Proof. See [Reference Beran and Sämann8, theorem 4.7].

Proof. See [Reference Beran and Sämann8, Appendix A].

Proof. See [Reference Beran and Sämann8, theorem 4.12] or a more complete picture in [Reference Beran, Kunzinger and Rott11, theorem 5.1].

Proof. This follows from [Reference Beran, Harvey, Napper and Rott9, theorem 3.6] and [Reference Beran, Kunzinger and Rott11, theorem 5.1 and proposition 4.18].

Proof. See [Reference Beran and Sämann8, proposition 4.14].

Proof. In ${\mathrm{AdS}} $, the $c+\frac{\pi}{2}$-level set of $\tau(p_-,\cdot)$ is given by ${\mathrm{AdS}} \cap\{(s_1,s_2,z):s_1=\sin(c)\}$. All timelike geodesics through $p_-,p_+$ are given by $t\mapsto(\sin(t),\cos(t)\cosh(x),\cos(t)\sinh(x))$ (the x parametrizes the angles at p), and they are distance realizing and thus pass at right angles through the level sets.

So we define $f:{\mathrm{AdS}} ^\prime \to X$ by setting

\begin{equation*} f(t,x)=(\sin(t),\cos(t)\cosh(x),\cos(t)\sinh(x))\,. \end{equation*}

This is an isometry: The vertical unit speed distance realizers $t\mapsto (t,x_0)$ map to the unit speed distance realizers $t\mapsto(\sin(t),\cos(t)\cosh(x_0),\cos(t)\sinh(x_0))$ (which cover all ${\mathrm{AdS}} ^\prime$ resp. X), and the horizontal spacelike geodesic $x\mapsto (t_0,x)$ has speed $\cos(t_0)$ and maps to $x\mapsto(\sin(t_0),\cos(t_0)\cosh(x),\cos(t_0)\sinh(x))$ which is also a spacelike geodesic of speed $\cos(t_0)$ (these are parametrizations of the level sets of $\tau(p_-,\cdot)$ and thus cover all ${\mathrm{AdS}} ^\prime$ resp. X), and the geodesics of the first family intersect the ones in the second family at right angles as described before.

That X is globally hyperbolic can be seen in the warped product picture.

For realizing timelike triangles satisfying the size bonds for $K=-1$, let $a+b\leq c \lt \pi$. Choose a realization with a timelike triangle $\Delta(p,q,r)$ in AdS. By applying a suitable isometry, we can w.l.o.g. assume that $p=(\cos(-c/2),\sin(-c/2),0),\,r=(\cos(c/2),\sin(c/2),0) \in X$ (these have the right τ-distance). Then $q\in X$ by causal convexity of X, so $\Delta(p,q,r)$ is realized in X, or equivalently in ${\mathrm{AdS}} ^\prime$ where $p,r$ have the same x-coordinate (‘vertical’).

For the inextendibility of X, we indirectly assume an extension X ^ʹ of X was still globally hyperbolic. Connect $p_-$ to $p_+$ by a timelike distance realizer γ. First, if there is a point $p\in I^+(\gamma)\setminus I^-(\gamma)$, we find a t such that $\gamma(t)\ll p$, and a timelike distance realizer α_t connecting them. Inside a localizing neighbourhood of p, we have a point $p\ll p_+$ and set $\varepsilon=2\tau(p,p_+) \gt 0$. α_t is initially contained in $I(\gamma)$, so we find a point $\alpha_t(s_t)\in\partial I(\gamma)$, i.e. in $\partial I^-(\gamma)$.

By lemma 3.15, we get that for all ɛ > 0 there is δ > 0 small enough such that $\tau(\gamma(r),\alpha_t(s_t-\delta))\to\pi-\varepsilon$ as $r\to-\frac{\pi}{2}$. In particular, $\tau(\gamma(r),p_+) \gt \pi$. By the synthetic Bonnet–Myers theorem 1.6, we know that this forces $\tau(\gamma(r),p_+)=+\infty$, contradicting finiteness of τ (see [Reference Kunzinger and Sämann26, theorem 3.28]).

The situation where there is a point $p\in I^-(\gamma)\setminus I^+(\gamma)$ is analogous.

Finally, assume that all points in $X \setminus I(\gamma)$ are in neither $I^+(\gamma)$ nor in $I^-(\gamma)$. Take any such point p, then by strong causality, we have $p_-\ll p\ll p_+$. Then also $p_-,p_+$ are neither in $I^+(\gamma)$ nor in $I^-(\gamma)$. Thus, $I(p_-,p_+)\subseteq X \setminus I(\gamma)$ is an open neighbourhood of p. This works for any p, so both $I(\gamma)$ and $X \setminus I(\gamma)$ are open, contradicting connectedness of X.

Proof. This follows from transforming the geodesics in AdS to warped product coordinates.

Proof. W.l.o.g. assume $x(\alpha(\lambda))\geq0$ for all λ. Note that $\tau((t_0,0),(t,x))$ is monotonically decreasing in $|x|$, and thus the worst case is attained when α is null, i.e. we can assume $\alpha(t)=\left(t,\int_{t_1}^t\frac{1}{\cos(t)}{\mathrm{d}} t\right)$.

Now we calculate $\lim_{t\to a}\tau((t_0,0),\alpha(t))$: For the calculation, we switch to the three-dimensional picture in definition 2.2, and there we can use the formula

\begin{equation*} \tau(p,q)=\arccos(-\left \lt p,q\right \gt ) \end{equation*}

where the inner product is in the ambient semi-Riemannian vector space, see [Reference Cunningham, Rideout, Halverson and Krioukov22, (2.7)]. The point $\lim_{t\to a}\alpha(t)$ exists and is given by the intersection of the two null geodesics:

\begin{equation*} \{(0,1,0)+s(1,0,1):s\in\mathbb{R}\} \;\cap\; \{(\cos(t_1),\sin(t_1),0)+s(-\sin(t_1),\cos(t_1),1) :s\in\mathbb{R}\} \end{equation*}

yielding the point

\begin{equation*} \lim\alpha(t)=\left(\frac{1-\sin(t_1)}{\cos(t_1)},1,\frac{1-\sin(t_1)}{\cos(t_1)}\right) \end{equation*}

so

\begin{equation*} \tau((\cos(t_0),\sin(t_0),0),\lim\alpha(t))=\arccos\left(\frac{\cos(t_0)(1-\sin(t_1))}{\cos(t_1)}+\sin(t_0)\right) \end{equation*}

where we used the inner product formula for τ.

as $t_0\to-\frac{\pi}{2}$, this goes to π:

\begin{equation*} \arccos\left(-1\right)=\pi \end{equation*}

Proof. This is a part of [Reference Beran, Ohanyan, Rott and Solis7, theorem 2.23].

