Sunday, 8 October 2006

at.algebraic topology - Ring of closed manifolds modulo fiber bundles

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



The starting observation is that if FtoExrightarrowpBFtoExrightarrowpB is a smooth fiber bundle then



0totextker(p)toT(E)topast(T(B))to00totextker(p)toT(E)topast(T(B))to0



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



But this is standard: we can take the even-dimensional spheres S2nS2n. First, observe that because S2nS2n is simply connected, it has no nontrivial covering spaces, and in addition every real vector bundle over S2nS2n is orientable, hence has well-defined Euler classes (after picking an orientation). Second, the Euler class e(T)e(T) of the tangent bundle is 22 times a generator of H2n(S2n)H2n(S2n), and in particular does not vanish. Since the Euler class is multiplicative with respect to direct sum, if T=T1oplusT2T=T1oplusT2 is a nontrivial splitting of the tangent bundle then e(T)=e(T1)e(T2)e(T)=e(T1)e(T2). But the cohomology groups that e(T1)e(T1) and e(T2)e(T2) live in both vanish for S2nS2n; contradiction. Hence the tangent bundle of S2nS2n admits no nontrivial splittings, and so S2nS2n 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 S2nS2n with something else. But fiber bundle maps p:EtoBp:EtoB are open, so the image of S2nS2n 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 EE breaks up, as a fiber bundle, as a disjoint union of S2nS2n and whatever else, and we can restrict our attention to S2nS2n again without loss of generality. In other words, in the defining relations we can assume that EE and BB are both connected without loss of generality.)

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