# 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 '10 at 22:20
what is -[M] ? – Paul Aug 25 '10 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 '10 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 '10 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 '10 at 6:04

Consider the variation where we ask for smooth manifolds and smooth fiber bundles. Then I claim that $R$ is not finitely generated.

The starting observation is that if $F \to E \xrightarrow{p} B$ is a smooth fiber bundle then

$$0 \to \text{ker}(p) \to T(E) \to p^{\ast}(T(B)) \to 0$$

gives a splitting of the tangent bundle of $E$. There are cohomological obstructions to such splittings existing in general, which we can compute. The upshot is that if $E$ is a simply connected (so it has no nontrivial covers) closed smooth manifold whose tangent bundle has no nontrivial subbundles, then $[E]$ does not participate in any of the interesting relations defining $R$, and in particular cannot lie in the subring of $R$ generated by manifolds of dimension smaller than $\dim E$. Hence to show that $R$ is not finitely generated it suffices to write down a sequence of such $E$ of arbitrarily large dimension.

But this is standard: we can take the even-dimensional spheres $S^{2n}$. First, observe that because $S^{2n}$ is simply connected, it has no nontrivial covering spaces, and in addition every real vector bundle over $S^{2n}$ is orientable, hence has well-defined Euler classes (after picking an orientation). Second, the Euler class $e(T)$ of the tangent bundle is $2$ times a generator of $H^{2n}(S^{2n})$, and in particular does not vanish. Since the Euler class is multiplicative with respect to direct sum, if $T = T_1 \oplus T_2$ is a nontrivial splitting of the tangent bundle then $e(T) = e(T_1) e(T_2)$. But the cohomology groups that $e(T_1)$ and $e(T_2)$ live in both vanish for $S^{2n}$; contradiction. Hence the tangent bundle of $S^{2n}$ admits no nontrivial splittings, and so $S^{2n}$ is not the total space of any nontrivial smooth fiber bundle of closed smooth manifolds.

(Maybe this argument can be rescued in the topological setting using tangent microbundles?)

(Strictly speaking this argument's not quite complete: we also need to show that there aren't any interesting bundles with total space the disjoint union of $S^{2n}$ with something else. But fiber bundle maps $p : E \to B$ are open, so the image of $S^{2n}$ under such a map is a connected component of the base, and we can restrict our attention to this connected component without loss of generality. Then $E$ breaks up, as a fiber bundle, as a disjoint union of $S^{2n}$ and whatever else, and we can restrict our attention to $S^{2n}$ again without loss of generality. In other words, in the defining relations we can assume that $E$ and $B$ are both connected without loss of generality.)

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It's worth noting that the definition of $R$ can be rearranged a bit as follows: first, on the set of homeo/diffeomorphism classes of manifolds, there is already a natural semiring structure given by disjoint union and product. Disjoint union is cancellative, so formally adjoining inverses to get a ring doesn't identify anything that shouldn't be identified. Now we add relations $[F \times B] = [E]$ for fiber bundles $F \to E \to B$ (where, as mentioned above, we can assume WLOG that $B$ and $E$ are connected). This makes it clear that the relation imposed by $F \to E \to B$ only matters... – Qiaochu Yuan Nov 3 '14 at 0:03
...in dimensions $\dim E$ and up, and only through the influence of manifolds which can be written in some way as fiber bundles. – Qiaochu Yuan Nov 3 '14 at 0:06
Very nice! I am sure that this can be done topologically. At least there are similar results without any smoothness assumption, e.g., Browder showed that any fibration with the sphere as a total space (and base and fibre connected finite polyhedra) must have $S^1,S^3$ or $S^7$ as fibre (up to homotopy). The proofs are using the Steenrod algebra, $H$-spaces structure and no assumption on smoothness etc. – Andreas Thom Nov 3 '14 at 12:22
I am accepting this as an answer, since you proved that the ring is not finitely generated - and the remaining questions were a bit too vague. – Andreas Thom Nov 7 '14 at 11:02
If $\mathbb{RP}^{2n} \to B$ is a locally trivial fiber bundle (with fiber $F$), then so is $S^{2n} \to B$. The fiber is some two-sheeted cover of $F$. Your argument gives that either $B$ or $F$ must have dimension $2n$, that's enough. – Andreas Thom Nov 7 '14 at 20:54

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 '10 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 '10 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 '10 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 '10 at 4:09