Let $M$ be a closed connected manifold and fix a basepoint $q \in M$ and a Riemannian metric on $M$. Let $F(M)$ denote the orthonormal frame bundle of $M$. This is a principal $O(n)$-bundle over $M$ ($n = \dim M$). The homotopy sequence of this bundle reads $$\dots \to \pi_2(M,q) \to \pi_1(O(n),I) \to \pi_1(F(M),F_0) \to \pi_1(M,q) \to \pi_0(O(n),I)\to\dots$$ where $F_0$ is a fixed frame over $q$. Let $\pi_1^{\text{or}}(M,q)$ be the kernel of the penultimate arrow; this is the subgroup of $\pi_1M$ represented by "orientable" loops. Then we have the group extension $$0 \to A \to \pi_1(F(M),F_0) \to \pi_1^{\text{or}}(M,q) \to 1$$ where $A$ is the quotient of $\pi_1(O(n),I)$ by the image of $\pi_2(M,q)$. Assume $A \simeq \mathbb Z_2$ (which happens for $n \geq 3$ and $w_2(M) = 0$, I think). Then this is a central group extension and therefore we have the associated class in the second group cohomology $H^2(\pi_1^{\text{or}}(M,q);\mathbb Z_2)$. What can be said about this class in terms of the topology of $M$? Does it have to do anything with Stiefel-Whitney classes? When is it trivial?

It seems that when $M$ is orientable and Spin, then this class is indeed trivial; moreover different splittings of this extension correspond to different Spin structures.

A reformulation of this is as follows (I'm adding this just to give an additional perspective): the above fiber bundle gives rise to the associated based loop space fibration $$\Omega SO(n) \to \Omega F(M) \to \Omega^{\text{or}}M$$ and the group extension is obtained by looking at the last terms of the corresponding homotopy sequence.

And lastly, it looks like this should be understood in the broader context of principal bundles and characteristic classes. If someone can point me in the direction of understanding this or give a reference, that'd be great.

  • $\begingroup$ First comment: I think your guess that this has something to do with Stiefel-Whitney classes is right. Choosing a Riemannian metric is tantamount to choosing a reduction of structure group from $GL_n(\mathbb{R})$ to $O(n)$ for the tangent bundle. The $O(n)$-characteristic classes of the frame bundle and the $GL_n(\mathbb{R})$-characteristic classes of the tangent bundle are the same because $O(n)$ is a deformation retract of $GL_n(\mathbb{R})$. $\endgroup$ Commented Jun 12, 2014 at 8:05
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    $\begingroup$ Second comment: I assume the simpler situation where the manifold is oriented (so that I do not have to talk about $\pi_1^{\textrm{or}}$). Then it seems to me that the central extension is the pull-back of the covering $B\textrm{Spin}(n)\to B\textrm{SO}(n)$ along the classifying map $M\to B\textrm{SO}(n)$ of the frame bundle. Triviality of the extension should then characterize the existence of a spin structure on $M$. So I believe that the image of the class of the extension under the restriction map $H^2(B\pi_1 M)\to H^2(M)$ is the second Stiefel-Whitney class. $\endgroup$ Commented Jun 12, 2014 at 8:08
  • $\begingroup$ Ctd: However, the class in $H^2(B\pi_1M)$ could have some more precise information that is not visible on the manifold. I am not sure about this, though, because $B\pi_1M$ can be obtained from $M$ by glueing in cells of dimension $\geq 3$. $\endgroup$ Commented Jun 12, 2014 at 8:12

1 Answer 1


Let $M$ be an orientable manifold, with chosen base point $q$ and chosen Riemannian metric. The extension of the question arises from the principal bundle $SO(n)\to F(M)\to M$.

