Classify manifolds up to diff homeomorphism v.s. the word problem for groups

Question

It is not possible to classify manifolds up to diffeomorphism or homeomorphism in dimensions ≥ 4 – because the word problem for groups cannot be solved. But it is possible to classify manifolds up to cobordism.

question 1: How does the word problem for groups affects classifying manifolds $M$ up to diffeomorphism or homeomorphism? Is that trouble of classifying manifold related to that the homotopy "group" ad winding numbers of sub-manifolds inside $M$ cannot be counted due to the word problem for groups?

question 2: Can the word problem for groups be solved? If so, can we classify manifolds up to diffeomorphism or homeomorphism in any dimensions, including dimensions ≥ 4?

Answer 1

There is a theorem due to A.A. Markov regarding this question (the son of the much more famous A.A. Markov) available in the proceedings of the 1958 ICM. I'll try and give a modern proof in the smooth category in dimension four.

Theorem (Markov): For any finite presentation $P$, there exists a pair of smooth closed orientable 4-manifolds $M_1$ and $M_2$ which are diffeomorphic if and only if $P$ presents the trivial group.

Proof: Assume $P$ has $m$ generators $a_1, \dots , a_m$ and $n$ relations $r_1, \dots ,r_n$. Consider $X$ the connect sum of $m$ copies of $S^1 \times S^3$. The fundamental group of $X$ is the free group on $m$ generators so we can represent the relations $r_1, \dots, r_n$ as elements of $\pi_1(X)$. By standard tranversality theory (for instance see Guillemin and Pollack), we can represent $r_1, \dots, r_n$ as disjoint simple closed curves. After surgering out neighborhoods of these curves (replacing their tubular neighborhoods $S^1 \times D^3$ each with an $D^2 \times S^2$) we get a smooth closed 4-manifold $X'$ where $P$ presents $\pi_1(X')$. After perhaps connecting summing with $\Bbb CP^2 \# -\Bbb CP^2$ we can assume $X'$ has indefinite and odd intersection form, hence its intersection form is determined by its signature, rank and parity.

A theorem of Wall implies then that if $X'$ is simply connected $X'$ is h-cobordant to $k_1\Bbb CP^2 \# k_2(-\Bbb CP^2)$ for some $k_1$ and $k_2$, and another theorem of Wall implies that for any two h-cobordant 4-manifolds $X_1$ and $X_2$, there exists an integer $N$ so that $X_1 \# N(S^2 \times S^2)$ is diffeomorphic to $X_2 \# N(S^2 \times S^2)$. In the case we are in we can actually find an explicit $N$ using the data of the presentation ($N$ just needs to be larger than the Morse number of $X'$), but I won't prove that here. Finally we let $M_1= X' \# N(S^2\times S^2)$ and $M_2=k_1\Bbb CP^2 \# k_2(-\Bbb CP^2) \# N(S^2 \times S^2)$.