Must the action of a torus on a manifold have a fixed point if each circle subaction has a fixed point? Let $T = (S^1)^k$ act smoothly on a compact manifold $M$ via $\rho : T \times M \to M$.
Suppose for every cocharacter/group map $\sigma : S^1 \to T$, the circle subaction $\rho \circ \sigma : S^1 \times M \to M$ has a fixed point.
Does there exist a fixed point of $\rho$ on $M$?
I'm interested in the case where $M$ is a symplectic manifold and $\rho$ is a Hamiltonian torus action.
The moment map of $\rho \circ \sigma$ will have a maximum, and this will be a fixed point of $\rho \circ \sigma$. Indeed, this is the origin of my assumption about circle subactions.
For $k=2$, I wonder whether we can show existence by contradiction.
Assume no fixed point of $\rho$ exists, and let $p_\sigma \in M$ be a fixed point of $\rho \circ \sigma$ for all $\sigma : S^1 \to T$. The set $\{ p_\sigma \}$ is an infinite set, and thus has a limit point in $M$. Can we show that the points $p_\sigma$ must be isolated (giving a contradiction)?
Any assistance or ideas would be welcome.
 A: Regarding a Hamiltonian torus action on a (compact connected) symplectic manifold, your question is answered positively by Atiyah--Guillemin--Sternberg's convexity theorem, which states that the image of the moment map is a (compact) convex polytope whose vertices have as preimages precisely the fixed point set of the torus action.
Besides the original research papers, the proof of Atiyah can be found in McDuff's and Salamon's Introduction to Symplectic Topology, whereas Guillemin's and Sternberg's proof is somewhat scattered in their book Symplectic Techniques in Physics. Atiyah's proof is more straightforward and relies on Morse-Bott theory; As you anticipated, the Hamiltonian functions associated to the 1-parameter subgroups of the torus play a key role. Guillemin's and Sternberg's proof also involves bits of Morse-Bott theory, but their proof relies on more general facts about smooth actions which allow to answer your question affirmately in the smooth setting too (at least when $k=2$):
Proposition (27.4 in their book). Let a compact group $G$ act smoothly on a compact manifold $M$. Then at most finitely many conjugacy classes of subgroups of $G$ appear as isotropy groups of points of $M$ for this action.
Here, the conjugation equivalence on subgroups of $G$ is given by $A \sim B$ iff there exists $g \in G$ such that $A = gBg^{-1}$. (Remark: This result is apparently a special case of a theorem of Mostow for continuous actions.) It follows from this proposition that when $G$ is abelian, i.e. a torus, subgroups coincide with conjugacy classes, hence only finitely many subgroups of $G$ can be isotropy groups of points of $M$ for the given torus action.
For $k > 1$, the torus $T^k$ admits infinitely many (conjugation inequivalent) circle subgroups. Your assumption on the torus actions however implies that each one of these circle subgroups fixes at least a point in $M$. In view of the proposition, most of those circle subgroups are only proper subgroups of the isotropy groups of the points they fix. Hence there are two nonequivalent circle subgroups belonging to the isotropy group of some point, implying that this isotropy group contains at least a 2-dimensional torus. When $k=2$, this proves that every smooth $T^2$-action on a compact manifold has a fixed point. I expect that pursuing this sort of argument would establish the corresponding result for any $k > 2$.
