I'm just curious from Berger's classification of Riemannian holonomy, how do Spin(7) manifolds intersect the other types of Riemannian manifolds?
In particular, what is the intersection of Spin(7) and U(4), seen as subgroups of SO(8)?
6
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4$\begingroup$ I would say it is $SU(4)$, since $SU(4)\subset Spin(7)$. In fact, $SU(4)$ is another of the few holonomy groups admissible for an irreducible Riemannian manifold of dimension eight. $\endgroup$– BilateralCommented Mar 18, 2015 at 18:47
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6$\begingroup$ Of course, this depends on which conjugates of $\mathrm{Spin}(7)$ and $\mathrm{U}(4)$ in $\mathrm{SO}(8)$ you choose to intersect. For instance, is not true that all the conjugates of $\mathrm{SU}(4)$ lie inside some fixed conjugate of $\mathrm{Spin}(7)$ $\endgroup$– Robert BryantCommented Mar 18, 2015 at 19:07
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$\begingroup$ @Bilateral and Robert Bryant: thank you both. $\endgroup$– seubCommented Mar 18, 2015 at 19:17
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$\begingroup$ So Spin(7) does not contain U(4) right? And if I am allowed to ask more: I suppose there are Ricci-flat 8-manifolds that are not Spin(7)? $\endgroup$– seubCommented Mar 18, 2015 at 19:22
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2$\begingroup$ The maximal abelian subgroups of $\mathrm{Spin}(7)$ all have dimension $3$, so it cannot contain even the maximal torus of $\mathrm{U}(4)$, let alone the whole thing. Yes, there are Ricci-flat Riemannian $8$-manifolds with holonomy not contained in $\mathrm{Spin}(7)$. I'm not aware of a compact one, though. $\endgroup$– Robert BryantCommented Mar 18, 2015 at 20:26
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