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Let $G$ be a (connected) Lie group, $X,Y$ homogeneous $G$-manifolds, and $p:Y\to X$ a $G$-equivariant covering space. Now, above a point $x\in X$ there are two actions on $S=p^{-1}(x)$:

First, one has monodromy with $\pi_1(X)$ acting on $S$. Since the map is $G$-equivariant, the image of $\pi_1(G)\to\pi_1(X)$ given by $G\to X, g\mapsto gx$ acts trivially. Plugging this into the homotopy long exact sequence of the fibration $C_G(x)\to G\to X$, we get that this action canonically factors through $\pi_0(C_G(x))=C_G(x)/C_G(x)^0$.

Second, one also has $C_G(x)$ acting on $S$ simply by permuting $S$. The connected component $C_G(x)^0$ acts trivially since it's connected and $S$ is discrete, so this action factors through $C_G(x)/C_G(x)^0$.

This gives two natural actions of $C_G(x)/C_G(x)^0$ on the fiber $S$, one constructed globally and one locally. Can one, under some suitable conditions, conclude that these are the same?

someone
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  • Just to be sure, what exactly do you mean by a $G$-equivariant covering space? Do you mean a representation of $G$ on deck transformations $G\rightarrow \text{Deck}(p)$, or that the $G$-actions on $X$ and $Y$ respect the covering space structure ($p(x\cdot g)=p(x)\cdot g$)? – Diana Pestana May 03 '25 at 03:12
  • @DianaPestana That the covering map is $G$-equivariant, that is, the latter – someone May 04 '25 at 08:30

1 Answers1

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If you consider that the fibers $S=p^1(\{x\})$ in a covering space are discrete, which is usually considered to be part of the definition. Then, for $G$ a connected(path-connected) Lie Group, if you let $X$ be a $G$-manifold and $p:Y\rightarrow X$ a covering space, then there is only one possible continuous action of $G$ on $Y$ which makes this into a $G$-covering space, that is, making $p$ $G$-equivariant: $$ p(y\cdot g)=p(y)\cdot g \qquad \forall g\in G, y\in Y $$

This $G$-action on $Y$ is given by the monodromy, it exists by the homotopy/path lifting theorem (which works for any fiber-bundle or Serre Fibration), and is also unique by the Unique Lifting Theorem (which works only for covering spaces). You might want to check on the latter, it's shown and proven in Gregory L. Naber's Topology, Geometry and Gauge Fields: Foundations as theorem 1.5.12 pg. 82 of 2nd edition.

You may see this by supposing points $x\in X,y\in Y$ such that $p(y)=x$. Then a path in $G$ starting at the identity and ending in $g$ would yield a path in $X$ starting in $x$ and a path in $Y$ starting at $Y$, by the continuous $G$-actions. Because the covering space is $G$-equivariant, the path in $Y$ must be a lift of the path in $X$. By the Unique Lifting Theorem the path on $Y$ is unique. If you consider two $G$-actions on $Y$ they must lift to the same path, so they must end that the same point $y\cdot_1 g=y\cdot_2 g$, and thus the both $G$-actions must agree on their path-connected components. If G is a connected Lie Group, the it's path connected, and as such all $G$-actions on $Y$ must be the same.

This in turn implies that both your $G$-actions on $Y$ must be the same, so both actions of $C_G(x)\big/C_g(x)^0$ on the fibers $S$ will always be the same: the trivial action, as the stabilizer of $X$ will also be the stabilizer on $Y$. To express this informally in words similar to yours, the discreteness of fibers implies there is no "local action" only the unique "global action".

Some comments about when these $G$-actions may not agree:

The question however becomes much more interesting if you consider non-discrete fibers, in which case it would not be called a $G$-equivariant covering space, perhaps a better term is $G$-equivariant fiber space. For this case the $G$-action on $Y$ is not unique, as the Unique Lifting Theorem no longer stands. These $G$-actions may not even be path independent, for smooth fiber bundles and smooth actions this would be a question about connections, where the monodromy is path-independent if and only if the connection is flat. This case would possibly cover your intuition of a action which has a local component.

(also if the fibers are discrete then the connection is unique and flat)

If $G$ is not path-connected, then the monodromy is only a action for $G^0\subseteq G$ the path-connected component of $G$, on which this action must agree with the $G$-action on $Y$. But there may be multiple $G$-actions on $Y$, differing basically by how the components $G\big/G^0$ act.

Also if you consider a $G$-action which is not a continuous group action, then there may also be multiple $G$-actions on $Y$. This happens because a path on $G$ may not necessarily yield a path on $Y$ by the action...

Diana Pestana
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