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I have been searching the literature for a construction of a simply connected spin manifold of dimension 8 with A-genus 1. I am not sure, but I think this is called a Bott manifold. Can anybody help me with a reference? Are there in the literature index computations for this type of manifold? Thank you for your help!

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One construction is somehow hidden in Kervaire-Milnors work on homotopy spheres. A textbook reference is Kosinski: ''Differential manifolds''. In section IX.8 (Theorem 8.7), you find the statement that there exists an 8-dimensional manifold which is almost parallelizable (i.e. parallelizable away from a point) whose signature is $8 \cdot 28$. Because this $M$ is almost parallelizable, $p_1 (TM)=0$, and from the formulae for the $\hat A$-class and the $L$-class, you get that $\hat{A} =1$. Typically, one wants that the signature is zero, and this you can achive by connected sum with $8 \cdot 28$ copies of $\overline{HP^2}$.

How is $M$ constructed? You take the $E_8$-plumbing manifold $V$. It is a $3$-connected $8$-manifold which is parallelizable, which has signature $8$ and whose boundary is a homotopy sphere. In fact, $\partial V$ generates the group $\Theta_7 \cong Z/28$ of exotic $7$-spheres. Now you form the boundary connected sum of $28$ copies of $V$; the boundary of the resulting manifold is the standard $S^7$, and you glue in a copy of $D^8$ to obtain $M$.

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  • $\begingroup$ Thank you! Is it known how does this manifold relate to $K^2/4$ used by Laures? In fact, what precisely is meant by "$/4$"? Also, a very similar construction is given in Hirzebruch-Berger-Jung (page 92), is this the same? $\endgroup$ May 28, 2014 at 6:00
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    $\begingroup$ The spin bordism group of 8-dimensional manifolds is $Z \oplus Z$, detected by the $\hat{A}$-genus and the signature. A basis is given by a Bott manifold $B$ and $HP^2$. The K3-surface K has $\hat{A} (K^2) =4$ and $\sign (K^2)=16^2$. Therefore $[K^2] = 16^2 [HP^2]+ 4 [B]$, and this yields Laures formula (note that $[K^2]$ is divisible by 4). $\endgroup$ May 28, 2014 at 9:43
  • $\begingroup$ I see. What I'm ignorant about is whether actually $[K^2]/4$ can be realized by an explicit construction (I don't know, maybe $K^2$ is a 4-fold covering of some other manifold, or something :) ). $\endgroup$ May 28, 2014 at 9:48
  • $\begingroup$ Another question - can $B$ be realized as a homogeneous space $G/H$ for some Lie groups $H<G$? In general, are there some algebraic conditions known for a manifold to be so realizable? $\endgroup$ May 28, 2014 at 9:50

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