Proof. Suppose there is a point $p \in I(\gamma)$ such that the claim does not hold at p, so w.l.o.g. there is a sequence $p_n \to p$, $t_n \to +\frac{\pi}{2}$ and maximal timelike curves σ_n from p_n to $y_n:=\gamma(t_n)$ such that σ_n converge to a future directed null ray σ. Choose some $x \in \gamma \cap I^-(p)$ and let µ_n be maximal timelike curves from x to p_n (assuming n to be large enough, $x\ll p_n$). Applying the limit curve theorem, it is easily seen that (up to a choice of subsequence) the µ_n converge locally uniformly to a maximizing limit causal curve µ from x to p. µ is timelike since $x \ll p$. Denote by γ_n the piece of γ that runs between x and y_n. Set $a_n:=L_{\tau}(\mu_n)$, $b_n:=L_{\tau}(\sigma_n)$ and $c_n:=L_{\tau}(\gamma_n)$. Let $\beta_n:=\measuredangle_x(\mu_n,\gamma_n)$ and $\theta_n:=\measuredangle_{p_n}(\mu_n,\sigma_n)$. Then $a_n \to a:=\tau(x,p)$ and by the continuity of angles (cf. proposition 2.15), $\beta_n \to \beta$, where β is the angle between µ and γ. Consider the comparison triangles for $\Delta(x,p_n,y_n)$ in AdS and call its sides $(\overline{\mu}_n,\overline{\sigma}_n,\overline{\gamma}_n)$, then the angle $\overline{\beta}_n$ between $\overline{\mu}_n$ and $\overline{\gamma}_n$ satisfies $\overline{\beta}_n \leq \beta_n$ and similarly $\overline{\theta}_n \geq \theta_n$. Since $\beta_n \to \beta$, there is some C > 0 such that $\overline{\beta}_n \leq C$ for all n. The hyperbolic law of cosines in ${\mathrm{AdS}} $ (see lemma 2.7) gives

\begin{align*} &\cos(b_n)=\cos(a_n)\cos(c_n)+\sin(a_n)\sin(c_n)\cosh(\overline{\beta}_n),\\ &\cos(c_n)=\cos(a_n)\cos(b_n)-\sin(a_n)\sin(b_n)\cosh(\overline{\theta}_n). \end{align*}

Using the first equation, noting that $\overline{\beta}_n$ stays bounded and that a_n and c_n stay bounded away from zero and π, we get that also b_n stays bounded away from zero and π. This means that also $\overline{\theta}_n$ stays bounded.

From the monotonicity condition, we get that

\begin{equation*} \tilde{\measuredangle}_{p_n}(\sigma_n(s),\mu_n(t)))\leq\tilde{\measuredangle}_{p_n}(x,y_n)=\bar{\theta}_n \end{equation*}

for each s and t, so this is bounded too. We will see that this is incompatible with σ_n ‘getting more and more null’: Note that we get the following estimate for some constant C ^ʹʹ:

\begin{align*} C^{\prime\prime} &\geq \cosh( \tilde{\measuredangle}_{p_n}(\sigma_n(s),\mu(t))) \\ &= \frac{\cos(\tau(\mu_n(t),p_n))\cos(\tau(p_n,\sigma_n(s)))-\cos(\tau(\mu_n(t),\sigma_n(s)))}{\sin(\tau(p_n,\sigma_n(s))\sin(\tau(\mu_n(t),p_n))}. \end{align*}

Since the denominator goes to 0 for $n \to \infty$ (as $\tau(p_n,\sigma_n(s)) \to \tau(p,\sigma(s)) = 0$ and $\tau(\mu_n(t),p_n) \to \tau(\mu(t),p) \gt 0$), the enumerator has to go to 0 as well; in particular, this means

\begin{align*} \tau(\mu(t),\sigma(s)) = \tau(\mu(t),p) \end{align*}

for all $s,t$. This implies that running along µ from $\mu(t)$ to p and then along σ to $\sigma(s)$ gives a maximizer, but this curve has a timelike and a null piece, a contradiction to [Reference Kunzinger and Sämann26, theorem 3.18].

Proof. Let $\alpha_n:[0,a_n]\to X;\,p\leadsto c(t_n)$ be the sequence converging to α, it is parametrized in d-arclength parametrization. As d is proper, we have that $a_n\to+\infty$ (otherwise by properness, $c(t_n)$ would converge to some P ₁, and if the whole curve didn’t converge to P ₁ other subsequence would converge to some other point P ₂ within a local causal closed neighbourhood of P ₁, and we get $P_1\leq P_2\leq P_1$, contradicting causality of X). Now we can apply the Limit curve theorem for inextensible curves [Reference Kunzinger and Sämann26, theorem 3.14] to get that α is inextensible.

Proof. We set $s_-=\sup(\gamma^{-1}(I^-(p)))$ and $s_+=\inf(\gamma^{-1}(I^+(p)))$. Then the set of s where $\gamma(s)$ is timelike related to p is $(-\infty,s_-)\cup(s_+,+\infty)$, see figure 5. We assume $p\ll y_2$, the other case $y_2\ll p$ can be reduced to this one by flipping the time orientation of the space.

Figure 5.

The domain where the y_i can lie in is doubled.

We create comparison situations for an Alexandrov situation: We take the triangles $\bar{\Delta}_{12}=\Delta(\bar{p},\bar{y}_1,\bar{y}_2)$ and $\bar{\Delta}_{23}=\Delta(\bar{p},\bar{y}_2,\bar{y}_3)$ as in the statement. We also create a comparison triangle $\tilde{\Delta}_{13}=\Delta(\tilde{p},\tilde{y}_1,\tilde{y}_3)$ for $(p,y_1,y_3)$ and get a comparison point $\tilde{y}_2$ for y ₂ on the side $y_1y_3$. Then we can apply curvature comparison to get $\bar{\tau}(\tilde{p},\tilde{y}_2) \geq \tau(p,y_2)=\bar{\tau}(\bar{p},\bar{y}_2)$. By lemmas 2.16 and 2.17 (if $p\ll y_1\ll y_2\ll y_3$, we use lemma 2.17 and if $y_1\ll p\ll y_2\ll y_3$ we use lemma 2.16), this means the situation is convex, i.e.

