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Let $(X,g)$ be a Riemannian manifold with holonomy group $Hol(X,g)$. Suppose that a finite group $G$ acts on $X$ freely and the metric $g$ is invariant under $G$. What can one say about the the holonomy group of the Riemannian manifold $(X/G,\tilde{g})$, where $\tilde{g}$ is the metric induced by $g$?

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All of the loops upstairs project to loops downstairs, with the same holonomy, so the holonomy group of the quotient contains that of the original manifold. For example, the holonomy group of the sphere is the rotation group, while that of real projective space is the orthogonal group; you can draw the pictures for the 2-sphere and see the effect of parallel transport and antipodal map.

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    $\begingroup$ ...and, furthermore, the holonomy group of the quotient contains the original one as a finite index subgroup (index is at most $|G|$). $\endgroup$ – Misha Apr 27 '13 at 10:05
  • $\begingroup$ Ben's example with the real projective space only works as described if the dimension is even. In odd dimension the real projective space is orientable and thus the holonomy is the special orthogonal group. $\endgroup$ – Bernd Ammann Jun 20 '17 at 18:56
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In the presence of a parallel spinor and irreducible restricted holonomy, the full holonomy groups were classified by Mackenzie Wang together with previous work by McInnes. However these methods do not apply to the reducible case. References are M. Wang, On non-simply connected manifolds with non-trivial parallel spinors, Ann. of Global Anal. Geom. 13 (1995), no. 1, 31–42 and B. McInnes, Methods of holonomy theory for Ricci-flat Riemannian manifolds, J. Math. Phys. 32, (1991), 888–896.

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