9
$\begingroup$

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.

$\endgroup$
12
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ That answer was out of the world. Thanks a lot :-) $\endgroup$ – Sandeep Thilakan Apr 13 '14 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.