Suppose $M$ is a compact, connected, orientable manifold ($\dim M=m$) with trivial tangent bundle and let $j \colon M \to \mathbb R^n$ be an embedding. Suppose we choose a trivialization of $TM$. Then we obtain a stable normal framing of the normal bundle $\nu(j)$ by

$$ \varepsilon^m \oplus \nu(j) \cong TM\oplus \nu(j) \cong \varepsilon^n|_M. $$

Any other embedding $j'\colon M\to \mathbb R^{n'}$ is isotopic to $j$ if we consider $j$ and $j'$ as maps into some $\mathbb R^l$ for $l$ big enough. The isotopy provides a cobordism $W$ between $j(M)$ and $j'(M)$ which is parallelizable and hence it has a stable normal framing which restricts to the stable normal framings of $\nu(j)$ and $\nu(j')$ at the boundary of $W$. Thus $M$ together with the chosen trivilization represents a well-defined element in the bordism group of stably framed $m$-dimensional manifolds $\Omega_m^{\text{fr}}$.

Question: We have $\Omega_1^{\text{fr}} \cong \pi_1^S\cong \mathbb Z_2$. Do the two classes of $\Omega_1^{\text{fr}}$ depend on the orientation of the circle? If no, how do I distinguish these two classes by simply framing the tangent bundle of a circle?

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    $\begingroup$ I think you need to frame the stable tangent bundle to get all framings of the stable normal bundle. Anyway: two test if two framings of the same circle are framed cobordamt you can do the following: use the two framings to cook up a map from the circle into $SO(n-1)$ . This can be non trivial map of fundamental groups, or not. If it is not, the framings are different. $\endgroup$
    – Thomas Rot
    Aug 28, 2018 at 16:21
  • $\begingroup$ Concerning your first sentence: This means that I get only some (not all) stable normal framings if I trivialize the tangent bundle? $\endgroup$ Aug 29, 2018 at 6:52
  • $\begingroup$ Yes I think that is true. The following notes of Andy Putman www3.nd.edu/~andyp/notes/HomotopySpheresLowDimTop.pdf can be helpful. $\endgroup$
    – Thomas Rot
    Aug 29, 2018 at 7:37

1 Answer 1


Any two framings on a sphere in $n$-dimensional space, say for $n$ large, differ by a homotopy class of map to $SO(n-1)$ (as $SO(n-1)$ acts freely and transitively on the set of frames at a given point). The two classes of $\Omega^{fr}_1$ correspond to framed bordisms of the circle in large dimension Euclidean space, with one corresponding to the framing that extends over the disk, and the other corresponding to the Lie group framing on the tangent bundle, so one can distinguish the two exactly by seeing whether the framings extend over the disk. For a fixed framing, the two framings will differ by an element of $\pi_1SO(n-1) = Z/2$ hence the isomorphism above with the first framed bordism group.

A word of caution: one cannot use the exact same argument for higher dimensional spheres. The isomorphism above really depends on the fact that the $J$-homomorphism is injective in dimensions 1,2 mod 8, so that indeed, the framed bordism classes can be described by the actual framings on the spheres in those dimensions. In general, the $J$-homomorphism is precisely the map that, given a framing, takes a framed sphere in large Euclidean space to its framed bordism class via the Pontryagin-Thom construction.

  • $\begingroup$ @ThomasRot apologies; for circles in Euclidean space of dimension $n$, it is SO(n-1). Will edit accordingly. $\endgroup$ Aug 29, 2018 at 14:26

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