Are manifold subsets submanifolds?

Question

New question: Can manifold subsets always be made into submanifolds?

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

A. Regular/embedded submanifolds are manifolds. My question is about the converse.

In algebra:

B. Subset groups are equivalent to subgroups (at least with the same law, but I believe "same identity" is not required because they'll just turn out to have the same identity anyway).
C. Rings not so much: For (commutative unital) rings, if $B$ is a ring and if $A \subseteq B$ and $A$ is a subring of $B$, then $A$ is a ring (with the same laws and identity as $B$ because this is how subring is defined anyway). However conversely, if they are both rings (NOT necessarily with the same laws or identity), then $A$ is not necessarily a subring of $B$.
- D. For example, $B$ has an idempotent element $e$ besides identity, and $A$ is the principal ideal generated by $e$, where we have $e$ as the identity of $A$ but not of $B$ (Algebra by Michael Artin Proposition 11.6.2). I think the laws of $A=(e)$ are the same as the one of $B$, and the only thing lacking for $A$ to be a subring of $B$ is that $A$ has a different identity from $B$ (I understand that $A=(e)$ has a different identity from $B$ if and only if $A$ doesn't contain the identity of $B$).
E. Based on what I think is the issue in (D) and based on my guess that manifolds have no such analogue for "identity", I expect manifold subsets to be regular/embedded submanifolds.
- Update: Based on Eric Wofsey's answer, I guess since there are indeed ways, that subset rings are not subrings, besides not sharing identity. I guess the ways are to do with the laws $+$ and $\times$ differing between $A$ and $B$, kind of like in the above parenthetical remark for groups.

Question: Let $A$ and $B$ be manifolds with respective dimensions $a$ and $b$. If $A \subseteq B$ (given the subspace topology because apparently people don't just assume this), then is $A$ a regular/an embedded $a$-submanifold of $B$?

I'll just attempt to prove embedded (I won't prove regular directly). Please verify.

$A$ is the image of the inclusion map $\iota: A \to B$. I will show $\iota$ is an embedding, with this definition (Using this equivalent definition would be circular since such definition says "smooth submanifold" and not "smooth manifold"):

Smooth: An inclusion between two smooth manifolds is smooth.
- Edit: I guess this is the problem. I can't quite use Theorem 11.14, but i think one can somehow modify the proof of Theorem 11.14 to prove "If N is a (smooth) manifold subset of M, then the inclusion $i: N \to M, i(p) = p$, is an embedding"
Immersion: Inclusions are the prototype of immersions.
- Edit: Oh, at least for Euclidean spaces.
Topological embedding: The restriction $\tilde{\iota}: A \to \iota(A)=A$ is identity on $A$, a homeomorphism of $A$ (because of subspace topology).

