Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Question: Consider the proper class of all $n$-dimensional smooth manifolds. If we take the equivalence classes where two manifolds are identified if there exists a diffeomorphism between them, is this collection of equivalence classes a set?

Remark: I do not assume my manifolds to be Hausdorff nor second countable. If the answer depends on those assertions (Edit: it most definitely does) I would like to hear about the difference.

Remark 2: As Omar pointed out, there may be a problem with the various long lines. To make the question slightly more tractable, I would be (mostly) satisfied if there is a statement even ignoring the smooth structure and consider the case of topological manifolds and homeomorphisms.

Motivation: for something that I am working on I need to consider the collection of all $n$-dimensional smooth manifolds satisfying "property $X$". Unfortunately property $X$ is diffeomorphism invariant, so most definitely this collection is not a set, which invalidates many constructions (I want to build a manifold out of this collection; if the collection is a proper class then even on the set level the thing that I constructed will be a proper class, instead of a set) or at least forces me to rethink how this constructions ought to go. Fortunately for my argument it suffices that I have one object in each diffeomorphism class in my collection.

share|improve this question
4  
With out second countability, probably the long lines of different lengths are all pairwise non-diffeomorphic. If you assume Hausdorff and second countable, you can embed each manifold in some $\mathbb{R}^n$ so you get the crude bound of at most $\sum_{n \in \mathbb{N}} 2^{2^{\aleph_0}} = 2^{2^{\aleph_0}}$ manifolds. –  Omar Antolín-Camarena Oct 4 '13 at 12:20
7  
Isn't this already false in dimension $0$? I mean, if you start with any set and equip it with the discrete topology, you get a $0$-dimensional manifold. In dimnesion $n$, you can instead take the product of that discrete space with any fixed manifold. –  Tobias Fritz Oct 4 '13 at 12:25
5  
@OmarAntolín-Camarena, there is no such thing as "long lines of different lengths". There is only one long line up to homeomorphism (and it has a set many smooth structures). –  André Henriques Oct 4 '13 at 12:26
4  
Restricting to 2nd countable, metrizable smooth manifolds, isn't it enough to look at submanifolds of $\mathbb R^{2n+1}$ which I think do form a set. Then one has the equivalence relation on the set given by (nonambient) diffeomorphism, and what you care about is the set of equivalence classes. –  Igor Belegradek Oct 4 '13 at 12:36
8  
@Omar, your longer lines are not manifolds, since the point $(\omega_1,0)$ in your order is problematic, and has no neighborhood like $\mathbb{R}$. –  Joel David Hamkins Oct 4 '13 at 12:57

3 Answers 3

up vote 11 down vote accepted

If we assume the Hausdorff separation axiom, connectedness, but not second countability then each such manifold has cardinality at most continuum. Therefore the collection of equivalence classes of such manifolds is a set and not a proper class. First of all, $U$ must be path connected since every connected locally path connected space is path connected. Fix $x_{0}\in U$. Then every point of $U$ is reachable from $x_{0}$ by a path. I now claim that there can be at most continuumly many choices of paths $f:I\rightarrow U$ with $f(0)=x_{0}$. To each $x\in U$, we associate an open set $U_{x}$ locally homeomorphic to $\mathbb{R}^{n}$. Let $T$ be the following tree of height $\omega_{1}$. The nodes of $T$ will branch into continuumly many parts at each point of $T$, so the cardinality of $T$ is at most continuum. The objects of $T$ are tuples $((r_{\alpha})_{\alpha\leq\lambda},(y_{\alpha})_{\alpha\leq\lambda},(f_{\alpha}))$ where $(r_{\alpha})_{\alpha\leq\lambda}$ is an increasing sequence in $[0,1]$, $(y_{\alpha})_{\alpha\leq\lambda}$ is a sequence of points in $U$ where $y_{\alpha+1}\in U_{y_{\alpha}}$ and $f_{\alpha}:[r_{\alpha},r_{\alpha+1}]\rightarrow U_{y_{\alpha}}$ is continuous. Now for each element $((r_{\alpha})_{\alpha\leq\lambda},(y_{\alpha})_{\alpha\leq\lambda},(f_{\alpha}))$, we associate a path $\bigcup_{\alpha}f_{\alpha}$ if possible (there could be discontinuities at limit ordinals). Clearly every path is determined by an element of this tree, so there could be at most continuumly many paths. Therefore $U$ has cardinality at most continuum.

