Vtyjkyuk

Suppose I have a connected Riemannian manifold $X$ and a covering $pi:Yto X$ with the pulled back metric on $Y$, making $pi$ into a local isometry between Riemannian manifolds.

Suppose we have a closed loop in $Y$ based at a point $yin Y$. This will give rise to an element of the holonomy group $mathrmHol_y(Y)$, which after we fix a basis of $T_yY$ becomes an element of $mathrmGL(n)$ (or $mathrm O(n)$ really).

If we project the path to $X$, we get a closed loop again, hence an element of $mathrmHol_xX$, where $x = pi(y)$, and by using $pi$ to map the basis of $T_yY$ to one of $T_xX$ this becomes an element of $mathrmGL(n)$ as well, and we can talk about equality in a sensible way.

Since all along the paths the geometry is identical (around every point there is a neighbourhood on which $pi$ induces an isometry) it seems clear that they must induce the same element in holonomy.

This would mean that the holonomy of $Y$ is a subgroup of the holonomy of $X$.

This inclusion can be strict, if $X$ is not orientable for example, its holonomy group might be $mathrm O(n)$, but if we take $Y$ to be its oriented double cover its holonomy must be contained in $mathrmSO(n)$.

Can anything more precise be said about the relation between the holonomy groups of $X$ and of $Y$? I would expect the fundamental group or rather the group of deck transformations to play a role, but I don't know in what form.

Thanks!

$rm Hol(Y)$ is indeed a subgroup of $rm Hol(X)$. The fundamental group introduces connected components to the holonomy; that is, there is a surjection $pi_1(X)torm Hol(X)/rm Hol(X)^circ=rm Hol(X)/rm Hol(tildeX)$ (with $tildeX$ being the universal cover). I'm not aware that this can be made more precise in general (if you read German, have a look at Helga Baum's book "Eichfeldtheorie", Section 5.1; can't think of a good English reference now, but something on this is probably in Kobayashi & Nomizu, Book I).

Example: For a complete flat manifold $X$, the fundamental group is a group of affine transformations on $tildeX=mathbbR^n$. The (linear) holonomy group is then given by the projection to the linear parts of the fundamental group
(e.g. Joe Wolf's book "Spaces of constant curvature", Chapter 3).

Hope this helped.