My book is An Introduction to Manifolds by Loring W. Tu.

As can be found in the following bullet points

we have that

  1. Tu's manifolds with or without boundaries do not necessarily have (uniform) dimensions.

  2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).

Question: For Definition 22.6 (see here and here), Tu says that "A manifold with boundary has dimension at least 1". Should this instead be "A manifold with boundary has dimension at least 1 if it has a dimension and if it has nonempty boundary" or "An $n-$manifold with boundary with non-empty boundary has $n \ge 1$" (Notice that the prefix "$n-$" precisely gives the manifold with boundary a dimension)?

Embedding photos:

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    $\begingroup$ Of course, this entirely depends on how you defined things and, once you have fixed definitions, the answer will be trivial. $\endgroup$ – AnonymousCoward May 10 '19 at 12:37
  • $\begingroup$ Also note that imgur is not accessible in every country. It is better to embed images directly into your post. $\endgroup$ – AnonymousCoward May 10 '19 at 12:52
  • $\begingroup$ @AnonymousCoward You mean one might not access imgur directly but can be embedded imgur images? $\endgroup$ – user636532 May 10 '19 at 13:11
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    $\begingroup$ Oh, they embed from imgur now? this is a shame. $\endgroup$ – AnonymousCoward May 10 '19 at 13:34
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    $\begingroup$ If you are having so many issues with the definitions in a given textbook, it might be a good time to pick up another book. There are so many introductory smooth manifold texts to choose from. The world is your oyster. $\endgroup$ – AnonymousCoward May 10 '19 at 13:35

I think Tu’s statement is fine:

A manifold, by definition, always has a dimension. Where are the charts going?

Usually when we say that a manifold “has boundary,” we mean that it has nonempty boundary.

After looking at some of Tu’s (non-standard!) definitions, I think you’re correct. An accurate statement might be

If an $n$-dimensional manifold has nonempty boundary, then $n\ge 1$.

  • $\begingroup$ I'll add these to the post to make the post hopefully self-contained: 1. Tu's manifolds do not necessarily have dimension. 2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries). $\endgroup$ – user636532 May 10 '19 at 12:27
  • $\begingroup$ Thanks Santana Afton! About non-standard, Tu said himself on stackexchange that these must be allowed: here. Your response? Haha $\endgroup$ – user636532 May 10 '19 at 13:11
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    $\begingroup$ @SeleneAuckland When learning about these objects, the important part is understanding the core of what they are and how they interact with one another. These edge cases can be important — keep them in mind — but don’t dwell on them at the moment. $\endgroup$ – Santana Afton May 10 '19 at 13:19
  • $\begingroup$ Good advice. Thanks! $\endgroup$ – user636532 May 10 '19 at 13:20
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    $\begingroup$ @SeleneAuckland I think it’s a tricky issue. It depends on what you think the purpose of an introductory text is, what lessons an author wants a reader to learn from the text, etc. There’s a much larger conversation here about standard definitions vs. “proper” definitions, how mathematics “ought” to behave, teaching students, and so on. I personally disagree, but I’m not sure how much weight that has. $\endgroup$ – Santana Afton May 10 '19 at 13:31

Assuming sensible definitions, an alternative solution is to change the statement to the following:

A connected manifold with non-empty boundary has dimension at least 1

Edit: I rejected the suggested edit to change "manifold with non-empty boundary" to "manifold with boundary with non-empty boundary" because it does not add new information. A manifold with non-empty boundary must be a manifold with boundary, or your definitions are nonsense.

  • $\begingroup$ Oh because connected manifolds with or without boundary, under Tu's dimension-less definitions, necessarily have dimensions? $\endgroup$ – user636532 May 10 '19 at 12:57
  • $\begingroup$ Also, the rest of the sentence talks about "a discrete set of points". I don't think connected is what was intended. Why don't we just say "An $n-$manifold with boundary with non-empty boundary has $n \ge 1$" ? $\endgroup$ – user636532 May 10 '19 at 12:58
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    $\begingroup$ I don't know what Tu's definition is, but for any sensible definition of manifold, a connected manifold will have a well defined dimension. I also cannot load imgur posts and cant see anything about discrete sets of points directly in your post. $\endgroup$ – AnonymousCoward May 10 '19 at 13:07
  • $\begingroup$ The statement in Section 22: "A manifold with boundary has dimension at least 1 since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary." The definition of manifolds (without boundary) in Section 5: here $\endgroup$ – user636532 May 10 '19 at 13:10

I would not say that a manifold could be dimensionless. A manifold consists of connected components, each of which has a dimension. As for the statement in question, a more accurate phrasing would be

"If an n-manifold has nonempty boundary, then $n \ge 1$"


"A connected manifold with nonempty boundary has dimension at least 1"

as was pointed out above by various commentators.

  • $\begingroup$ Thanks Prof Tu! $\endgroup$ – user636532 May 28 '19 at 23:36

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