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Let $M=G/H$ be a homogeneous manifold, with $G$ connected Lie group. Suppose that $\widetilde{M}$ is a covering of $M$.

QUESTION: is there a general prescription to obtain a Lie group $\widetilde{G}$, starting from $G$, in such a way that $\widetilde{M}=\widetilde{G}/\widetilde{H}$?

Using the case of $M=G=S^1$ and $H=1$ as a toy model (and, for instance, $\widetilde{M}=\widetilde{G}=\mathbb{{R}}$) we see that $\widetilde{G}$ has two remarkable properties:

  1. it contains the group $\Gamma=\mathbb{{Z}}$ of "gauge symmetries" of $\widetilde{M}\to M$;

  2. its factor by $\Gamma$ returns the original group $G$.

So, I guess that the two properties above are enough to characterise $\widetilde{G}$ but I'm not able to prove it. I'm sure it's a well-known result, but I can't find any reference (I could not get which book is this "Bredon" mentioned here: lifting group actionlifting group action). In the case that my guess is correct, I'd like to understand if there is a constructive way to obtain $\widetilde{G}$, e.g., by realising the Lie algebra $\frak{g}$ as vector fields on $M$, lifting them to $\widetilde{M}$, and then take the group generated by their flows.

Let $M=G/H$ be a homogeneous manifold, with $G$ connected Lie group. Suppose that $\widetilde{M}$ is a covering of $M$.

QUESTION: is there a general prescription to obtain a Lie group $\widetilde{G}$, starting from $G$, in such a way that $\widetilde{M}=\widetilde{G}/\widetilde{H}$?

Using the case of $M=G=S^1$ and $H=1$ as a toy model (and, for instance, $\widetilde{M}=\widetilde{G}=\mathbb{{R}}$) we see that $\widetilde{G}$ has two remarkable properties:

  1. it contains the group $\Gamma=\mathbb{{Z}}$ of "gauge symmetries" of $\widetilde{M}\to M$;

  2. its factor by $\Gamma$ returns the original group $G$.

So, I guess that the two properties above are enough to characterise $\widetilde{G}$ but I'm not able to prove it. I'm sure it's a well-known result, but I can't find any reference (I could not get which book is this "Bredon" mentioned here: lifting group action). In the case that my guess is correct, I'd like to understand if there is a constructive way to obtain $\widetilde{G}$, e.g., by realising the Lie algebra $\frak{g}$ as vector fields on $M$, lifting them to $\widetilde{M}$, and then take the group generated by their flows.

Let $M=G/H$ be a homogeneous manifold, with $G$ connected Lie group. Suppose that $\widetilde{M}$ is a covering of $M$.

QUESTION: is there a general prescription to obtain a Lie group $\widetilde{G}$, starting from $G$, in such a way that $\widetilde{M}=\widetilde{G}/\widetilde{H}$?

Using the case of $M=G=S^1$ and $H=1$ as a toy model (and, for instance, $\widetilde{M}=\widetilde{G}=\mathbb{{R}}$) we see that $\widetilde{G}$ has two remarkable properties:

  1. it contains the group $\Gamma=\mathbb{{Z}}$ of "gauge symmetries" of $\widetilde{M}\to M$;

  2. its factor by $\Gamma$ returns the original group $G$.

So, I guess that the two properties above are enough to characterise $\widetilde{G}$ but I'm not able to prove it. I'm sure it's a well-known result, but I can't find any reference (I could not get which book is this "Bredon" mentioned here: lifting group action). In the case that my guess is correct, I'd like to understand if there is a constructive way to obtain $\widetilde{G}$, e.g., by realising the Lie algebra $\frak{g}$ as vector fields on $M$, lifting them to $\widetilde{M}$, and then take the group generated by their flows.

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How to "lift" a transitive group action on a manifold?

Let $M=G/H$ be a homogeneous manifold, with $G$ connected Lie group. Suppose that $\widetilde{M}$ is a covering of $M$.

QUESTION: is there a general prescription to obtain a Lie group $\widetilde{G}$, starting from $G$, in such a way that $\widetilde{M}=\widetilde{G}/\widetilde{H}$?

Using the case of $M=G=S^1$ and $H=1$ as a toy model (and, for instance, $\widetilde{M}=\widetilde{G}=\mathbb{{R}}$) we see that $\widetilde{G}$ has two remarkable properties:

  1. it contains the group $\Gamma=\mathbb{{Z}}$ of "gauge symmetries" of $\widetilde{M}\to M$;

  2. its factor by $\Gamma$ returns the original group $G$.

So, I guess that the two properties above are enough to characterise $\widetilde{G}$ but I'm not able to prove it. I'm sure it's a well-known result, but I can't find any reference (I could not get which book is this "Bredon" mentioned here: lifting group action). In the case that my guess is correct, I'd like to understand if there is a constructive way to obtain $\widetilde{G}$, e.g., by realising the Lie algebra $\frak{g}$ as vector fields on $M$, lifting them to $\widetilde{M}$, and then take the group generated by their flows.