4
$\begingroup$

Any two Riemannian metrics can easily be deformed into each other, only obtaining positive definite metrics in between. However, for metrics of other signatures this might not be possible.

Which Lorentzian metrics does the two-torus $\Bbb T^2$ admit, up to continuous deformations via Lorentzian metrics? In other words, I'm interested in $\pi_0(\mathbf{LMet}(\Bbb T^2))$.

Of course there is the standard metric $g:=dx^2-dy^2$, but there are also metrics whose lightcones turn around several times when one wanders along a non-nullhomotopic circle, explicitly given by $g\circ (M\otimes M)$ where $M\in C(\Bbb T^2, End(T\Bbb T^2))$, and hence can not be deformed into the standard metric, thus yielding an injection $\langle T^2, SO(2)\rangle \to \pi_0(\mathbf{LMet}(\Bbb T^2))$. Are these already all different metrics?

$\endgroup$

1 Answer 1

9
$\begingroup$

Yes those are already all different metrics. Since the tangent bundle of the 2-torus $\mathbb{T}^2$ is trivial you have a correspondence between the set of homotopy classes of maps $\mathbb{T}^2\to\mathbb{RP}^1$ and $\pi_0(\mathbf{LMet}(\mathbb{T}^2))$. Since $\mathbb{T}^2=S^1\times S^1$ and $\mathbb{RP}^1\simeq S^1$ are $K(\mathbb{Z}^2,1)$ and $K(\mathbb{Z},1)$ respectively, you get $\pi_0(\mathbf{LMet}(\mathbb{T}^2))\simeq\hom(\mathbb{Z}^2,\mathbb{Z})$.

In general there is a bijection between the set of path components of metrics of signature $(p,q)$ on a given vector bundle $E\to X$ and the set of homotopy classes of rank $p$ subbundles of $E$. When $E$ is not trivial the classification is a difficult task. For $p=1$ interesting results are in this paper by Ulrich Koschorke, "Homotopy classification of line fields and of Lorentz metrics on closed manifolds", Math. Proc. Camb. Phil. Soc. (2002), 132, issue 02, p.281-300.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.