Weak Whitney Immersion Theroem

Lee Smooth Manifolds

Theorem: Every smooth $n$-manifold without boundary admits a smooth immersion into ${\mathbb{R}}^{2n}$.

Proof) Let $M$ be an arbitrary smooth $n$-manifold without boundary. Note that, by the Whitney Embedding Theorem, $M$ can be properly embedded into ${\mathbb{R}}^{2n+1}$. We use the same core mechanism as in the Whitney embedding theorem, except that instead of passing on the entire tangent bundle $TM$ to define a projection vector, we pass the unit tangent bundle $UM$ to squeeze out an extra dimension.

We first show the unit tangent bundle $UM$ is a well-defined $2n-1$ dimensional submanifold of $T{\mathbb{R}}^{2n+1}$. More generally, we show that if $N\subset {\mathbb{R}}^n$ is an embedded $m$-dimensional submanifold, the unit tangent bundle $UN\subset T{\mathbb{R}}^n$ defined as

$$UN=\{(x,v)\in T{\mathbb{R}}^n:x\in N,v\in T_{x}N,|v|=1\}$$

is an embedded $2m-1$ dimensional submanifold of $T{\mathbb{R}}^n\approx {\mathbb{R}}^n\times {\mathbb{R}}^n$.

First note that the submanifold tangent bundle $TN$ is a $2m$-dimensional embedded submanifold of $T{\mathbb{R}}^n$. If we let $i:N\to {\mathbb{R}}^n$ be the inclusion map, $di:TN\to T{\mathbb{R}}^n$ has the representation $(x_1,x_2,\dots ,x_m,v_1,\dots ,v_m)\to (x_1,\dots ,x_m,0,\dots ,0,v_1,\dots ,v_m,0,\dots ,0)$ in slice coordinates, clearly indicating $di$ is a smooth immersion. $di$ is also clearly closed, and thus is a smooth embedding.

It suffices to show $UN$ is a smooth embedded hypersurface of $TN$. If we define $f:TN\to \mathbb{R}$ as $f(x,v)=|v{|}^2$, then $UN$ is a regular level set of $f$, completing our proof. We now return to the main proof, discarding this temporary notation.

It suffices to show there is some vector $v\in {\mathbb{R}}^{2n+1}\setminus {\mathbb{R}}^{2n}$ such that the projection ${\pi }_v{|}_M$ is a smooth (and not necessarily injective!) immersion into ${\mathbb{R}}^{2n}$; this is equivalent to no element of $UM$ being parallel to $v$. If we define $\tau :UM\to {\mathbb{RP}}^{2n}$ by $\tau (x,v)=[v]$, it suffices to show there is some $v\in {\mathbb{R}}^{2n+1}\setminus {\mathbb{R}}^{2n}$ such that $[v]\notin \tau (UM)$. As $UM$ is $2n-1$ dimensional, by Sard’s Theorem $\tau (UM)$ has empty interior in ${\mathbb{RP}}^{2n}$, meaning ${\mathbb{R}}^{2n}\setminus \tau (UM)$ is dense. Thus, such a $v$ must exist.