Fix a differentiable non-compact manifold $M$. Denote by $\mathrm{Lor}(M) := \{\text{Lorentzian metrics on $M$}\}.$ One can define a topology on this set via: fix any open covering $\mathcal{A}$ on $M$. For each $g \in \mathrm{Lor}(M)$ and for each positive continuous function $r : M\to ]0,\infty[,$ one defines: $$\mathcal V(g,r) := \{h \in \mathrm{Lor}(M) : \forall p \in M, |\nabla^kg_{ij}(p) - \nabla^kh_{ij}(p)| < r(p)\},$$ where $g_{ij}, h_{ij}$ are the coordinates on some local chart with open domain contained on some open of $\mathcal{A}$. (I am sorry being a bit informal at this point, but I think it is clear what I mean). Also, $\nabla^k$ is supposed to denote any $k$-derivative of the metric. This is a sub-basis for the $C^k$-topology on $\mathrm{Lor}(M)$.

My first question is: is this topology somehow metrizable?

Further, if $M$ was compact (and $\chi(M) = 0)$, then $\mathrm{Lor}(M)$ is not empty, by imposing the $L^2$ metric (for some Riemannian metric) on $\mathrm{Lor}(M)$ induces the same topology as that I have defined? I mean, if we are interested on statements like:

(M,g) is locally causal if $g$ is close (on the sense of $C^r$-topology) to a causal metric $h$ on $M$,

Does this can be interchanged by

(M,g) is locally causal if $g$ is close to a causal metric $h$ on $M$ on the $L^2$-norm.

I am also sorry it these questions don't make sense at all, at the end, my question is: in general, introducing a Riemannian metric for comparing Lorentzian metric is somehow inappropriate, in the sense it leads to lost of some information?

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    To supplement Pedro's excellent answer, you might also be interested in the article "The space of Lorentz metrics" by David E. Lerner CMP 32 19 (1973), where he discusses the stability of various causal properties with respect to the Whitney topologies. – Igor Khavkine Oct 24 at 8:01
up vote 12 down vote accepted

First of all, there is a bunch of basic things that you need to write in a slightly clearer way. If you try to topologize the set of Lorentzian metrics as you did, you need first:

  1. Restrict to the subset of Lorentz metrics of a given (say, $C^k$, $0\leq k\leq\infty$) regularity, otherwise your definition for the basic neighborhoods $\mathcal{V}(g,r)$ makes no sense.
  2. Once you did the above (denote the resulting set by, say, $\mathrm{Lor}_k(M)$), replace $|\nabla^k g_{ij}(p)-\nabla^k h_{ij}(p)|$ by the sum $\sum_{0\leq j\leq k}|\nabla^j g_{ij}(p)-\nabla^j h_{ij}(p)|$ in the definition of $\mathcal{V}(g,r)$ for $k$ finite. If $k=\infty$, you must include such $\mathcal{V}(g,r)\doteq\mathcal{V}_k(g,r)$ for all $k\geq 0$ (or at least for all $k$ in an infinite subset of $\mathbb{N}\cup\{0\}$).

This is the bare minimum. Ideally, it would be better to do things in a coordinate-free way: denote by $\nabla^k g$ the iterated covariant derivative of order $k$ of $g$ w.r.t. some (say, torsion-free) covariant derivative operator $\nabla$ on $M$ and define the pointwise Euclidean norm $|T|$ of a tensor field $T$ by lifting some reference Riemannian metric $e$ on $M$ to the corresponding tensor bundle. This provides a way to write a fiberwise scalar product on the jet bundle of order $k$ of the fiber bundle of Lorentzian metrics over $M$ (more generally, on the jet bundle of order $k$ of the vector bundle of covariant tensors of rank 2 over $M$). One can write these things in more detail, but this is more or less standard. Anyway, I think you get the gist.

With the trivialities out of the way (at this point, if something about them is not yet clear, please do let me know), we can begin to address your question proper. What you wrote above (with the tacit understanding that the amendments 1. and 2. above are included) are the basic neighborhoods for the $C^k$ Whitney topology of $\mathrm{Lor}_k(M)$. In fact, this topology does not depend on the choice of a reference Riemannian metric $e$ as above (the reason will be explained below).

If $M$ is compact, this topology is even normable if $k$ is finite (as a subset of the normable vector space of $C^k$ covariant tensor fields of rank two) and still metrizable if $k=\infty$ for then it coincides with the compact-open $C^k$ topology. However if $M$ is non-compact (as you assumed, since you seem to be ultimately interested in causality theory for Lorentzian metrics and this theory is nontrivial only for non-compact $M$) this topology is non-metrizable for all $k$ (even $k=0$). In fact, this topology is not even first-countable in this case.

This is easier to visualize in the case of $C^k$ scalar fields (i.e. $C^k$ real-valued functions) on $M$ instead of Lorentzian metrics: the connected component of $f\equiv 0$ in the $C^k$ Whitney topology of $C^k(M)$ is the space $C^k_c(M)$ of $C^k$ functions with compact support on $M$ with the usual inductive limit (locally convex vector space) topology. This topology is not first-countable, hence non-metrizable. More generally, the connected component of any $C^k$ function $f$ in the $C^k$ Whitney topology of $C^k(M)$ is precisely $f+C^k_c(M)$. One sees from this remark that the Whitney topologies get so fine when $M$ is non-compact, they become extremely disconnected. A similar fact holds for the $C^k$ Whitney topology in $\mathrm{Lor}_k(M)$ - the metrics $h$ in the connected component of $g\in\mathrm{Lor}_k(M)$ in this topology differ from $g$ only inside some compact subset of $M$ (depending on $h$). This remark also makes it clear why the choice of the Riemannian metric $e$ is not relevant to the definition of the $C^k$ Whitney topology, despite the fact that $M$ is not compact.

For (many!) details on the Whitney topologies, you may want to check The Convenient Setting of Global Analysis by Andreas Kriegl and Peter W. Michor (AMS, 1997), specially Chapter IX (Manifolds of Mappings). The paper by Lerner quoted by Igor in his comment relates the Whitney topologies to structures which are natural to Lorentz metrics - e.g. conformal classes of Lorentzian metrics with representatives being $C^0$ Whitney-near to each other amounts to their light cones being close to each other, metrics which are $C^1$ Whitney-near to each other have their geodesics near to each other in some sense, and so on.

  • Thank you for the answer, I will just ask kindly to you be less rude, perhaps, of course my question should be improved but it seems you understood it and a little of kindness is not bad at all. Thank you again. – L.F. Cavenaghi Oct 24 at 17:07
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    I apologize if the beginning of my answer sounded harsh. My point was that MO is a forum for research-level math questions, which therefore must be written as precisely as possible. More importantly, yes, I understood what you meant in your question but that might not be the case for others if you don't write the precise definition for the Whitney topologies, especially for the subject tag you chose, for which the notion is not exactly common knowledge. Anyhow, again I apologize for the tone. – Pedro Lauridsen Ribeiro Oct 24 at 17:17
  • That's ok Pedro, I am very thankful for your great answer and criticism, I totally agree I should have written the question in a more precise way, but, in my defense, I was just introduced to such theme yesterday and it was pretty fresh and not very clear on my head, but I totally agree, it should be more clear and precise. Thank you again, it will not happen on other questions. – L.F. Cavenaghi Oct 24 at 17:56

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