Definition of smooth action

differential-topologyexamples-counterexamplesgroup-actionslie-groupssmooth-manifolds

Let $G$ be a Lie group and let $M$ be a smooth manifold.
In chapter 7 of Lee's Introduction to Smooth Manifolds, a smooth (left) action of $G$ on $M$ is defined to be a smooth map $\theta:G \times M \rightarrow M$ written as $(g,p) \mapsto g \cdot p$, such that $g \cdot (h \cdot p) = gh \cdot p$ and $1_G \cdot p = p$ for all $g,h \in G$ and $p \in M$. We have the following

Proposition. Let $\theta:G \times M \rightarrow M$. Then the map $\Phi:G \rightarrow \text{diff}(M)$ given by $\Phi(g)(p) := g\cdot p$ is a group homomorphism.

Proof. Fix $g \in G$. Then $\Phi(g):M \rightarrow M$ is smooth because it is the composition of smooth maps
$$M\xrightarrow{p \mapsto (g,p)} G \times M\xrightarrow{\theta} M.$$
Moreover, $\Phi(g)$ is a diffeomorphism because $\Phi(g^{-1})$ is its inverse. To see this, fix $p \in M$. Then
$$(\Phi(g) \circ \Phi(g^{-1}))(p) = g \cdot (g^{-1} \cdot p) = e \cdot p = p,$$
and similarly $(\Phi(g^{-1}) \circ \Phi(g))(p) = p$. Finally, $\Phi$ is a group homomorphism because given $g,h \in G$ and $p \in M$, we have
$$(\Phi(g) \circ \Phi(h))(p) = g \cdot (h \cdot p) = gh \cdot p = \Phi(gh)(p),$$
so $\Phi(gh) = \Phi(g) \circ \Phi(h)$. $\square$

Now, suppose we instead begin with a group homomorphism $\Phi:G \rightarrow \text{diff}(M)$. We can then obtain a group action $\theta:G \times M \rightarrow M$ via the assignment $g \cdot p := \Phi(g)(p)$. To see that this is indeed a group action, observe that given $g,h \in G$ and $p\in M$, we have
$$g \cdot (h \cdot p) = (\Phi(g)\circ\Phi(h))(p) = \Phi(gh)(p) = gh \cdot p, \qquad 1_G \cdot p = \Phi(1_G)(p) = \text{Id}_M(p) = p.$$

Question. Is $\theta:G \times M \rightarrow M$ a smooth action? That is, is $\theta$ a smooth map from the product manifold $G \times M$ into $M$? If not, then what is an example of a group homomorphism $\Phi:G \rightarrow \text{diff}(M)$ such that the induced action $G \times M \rightarrow M$ is not smooth?

I suspect that the answer is no, but I cannot think of a counterexample.

Best Answer

Let $G=S^1=SO(2)$, the group of rotations around the origin in the Euclidean plane. Every rotation is either rational (has finite order) or is irrational (has infinite order).

  1. Verify that $G$ is isomorphic (as an abstract group) to the product $G_1\times G_2$, where each nontrivial element of $G_1$ is an irrational rotation and each element of $G_2$ is a rational rotation.

  2. Verify that both $G_1, G_2$ are dense in $G$. (It suffices to check that one of these subgroups is dense if you have trouble proving density of both.)

  3. Verify that each continuous action $\theta: G\times S^1\to S^1$ is determined by its restriction to $G_1$ and is also determined by its restriction to $G_2$. (Assuming that you proved density of both subgroups.)

  4. Suppose you have managed to prove that $G_2$ is dense in $G$. Find a group homomorphism $\Phi: G\to G< Diff(M)$ which is the identity when restricted to $G_2$ ($\Phi(g)=g$ for all $g\in G_2$) and is not the identity on $G_1$. Conclude that $\Phi$ does not define a continuous $G$-action on $M=S^1$. Do the same, swapping the roles of $G_1, G_2$, if you proved that $G_1$ is dense in $G$.

Related Question