Weak Whitney Immersion Theorem

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Theorem

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


Proof

Let $M > 2 k + 1$ 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:

$\map h {x, y, t} = \map t {\map f x - \map f y}$

Define a map $g: \map T X \to \R^M$ by:

$\map g {x, v} = \map {\d f_x} v$

where $\map T X$ is the tangent bundle of $X$.

Since $M > 2 k + 1$, the Morse-Sard 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 $a, H$.


Suppose:

$\map {\paren {\pi \circ f} } x = \map {\paren {\pi \circ f} } y$

for some $x, y$.

Then:

$\map f x - \map f y = t a$

for some scalar $t$.

Aiming for a contradiction, suppose $x \ne y$.

Then as $f$ is injective:

$t \ne 0$

But then $\map h {x, y, 1/t} = a$, contradicting the choice of $a$.

By Proof by Contradiction:

$x = y$

and so $\pi \circ f$ is injective.


Let $v$ be a nonzero vector in $\map {T_x} X$ (the tangent space of $X$ at $x$) for which:

$\map {\d \paren {\pi \circ f}_x} v = 0$

Since $\pi$ is linear:

$\map {\d \paren {\pi \circ f}_x} = \pi \circ \d f_x$

Thus:

$\map {\pi \circ \d f_x} v = 0$

so:

$\map {\d f_x} v = t a$

for some scalar $t$.

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

Hence:

$\map g {x, 1/t} = a$

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


Also see

  • Whitney Immersion Theorem, a strictly stronger result: every $k$-manifold admits of a one-to-one immersion in $\R^{2 k}$, and an immersion (not necessarily one-to-one) in $\R^{2 k - 1}$.


Source of Name

This entry was named for Hassler Whitney.