# Holonomy group of quotient manifold

Let $(M,g_M)$ be a compact Riemannian manifold with holonomy group $Hol(M,g_M)$. Suppose that a finite group $G$ acts on $M$ freely and preserves the metric $g$.

What can one say about the holonomy group $Hol(M/G,g_{M/G})$ of $M/G$ equipped with the induced metric $g_{M/G}$?

Are there any good conditions to guarantee $Hol(M,g_M)\cong Hol(M/G,g_{M/G})$?

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Why do you think they would be any different? – Bombyx mori Jan 1 '13 at 1:12
Without any condition, they must be different. For example the fundamental group of a manifold affect its holonomy group. – M. K. Jan 1 '13 at 2:11
@user32240: For a concrete example of what M.K. said, consider that $S^2$ has holonomy group $SO(2)$ while $\mathbb{R}P^2$ has holonomy group $O(2)$. – Jason DeVito Jan 1 '13 at 2:30
Here's an almost trivial thought: If $M^{2n+1}$ is compact with $SO(2n+1)$ holonomy and positive sectional curvature, then $M/G$ will also have $SO(2n+1)$ holonomy. This follows from Synge's theorem which states that a positively curved compact odd dimensional manifold is orientable. The analogous statement for $M^{2n}$ is that $M/G$ has $O(2n)$ holonomy. – Jason DeVito Jan 1 '13 at 2:38
Similar to Jason's comment, but if $M$ is a K\"ahler manifold it has holonomy $U(n)$, and we know that $M/G$ is also a K\"ahler manifold, then $Hol(M/G)\subset U(n)$ by the holonomy principle. But $U(n)=Hol(M)\subset Hol(M/G)$ (at least this is certainly true if $M$ is simply connected, in which case $Hol(M) = Hol_{0}(M/G)$) and thus we have $Hol(M/G) = U(n) = Hol(M)$ – Daniel Mckenzie Apr 2 '14 at 22:01