(4.1)\begin{equation} \tilde{\measuredangle}_{y_2}(p,y_1)\leq \tilde{\measuredangle}_{y_2}(p,y_3)\,. \end{equation}

If $p\ll y_1$ we also get

(4.2)\begin{equation} \tilde{\measuredangle}_{y_1}(p,y_2)\leq\tilde{\measuredangle}_{y_1}(p,y_3) \end{equation}

and if $y_1\ll p$, inequality (4.2) flips.

Note that inserting back $y_2=\gamma(t_2)$ and $y_3=\gamma(t_3)$ in (4.2), varying $t_2,t_3$ and replacing y ₁ by y ₂ gives that $\tilde{\measuredangle}_{y_2}(p,\gamma(t))$ is monotonically increasing in t for $t \gt t_2$. Similarly, the time-reversed situation gives that for all t with $p\ll \gamma(t)$, $\tilde{\measuredangle}_{y_2}(p,\gamma(t))$ is monotonically increasing in t and similarly for t with $\gamma(t)\ll p$. Note that inserting the definitions of y ₁ and y ₃ in (4.1) and varying $t_1,t_3$ gives us the last piece to say $\tilde{\measuredangle}_{y_2}(p,\gamma(t))$ is monotonically increasing on its whole domain. We revisit (4.1): $\tilde{\measuredangle}_{y_2}(p,\gamma(t_1))\leq \tilde{\measuredangle}_{y_2}(p,\gamma(t_3))$. Note that the left hand side is decreasing with decreasing t ₁ and the right hand side is increasing with increasing t ₃.

Claim: For each t ₁ and t ₃, equality holds in (4.1).

Then: The comparison situation is straight, i.e. $\bar{y}_1,\bar{y}_2,\bar{y}_3$ lie on a distance realizer, and by triangle equality along distance realizers we have $\tau(y_1,y_3)=\tau(y_1,y_2)+\tau(y_2,y_3)=\bar{\tau}(\bar{y}_1,\bar{y}_2)+\bar{\tau}(\bar{y}_2,\bar{y}_3)=\bar{\tau}(\bar{y}_1,\bar{y}_3)$, i.e. $\bar{p},\bar{y}_1,\bar{y}_3$ is a comparison triangle for $p,y_1,y_3$.

Proof: For any $t_1,t_3$ we can draw $\bar{\Delta}_{12}$, $\bar{\Delta}_{23}$ simultaneously in AdS. We indirectly assume there are $t_1,t_3$ such that the inequality (4.1) is strict. We first want to show that there are such $t_1,t_3$ where $\bar{\Delta}_{12}$, $\bar{\Delta}_{23}$ can be drawn simultaneously in the warped product picture ${\mathrm{AdS}} ^\prime$. As a first case, let inequality (4.1) always be strict. Then we take the limit as $t_1\nearrow t_2$ and $t_3\searrow t_2$, as the sides are monotonous the limits of the sides are both finite, and thus for $t_1,t_3$ close enough to t ₂, we have that $\bar{y}_1,\bar{y}_3$ are realizable in ${\mathrm{AdS}} ^\prime$ (by triangle equality of angles in AdS, we take an angle a little bit bigger than the limits, then small distances can be realized at that angle or lower from a prescribed side $\bar{p}\bar{y}_2$). The other case is where inequality (4.1) has equality for $t_1,t_3$ close to t ₂, but one of the sides changes when taking t ₁ resp. t ₃ further away from t ₂. W.l.o.g., say the right side changes, so there is a $t_3^0 \gt t_3$ such that $\tilde{\measuredangle}_{y_2}(p,\gamma(t_3^0)) \gt \tilde{\measuredangle}_{y_2}(p,\gamma(t_3))$. Note that as long as (4.1) has equality, we have that $\tau(\bar{y}_1,\bar{y}_3)=\tau(\bar{y}_1,\bar{y}_2)+\tau(\bar{y}_2,\bar{y}_3) \lt L(\gamma)=\pi$. As $\tilde{\measuredangle}_{y_2}(p,\gamma(t_3))$ is continuous in t ₃, there also exists a $t_3^0$ such that $\tilde{\measuredangle}_{y_2}(p,\gamma(t_3^0)) \gt \tilde{\measuredangle}_{y_2}(p,\gamma(t_3))$ and $\tau(\bar{y}_1,\bar{y}_3) \lt \pi$. In particular, it is realizable, with $\bar{y}_1^0,\bar{y}_3^0$ on a vertical line, and all of $\bar{y}_1^0,\bar{y}_3^0,\bar{p}$ to the left of the vertical line with x-coordinate $x(\bar{y}_2)$. In any case, we now fix this realization, i.e. fix the points $\bar{y}_1^0,\bar{y}_2,\bar{y}_3^0$ and $\bar{p}$.

Now consider $t_1^* \lt t_1^0$ and $t_3^* \gt t_3^0$. On the side $\bar{y}_2\bar{p}$, we can construct comparison points for $\gamma(t_1^*)$ and $\gamma(t_3^*)$, call them $\bar{y}_1^*,\bar{y}_3^*$, if they are still contained in ${\mathrm{AdS}} ^\prime$. Note that $\bar{y}_1^*,\bar{y}_3^*$ vary continuously with $t_1^*,t_3^*$, so if $\bar{y}_1^*,\bar{y}_3^*$ are not defined for some $t_i^*$ we get a $t_1^{\min},t_3^{\max}$ such that it is defined before that value, and $\bar{y}_1^*$ or $\bar{y}_3^*$ converge to infinity in the limit to $t_1^{\min}$ resp. $t_3^{\max}$, otherwise set $t_1^{\min}=-\frac{\pi}{2}$ and / or $t_3^{\max}=\frac{\pi}{2}$, then similar limits hold. As $\tilde{\measuredangle}_{y_2}(p,\gamma(t_3^*)) \gt \tilde{\measuredangle}_{y_2}(p,\gamma(t_3^0))$, we get that $\bar{y}_3^*$ is to the left of the (extended) geodesic connecting $\bar{y}_2t_3^0$, and as $\tau(\bar{y}_2,\bar{y}_3^*) \gt \tau(\bar{y}_2,\bar{y}_3^0)$, it has much more negative x-coordinate. As the (extended) geodesic connecting $\bar{y}_2t_3^0$ is going to the left, it has x-coordinate converging to $-\infty$ towards the boundary, and thus so has $\bar{y}_3^*$, similarly $\bar{y}_1^*$. In particular, as $\bar{y}_1^*,\bar{y}_3^*$ depend continuously on $t_1^*,t_3^*$, there are parameters such that $\bar{y}_1^*,\bar{y}_3^*$ have the same x-coordinate as p, thus by triangle equality along distance realizers and strict triangle inequality we have that

\begin{equation*} \tau(\bar{y}_1^*,\bar{p})+\tau(\bar{p},\bar{y}_3^*)=\tau(\bar{y}_1^*,\bar{y}_3^*) \gt \tau(\bar{y}_1^*,\bar{y}_2)+\tau(\bar{y}_2,\bar{y}_3^*)\,, \end{equation*}

in contradiction to the fact that $\gamma(t_1^*),\gamma(t_2),\gamma(t_3^*)$ lie on a distance realizer. Thus we get the claim. See figure 6 for a visualization of the construction.

