Suppose $M= \mathbb R \times M\_0$ with a Lorentzian metric $g(t,x)=-dt^2+ g\_0(t,x)$ where
$M\_0$ is a compact manifold with a smooth boundary and $g\_0$ is a family of smooth Riemannian metrics on $M\_0$ depending smoothly on $t$. Assume furthermore that $M$ is nontrapping, that is to say, all inextendible null geodesics eventually exit $M$. Now let $p \in M^{\textrm{int}}$ and let $L^+(p)$ denote the future null cone emanating from the point $p$. Let us assume that the null cone $L^+(p)$ is smooth (away from $p$) and that it intersects $\mathbb R\times \partial M\_0$ on a surface $S$ of codimension 2. My question is whether $S$ can be realized as the boundary of a spacelike smooth hypersurface $\Sigma$ in $M$.



A question on light cones in Lorentzian manifolds with timelike boundary
**Edit:** Right now I will *not* give a complete argument in what follows (the previous version of the answer missed the part of the question asking for $S$ to be the boundary of a spacelike hypersurface in $M$), I will add the missing ingredients and/or additional hypotheses (if needed) as time allows in a future edit.


If by "future null cone" you mean the (achronal) boundary $\partial I^+(p)$ of the chronological future $I^+(p)$ of $p$, then one can show that $S$ is a smooth, acausal hypersurface of $\partial M$ even without the nontrapping condition, thanks to your smoothness assumption. What may fail is that tangent vectors will be either spacelike or null, and the latter case cannot be excluded without additional assumptions on $\partial M$. For example, take $M$ as the Minkowski space-time minus an open cylinder around the time axis - $S$ will have a null tangent vector at any point where a null generator of $\partial I^+(p)$ meets $\partial M$ tangentially. Of course, this example has trapped null geodesics, but even with the nontrapping assumption the null generators could in principle meet $\partial M$ tangentially, causing $S$ to have null tangent vectors at the meeting point.


The argument goes as follows. Since $\partial I^+(p)$ is assumed to be smooth away from $p$ and this point is assumed to be in the interior of $M$, it turns out that $\partial I^+(p)$ meets $\partial M=\mathbb{R}\times\partial M\_0$ *transversally*. This is due to the fact that timelike boundary curves in $\partial M$ such as $t\mapsto(t,p\_0)$, $p\_0\in\partial M\_0$ are *never* tangent to $\partial I^+(p)$ (tangent vectors to a lightlike hypersurface are either lightlike or spacelike). As a consequence, $S=\partial I^+(p)\cap\partial M$ is indeed a smooth hypersurface of $\partial M$ (hence with codimension two in $M$) if nonvoid, and $TS$ has no timelike vectors. Moreover, also due to the assumed smoothness of $\partial I^+(p)$ away from $p$, each $q\in S$ belongs to *exactly one* null generator $\gamma\_q$ of $\partial I^+(p)$. If $\gamma\_q$ meets $\partial M$ transversally at $q=\gamma\_q(t)$, then $\dot{\gamma}\_q(t)$ is not tangent to $S$ at $q$ and therefore $T\_q S$ is spacelike. If, on the other hand, $\gamma\_q$ meets $\partial M$ tangentially at $q$, then $\dot{\gamma}\_q(t)$ is tangent to $S$ and therefore $T\_q S$ is null.


Notice that, regardless of the fact that $S$ may have null tangent vectors, we still have that $S$ is *acausal* since, as pointed above, each point of $S$ belongs to exactly one generator of $\partial I^+(p)$ and any subset of the latter with that property is necessarily acausal, because the only possible causal curve connecting two points of $\partial I^+(p)$, if any, must be a null generator thereof.


To rule out $S$ having null tangent vectors, a sufficient condition is that $\partial M$ is *totally geodesic*. More generally, one could impose a sort of "convexity" assumption on $\partial M$ with respect to null geodesics.


**Edit:** Let us assume from now on that $S$ is indeed spacelike. It remains to show the existence of a spacelike hypersurface $\Sigma\subset M$ such that $\partial\Sigma=S$. We will sketch below a partial argument assuming that $M\_0$ is *connected* and $g$ is *ultrastatic* (i.e. $g\_0(t,x)=g\_0(x)$ does not depend on $t$). Let $p=(t\_0,p\_0)$ and let $\gamma\_0:[0,1]\rightarrow M\_0$ be a smooth curve connecting $p\_0=\gamma\_0(0)$ to $r\_0=\gamma\_0(1)$. Given $$\alpha>\sup\_{s\in[0,1]}\sqrt{g\_0(\gamma\_0(s))(\dot{\gamma}\_0(s),\dot{\gamma}\_0(s))}\ ,$$ we have that the curve $$\gamma(s)=(t\_0+\alpha s,\gamma\_0(s))\ ,\quad s\in[0,1]$$ is *timelike* and links $p$ to the point $r=(t\_0+\alpha,r\_0)\in\partial M$. This shows that the timelike curve $t\mapsto (t,r\_0)$ must cross $\partial I^+(p)$ at some $t\in(t\_0,t\_0+\alpha)$, which must be unique since $S$ is acausal. This shows that $\partial I^+(p)$ crosses the timelike curve $t\mapsto (t,q\_0)$ exactly once for all $q\_0\in\partial M\_0$, thus implying that $S$ is the graph of a smooth function $\phi:\partial M\_0\rightarrow\mathbb{R}$ - namely, if $q\in S$ then $q=(\phi(q\_0),q\_0)$ for a unique $q\_0\in\partial M\_0$. Since $S$ is spacelike, one must have $g\_0^{-1}(q\_0)(d\phi(q\_0),d\phi(q\_0))<1$. The missing piece is to extend $\phi$ to a smooth function $\Phi:M\_0\rightarrow\mathbb{R}$ such that $g\_0^{-1}(q\_0)(d\Phi(q\_0),d\Phi(q\_0))<1$ - if that is accomplished, then the graph of $\Phi$ is the desired spacelike hypersurface $\Sigma$ with $\partial\Sigma=S$. It is straightforward to extend $\phi$ to a collar neighborhood of $\partial M\_0$ in $M\_0$ while keeping the derivative bound above. Using this to extend it further to a smooth function $\Phi$ on $M\_0$ using a partition of unity is also straightforward - the problem is to do it in such a way so as to keep the derivative bound. I will complete the details in a future edit.


Connectedness of $M\_0$ seems necessary to achieve the result, for otherwise it is hard to tell how to extend $S$ into the interior of $M$. As for ultrastaticity of $g$, it may be possible to weaken it to a form of global hyperbolicity suitable for Lorentzian manifolds with boundary.


