I read here that a major ingredient in Whitney's strong embedding theorem and later Smale's celebrated h-cobordism theorem is the Whitney trick.

Can someone give an intuitive description of the trick? To be more specific, I would be happy to know why and where the trick was applied ? Thanks.


1 Answer 1


I hope this answer might still be useful after almost two months.

These notes contains some nice pictures of examples of the Whitney trick.

But here is a very low-dimensional example which really captures the whole idea. Suppose you have two oriented simple closed curves $c_1,c_2$ in the sphere $S^2$. You should visualize this by drawing two such curves on paper which cross-cross each other, without either of them every crossing itself. We know, of course, that you can pull $c_1,c_2$ apart by an isotopy: just shrink $c_1$ down until it is close to the north pole and shrink $c_2$ down until it is close to the south pole.

But suppose that you wanted to pull $c_1,c_2$ apart by some kind of local move, rather than such a huge global move like shrinking big curves down to little curves. Here is a step-by-step procedure to do that.

  1. Jiggle $c_1$ until it intersects $c_2$ transversely.
  2. Choose orientations of $c_1,c_2$.
  3. Assign $+$ and $-$ signs to each point of the intersection $c_1 \cap c_2$, according to whether $c_1,c_2$ cross each other in a "right hand" or a "left hand" manner.
  4. Find arcs $\alpha_1 \subset c_1$ and $\alpha_2 \subset c_2$ so that these arcs have the same endpoints $x_-,x_+$, and otherwise $\alpha_1$ is disjoint from $c_2$ and $\alpha_2$ is disjoint from $c_1$. So $\alpha_1 \cup \alpha_2$ forms a circle in $S^2$.
  5. Find an embedded 2-dimensional disc $D$ in $S^2$ whose boundary is $\alpha_1 \cup \alpha_2$ and whose interior is disjoint from $c_1$ and $c_2$.
  6. Now you just push $c_1$ across the disc $D$, removing the two intersection points $x_-,x_+$, reducing the cardinality of $c_1 \cap c_2$ by $2$.

The general Whitney trick in higher dimensions is quite similar, except that finding the embedded 2-dimensional disc $D$ in Step 5 is harder and sometimes impossible. In general, rather than being in $S^2$, one is in an $n$-dimensional manifold $M$. Rather than $c_1,c_2$ being circles, they are submanifolds of $M$ whose dimensions add up to equal the dimension of $M$. But in step 4, $\alpha_1,\alpha_2$ are still arcs, and their union is still a circle.

Here's what happens in Step 5: if the circle $\alpha_1 \cup \alpha_2$ is not homotopic to a constant then the embedded 2-dimensional disc $D$ simply does not exist. But even if that circle is homotopic to a constant, for instance if the manifold $M$ is simply connected, finding $D$ is still hard and sometimes still impossible. But using transversality one can always find $D$ if $M$ is a simply connected manifold of dimension five or higher, and this is an important step in the proof of the $h$-cobordism theorem that you asked about.

  • 5
    $\begingroup$ That answer was out of the world. Thanks a lot :-) $\endgroup$ Apr 13, 2014 at 22:08
  • $\begingroup$ Wonderfully clear explanation! $\endgroup$ Sep 16, 2020 at 17:25

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