Figure 6.

We assume that the path $\bar{y}_1^0\bar{y}_2\bar{y}_3^0$ is longer than the path $\bar{y}_1^0\bar{p}\bar{y}_3^0$. If the angle at $\bar{y}_2$ is not straight, we extend some lines and as points further from $\bar{p}$ are more on the left, we find some $t_1^*$ and $t_3^*$ such that $\bar{y}_1^*\bar{p}\bar{y}_3^*$ are collinear, but then the path $\bar{y}_1^0\bar{y}_2\bar{y}_3^0$ is shorter than $\bar{y}_1^0\bar{p}\bar{y}_3^0$, which then yields a contradiction to γ being maximizing.

Proof. First, we check that the comparison angle $\tilde{\measuredangle}_x(p,\gamma(s))$ is constant in s: For s ₁ and s ₂ for which this is defined, we have three parameters on γ involved: $s_1,s_2,t_0$. The previous result (proposition 4.7) tells us we can construct a comparison situation for all three triangles at once. As the comparison situations have the comparison angle $\tilde{\measuredangle}_x(p,\gamma(s_1))$ resp. $\tilde{\measuredangle}_x(p,\gamma(s_2))$ as the angle in $\bar{x}$, they are equal.

Now we look at the angle $\measuredangle_x(\alpha,\gamma_\pm)$: We assume $p\ll x$. We already know $\tilde{\measuredangle}_x(\alpha(s),\gamma(t))$ is constant in t. We now have to look at its dependence on s: By theorem 2.12, $\tilde{\measuredangle}_x^{\mathrm{S}}(\alpha(s),\gamma(t))$ is monotonically increasing in s. Note now that for $t \lt t_0$, the sign of this angle is $\sigma=-1$, and for $t \gt t_0$, the sign of this angle is $\sigma=+1$. So choose some $t_- \lt t_0$ and $t_+ \gt t_0$ for which all the necessary angles exist (i.e. $\gamma(t_-)\ll p$ and $t_+ \gt t_0$). Then the above applied to $\alpha(s)$ instead of p gives that $\tilde{\measuredangle}_x(\alpha(s),\gamma(t_-))=\tilde{\measuredangle}_x(\alpha(s),\gamma(t_+))$ is both a monotonically decreasing and increasing function in s. Thus it is constant, and $\tilde{\measuredangle}_x(\alpha(s),\gamma(t))=\measuredangle_x(\alpha,\gamma_-)=\measuredangle_x(\alpha,\gamma_+)$, which includes the desired equalities.

Proof. We distinguish which sides the q_i lie on: Note we assumed one of them is on the side $x_1x_2$. We only prove the case where q ₁ is on the side $\gamma_+$ connecting $x_1x_2$ (say $q_1=\gamma_+(s)$) and q ₂ is on the side β connecting $x_1,p$ (say $q_2=\beta(t)$), the proof of the other cases is easily adapted.

Now the comparison triangle $\tilde{\Delta}$ for $\Delta(x_1,q_1,q_2)$ has angle $\tilde{\measuredangle}_{x_1}(q_1,q_2)=\measuredangle_{x_1}(\gamma_+,\beta)$ at the point corresponding to x ₁, and the comparison triangle for $\Delta=\Delta(x_1,p,x_2)$ has the same angle $\tilde{\measuredangle}_{x_1}(p,x_2)=\measuredangle_{x_1}(\gamma_+,\beta)$ at the point corresponding to x ₁. As the comparison points $\bar{q}_1$ and $\bar{q}_2$ lie on the sides enclosing the angle, also $\Delta(\bar{x}_1,\bar{q}_1,\bar{q_2})$ has the angle $\measuredangle_{x_1}(\gamma_+,\beta)$ at $\bar{x}_1$. Now the triangles $\tilde{\Delta}$ and $\Delta(\bar{x}_1,\bar{q}_1,\bar{q_2})$ have two sides and the angle between them equal, and thus also the opposite side is equal, establishing $\tau(q_1,q_2)=\bar{\tau}(\bar{q}_1,\bar{q}_2)$. A continuity argument (moving q ₁ a bit into the past) then gives that $q_1\leq q_2$ if and only if $\bar{q}_1\leq\bar{q}_2$.

Proof. Let $x\in I(\gamma)$. We dive into the limits: We have that for t ₁ small enough and t ₂ large enough that

\begin{equation*} \tau(\gamma(t_1),\gamma(t_2))-\pi\geq \underbrace{\tau(\gamma(t_1),x)-\frac{\pi}{2}}_{\to-b_-(x)}+\underbrace{\tau(x,\gamma(t_2))-\frac{\pi}{2}}_{\to-b_+(x)}\,, \end{equation*}

and that the left hand side converges to 0 as $t_1\to-\frac{\pi}{2}$ and $t_2\to\frac{\pi}{2}$.

Proof. By construction, $\alpha:[0,b) \to X$ arises as a locally uniform limit of timelike maximizers $\alpha_n:[0,a_n] \to X$ from x to $z_n:=\gamma(t_n)$, where $t_n \to +\frac{\pi}{2}$. For each compact subinterval of the domain of α, we can apply lemma 4.3 to get $L(\alpha|_{[0,c]})=\lim_nL(\alpha_n|_{[0,c]})\leq \lim_n L(\alpha_n)$, thus $L:=L_{\tau}(\alpha) \leq \frac{\pi}{2}-t$. Now indirectly suppose that $L:=L_{\tau}(\alpha) \lt \frac{\pi}{2}-t$. By continuity of angles (see proposition 2.15), $\omega_n:=\measuredangle_x(\alpha,\alpha_n) \to 0$ for $n \to \infty$.

