# Is the Euler characteristic of aspherical connected 2-complexes at most 1? (No!) What can be said about subcomplexes of 2-complexes deformation retractible onto graphs.

I have several related questions, i do not know which one is more important to me, i think it would depend on their answers.

1. Is it true that the Euler characteristic of a finite connected aspherical simplicial 2-complex cannot be greater than 1?

2. If $A$ is a finite simplicial 2-complexe that retracts by deformation onto a graph (1-complex), is it true that every subcomplex of $A$ is aspherical?

3. (This is the question that i am most likely interested in.) If $A$ is as above, is it true that every connected subcomplex $B$ of $A$ is of Euler characteristic at most 1, and if the Euler characteristic of $B$ is 1, then $B$ is contractible?

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This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ deformation retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. I do not see how $\chi(X)=1$ could imply that $X$ is contractible.

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Thanks a lot Andreas, i should have thought of it! I am however still mostly interested in question 3. – Alexey Muranov Jan 30 '12 at 18:11
I understood that there would be no negative answer to 2. You have answered the question from the title, so i would have to mark it as answered, but i am afraid that then people wouldn't look at question 3, so i cheated and changed the title :). – Alexey Muranov Jan 30 '12 at 18:17
No problem. Question 3 is interesting. – Andreas Thom Jan 30 '12 at 18:29
I've decided to post Question 3 separately: mathoverflow.net/questions/87342/… – Alexey Muranov Feb 2 '12 at 16:31

About Question 3, thanks to Andreas' answer it is enough to prove that there is no acyclic 2-complex $X$ with $\pi_1(X)=D\pi_1(X)\neq\lbrace0\rbrace$.

Indeed it would mean that $\chi(B)=1 \implies$ $B$ is weakly contractible $\overset{Whitehead}{\implies}$ $B$ is contractible.

I have no examples in mind, all I can say is that making a loop into a commutator amounts to make it bound the complementary of a disk in a torus, this is how I would go for a counterexample.

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Hi Arnaud, what is $D\pi_1(X)$? – Alexey Muranov Jan 31 '12 at 12:16
Unfortunately there exists acyclic 2-complexes which are not contractible, so this approach won't work. – Jesper Grodal Jan 31 '12 at 15:09
@Jesper Grodal: can you point me to an example of an acyclic non contractible 2-complex, please? By the way, i do not see why the subcomplex $B$ will be acyclic, i only see that $H_1(B)$ will be finite. – Alexey Muranov Feb 1 '12 at 9:12
Sorry about the silly question: of course, if it is acyclic but $\pi_1$ is nontrivial, it is not contractible. I do not see though an easy example. – Alexey Muranov Feb 1 '12 at 11:59
It has been explained to me by a colleague why $B$ will be acyclic: $H_1(B)$ with coefficients in any $\mathbb Z_p$ has to be trivial, not only with coefficients in $\mathbb Q$. – Alexey Muranov Feb 2 '12 at 15:22