A SO(3) group, Heisenberg Hamiltonian

[LagrangeEuler]

We have commutation relation $[J_j,J_k]=i \epsilon_{jkl}J_l$ satisfied for $2x2$, $3x3$, $4x4$ matrices. Are in all dimensions these matrices generate $SO(3)$ group? I am confused because I think that maybe for $4x4$ matrices they will generate $SO(4)$ group. For instance for Heisenberg Hamiltonian
H=-\frac{J}{2}\sum_{n,m}\vec{S}_n \cdot \vec{S}_m depending on spin of the sistem is this Hamiltonian always $SO(3)$ invariant or no? Or it is SO(n) symmetry in general?

 

[jambaugh]

They together close to form the Lie algebra of the SO(3) group. But SO(3) (and so(3) the Lie algebra) can have various dimensional representations. So yes, it depends on the "spin of the system" which is another way of specifying which irreducible representation of SO(3) (or rather its corresponding spin group) you're considering.

Note: The dimension of the group is number of independent parameters which equates to the number of (linearly) independent generators. Hence here the general group element is: $G(\Theta) =\exp( i\theta^k J_k)$. The group here is 3 dimensional... but which 3-dim group? That's what the commutator relations determine. Then what dimensional matrices these generators are describes the dimension of the representation.

[Edit] Let me add, for SO(n) the n is the dimension of the fundamental representation, namely the spin 1 or vector representation.