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I have an attempt at solving the following problem. This is not so much a question asking for a solution in general, but more on how to complete my own.

Let $M^n$ be a compact smooth manifold. Show that there exists an embedding of $M$ into $\mathbb{R}^N$ for some $N$.

(Recall that an embedding of smooth manifolds is a topological embedding such that each differential is injective.)

My attempt:

Since $M$ is compact, we can cover it by finitely many coordinate neighbourhoods $U_i$ for $i=1,\dotsc,k$, where $\varphi_i : U_i \to \mathbb{R}^n$ are the corresponding charts. Choose a subordinate (smooth) partition of unity $\psi_i : M \to \mathbb{R}$. Then the functions $$ f_i := \psi_i \cdot \varphi_i$$ are smooth, where we interpret $\varphi$ as being zero outside of the support of $\psi$. Now define $$ F : M \to \mathbb{R}^{n\cdot k + k} : x\mapsto (f_1(x),\dotsc,f_k(x),\psi_1(x),\dotsc,\psi_k(x)).$$ My hope was that $F$ is an embedding of smooth manifolds. For this, we verify:

  • Injectivity. This is why the $\psi$'s were stuck at the end of the map. If all the $\psi(x)$'s are the same, then all the $\varphi(x)$'s are the same, but these are local diffeomorphisms, in particular bijections.
  • Smoothness. Trivial.
  • Topological embedding. Immediate since $M$ is compact and $F$ is injective and continuous.
  • Injective differentials. This is my issue.

Are all the differentials injective for $F$ as defined above? Or would we need more assumptions on the covering or of the partition of unity for this to work (or for this to work more easily)?

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  • $\begingroup$ This is whitney lemma $\endgroup$
    – Tsemo Aristide
    Commented Dec 12, 2018 at 18:26
  • $\begingroup$ @TsemoAristide Which lemma are you referring to? If it's just about embedding a compact manifold in $\mathbb{R}^N$, then like I stated, that's not the point of me raising this question. $\endgroup$
    – SvanN
    Commented Dec 12, 2018 at 18:27
  • $\begingroup$ en.m.wikipedia.org/wiki/Whitney_embedding_theorem $\endgroup$
    – Tsemo Aristide
    Commented Dec 12, 2018 at 18:29
  • $\begingroup$ @TsemoAristide Thanks for the reference, but again, I am asking this question about my own attempted solution, not a general proof. $\endgroup$
    – SvanN
    Commented Dec 12, 2018 at 18:30
  • $\begingroup$ Read the proof of Whitney's theorem, it starts like your idea, then by using sard lemma it can be shown that $M$ can be embedded in $\mathbb{R} ^{2n+1}$ where $n$ is the dimension of $M$. $\endgroup$
    – Tsemo Aristide
    Commented Dec 12, 2018 at 18:43

1 Answer 1

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Let $x \in M$ be such that $T_x F$ is not injective. Then there exists some nonzero $X \in T_x M$ such that $T_xF(X)=0$.

Let $i$ be such that $\psi_i(x)>0$, then $d_x \psi_i(X)=0$ (It is a component of $F$) and $T_x(\psi_i\phi_i)(X)=0$. Now, since $d_x\psi_i(X)=0$, $0=T_x(\psi_i\phi_i)(X) = \psi_i(x)T_x\phi_i(X)$.

Thus, $T_x\phi_i(X)=0$, a contradiction. So your differentials are injective.

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  • $\begingroup$ How does it follow that $\nabla_x \psi_i =0$? (And I’m presuming that by $\nabla \psi$ you mean the differential of $\psi$?) $\endgroup$
    – SvanN
    Commented Dec 12, 2018 at 20:24
  • $\begingroup$ Edited, I hope I answered your concern. $\endgroup$
    – Aphelli
    Commented Dec 12, 2018 at 20:31
  • $\begingroup$ You did, yes :) Thank you! $\endgroup$
    – SvanN
    Commented Dec 12, 2018 at 20:49

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