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The Pontryagin-Thom construction allows one to identify the stable homotopy groups of spheres with bordism classes of stably normally framed manifolds. A stable framing of the stable normal bundle induces a stable framing of the stable tangent bundle.

This means that a framed manifold (one whose tangent bundle is trivial, e.g. a Lie group) represents an element of the stable homotopy groups of spheres.

So some elements are represented by honestly framed manifolds (as opposed to stably framed).

What is known about such elements? Is every element of the stable homotopy groups of spheres represented by an honestly framed manifold (i.e. with a trivial tangent bundle)?

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    $\begingroup$ At least for Lie group framings there are some articles: core.ac.uk/download/pdf/82048719.pdf, projecteuclid.org/euclid.jmsj/1468956166 $\endgroup$ – Panagiotis Konstantis Jun 26 at 6:19
  • $\begingroup$ Am I confused or doesn't every tangential framing on $S^1$ give the trivial stable homotopy class? $\endgroup$ – Lennart Meier Jun 26 at 7:17
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    $\begingroup$ No, if you take the Lie group framing you obtain the generator of $\Omega_1^{\rm{fr}}$. If you would like to obtain the trivial element in this framed bordism group, then you need to trivialize the stable tangent bundle accordingly. With other words: Not every stable framing is induced by a framing of the tangent bundle. $\endgroup$ – Panagiotis Konstantis Jun 26 at 7:24
  • $\begingroup$ Suppose $S^1 \subset \mathbb R^n$ and suppose furthermore you have the Lie group framing on $S^1$ than you obtain a framing on the normal bundle of $S^1$. But this framing has to be in such way that if you compare it to the standard background framing of $\mathbb R^n$ you obtain a map $S^1 \to \mathrm{SO}(n)$ which has to be the trivial homotopy class (since you want to obtain the standard framing of $\mathbb R^n$). Pontryagin showed that the stable homotopy class is given by this homotopy class $S^1 \to \mathrm{SO}(n)$ in $\pi_1(\mathrm{SO}(n))$ plus the generator of that group $\endgroup$ – Panagiotis Konstantis Jun 26 at 8:14
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    $\begingroup$ All the useful comments were deleted. Maybe they were wrong, but at least they were trying. You have to take risks! Did Ryan's comment lead to the Kervaire invariant 1 problem? Is that what @archipelago was talking about? $\endgroup$ – Ben Wieland Jun 26 at 18:16
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I think all elements are representable by honestly framed manifolds.

Let $M$ be a closed $d$-manifold with a stable framing, and consider the obstructions to destabilising a stable framing. Asumng $M$ is connected, which we can arrange by stably-framed surgery, there is a single obstruction, lying in $H^d(M ; \pi_d(SO/SO(d)))$.

If $d$ is even then $\pi_d(SO/SO(d)) = \mathbb{Z}$ and this obstruction may be identified with half the Euler characteristic of $M$. (As $M$ is stably framed, its top Stiefel--Whitney class vanishes and so its Euler characteristic is even.) We can change $M$ to $M \# S^p \times S^{2n-p}$ by doing a trivial surgery in a ball, and the stable framing extends over the trace of such a surgery. By taking $p$ to be 1 or 2 we can therefore change the Euler characteristic by $\mp 2$: thus we can change $M$ by stably framed cobordism until its Euler characteristic is 0, whence the stable framing destabilises to an actual framing.

If $d$ is odd then then $\pi_d(SO/SO(d)) = \mathbb{Z}/2$ and the obstruction is obscure to me (it is realised by the stable framing induced by $S^d \subset \mathbb{R}^{d+1}$, and is non-trivial even in Hopf invariant 1 dimensions where $S^d$ does admit a framing). I can't see an elementary argument for $d$ odd, but I think it is nontheless true by the following.

Let $d=2n+1$ with $d \geq 7$ (lower dimensions can be handled manually). Consider the manifold $$W_g^{2n} = \#g S^n \times S^n.$$ This has a stable framing by viewing it as the boundary of a handlebody in $\mathbb{R}^{2n+1}$. By doing some trivial stably-framed surgeries as above (with $p=2,3$ say, to keep it simply-connected), we can change it by a cobordism to a manifold $X$ having an honest framing $\xi$. I wish to apply [Corollary 1.8 of Galatius, Randal-Williams, ``Homological stability for moduli spaces of high dimensional manifolds. II"], to $(X, \xi)$. There is a map $$B\mathrm{Diff}^{fr}(X, \xi) \to \Omega^{\infty+2n} \mathbf{S}$$ given by a parameterised Pontrjagin--Thom construction. Now there is a step that I would have to think about carefully, but I think that the choices made can be arranged so that $(X,\xi)$ has genus $g$ in the sense of that paper, and so taking $g$ large enough the map above is an isomorphism on first homology. But this has the following consequence: any element $x \in \pi_{2n+1}(\mathbf{S})$ is represented by the total space of a fibre bundle $$X \to E^{2n+1} \overset{\pi}\to S^1$$ with a framing of the vertical tangent bundle (and the Lie framing of $S^1$).

(Again, I'm sure there must be a more elementary way of seeing this.)