Now note that for any t_n with $n \geq N$, $\partial J^-(z_n) \cap \alpha = (J^-(z_n)\setminus I^-(z_n)) \cap \alpha$ is non-empty: Certainly, $x \in I^-(z_n)$, so if this intersection were empty, then α would be imprisoned in the compact set $J^+(x) \cap J^-(z_n)$, which cannot happen by proposition 4.6. So we find a point $y_n \in \alpha$ that is null-related to z_n, i.e. $y_n \lt z_n$ and $\tau(y_n,z_n) = 0$.

We now consider the triangle given by the vertices $x,y_n,z_n$. Two of the sides have natural curves: From x to y_n set ν_n to be the part of α from x to y_n, and from x to z_n take α_n. We name the side-lengths: $c_n=\tau(x,z_n)=L(\alpha_n)$, $a_n=\tau(x,y_n)=L(\nu_n)$, and $b_n=\tau(y_n,z_n)=0$. We know a lot about this triangle as $n\to+\infty$:

• • By the definition of $t=b_+(x)$, $c_n\to \frac{\pi}{2}-t$. Note this is the longest side-length and c_n is bounded away from π as $b_+(x)\geq -b_-(x) \gt -\frac{pi}{2}$.

• • $b_n=0$ by the choice of y_n.

• • $a_n=\tau(x,y_n) \lt L(\alpha)$, so $\limsup_{n} a_n \leq L \lt \frac{\pi}{2}-t$.

• • Our curvature assumption gives that $\bar{\omega}_n:=\tilde{\measuredangle}_x(y_n,z_n)\leq \omega_n\to 0$.

But this cannot be: we claim we can take a comparison situation $\Delta(\bar{x},\bar{y}_n,\bar{z}_n)$ which converges as $n\to+\infty$: Fix $\bar{x}$, choose $\bar{z}_n$ along a fixed distance realizer $\bar{\alpha}_n$ (independent of n!) through $\bar{x}$, take $\bar{y}_n$ on a distance realizer $\bar{\alpha}$ (dependent on n!) through $\bar{x}$ at the correct angle $\measuredangle_{\bar{x}}(\bar{\alpha}_n,\bar{\alpha})=\bar{\omega}_n$. Then $\bar{x}$ and $\bar{z}_n$ automatically converge, and $\bar{y}_n$ does so too as $\bar{\omega}_n$ converges.

Note the limiting situation is degenerate, as $\bar{\omega}_n\to0$, c_n is bounded away from 0 and $a_n \gt 0$ is increasing, so bounded away from 0 as well. Thus, the triangle inequality of angles has equality in the limit, so

\begin{equation*} 0=\lim_n (c_n-a_n-b_n) = \left(\frac{\pi}{2}-t\right) - L - 0 \end{equation*}

which contradicts $L \lt \frac{\pi}{2}-t$.

Proof. Let $\sigma_n^+$ and $\sigma_n^-$ be two sequences of timelike maximizers from p to $\gamma(r_n)$ and $\gamma(-r_n)$, respectively, such that $\sigma^+$ and σ ⁻ arise as limits of these sequences as $r_n \to \infty$. To show that σ is a line, it is sufficient to show that for any $s \lt 0 \lt t$, $\tau(\sigma(s),\sigma(t)) = L_{\tau}(\sigma|_{[s,t]})$. To see this, let $q_+:=\sigma(t)$ and $q_-:=\sigma(s)$, and $q_+^n:=\sigma_n^+(t)$, $q_-^n:=\sigma_n^-(s)$. Then $q_{\pm} = \lim_n q_{\pm}^n$. Consider the triangle $\Delta(q_-^n,p,q_+^n)$ with sides $\sigma_n^-$, $\sigma_n^+$ and a part of γ. Consider a comparison triangle with points $\overline{q_{\pm}^n}$ corresponding to $q_{\pm}^n$. Sending $n \to \infty$, we see that the stacked comparison triangles in ${\mathrm{AdS}} ^\prime$ converge to a vertical line (here we use proposition 4.7 and that γ has length π); hence, our curvature bound gives

\begin{align*} \tau(q_-,q_+) = \lim_{n \to \infty} \tau(q_-^n,q_+^n) \leq \lim_{n \to \infty} \overline{\tau}(\overline{q_-^n},\overline{q_+^n}) = \lim_n L_{\tau}(\sigma_n^-\sigma_n^+|_{[s,t]})= L_{\tau}(\sigma|_{[s,t]}), \end{align*}

which is what we wanted to show, as the other inequality is trivial.

$a_+-a_-=\pi$ also follows as the stacked comparison triangles in ${\mathrm{AdS}} ^\prime$ converge to a vertical line. For the future and past Busemann functions, we also look at this picture: we get a vertical line $\bar{\gamma}:(-\frac{\pi}{2},\frac{\pi}{2})\to{\mathrm{AdS}} ^\prime$ and $\bar{p}\in{\mathrm{AdS}} ^\prime$ such that any $\Delta(\bar{\gamma}(t_1),\bar{p},\bar{\gamma}(t_2))$ is a comparison triangle. In particular, $b_+(p)=\frac{\pi}{2}-\lim_{t_2\to\frac{\pi}{2}}\bar{\tau}(\bar{p},\bar{\gamma}(t_2)) = t(p) =\cdots= -b_-(p)$ is the t-coordinate.

Proof. We consider the sets $I=\{(s,t):\alpha(s)\ll\beta(t)\}$ (open) and $J=\{(s,t):\alpha(s)\leq\beta(t)\}=\bar{I}$ (as X is globally causally closed). They are non-empty: as $\beta\cap I^+(\alpha)\neq\emptyset$, we find $\alpha(s_0)\ll\beta(t_0)$, so $(s_0,t_0)\in I$. We claim that the first projection of I is full, i.e. $pr_1(I)=\{s:\exists t:\alpha(s)\ll\beta(t)\}=(-\frac{\pi}{2},\frac{\pi}{2})$. We know s ₀ is in the left hand side, and with it all smaller values $s \lt s_0$. So let indirectly $s_n\nearrow s_+ \lt \frac{\pi}{2}$ and $t_n\nearrow \frac{\pi}{2}$ such that $(s_n,t_n)\in I$, but $s_+\not\in pr_1(I)$. Then by continuity of τ, $\tau(\alpha(s_n),\beta(t_n))\to0$. Looking at the formula for $c_{\alpha\beta}(s_n,t_n)$, one sees that the enumerator stays away from 0, whereas the denominator approaches 0. Thus $c_{\alpha\beta}(s_n,t_n)\nearrow+\infty$ which cannot be as it is constant, so $pr_1(I)=(-\frac{\pi}{2},\frac{\pi}{2})$.

