# Proof of Geroch's quiz part I : Manifolds, metrics and orientation

Here's a list of proofs of Geroch's quiz on spacetimes from the previously mentionned paper. Most of which done using counterexamples. For a brief reminder :

Take all spacetimes to be four-dimensional, and, for assertions involving causal structure, time-oriented. All manifolds are to be Hausdorff, paracompact, connected, and without boundary.

### 1. A manifold admits a Lorentz metric if and only if its universal covering manifold does. (FALSE)

As is well known (from the mysterious proof of Steenrod's Topology of fiber bundles), given Geroch's definition of a manifold, there's only one case for which a manifold may fail to admit a Lorentz metric : if it is both compact and has an Euler characteristic different from zero. So what we are looking for is a compact manifold of non-zero Euler characteristic which has a universal cover that either does not, or is non-compact.

A very simple example is the surface of genus two $S_2 = T^2 \# T^2$, which has Euler characteristic $2$, but has the hyperbolic plane $\mathbb{H}^2$ as its universal cover, which is non-compact. But for our case, we have to assume $4$-dimensional manifolds. Fortunately, as the covering space of a product of spaces is the product of their respective covering spaces, we have that the product of two such surfaces $S_2 \times S_2$ (with Euler characteristics $\chi(S_2 \times S_2) = \chi(S_2) \times \chi(S_2) = 4$) is $\mathbb{H}^2 \times \mathbb{H}^2$, also non-compact. Hence we have a proper counterexample.

### 2. Every flat spacetime admits a nonzero Killing field. (FALSE)

Not to get confused (this wouldn't work with some notion based on local properties like simply a vector field obeying the Killing equation), let's have a quick reminder of what a Killing vector field is :

A *Killing vector field* $K$ is a vector field generating a one-parameter group of diffeomorphism $\phi_t$, in other words, for any $X, Y \in T_pM$,

Given Minkowski space, we have the full PoincarĂ© group as the set of all isometries on the manifold. A simple way of removing those isometries is by simply removing points : removing a single point from the manifold, all translation isometries already vanish (for instance removing $0$, the translation isometries of $x,y,z$ with parameter $t = 1$ no longer work for $(-1,0,0)$, $(0,-1,0)$, and $(0,0,-1)$). For $4$ dimensions, we simply have to remove a point along every axis as well to prevent rotational Killing vector fields.

### 3. Call two Lorentz metrics on M equivalent if there is a diffeomorphism from M to M which sends one to the other. Then each of two equivalent metrics can be continuously deformed to the other. (FALSE)

### 4. For $M$ a, a manifold, and $p$, a point of $M$, $M$ with $p$ removed is not diffeomorphic with $M$. (FALSE)

The counterexample is fairly simple to come up with once you realize that the big obstacle is just the lack of bijections between two finite sets of different cardinalities. Just have the manifold already contain a countable infinity of points removed :

$$M = \mathbb{R}^4 \setminus \{ (t,x,y,z) | t = y = z = 0 \wedge x \in \mathbb{N}_{> 0} \}$$Then simply remove $(0,0,0,0)$, $M' = M \setminus \{ (0,0,0,0) \}$. The diffeomorphism $\phi : M \to M'$ is then just

$$\phi(t,x,y,z) = (t,x-1,y,z)$$### 5. If spacetime $M, g_{ab}$ has vanishing Weyl tensor, then $\Omega^2 g_{ab}$ is flat for some $\Omega$. (FALSE)

### 6. If, for $C$ a closed subset of $M$, $M \setminus C$ admits a Lorentz metric, then so does $M$. (FALSE)

Fairly easy to do by considering the usual counterexample of Lorentzian manifolds : $S^2$ minus a disk is homeomorphic to $\mathbb{R}^2$, and the same is true of $S^4$ and $\mathbb{R}^4$. As $S^4$ admits no Lorentz metric, this is false.