Do you assume any relationship between the smooth manifold structures on $A$ and $B$ at all? If not, then merely knowing that $A$ is a subset of $B$ tells you extremely little. Any set of cardinality $2^{\aleph_0}$ can be given a smooth manifold structure of any positive dimension... — Eric Wofsey, Jul 20 '19 at 05:26
In any case, you have not given any justification for most of the claims in your "proof". — Eric Wofsey, Jul 20 '19 at 05:29
@EricWofsey Thanks. I added a link for (2). For (3), I think this is known from elementary topology. For (1), I guess this is the main issue. Is it related to one of the issues here? — , Jul 20 '19 at 05:43
@EricWofsey Thanks for the edit! I cannot believe I put algebraic-geometry instead of differential-geometry. — , Jul 20 '19 at 05:50
You‘re saying „If they are both rings, then...“. Unfortunately, that‘s wrong, subsets that are rings are subrings by definition. Note that ideals are not rings (but rather submodules)... — Qi Zhu, Jul 20 '19 at 05:54
@Kezer It's been 5-6 months since I've done any algebra, but I remember an ideal of a unital commutative ring is still, or at leastcan be (I forgot), a ring but with a different identity. In such a case, the ideal is not a subring because they don't have the same identity. At least my textbook defines subrings to have identities and to have the same identities as the original ring. Do you disagree with the parenthetical remark, on rings, of Eric Wofsey's answer below? (I understand Eric Wofsey agrees with me about rings but for a different reason) — , Jul 20 '19 at 06:07
You can make almost anything (perhaps even everything?) into a ring by choosing everything suitably, so just changing operations doesn't make much sense. An ideal by definition inherits the structure of the ring $R$; an ideal is not just the set $I$ but rather $(I,+,\cdot)$ combined with the inherited operations. (And I don't see Eric Wofsey talking about ideals, he says that subsets don't have to be subrings which of course is true.) — Qi Zhu, Jul 20 '19 at 09:37
@Kezer You said "subsets that are rings are subrings by definition." What about the example of the principal ideal $(e)$ that I edited to include? Thanks by the way! — , Jul 20 '19 at 09:44
Once again, ideals are NOT rings... Ideals $I$ are never rings unless $I = (0)$ or $I = R$. There is a huge difference between these two structures. Substructures of rings are subrings. Ideals are submodules of $R$ as an $R$-module - a completely different (and by the way categorically much nicer) structure. This is universal in mathematics: A substructure of a structure is just a subset with the same structure (subgroups are subsets that are groups, subrings and subsets that are rings, etc.). — Qi Zhu, Jul 20 '19 at 10:17
@Kezer Right yeah: I was wrong earlier to say that ideals are (commutative unital) rings with different identities because they not have identities and thus are not rings. I think they are sometimes rings and sometimes not rings. For example, in the case of the $e$ above, I think that $(e)$ is a ring WITH A DIFFERENT IDENTITY: Is THIS ONE also wrong? Artin might say "Ideals $I$ are never subrings unless $I=(0)$ or $I=R$". I believe Artin asserts that $(e)$ is a ring. You might have a different definition. Do you disagree with Artin? — , Jul 20 '19 at 11:34
Do you understand Artin's Statement? He says $(e)$ is a subring only if $e = 1$. Do you know what $(1)$ is? It's precisely the ring itself. (I have to correct myself, $I = (0)$ only works if $I = R$.) Please try to read and understand my statements. I've clearly asserted that it makes no sense to look at different identities. — Qi Zhu, Jul 20 '19 at 16:32
@Kezer I know $(e)$ is not a subring because we're not given $e=1$ (Oh I didn't realize $e=0$ didn't work. I've forgotten this already). However, $(e)$ is a ring (not necessarily with the same laws or identity, in particular $(e)$ has a different identity) that happens to be a subset of another ring. I believe there no contradiction here. Artin does not define "subring" as equivalent to "subset that is a ring (not necessarily with the same laws or identity)". You said "it makes no sense to look at different identities". Okay, but what if we did? I'll edit to clarify in the post. Thanks. — , Jul 21 '19 at 05:42
@Kezer Note: I'm talking about 11.6.2b (which I believe says ring subset of $S$) not 11.6.2c (which says not a subring of $S$ for $e \ne 1$). — , Jul 21 '19 at 11:22
@Kezer Do you disagree with Eric Wofsey please? Thanks by the way! — , Jul 21 '19 at 11:25
@EricWofsey is what you talk about in your 1st comment ('can be given a smooth manifold structure of any positive dimension...') precisely this ? — BCLC, Apr 28 '21 at 01:32

Eric Wofsey · Accepted Answer · 2019-07-20T05:58:31.823

5

No, this is very very false. For instance, let $B$ be $\mathbb{R}$ with its usual smooth manifold structure, and let $A$ be $\mathbb{R}$ with a smooth manifold structure given by picking a bijection $\mathbb{R}\to\mathbb{R}^2$ and pulling back the usual smooth manifold structure on $\mathbb{R}^2$. Then $A$ is certainly not an embedded submanifold of $B$, since it has larger dimension. Indeed, the inclusion map $A\to B$ cannot even be continuous.

Even if you assume $A$ has the subspace topology, it is still very false. For instance, in the example above, you can instead pick a homeomorphism $\mathbb{R}\to\mathbb{R}$ that is not a diffeomorphism and pull back the usual smooth manifold structure of $\mathbb{R}$ to a new one and call it $A$. Then the inclusion map $A\to B$ will be a homeomorphism but not a diffeomorphism.

The key thing to understand here is that being a manifold is not a property of a set. It's an additional structure you can put on a set. All that that $A\subseteq B$ tells you is that every element of $A$ happens to be an element of $B$; it tells you nothing at all about their manifold structures, which could be totally unrelated. (The same thing happens with rings: if $A$ and $B$ are rings with $A\subseteq B$, then there is no reason at all to think that $A$ is a subring of $B$, because the ring operations of $A$ are probably totally different from those of $B$.) Being a smooth manifold is similarly not a property of a topological space, but an extra structure you can put on it.

As for your proposed proof, all three of your claims are wrong as shown by the example above. You gave no justification for claim 1 or claim 2 ("inclusions are the prototype of immersions" is just a vague slogan that has no meaning in a proof). For claim 3, to prove $\iota$ is an embedding you need to prove it is a homeomorphism from $A$ to $\iota(A)$ with the subspace topology from $B$, and you have no reason to believe that topology is the same as the given topology on $A$.

