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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 $X$?

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    $\begingroup$ It should be enough to pick a contractible manifold with non-contractible diffeomorphism group (e.g. $\mathbb{R}^n$) $\endgroup$
    – Denis Nardin
    Commented Mar 13, 2019 at 21:41
  • $\begingroup$ Q 1 : $S^1$ acting on $S^1$ by complex multiplication and by identity $\endgroup$
    – InfiniteLooper
    Commented Mar 13, 2019 at 21:43
  • $\begingroup$ @Bleuderk That answers Q2 also (by taking the other action to be trivial) $\endgroup$
    – Denis Nardin
    Commented Mar 13, 2019 at 21:45
  • $\begingroup$ @DenisNardin You mean by taking action on the whole $\mathbb C$ ? $\endgroup$
    – InfiniteLooper
    Commented Mar 13, 2019 at 21:50
  • $\begingroup$ @Bleuderk Oh yes, sorry. For some reason I thought you wrote $S^1$ acting on $\mathbb{C}$... $\endgroup$
    – Denis Nardin
    Commented Mar 13, 2019 at 21:55

1 Answer 1

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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.

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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Mar 13, 2019 at 22:26
  • $\begingroup$ @YCor Sorry, the addition of $G=S^1$ was later. Will amend my answer (pretty much the same thing works, of course) $\endgroup$
    – Denis Nardin
    Commented Mar 13, 2019 at 22:47
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    $\begingroup$ 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...) $\endgroup$
    – Denis Nardin
    Commented Mar 13, 2019 at 22:55

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