 # Celebratio Mathematica

## Martin Scharlemann

### Markov’s theorem on the nonrecognizablility of 4-manifolds:an exposition

#### by Rob Kirby

In the mid 1950s A. A. Markov proved [e6] that there could be no al­gorithm to dis­tin­guish all pairs of closed, com­pact, smooth 4-man­i­folds. His proof star­ted with the fact that there is no al­gorithm to de­cide wheth­er a fi­nitely presen­ted group $$G$$ is trivi­al, a fact us­ing ideas of Gödel and proved by P. S. Novikov [e1], [e2] and W. W. Boone [e5] (also see [e8] and [e9]).

Markov’s proof is in Rus­si­an, and not too widely known in the West. Marty Schar­le­mann men­tioned to me a few years ago that, as a gradu­ate stu­dent in the early 70s, he had wanted to un­der­stand Markov’s proof but knew no Rus­si­an. So he looked up the pa­per in Math­em­at­ic­al Re­views and found the ac­count of the proof there un­con­vin­cing, which led him to work out what he thought Markov’s ar­gu­ment must have been. To­geth­er we mined his memory and sor­ted out the proof which is ex­pos­ited here. To the best of my know­ledge this simple proof is not writ­ten up else­where (but see ([e10], p. 149) where the proof is giv­en as an ex­er­cise with help from earli­er ex­er­cises but con­tains no hint as to adding $$g$$ ex­tra 2-handles; also see [e11]).

Defin­i­tion: A man­i­fold $$X$$ be­long­ing to a col­lec­tion $$C$$ of man­i­folds is said to be re­cog­niz­able if there ex­ists an al­gorithm which can de­cide wheth­er or not any man­i­fold $$Y$$ in $$C$$ is dif­feo­morph­ic to $$X$$.
The 4-man­i­fold $X^4 = \mathop{\#}\nolimits^k S^2 \times S^2$ for some $$k > 0$$ is not re­cog­niz­able in the col­lec­tion $$C$$ of all smooth, com­pact, closed 4-man­i­folds.
There does not ex­ist an al­gorithm to dis­tin­guish com­pact, smooth, closed 4-man­i­folds.
Giv­en a fi­nitely presen­ted group $$G$$, we can con­struct a smooth 5-man­i­fold $$Y^5_G$$ such that $\pi_1 (Y_G) = \pi_1(\partial Y_G) = G$ and $$Y_G$$ has no handles of in­dex $$> 2$$.

Proof.  Sup­pose $$G$$ has $$g$$ gen­er­at­ors $$g_1$$, …, $$g_g$$ and $$r$$ re­la­tions $$r_1$$, …, $$r_r$$. We can con­struct a 5-man­i­fold $$Y^5_G$$ by adding $$g$$ 1-handles to $$B^5$$ (ob­tain­ing $$\mathop{\#}\nolimits^g (S^1 \times B^4)$$) and then adding $$r$$ 2-handles ac­cord­ing to the re­la­tions $$r_1$$, …, $$r_r$$. Be­cause the at­tach­ing circles lie in a 4-man­i­fold, these iso­topy classes are de­term­ined by their ho­mo­topy classes in $$\mathop{\#}\nolimits^g S^1 \times S^3$$ which in turn are de­term­ined by the re­la­tions.

However a 2-handle $$B^2 \times B^3$$ is at­tached by an em­bed­ding of $$S^1 \times B^3$$ and giv­en the em­bed­ding of $$S^1 \times 0$$; there are then two ways to ex­tend that em­bed­ding since $\pi_1(\mathrm{SO}(3)) = \mathbb{Z}/2.$ For ex­ample for the un­knot in $$\partial B^5$$, the two fram­ings give $$S^2 \times B^3$$ and the non­trivi­al $$B^3$$ bundle over $$S^2$$, namely $$S^2 \mathbin{\tilde{\times}} B^3$$. Let $$Y_G$$ be defined by any one choice of fram­ings of the 2-handles.

Two dif­fer­ent present­a­tions of $$G$$ dif­fer by Tiet­ze moves, but these moves cor­res­pond to the birth or death of a 1-2 handle pair and to handle slides, and these do not change the 5-man­i­fold $$Y_G$$.

