Diffeomorphisms between $ \mathbb{R}^{p, q} $ and $ \mathbb{R}^{p+q} $

Question

suppose an euclidean space $ \mathbb{R}^{n} $ with a quadratic form $ Q: \mathbb{R}^{n} \rightarrow \mathbb{R} $ such that

$$ \forall X \in \mathbb{R}^n, \, \ Q(X) = \sum_{i=1}^p X_i^2 - \sum_{j=1}^q X_j^2 $$

with $ p + q = n $

then this space is the pseudo-euclidean space $ \mathbb{R}^{p,q} $ of signature $ (p,q) $ which can be made a pseudo-Riemannian manifold by setting $ Q $ to be a pseudo-Riemannian metric.

is there any way to prove $ \mathbb{R}^{p,q} \simeq \mathbb{R}^n $ as manifolds ?

my intuition tells this may be possible but there's the extra step of requiring that the quadratic form is preserved right? i am really lost

Answer 1

You’re overthinking things. As sets, topological spaces, $C^k$ manifolds (for all $k\in \{0,1,\dots,\infty,\omega\}$), and vector spaces, we have $\Bbb{R}^{p,q}:=\Bbb{R}^{p+q}$, by definition. The only difference between the two is when you decide to endow each with a geometry, namely a semi-inner product, of signature $(p,q)$ in the first case and positive-definite in the second case. In other words, for the purposes of diffeomorphisms, the identity map is a diffeomorphism, and you should completely ignore the roles of the respective quadratic forms.