# More four-dimensional counterexamples

To follow up on A four-dimensional counterexample?, I am probably being dense, but are there examples of spaces which are homotopy equivalent to bundles of surfaces over surfaces (or three-manifolds over the circle, or circle over three-manifold) and yet are not such. Same question if you change "bundle" to "product". In Hillman's book

he seems very careful to sidestep this question and talk about homotopy equivalence only...

I am interested primarily in spaces where the fiber and the base are $K(\pi, 1)$ spaces (so not spheres) and are oriented (if that makes any difference).

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Borel's conjecture predicts that a homotopy equivalence of closed aspherical manifolds is homotopic to a homeomorphism. I would be very surprized if the conjecture were not true for surface bundles over surfaces. –  Igor Belegradek Jul 4 '11 at 3:58
Incidentally, if both the fiber and the base have zero Euler characteristic, then I think the fundamental group of the total space has subexponetial growth in which this case the Borel's conjecture is true. –  Igor Belegradek Jul 4 '11 at 4:17
Cool! I would assume that "Borel's conjecture" would be based on some set of cases where Borel actually knew this to be true... Also, what's the reference for the "subexponential growth" case? –  Igor Rivin Jul 4 '11 at 5:30
@Igor: "if both the fiber and the base have zero Euler characteristic, then I think the fundamental group" --- what about $T^2$-bundle over $S^1$? –  Anton Petrunin Jul 4 '11 at 8:01
@Anton: thanks, I wrote this in haste. In fact, I do not even know the 4-dimensional Borel conjecture for all subexponential growth groups, but all virtually nilpotent groups are okay. I will give references later today. –  Igor Belegradek Jul 4 '11 at 19:57

Basic ingredients are topological surgery, and computations of $L$-groups and $K$-groups. Surgery works perfectly in dimensions $\ge 5$, but in dimension $4$ one is limited to fundamental groups of subexponential growth, see this paper by Kruskal-Quinn.