The map in­duced by in­clu­sion, $\pi_1 (\partial Y_G) \to \pi_1(Y_G),$ is an iso­morph­ism. It is an epi­morph­ism be­cause any loop in $$Y_G$$ can be pushed off the 2-com­plex spine of $$Y_G$$ and thus in­to $$\partial Y_G$$. It is a mono­morph­ism for the same reas­on; if a loop in $$\partial Y_G$$ is ho­mo­top­ic­ally trivi­al in $$Y_G$$, then that 2-disk can also be pushed off the spine and in­to $$\partial Y_G$$.  ◻

Now define $$Y_g$$ to be $$Y_G$$ with $$g$$ trivi­al 2-handles (at­tached to the un­link in a small 4-ball in $$\partial B^5$$ which avoids all oth­er at­tach­ing maps, and with the fram­ing on the at­tach­ing circle that gives $$B^2 \times B^3$$ ). Note that $Y_g = Y_G \mathop{\#}\nolimits^g S^2 \times B^3.$

The proof of the the­or­em will fol­low from the next lemma.

$$G$$ is trivi­al if and only if $\partial Y_g = \mathop{\#}\nolimits^r S^2 \times S^2.$

Proof.  The if part fol­lows from $G = \pi_1(Y_g) = \pi_1 (Y_G) = \pi_1 (\partial Y_G) = 0.$

For the only if part, note that $\pi_1(\partial Y_G) = G =0.$ Then the at­tach­ing circles of the $$g$$ trivi­al 2-handles can be ho­mo­toped and thus iso­toped in $$\partial Y_G$$, to geo­met­ric­ally can­cel the $$g$$ 1-handles. (Note that we can­not do this with the ori­gin­al $$r$$ 2-handles be­cause they do not lie in a simply con­nec­ted 4-man­i­fold, but rather in a con­nec­ted sum of products $$S^1 \times S^3$$, where­as the ex­tra $$r$$ 2-handles may be thought of as be­ing ad­ded to $$\partial Y_G$$.) Do the can­cel­la­tion, and what re­mains of $$Y_g$$ is an un­link of $$r$$ com­pon­ents to which are at­tached $$r$$ 2-handles, so that $$Y_g$$ is dif­feo­morph­ic to $$\mathop{\#}\nolimits^r S^2 \times B^3$$ with bound­ary $$\mathop{\#}\nolimits^r S^2 \times S^2$$ or the twis­ted case. However, go­ing back to ori­gin­al $$r$$ 2-handles, we could have chosen their fram­ings so that we avoid the twis­ted case.  ◻

The proof of the The­or­em now fol­lows be­cause giv­en a fi­nitely presen­ted group $$G$$, con­struct $$\partial Y_G$$. If we had an al­gorithm to re­cog­nize $$\partial Y_G$$ as a con­nec­ted sum of $$S^2 \times S^2$$s, then we would have an al­gorithm to de­cide wheth­er $$G$$ is the trivi­al group.

It is still an open ques­tion as to wheth­er the 4-sphere is re­cog­niz­able, that is, if there is an al­gorithm which in fi­nite time will de­cide wheth­er a ho­mo­logy 4-sphere $$H$$ is dif­feo­morph­ic to $$S^4$$. Pos­sibly the de­scrip­tion of $$H$$ as a tri­sec­tion de­term­ined by a dif­feo­morph­ism of a 3-di­men­sion­al handle­body will lead to such an al­gorithm.

Note that there is no al­gorithm to re­cog­nize the $$n$$-sphere for $$n\geq 5$$. Ker­vaire [e7] showed that if a fi­nitely presen­ted group $$G$$ is su­per­per­fect, that is, if $H^1(G) = H^2(G) =0,$ then it is the fun­da­ment­al group of a ho­mo­logy $$n$$-sphere $$H$$, $$n > 4$$. If the $$n$$-sphere can be re­cog­nized, then $$H = S^n$$ if and only if $$\pi_1(H) = 1$$ (by the Poin­caré the­or­em). However the class of su­per­per­fect groups, by the Ad­jan–Ra­bin the­or­em [e3] [e4], can­not have an al­gorithm to re­cog­nize the trivi­al group.