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This is an exercise from Lee's Riemannian geometry. I'm not clear about how to use the hint to solve this exercise exactly.

Exercise 2.3. Suppose $M \subset \widetilde{M}$ is an embedded submanifold.

(a) If $f$ is any smooth function on $M$, show that $f$ can be extended to a smooth function on on a neighborhood of $M$ in $\widetilde{M}$ whose restriction to $M$ is $f$. [Hint: Extend $f$ locally in slice coordinates by letting it be independent of $\left(x^{n+1}, \ldots, x^m\right)$, and patch together using a partition of unity.]

I don't know how to construct $f^\alpha$ in each open sets $U^\alpha$ of open cover $\{U^\alpha\}$. Please enlighten me.

Edit

Here is my solutions according to the first answer.

$\forall p \in M$, let $\{ \tilde{U_p},\tilde{\phi_p} \}$ be a chart in $\tilde{M}$ centered at $p$ with slicing property. Then define $\tilde{f}_p:\tilde{U} \rightarrow \mathbb{R}$ as

$$\tilde{f}_p \tilde{\phi_p}^{-1} = f \tilde{\phi_p}^{-1}(x^1,\cdots,x^n,0,\cdots,0).$$ Then $\tilde{f}$ coincides with $f$ on $ \tilde{U}_p \cap M$.

Let $\tilde{U}= \bigcup\limits_{p\in M} \tilde{U}_p$. By a smooth bump function on $\tilde{U}$ supported in $\tilde{U}_p$, each $\tilde{f}_p:\tilde{U}_p \rightarrow \mathbb{R}$ can be extended to a smooth function $f_p$ on $\tilde{U}$. Since $f_p$ will coincide with $\tilde{f}_p$ in a possibly smaller neighborhood of $p$, denote this neighbourhood as $U_p \subset \tilde{U}_p$. Then $\tilde{f}_p : \tilde{U} \rightarrow \mathbb{R}$ coincides with $f$ in $U_p \cap M$.

By manifoldness of $U= \bigcup\limits_{p\in M} U_p$, there exists a partition of unity $\{\alpha_p\}$ subordinate to the open cover $\{{U}_p\}$ of $U$. Let $g=\sum \alpha^p f_p$ (here $f_p= f_p|_U$). Since each $f_p$ coincide with $f$ on $U_p \cap M$, $g$ is the extension of $f$.

gsoldier
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1 Answers1

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A hint which slightly elaborates on the hint: fix a point $p\in M$, and consider a chart $(U,\phi)$ of $\widetilde{M}$ around $p$, which has the slice property, so $\phi[U\cap M]=\phi[U]\cap \left(\Bbb{R}^n\times \{0\}\right)$. Consider the function $F:U\to\Bbb{R}$ defined such that for all $x=(x^1,\dots, x^m)\in \phi[U]$, \begin{align} (F\circ \phi^{-1})(x^1,\dots, x^m):=(f\circ \phi^{-1})(x^1,\dots, x^m, 0,\dots, 0). \end{align} This definition makes sense because setting the last few coordinates to zero means the point $\phi^{-1}(x^1,\dots, x^m,0,\dots, 0)$ lies in $U\cap M$, and hence be fed into $f$.

This is a suitable smooth local extension. Why? Use partitions of unity etc to complete the argument.

peek-a-boo
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  • Thanks for answering. I have wrote a proof according to your hint. But I am not sure about my details. Could you please check my proof? Especially the part about combining local extension by partition of unity. Thanks a lot! – gsoldier Feb 21 '23 at 15:00
  • @gsoldier sure, but there’s no need for the intermediate bump function. If you take a partition of unity of the original open set, then you can just sum it up (and since each function in the partition of unity has compact support inside the $\tilde{U}_p$, it follows that when you multiply by $\tilde{f}_p$, you can automatically smoothly extend it to be zero outside $\tilde{U}_p$). i.e you did one extra step for no significant reason. The whole point of partitions of unity is to do it all in one fell swoop. – peek-a-boo Feb 22 '23 at 05:23