## Ring of closed manifolds modulo fiber bundles

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.

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?

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 $S^1$.

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If you are interested in the algebraic category, I suggest Bridgeland's Introduction to Motivic Hall Algebras (arxiv.org/abs/1002.4372) which develops similar rings for varieties, schemes, and stacks. – David Steinberg Aug 25 2010 at 22:20
what is -[M] ? – Paul Aug 25 2010 at 23:14
@Sean : Can't all of them be the answer? All that happens is that in Andreas's ring, all the total spaces you are talking about get identified. – Andy Putman Aug 26 2010 at 4:13
@Sean, read damiano and Agol's answers, they're addressing your concern. The ring is defined via two equivalence relations, first you take homeomorphism types of manifolds, then you form the free commutative ring on the homeormophism types of the manifolds, then you mod out by the ideal generated by all $[M]+[N]-[N \sqcup M]$ and $[E]-[M][N]$. – Ryan Budney Aug 26 2010 at 5:17
@Paul: $-[M]$ is the just the formal additive inverse of $[M]$. ̯@Sean: I think that the notion of a ring generated by variables subject to relations is standard. Of course it can happen that variables get identified. More formally, it is the quotient of the (a priori non-commutative) polynomial ring with variables indexed by homeomorphism classes modulo the relations $[M] + [N] - [M \cup N]$, $[E] - [B][F]$ and $[pt]=1$. – Andreas Thom Aug 26 2010 at 6:04

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.

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.

Therefore, unless I made a mistake, in the orientable case we have found a non-zero element in the kernel of $\chi$.

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 Thanks. Right, since it is really a graded ring, the sum of the sphere and the surface of genus $2$ seems to be non-trivial in $R$. – Andreas Thom Aug 26 2010 at 5:55

If you believe Thurston's virtual fibering conjecture, 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.

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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 here.

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 In fact there are a number of multiplicative genera which give ring homomorphisms from this ring to the integers (or something) - though they usual require some additional structure (e.g. unitarity for the Todd genus). Thus, one should perhaps consider a variants of this ring for different flavors of manifolds (unoriented, oriented, spin, "string"), where then by definition there is a universal genus sending M to [M]. Perhaps that was the motivation for this question? – Dev Sinha Aug 26 2010 at 21:29 Right, the ring $R$ is some sort of universal target for a multiplicative genus and there are many variations of it. Does that help to compute it? – Andreas Thom Aug 27 2010 at 15:19 That does help identify many interesting quotients of it, but this is just rephrasing what has been noticed so far. – Dev Sinha Sep 2 2010 at 4:09