# Does the gluing procedure in Robert Wald’s book *General Relativity* yield a Hausdorff spacetime?

Before I state my problem, let me provide some definitions pertaining to the Cauchy Problem in General Relativity.

Definition 1: A triplet $(\Sigma,h,k)$ is called an initial data set if $(\Sigma,h)$ is a Riemannian $3$-manifold, $k$ is a symmetric $2$-form on $\Sigma$, and both $h$ and $k$ satisfy the so-called Einstein constraint equations for vacuum (we do not need to know what these equations are). $\quad \spadesuit$

Definition 2: Let $(\Sigma,h,k)$ be an initial data set. A triplet $(M,g,i)$ is called a vacuum development of $(\Sigma,h,k)$ if

• $(M,g)$ is a spacetime manifold (i.e., a Lorentzian $4$-manifold) that satisfies the vacuum Einstein Field Equations;

• $i$ is a smooth embedding of $\Sigma$ into $M$ such that ${i^{*}}(g) = h$ and ${i^{*}}(\Pi) = k$, where $\Pi$ is the second fundamental form (i.e., the extrinsic curvature) of $\Sigma$ in $M$; and

• $i[\Sigma]$ is a smooth Cauchy surface in $M$. $\quad \spadesuit$

The Cauchy Problem in General Relativity is to prove the existence of a vacuum development for every initial data set. The proof proceeds by working locally on small patches of the initial data set and ends by performing some gluing to obtain a global solution.

It is a well-known result from the theory of hyperbolic PDE’s that if $U$ is a local coordinate patch in $\Sigma$ (i.e., $U$ is homeomorphic to an open subset of $\mathbb{R}^{3}$), then we can find a vacuum development $(M_{U},g_{U},i_{U})$ of the initial data set $(U,h|_{U},k|_{U})$ such that $M_{U}$ is an open neighborhood of $U \times \lbrace 0 \rbrace$ in $U \times \mathbb{R}$ and ${i_{U}}[U] = U \times \lbrace 0 \rbrace$.

Knowing now that local vacuum developments exist, we have to glue them together in order to obtain a global-in-space vacuum development of $(\Sigma,h,k)$. Once the Hausdorff property is established, paracompactness is for free by a 1968 result of Robert Geroch, which states: If a smooth $4$-manifold is Hausdorff and admits a smooth Lorentz metric, then it is automatically paracompact.

In Robert Wald’s book General Relativity, the following method for gluing is proposed:

1. By the paracompactness of $\Sigma$ as a $3$-manifold, we can find a locally finite open cover $\mathcal{U}$ of $\Sigma$ consisting of coordinate patches. As mentioned above, for each $U \in \mathcal{U}$, we have a vacuum development $(M_{U},g_{U},i_{U})$ of $(U,h|_{U},k|_{U})$.

2. For each $p \in \Sigma$, define $\mathcal{U}_{p} := \lbrace U \in \mathcal{U} ~|~ p \in U \rbrace$. We have from (1) that $\mathcal{U}_{p}$ is a finite collection for each $p \in \Sigma$.

3. For each $p \in \Sigma$, let $W_{p} \subseteq \Sigma$ be a neighborhood of $p$ such that $W_{p}$ is contained in all the open sets belonging to $\mathcal{U}_{p}$, i.e., $W_{p} \subseteq \bigcap \mathcal{U}_{p}$. Using a result known as ‘local geometric uniqueness’, we can then find a vacuum development $(M_{p},g_{p},i_{p})$ of $(W_{p},h|_{W_{p}},k|_{W_{p}})$ such that there exists an isometric embedding $j_{p,U}: (M_{p},g_{p}) \to (M_{U},g_{U})$ for each $U \in \mathcal{U}_{p}$.

4. Using the embeddings $j_{p,U}$ to make identifications, we can glue the $(M_{p},g_{p})$’s together in order to obtain a vacuum development of $(\Sigma,h,k)$. Although Wald does not clearly describe this step, this is how I interpret it. For each $U \in \mathcal{U}$, consider the union $\tilde{M}_{U} := \displaystyle \bigcup_{p \in U} {j_{p,U}}[M_{p}] \subseteq M_{U}$. We wish to glue the $(\tilde{M}_{U},g_{U}|_{\tilde{M}_{U}})$’s together by identifying points as follows. Given $U_{1},U_{2} \in \mathcal{U}$, $x_{1} \in \tilde{M}_{U_{1}}$ and $x_{2} \in \tilde{M}_{U_{2}}$, we say that $x_{1} \sim x_{2}$ if and only if there exist a $p \in U_{1} \cap U_{2}$ and a $q \in M_{p}$ such that ${j_{p,U_{1}}}(q) = x_{1}$ and ${j_{p,U_{2}}}(q) = x_{2}$. Finally, define $\displaystyle M := \bigsqcup_{U \in \mathcal{U}} \tilde{M}_{U} \bigg/ \sim$.

Problem: How do we show that $M$ is Hausdorff?

This has eluded my best efforts for the past month, so I would appreciate it if anyone on MathOverflow could offer any insight. Thank you very much!

