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  1. What is the definition of Lie group framing, in simple terms?

  2. Is the Lie group framing of spheres a particular type of Lie group framing? (How special is the Lie group framing of spheres differed from the generality of Lie group framing?)

  3. Examples of Lie group framing includes:

  • $\Omega_3^\text{fr}=\mathbf{Z}/{24}$ generated by a 3-sphere $S_3=\operatorname{SU}(2)$ with the Lie group framing.

  • $\Omega_6^\text{fr}=\mathbf{Z}/{2}$ generated by $S_3 \times S_3$ with the Lie group framing.

How do we construct the explicit class of $\mathbf{Z}/{24}$ and $\mathbf{Z}/{2}$ class of Lie group framing above?

"Framed bordism" at the Manifold Atlas Project

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  • $\begingroup$ My understanding of the term is that left multiplication on a Lie group gives a trivialization of its tangent bundle. That's typically called "the" Lie group framing, although there is another one due to right multiplication. Since the $3$-sphere is a Lie group, it has its left-invariant framing, which is essentially unique up to the framing of the tangent space of the identity. Similarly, the product of two Lie groups is a Lie group. $\endgroup$
    – Ryan Budney
    Commented Nov 27, 2023 at 17:46
  • $\begingroup$ Thanks @RyanBudney is this paper related - Lie groups as framed manifolds - Erich Ossa -sciencedirect.com/science/article/pii/0040938382900131 $\endgroup$
    – zeta
    Commented Nov 27, 2023 at 18:49
  • $\begingroup$ Yes, that paper is using the same basic idea. $\endgroup$
    – Ryan Budney
    Commented Nov 27, 2023 at 22:15
  • $\begingroup$ could you write up an answer? please thanks! $\endgroup$
    – zeta
    Commented Nov 28, 2023 at 7:54

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Lie group framing is a reference to the group action. A Lie group $G$ acts on the left of $G$ by the map

$$(g,h) \longmapsto gh.$$

Similarly there are actions on the right, and conjugation actions, etc. The left-invariant vector fields on $G$ correspond to the elements of $T_e G$. So any basis for $T_e G$ extends to a unique left-invariant framing of the tangent bundle $TG$.

So for example, the $3$-sphere has left-invariant vector fields

$$S^3 \ni g \longmapsto (g,gv) \in TS^3$$

where $v \in T_1 S^3$, i.e. $v$ is orthogonal to $1 \in \mathbb R$ as a vector in $\mathbb R^4$, and we are using quaternionic multiplication.

This gives the framing of $TS^3$,

$$S^3 \ni g \longmapsto (g, gi, gj, gk) \in FS^3.$$

$FS^3$ is the frame bundle to $TS^3$, i.e. it consists of a point $g \in S^3$ together with three linearly independent vectors in $T_g S^3$.

You can visualize this framing in terms of the Hopf fibration.

enter image description here

The first vector field $(g,gi)$ is tangent to the fibers of the Hopf fibration, and the other two are normal to the fibration, where the normal vector fields wind about the fibers.

I think if you use right multiplication instead of left multiplication for all the definitions you get the opposite twisting of the normal vector fields.

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  • $\begingroup$ Thanks +1 - do you have comments about How do we construct the explicit class of 𝐙/24 and 𝐙/2 class of Lie group framing above? $\endgroup$
    – zeta
    Commented Dec 14, 2023 at 0:35
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    $\begingroup$ I think the $\Omega_3^{fr}$ is $\pi_8 S^5$, so you're talking about a 3-sphere embedded in $\mathbb R^8$ with a framing of its normal bundle. In this dimension all embedded $3$-spheres are isotopic, so it's the standard linearly embedded sphere, and the framings are parametrized by $\pi_3 SO_5 \simeq \pi_3 SO_4 \simeq \mathbb{Z}^2$. The language "Lie group framing" appears to have been a choice of terminology by Martin Olbermann, the person that wrote that Manifold Atlas page you link to. It might require a conversation with him to verify, but what I suspect he means is that there is... $\endgroup$
    – Ryan Budney
    Commented Dec 14, 2023 at 1:00
  • $\begingroup$ Thanks so much, I also asked this mathoverflow.net/q/460345/505775 maybe you know the differences? $\endgroup$
    – zeta
    Commented Dec 14, 2023 at 1:02
  • $\begingroup$ a homomorphism $S^3 \to SO_4 \to SO_8$ that acts transitively on this linearly embedded $S^3$ in $\mathbb R^8$, so the normal framing comes from taking a basis for the normal bundle and translating it around. $\endgroup$
    – Ryan Budney
    Commented Dec 14, 2023 at 1:02

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