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I was doing Lee's smooth manifold exercise,in problem 10-14 needs to prove the following result:

Let $S\subset M$ be some smooth immersed submanifold,prove that $TS$ is subbundle of $TM|_S$.

It seems we need to use the local frame criterion,which needs to find a neigborhood such that exist k smooth local section $\sigma_1,...,\sigma_k :U\to TM|_S$ as frame for $T_pS$ for each $p \in U$.

To prove this observe that $i:S^k\to M^n$ is smooth immersion,I have idea how to prove it when $S$ is embedded submanifold,which can be shown as follows:

First for $p\in S$,there exist slice chart around it that is $(U\cap S,\pi\circ\varphi = (x^1,...,x^k))$ as chart for $S$,where $(U,\varphi =(x^1,...,x^n))$ is the corresponding chart for $M$.

We know chart $(U,\varphi =(x^1,...,x^n))$ provide a local trivialization for $TM$,we want to find the local section for $TM|_S$(needed for the local section criterion) which is obvious given by this trivialization.

Now construct the local section as follows:

$$\sigma_i : U\cup S \to (U\cap S) \times \Bbb{R}^n \to \pi^{-1}(U\cap S) \to TM|_S$$

Such that this map maps $p \mapsto (p,e_i) \mapsto \frac{\partial}{\partial x^i}|_p $

Our guess may be $\sigma_1,...,\sigma_k$ span the $TS$ but I can't make it clear.

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I wrote this a while ago, and now I kind of having difficulty to follow my proof. But I'm sure I convinced by this proof at the time I wrote this. Let me know if something unclear.

Setup : Let $ \pi|_S : TM|_S \to S$ be the ambient tangent bundle, $\iota : S \hookrightarrow M$ is an immersed submanifold of dimension $k$ of a $n$-dimensional smooth manifold $M$. For any $p \in S$, we have an $k$-dimensional subspace $d\iota_p (T_pS) \subseteq T_pM$. We want to show that $\bigcup_{p \in S} d\iota_p(T_pS) \subseteq TM|_S$ is a smooth subbundle of rank-$k$. To do this, it is enough to show that for any $p \in S$ there is a neighbourhood $V$ of $p$ and smooth local sections $\sigma_1,\dots,\sigma_k :V \to TM|_S$ such that at any $q \in V$, $\sigma_1(q),\dots,\sigma_k(q)$ form a basis for subspace $d\iota_q(T_qS)$.

Proof : Since $S \subseteq M$ is an immersed submanifold, then $S$ locally embedded. So for any $p \in S$, there exists a neighbourhood $V \subseteq S$ of $p$ such that $V$ is embedded submanifold of $M$. By shrinking $V$ if necessary, we may assume that $V \subseteq S$ such that $\iota(V)=V \subseteq U$, so $V = V \cap U$ is a single slice in $U$, where $(U,\varphi)$ is a slice chart for $V$. That is $$ \varphi(V) = \{(x^1,\dots,x^n) \in \varphi(U) : x^{m+1} = \cdots= x^n = 0 \} \subseteq \varphi(U). $$ By Theorem 5.8, we have smooth (global) chart $(V,\psi)$ where the coordinate functions $\psi (V) = \{ (x^1,\dots,x^m)\} \subseteq \mathbb{R}^m$ are exactly the nonvanishing coordinates of image $\varphi(V)$ in $\varphi(U)$. Hence the representation of $\iota : S \hookrightarrow M$ in smooth charts $(V,\psi)$ and $(U,\varphi)$ is \begin{equation}\tag{$\star$} (x^1,\dots,x^m) \mapsto (x^1,\dots,x^m,0,\dots,0). \end{equation} Now for $i=1,\dots,m$, the coordinate vector fields $\tau_i : V \to TS$ defined as $\tau_i(q) = \frac{\partial}{\partial x^i}\big|_q$ are smooth local frame (since the component functions are constant, hence smooth) for $TS$ over $V$. Define local sections $\sigma_i : V \to TM|_S$ defined as $\sigma_i(q) = d\iota_q (\tau_i(q))$, for $i =1,\dots,m$, where $d\iota_q : T_qS \to T_qM$ is the differential of $\iota : S \hookrightarrow M$ at $q\in S$. From ($\star$), we can compute $\sigma_i$ as $$ \sigma_i(q) = d\iota_q \circ \tau_i(q) = d\iota_q \Big(\frac{\partial}{\partial x^i}\bigg|_q \Big) = \frac{\partial}{\partial x^i}\bigg|_q, $$ which is form a basis for $d\iota_q(T_qS) \subseteq T_qM$.

We have to show that these sections are smooth sections on $TM|_S$. Since $\sigma_i (V) \subseteq \bigcup_{q \in V} T_qM = (\pi|_S)^{-1}(V) = TM|_V$ and $TM|_V$ is open in $TM|_S$, then we only need to check the smoothness in $TM|_V$. For the local trivialization of $TM$ over slice chart $U$ as above is $\Phi : \pi^{-1}(U) \to U \times \mathbb{R}^n$, defined as $$ \Phi \Big(v^i \frac{\partial}{\partial x^i}\Big|_p \Big) = (p,(v^1,\dots,v^n)). $$ So the local trivialization for $TM|_V$ is just the restriction of $\Phi$ to $\pi^{-1}(U \cap V)$. Denote this as $\Psi = \Phi|_U : \pi^{-1}(U \cap V) \to (U \cap V) \times \mathbb{R}^n$. Therefore $$ \Psi \circ \sigma_i (q) =\Phi(\sigma_i(q)) =(q,e_i), $$ which is obviously smooth. Hence $\sigma_i$ are the smooth local sections tha we seek. Since we can do this for every point $p \in S$, therefore $TS$ identified as its image $d\iota(TS) \subseteq TM|_S$ is a smooth subbundle.

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  • $\begingroup$ thanks for your nice solution very clear. $\endgroup$
    – yi li
    Commented Apr 30, 2021 at 6:51
  • $\begingroup$ Problem 10-14 actually states that both $M$ and $S$ might have nonempty boundaries, in which case slice charts are not available. Do you have a proof for this case? $\endgroup$
    – Jeff Rubin
    Commented Sep 23, 2023 at 15:59
  • $\begingroup$ @JeffRubin Tbh, I already forgot how to work around this issue. You can try to find a result in Lee's book about the slice chart, or post a question for this. $\endgroup$ Commented Sep 24, 2023 at 10:05

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