# What happens to M2 and M5 brane solutions upon orbifolding?

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M-theory has M2 and M5 brane solutions. Suppose M-theory is compactified on $\mathbb{R}^{10} \times S^{1}/\mathbb{Z}_2$, what happens to the M2 and M5 brane solutions? How does one define the near horizon limit in this case?

The M2 brane in d = 11 has a near horizon limit of $AdS_4 \times S^7$. Upon orbifolding by $\mathbb{Z}_2$, would this translate to $AdS_4 \times S^{7}/\mathbb{Z}_2$?

The physical picture of Horava-Witten theory corresponds to an M2 brane somehow stretched between end of the world 9-branes at the orbifold singularities. Does this mean that there are now two near-horizon limits? [Here is a link to the picture being referred to.]

This post imported from StackExchange Physics at 2015-06-24 12:05 (UTC), posted by SE-user leastaction

asked Jun 22, 2015
edited Jun 24, 2015

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