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One thing I think is interesting about Spivak's book A Comprehensive Introduction to Differential Geometry is that Spivak uses metric spaces instead of topological spaces when defining an abstract manifold. For example, on p. 19:

A manifold-with-boundary is a metric space $M$ with the following property: If $x \in M$, then there is some neighborhood $U$ of $x$ and some integer $n \geq 0$ such that $U$ is homeomorphic to either $\mathbb R^n$ or $\mathbb H^n$.

Munkres takes a similar approach in the final chapter of Analysis on Manifolds.

In the preface to the first edition of A Comprehensive Introduction to Differential Geometry, Spivak states:

An acquaintance with topological spaces is even better, since it allows one to avoid the technical troubles which are sometimes relegated to the Problems, but I tried hard to make everything work without it.

What are the "technical troubles" that Spivak is referring to here? And why did Spivak have to try hard to make everything work out with this approach?

It seems like building the theory using the metric space definition would be no more difficult than building the theory using the topological space definition. I don't see what extra difficulties would arise.

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    $\begingroup$ Another question is why would one bother using metric spaces, since the definition of a topological space is not more difficult. $\endgroup$ – Spenser Apr 8 '16 at 17:18
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    $\begingroup$ One wants, for most purposes, to restrict manifolds to paracompact ones, and demanding that they be metric is equivalent to that. $\endgroup$ – Mariano Suárez-Álvarez Apr 8 '16 at 18:04
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This arises mostly when you're constructing new manifolds. It's easy to define a quotient space; it's much harder to define a quotient metric space. (If the torus is $[0,1] \times [0,1]$ with appropriate edge identifications, how do you define the metric on the quotient?)

This particular example is a bit disingenuous (see quotient metric space for how to get a metric on precisely this object) but the spirit is there.

It's often much easier to specify a topology than to specify a metric, and you will often want to build new manifolds. Of course, as noted in the comments in anomaly's answer, topological manifolds (meaning locally Euclidean, second countable, Hausdorff) are always metrizable; it's just that we don't want to have to do so much work to metrize them!

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There are two major problems: If you remove the condition that $M$ is a metric space, then $M$ may not even be Hausdorff; the canonical example is $\mathbb{R}$ with a doubled origin. Furthermore, even if $M$ is Hausdorff, it may not be metrizable.

In the topological setting, the usual requirement is that $M$ should be locally homeomorphic to some $\mathbb{R}^n$ (or $\mathbb{H}^n$, if we allow manifolds with boundary), and also that $M$ should be Hausdorff and second-countable. Hausdorff is generally a nice property have. Second-countability implies paracompactness and thus metrizability, and also implies that the manifold embeds in some $\mathbb{R}^N$ by a standard partition of unity argument.

I'm not familiar with how far Spivak gets into the subject, but dropping paracompactness does cause some technical problems. See, for example, Milnor and Stasheff's book on characteristic classes, which takes painful measures to avoid paracompactness for reasons that are really never explained in the text. In dealing with noncompact manifolds (compact manifolds are clearly paracompact), you can run into issues like the lack of an orientation class in the de Rham cohomology, failures of the existence of the normal bundle, etc. In particular, it can also prevent the existence of a Riemannian metric, which is probably what Spivak is referring to.

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  • $\begingroup$ I do not really consider these problems, since you immediately add them to your list of axioms. (I do agree that both assumptions are more-or-less essential to the theory.) $\endgroup$ – user98602 Apr 8 '16 at 17:20
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    $\begingroup$ Sure: I'm saying that the usual definition of a topological manifold $X$ is a locally Euclidean, Hausdorff, second-countable space; that turns out to make $X$ metrizable, so it's more or less equivalent to Spivak's definition. ("More or less" only because the choice of metric isn't canonical.) The last paragraph deals with what breaks if you just take a locally Euclidean space. $\endgroup$ – anomaly Apr 8 '16 at 17:26
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    $\begingroup$ This sounds like you're describing difficulties that arise when you remove the condition that $M$ is a metric space. That's interesting / useful to me too, but Spivak seems to say that he had to "try hard" to make the theory work with the metric space assumption. So I'm wondering how the metric space assumption does anything other than make life easier. $\endgroup$ – eternalGoldenBraid Apr 8 '16 at 17:35
  • $\begingroup$ @eternalGoldenBraid Anomaly's comment is that it doesn't do anything other than make life easier. If you read a book like Lee, where he does not assume things to be metric spaces, you'll see that to him a topological manifold is a locally euclidean second countable Hausdorff space, and as anomaly says, such things are metrizable. So the theories are equivalent. You just have to find the metrics. $\endgroup$ – user98602 Apr 8 '16 at 17:40
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    $\begingroup$ I'm not sure what Spivak means by that, since removing the metrizability assumption means that you're working in a superset of the metrizable case that contains some pathological spaces (the line with two origins, the long line, etc.) Now, sometimes you want those spaces; for example, infinite-dimensional spaces often come up in algebraic topology. My guess, though, is that Spivak doesn't want to deal with the conditions that make a space metrizable, and he doesn't want to have to explicitly define the metric you'd get on, say, a quotient space or badly embedded subspace. $\endgroup$ – anomaly Apr 8 '16 at 17:43
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The main reason for not including a metric in the definition of differential manifold is that it is an irrelevant artificial structure which is immediately abandoned when one defines an atlas: obviously transition maps between charts are smooth but definitely not isometric!
In the same vein the category of smooth manifolds and that of metric spaces are not related by natural functors in either direction.
A more technical issue is that some manifolds are plain not metrizable, even in dimension one: Spivak himself describes an example, the long line (Appendix A, corollary 6).
It is true that such manifolds are not very common, but it is very unaesthetic to exclude them from the very definition, especially since you might start with metrizable manifolds and after operating on them with standard constructions end up with a non-metrizable one.

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  • $\begingroup$ An interesting answer, but I don't think it addresses the actual question that OP posed at all. $\endgroup$ – Marc van Leeuwen Apr 9 '16 at 7:39
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    $\begingroup$ Dear @Marc: The OP writes "It seems like building the theory using the metric space definition would be no more difficult than building the theory using the topological space definition" Since I give reasons (transition functions not isometric, long line) for not using the metric space definition, your criticism that I "don't address the actual question that OP posed at all" seems to me not completely justified. $\endgroup$ – Georges Elencwajg Apr 9 '16 at 8:17

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