# Homotopy of group actions

Let $$G$$ be a topological group and $$X$$ be a topological space. Let $$\alpha$$, $$\beta:G\times X\to X$$ be two group actions. We say that these two actions are homotopic actions if there is a continuous path of actions connecting $$\alpha$$ to $$\beta$$. This means that there is a continuous function $$\Gamma:[0,1]\times G\times X \to X$$ such that each $$\Gamma_t$$ is a group action and we have $$\Gamma_0=\alpha, \Gamma_1=\beta$$.

Question 1: What is an example of the following situation: There are two group actions $$\alpha$$, $$\beta$$ by $$G=S^1$$ on a manifold $$M$$ which are not homotopic actions but for every $$g\in G$$, the two homeomorphisms $$\alpha_g$$, $$\beta_g$$ of $$M$$ are homotopic maps?

Question 2: What is an example of the following situation: There are two group actions $$\alpha$$, $$\beta$$ by $$G=S^1$$ on a manifold $$M$$ which are not homotopic actions but $$\alpha, \beta$$ are homotopic maps as maps from $$G\times X$$ to $$X$$?

• It should be enough to pick a contractible manifold with non-contractible diffeomorphism group (e.g. $\mathbb{R}^n$) – Denis Nardin Mar 13 at 21:41
• Q 1 : $S^1$ acting on $S^1$ by complex multiplication and by identity – Bleuderk Mar 13 at 21:43
• @Bleuderk That answers Q2 also (by taking the other action to be trivial) – Denis Nardin Mar 13 at 21:45
• @DenisNardin You mean by taking action on the whole $\mathbb C$ ? – Bleuderk Mar 13 at 21:50
• @Bleuderk Oh yes, sorry. For some reason I thought you wrote $S^1$ acting on $\mathbb{C}$... – Denis Nardin Mar 13 at 21:55

It's enough to pick a contractible manifold $$M$$ with two non-homotopic actions. For example, let us pick $$M=\mathbb{R}^2$$ with two actions of $$S^1$$, the trivial one and the one given by rotations. These two actions are not homotopic. In fact for every action $$\rho$$ of $$S^1$$ on $$\mathbb{R}^2$$ we can consider the map $$\mathbb{R}^2\times S^1\to GL_2^+(\mathbb{R})$$ sending $$(x,\lambda)$$ to the differential of $$\rho(\lambda)$$ at $$x$$. This map is clearly homotopy invariant under the action, and its degree as a map $$S^1\to S^1$$ is 0 for the trivial action and 1 for the defining action.
However all continuous maps with target $$\mathbb{R}^2$$ are homotopic.
• This answers the second question. For the first, just take $n\ge 2$ and choose the nontrivial action using a element of order 2 and determinant 1. – YCor Mar 13 at 22:26
• @YCor Sorry, the addition of $G=S^1$ was later. Will amend my answer (pretty much the same thing works, of course) – Denis Nardin Mar 13 at 22:47
• Also, if I'm not mistaken, question 1 is true for any action, since $S^1$ is a connected Lie group (so for every $g$, $\alpha_g$ is homotopic to the identity...) – Denis Nardin Mar 13 at 22:55