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Let $Y$ be a contractible finite simplicial 2-complex.
Let $X$ be an acyclic subcomplex of $Y$ (i.e. $X$ connected, $H_1(X)=0$, $H_2(X)=0$).
Is $X$ contractible? (Equivalently, is $\pi_1(X)$ trivial?)

I asked this question among others here before, but it remained unanswered.

Apparently this question is equivalent to the following one.

If $m < n$, $G = \langle a_1,\dotsc,a_m \mid r_1 = \dotsb = r_m =1 \rangle$ and $J=\langle a_1,\dotsc,a_n \mid r_1 = \dotsb = r_n =1 \rangle$ are two groups given by (balanced) presentations, the presentation of $G$ is a sub-presentation of the presentation of $J$, and $J$ is trivial, is $G$ trivial? (It can be shown that the abelianization of $G$ is trivial.)

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I first formulated the question for the case when $Y$ retracts by deformation onto a graph, but then realized that it is equivalent to assuming that $Y$ is contractible (it can be made contractible by attaching 2-cells). –  Alexey Muranov Feb 3 '12 at 17:31
    
I just want to make a comment: There exists an acyclic group $G$ of geometric dimensional 2 (Thm1.2 in math.nus.edu.sg/~matberic/Berrick_acydichotomy_sent10416.pdf). Take $X=BG$. Do the plus construction to X to get a complex Y. Although $Y$ is contractible, it is a 3-dimensional complex. –  yeshengkui Feb 5 '12 at 7:34
    
Concerning your update: So we are to assume that the $r_1$ etc. in $G$ are the same as those in $J$? So the first $m$ relations of $J$ only involve $a_1$ through $a_m$?? –  Steve D Feb 5 '12 at 8:20
    
@Steve D: yes, this is what i mean. –  Alexey Muranov Feb 5 '12 at 16:08
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2 Answers

up vote 7 down vote accepted

According to Sergei Ivanov, On balanced presentations of the trivial group, Invent. Math. 165, 525--549 (2006), for $n=m+1$, this is a particular case of Kervaire–Laudenbach conjecture which is also open, and according to one of Klyachko's results, the negative answer in this special case would imply the negative answer to Whitehead's asphericity conjecture.

It seems however that nothing is known in general, not even equivalence with the Kervaire–Laudenbach conjecture.

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One can show the following:

Theorem: If the group given by the subpresentation is hyperlinear (in particular if it sofic), then it must be trivial.

The argument is based on an old Theorem of Gerstenhaber-Rothaus. I can only give a sketch here: Denote the group given by the subpresentation by $G$. We want to show that $G$ embeds into the group given by the full presentation to get a contradiction. Now, as $G$ is hyperlinear, one can satisfy the group multiplication on a finite subset of $G$ by unitaries up to any desired precision in the normalized Hilbert-Schmidt norm. Equivalently, $G$ embeds into a metric ultraproduct of unitary groups. Now, if unitaries $g_{m+1},\dots,g_n$ can be found which satisfy the relations $r_{m+1},\dots,r_n$, then the embedding of $G$ can be extended to homomorphism into the ultraproduct defined on $g_1,\dots,g_n$. In particular, $G$ embeds into the group given by the full presentation. Now, the fact that $g_{m+1},\dots,g_n$ can be found relies on the solvability of the equations $r_{m+1},\dots,r_n$ and can be proved along the lines of Gerstemhaber-Rothaus using degree theory and Hopf's computation of the cohomology of unitary groups.

Gerstenhaber-Rothaus proved some generalized form of the Kervaire-Laudenbach Conjecture for residually finite groups in

M. Gerstenhaber and O.S. Rothaus, The solution of sets of equations in groups, Proc. Nat. Acad. Sci. U.S.A. 48 (1962), 1531–1533.

The above argument is essentially only the observation that their argument extends without any problems to hyperlinear groups.

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Thanks Andreas, i will think about it. I still do not know if the fact that the presentation is balanced can be used somehow. –  Alexey Muranov Feb 5 '12 at 20:04
    
My argument only works if the presentation is balanced. –  Andreas Thom Feb 5 '12 at 21:49
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