Existence of parametrization with dense image for an invariant surface in $\Bbb R^4_1$.

Question

I am trying to read the paper Boost invariant marginally trapped surfaces in Minkowski $4$-space, by Haesen and Ortega. Basically, they work with metric signature $(-+++)$ in $\Bbb R^4_1$, consider the group of boosts $$G = \{ R^h_\theta\oplus {\rm Id}_2 \mid \theta \in \Bbb R \},$$where $R^h_\theta \in {\rm O}_1^{+\uparrow}(2,\Bbb R)$ denotes hyperbolic rotation by $\theta$, and they claim that if $S \subseteq \Bbb R^4_1$ is a connected spacelike surface which is invariant by the action of $G$ in $\Bbb R^4_1$, then $S$ admits a parametrization $X\colon I \times \Bbb R \to \Sigma_\alpha \subseteq S$ of the form $$X(s,\theta) = B_\theta(\alpha(s))$$for some open interval $I\subseteq \Bbb R$, where the image $\Sigma_\alpha$ is an open dense subset of $S$.

I can't check that such a parametrization is always available, let alone prove that the image is dense in $S$.

They mention that the set of fixed points of the action is $\{(0,0)\} \times \Bbb R^2$ very near of that claim, so I think this should be useful. Please help.

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Edit 1: I think I can convince myself why the parametrization has that form. If you take $(x,y,z,w) \in S$, since $S$ is invariant, we can write $(x(s),y(s)) = (\alpha_1(s)\sinh \theta, \alpha_1(s)\cosh\theta)$, assuming that $S \subseteq \mathcal{R}$. Since $\dim S = 2$, $z$ and $w$ can only depend on one parameter, which obviously cannot be $\theta$. Meaning $(z(s),w(s))=(\alpha_3(s),\alpha_4(s))$. So every point has the form $B_\theta(\alpha(s))$. But I think this is a bit too much handwaving and tells me nothing about the image being dense.

Answer 1

Yes, their claim is true and you should absolutely learn more about bundles (vector bundles, principal bundles, connections, curvature, etc. I just do not see how you can study the modern GR without this).

Here are steps of the proof. You first observe that the action of $G$ on the region $Q=\{(x,y,z,w): |x|>|y|\}$ is proper: For every compact $K\subset Q$ the subset $$ G_K:= \{g\in G: gK\cap K\ne \emptyset\}\subset G $$ is compact. To prove this it suffices to analyze the 2D case, i.e. the action of $G$ on the intersection of $Q$ with the xy-plane and observe that the restriction of the Lorentzian metric to the quadrants $|x|>|y|$ in the xy-plane is definite (invariant under $G$, of course); then the claim follows e.g. from the Arzela-Ascoli theorem (since any group of isometries forms an equicontinuous family).

Since they assume that $S$ is contained in $Q$, it follows that the $G$-action on $S$ is also proper.

The next step is to quote a theorem about $G$-bundles: Let $X$ be a smooth manifold, $G\times X\to X$ a free smooth proper action of a Lie group (free means that for each $x\in X$, $gx=x$ implies that $g=e$) determines a principal $G$-bundle $X\to B=G/X$ (in particular, $B$ is a smooth manifold of dimension $dim(X)-dim(G)$).

Specializing to out case, $G\cong ({\mathbb R}, +)$, $X$ is a connected surface $S$, hence, $B$ is a connected 1-dimensional manifold, i.e. $B$ is either diffeomorphic to $S^1$ or to ${\mathbb R}$. (I do not think the former case can occur, but let's not worry about this.) In particular, the group $G$ is contractible. It is a general fact about bundles that a fiber bundle with contractible fibers has a section, so we obtain a section $s: B\to X$. By the definition of a section of a principal bundle, $g s(B)\cap s(B)\ne \emptyset$ if and only if $g=e$. Hence, we obtain an injective differentiable map $G\times B\to X$, $(g,b)\mapsto g s(b)$. By the equality of the dimensions of the domain and the range, this map is open, hence, a diffeomorphism to its image. It is also clear that this map is surjective (since the bundle is principal, $G$ acts transitively on its fibers). Hence, the bundle $X\to B$ is trivial. This diffeomorphism also gives you the required parameterization of your surface $S$, since you can parameterize the base $B$, and, hence, $s(B)\subset S$, by $\alpha: I\to s(B)$, where $I$ is an interval in ${\mathbb R}$ and the image of $\alpha$ is either the entire $s(B)$ (if the latter is not diffeomorphic to the circle) or an open and dense subset of $s(B)$ (if the latter is diffeomorphic to the circle).

Answer 2

I came up with something more pedestrian: following the paper's notation, put $\mathcal{P} = \{(x,0,z,w) \mid x>0\}$.

For each $p \in S$ there is a unique $\theta(p) \in \Bbb R$ such that $B_{\theta(p)}(p) \in S \cap \mathcal{P}$. This also defines a continuous map $\theta\colon S \to \Bbb R$. Now, $f\colon S \to \Bbb L^4$ given by $f(p) = B_{\theta(p)}(p)$ is continuous and its image is precisely $S \cap \mathcal{P}$. Since the dimension of the intersection $S \cap \mathcal{P}$ is $1$ and $S$ connected now implies $S \cap \mathcal{P}$ connected (path-connected actually), we have that $S \cap \mathcal{P}$ is diffeomorphic either to $\Bbb R$ or $\Bbb S^1$. Take a global parametrization $\alpha$ and define $X$ by that formula, done. If $S \cap \mathcal{P}$ is diffeomorphic to $\Bbb R$, then $\Sigma_\alpha$ will be all of $S$. In the other case, to consider parametrizations only with open domains, take a parametrization of $S \cap \mathcal{P}$ omitting a single point - in this case $\Sigma_\alpha$ is a proper open dense subset of $S$.

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By the way, it seems that the latter case is impossible. If this were the case, the compactness of $S \cap \mathcal{P}$ would give up a timelike vector tangent to $S \cap \mathcal{P}$, since we can see $S \cap \mathcal{P}$ as a curve in $\Bbb L^3$.