Does Heisenberg ferromagnet have infinite number of phases below the critical temperature? This is an upshot of the question here. The up-aligned and the down-aligned spin configurations are assumed to be two distinct phases in case of an Ising ferromagnet. But for Heisenberg ferromagnet, there are infinitely many spin configurations distinguished by different orientations of the spins in full 3-dimensional space. 


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*Does it mean a Heisenberg ferromagnet has an infinite number of phases? If not, why?

*If yes, what is the nature of phase transition between any two such phases below $T_c$? Is it a first order or a continuous transition? Note that, in the case of Ising ferromagnet, the transition between up-aligned ferromagnetic phase and the down-aligned phase is first order. 
 A: *

*Yes, but only in the same sense that there are infinitely many directions for the hydrogen m=1 state to point it's angular momentum in. This degeneracy is from global rotational symmetry. They are not distinct phases, simply rotating the entire magnet moves you from one to the other. 

*All of these states have the same energy,  so there is no concept of phase transitions between them. It costs zero energy to go between them. This is related to the fact that symmetry breaking of the ferromagnet comes with a (gapless) Goldstone mode 
Edit: So OP is adamant about definining these different states as different phases, fine its up to you what you define as "phase", but it disagrees with many classic texts of phase transitions (cf. Phase transitions by Goldenfeld). In that case, yes there are as many "phases" as there are on the surface of a sphere.
As I mentioned in the comments, there is only a first order transition if you pass through $H=0$ going from one of these states to another. This means the "phase" boundary between these states is a single point in 3D configurational $H$ space. This is why I would say that there is no distinction between these phases, you can get from one to the other without passing through any transition (1st, 2nd order or otherwise).
You then ask why can't it be that there is a no line or plane in $H$ that separates states. I return this question with: why would there be? Surely you agree that all of these states are degenerate when $H$ points in the appropriate direction. Because the energy goes smoothly as $\mathbf{M} \cdot \mathbf{H}= M H \cos(\theta)$, you then can see there is no energy barrier going from one state to the other if $\theta$ is changed continuously.
