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Let $(M,g)$ be a 2 dimensional pseudo-Riemannian manifold that is topologically a disc. Is it possible to construct a global coordinate system in which the metric is conformally flat? I.e. coordinates $(t,x)$ which cover the whole manifold such that the line element takes the form

$ds^2=\Omega^2(t,x)(-dt^2 + dx^2)$

for some conformal factor $\Omega$.

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Yes, every disk (topologically) is conformally flat. String theorists would say that the disk has "no moduli".

If it were not possible to flatten a disk to the standard round form by Weyl transformations, there would have to exist a family of inequivalent solutions. The family would be labeled by some parameters, "moduli". It's pretty clear that they would have to be continuous. So there would have to exist new "disks" that are conformally inequivalent to the standard disk but that are infinitesimally close to it.

It's impossible because if one defines a slightly perturbed disk by describing its boundary via polar coordinates $$ r(\phi) = r_0 + \epsilon(\phi), $$ where we also use $z=r \exp(i\phi)$, it is always possible to find a holomorphic function of $z$ such that the curve given by the equation above is mapped to the standard $r=r_0$ round form. Just Fourier-expand $\epsilon(\phi)$ and redefine $z$ by a Laurent series whose coefficients are the Fourier coefficients I just mentioned. One can give details. But in the formula $$z'=z+\sum_{n=0}^\infty \delta_n z^n$$ one has enough coefficients $\delta_n$ to adjust all the Fourier modes of $\epsilon(\phi)$. They're essentially the same things. Because every two (topologically) disks may be connected by continuous transformations (not necessarily preserving anything beyond topology), it's possible to divide the proof of their conformal equivalence to the pairs of nearby manifolds.

Moduli and conformal isometries of Riemann surfaces

Much more generally, if there is a Riemann surface with $g$ handles and $b$ boundaries and $c$ crosscaps (boundaries whose opposite points are identified, and therefore help to create an unoriented but possibly closed manifold), the number of real moduli $\mu$ can be calculated from the equation $$\frac{1}{3} ( s - \mu ) = \chi = 2 - 2g - b - c $$ where $\chi$ is the Euler character, calculable by the explicit formula above, and $s$ is the real dimension of the conformal isometry group of the manifold; usually $s$ vanishes for a sufficiently high value of $g$ etc. It's the Riemann-Roch index theorem which may be extended to the Atiyah-Singer theorem etc.

So for example, the disk has an $SU(1,1)$ isometry group - isomorphic to the $SL(2,R)$ conformal isometry of its conformally equivalent surface, the half-plane - which has $s=3$ and the formula above becomes $$ \frac{1}{3} ( 3 - 0 ) = 2-1 = 1 $$ For a sphere, $s=6$ and $\chi=2$ since $g=b=c=0$. For a torus, $s=2$ from the $U(1)^2$ group and $\mu=2$ as well because $\chi=2-2=0$. Similarly for a Möbius strip, $b=c=1$ and $\chi=0$, matching $s=\mu=1$. For a Klein bottle, $b=0$, $c=2$, and $s=\mu=1$ as well. All other topologies of Riemann surfaces have $s=0$ (no isometries) so the formula above simplifies to $\mu = 6g+b+c - 6$ for the number of real moduli.

Indefinite signature

If your word "pseudo-Riemannian" really meant that you have an indefinite signature, like in special relativity, then it is not correct to say that such manifolds ever have the "topology of a disk". In particular, seemingly compact sets of points cut into the Minkowski space are not open sets - a pretty basic problem. It's because open sets require that together with every point, you must include a whole neighborhood of all points whose distance is smaller than $\epsilon$.

That's not possible in a Minkowski-like space because all lightlike-separated points - that can be arbitrarily far in the coordinate space - have proper distance from the original point equal to zero, so it is smaller than $\epsilon$. Topologies work differently in the Lorentzian-signature manifolds and they're really very messy.

In string theory - following the procedures in physics in general - the relevant calculations are always done with the Euclidean, positive-definite-signature manifolds, and only the final results are extrapolated back to the Minkowski-signature spacetime. That's perfectly OK and one may show that the resulting theory satisfies all required consistency conditions. This "messy Lorentzian topologies" case is a textbook example of the fact that the calculations in the Wick-rotated framework are more well-behaved than the calculations that someone would naively try to do directly in the original, indefinite-signature spacetime.

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When I say pseudo-Riemannian I do mean the usual Lorentzian signature (- +) or (+ -) depending on your conventions. I know that all surfaces are conformally flat however what I'm interested in here is a specific coordinate system in which the metric is globally conformally flat. What I mean by topologically a disk is that $M$ is toplogically $\Sigma \times \mathbb{R}$ and $\Sigma$ is topologically trivial. –  Dionigi May 18 '11 at 12:30
    
Apologies, I don't understand in what sense is $\Sigma \times {\mathbb R}$ a disk: isn't it a noncompact manifold? –  Luboš Motl May 18 '11 at 13:36
    
Well the open disk is not compact actually. Examples: You can think of $\mathbb{M}^2$ topologically as $\mathbb{R}\times\mathbb{R}$ and the ''cylinder'' spacetime as $S^1\times\mathbb{R}$. The cylinder spacetime is then not topologically a disk since it admits non contractible closed curves (those wrapping around the $S^1$), it is in fact topologically a cylinder. $\mathbb{M}^2$ instead is, what I called, topologically a disk since there are no non-contractible closed curves (I'm assuming there are no singular points). –  Dionigi May 18 '11 at 14:07
    
So I guess what I mean by topologically a disk is that it is contractible. –  Dionigi May 18 '11 at 14:11
    
Dear @Dionigi, the Lorentzian-signature "disks" defined as any contractible manifolds can't be mapped to each other. Imagine that the shape is described by an equation involving $x^+,x^-$, the light-like coordinates. The conformal transformations in this case are separate reparametrizations of $x^+$ and of $x^-$. This is clearly not enough to relate all contractible shapes. For example, a null boundary of a "diamond" will always stay null under conformal transformations, and manifolds with piecewise spacelike and then timelike boundaries etc. will always have these pieces. –  Luboš Motl May 18 '11 at 15:18
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