Let $M$ be a real manifold with an action of a two-torus $T^2$. Let $p$ be a point in $M$ and $H_p$ the stabiliser of $p$ in $T^2$. If $\dim H_p = 1$, then $H_p$ is a circle.
I have a topological argument to prove this fact in a specific context, but I came up with a second, more general argument using the exponential map, I would like to double check it or know if it can be improved.
First of all $H_p$ is compact, because it is closed in $T^2$, and Abelian, as a subgroup of an Abelian group. So the claim is that $H_p$ is connected.
Since $H_p$ has dimension one, take a generator $U$ of its Lie algebra. Consider the one-parameter group $$E := \{ \exp (tU):t \in \mathbb{R}\} \subset T^2.$$ Of course $E \subseteq H_p$. Now the exponential map is known to be surjective on compact and connected groups, so $\exp \colon \mathfrak{t}^2 \to T^2$ is surjective. Then, given $g \in H_p \subset T^2$, there exists $W \in \mathfrak{t}^2$ such that $\exp (W) = g$. Assume $W \not \in \mathbb{R}U$. Then $U$ and $W$ generate $\mathfrak{t}^2$, in particular they commute. Therefore $$ \exp(W+U)p = \exp(W)\exp(U)p = \exp(W)p = gp = p. $$ This is equivalent to saying $W+U$ is in the Lie algebra of $H_p$, which is spanned by $U$, and then $W$ is spanned by $U$, contradiction. So $W$ must be in the Lie algebra of $H_p$, namely there is a number $s$ such that $W = sU$, and $g = \exp (W) = \exp (sU) \in E$. This proves $H_p \subseteq E$, which is to say $H_p$ is the image of the line $t \mapsto tU$, which is path-connected and thus connected. But the exponential map is in particular continuous, so $H_p$ is connected. We found that $H_p$ is compact, connected and Abelian, so it is a circle.