Question

By the Pati-Salam group I refer to SU(2) x SU(2) x SU(4). It can be obtained as the group of isometries of the 8 dimensional manifold $S^3 \times S^5$, but I wonder if this is the only 8 dimensional manifold having this group of isometries.

This particular manifold is interesting because a quotient by any U(1) will produce a 7 dimensional manifold whose isometry group is the unbroken standard model group, as pointed out by Witten time ago. But my particular curiosity comes because Non Commutative Geometry gets the Pati Salam group from a different setup: the finite algebra $M_2(H) \oplus M_4(C)$, related perhaps to deformations of even spheres.

Answer 1

$S^3 \times S^5$ has isometry group $SO_4(\mathbb{R}) \times SO_6(\mathbb{R})$, which has $SU(2) \times SU(2) \times SU(4)$ as a four-fold cover. Since it appears that you aren't worrying too much about central terms, we can replace $S^3$ with $\mathbb{R}P^3$, $S^5$ with $\mathbb{R}P^5$, or take a quotient by a diagonal group of order 2.

I'm pretty sure these are the only connected choices, because we can characterize homogeneous orbits by the stabilizers of points. In this case, you need a closed subgroup of Pati-Salam of dimension at least 13 whose intersection with each factor group is not the whole factor. There just aren't that many subgroups of suitably large dimension: we need a diagonally embedded $SU(2)$ (possibly with a central translate) to get dimension at least 3 in the first factors, and we need Spin(5) in the last factor to get dimension at least 10. This forces the orbits to be connected components.