Is it possible to make a pseudo-euclidean space $ \mathbb{R}^{p, q} $ into a differentiable manifold?
My intuition tells me that yes since in special relativity we deal with the flat Minkowski space $ \mathbb{R}^{1, 3} $ however i tried to approach the problem of proving that for any given pseudo-euclidean space we can make it a differentiable manifold and i am stuck.
From what i understand, the way we prove this is by defining two charts $ (U, x) $ and $ (V, y) $ on local regions U and V in $M$ with two maps $ x: U \rightarrow A $ and $ y: V \rightarrow B $ where A and B are subsets of $ \mathbb{R}^{\dim(M)} $ and then looking at the transition maps $ y(x^{-1}(p)) $ to see if they are of class $ C^k $ and if they are for any pair of charts in M then M is a differentiable manifold
I tried to do this by defining a subset of $ \mathbb{R}^{1,2} $ as a collection of intervals.
$ U = ( (0;1); (1;2); (3; 4)) $
Which would mean my local region U would be from 0 to 1 in x, from 1 to 2 in y and from 3 to 4 in z.
And then i defined a map to be
$ \phi(p) = a_i p_i $
With $ a \in \mathbb{R} $ and then i did the same for V but with different intervals and different constants.
However when i evaluate the derivatives of the transition maps this is not $ C^k $
Help, i've literally never done this kind of exercises and now i'm stuck. I don't know how to face this problem
OK, let's try to make sense of your question. First of all, what is ${\mathbb R}^{p,q}$? Let $n=p+q$. Then ${\mathbb R}^{p,q}$ is the vector space ${\mathbb R}^{n}$ equipped with the quadratic form $$ Q({\mathbf x})= x_1^2+...+x_p^2 - x_{p+1}^2- ... - x_n^2. $$ In particular, as a topological space, ${\mathbb R}^{p,q}$ is just ${\mathbb R}^{n}$. (At this point it is critical that you know what the words "topological space" mean. I assume that you do.)
What does it mean to make ${\mathbb R}^{p,q}$ into a differentiable manifold? You want to accomplish two things:
Find a (maximal) smooth atlas on the topological space ${\mathbb R}^{n}$ such that the transition maps are smooth. (The "definition" that you wrote in your question is not quite the right one.) This will give ${\mathbb R}^{n}$ a structure of a differentiable manifold, I will call this manifold $M$.
Make sure that the quadratic form $Q$ becomes a semi-Riemannian metric on $M$ (of the signature $(p,q)$). (I assume that you know what these words mean.)
Now, let's do it. As an atlas ${\mathcal A}$ on ${\mathbb R}^{n}$ I will take a single chart $(U,\phi)$, where $U={\mathbb R}^n$ (the entire space) and $\phi$ the identity map, $\phi({\mathbf x})={\mathbf x}$. Let's check the transition maps: Since there is only one chart, all what we have to check is that $\phi\circ \phi^{-1}$ and $\phi^{-1}\circ \phi$ are smooth. But this is clear since both are just the identity map, so there is nothing to check here. Lastly, I will take a maximal smooth atlas containing ${\mathcal A}$ (this step is just a bookkeeping). The result is a differentiable manifold structure on ${\mathbb R}^n$. This concludes Part 1.
For Part 2 you have to identify the tangent spaces $T_xM$ to our manifold $M$. Since $M$ is just ${\mathbb R}^n$, tangent vectors, elements of $T_xM$, are simply directed line segments $\overrightarrow{xy}$ with the tail at $x$ and the head at various points $y\in M$. I assume that in your vector calculus class you have learned how to identify the space of such segments with ${\mathbb R}^n$ itself (by computing the difference of coordinates of the points $x$ and $y$). With this identification, we define a quadratic form $Q_x$ on each $T_xM$ simply as our original form $Q$. If you want to be more explicit, for $x=(x_1,...,x_n)$, $y=(y_1,...,y_n)$, we have $$ Q_x(\overrightarrow{xy})= Q(y_1-x_1,,....,y_n-x_n)= (y_1-x_1)^2 + ... + (y_p-x_p)^2 - (y_{p+1}-x_{p+1})^2 - ... - (y_n-x_n)^2. $$ The function $(x,y)\mapsto Q_x(\overrightarrow{xy})$ is clearly smooth, hence, making our collection of quadratic forms $Q_x$ a semi-Riemannian metric on our manifold $M$.