Suppose that we are given a topological space $X$: assume for simplicity that $X$ is compact we want to adress the following question:

Is it true that one can find a manifold $M$ which is homotopy equivalent to $X$?

Necessary condition $X$ must satisfy Poincare duality, namely there must be a class in the top homology of $X$ such that the cap product with this class induces isomorphism between homology and cohomology. As one could expect this is not enough. So let us assume that $X$ indeed satisfy Poincare duality. Then one can associate to $X$ the so called Spivak normal fibration. This fibration is classified by the map $f:X \to BG$ where $G$ is a space of self-homotopy equivalences of the sphere.
Second assumption: Let us assume that $X$ is simply connected. Then one can find a manifold homotopy equivalent to $X$ which is:
-smooth iff the map $f$ lifts to a map $X \to BO$
-piecewise linear iff the map $f$ lifts to a map $X \to BPL$
-topological iff the map $f$ lifts to a map $X \to BTop$.
(Here as far as my knowledge goes, $Top$ is the space of homoeomorphisms of the sphere, $PL$ the space of piecewise linear homeomorphisms of the sphere and so on-but please correct me if I'm wrong). Problem of lifting maps leads to the obstruction theory which gives classes in cohomology $H^{n+1}(X,\pi_n(F))$ where $F$ is a fiber of our fibration which we would like to lift (in our case $F$ is equal to $G/O,G/PL,G/Top$ respectively).

Question 1 It is general knowledge of the obstruction theory that it works better in the simply connected case: in the non simply connected case one has to deal with local coefficients. In our problem of finding a manifold in the given homotopy type in the non simply connected case there is a further obstruction: is it right to think that the presence of this obstruction is due to the fact that one has to work with local coefficients?

Question 2 What is known about the homotopy groups of the fibers $G/O$ (resp. $G/PL$, $G/Top$)? Is it somehow possible to interpret the various theorems of the form ,,up to some dimension/in some dimension two notions of manifolds (e.g. smooth, PL, Top) coincide'' in this language, i.e. as vanishing of the homotopy groups of the fiber up to some dimension?

Question 3 Due to the theorem of Sullivan every manifold of dimension different than $4$ has unique Lipschitz atlas: how does it fit into this picture?

My question is very broad, it is rather some kind of a big picture: nevertheless I hope that it would be interesting to someone who is less familiar with this theory.

  • 2
    $\begingroup$ The existence of a manifold structure on a Poincare complex is described by the surgery exact sequence, see en.wikipedia.org/wiki/Surgery_exact_sequence which works regardless of the fundamental group. In the simply-connected case the L-groups are known and quite simple, but this is also true for other classes of fundamental groups. The sequence doesn't use local coefficients in the sense they are used in the obstruction theory. Incidentally, the obstruction theory works reasonably well with local coefficients - no problem here. $\endgroup$ Jul 22, 2018 at 22:43
  • $\begingroup$ "$X$ must satisfy Poincare duality, namely there must be a class in the top homology of $X$ such that the cap product with this class induces isomorphism between homology and cohomology". So you are at least assuming that $X$ is compact. In fact, there are plenty of open manifolds that do not satisfy Poincaré duality. $\endgroup$ Jul 23, 2018 at 7:24
  • $\begingroup$ I've edited my post accordingly, indeed I'm interest in the compact case. $\endgroup$
    – truebaran
    Jul 23, 2018 at 8:26

1 Answer 1


Let's talk first about the smooth and simply connected case. As you say, Poincare duality for $X$ yields a spherical fibration, or map $X\to BG$. A lifting $X\to BO$ is necessary but not sufficient for a manifold structure. Such a lifting, or "normal structure", determines an element of the surgery obstruction group ($L$-group). A necessary and sufficent condition (assuming the dimension is not too small) for existence of a manifold structure compatible with the given normal structure is the vanishing of this surgery obstruction.

If we say PL or topological instead of smooth, then the relevant normal structure is a lift to $BPL$ or $BTop$ instead of $BO$.

In the non-simply-connected case there are two important differences. First, the surgery obstruction groups are different. (But they depend only on $\pi_1(X)$ and the orientation character $\pi_1(X)\to \lbrace\pm 1\rbrace$ and the mod $4$ class of the dimension.) Second, in order for any of this to work you need to assume that $X$ satisfies Poincare duality in a strong sense (cap product isomorphisms between cohomology with twisted coefficients and homology with twisted coefficients).

In the following sense it is correct to say that the extra complications in the non-simply-connected case are because of local coefficents: In the simply-connected case Poincare duality (say, in the case when the dimension is a multiple of $4$) produces a quadratic form over $\mathbb Z$, and the $L$-groups are what they are because of some stable classification of such forms, and this is a powerful enough invariant because for simply connected spaces homology has a powerful influence on homotopy theory (sorry for the vagueness). But in the general case one needs twisted homology (with coefficients in $\mathbb Z[\pi_1(X)]$-modules) to get that power, so one needs to think about twisted duality and about quadratic forms over $\mathbb Z[\pi_1(X)]$.

  • $\begingroup$ Tom, correct me if I am wrong: in the odd dimensional, simply connected case a lift of the map $X \rightarrow BG$ to $X \rightarrow BO$ is enough to conclude $X$ has the homotopy type of a smooth manifold. Is it also true in this case that a lift of $X \rightarrow BG$ to $X \rightarrow BTOP$ is the only obstruction to being the homotopy type of a topological manifold? $\endgroup$ Aug 19, 2020 at 21:32
  • 1
    $\begingroup$ Yes, there is a parallel story involving topological instead of smooth manifold structures, $G/TOP$ instead of $G/O$, and the same surgery obstruction groups as in the smooth case (which in odd dimensions and for simply-connected spaces are trivial). $\endgroup$ Aug 20, 2020 at 10:31

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