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    $\begingroup$ It seems to me that the obstruction in odd dimension could be the semi-characteristic of the manifold. Bredon and Kosinski showed in (jstor.org/stable/1970531) that a stably framed manifold of dimension $n$ admits $n$ linear independent vector fields if only if the ($\mathbb Z_2$-valued) semi-characteristic vanishes. $\endgroup$ – Panagiotis Konstantis Jun 26 at 20:52
  • $\begingroup$ Forgive me if I am just stating the obvious here, but wouldn't the GMTW-theorem for cobordisms with honest framings of the tangent bundle immediately show that all are represented by manifolds which have tangent bundle trivial after adding a single trivial bundle. Obviously not as strong as the result you prove, but at least a sanity check. $\endgroup$ – Connor Malin Jul 4 at 13:17
  • $\begingroup$ Ah, after reading @archipelago's answer I realize that since tangent bundles are the same rank as the dimension of the manifold, you are stably trivial if and only if adding a single trivial bundle makes you trivial. $\endgroup$ – Connor Malin Jul 4 at 13:36
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Repeating the first part of Oscar's answer and elaborating on comments by Chris and Panagiotis, here is a down-to-earth argument in all cases:

These cases $n=1,3,7$ are fine, since in these cases the stable stems are generated by $S^1$, $S^3$, $S^7$ with the unstable framing induced by the multiplication in the unit complex numbers, quaternions, or octonions.

In the other cases, we use that the obstruction to destabilising a given stable framing $F$ of an oriented closed manifold $M^n$ lies in $H^n(M,\pi_n(SO/SO(d))$, which is isomorphic (in a preferred way) to $\mathbb{Z}$ if $n$ is even and to $\mathbb{Z}/2$ if $n$ is odd. It is not too hard to see that, with respect to this isomorphism, the obstruction is given by the semi-characterstic: half the Euler characteristic for $n=2d$ and $\sum_{i=0}^d\mathrm{dim}(H_i(M,\mathbb{Z}/2))\text{ mod }(2)$ for $n=2d+1$ and $n\neq1,3,7$. In particular, the obstruction to destabilising is independent of $F$ which is somewhat surprising.

Originally this was proved by to Bredon and Kosinksi [1] who used a more geometric description of this obstruction: it is the degree (mod $2$ if $n$ is odd) of the Gauss map $M\rightarrow{S^n}$ induced by the stable framing $TM\oplus \varepsilon\cong \varepsilon^{n+1}$ (take the image of the canonical vector field in the trivial line bundle and normalize).

Now observe that, as Oscar explained, by doing a couple of trivial surgeries in a ball corresponding to taking connected sums with $S^1\times S^{n-1}$ or $S^2\times S^{n-2}$ and extending the stable framing, any stably framed bordism class in even dimensions contains a representative with trivial Euler-characteristic. The same works with the semi-characteristic in odd dimensions (here at most one surgery is necessary), so by the discussion above every stably framed bordism class has a representative whose stable framing can be destabilised.

[1] G.E. Bredon and A. Kosinski, Vector fields on $\pi$-manifolds. Annals of Math. 84, 85– 90 (1960).

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  • $\begingroup$ Could you explain why your last paragraph is true? $\endgroup$ – Panagiotis Konstantis Jun 27 at 3:44
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    $\begingroup$ Oscar explained it in his answer in the even-dimensional case and the same works in odd dimensions. I added it to the answer. $\endgroup$ – archipelago Jun 27 at 10:12
  • $\begingroup$ Thanks for this summary and for an easier argument in the odd dimensional case. $\endgroup$ – Chris Schommer-Pries Jun 27 at 17:36
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$k \cdot[\mathrm{point}]\in \pi_0^s$ is represented by an honestly framed 0-manifold if and only if $k \geq 0$.

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In terms of homotopy theory, the questions is that when an element of $\pi_n^s$ pulls back to an element of $\pi_{n+i}S^i$ for some $i$ and the answer to this question is positive by Freudenthal's suspension theorem as there is an epimorphism $$\pi_{2n+1}S^{n+1}\longrightarrow \pi_n^s$$ unless you put more restrictions, e.g. ask for some specific $i$. The point is that there could be two different pull backs whose normal bundles are not isomorphic, only stably isomorphic. An example is given by $\eta_3\in\pi_8^3$ which equals to $\eta\sigma=\sigma\eta$ in $\pi_*^s$. However, one can do some unstable computations to show that one pulls back one step further than the other. Hence, as unstable elements they are not the same really, but map to the same element. I think it could be interesting to sort this out in terms of bordism theory and I don't know if such specific examples are considered somewhere in the literature.

ADDED I think the answer still is positive. I think manifolds with tangential structures are understood in terms of Madsen-Tillmann spectra using the Madsen-Tillmann-Weiss map; experts can comment more on this and correct me if this is wrong or vague. In the case of trivialisation of the tangent bundle of $m$ dimensional manifolds, the relative spectrum is $\mathbb{S}^{-m}=\Sigma^{-m}S^0$. The general result of Galatius-Madsen-Tillmann-Weiss provides an interpretation of $\pi_i\Omega^\infty\mathbb{S}^{-m}$ in terms of specific submersions (I guess). Now, the point is that $\pi_i^s\simeq\pi_{i-m}\mathbb{S}^{-m}$ for any $m>0$ and I think again using Freudenthal's theorem one can see the answer is positive.

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    $\begingroup$ No, the question is whether the class is represented by a manifold with trivial tangent bundle, not trivial normal bundle. $\endgroup$ – Chris Schommer-Pries Jun 25 at 19:54
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    $\begingroup$ Your edit does not answer the question: roughly speaking, MT-spectra encode bordism groups of manifolds equipped with a vector bundle with some tangential structure that is stably equivalent to the tangent bundle, not tangential structures on the tangent bundle itself. $\endgroup$ – archipelago Jun 26 at 14:48

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