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In Witten's 1989 QFT and Jones polynomial paper, he said

Although the tangent bundle of a three manifold can be trivialized, there is no canonical way to do this.

So if I understand correctly, Witten meant:

3-manifold has no canonical framing.

But in Atiyah's paper "On framings for 3-manifolds" , Atiyah said in his Proposition 1):

every oriented 3-manifold has a canonical 2-framing.

I am puzzled by why 3-manifold has no canonical framing but 3-manifold has a canonical 2-framing. Could someone explain the reason and intuition as simple as possible?

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    $\begingroup$ Presumably the short answer is Witten did not know of a canonical framing at the time, and then Atiyah showed him one. $\endgroup$
    – Ryan Budney
    Commented Dec 7, 2023 at 6:36
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    $\begingroup$ And of course, the two mean different things by "framing". $\endgroup$
    – Ryan Budney
    Commented Dec 7, 2023 at 6:43

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The answer is a paper aptly titled Canonical framings for 3-manifolds by Rob Kirby and Paul Melvin.

Here is a 2012 email correspondence between me and Kirby concerning that paper (glad I didn't walk over to his office otherwise I may have forgotten verbally):

  • Chris: For a spin 3-manifold M, the spin structure s+s on TM+TM is independent of the choice of spin structure s on TM. Could you explain why?
  • Kirby: Here is a down to earth way to think about it. A spin structure is a trivialization over the 1-skeleton which extends (you don't care how) over the 2-skeleton. There are two ways to trivialize over a circle, corresponding to \pi_1 of S(n) n>2, and in the case of n=2 or 1 we work mod 2. Then on s+s, if you change the trivialization of s over a circle, then you double that change for s+s, which mod 0 is no change.
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This is maybe not the answer you're looking for, but it's certainly too long for a comment.

First, there's no contradiction here, and nothing too strange. A 2-framing on $M$ is a framing of $TM\oplus TM$, which a priori doesn't have much to do with $TM$ (and moreover it comes with a splitting, which $TM$ doesn't).

Anyway, let me give an answer by analogy. Consider the case of $M = S^1$. Then $TM$ is trivial, but not canonically so: in this case a framing is the same thing as an orientation, or as a nowhere-vanishing section. $TM\oplus TM$ is also trivial, and either trivialisation of $TM$ induces a "diagonal" trivialisation of $TM \oplus TM$. (Note that Atiyah talks about the diagonal embedding $\Delta\colon SO(3) \to SO(3) \times SO(3)$, which then embeds in $SO(6)$ via an embedding $\iota$.) Moreover, the two induced trivialisations differ by multiplication by $-1$, which on a rank-2 bundle is isotopic to the identity. So the two trivialisations induced are equivalent (as trivialisations of a rank-2 bundle).

EDIT (What follows is slightly speculative.) This should boil down to showing that for every compact oriented 3-manifold $M$ and every map $\tau \colon M \to SO(3)$ (which measures a difference in trivialisations) the composition of maps $\iota\circ\Delta\circ\tau$ is homotopic to a constant map. This is non-trivial even in the case of $M = S^3$, since $\pi_3(SO(6)) = \mathbb{Z}$. What mme (is saying that Atiyah) is saying is that, for a 3-manifold $M$, the composition $\iota\circ\Delta\circ\tau$ gives an identification of the induced framings of $TM\oplus TM$ with $\mathbb{Z}$, and in particular there is a preferred framing (the one that corresponds to $0 \in \mathbb{Z}$). See mme's comments below for more details. (Thanks, mme!)

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  • $\begingroup$ I don't think the last claim is true. Take $\tau$ to be the identity to see you're asking if the diagonal embedding $SO(3) \to SO(6)$ is null. There are many ways to see that this is false, but here's one: the standard embedding induces an isomorphism on $\pi_3$, and the embedding as the bottom-right block is homotopic to that. Your map is the product of those, so it suffices to show that multiplication induced the sum map on $\pi_3$, which is the Eckmann-Hilton argument. Thus the induced map on $\pi_3$ is injective (it's multiplication by 2, if you like). $\endgroup$
    – mme
    Commented Dec 7, 2023 at 13:45
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    $\begingroup$ Looking at Atiyah's paper, he's not claiming that there's one framing up to homotopy (which appears to be the goal of your last paragraph), he's claiming the framings carry a canonical bijection to the integers, this bijection arising from the Hirzebruch signature formula. $\endgroup$
    – mme
    Commented Dec 7, 2023 at 14:19
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    $\begingroup$ Oh, nice, thanks Mike! I trusted the OP had the question right... So there's no canonical 2-framing, either, just a canonical bijection with $\mathbb{Z}$, which maybe isn't too far from what is in my answer? (I will edit, or maybe even delete the answer, when I'm sure I got what is needed.) $\endgroup$
    – Marco Golla
    Commented Dec 7, 2023 at 14:49
  • $\begingroup$ Please don't delete the answer! It provides great clarity on the difference between framing $TM$ vs $TM \oplus TM$. $\endgroup$
    – Alon Amit
    Commented Dec 7, 2023 at 15:42
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    $\begingroup$ @MarcoGolla Well, $0 \in \Bbb Z$ is a pretty canonical element, so I'm satisfied with Atiyah calling this a "canonical framing". (More precisely, the bijection is determined by sending a 2-framing $\alpha$ to the integer $\sigma(Z) - p_1(Z, \alpha)$, where $Z$ is an oriented 4-fold bounding $Y$; the `canonical framing' for which this is zero is the one for which Hirzebruch's signature formula is still true.) $\endgroup$
    – mme
    Commented Dec 7, 2023 at 15:56

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