Homotopy Equivalence – How to Tell if a Space is Homotopy Equivalent to a Manifold

at.algebraic-topologysmooth-manifolds

Is there some criterion for whether a space has the homotopy type of a closed manifold (smooth or topological)? Poincare duality is an obvious necessary condition, but it's almost certainly not sufficient. Are there any other special homotopical properties of manifolds?

Best Answer

In surgery theory (which is basically a whole field of mathematics which tries to answer questions as the above), the next obstruction to the existence of a manifold in the homotopy type is that every finite complex with Poincaré duality is the base space of a certain distinguished fibration (Spivak normal fibration) whose fibre is homotopy equivalent to a sphere. (In order to get a unique such fibration, identify two fibrations if they are fiber homotopy equivalent or if one is obtained from the other by fiberwise suspension.)

For manifolds, this fibration is the spherization of the normal bundle, so the Spivak normal fibration comes from a vector bundle. This is invariant under homotopy equivalence. Thus the next obstruction is: the Spivak normal fibration must come from a vector bundle.

If I remember right, then it was Novikov who first proved that for simply-connected spaces of odd dimension at least 5, this is the only further obstruction.

In general, there is a further obstruction with values in a group $L_n(\pi_1,w)$ which depends on the fundamental group, first Stiefel-Whitney class and the dimension. See Lück's notes on surgery theory at https://www.him.uni-bonn.de/lueck/data/ictp.pdf