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:
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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.
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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.
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$f=g$ 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.
Best Answer
One (standard) approach to defining manifolds is to start with a topological space $M$. We say that this topological space is a (topological) manifold if it is (second countable Hausdorff and) is locally homeomorphic to $\mathbb{R}^n$, i.e. every point $p$ in $M$ has an open neighborhood homeomorphic to an open set in $\mathbb{R}^n$. The invariance of domain theorem then implies that 1) for a given $p$ only one $n$ can work and 2) for all $p$ in a given connected component of $M$ this $n$ is the same. So, if $M$ is connected, $n$ is uniquely determined. If not, $n$ is uniquely determined on each component. Most authors then put the requirement that the $n$s for all the components are also equal into the definition of a manifold as well. Such a topological space is then called an $n$-manifold.
Note also that for a topological space being a topological manifold is not a structure, it's a property. It is meaningless to talk about given topological space having two "topological manifold structures". The topological space either is a topological manifold or isn't.
All the extra structures (smooth, complex et cetera) are then on top of this, so of course, they are also all of the same dimension $n$ -- the dimension of the underlying topological manifold $M$.
On the other hand, if you start with a set without any topology then of course any set of the cardinality of the continuum can be topologized to become homeomorphic to any given manifold (and thus can be made the underlying set of a manifold of any dimension bigger than $0$, but for a very silly reason!).