# $T_pS \subseteq T_pM$: Are tangent spaces of submanifolds subsets (and not embedded) of tangent spaces of the original manifold?

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

From Definition 8.1 and Remark 8.2 (and definitions from Section 2. see below), we have that

• A. $$T_pM = T_pU$$

• B. and $$C_p^{\infty}M = C_p^{\infty}U$$, where (B) implies (A).

• I believe both equalities are really equalities are not isomorphisms (see this question, my previous question and my other previous question).

This question says the differential of inclusion map of smooth manifolds is an inclusion map of tangent spaces.

1. Question 1. If A and B are indeed equalities, then is it really that $$T_pS \subseteq T_pM$$ (this may or may not be vector subspace, but I think it is) for $$S \subseteq M$$ a regular/embedded submanifold and not merely that $$T_pS$$ is (vector space) isomorphic to a vector subspace of $$T_pM$$ (so $$T_pS$$ is embedded, in the vector space sense, but probably equivalent to the manifold sense, in $$T_pM$$)?

2. Question 2. If A and B are indeed equalities and the answer to (1) is yes, then, then how do we see this from Definition 8.1 and Remark 8.2 with $$C_p^\infty S$$ and $$C_p^{\infty} M$$?

• I think I can kind of see this geometrically with $$S=S^1$$, $$M=\mathbb R^2$$ and $$p=(1,0)$$ with $$T_pS^1$$, as the vertical line isomorphic to $$\mathbb R$$ with $$p$$ as origin, being a vector subspace of $$T_p\mathbb R^2$$, as the plane isomorphic to $$\mathbb R^2$$ with $$p$$ as origin

• but I don't quite see how a map $$D:C_p^{\infty}S \to \mathbb R$$ is also a map $$D:C_p^{\infty}M \to \mathbb R$$.

• I think we somehow do some kind of smooth extensions, but most of the smooth extensions I've encountered are from open submanifolds or open subsets are not arbitrary regular submanifolds.

• Maybe $$C_p^{\infty} S \supseteq C_p^\infty M$$ or something.

3. Question 3. If (A) is merely an isomorphism and not equality, then why exactly? Is it actually that (B) is also merely an isomorphism and not equality, and if so, then why exactly?

Note: Other definitions of tangent space of submanifold seem to be explicit in being subsets of tangent spaces (at the same point) of the original manifold. See the (embedded) submanifold part in this question (on immersed submanifolds).

Also: Exercise 11.1 seems to implicitly and casually assume that $$T_pS^n \subseteq T_p \mathbb R^{n+1}$$ (and so proceeds to discuss a condition on how certain elements of $$T_p \mathbb R^{n+1}$$ are also elements of $$T_pS^n$$)

Section 2 definitions: page 11, page 12, page 13, page 14

I believe both equalities are really equalities are not isomorphisms

This is not quite true; it might help to write out the definitions. Given a manifold $$M,$$ we can form the vector space $$C^{\infty}(M)$$ of smooth functions $$f : M \rightarrow \Bbb R.$$ Then we define, $$C^{\infty}_p(M) = \left\{ [(f,V)]_M : V \subset M \text{ open containing } p, \ f \in C^{\infty}(U)\right\} / \sim_M,$$ where we quotient by a suitable equivalence relation $$\sim.$$ From this we see for $$U \subset M$$ open there's an obvious isomorphism, $$C_p^{\infty}(U) \rightarrow C_p^{\infty}(M) \text{ sending } [(f,V)]_U \mapsto [(f,V)]_M,$$ which makes it looks like the two spaces are equal. But not the equivalence classes are not equal as sets, because even if we start with $$[(f,V)]_U \in C^{\infty}_p(U),$$ by multiplying by a suitable cutoff we can find $$g \in C^{\infty}(M)$$ such that $$(g,M) \sim_M (f,V).$$ Evidently if $$U \neq M,$$ then $$(g,M) \not\in [(f,V)]_U.$$

That being said, because we have a "canonical" isomorphism $$C_p^{\infty}(U) \cong C_p^{\infty}(M)$$ we can basically pretend these sets are equal and drop the subscripts, where we implicitly use this identification each time. This is something that's done quite often, because there's really no harm in pretending these objects are actually equal since we have fixed a choice of identification.

Note this also gives a canonical isomorphism $$T_pM \cong T_pU$$ using the above identification, but again they aren't equal as sets. People still write $$T_pM=T_pU$$ however, because they are essentially the same.

It's important to note also that there's a lot of ways to define all of these notions, and one can show they are all equivalent in a suitably canonical sense. This is partly why we might not be too precise, because that both convention-dependent and provably not going to cause any ambiguities.

I think this more or less should answer questions 2 and 3, so let me now address your first question:

Note that if $$S \subset M$$ is an embedded submanifold, the inclusion map $$\iota : S \hookrightarrow M$$ induces a map $$\mathrm{d}\iota_p : T_pS \rightarrow T_pM$$ for all $$p \in M,$$ and this map is always an injection. Again because the fact that $$S \subset M$$ gives us a canonical choice of identification, there's not much harm in thinking of $$T_pS \subset T_pM.$$

Moreover technical difficulties aside, geometrically you want to to think of $$T_pS$$ being a subset of $$T_pM;$$ intuitively the tangent space of a manifold at a point encodes all the possible directions in which you can travel. So if you're in a submanifold $$S,$$ the set of directions you can travel can naturally as a subset of directions you can travel in the ambient space $$M.$$ This is especially instructive in the case when $$M = \mathbb R^n,$$ where we have classical, equivalent definitions of the tangent space which encodes this intuition (and where some authors actually define $$T_pS$$ as a subset of $$\Bbb R^n$$).

• Thanks. About the equalities vs isomorphisms, 1. is this attempt wrong? (no one has yet answered that) 2. Are you sure your definitions of $C_p^{\infty}M$ are the same as the one in the textbook? I'll edit to include the definitions from Section 2. – Selene Auckland Jun 21 at 13:35
• @Selene Auckland The definition I wrote is me explicitly writing out definition 8.1 in set theoretic language (note the subscript $M$ is purely notational, to prevent confusion). There's probably other equivalent ways to do this, but this is how I naturally interpret it. From this it becomes clear that asking if the sets are equal doesn't make sense, because they contain different elements. – ktoi Jun 21 at 13:46