Considering a static spacetime of the metric form \begin{equation} \mathrm{d}s^{2}=-V^{2}\mathrm{d}t^{2}+h_{ij}\mathrm{d}x^{i}\mathrm{d}x^{j} \end{equation} with a timelike killing field $\xi^{\mu}=(\partial_{t})^{\mu}$ we can choose a function space on each constant time hypersurface $\Sigma$ as$\mathcal{H}=\left\{ f\,\mid\,\parallel f\parallel<\infty\right\} $with the Sobolev norm $\parallel f\parallel$being given by \begin{eqnarray*} \parallel f\parallel^{2} & = & \frac{q^{2}}{2}\underset{\Sigma}{\int}\mathrm{d}\Sigma V^{-1}f^{*}f+\frac{1}{2}\underset{\Sigma}{\int}\mathrm{d}\Sigma Vh^{ij}D_{i}f^{*}D_{j}f. \end{eqnarray*} In (negative mass) Schwarzschild coordinates this norm reads reads \begin{equation} \parallel f\parallel^{2}=\frac{q^{2}}{2}\underset{\Sigma}{\int}\mathrm{d}\mu\mathrm{d}rR^{n}V^{-2}\mathcal{\mid}f\mid^{2}+\frac{1}{2}\underset{\Sigma}{\int}\mathrm{d}\mu\mathrm{d}rR^{n}V^{2}\mid f'\mid^{2} \end{equation} with $\mathrm{d}\mu$ being the volume element of unit $n$-sphere ($\mathrm{d}\Sigma=\mathrm{d}\mu\mathrm{d}rV^{-1}R^{n}$).

My question is: How does this norm read for positive mass Schwarzschild spacetime? $V^2$ is assumed positive above (which is ok for $V=(1+2M/r)$ (M positive). But for $V=(1-2M/r)$ there will be the point of a change in sign so that the above norm cannot hold anymore.How does the norm then look like???


  • $\begingroup$ I do not understand well: Are you dealing with one of the two external static regions of Schwarzschild-Kruskal space or with the whole manifold? $\endgroup$ Feb 26, 2014 at 18:38
  • $\begingroup$ If I worked with $r>2M$, $V$ would be positive - so in this case the norm is still positive definite and ok. But for $r<2M$ the problem occures that the norm can be negative...so I must find another norm. $\endgroup$
    – user40852
    Feb 27, 2014 at 8:28
  • $\begingroup$ also, what is $\mu$? $\endgroup$ Feb 27, 2014 at 14:39

1 Answer 1


Notice that if $r<2M$, the coordinate $t$ turns out to be spacelike, so surfaces at $t$-constant are not appropriate for imposing, for instance, Cauchy data. (I guess that your Sobolev norms are used to deal with the Cauchy problem).

If you are dealing with the whole Kruskal manifold you should fix a spacelike Cauchy surface instead, which for instance coincides to a $t$-constant surface in both the static wedges.

If you use the $t=0$-surface, the singularity region (where the measure you exploit to construct your norm is singular) is made of the bifurcation sub manifold $\cal B$ (a $2$-sphere) localized at $r=2M$ and you never encounter the region $r<2M$ on this spacelike Cauchy surface. The submanifold $\cal B$ has obviously zero measure.

In http://en.wikipedia.org/wiki/File:Kruksal_diagram.jpg the Cauchy surface I am mentioning is represented by the $x$ axis. $\cal B$ is the intersection of the $x$ axes and the $y$ one. $r<2M$ holds in both the open regions indicated by II and IV, so you see that the Cauchy surface never meets them.

To construct a well defined Sobolev norm it is sufficient to change the measure in a neighbourhood of ${\cal B}$ to make it finite. I do not know if there is a canonical way to do it.

(Together with a pair of colleagues I used a similar technology to rigorously construct the so called Unruh vacuum for zero mass particles, proving that the state verifies the Hadamard condition at short distances so that it can be used for renormalization procedures (Adv. Theor. Math. Phys. 15, vol 2, 355-448 (2011) ))

  • $\begingroup$ Thanks a lot! I will see whether I can understand what you told me. To give you the context: arxiv.org/abs/gr-qc/9907009 I wonder why they only stick to the negative mass Schwarzschild case - the positive mass one is the one that should be addressed, right?! $\endgroup$
    – user40852
    Feb 27, 2014 at 20:01
  • $\begingroup$ Sorry, I am too busy with my teaching activity presently. I cannot read the paper you quoted. $\endgroup$ Feb 28, 2014 at 9:10

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