# Celebratio Mathematica

## Hassler Whitney

### The Whitney trick

#### by Rob Kirby

The Whit­ney trick is a meth­od for re­mov­ing points of in­ter­sec­tion between two sub­man­i­folds. It can be seen in its most ele­ment­ary form in Fig­ure 1, in which it is ob­vi­ous that the two points of in­ter­sec­tion can be re­moved by an iso­topy (a 1-para­met­er fam­ily of em­bed­dings) of the arc labeled $P^p$ which pulls the arc across the disk $D$. (Note that $x$ and $y$ have op­pos­ite signs.) More gen­er­ally the Whit­ney trick is used to re­move a pair of in­ter­sec­tions, $x$ and $y$, between two man­i­folds $P^p$ and $Q^q$ which are em­bed­ded in an am­bi­ent man­i­fold $M^{p+q}$. To see how this is done, we first con­struct a mod­el, then show how to em­bed it in $M$ (if pos­sible), and then sketch some ap­plic­a­tions of the Whit­ney trick.

The mod­el is merely a sta­bil­iz­a­tion of the ex­ample in Fig­ure 1. We cross the plane in which $D$ is em­bed­ded with $\mathbb R^{(p-1) + (q-1)}$ so that the am­bi­ent space is just $\mathbb R^{p+q}$, and then we cross the curve which in­cludes $\alpha$ by $\mathbb R^{p-1}$ to get an $p$-di­men­sion­al man­i­fold $P$, and sim­il­arly cross with $\mathbb R^{q-1}$ to get an $q$-man­i­fold $Q$. These two man­i­folds still meet in two points $x$ and $y$, which are con­nec­ted in $P$ by the ori­gin­al arc $\alpha$ and in $Q$ by the ori­gin­al arc $\beta$. Note that the two arcs still bound a 2-di­men­sion­al disk $D$, and that $D$ lies in­side a lar­ger open disk $\Delta$ in the plane. Also note that $\Delta$ has a nor­mal $(p-1)+(q-1)$-plane bundle which splits as the dir­ect sum (also called “Whit­ney sum”) of a $(p-1)$-plane bundle which co­in­cides along $\alpha$ with the nor­mal bundle of $\alpha$ in $P$, and an $(q-1)$-plane bundle which co­in­cides along $\beta$ with the nor­mal bundle of $\beta$ in $Q$.

The plane iso­topy de­scribed in Fig­ure 1 eas­ily ex­tends to an iso­topy tak­ing place in the plane crossed with the $p-1$ co­ordin­ates of $P$, as drawn for $p=2$ in Fig­ure 2 [e4]; noth­ing hap­pens with the oth­er $q-1$ co­ordin­ates.

Now this mod­el must be em­bed­ded in $M^{p+q}$ so that the ac­tu­al man­i­folds $P$ and $Q$ and two points of in­ter­sec­tion $x$ and $y$ cor­res­pond to the man­i­folds and points in the mod­el.

If both $P$ and $Q$ are con­nec­ted, then the arcs $\alpha$ and $\beta$ ex­ist, and if $P$ and $Q$ are simply con­nec­ted (as they of­ten are in ap­plic­a­tions), then the arcs are unique up to ho­mo­topy. If $M$ is simply con­nec­ted, then the disk $D$ can be mapped in­to $M$. If not, then $x$ must be con­nec­ted by an arc (unique up to ho­mo­topy if $P$ is simply con­nec­ted) to a base point $x_0 \in P$ which is con­nec­ted by an arc to a base point $z \in M$. Sim­il­arly with arcs to a base point $y_0 \in Q$. It fol­lows that $x$ then de­term­ines an ele­ment of $\pi_1(M)$ by run­ning from $z$ to $x_0$ to $x$ to $y_0$ and back to $z$. Now if $x$ and $y$ both rep­res­ent the same ele­ment of $\pi_1(M)$, then we can still map a disk $D$ in­to $M$. (This is im­port­ant in prov­ing the $s$-cobor­d­ism the­or­em.)

Once $D$ is mapped in­to $M$, we can em­bed it if the di­men­sion of $M$, $p+q$, is five or more. Fur­ther­more, if each of $p$ and $q$ is three or more, then the em­bed­ding of $D$ can be chosen to miss $P$ and $Q$ ex­cept along its bound­ary.

Now that $D$ is em­bed­ded miss­ing $P$ and $Q$, it re­mains to find the em­bed­ding of the nor­mal bundle of $D$. The nor­mal $(p+q-2)$-bundle to $D$ (in fact, $\Delta$) in $M$ can be split along $\alpha$ as the nor­mal $(p-1)$-bundle to $\alpha$ in $P$ dir­ect sum the or­tho­gon­al $(q-1)$-bundle. That split­ting ex­tends across $\Delta$. The only prob­lem re­main­ing is that this $(p-1)$-plane bundle may not co­in­cide with the $(p-1)$-plane bundle which is the nor­mal bundle to $\beta$ in $Q$.

The prob­lem re­duces to an arc of $(p-1)$-planes in $\mathbb R^{(p-1) + (q-1)}$ which we want to iso­tope to the trivi­al arc, re­l­at­ive to the en­d­points. Note that the trivi­al arc, as in the mod­el, cor­res­ponds to $x$ and $y$ hav­ing op­pos­ite signs, so this is ne­ces­sary. Now, this is pos­sible be­cause the fun­da­ment­al group of the Stiefel man­i­fold of $(p-1)$-planes in $\mathbb R^{p+q-2}$ is trivi­al when $p > 2$ (see [e3], p. 202). For more de­tails, see the ex­cel­lent de­scrip­tion in [e4].

Whit­ney de­veloped the Whit­ney trick in or­der to em­bed $P^p$ in $\mathbb R^{2p}$ [e1]. For $p=2$, this is easy. In high­er di­men­sions, $P$ only im­merses in $\mathbb R^{2p}$ (by gen­er­al po­s­i­tion), so for each double point, Whit­ney in­tro­duces in loc­al fash­ion an­oth­er double point of op­pos­ite sign (some thought is needed if $P$ is non-ori­ent­able), and then uses the Whit­ney trick to re­move both points of in­ter­sec­tion.

A later, and cru­cial, use of the Whit­ney trick is in Smale’s proof of the $h$-cobor­d­ism the­or­em [e2].