Topology of $\mathbb{R}^{n+1}$ versus $\mathbb{R}^{n,1}$

Question

What are the topology of these two noncompact manifolds $M_1=\mathbb{R}^{n+1}$ versus $M_2=\mathbb{R}^{n,1}$, their differences? The inner product of the vectors for $M_1$ relies on the Euclidean metric: $$ g_{ab}=\delta_{ab}, \quad a,b \in 1,2,...,n+1. $$ The inner product of the vectors for $M_2$ relies on the (pseudo) Minkowski metric: $$ g_{1,1}=-1, \quad g_{ab}=+\delta_{mn}, \quad a,b \in 2,...,n+1. $$ All off diagonal components are zeros.

1. Homotopy group:

• I believe both have the $\pi_0(M_1)=\pi_0(M_2)=0$. So only one piece.

• I believe both have the $\pi_1(M_1)=\pi_1(M_2)=0$. So only contractible cycles.

• how about higher homotopy groups $\pi_k(M_1)$, or $\pi_k(M_2)$?

2. Homology groups:

• how about higher homotopy groups $H_k(M_1)$, or $H_k(M_2)$?

3. Are there some topology ways to distinguish $\mathbb{R}^{n+1}$ versus $\mathbb{R}^{n,1}$? What are those topological differences between the $M_1=\mathbb{R}^{n+1}$ versus $M_2=\mathbb{R}^{n,1}$?

Answer 1

Short Answer: They're the same thing.

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Just echoing what has alredy been mentioned in the comments. A smooth manifold is a triple of information $(M,\mathcal{T}, \mathcal{A})$, where $M$ is a set, $\mathcal{T}$ is a (Hausdorff, second countable) topology for $M$, and $\mathcal{A}$ is a maximal smooth atlas. A pseudo-Riemannian manifold is then a quadruple of information $(M,\mathcal{T}, \mathcal{A}, g)$, where $g$ is a smooth $(0,2)$ tensor field on $M$ which is pointwise a symmetric, bilinear, non-degenerate functional. As you can see, the definition of the topology $\mathcal{T}$ comes BEFORE we even talk about the tensor field $g$. Said differently, changing $g$ DOES NOT affect the topology.

When we speak of "the Riemannian manifold $\Bbb{R}^{n+1}$", we mean

• We consider the set $\Bbb{R}^{n+1}$, together with the "usual" topology $\mathcal{T}_{n+1}$ (i.e $U\in\mathcal{T}_{n+1}$ if and only if for each $p\in U$ there is an $r>0$ such that for all $q\in\Bbb{R}^{n+1}$ with $\sqrt{\sum_{i=1}^{n+1}(q^i-p^i)^2}<r$, we have $q\in U$). Then we consider the unique maximal atlas which contains the identity chart $(\Bbb{R}^{n+1},\text{id}_{\Bbb{R}^{n+1}})$. The Riemannian structure is obtained by considering the tensor field $\delta:=\sum_{i=1}^{n+1}(dx^i)\otimes (dx^i)$.

When we speak of "the Lorentzian manifold $\Bbb{R}^{n,1}$", we mean

• We consider the same set $\Bbb{R}^{n+1}$, the same topology $\mathcal{T}_{n+1}$, the same maximal smooth atlas containing the identity chart. The only difference is we're considering a different 'geometry', i.e given by the tensor field $\eta:=-(dx^1)\otimes (dx^1)+\sum_{i=2}^{n+1}(dx^i)\otimes (dx^i)$.

But like I've already said, the tensor field you choose to consider, $\delta$ vs $\eta$ makes no difference at all to the topology.