I was constructing an explicit example of a torus action in order to compute the Sullivan model of the associated Borel fibration, and I came up with the following action: $T^2 \times T^2 \to T^2$ given by $((\alpha, \beta), (z, w)) \mapsto (\alpha \cdot z, \beta \cdot w)$. Here I denote $T^2 = S^1 \times S^1$ the torus.
- Is this a (topological) action? In my opinion it should be, beacuse it is continuous (in fact, differentiable).
- Is this action free (i.e. every isotropy group is trivial)? I believe so, because the restriction to actions $S^1 \times S^1 \to S^1$, $(\alpha, z) \mapsto \alpha \cdot z$ is free.
Could someone please tell me if this is correct? I have the feeling it is, but I'm not an expert in this area so I really can't tell it for sure...
Yes, more generally, $T^n=S^1\times\cdots \times S^1$ being a product of Lie groups is a Lie group (group law being the one on the product, i.e component wise). And in general, every Lie group $G$ can act on itself from the left by group multiplication $G\times G\to G$, so the action map is the same as the multiplication operation, which is why the action is smooth and also free (for each $x\in G$, the existence of multiplicative inverses in groups implies that $g\cdot x=x$ implies $g=e$, so the stabilizer/isotropy subgroup in $G$ of each point $x\in G$ is just $\{e\}$).