Proof. We define f as in the definition of parallel AdS-lines with distance c: $f(\alpha(s)):=(s,0)$, $f(\beta(s)):=(s,c)$ in ${\mathrm{AdS}} ^\prime$. We will prove that f is ≤-preserving if and only if $c_{\alpha+}^N$ and $c_{\beta+}^N$ are constantly c, and under the assumption that this is the case f is τ-preserving if and only if $c_{\alpha\beta}$ and $c_{\beta\alpha}$ are constantly c wherever defined.

As α and β are future directed τ-arclength parametrized AdS-lines, it is clear that f is ≤- and τ-preserving along α and along β, i.e. we only have to check the conditions on $f(\alpha(s)){\bar\leq} f(\beta(t))$ and their τ-distance, and both also for $\alpha,\beta$ reversed.

So first, for ≤-preserving: For a fixed s, we describe the set of t such that $\alpha(s)\leq\beta(t)$ resp. $f(\alpha(s)){\bar\leq} f(\beta(t))$, the situation switching α and β is analogous (using $c_{\beta+}^N$ instead of $c_{\alpha+}^N$). By transitivity of ≤ resp. $\bar{\leq}$, these sets will both be an interval of the form $[t_{min},+\frac{\pi}{2})$ or $(t_{min},+\frac{\pi}{2})$, resp. $[\tilde{t}_{min},+\frac{\pi}{2})$ (always closed). Note that inserting $\tilde{t}_{min}$ instead of t into the definition of $c_{\alpha+}^N(s)$ yields c by the geometric definition. As the formula of $c_{\alpha+}^N(s)$ is strictly increasing in t, we have that $t_{min}=\tilde{t}_{min}$ if and only if $c=c_{\alpha+}^N(s)$. Thus $\{t:\alpha(s)\leq\beta(t)\}=[\tilde{t}_{min},+\frac{\pi}{2})$ if and only if $c_{\alpha+}^N(s)=c$ and the infimum in the definition of $c_{\alpha+}^N(s)$ is achieved. The same argument works out for α and β exchanged, thus proving that f is ≤-preserving if and only if $c_{\alpha+}^N$ and $c_{\beta+}^N$ are constant and equal to c, and the infima appearing are minima.

For τ-preserving, we assume that f is already ≤-preserving. We will only prove $\tau(\alpha(s),\beta(t))=\bar{\tau}(f(\alpha(s)),f(\beta(t)))$ for $\alpha(s)\leq\beta(t)$, the case where they are causally unrelated is covered by ≤-preserving, and the case where α and β are interchanged follows analogously. Note that inserting $\bar{\tau}(f(\alpha(s)),f(\beta(t)))$ for $\tau(\alpha(s),\beta(t))$ into the definition of $c_{\alpha\beta}(s,t)$ yields c by the geometric definition. As the formula of $c_{\alpha\beta}(s,t)$ is strictly decreasing in $\tau(\alpha(s),\beta(t))$, we have that $\tau(\alpha(s),\beta(t))=\bar{\tau}(f(\alpha(s)),f(\beta(t)))$ if and only if $c_{\alpha\beta}(s,t)=c$. The same argument works out for α and β exchanged, thus proving that if f is ≤-preserving, f is τ-preserving if and only if $c_{\alpha\beta}$ and $c_{\beta\alpha}$ are constant and equal to c.

For the additional statement, fix s. By global causal closedness, the infimum in the formula of $c_{\alpha+}^N$ is automatically a minimum, and it is non-void by lemma 4.18. For the value of $c_{\alpha+}^N(s)$, let t ₀ be the minimum. Then for any $t \gt t_0$ we have $\alpha(s)\ll\beta(t)$, and by continuity of τ we have that $\tau(\alpha(s),\beta(t_0))=0$. Now note that plugging this into $c_{\alpha\beta}(s,t_0)$ gives the formula for $c_{\alpha+}^N$.

Proof. We set $f_+(s,t)=\tau(\alpha(s),\beta(t))$ and $f_-(s,t)=\tau(\beta(s),\alpha(t))$. They are both monotonically increasing in t and continuous. We want to describe the set where $f_+ \gt 0$ resp. $f_- \gt 0$. We set $t_s^+=\inf \{t:f_+(s,t) \gt 0\}$, then $\lim_{t\searrow t_s^+} f_+(s,t)=0$. Thus in the law of cosines (lemma 2.7), we get $\lim_{t\searrow t_s^+}\cosh(\tilde{\measuredangle}_x(\alpha(s),\beta(t)))=1 = \lim_{t\searrow t_s^+} \frac{\cos(f_+(s,t)) - \cos(s) \cos(t)}{\sin(s) \sin(t)}$, so $s=t_s^+$. Analogously, we get $s=\inf \{t:f_-(s,t) \gt 0\}$, in total $f_\pm(s,t) \gt 0$ for s < t.

That means that whenever $s_- \lt t \lt s_+$ we have that $\alpha(t)\in I(\beta(s_-),\beta(s_+))$. By strong causality, the $I(\beta(s_-),\beta(s_+))$ form a neighbourhood basis of the point $\beta(t)$, and $\alpha(t)$ is inside of all of these neighbourhoods. As X is Hausdorff, we get that $\alpha(t)=\beta(t)$ for all $t\in(0,b)$, so $\alpha=\beta$.

Proof. We indirectly assume there are two AdS-parallel lines to α through p, namely $\beta:(-\frac{\pi}{2},\frac{\pi}{2})\to X$ and $\tilde{\beta}:(-\frac{\pi}{2},\frac{\pi}{2})\to X$, with $p=\beta(t_0)=\tilde{\beta}(\tilde{t}_0)$. We know that $t_0=\tilde{t}_0$ and that both $\alpha,\tilde{\beta}$ have the same distance, as there is only one way (up to isometry) of realizing α vertically in ${\mathrm{AdS}} ^\prime$ and putting p at the right distance.

We construct a comparison situation for all three lines at once: We choose the parallel realization f for α and β given by $f(\alpha(s))=(s,0)$ and $f(\beta(s))=(s,c)$, and similarly we choose $\tilde{f}$ for α and $\tilde{\beta}$ given by $\tilde{f}(\tilde{\beta}(s))=(s,-c)$ and $\tilde{f}(\alpha(s)) = (s,0)$, i.e. we realize β and $\tilde{\beta}$ on opposite sides of α.

