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A group $G$ is Hopfian if every epimorphism $G\to G$ is an isomorphism. A smooth manifold is aspherical if its universal cover is contractible. Are all fundamental groups of aspherical closed smooth manifolds Hopfian?

Perhaps the manifold structure is irrelevant and makes examples harder to construct, so here is another variant that may be more sensible. Let $X$ be a finite CW-complex which is $K(\pi,1)$. If it helps, assume that its top homology is nontrivial. Is $\pi=\pi_1(X)$ Hopfian?

Motivation. Long ago I proved a theorem which is completely useless but sounds very nice: if a manifold $M$ has certain homotopy property, then the Riemannian volume, as a function of a Riemannian metric on $M$, is lower semi-continuous in the Gromov-Hausdorff topology. (And before you laugh at this conclusion, let me mention that it fails for $M=S^3$.)

The required homotopy property is the following: every continuous map $f:M\to M$ which induces an epimorphism of the fundamental group has nonzero (geometric) degree.

This does not sound that nice, and I tried to prove that some known classes of manifolds satisfy it. My best hope was that all essential (as in Gromov's "Filling Riemannian manifolds") manifolds do. I could not neither prove nor disprove this and the best approximation was that having a nonzero-degree map $M\to T^n$ or $M\to RP^n$ is sufficient. I never returned to the problem again but it is still interesting to me.

An affirmative answer to the title question would solve the problem for aspherical manifolds. A negative one would not, and in this case the next question may help (although it is probably stupid because I know nothing about the area):

Question 2. Let $G$ be a finitely presented group and $f:G\to G$ an epimorphism. It it true that $f$ induces epimorphism in (co)homology (over $\mathbb Z$, $\mathbb Q$ or $\mathbb Z/2$)?

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For question 2, $f$ induces an epi on abelianizations, i.e. on $H_1(G;Z)$. The universal coefficient theorem shows it remains epi over $Z/2$ or $Q$. –  Paul Apr 9 '10 at 12:30
    
I'm interested in higher homology groups. –  Sergei Ivanov Apr 9 '10 at 12:33
    
I think the question "do closed aspherical manifolds have Hopfian fundamental group?" is a well-known open problem. The difficulty in Question 2 is that non-Hopfian finitely presented groups are not so easy to construct; it is a good question! –  Igor Belegradek Apr 9 '10 at 16:44
    
I suspect that the closest known such group is the example of a non-Hopfian CAT(0) group that Wise constructed in his thesis. –  HJRW Apr 9 '10 at 16:55
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One partial result: if a closed manifold X admits a metric of negative sectional curvature bounded away from 0 then it follows from a theorem of Sela that $\pi_1(X)$ is Hopfian. –  HJRW Apr 9 '10 at 17:28

3 Answers 3

up vote 15 down vote accepted

The Baumslag-Solitar group $B(2,3)=\langle a,b\vert ba^2b^{-1}=a^3\rangle$ is not Hopfian. But it has a natural $K(\pi,1)$ given by the double mapping cylinder of $S^1 \rightrightarrows S^1$ where the maps are $z\mapsto z^2$ and $z\mapsto z^3$. This is a finite CW complex.


Edit: The double mapping cylinder can be constructed like this. Take a circle $S^1$ and a cylinder $S^1\times I$. Glue one end of the cylinder to the circle by the degree 2 map $z\mapsto z^2$, and glue the other end of the cylinder to the circle by the degree 3 map $z\mapsto z^3$. The $a$ in the presentation above is the loop around the circle, while the $b$ is the loop that goes along the cylinder (whose ends have been brought together, forming a loop).

To see that this is a $K(\pi,1)$, you can check that its universal cover is the product $T_5\times \mathbb{R}$ of the infinite $5$-regular tree $T_5$ with the line $\mathbb{R}$. (Think about what this CW complex looks like locally: away from the circle where we glued everything together, it's locally a $2$-manifold. At the circle, we have $2+3=5$ half-planes meeting along their edges.) There is a picture of this universal cover on page 3 of Farb-Mosher, "A rigidity theorem for the solvable Baumslag-Solitar groups", Inventiones 131 2 (1998), 419-451.

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Thank you for explanation. –  Sergei Ivanov Apr 9 '10 at 13:54
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Can you construct a manifold like that? –  Anton Petrunin Apr 15 '10 at 22:02

A couple of remarks on this question:

It is true for closed aspherical manifolds of dim $\leq 3$, since they have residually finite fundamental group.

If one could find an aspherical 4-manifold $W$ with boundary, such that $\pi_1 W$ is non-Hopfian, and a degree >1 map $f:W\to W$ which induces a self-covering $f: \partial W \to \partial W$, then one might be able to use a reflection trick of Davis to find a closed aspherical manifold with the same property. Since there exist 3-manifolds which are not co-Hopfian (in fact, which are finite-sheeted covers of themselves), this approach might work.

I attempted this approach by thickening up the Baumslag-Solitar group presentation complex to a 4-manifold with boundary, but the boundary wasn't co-Hopfian, so I couldn't get the self-mapping of the 2-skeleton to extend.

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I just wish to point out that Hopfian spaces/manifolds are studied in topology, e.g. below is mathscinet review (written by Ross Geoghegan) of Hausmann's paper "Geometric Hopfian and non-Hopfian situations":

"Let $M$ be a closed oriented smooth $n$-manifold and let $f:M\to M$ be a map of degree $d$. The author addresses Hopf's conjecture that if $d=1$, or $d=-1$ then $f$ is a homotopy equivalence. He proves the conjecture for the case in which $M$ has virtually nilpotent fundamental group. He also gives partial answers to related questions. Example: For any $d$ is not $1$ or $-1$ and any $n\ge 6$ there exist $M$ and $f$ as above such that $f$ induces an epimorphism with nontrivial kernel on the fundamental groups."

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