This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
By definition, a submanifold of a manifold \(M\) is an injective immersion. Write this as \(\Phi:P\rightarrow M\). Restricting the codomain of \(\Phi\) yields a bijection \(\bar{\Phi}:P\rightarrow \Phi(P)\), so we may transfer the topology of \(P\) directly onto \(\Phi(P)\). Through this process, a submanifold of \(M\) can also be regarded as the inclusion of a subset \(\Phi(P)\hookrightarrow M\). In this post we examine this perspective in more detail.
Submanifolds and \(C^\infty\) Functions
First, fix a manifold \(M\) and a submanifold \(\Phi:P\rightarrow M\). If another \(C^\infty\) function \(F:N\rightarrow M\) satisfies \(F(N)\subseteq\Phi(P)\), then using \(\bar{\Phi}\) as above we can define a new injective function \(F_0:N\rightarrow P\) by the formula
\[F_0=\bar{\Phi}^{-1}\circ F\]It is natural to ask whether \(F_0\) defined in this way is an immersion.
If we view a submanifold simply as a subset of the original manifold, this question becomes: whenever the image of an arbitrary \(C^\infty\) function \(F:N\rightarrow M\) satisfies \(F(N)\subseteq P\subseteq M\), is the restriction of \(F\) to codomain \(P\) again \(C^\infty\)? However, this question is less trivial than it appears, and the answer is not always affirmative.
Example 1 Suppose both manifolds \(N\) and \(P\) are \(\mathbb{R}\) with their natural manifold structures, and let \(M=\mathbb{R}^2\) also carry its natural manifold structure. Define two submanifolds \(\Phi:N\rightarrow M\) and \(\Psi:P\rightarrow M\) by the two figures below
and
Then \(\Phi(N)=\Psi(P)\), and therefore the bijection \(\bar{\Psi}^{-1}\circ\Phi\) from \(N\) to \(P\) is well defined.
Now consider a sufficiently small open neighborhood \(U\) of the origin in \(N\), and look at the image of \(U\) under \(\bar{\Psi}^{-1}\circ\Phi\). Then \((\bar{\Psi}^{-1}\circ\Phi)(U)\) is not an open set in \(P\).
Hence \((\bar{\Psi}^{-1}\circ\Phi)^{-1}\) is not continuous, so this function is not even \(C^\infty\).
The following proposition shows, however, that the obstruction in the preceding example is purely topological.
Proposition 2 Let \(\Phi:P\rightarrow M\) be a submanifold of a manifold \(M\). If an arbitrary \(C^\infty\) function \(F:N\rightarrow M\) satisfies \(F(N)\subseteq\Phi(P)\), then for the function \(F_0:N\rightarrow P\) defined by
\[F_0=\bar{\Phi}^{-1}\circ F\]the following hold:
- If \(F_0\) is continuous, then \(F_0\) is \(C^\infty\).
- If \(\Phi\) is an embedding, then \(F_0\) is continuous.
Proof
The second claim is immediate from the definition, so it suffices to prove the first.
Assume \(F_0\) is continuous; we show that \(F_0\) is \(C^\infty\). That is, for arbitrary \(x\in N\) we must exhibit a coordinate system \((U,\varphi)\) centered at \(x\) such that the restriction of \(F_0\) to \(U\) is \(C^\infty\). Since we have assumed that \(F_0\) is a continuous bijection, to show this it suffices to exhibit, for arbitrary \(y\in P\), a coordinate system \((V,\psi)\) containing \(y\) such that the restriction of \(\psi\circ F_0\) to the open set \(F_0^{-1}(V)\) is \(C^\infty\).
Let \(y\in P\) be given, and choose a coordinate system \((W,z^1,\ldots, z^m)\) of \(M\) containing \(\Phi(y)\). Then by §Submanifolds and the Inverse Function Theorem, ⁋Corollary 10 (Rank theorem, Immersion case), restricting a suitable subset of \(\{z^k\circ\Phi\mid 1\leq k\leq m\}\) to a suitable open neighborhood \(V\) yields a coordinate system at the point \(y\in P\).
Denote these by \(\{z^1\circ\Phi,\ldots,z^p\circ\Phi\}\). Without loss of generality, assume \(\gamma=(z^1,\ldots, z^m)\) is a surjection onto \(\mathbb{R}^m\); then the above claim is equivalent to saying that \((V,\pi\circ\gamma\circ\Phi)\) is a coordinate system at \(y\) via the projection \(\pi:\mathbb{R}^m\rightarrow\mathbb{R}^p\). Now
\[(\pi\circ\gamma\circ\Phi)\circ F_0|_{F_0^{-1}(V)}=\pi\circ\gamma\circ F|_{F_0^{-1}(V)}\]and the right-hand side is a composition of \(C^\infty\) functions, hence \(C^\infty\).
Equivalence of Submanifolds
Definition 3 Let a manifold \(M\) and two submanifolds \(\Phi_1:N_1\rightarrow M\), \(\Phi_2:N_2\rightarrow M\) be given. These two are called equivalent if there exists a diffeomorphism \(\theta:N_1\rightarrow N_2\) such that \(\Phi_1=\Phi_2\circ\theta\).
That the relation thus defined induces an equivalence relation on the set of all submanifolds \((N,\Phi)\) of \(M\) is obvious. Choose an arbitrary equivalence class \([(N,\Phi)]\). Then we may consider the pair
\[A=\Phi(N)\subseteq M, \qquad \iota:A\hookrightarrow M\]just as at the outset. Here \(A\) is a manifold endowed with the differentiable and topological structures of \(N\) via the bijection \(\bar{\Phi}:N\rightarrow A\). Then \(\bar{\Phi}\) is a diffeomorphism between \(N\) and \(A\), so \(\bar{\Phi}^{-1}\) is also a diffeomorphism and therefore the composition
\[\iota=\Phi\circ\bar{\Phi}^{-1}\]is an immersion from \(A\) into \(M\). That is, \((A,\iota)\) is a submanifold of \(M\). Moreover, the above identity shows that \((N,\Phi)\) and \((A,\iota)\) are equivalent.