share|improve this answer
    
Could you explain why without assuming second countability your manifolds have bounded cardinality? –  Herman Stel Oct 5 '13 at 3:53
1  
You want to also demand that the functions $\alpha\mapsto r_\alpha$ and $\alpha\mapsto y_\alpha$ are continuous. Otherwise, for $\alpha$ limit, $y_\alpha$ is allowed to be any element of $U$, and you lose the cardinality bound. This is also where the Hausdorff assumption is used (it tells you the limit $y_\alpha$ are uniquely determined by the successor $y_\alpha$). –  Eric Wofsey Oct 5 '13 at 12:06

Already in dimension $0$, the collection of all manifolds in the sense of the OP is a proper class: any set, equipped with the discrete topology, is a $0$-dimensional manifold. In dimension $n$, one can instead take the product of such a discrete space with any fixed manifold, so that the collection of all $n$-manifolds is still a proper class.

(In the generic case, these manifolds are Hausdorff, but not second countable.)

As pointed out by Omar Antolín-Camarena in the comments, the question becomes more interesting if one also assumes connectedness.

share|improve this answer
2  
But as Willie Wong points out in the comments a connected non-Hausdorff manifold of dimension $n>0$ can have arbitrarily large cardinality. –  Tom Goodwillie Oct 4 '13 at 13:16

I claim that the class $\mathcal{K}$ of weak homotopy equivalence classes of connected Hausdorff smooth 2-dimensional manifolds is not a set. You know you can make compact second countable surfaces by sewing any finite number of donuts together, and if you don't demand compactness or second countability you can just keep on sewing.

First of all, the class $\text{Card}$ of all cardinals is not a set. I construct an "function" $\text{Ord}\rightarrow \mathcal{K}$ whose restriction to $\text{Card}$ is "injective" ($\text{Ord}$ denotes the class of ordinals!). Let $\alpha$ be an ordinal. Let's first deal with $\alpha=0$. Then we assign to $\alpha$ the empty manifold $M_0$. If $\alpha=1$ we assign to $\alpha$ the torus $M_1$. If $\alpha=\beta+1$ is some other successor ordinal, take a torus and chop off a disk, and take $M_\beta$ and chop off a disk to obtain $M'_\beta$, now glue them together along the boundary. If $\lambda\neq 0$ is a limit ordinal do the following. Assume that for every limit ordinal $0\neq \mu<\lambda$ we have that the underlying topological space of $M_\mu$ is the union of $M'_{\beta}$ for $\beta<\mu$ and the smooth atlas of $M_\mu$ is the set of all charts of $M'_\beta$ for $\beta<\mu$ whose domain (or codomain, whichever your convention is) is diffeomorphic to $\mathbb{R}^2$ (so we are excluding charts hitting the boundary of $M'_\beta$). Then define the underlying topological space of $M_\lambda$ to be the union of the $M'_\alpha$ for $\alpha<\lambda$ and take as an atlas for $M_\lambda$ the set of all charts of all $M'_\alpha$ for $\alpha<\lambda$ that miss the boundary of $M'_\alpha$. Set $M'_\lambda$ to be $M_\lambda$ with a disk removed.

This gives a "function" $\text{Ord}\rightarrow \mathcal{K}:\alpha\mapsto M_\alpha$. "It's restriction to the class of cardinals is injective" because if $|\alpha|\neq |\beta|$ then $\pi_1\left(M_\alpha\right)\not\cong \pi_1\left(M_\beta\right)$. If you want to get rid of the """'s, use a Grothendieck universe (a modest large cardinal), if you allow yourself to. If you don't the proof is still valid.

It seems likely this can be done for any $n$ and the class of suitably connected manifolds, as was pointed out for $0$-dimensional manifolds by Tobias. For $1$-dimensional manifolds, take $\underline{\alpha}\times \mathbb{R}$ for each ordinal $\alpha$ ($\underline{\alpha}$ is the discrete topological space whose underlying set is $\alpha$). Again the restriction to $\text{Card}$ is injective, since $\pi_0$ tells them apart.

EDIT: The constructing shown here does not work because there exists an $n$ for which $M'_n\not\subseteq M'_\omega$, as Eric points out in a comment below.

share|improve this answer
    
You can't just say $M_\lambda$ is the union of the $M_\alpha'$, because the $M_\alpha'$ might not be nested (later steps might chop out part of them). Indeed, $M_\omega'$ must necessarily chop out a disk that intersects some $M_n'$. You then run into trouble if at limit steps, the disks you've chopped out accumulate at a point. –  Eric Wofsey Oct 5 '13 at 11:57
    
@Eric: You are right! Thank you. –  Herman Stel Oct 5 '13 at 12:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.