# 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].