Conversely, for a subset \(A\subseteq M\) equipped with a manifold structure, if the inclusion \(\iota:A\hookrightarrow M\) is an immersion then \((A,\iota)\) is always a submanifold of \(M\). This means that for arbitrary \(x\in A\), the differential \(d\iota_x:T_xA\rightarrow T_xM\) is injective, so \(d\iota_x\) induces a bijection between \(T_xA\) and \(d\iota_x(T_xA)\). Setting aside a slight notational issue, we may identify \(T_xA\) with \(d\iota_x(T_xA)\).
Uniqueness of Submanifolds
The pair \((A,\iota)\) defined in the preceding section is determined uniquely for each equivalence class \([(N,\Phi)]\). First, since \(\theta\) in Definition 3 is a diffeomorphism, it is in particular bijective, and therefore
\[\Phi_2(N_2)=\Phi_1(\theta(N_2))=\Phi_1(N_1)\]holds, so \(A\) is uniquely determined. On the other hand, for the subset \(A\) of \(M\) thus determined and the inclusion \(\iota:A\hookrightarrow M\) to belong to \([(N,\Phi)]\), there must exist a diffeomorphism \(\theta\) satisfying \(\iota=\Phi\circ\theta\); taking \(\bar{\Phi}^{-1}\) on the left of both sides yields \(\theta=\bar{\Phi}^{-1}\), so the manifold structure on \(A\) is necessarily the one defined above.
By contrast, given an arbitrary subset \(A\subseteq M\) with inclusion \(\iota:A\hookrightarrow M\), a manifold structure on \(A\) making \((A,\iota)\) a submanifold need not be unique. For instance, as in Example 2, if two submanifolds \((N_1,\Phi_1),(N_2,\Phi_2)\) of \(M\) are not diffeomorphic to each other yet satisfy \(\Phi_1(N_1)=\Phi_2(N_2)\), then the two manifold structures on \((A,\iota)\) obtained from \([(N_1,\Phi_1)]\) and \([(N_2,\Phi_2)]\) via the above process must be distinct.
Nevertheless, if additional conditions are imposed on \((A,\iota)\), the submanifold structure on it may be determined uniquely. The theorems to be introduced in the next post satisfy even this stronger uniqueness, and for that purpose the following two propositions are useful.
Proposition 4 Fix a subset \(A\) of a manifold \(M\) and a topology \(\mathcal{T}\) on \(A\). Then there exists at most one differentiable structure making \((A,\iota)\) a submanifold of \(M\).
Proof
This is immediate from the first claim of Proposition 2.
Proposition 5 Let a subset \(A\) of a manifold \(M\) be given. If, when \(A\) is viewed as a subspace of \(M\), there exists a differentiable structure making \((A,\iota)\) a submanifold of \(M\), then this differentiable structure together with the subspace topology constitutes the unique manifold structure that can be given to \((A,\iota)\).
Proof
First, applying Proposition 4 to the subspace topology \(\mathcal{T}\), the differentiable structure making \((A,\mathcal{T},\iota)\) a submanifold of \(M\) is unique. Denote it by \(\mathcal{A}\). Now suppose a topology \(\mathcal{T}'\) and a differentiable structure \(\mathcal{A}'\) making \((A,\iota)\) a submanifold of \(M\) are given, and consider the following diagram.
Here \(\iota\) and \(\iota'\) are both \(A\hookrightarrow M\), but we have denoted them by different names for distinction. By definition \((A,\mathcal{T},\mathcal{A})\) is an embedded submanifold of \(M\), so the vertical \(\iota\) is an embedding; hence \(\operatorname{id}\) is \(C^\infty\) by Proposition 2. Moreover, by the chain rule
\[d\iota'=d\iota\circ d(\id)\]holds, and since \(d\iota'\) is injective at every point, \(d(\id)\) is also injective at every point. Therefore it suffices to show that \(d(\id)\) is surjective at every point.
Suppose for contradiction that there exists a point \(a\) at which \(d(\id)\) is not surjective. Then considering the dimension of the tangent space at this point,
\[\dim(A,\mathcal{T}',\mathcal{A}')<\dim(A,\mathcal{T},\mathcal{A})\]holds. Let \(d\) be the dimension of \((A,\mathcal{T},\mathcal{A})\) and \(d'\) be the dimension of \((A,\mathcal{T}',\mathcal{A}')\).
Let \((U,\varphi)\) be a coordinate system of \((A,\mathcal{T},\mathcal{A})\). Without loss of generality we may assume that the image of \(\varphi\) is \(\mathbb{R}^d\), and since \(\id\) is surjective, the image of the composition \(\varphi\circ\id\) is also \(\mathbb{R}^d\).
On the other hand, since \((A,\mathcal{T}',\mathcal{A}')\) is a manifold, we can cover it by countable coordinate systems \((V,\sigma)\) each homeomorphic to \(\mathbb{R}^{d'}\). But \(\varphi\circ\id\circ\sigma^{-1}\) is \(C^\infty\), and as \(C^1\) functions they all map measure-zero sets to measure-zero sets, so it is a contradiction that their images could be \(\mathbb{R}^d\).
References
[War] Frank W. Warner. Foundations of Differentiable Manifolds and Lie Groups, Graduate texts in mathematics, Springer, 2013
[Lee] John M. Lee. Introduction to Smooth Manifolds, Graduate texts in mathematics, Springer, 2012
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