There is another bundle giving rise to the same extension, namely $\mathbb{RP}^\infty\to \tilde{M}\to M$, and there are three equivalent ways of constructing it. On the one hand, it can be obtained from stabilization to $SO(\infty)$ plus change of fibre along the action $SO(\infty)\times\mathbb{RP}^\infty\to\mathbb{RP}^\infty$. On the other hand, it can be obtained by passing to the stable tangent bundle and then taking the associated $\mathbb{RP}^\infty$-bundle. Third, the $\mathbb{RP}^\infty$-bundle is the pullback of the universal $\mathbb{RP}^\infty$-bundle $\mathbb{RP}^\infty\to\{\operatorname{pt}\}\to K(\mathbb{Z}/2,2)$ along the classifying map $M\to BSO(n)\to K(\mathbb{Z}/2,2)$. Under the identification $[M,K(\mathbb{Z}/2,2)]\cong H^2(M,\mathbb{Z}/2)$, this is the Stiefel-Whitney class $w_2$. (Note that the classifying map of the frame bundle can be identified with the classifying map for the tangent bundle, the frame bundle arises from choice of structure group reduction from $GL_n(\mathbb{R})$ to $SO(n)$ applied to the tangent bundle. This choice of reduction of structure group is the same as a choice of a Riemannian metric.)

[Edit:] To see that the $\mathbb{RP}^\infty$-bundle yields the same extension on fundamental groups, the first construction (via change-of-fiber) is the most helpful. The change of fiber induces a map $F(M)\to \tilde{M}$, such that the map induced on homotopy fibers is $SO(n)\to SO(\infty)\to \mathbb{RP}^\infty$. Therefore, we get a commutative ladder of the long exact homotopy sequences associated to $SO(n)\to F(M)\to M$ and $\mathbb{RP}^\infty\to\tilde{M}\to M$. The change-of-fiber has the effect of killing all homotopy groups of $SO(n)$ in degrees $\geq 2$, but induces an isomorphism on $\pi_1$; therefore, the part of the exact sequence relevant for the extension question is therefore unchanged when changing the fiber from $SO(n)$ to $\mathbb{RP}^\infty$.

I think, obstruction theory shows that a splitting of the extension is the same as a section of the bundle $\mathbb{RP}^\infty\to \tilde{M}\to M$, because $\mathbb{RP}^\infty\cong K(\mathbb{Z}/2,1)$. This way, one could see that the extension splits if and only if $w_2=0$.

In the general case where $M$ is not necessarily orientable, then $\pi_1^{\operatorname{or}}(M)$ is the fundamental group of the orientation cover, the pullback of $BSO(n)\to BO(n)$ along the classifying map $M\to BO(n)$. I think the above arguments can be applied to the orientation cover and then characterize the splitting of the extension as triviality of $w_2$ for the orientation cover.

  • $\begingroup$ If I understand correctly, the idea is: replace the "complicated" $SO(n)$-fibration $F(M) \to M$ by the "simpler" $\mathbb R P^\infty$-fibration, which yields the same group extension and therefore is classified by the same class; since we know how the $\mathbb R P^\infty$-fibration comes about, we know what this class is, namely it's $w_2$. The fibration is trivial iff the classifying map is nullhomotopic iff $w_2(L) = 0$. What I don't understand is how to see that the group extension is the same in this case. Could you elucidate this? $\endgroup$ Commented Jul 23, 2014 at 16:03
  • $\begingroup$ Another point is that $H^2(\pi_1(L);\mathbb Z_2)$ embeds into $H^2(L;\mathbb Z_2)$. In case $L$ is orientable, it seems that the class of the extension really maps to $w_2(L)$ under this embedding. $\endgroup$ Commented Jul 23, 2014 at 16:05
  • $\begingroup$ The point is that we do not just replace one fibration by another one. By construction, they are more closely related, they actually give a commutative ladder of long exact homotopy sequences. From this you can see that the extensions will be the same. I will edit the question to make this clearer. $\endgroup$ Commented Jul 23, 2014 at 18:54
  • $\begingroup$ I added some explanation to show that the two extensions of fundamental groups, arising from the $SO(n)$-bundle and the $\mathbb{RP}^\infty$-bundle, are equivalent. I hope it is clearer now. $\endgroup$ Commented Jul 26, 2014 at 17:45

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