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Given closed $3$-manifolds $M$ and $N$ and an element $\alpha\in H^1(M;\mathbb{Z}_2)$, when does there exist a map $f:M\to N$ such that $\alpha=f^*(w_1(N))$?

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2 Answers 2

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This basically boils down to

  1. Understanding what the class $w_1$ represents, and
  2. Doing obstruction theory.

This is generally not an easy question to answer, since obstruction theory requires you to have a clear understanding of some cell decomposition of the domain $M$, as well as the homotopy groups of the target $N$. Typically one cannot say much for general manifolds, but here is one example of the type of thing that this theory implies:


Suppose $N$ is simply connected, then $H^1(N;\mathbb Z/2)=0$ and thus the only possible $\alpha$ is the zero class, and this can be realized by a constant map $f:M\to\{n_0\}\subseteq N$.


Aside from obstruction theory, there aren't any theories that I know of that produce continuous maps. Also you have written $w_1(N)$, which typically refers to the 1st SW class of the tangent bundle. This indicates that you are working with smooth manifolds (though it is possible to make sense of $w_1(N)$ in the topological case). I mention this only to emphasize that obstruction theory is only attempting to build for you a continuous map. If that is fine for you, great! If not, then you would further need to use smoothing theory to see if any of the resultant maps can be made smooth.

I recommend Milnor and Stasheff's 'Characteristic Classes' for further intuition about $w_1$, and I like the treatment of obstruction theory given in Davis and Kirk's 'Lecture Notes in Algebraic Topology'

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    $\begingroup$ Any continuous map between smooth manifolds is homotopic to a smooth map, so it suffices to consider continuous maps for the problem $\endgroup$ Commented Jun 12, 2022 at 8:03
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    $\begingroup$ And every topological 3-manifold admits a unique smooth structure, so $w_1$ is well-defined. $\endgroup$ Commented Jun 12, 2022 at 11:40
  • $\begingroup$ moreover, Stiefel-Whitney classes are homotopy invariant, so in fact the problem is reduced to the homotopy cathegory $\endgroup$ Commented Jun 12, 2022 at 14:49
  • $\begingroup$ on the other hand, the Stiefel-Whitney classes of a manifold have some special properties, e.g. $w_1(N)^2=w_2(N)$, which maybe can help $\endgroup$ Commented Jun 12, 2022 at 14:52
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For manifolds $M,N$ of dimension $<3$ I know the complete answer for the similar question.


If $M,N$ are closed $1$-manifolds, obviously, $f$ exists if and only if $\alpha=0$.


If $M,N$ are closed $2$-manifolds, $f$ exists if and only if any of the following four conditions hold:

  • $\alpha=0$;
  • $w_1(N)\ne0$ and $\alpha^2=0$;
  • $w_2(N)\ne0$, $\alpha=w_1(M)$ and $\chi(M)\le\chi(N)$;
  • $w_2(N)\ne0$, $\alpha^2\ne0$ and $\chi(M)<\chi(N)$.

In the proof I used the technique of simple closed curves on surfaces, inspired by Farb-Margalit.


Maybe, there is some similar methods and arguments in case of $3$-manifolds?

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