# Does strong causality imply the existence of convex normal neighborhoods that are causally convex?

Let $(M,g)$ be a spacetime. An open set $A\subset M$ is causally convex if no causal curve intersects $A$ in a disconnected set. If $(M,g)$ is strongly causal, then for any point $p\in M$ and neighborhood $U$ of $p$, there is a causally convex set $A$ such that $p\in A\subset U$. It might be convenient to assume that $A$ can be taken to be a convex normal neighborhood (CNN), i.e. $A$ can be covered by normal coordinates centered at any $q\in A$ and that any $r,s\in A$ can be connected by a unique geodesic in $A$. Is this possible in general?

The existence of such sets is nonchalantly declared on page 195 of Hawking and Ellis, and on page 60 of Beem et al.

I attemped a proof by a standard argument: Take a CNN $U$ of $p$, and a causally convex set $A$ with $p\in A\subset U$. Then take another CNN $V$ of $p$ with $V\subset A$. Then $V$ should be causally convex or some small modification is. This seems to make sense, because causal convexity is supposed to protect from global "almost" causal violations. But in the case of $\Bbb R^2$ Minkowski space we can see that this program fails. A CNN is any set that is convex in the linear algebra sense, in particular a square with "the flat part down." But this is not causally convex because you can have a timelike curve that passes through the bottom corner (+time pointing up), then leaves through the side, and comes back in near a corner at the top. But if you rotate so it's a diamond, you get a causally convex CNN.

So maybe a causal diamond like $I^+(r)\cap I^-(s)$ with $r\ll p$ and $p\ll s$ could work, where $r$ and $s$ are "close enough" to $p$. I'm not sure how the details play out.

• 0ßelö7: I find your question interesting as far as it seeks to make a quantitative determination (whether some particular causal curve between two suitable given events is "of extremal Lorentzian arc length") based only on given causal and topological relations. (Good luck! ;) p.s. "An open set $A \subset M$ is causally convex if no causal curve intersects A in a disconnected set" -- For a moment I found this wording confusing (even though it's original Beem/Ehrlich/Easley). But clearly, it means: "The intersection set is not disconnected."; not: "What's left of set $A$ is not disconnected." – user12262 Sep 19 '17 at 6:19

Every spacetime $(M,g)$ admits arbitrarily small globally hyperbolic neighborhoods at every point, which in a strongly causal spacetime can be chosen to be causally convex, see Theorem 2.14 in Minguzzi and Sanchez: the causal hierarchy of spacetimes arXiv:gr-qc/0609119. Notice that global hyperbolicity implies causal geodesic connectedness.

The problem is that they are not convex normal neighborhoods (they might be non convex with respect to spacelike geodesics). The problem of existence of arbitrarily small globally hyperbolic convex normal neighborhoods (which in a strongly causal spacetime could be chosen to be causally convex) is more difficult and requires a deeper analysis of the exponential map (in HE and Beem et al book they do not seem assume the existence of such neighborhoods). The problem has been solved in the affirmative in my paper "Convex neighborhoods for Lipschitz connections and sprays" {arXiv}:1308.6675, see Sec. 1.6 Corollary 2.

• Hi Ettore, welcome to PSE... After our discussion in Florence on this point we had more than one year ago, I pointed out your paper commenting/answering an analogous question in PSE. – Valter Moretti Aug 5 '18 at 9:46

If you pick some causally convex set $A$ within a Cauchy normal neighbourhood $U$, consider the following subset :

In the tangent space, take the domain $A' = \exp^{-1}_p(A)$. Pick two points $a,b$ such that in Minkowski space $a \ll p \ll b$. Then consider the subset $A'' = I^+(a) \cap I^-(b)$. We have

$$I^+(a) = \{ x | (x_t - a_t) > 0,\ (x_t - a_t)^2 > (\vec x - \vec a)^2 \}$$ $$I^-(b) = \{ x | (x_t - b_t) < 0,\ (x_t - b_t)^2 > (\vec x - \vec b)^2 \}$$

