Ring of closed manifolds modulo fiber bundles

Question

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 $S^1$.

Answer 1

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.)

Answer 2

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.

Answer 3

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 $\chi$.

Answer 4

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.