If we have a system, with total spin angular momentum given by $S$, then we have spin multiplicity equal to, $2S+1$. This spin multiplicity basically tells us the different spin states, this system can have, based on the different values of $m_s$.
However, if we have also been given total angular momentum $L$, and we have $L\ge S$, then we know that $2S+1$ also gives us the multiplicity of J if we write the states in the coupled basis.
My question is, in uncoupled basis, the triplet states are defined by the different values of $m_s$. For example, $|L,S,m_s\rangle = |1,1,-1\rangle,|1,1,0\rangle,|1,1,1\rangle$ is an example of a triplet defined by $m_s$.
However, in coupled basis representation, $|L,S,J\rangle = |1,1,0\rangle,|1,1,1\rangle,|1,1,2\rangle$ is an example of a triplet defined by $J$.
There is no one-to-one correspondence between the states, but they can be related using Clebsch Gordon coefficients I think.
So, my question is, do we define triplets on the basis of $m_s$ in uncoupled basis or when we ignore angular momentum, and by using $j$ in the coupled basis, in spin-orbit coupling for example ? In spin-orbit coupling, we ignore $m_s$ because it's not a good quantum number, and yet we get the triplet state for $S=1$, but because of values of J. Are total angular momentum triplets and spin triplets the same?
If I'm wrong, can someone tell me the correct explanation?