We calculate the angle $\omega:=\measuredangle_p(\beta|_{[t_0,\frac{\pi}{2})},\tilde{\beta}|_{[t_0,\frac{\pi}{2})})$: By proposition 4.8, this is equal to any comparison angle $\tilde{\measuredangle}_p(\beta(s),\tilde{\beta}(t))$ as long as $\beta(s)\ll\tilde{\beta}(t)$ or conversely.

We know by lemma 4.18 that given $r\in(-\frac{\pi}{2},\frac{\pi}{2})$, we find an $s\in (-\frac{\pi}{2},\frac{\pi}{2})$ such that $\beta(r)\ll\alpha(s)$ and a $t\in (-\frac{\pi}{2},\frac{\pi}{2})$ such that $\alpha(s)\ll\tilde{\beta}(t)$. In particular, for all $r\in(-\frac{\pi}{2},\frac{\pi}{2})$, there is a $t\in(-\frac{\pi}{2},\frac{\pi}{2})$ such that $\beta(r)\ll\tilde{\beta}(t)$. We now look at the law of cosines to estimate the comparison angle $\omega=\tilde{\measuredangle}_p(\beta(r),\tilde{\beta}(t))$: $\cosh(\omega)=\frac{\cos(\tau(\beta(r),\tilde{\beta}(t)))-\cos(r-t_0)\cos(t-t_0)}{\sin(r-t_0)\sin(t-t_0)}\leq \frac{1-\cos(r-t_0)\cos(t-t_0)}{\sin(r-t_0)\sin(t-t_0)}$ which converges to 1 as s and t both approach the maximal value $\frac{\pi}{2}$. In particular, ω = 0. We can now apply the strong causality trick (lemma 4.20) to get $\beta=\tilde{\beta}$, so there is only one line parallel to α through p.

Proof. By construction, α arises as the limit of timelike maximizing segments $\alpha^+_n$ from $p:=\alpha(t_0)$ to $\gamma(r_n)$ and $\alpha^-_n$ from p to $\gamma(t_n)$, where $r_n \to -\frac{\pi}{2}$ and $t_n\to\frac{\pi}{2}$. We will use the stacking principle (cf. proposition 4.7) to show that α and γ are parallel. Indeed, the triangles corresponding to $\Delta(\gamma(r_n),p,\gamma(t_n))$ stack in ${\mathrm{AdS}} ^\prime$ (see figure 8), and we have that the sides corresponding to $\alpha^+_n$ and $\alpha^-_n$ in ${\mathrm{AdS}} ^\prime$ converge to the vertical line through $\bar{p}$.

Figure 8.

Stacking comparison triangles. Every three subsequent $\bar y_k$ together with $\bar{p}$ form two triangles as in proposition 4.7.

We now argue that the map sending $\alpha((-\frac{\pi}{2},\frac{\pi}{2}))$ and $\gamma((-\frac{\pi}{2},\frac{\pi}{2}))$ to the corresponding limit lines $\bar{\alpha}$ and $\bar{\gamma}$ is a parallel realization.

For any $t,s$, consider $\tau(\alpha(t),\gamma(s)) = \lim_n \tau(\alpha_n(t),\gamma(s))$. For n so large that $-r_n \lt s \lt r_n$, $\gamma(s)$ is part of the triangle $\Delta(\gamma(-r_n),p,\gamma(r_n))$. From corollary 4.9, we conclude that $\alpha_n(t) \ll \gamma(s)$ if and only if $\overline{\alpha}_n(t) \ll \overline{\gamma}(s)$, and $\tau(\alpha_n(t),\gamma(s)) = \overline{\tau}(\overline{\alpha}_n(t),\overline{\gamma}(s))$, where the bars denote the corresponding points on the comparison side. This shows that the realization of α as $\bar{\alpha}$ and γ as $\bar{\gamma}$ is a parallel realization.

Proof. Let $\tilde{\alpha}$ be the asymptotic line through $\alpha(s)$. Then the previous result shows that both γ and α as well as γ and $\tilde{\alpha}$ are parallel and they meet at $\alpha(t_0)=\tilde{\alpha}(t_0)$. Thus, as parallel lines are unique (see lemma 4.21), we have that $\alpha=\tilde{\alpha}$.

Proof. Since asymptotic lines to γ are parallel to γ by lemma 4.22, this readily follows.

Proof. We dive into the construction of asymptotes: We first assume $x\ll y$. We set $s_0=b^+(x)$ and $t_0=b^+(y)$, then there are maximizers α_n from x to $\gamma(T_n)$ parametrized in τ-arclength such that $\alpha_n(s_0)=x$ and β_n from y to $\gamma(T_n)$ parametrized in τ-arclength such that $\beta_n(t_0)=y$, which converge (pointwise) to the upper parts of α and β in τ-arclength parametrization for $T_n \to \infty$, respectively. Note that $\alpha_n(s_0+\tau(x,\gamma(T_n)))=\beta_n(t_0+\tau(y,\gamma(T_n)))=\gamma(T_n)$.

We try to prove the c-criterion (lemma 4.19): Clearly, due to our assumptions on X, we do not need to consider the null functions. First, we look at $c_{\alpha\beta}(s_0,t_0)$ and $c_{\alpha\beta}(s^\prime,t^\prime)$ or $c_{\beta\alpha}(s^\prime,t^\prime)$ for $s_0\leq s^\prime$, $t_0\leq t^\prime$: We require $a:=x=\alpha(s_0)\ll b:=y=\beta(t_0)$. We get points converging to the primed parameters: $a^\prime_n:=\alpha_n(s^\prime)\to a^\prime:=\alpha(s^\prime)$ and $b^\prime_n:=\beta_n(t^\prime)\to b^\prime:=\beta(t^\prime)$ (note that $\alpha_n(s_0)=a$ and $\beta_n(t_0)=b$ anyway). We get a timelike triangle $\Delta_n=\Delta(a, b, c_n:= \gamma(T_n))$ containing the points $(a^\prime_n,b^\prime_n)$. We form a comparison situation for this in ${\mathrm{AdS}} ^\prime$: $\bar{\Delta}_n=\Delta(\bar{a},\bar{b}, \bar{c}_n)$ with comparison points $\bar{a}^\prime_n,\bar{b}^\prime_n$. As X has global curvature bounded below by $K=-1$, we get that $\tau(a^\prime_n,b^\prime_n)\leq\bar{\tau}(\bar{a}^\prime_n,\bar{b}^\prime_n)$ and $\tau(b^\prime_n,a^\prime_n)\leq\bar{\tau}(\bar{b}^\prime_n,\bar{a}^\prime_n)$.

