# M-theory compactified on $S^4$

I am trying to understand what happens to the 32 supersymmetries of M-theory when it is compactified on $S^4$, with the other seven dimensions being flat, e.g., $\mathbb{R^7}$, $\mathbb{R^6} \times S^1$, or $\mathbb{R^5} \times S^1 \times S^1$. From what I have read, $S^4$ admits Killing spinors, so I believe some number of supersymmetries should be preserved in the 11-dimensional spacetime of M-theory. My question is, how many supersymmetries are actually preserved?

In this case, we are interested in supersymmetric backgrounds $\mathrm{AdS}_7 \times X^4$, with $X$ possibly an orbifold. Bär's construction, together with the non-existence of irreducible five-dimensional holonomy representations, imply that the only (complete) four-dimensional manifold admitting real Killing spinors is the round sphere $S^4$, hence any other supersymmetric background must be an orbifold of $S^4$ by a finite subgroup of $\mathrm{SO}(5)$, lifting isometrically to a subgroup $\Gamma < \mathrm{Sp}(2)$. The space of Killing spinors is again identified with the $\Gamma$-invariant parallel spinors on $\mathrm{R}^5$ which is the irreducible spinor representation $\Delta$ of $\mathrm{Sp}(2)$. This representation is quaternionic, whence the space of $\Gamma$-invariant spinors is a quaternionic subspace: if a spinor is invariant, so is its quaternion line, by the quaternion-linearity of the action of $\mathrm{Sp}(2)$. Since $\dim_\mathbb{H}\Delta =2$, necessarily $0\leq \dim_\mathbb{H}\Delta^\Gamma \leq 2$. Hence if we demand some supersymmetry, either $\Gamma = \{1\}$ and we have $X = S^4$, or else the orbifold is half-BPS. In this case $\Gamma$ is contained in an $\mathrm{Sp}(1)$ subgroup of $\mathrm{Sp}(2)$, leaving a nonzero vector invariant in the fundamental representation of $\mathrm{Sp}(2)$. Up to automorphisms, we see that $\Gamma$ is one of the ADE subgroups in Table 1, but this time embedded in $\mathrm{Sp}(2)$ in such a way that if $u \in \Gamma < \mathrm{Sp}(1)$ and $(x,y)\in\mathbb{H}^2$, then $u \cdot (x,y) = (ux,y)$.
The action of that $\mathrm{Sp}(1)$ subgroup of $\mathrm{SO}(5)$ on $S^4$ is given by restricting the action on $\mathrm{RR}^5$. This is given as follows. First of all, there is a vector which is fixed, call it $\nu$. If we identify the four-dimensional subspace perpendicular to $\nu$ with $\mathbb{H}$, then the action of $\mathrm{Sp}(1)$ is by left quaternion multiplication. Finally, using the arguments described in Section 7.1 and in particular Table 11, it is a simple exercise to decompose such orbifolds $S^4/\Gamma$, except for $\Gamma = \mathbb{E}_8$, into a sequence of cyclic quotients. This should become useful if and when we understand the six-dimensional superconformal field theory dual to $\mathrm{AdS}_7 \times S^4$.
• I think it should be added that $S^4$ admits four Killing (Dirac) spinors. – Meer Ashwinkumar Sep 13 '16 at 6:58