I am currently trying to understand the computation of an x-ray for a $T^2$-action of $M_1 = U_3 \lambda$, where $\lambda = i \operatorname{diag}(\lambda _1,\lambda _2,\lambda _3) \in \mathfrak{u}^*(3)$ and $T^2 \subset T^3$ and $T^2 =U(1)^2 \times \mathrm{id}$, to be found in C. Woodwards paper Multiplicity-free hamiltonian torus actions need not to be Kähler.
The remark in his computation I have trouble following is the following: If $X \in \chi$ has isotropy group $H$, then $X$ is a component of the fixed point set of $H$, that is, a compact submanifold, and therefore $X$ must contain $T$-fixed points.
Here $\chi$ denotes the set of connected components of orbit-type strata, and $H$ is a subgroup of $T=T^2$. I assume that I am missing some background here. I understand that $X$ must be $H$-fixed, but how does he get that the closure must be $T$-fixed?
Here is the proof that $X$ contains $T^{2}$ fixed points.
The subset $X$ is preserved by the $T^{2}$ action. This is because if $p$ is fixed by $H$, then so is the point $g.p$ for any $g \in T^2$, since $T^{2}$ is commutative. The other remark is that $X$ is a symplectic submanifold, this is lemma 5.53 of "Introduction to Symplectic Topology" (second edition).
After this, one needs to use a general fact, namely that if a symplectic submanifold is preserved by a Hamiltonian torus action, then the restricted action is a Hamiltonian action (for the restriction of the symplectic form). This is because the Hamiltonian equations for the submanifold are a subset of the Hamiltonian equations for the original manifold, you may check for yourself.
Lastly, one needs that any Hamiltonian action on a compact symplectic manifold has fixed points. This is well-known. In fact more is true, namely that the image of the moment map is a convex polygon and the pre-images of the vertices of this polygon are fixed by $T^{2}$ (implying that in this case there are at least $3$ fixed points). This is Theorem 5.47 of "Introduction to Symplectic Topology" (second edition).