Ring of closed manifolds modulo fiber bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:22:06Z http://mathoverflow.net/feeds/question/36697 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36697/ring-of-closed-manifolds-modulo-fiber-bundles Ring of closed manifolds modulo fiber bundles Andreas Thom 2010-08-25T21:40:33Z 2010-08-26T12:13:10Z <p>Let $R$ be the ring which is generated by homeomorphism classes $[M]$ of compact closed manifolds (of arbitrary dimension) subject to the relations that $$[F]\cdot [B] = [E]$$ if there exists a fibre bundle $F \to E \to B$, and $$[M] + [N] = [M \cup N]$$ if $M$ and $N$ are of the same dimension. Clearly, $[pt]$ behaves as a unit and we can write $[pt]=1$. Moreover, since $[F] \cdot [B] = [F \times B] = [B \times F] = [B] \cdot [F]$, we see that $R$ is a commutative ring.</p> <p>It is clear that the Euler characteristic defines a homomorphism $\chi : R \to {\mathbb Z}$. What else can we say about the ring $R$ ? What can we say if everything is required to be oriented and/or smooth etc.? Is the ring $R$ finitely generated?</p> <p>Example: Since $S^1$ is a double cover of itself, we get $[S^0] \cdot [S^1] = [S^1]$, but $[S^0] = 2$ and hence $[S^1]=0$. In particular, the classes of all mapping tori of homeomorphisms vanish in $R$ since they are fiber bundles over $S^1$.</p> http://mathoverflow.net/questions/36697/ring-of-closed-manifolds-modulo-fiber-bundles/36705#36705 Answer by damiano for Ring of closed manifolds modulo fiber bundles damiano 2010-08-25T23:54:40Z 2010-08-25T23:54:40Z <p>The ring $R$ is graded by dimension, and it is trivial in dimension one, by the observation in the question. In dimension two, the connected orientable surfaces of genus at least two are all topological covers of the surface of genus two. In particular, the class of the 2-sphere and the class of the orientable surface of genus two represent in $R$, up to multiples, all orientable two manifolds. Using orientable double covers, we might also deal with the non-orientable ones, but I am not going to think about non-orientable surfaces.</p> <p>Observe that the sum of the two sphere and the surface of genus two has vanishing Euler characteristic: this is the first candidate for something with trivial Euler characteristic that might be non-zero! In fact, neither of these surfaces fibers over a circle (Euler characteristic is non-zero), and neither is a non-trivial cover of an orientable surface (Euler characteristic of a putative base space would have to be odd). Thus, there seems to be no possibility for a relation between these two surfaces.</p> <p>Therefore, unless I made a mistake, in the orientable case we have found a non-zero element in the kernel of $\chi$.</p> http://mathoverflow.net/questions/36697/ring-of-closed-manifolds-modulo-fiber-bundles/36715#36715 Answer by Agol for Ring of closed manifolds modulo fiber bundles Agol 2010-08-26T03:52:33Z 2010-08-26T03:52:33Z <p>If you believe <a href="http://en.wikipedia.org/wiki/Virtually_fibered_conjecture" rel="nofollow">Thurston's virtual fibering conjectur</a>e, then hyperbolic 3-manifolds represent torsion in your ring. Also, Seifert fibered spaces are torsion. There are graph manifolds which do not virtually fiber, so I'm not sure about that case. </p> http://mathoverflow.net/questions/36697/ring-of-closed-manifolds-modulo-fiber-bundles/36752#36752 Answer by Igor Belegradek for Ring of closed manifolds modulo fiber bundles Igor Belegradek 2010-08-26T12:13:10Z 2010-08-26T12:13:10Z <p>If you restrict to simply-connected smooth manifolds, the signature becomes multiplicative under fiber bundles. In general it is multiplicative mod $4$ as proved by Hambleton-Korzeniewski-Ranicki, see <a href="http://www.math.mcmaster.ca/ian/published/hkr_2007.pdf" rel="nofollow"> here. </a></p>