Section 7 (you have to scroll down a bit to get the corollary for the semi-Riemannian case) in

Christian Bär, Paul Gauduchon, and Andrei Moroianu, Generalized cylinders in semi-Riemannian and Spin geometry, Math. Z. 249 (2005), no. 3, 545--580.

where the theorem is proved for the hypersurface case in arbitrary dimensions. (So not just the 2D version as referred to in the OP.)

A few caveats:

  1. The theorem proves the existence of a local embedding around every point.
  2. By patching you get the existence of a global immersion provided that the hypersurface is simply connected. (This part is standard and the authors just refer to Kobayashi-Nomizu.)
  3. Coming back to the question as asked above, as a side remark note that a closed Riemann surface cannot be isometrically embedded in $\mathbb{R}^{1,2}$, for the reason that any closed hypersurface of $\mathbb{R}^{1,2}$ must have a point where the tangent plane is time-like. The same is true for embeddings in to anti-de-Sitter. This is of course different from the Euclidean setting.
$\endgroup$
1
  • $\begingroup$ Perfect! Thank you very much! \\ Thanks also for the comments --- concerning point 3., usually this theorem is used to isometrically embedded the universal cover of closed surfaces, that can be still very interesting. $\endgroup$ Commented Mar 22, 2016 at 23:12

You must log in to answer this question.

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