### 7. For $S$, a two-dimensional manifold, and $p$, a point of $S$, $S \setminus p$ is not simply connected. (FALSE)

Consider $S^2$ for the manifold, $S^2 \setminus \{ p \} \cong \mathbb{R}^2$, which is simply connected.

### 8. The product of two simply connected manifolds is simply connected. (TRUE)

From standard results, we have that $\pi_1(X \times Y) = \pi_1(X) \times \pi_1(Y)$, so that in our case, the product of the trivial groups $\{ i_X \}$ and $\{ i_Y \}$ is the also trivial group $\{ (i_X, i_Y) \}$.

### 9. If, for $S$ and $S'$, three-dimensional manifolds, $S \times \mathbb{R}$ and $S' \times \mathbb{R}$ are diffeomorphic then $S$ and $S'$ are diffeomorphic. (FALSE)

Not a whole lot of simple counterexamples unfortunately, but it is famously true of the Whitehead manifold.

### 10. No compact spacetime is extendible. (TRUE)

If a compact manifold $M$ is extendible, there exists some inclusion $\iota : M \to M'$, $M'$ some connected manifold, such that $\iota(M) \neq M'$, and where $\iota(M)$ is an open submanifold isometric to $M$. From the isometry, $\iota(M)$ is compact and therefore closed. It is therefore closed and open, which means that since $M'$ is connected, it can only be $M'$ or $\varnothing$.

### 11. Every simply connected $4$-manifold which is a product admits a Lorentz metric, except $S^2 \times S^2$. (TRUE)

While $4$-manifolds aren't so easily classified, things are much simpler in this case. We can only consider the following two cases : either the product of two $2$-manifolds, or the product of a $3$-manifold with a $1$-manifold (the case of four $1$-manifolds can be reduced to either).

From the product property of the fundamental group, if the $4$-manifold is simply connected, then so are its components here. And the same is true of compactness : it will only be compact if both are. This simplifies things considerably : $\mathbb{R}$ is not compact and $S$ is not simply connected, hence we can only have the product of two $2$-manifolds. Of these, only the sphere is both compact and simply connected, from the usual theorem of classification of surfaces. Hence this statement is correct.

### 12. A spacetime is extendible if and only if its universal covering spacetime is. (FALSE)

### 13. A non-simply connected manifold which admits a Lorentz metric admits one which is neither time-orientable nor space-orientable. (FALSE)

### 14. A flat spacetime is time-orientable. (FALSE)

A seldomly brought up theorem is that, given a spacetime of the form $\mathbb{R} \times \Sigma$, $\mathbb{R}$ being a timelike coordinate and $\Sigma$ a spacelike hypersurface, and considering the time-reversal operator $T(t, \sigma) = (-t, \sigma)$ and some involution $I$ on $\Sigma$, then $(\mathbb{R} \times \Sigma) / (T \times I)$ fails to be time-orientable (this can be checked by considering some closed curve passing through two points $p, q \in \mathbb{R} \times \Sigma$ such that $p \sim q$ and checking what happens to a timelike vector transported on it).

Given this, a simple example of a flat non-time-orientable spacetime is the Minkowski cylinder $\mathbb{R} \times S$,

$$ds^2 = -dt^2 + d\theta^2$$with the involution $I(t,\theta) = (t, \theta + \pi)$.

### 15. If each of two spacetimes can be embedded isometrically as an open subset of the other, then the spacetimes are isometric. (FALSE)

An example of such a pair of spacetimes is to consider two upper half Minkowski planes, $\mathbb{R}^2_{t > 0}$, but the second one has a point $(t_1, x_1)$ removed. Then we can consider the isometry of simply the identity map with a point removed for going from the first to the second, and the identity translated by $t_1$ for the second to the first.

This theorem is also apparently true if the manifolds are compact and oriented.

### 16. The sum of two complete vector fields is complete. (FALSE)

Posted on

*2019-03-01 15:13:14*