Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $Hol(M,g_M)\cong Hol(M/G,g_{M/G})$?

share|improve this question
    
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 at 22:01

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.