edited Jul 20 '19 at 05:58

answered Jul 20 '19 at 05:44

Eric Wofsey

330,363

Thanks! About the stuff: Claim 3: Okay this is weird. I thought subspace topology is always assumed unless otherwise stated. This is what I gathered from elementary topology. I think we will go insane if we have to say "subspace" topology every time. Is that the only problem with Claim 3? – Jul 20 '19 at 05:52
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Yes, claim 3 becomes true if you assume $A$ has the subspace topology. This is generally implied if you are just starting with a topological space and taking a subset of it, but that's not what's going on in your question: you started with two separate manifolds $A$ and $B$, and then added the assumption that $A$ is a subset of $B$. Given that you are explicitly not assuming $A$ is a submanifold of $B$, I would think most readers would think you aren't assuming it is a subspace either. – Eric Wofsey Jul 20 '19 at 05:55
Claim 2: Okay, so I remembered wrong. The prototype was for Euclidean space (I read again), but I understand the differential at any point of an inclusion (a smooth inclusion, otherwise differential is not defined in this book) map of smooth manifolds is still inclusion. As far as I know every inclusion map of anything is injective. Therefore the differential at any point is injective and thus the original inclusion map is an immersion everywhere? – Jul 20 '19 at 05:59
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It is not true that the differential of an inclusion map of smooth manifolds is an inclusion (for one thing, it certainly isn't literally an inclusion at least for most ways of defining tangent spaces; at best you might hope for it to be an injection). – Eric Wofsey Jul 20 '19 at 06:01
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The question you linked seemed to be tacitly assuming that $X$ is a smooth submanifold of $Y$ (and the answer also makes this assumption). – Eric Wofsey Jul 20 '19 at 06:03
Claim 2: I had a feeling!. Do you disagree with this? Anyway, inclusion, if a smooth map of smooth manifolds (not just euclidean spaces), is an immersion? – Jul 20 '19 at 06:03
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No, consider $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^3$. It is smooth and injective but not an immersion. (It is not literally an inclusion, but any injection is "isomorphic" to an inclusion since you can transport the structures on the domain to a structure on its image.) – Eric Wofsey Jul 20 '19 at 06:08
Claim 2: Eric Wofsey, how is your $f$ an inclusion? My understanding of inclusion is $f(x)=x$. My question is if smooth inclusion maps of smooth manifolds are immersions. Oh man, this might be another one of those different definitions for a thing (in this case inclusion) again. – Jul 20 '19 at 06:12
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Let $A$ denote the image of $f$ (i.e., the set $\mathbb{R}$), equipped with the unique smooth structure that maps $f:\mathbb{R}\to A$ a diffeomorphism (push the smooth structure of $\mathbb{R}$ forward along $A$). Now the inclusion map $A\to\mathbb{R}$ is smooth and not an immersion (and it is basically the same as $f$; we have just renamed the elements of its domain along $f$ to make it an inclusion). – Eric Wofsey Jul 20 '19 at 06:14
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Again, the thing you need to keep in mind is that the sets involved never really matter. What matters is the structure we put on those sets. You can "replace" the elements of a set with new elements via a bijection and get a totally different set, but with essentially the same structure. In this way the distinction between an inclusion and an injection is artificial: every injection is an inclusion, up to renaming the elements of the sets involved. – Eric Wofsey Jul 20 '19 at 06:18
Is this exercise related? – Jul 20 '19 at 06:18
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Yes, what I called $A$ in my comment above is the same as $\mathbb{R}'$ in that exercise. – Eric Wofsey Jul 20 '19 at 06:20
Ok thanks. Wait wait, Eric Wofsey, what exactly am I missing then to make manifold subsets into regular/embedded submanifolds? I can't quite pinpoint exactly, but there have been a lot of times I've, and I think the book has too, just been implicitly assuming that when we have manifolds (with dimension) $A$ and $B$ and $A \subseteq B$, we have somehow that $A$ is a submanifold of $B$. There was probably an assumption I overlooked then. I have a feeling this is something that can be fixed simply (like assuming the subspace topology for Claim 3) – Jul 20 '19 at 06:22
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Normally, you just directly assume that $A$ is an embedded submanifold of $B$. There isn't really any shortcut to it. – Eric Wofsey Jul 20 '19 at 06:23
Ok thanks. I'll have to revisit in the weeks or months to come when I find something. – Jul 20 '19 at 07:08
Eric Wofsey, about the algebra, 1. what is the analogue of "ring operations" here please? I guess it's those adapted charts. 2. okay I actually didn't consider that thing about the operations, but it's possible to have the same operations but still not be a subring ONLY because of identity? From Algebra by Michael Artin Proposition 11.6.2 – Jul 20 '19 at 08:59
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Well, there's not an exact analogue, since the relationship between a submanifold and a manifold is more complicated than the relationship between a subring and a ring. You need a certain compatibility between the charts of the two manifolds, but it's not as simple as just saying "restrict the operations" like it is for rings.

Eric Wofsey

Jul 20 '19 at 17:10

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I would consider the identity element to be one of the operations of a ring, so "same operations" means also the same identity. But if you take "operations" to just mean $+$ and $\cdot$, then yes, you can have a subset with the same operations but different identity, and it's not a subring. (Here I assume "ring" means "unital ring", which it doesn't always.)

– Eric Wofsey Jul 20 '19 at 17:11

Are manifold subsets submanifolds?

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