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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.

BCLC
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    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 $n=4$. – symplectomorphic Apr 26 '21 at 05:43
  • @symplectomorphic Ah so $(\mathbb R^3, \text{whatever})$ for example is always diffeomorphic to any other $(\mathbb R^3, \text{whatever})$? – BCLC Apr 26 '21 at 05:44
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    Yes, in the case when “whatever” means a smooth structure. – symplectomorphic Apr 26 '21 at 05:46
  • @symplectomorphic thanks. – BCLC Apr 26 '21 at 05:46
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    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 $d$. – Rivers McForge Apr 26 '21 at 05:55
  • @RiversMcForge i was thinking of that. wasn't really sure people would be bothered. hehe. will edit. thanks. – BCLC Apr 26 '21 at 05:57

2 Answers2

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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!).

Max
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    '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 '21 at 05:46
  • 2 (part1) - '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 structures.' --> right yeah sooo...'topological structure' refers to the topology itself? eg $A$ is a set and then $\mathscr M$, $\mathscr N$ are topologies on $A$ and then if $(A, \mathscr M)$ and $(A, \mathscr N)$ each are topological $m/n$-manifolds (not sure if same dimension!) then 'topological structure' refers to the, resp, $\mathscr M$ and $\mathscr N$? --> Ah wait yeah in this case... – BCLC Apr 26 '21 at 05:53
  • 2 (part2) ...in this case I mean that $\mathscr F$ and $\mathscr G$ include the topological information i.e. they have the same topology. I'll edit this in post – BCLC Apr 26 '21 at 05:54
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    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 '21 at 05:59
  • thanks Max. 3 - what about cardinality less than continuum? 4 - (kinda restating (1)) wait so we can have like $(\mathbb R, \text{something})$ as like say $0$-dimensional or $2$-dimensional? If so, then how? I have a feeling there's some really trivial silly way to do this, and this question is precisely about those trivial and silly ways. – BCLC Apr 26 '21 at 06:06
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    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 '21 at 06:11
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    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 '21 at 06:12
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    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 '21 at 06:15
  • oh ok that makes complete sense. thanks Max. one more thing is it that 'every immersed submanifold can indeed become a manifold on its own even though it's not gonna be regular/embedded submanifold' ? – BCLC Apr 26 '21 at 06:38
  • note: the linked question is my motivation for asking this question. it's not some irrelevant thing. – BCLC Apr 28 '21 at 01:30
  • oh wait max something i realised. for the other question, it's relevant to take into account the topology of the immersed submanifold because if we just consider the set, then we just apply your answer? – BCLC Apr 28 '21 at 02:08
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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.

Azur
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  • 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 '21 at 01:29
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    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 '21 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 '21 at 12:24
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    No, just detailing the reasoning behind the "invariance of domain theorem" that they mentioned :) – Azur Apr 30 '21 at 12:34
  • ok thanks i'll have to upvote later on since limit reached. but anyhoo $\mathbb R^7$ can indeed be like 4-dimensional right? – BCLC Apr 30 '21 at 13:12
  • Well, no, because that would imply somehow finding a neighbourhood of a point in $\mathbb{R}^7$ which is homeomorphic to $\mathbb{R}^4$. This is impossible by the said "invariance of domain" statement – Azur May 01 '21 at 15:51
  • Azur, well it's not the $\mathbb R^7$ under its standard topology. that's kind of the point right? do you disagree with Max? – BCLC May 02 '21 at 08:11
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    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 02 '21 at 08:24
  • Well I figure it's just based on cardinality of continuum: If you have $(\mathbb R, \mathscr T_{std})$, then just do $(\mathbb R^7, \mathscr T_X)$ where $\mathscr T_X$ is a topology based on any bijection $f: \mathbb R \to \mathbb R^7$ s.t. '$(\mathbb R^7, \mathscr T_X)$ is a topological $1$-manifold' if and only if '$f$ is a homeomorphism'? i think that's the idea of this post – BCLC May 02 '21 at 08:38
  • I wanna say yes but I'll leave that for someone who knows more about it than I do :) – Azur May 02 '21 at 08:40