# Do we always get the same dimension regardless of the manifold/topological structure?

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:

1. 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.

2. 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.

3. $$f=g$$ 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.

• For context re (1), note that even though there are infinitely many different smooth structures on $\mathbb{R}^n$ when $n\neq 4$, there is only one smooth structure up to diffeomorphism for those $n$. The exotic structures (meaning structures not diffeomorphic to the usual one) occur only for $n=4$. Apr 26, 2021 at 5:43
• @symplectomorphic Ah so $(\mathbb R^3, \text{whatever})$ for example is always diffeomorphic to any other $(\mathbb R^3, \text{whatever})$?
– BCLC
Apr 26, 2021 at 5:44
• Yes, in the case when “whatever” means a smooth structure. Apr 26, 2021 at 5:46
• @symplectomorphic thanks.
– BCLC
Apr 26, 2021 at 5:46
• Notational quibble: until reading the answer, I thought $f$ and $g$ were functions, and couldn’t understand the question that was being asked. It might be better to use a variable name for those that is more typically associated with nonnegative integers, like $i, j, k, m, n,$ or $d$. Apr 26, 2021 at 5:55

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!).

• '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.' --> HUGE FACEPALM. THANKS. 1 - ok but fine if i were to ask for topological instead of smooth, then this is what your last paragraph answers e.g. we can have $(\mathbb R, \text{something})$ as topological $m$-manifold for $m \ne 1$ (but still a non-negative integer) ?
– BCLC
Apr 26, 2021 at 5:46
• I edited a bit more, hope it's clearer. Yes, once the topology is fixed, the resulting space either is or is not a (topological) manifold. If a given set is given different topologies each such topology either produces a manifold or not, and the resulting manifolds can be of different dimensions.
– Max
Apr 26, 2021 at 5:59
• I operate under the definition that manifolds are Hausdorff second countable. If so, sets of finite cardinality can be topologized to be 0-dimensional manifolds in the unique way (discrete topology). Sets of same cardinality as $\mathbb{R}$ can be topologized to be homeomorphic to any given positive-dimensional manifold. Sets of any other cardinality can not be topologized with a topology that makes them into topological manifolds.
– Max
Apr 26, 2021 at 6:11
• This is nothing but a "backwards" restatement of the fact that (Hsusdorf second countable) 0-dimensional manifolds have (underlying sets of) finite cardinality, and positive dimensional manifolds have (underlying sets of) cardinality of the continuum.
– Max
Apr 26, 2021 at 6:12
• I.e. take any positive-dimensional manifold $M$. Its underlying set has cardinality or $\mathbb{R}$ i.e. is in bijection with $\mathbb{R}$. Use this bijection to transfer the topology from $M$ to $\mathbb{R}$. Voila, you have $(\mathbb{R}, U)$ homeomorphic to $M$. Of course $U$ is very far from "the standard" topology on $\mathbb{R}$.
– Max
Apr 26, 2021 at 6:15

Dimension needs to be consistent over all smooth structures.

Indeed, you can prove that if $$U \subset \mathbb{R}^n$$ and $$V \subset \mathbb{R}^m$$ are homeomorphic, then $$n = m$$.

The proof (if you're familiar with homology theory), is by looking at the relative homology of the pair $$(U, U\backslash\{0\})$$. In particular, you can show that it is isomorphic to the reduced homology of the $$n$$-sphere, while the homology of $$(V, V\backslash\{0\})$$ is isomorphic to the reduced homology of the $$m$$-sphere. So asking for $$U \cong V$$ enforces $$m = n$$. (Let me know if you want me to get into more details on that)

In particular, assume that have two different smooth structures on your manifold, making it locally homeomorphic to respectively $$\mathbb{R}^m$$ and $$\mathbb{R^n}$$. Pick a point $$x$$, two charts $$U$$ and $$V$$ around it. Then, using the result above, $$m$$ and $$n$$ have to be the same.

• thanks Azur. 1 - is the relevant proof for smooth case perhaps what Max suggested: namely that dimension as smooth manifold comes from dimension as topological manifold? i mean, i know what invariance of domain/dimension is 2 - is perhaps the spirit of this question more like asking about the topological case where you can indeed have different dimensions?
– BCLC
Apr 28, 2021 at 1:29
• Well, I don't think this proof is specific to the smooth case, since we are not making use of the smooth structure on the manifold at any point (we are only working in one chart). In my last paragraph, you can replace "smooth structures" with "atlases" and the result would still hold - all we are using is the fact that the manifold is locally homeomorphic to Euclidean space, which is the definition of a topological manifold.
– Azur
Apr 30, 2021 at 12:08
• Sooo...you're saying for topological atlases/structures...that what...? I mean, are you saying something different from Max?
– BCLC
Apr 30, 2021 at 12:24
• No, just detailing the reasoning behind the "invariance of domain theorem" that they mentioned :)
– Azur
Apr 30, 2021 at 12:34
• Ooh, I hadn't seen the last sentence in their answer, my bad! And I was assuming $\mathbb{R}^7$ had the standard topology (in which case it can't be endowed with anything but a 7-dimensional topological structure). I reckon if you drop the standard topology that allows for weirder cases. I'll trust them with that because I've never seen such an example!
– Azur
May 2, 2021 at 8:24