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truebaran
truebaran
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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.

Suppose that we are given a topological space $X$: 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.

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.

Source Link
truebaran
truebaran
  • 9.3k
  • 5
  • 30
  • 88

Poincare duality spaces vs. manifolds via lifting maps, the obstruction theory and the role of simply connectedness

Suppose that we are given a topological space $X$: 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.