Weak Whitney Immersion Theorem

Theorem

Every $k$-dimensional manifold $X$ admits a one-to-one immersion in $\R^{2k+1}$.

Proof

Let $M>2k+1$ be a natural number such that $f: X \to \R^M$ is an injective immersion.

Define a map $h:X \times X \times \R \to \R^M$ by $h(x,y,t)=t((f(x)-f(y))$.

Define a map $g:T(X) \to \R^M$ by $g(x,v)=df_x (v)$

where $T(X)$ is the tangent bundle of $X$.

Since $M>2k+1$, Sard's Theorem implies $\exists a \in \R^M$ such that $a$ is in the image of neither function.

Let $\pi$ be the projection of $\R^M$ onto the orthogonal complement of $a, H$.

Since $f$ is injective, $\pi \circ f$ is injective as well, for suppose $(\pi \circ f)(x) = (\pi \circ f)(y)$.

Then $f(x)-f(y) = ta$ for some scalar $t$.

If $x \neq y$ then $t \neq 0$ because $f$ is injective, but then $h(x,y,1/t)=a$, contradicting the choice of $a$.

Now suppose $v$ is a nonzero vector in $T_x (X)$ (the tangent space of $X$ at $x$) for which $d(\pi \circ f)_x (v)=0$.

Since $\pi$ is linear, $d(\pi \circ f)_x = \pi \circ df_x$.

Thus $\pi \circ df_x(v)=0$, so $df_x(v)=ta$ for some scalar $t$.

Because $f$ is an immersion, $t \neq 0$.

Hence $g(x,1/t)=a$, contradicting the choice of $a$.

Hence $\pi \circ f:X \to H$ is an immersion.

$H$ is obviously isomorphic to $\R^{M-1}$.

Thus whenever $M>2k+1$ and $X$ admits of a one-to-one immersion in $\R^M, X$ also admits of a one-to-one immersion in $\R^{M-1}$.

Comments

This result is not the strongest possible. The Whitney Immersion Theorem is strictly stronger, stating that every k-manifold admits of a one-to-one immersion in $\R^{2k}$, and an immersion (not necessarily one-to-one) in $\R^{2k-1}$. This result can be used to prove the Whitney Embedding Theorem. The Whitney Immersion Theorem deals with the use of Sards, and the difficulty that causes in the above proof when $M=2k+1$.

Source of Name

This entry was named for Hassler Whitney.