In Minkowski space, this subset is convex. Any geodesic with boundary $i, f$ inside it, that is, a straight line $d$, will be entirely within that shape. The curve defined by the vector $f - i$ is, with $\tau \in [0,1]$,

$$x^\mu(\tau) = i + (f-i)\tau$$

We have $i_t, f_t \in (a_t, b_t)$, so it can be shown easily that $x^t(\tau) \in (a_t, b_t)$, and we have also

$$(i_t - a_t)^2 > (\vec i - \vec a)^2$$ $$(f_t - a_t)^2 > (\vec f - \vec a)^2$$ $$(i_t - b_t)^2 > (\vec i - \vec b)^2$$ $$(f_t - b_t)^2 > (\vec f - \vec b)^2$$

Then by expanding it, we can show that this is also true of $x^\mu(\tau)$.

[Left as an exercize]

Since every geodesic in the CNN is a straight line in the tangent space, then every point of a geodesic with endpoints in $\exp A''$ remains inside the region.

Now, any causal curve intersecting $A$ is intersecting it as a connected piece. Since homeomorphisms preserve connected components, this is also true in $A'$. We need to show that for any timelike curve $\gamma$ (since the exponential map maps timelike curves in Minkowski space to timelike curves in the manifold) in $A'$, the intersection of $A''$ and $\gamma$ is also connected. So we need to show that if a timelike curve leaves $A''$, it never re-enters it.

If the curve goes through a point $c \in A''$, being a causal curve, its future directed part will remain within $I^+(a)$, since $a \ll c$ and $c \leq d$ implies $a \ll d$, and likewise for the past directed part and $I^-(b)$.

Once the future directed curve will leave $A''$, that is, will leave $I^-(b)$, since it must remain in $I^+(a)$, it can never re-enter it, since in Minkowski space, we have $b \ll x \ll y$ implies $y \not \ll b$ (cf Zeeman's "Causality implies the Lorentz group"). The same argument applies for the past-directed curve. Hence once the curve leaves $A''$, it never re-enters it, so that it is causally convex in Minkowski space, and by homeomorphism, the same is true in the manifold.

Hence $\exp_p A''$ is both geodesically convex and causally convex.

Edit : If a spacelike geodesic $\gamma$ does not go through $p$, then at $\lambda = 0$, $\exp^{-1}_p(\gamma(0)) = m$, and the curve is the mapping of some curve $y^\mu(\lambda)$ in the tangent space, $\gamma(\lambda) = \exp_p (y^\mu(\lambda))$.

Since it is a geodesic, in Riemann normal coordinates around $p$, we can expand it as

$$y^\mu(\lambda) = y^\mu_0 + u^\mu \lambda + \mathcal{O}(\lambda^2)$$

For a small enough causal diamond around $p$, we have that the curve in the tangent space is arbitrarily close to a straight line. I'm not quite sure how to prove that this is enough to guarantee convexity from there, though. I'd say that in an open set a small enough deformation of a curve contained in that set is still contained in that open set.

• I do not understand why $\exp_p A''$ is geodesically convex. If $r,s \in \exp_p A''$ why the geodesic joining them must be included in $\exp_p A''$? It is generally false that a geodesic in $\exp_p A''$ is a straight line in the tangent space ($T_pM$?). It is true for the geodesics passing through $p$, but not for other geodesics in general... – Valter Moretti Jun 12 '17 at 12:19
• The image of the geodesic joining them $\exp^{-1}_p \vec{rs}$ will be a straight line in $T_pM$, which is contained entirely within $A''$ if $\exp^{-1}_p r$ and $\exp^{-1}_p s$ are (since $A''$ is just the intersection of two cones, which is convex for straight lines) – Slereah Jun 12 '17 at 12:35
• No it is not true – Valter Moretti Jun 12 '17 at 12:38
• Geodesics are straight lines in $T_pM$ when they pass through $p$. In our case you are considering a geodesic which does not pass through $p$ – Valter Moretti Jun 12 '17 at 12:39
• Well so much for that idea. Which part exactly isn't true? Is it the assumption that geodesics map to straight lines? – Slereah Jun 12 '17 at 12:39