We would now like to let $T_n \to+\frac{\pi}{2}$, so we need control over what the comparison triangle $\bar{\Delta}_n$ converges to. For this, we select which way to realize $\bar{\Delta}_n$ in ${\mathrm{AdS}} ^\prime$. We choose:

• • $\bar{a}=(s_0,0)$ constant in n, i.e. at the right t-coordinate and with x-coordinate 0,

• • $\bar{b}=(t_0,c_{\alpha\beta}(s_0,t_0))$ constant in n. Note this has the right τ-distance to $\bar{a}$ by definition of $c_{\alpha\beta}$, and it has the right t-coordinate.

• • $\bar{c}_n$ with $\bar{\tau}(\bar{a},\bar{c}_n)=\tau(a,c_n)$ and $\bar{\tau}(\bar{b},\bar{c}_n)=\tau(b,c_n)$.

  • • Note that $\bar{c}_n$ need not be realizable in ${\mathrm{AdS}} ^\prime\subseteq {\mathrm{AdS}} $, but in ${\mathrm{AdS}} $ it is.

  • • But still, the initial segments of the sides $\bar{\alpha}_n$ connecting $\bar{a}\bar{c}_n$ and $\bar{\beta}_n$ connecting $\bar{b}\bar{c}_n$ are contained in ${\mathrm{AdS}} ^\prime$.

  • • Note that the future tip $\bar{c}=(1,0,0)\in{\mathrm{AdS}} $ of ${\mathrm{AdS}} ^\prime$ satisfies $\bar{\tau}(\bar{a},\bar{c})=\lim_n\tau(a,c_n)$ and $\bar{\tau}(\bar{b},\bar{c})=\lim_n\tau(b,c_n)$.

  • • Assuming $c_{\alpha\beta}(s_0,t_0)\neq0$, the function $g:{\mathrm{AdS}} \to\mathbb{R}^2$, $g(\bar{p})=(\bar{\tau}(\bar{a},\bar{p}),\bar{\tau}(\bar{b},\bar{p}))$ is continuous near $\bar{c}$ and maps a neighbourhood of $\bar{c}$ to a neighbourhood of $(\frac{\pi}{2}-s_0=\lim_n\tau(a,c_n),\frac{\pi}{2}-t_0=\lim_n\tau(b,c_n))$. In particular, for n large, we can assume $\bar{c}_n$ lie in this neighbourhood and $\bar{c}_n\to\bar{c}$.

  • • Assuming $c_{\alpha\beta}(s_0,t_0)=0$, we have that $\bar{b}=(t_0,0)$, and as $\tau(a,c_n)$ stays bounded away from π, the triangle $\Delta(\bar{a},\bar{b},\bar{c}_n)$ converges (in AdS) to a triangle which satisfies the size bounds (thus realizable uniquely up to isometries in ${\mathrm{AdS}} $, giving one or two realizations given $\bar{a}$ and $\bar{b}$), and $\bar{\tau}(\bar{a},\bar{c}_n)\to \frac{\pi}{2}-s_0$ and $\bar{\tau}(\bar{b},\bar{c}_n)\to \frac{\pi}{2}-t_0$ makes the limit triangle degenerate, thus $\bar{c}_n$ converge to $\bar{c}$ (as this point makes $\bar{a},\bar{b},\bar{c}$ degenerate and is at the right distance).

  • • In particular, $\bar{\alpha}_n\to\bar{\alpha}$ and $\bar{\beta}_n\to\bar{\beta}$ in τ-arclength parametrization. Thus the initial segments of $\bar{\alpha}_n$ and $\bar{\beta}_n$ become vertical in the limit.

In the ${\mathrm{AdS}} ^\prime$ picture, the limit comparison sides have the form $\bar{\alpha}(s)=(s,0)$ and $\bar{\beta}(t)=(t,c_{\alpha\beta}(s_0,t_0))$. By continuity of τ and $\bar{\tau}$, our curvature assumption gives $\tau(a^\prime,b^\prime)\leq\bar{\tau}(\bar{a}^\prime,\bar{b}^\prime)$ in the limit (similarly with arguments flipped). Now we compare the definition of $c_{\alpha\beta}(s^\prime,t^\prime)$ resp. $c_{\beta\alpha}(s^\prime,t^\prime)$ with the c-functions for $\bar{\alpha}$ and $\bar{\beta}$ at the same parameters:

\begin{align*} c_{\alpha\beta}(s^\prime,t^\prime)&=\mathrm{arcosh}\left(\frac{\cos(\tau(a^\prime,b^\prime))-\sin(s^\prime)\sin(t^\prime)}{\cos(s^\prime)\cos(t^\prime)}\right),\\ c_{\bar{\alpha}\bar{\beta}}(s^\prime,t^\prime)&=\mathrm{arcosh}\left(\frac{\cos(\bar{\tau}(\bar{a}^\prime,\bar{b}^\prime)-\sin(s^\prime)\sin(t^\prime)}{\cos(s^\prime)\cos(t^\prime)}\right)= c_{\alpha\beta}(s_0,t_0), \end{align*}

whenever defined. The last equality holds because we know $\bar{\alpha}$ and $\bar{\beta}$ are AdS-parallel with distance $c_{\alpha\beta}(s_0,t_0)$. Note that $a^\prime \leq b^\prime$ implies $\bar{a}^\prime \leq \bar{b}^\prime$ by the curvature bound and a simple continuity argument, so whenever the first line is defined so is the second. Notice these equations only differ in the τ term, and we know $\tau(a^\prime,b^\prime)\leq\bar{\tau}(\bar{a}^\prime,\bar{b}^\prime)$, so we get $c_{\alpha\beta}(s^\prime,t^\prime)\geq c_{\alpha\beta}(s_0,t_0)$ whenever the former is defined.

Similarly, if $b^\prime \leq a^\prime$ we get $c_{\beta\alpha}(s^\prime,t^\prime)\geq c_{\alpha\beta}(s_0,t_0)$ for all $s^\prime\geq s_0$ and $t^\prime\geq t_0$. We can also do this if $a\gg b$, giving $c_{\alpha\beta}(s^\prime,t^\prime)\geq c_{\beta\alpha}(s_0,t_0)$ and $c_{\beta\alpha}(s^\prime,t^\prime)\geq c_{\beta\alpha}(s_0,t_0)$.