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I'm unsure whether this question is appropriate for mathoverflow, so feel free to criticize.

All manifolds are closed, smooth and have dimensions $n\ge 5$.

The Atiyah-Shapiro-Bott-Orientation gives a ring homomorphism $$\alpha\colon\Omega_*^{spin}\rightarrow KO^{-*}(pt),$$ from the spin-bordism ring to real K-theory, whose vanishing is a necessary condition for a spin manifold admitting a metric of positive scalar curvature. Recall $$KO^{-n}(pt)\cong \begin{cases} \mathbb{Z} &\mbox{if } n \equiv 0,4 (8)\\ \mathbb{Z}/2\mathbb{Z} & \mbox{if } n \equiv 1,2(8) \\ 0 &\mbox{if } n \equiv 3,5,6,7 (8). \end{cases}.$$ In dimensions $n\equiv0,4(8)$, $\alpha$ is just the $\hat{A}$-genus (respectively twice of it).

Considering the spin-cobordism class of homotopy spheres, the $\alpha$-invariant induces homomorphisms $$\beta_n\colon\Theta_n\rightarrow KO^{-n}(pt),$$ ($\Theta_n$ is the group of $n$-dimensional homotopy spheres), which are zero in dimensions $n\equiv 3,5,6,7 (8)$ for trivial reasons.

In the other dimensions, we have

  1. $\beta_n$ is zero in dimensions $n\equiv0,4(8)$, what is equivalent to the vanishing of the $\hat{A}$-genus for all homotopy spheres.
  2. $\beta_n$ is surjective in dimensions $n\equiv1,2(8)$.

I am trying to better understand these two claims. Lawson and Michelson simply write in their book "Spin Geometry", that (2) follows from "deep work of Adams and Milnor".

Since I am absolutely not an expert in this field, can someone elaborate a bit (more than "It follows from the work of Adams and Milnor.", what is not really helpful for me.) on what I really need to prove (2) and how one can place it in a wider context?

How can one prove (1) and is there a reference for it?

I searched the literature without good results, so simply giving references might also answer my question.

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If $M$ is a homotopy sphere of dimension $4k>0$, then the signature is clearly zero. By the Hirzebruch signature theorem, you get $0=\langle L_k (TM); [M] \rangle = b_k \langle p_k (TM); [M] \rangle$ for a certain number $b_k \neq 0$. Therefore, the Pontrjagin classes of $TM$ are all trivial, and hence the $\hat{A}$-genus is zero. This settles part (1).

Part (2) is harder. Adams proved in his $J(X)$ papers (it is one of the main results stated in the introduction to part IV) that the unit map from the sphere spectrum to $KO$ is surjective in homotopy of degrees $8k+1$ and $8k+2$. The unit map $\mathbb{S} \to KO$ factors as

$$\mathbb{S} \to MSpin \to KO$$

with the first map the unit and the second the Atiyah-Bott-Shapiro orientation. All that is needed to show this claim is that the $\alpha$-invariant of a point is $1$ (this is a statement about the homomorphism $\pi_0 (MSpin) \to \pi_0 (KO)$).

Therefore, there exists framed manifolds of dimension $8k+1$ and $8k+2$ whose $\alpha$-invariants (as spin manifolds) are nonzero.

Now use the work of Kervaire-Milnor. They prove in ''Groups of homotopy spheres'' that each odd-dimensional framed manifold is framed cobordant to a homotopy sphere, if the dimension is at least $5$. Take a framed manifold of dimension $8k+1$ with nonzero $\alpha$ and replace it by a homotopy sphere cobordant to it. Therefore, $\beta_n$ is surjective if $n = 8k+1$ (for $k=0$, this is also true).

In dimensions $8k+2$, a framed manifold is framded cobordant to a homotopy sphere if and only if its Kervaire invariant is zero. This is why I do not see a short argument in that case.

EDIT: let me remark that $\beta_2$ is not surjective: the unique spin structure on $S^2$ has $\alpha=0$; to get a nonzero $\alpha$, you need the torus.

EDIT II: in dimensions 8k+2, the argument (due to Milnor, I believe) is as follows. Take a homotopy sphere $\Sigma^{8k+1}$ with nonzero $\alpha$-invariant, and take the product $M=\Sigma \times S^1$, where $S^1$ has the spin structure with nonzero $\alpha$-invariant. Then $M$ has nonzero $\alpha$-invariant as well. Then do surgery on the embedded $x \times S^1$; the result is a homotopy sphere, and the $\alpha$-invariant is unchanged.

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  • $\begingroup$ Minor correction: "[...]whose $\alpha$-invariants (as spin manifolds) are NONzero." $\endgroup$ – archipelago Jan 3 '15 at 12:03
  • $\begingroup$ Thank you for your second editing, which was enlightening for me. $\endgroup$ – archipelago Feb 17 '15 at 22:48

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