• have you tried looking at the equivalent condition that the quotient map be open and $\sim$ be a closed subset of $\times^2\bigsqcup_{U\in\mathcal{U}}\tilde{M}_U$? – johndoe Jun 16 '13 at 10:05
• I think following johndoe's suggestion should get you to the answer. But just in case you wanted to see the answer written out, the Hausdorff property is checked explicitly in Ringström's book The Cauchy Problem in General Relativity (EMS, 2009). See proof of Theorem 16.6, p.179. – Igor Khavkine Jun 16 '13 at 16:03
• @johndoe: I’ve tried the approach that you suggested. The quotient map is indeed open, but proving that $\sim$ is a closed subset of $\displaystyle \left( \bigsqcup_{U \in \mathcal{U}} \tilde{M}_{U} \right) \times \left( \bigsqcup_{U \in \mathcal{U}} \tilde{M}_{U} \right)$ is the problem. – Leonard Jun 16 '13 at 18:51
• @Igor Khavkine: Thanks for the suggestion; I’ve read Hans Ringström’s book before. The Hausdorff property is evident in his approach because he constructs a global development within $\Sigma \times \mathbb{R}$, which is obviously Hausdorff, without having to form any quotient space. That is because the local developments agree on the overlaps. However, in Wald’s argument, the local developments do not agree on the overlaps, so one has to perform a coordinate transformation in order to get from one local development to another. Hence, there is a need to form a quotient space. – Leonard Jun 16 '13 at 19:05
• @Igor Khavkine: I’m mainly interested in seeing how Wald’s argument survives this final test. I asked him recently, but he told me that he couldn’t remember anything he had done on the Cauchy Problem, which was $25$ years ago! He’s an elderly gentleman indeed. – Leonard Jun 16 '13 at 19:12

Your construction won't work as stated. The problem is that $M_p$ can be too small. Let me give a slightly silly example of why it doesn't work.

Cover $\mathbb{R}^3$ with the two local coordinate charts $U = \lbrace x_1 < 1\rbrace$ and $V = \lbrace x_1 > -1\rbrace$, and prescribe on it trivial initial data. An admissible "local solution" has $M_U$ and $M_V$ being the following subsets of the Minkowski space $\mathbb{R}^{1,3}$:

\begin{align} M_U &= \lbrace x_1 < 1 - |t|\rbrace \newline M_V &= \lbrace x_1 > -1 + |t| \rbrace \end{align}

Let $W_p = \lbrace |x_1| < 1\rbrace$ for every $p \in U\cap V$. And let $$M_p = M_U \cap M_V \cap \lbrace |t| < \frac12\rbrace$$

So far all the objects are admissible ones that could, in principle, be the ones you picked out from steps 1 through 3.

By your step 4, you have two distinct points corresponding to what would've been the point $t = \frac12, x_1 = x_2 = x_3 = 0$, since the ones in $M_U$ and $M_V$ are not identified since that point technically lies outside of $M_p$ for every $p$. Those two points are not separated by neighborhoods in the space-time you constructed, which violates Hausdorff axiom.

If I am not mistaken, this would be the primary obstruction. Possible resolutions are to either make $M_p$ larger or allow us to make $M_U$ and $M_V$ smaller. (With the caveat that I haven't written down a full argument saying that this will work.) In either case you need to get yourself down to something analogous to Ringström's case where the images of gluing maps $j_{p,U}$ cover the neighborhood of intersection.

(Edit 19.06.2013) Ah, I found a reference where the argument is given in slightly more detail.

It depends on a slightly stronger local uniqueness theorem than you supposed. (See Hawking-Ellis, Large Scale Structure... section 7.5.) And some details are still missing. But it appears to be more easily filled-in. And considering the footnote on page 106, they certainly did worry about Hausdorff-ness in their construction.

• Hi Willie. I was hoping you would provide an answer, and you did! Anyway, I might have some further questions, so let me study your example first. Thanks, by the way. – Leonard Jun 17 '13 at 17:39
• The four steps that I’ve listed aren’t my idea, by the way. I simply took them from Wald. It thus seems that his argument can’t guarantee a Hausdorff spacetime. Many other books, such as Hawking’s and Ellis’ The Large Scale Structure of Space-Time, follow the same pattern: Local-in-space-and-time developments are glued together, via the identification of points, to obtain a global-in-space (though not necessarily global-in-time) development, but the Hausdorff property is never established. Curiously enough, the authors take great care to prove that a maximal development is Hausdorff. – Leonard Jun 17 '13 at 23:07
• However, the proof that a maximal development exists depends on the fact that a development exists in the first place, which must be proven to be Hausdorff! – Leonard Jun 17 '13 at 23:08
• "It thus seems that his argument can't guarantee a Hausdorff spacetime" This is not true. What we've discussed is that the particular interpretation of his (rather handwavy) argument suffers a flaw. But one can get around it if one modifies the gluing procedure so that $x_1 \sim x_2$ if, for every open neighborhood of $x_1\in V_1 \subset M_{U_1}$ and $x_2\in V_2 \subset M_{U_2}$, there exists $y_1\in V_1$ and $y_2\in V_2$ such that there exists $p$ in the base and $q$ in its local development where $j_1(q) = y_1$ and $j_2(q) = y_2$. This will trivially get around the difficulty in my answer. – Willie Wong Jun 18 '13 at 9:18
• With the above gluing procedure, we see that the "solution" we constructed is a Hausdorff topological space, but not a manifold. Since $(1/2,0,0,0)$ is not locally Euclidean: it has three sheets coming off of it. This "boundary set" will be acausal and smooth if you defined your initial notion of development suitably, and so probably you can start there and run the local local existence to enlarge... But it is probably much simpler to use in reverse the Globally Hyperbolicity condition and directly construct the development as a submanifold of $\Sigma \times \Real$ (so Ringstrom's argument). – Willie Wong Jun 18 '13 at 9:27