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We will say that a Hausdorff topological space $X$ is a smooth manifold if there is an open cover $(U_{\alpha})$ of $X$ and a corresponding collection of homeomorphisms $\varphi_{\alpha} : U_{\alpha} \to V_{\alpha} \subset \mathbb{R}^n$ such that on any overlap $U_{\alpha} \cap U_{\beta}$, the maps $$\varphi_{\beta} \circ \varphi_{\alpha}^{-1} : \varphi_{\alpha}(U_{\alpha} \cap U_{\beta}) \longrightarrow \varphi_{\beta}(U_{\alpha} \cap U_{\beta})$$ are smooth.

Note that I have omitted the assumption of paracompactness. I recall vaguely from my undergraduate years of my lecturer telling us that there are good reasons for omitting paracompactness from the definition of a manifold unless one wants to look at metric properties of $X$ or integrate.

His expertise was in complex geometry and homogenous spaces. I vaguely recall his justification for this being that every group can be declared a Lie group if one allows the more general definition since one is permitted to have an uncountable number of connected components.

I am not sure if I have correctly remembered this, but I was hoping someone could maybe elaborate on this thought further?

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    $\begingroup$ Sure, every discrete group is a zero-dimensional Lie group. There are examples of locally convex infinite-dimensional manifolds that aren't paracompact, but if one is happy to go there, then mere lack of paracompactness won't be scary. Note that the long line is a 1-dimensional manifold-without-paracompactness, and doesn't have a Riemannian metric inducing its topology. This is one reason why one might want to insist on paracompactness. $\endgroup$
    – David Roberts ♦
    Commented Nov 25, 2019 at 6:16
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    $\begingroup$ I removed the mathematical philosophy tag as I do not think it means what you think it means. $\endgroup$
    – David Roberts ♦
    Commented Nov 25, 2019 at 6:18
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    $\begingroup$ @YCor: Doesn't the long line admit a differentiable structure? $\endgroup$
    – abx
    Commented Nov 25, 2019 at 9:34
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    $\begingroup$ If one thinks he or she understand surfaces, then one thinks of manifolds as paracompact. The variety of non-paracompact surfaces is wild. $\endgroup$
    – Benoît Kloeckner
    Commented Nov 25, 2019 at 11:50
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    $\begingroup$ Your question appears to be just seeking opinions. i.e. is this a MathOverflow question? To give it some structure you should perhaps explain what you want to use the notion of manifold for. $\endgroup$
    – Ryan Budney
    Commented Nov 25, 2019 at 20:32

2 Answers 2

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every group can be declared a Lie group if one allows the more general definition since one is permitted to have an uncountable number of connected components.

A manifold is paracompact if and only if all of its connected components are second countable. So in particular, any discrete group is a paracompact Hausdorff smooth manifold. This manifold is second countable if and only if the group is countable, so if we want uncountable discrete groups to be Lie groups, we cannot require manifolds to be second countable, only paracompact.

On the other hand, removing the weaker assumption of paracompactness from the definition of a smooth manifold will immediately eliminate all theorems that use partitions of unity (examples: existence of Riemannian metrics, existence of connections, existence of embeddings in R^n, the Serre–Swan theorem), which in the context of differential geometry means the vast majority of nontrivial theorems. Unless one intends to make all these theorems false, it probably makes sense to make manifolds paracompact by definition.

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  • $\begingroup$ Are the existence of connections and Serre–Swan provably false for some non-paracompact manifold? $\endgroup$
    – Z. M
    Commented Jul 6 at 5:32
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    $\begingroup$ @Z.M: The tangent bundle of the long line is a counterexample to both statements: it does not admit a connection and its module of sections is not a finitely generated projective module. Indeed, if it did admit a connection, one could construct a metric by parallel transporting some metric on the fiber over some fixed point to all other points. If the module of sections of the tangent bundle of the long line was finitely generated and projective, we could induce a metric on it from its presentation as a direct summand of a trivial bundle, which does admit a metric. $\endgroup$
    – Dmitri Pavlov
    Commented Jul 6 at 14:18
  • $\begingroup$ It seems that the crucial failure is affineness. I wonder whether the higher sheaf cohomology of the structure sheaf $C^\infty(-)$ is non-vanishing for the long line as well? $\endgroup$
    – Z. M
    Commented Jul 6 at 15:41
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    $\begingroup$ @Z.M: Some examples of nontrivial bundles over nonparacompact spaces are given in arxiv.org/abs/1309.2524. Since every smooth function on the long line is eventually constant, I suspect the techniques of that article can be adapted to show that the higher cohomology of the structure sheaf of the long line is nonvanishing. $\endgroup$
    – Dmitri Pavlov
    Commented Jul 7 at 1:34
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Another good gift of paracompactness:

A Hausdorff $C^k$ manifold ($k\ge0$) is metrizable iff it is paracompact.

This is also true for infinite dimensional manifold modelled on a Banach space (because the underlying fact is that a topological space is metrisable iff it is Hausdorff, paracompact and locally metrisable: local metrics may be glued by means of a partition of unity).

Talking about Banach manifolds, recall, just in case, that

A second countable, regular, Banach manifold is paracompact.

But here Hausdorff in place of regular only gives an equivalent statement in finite dimension: it would be false for infinite dimensional Banach manifolds. (An enjoyable account of this in in R.Palais' paper Critical point theory and the minimax principle)

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