4
$\begingroup$

Main question is this: Suppos $M$ is a manifold and $G$ is a finite group. If there is a group homomorphism $\phi:\pi_1 M\to G$, is there a continuous map $f:M\to BG$, where $BG$ is a classifying space?

One may generalize this question to ask if the functor $\pi_1$ is full, but I think this is too good to be true.

This question arises from the following context. When I was reading some article, I read that there is a bijection between the isomorphism class of principal $G$-bundles over $M$ and the homotopy class of maps from $M$ to $BG$. I wanted to verify this, so I found out that (1) if two maps $f,g:M\to BG$ are in the same homotopy class, $f_*, g_*:\pi_1 M\to G$ are conjugate, (2) a map $M\to BG$ induces a principal $G$-bundle, which is a pull back of $EG\to BG$, (3) if $f_*, g_*$ are in the same conjugacy class, the principal $G$-bundles induced by each of them are isomorphic to each other, where the isomorphism is $p\mapsto hph^{-1}$ where $hf_*h^{-1}=g$, and (4) a principal $G$-bundle $N\to M$ induces a group homomorphism $\pi_1(M)\to G$, since every loops on $M$ induces a $G$-action on $M$. I can deduce that the homotopy class of maps from $M$ to $BG$ gives an isomorphism class of principal $G$-bundles over $M$, but I am stuck on the converse statement. I think it is enough to prove that $\pi_1(M)\to M$ gives a continuous map $M\to BG$, and I presume that there is something with classifying space.

Thanks in advance.

$\endgroup$
4
$\begingroup$

Yes, such a map $f$ does exist. In fact, it exists even when $M$ is a CW complex (and every smooth manifold has a CW complex structure) and when $G$ is an arbitrary group (not just a finite groups).

The proof is a straightforward obstruction theory argument. First pick a maximal tree in the 1-skeleton of $M$ and map that to the base point of $BG$. Next, each remanining 1-cell in $M$ represents an element of $\pi_1(M)$, map that edge to a loop in $BG$ representing the $\phi$ image of that element. Now extend over the 2-skeleton: the boundary of each 2-cell represents the trivial element of $\pi_1(M)$, so the image loop represents the trivial element of $\pi_1(BG)$, so you can extend the map continuously over that 2-cell. Now induct up the dimensions: assuming the map is defined on the $n$-skeleton for $n \ge 2$, for each $n+1$ cell the restriction of the map to the boundary of that cell is homotopically trivial, because $\pi_n(BG)$ is trivial, so the map can be extended continuously over the whole $n+1$ cell.

$\endgroup$
  • 1
    $\begingroup$ When you say "$G$ is an arbitrary group" it still has to be discrete, otherwise $\pi_n(BG)$ isn't necessarily trivial for $n > 1$. $\endgroup$ – William Feb 3 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.