TITLE: isometry group of the riemannian manifold $\mathbb{T}^2$?
QUESTION [7 upvotes]: What is the isometry group of the riemannian manifold $\mathbb{T}\times \mathbb{T}$ where $\mathbb{T}=\{z\in \mathbb{C}\ :\ |z|=1\}$ is the classical torus?

REPLY [6 votes]: I'd like to complement Jason's excellent and insightful answer by pointing out that the stabilizer of a point that he calculates is the dihedral group $D_8$ and the full isometry group is not $D_8\times \mathbb T^2$ as he guessed, but actually the semidirect product $D_8\rtimes_\varphi \mathbb T^2$ where $\varphi\colon D_8\to Aut(\mathbb T^2)$ is the obvious map. You can see that it's not a direct product by calculating on a simple example. Say take $\rho\in D_8$ to be reflection in the $y$-axis and $T$ to be the isometry moving things to the right by $.25$ units. Then $T\rho(0)= 0.25\neq \rho T(0)=.75$. 
It shouldn't be too hard to write an explicit isomorphism from $Isom(\mathbb T^2)$ to $D_8\rtimes_\varphi \mathbb T^2$.