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)?
Best Answer
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.