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Is there a finite dimensional closed manifold $M$ which is a $K(\pi,1)$, whose fundamental group is not word-hyperbolic, but which has a positive simplicial volume (ie "Gromov norm")?

(Added:) The answers of Jim and Richard are both excellent; another example is any closed, irreducible locally symmetric manifold (of non-positive curvature). But these examples are all CAT(0); I wonder if there is an example which is not CAT(0)? (of course then it is hard to see the example is a $K(\pi,1)$ . . .)

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Both Jim's and Richard's answers are good; I flagged Richard's "correct" because it is the easiest to see. –  Danny Calegari Nov 13 '09 at 0:55
    
Hmm, maybe Richard's method works if you double a cusped complex-hyperbolic manifold. The cusp is a nilmanifold, which in general won't be CAT(0) - does this work? –  Danny Calegari Nov 13 '09 at 17:07
    
That sounds like a good idea. Hm. –  Richard Kent Nov 13 '09 at 17:12
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5 Answers 5

up vote 8 down vote accepted

You can just take the double of a hyperbolic knot complement.

See Soma's paper The Gromov invariant of links, Invent. Math. 64 (1981) 445–454

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Yes: the product of 2 genus two surfaces is an example. Take a look at

Bucher-Karlsson, Michelle. The simplicial volume of closed manifolds covered by H^2 x H^2. J. Topol. 1 (2008), no. 3, 584 - 602. MR2417444.

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To get an example that is not CAT(0) you just double a finite volume complex hyperbolic manifold with cusps chopped off. The cusp cross section is pi_1-incompressible and its fundamental group is not virtually abelian (it is virtually nilpotent), and hence it cannot be CAT(0).

Actually, as I show in arXiv:math/0509504v3 ANY closed aspherical manifold can be realized as a codimension one incompressible submanifold of a closed aspherical manifold with nonzero simplicial volume. So there are tons of the kind of examples you ask for.

For some reason I cannot add a comment to DC's reply below, so I put it here. DC, you did not sound sure that you complex hyperbolic example works. :) Incidentally, for such a double there are two ways to show that its simplicial volume is nonzero. One is to use Gromov's result on gluing along amenable subsets (if my memory serves me this was proved in some detail by Kuessner but I could be wrong). What seems to me a better way is to note that the fundamental group of the double is hyperbolic rel cusp cross-section, and that apply recent paper of Mineyev-Yaman that in this case relative hyperbolicity implies nonvanishing of simplicial volume. Neither way is elementary. Examples in my paper mentioned above is more elementary.

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Dear Igor - actually, I already mentioned the example of a doubled cusped complex hyperbolic manifold as a comment to my own question :) Your arXiv paper seems to be exactly the answer to the best version of the question I could have formulated - and it's a great paper too! Thanks very much! –  Danny Calegari Nov 13 '09 at 19:27
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Just some musings:

If there were an atoroidal surface bundle over a surface whose group wasn't $\delta$-hyperbolic (so the fundamental group of the base would be a non-convex-cocompact purely pseudo-Anosov surface group in Mod(fiber)), then I'd expect the simplicial volume to be positive and that it wouldn't be CAT(0) (maybe my intuition is wrong about the CAT(0) thing).

But of course,

  1. This is just vague intuition.

  2. We don't even know if there are purely pseudo-Anosov surface groups.

  3. We don't even know if there are f.g. pseudo-Anosov non-convex-cocompact groups.

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Just Take an irreduzible, but not atorial, 3-manifold with the property that all (or at least One) of the pieces in the JSJ-decomposition are hyperbolic.

This is nonpositively curved by Leeb's thesis, but the fundamental Group is Not hyperbolic because of the abelian subgroups coming from incompressible tori. The simplicial Volume is the sum of the simplicial Volumina of the hyperbolic pieces.

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