Note: I ask this motivated by this other question: Are manifold subsets that are immersed submanifolds (regular/embedded) submanifolds?
Maybe a weird question, but:
Let $A$ be a set s.t. it is possible to endow $A$ with 2 different smooth (or topological or $C^k$ or possibly even holomorphic/complex/Kähler or whatever) manifold structures. Endow $A$ with 2 different smooth manifold structures $\mathscr F$ and $\mathscr G$ to get, resp, $(A,\mathscr F)$ and $(A,\mathscr G)$.
If $(A,\mathscr F)$ and $(A,\mathscr G)$ have respective dimensions $f$ and $g$, then is $f=g$?
- Edit: Note: Oh right so I really mean that $A$ is a set and not a topological manifold, so the way $(A,\mathscr F)$ and $(A,\mathscr G)$ are topological manifolds in the 1st place are that they are based on the same topological structure i.e. they are based on the same topology that makes the set $A$ into a topological space (and then this topological space is indeed a topological manifold).
What I have in mind:
So, like, there are many smooth (or whatever) manifold structures on $\mathbb R^n$, but I'm wondering if they all make $\mathbb R^n$ a smooth $n$-manifold. Perhaps there's some wild smooth manifold structure to make $\mathbb R^n$ locally $\mathbb R^{n-1}$ (i guess in the 1st place, such structure would be s.t. the topological structure makes $\mathbb R^n$ locally $\mathbb R^{n-1}$). I think of something like $\mathbb R^n$ and $\mathbb R^{n-1}$ as diffeomorphic/homeomorphic, but then it's not under the standard manifold or even topological structures.
I believe there's a rule that says for $(A,\mathscr F)$ to be a smooth $f$-manifold, we must have $f$ equal to the same dimension $h$ that allows us to say in the 1st place $(A,\mathscr F)$ is a topological $h$-manifold. But here...I'm not sure but I think I recall that the creation of a smooth manifold begins with a topological manifold and then this creation relies on same dimension.
$f=g$ if $(A,\mathscr F)$ and $(A,\mathscr G)$ are diffeomorphic/homeomorphic, but I recall $(A,\mathscr F)$ and $(A,\mathscr G)$ need not be diffeomorphic/homeomorphic.