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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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  • $\begingroup$ Why do you think they would be any different? $\endgroup$
    – Bombyx mori
    Jan 1, 2013 at 1:12
  • $\begingroup$ Without any condition, they must be different. For example the fundamental group of a manifold affect its holonomy group. $\endgroup$
    – M. K.
    Jan 1, 2013 at 2:11
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    $\begingroup$ @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 $O(2)$. $\endgroup$
    – Jason DeVito - on hiatus
    Jan 1, 2013 at 2:30
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    $\begingroup$ 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. $\endgroup$
    – Jason DeVito - on hiatus
    Jan 1, 2013 at 2:38
  • $\begingroup$ 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 $Hol(M/G) = U(n) = Hol(M)$ $\endgroup$
    – Daniel Mckenzie
    Apr 2, 2014 at 22:01

1 Answer 1

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Perhaps you can get an idea of the difficulty of your question by considering quotients of a flat torus. Have a look of the following papers:

http://terpconnect.umd.edu/~wmg/Bieberbach.pdf

http://mat.ug.edu.pl/~aszczepa/problemsfv.pdf

cheers, h.

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