# I Rank of the SO(3) algebra

1. Apr 14, 2016

### gentsagree

I am reading in my group theory book the well known commutation relations of the Lie algebra of SO(3), i.e. [J,J]=i\epsilon J.

What I don't understand is the statement that "from the relations we can infer that the algebra has rank 1".

Any ideas?

2. Apr 14, 2016

### ChrisVer

The rank of a Lie Group is the maximum number of its mutually commuting generators..In other words it's how many Casimir operators you can construct (Racah's theorem)... for SO(3) that's 1 because if you take $J_1, J_2, J_3$ they don't commute with each other. The Casimir operator is the $J^2 = J_1^2 +J_2^2 +J_3^2$ which commutes with the generators $J_i$: $[J_i, J^2]=0$

3. Apr 15, 2016

### gentsagree

Thank you very much, very clear. So, I suppose there is a way to find out how many Casimir operators I can construct, given an algebra with its commutation relations. I know SO(3) has J^2, but it is not clear to me how I know that it is the only Casimir there is.

Also, the rank of the Lie group corresponds to the rank of the Lie algebra, which is given by the dimension of the Cartan subalgebra (as you said, the number of mutually commuting group generators); so which is the one element of the dimension-1 Cartan subalgebra here? Is it correct to think that the Cartan is of dimension 1 since each generator commutes only with itself?

4. Apr 15, 2016

### ChrisVer

how many elements does the Cartan subalgebra of SU(3) have?

5. Apr 17, 2016

### gentsagree

Ok, what I said above was correct; I see now that from the commutation relations there are no commuting generators, but I can always pick one at random and see that it commutes trivially with itself. This is one way to see that the rank is 1. And from this I can infer there will be only one Casimir by Racah's theorem.

Why the reference to SU(3) ?

6. Apr 17, 2016

### ChrisVer

because I think that the Cartan subalgebra of SU(3) is not of dimension 1.

7. Apr 17, 2016

### gentsagree

Sure, the rank of SU(3) is 2, so is the dimension of the Cartan subalgebra. There one could see more clearly from the (modified) commutation relations that there are indeed two commuting generators.

8. Apr 17, 2016

### ChrisVer

ΟΚ then, it seems I misunderstood what you meant with this:

9. Apr 17, 2016

### gentsagree

I see. Thanks for the help anyway.

10. Apr 18, 2016

### gentsagree

Ok, I think I confused myself again. I thought I was thinking about this the right way, but I can't apply my reasoning to a slightly harder case: the proper orthochronous Lorentz group, SO(3,1), or equivalently, SL(2,C).

$$\begin{split} [J_{i}, J_{j}] &= i\epsilon_{ijk}J_{k} \\ [K_{i}, K_{j}] &= i\epsilon_{ijk}J_{k}\\ [J_{i}, K_{j}] &= i\epsilon_{ijk}K_{k} \end{split}$$

and my book says very confidently that from these I can read off that the rank of the algebra is 2. And there is no way I can think of in which I can see it. I don't see any commuting generators, apart from the trivial case where each specific generator, eg $J_{1}$, commutes with itself.

11. Apr 18, 2016

### ChrisVer

First a fast investigation: the generators of the sl(2,c) are for exaple the pauli matrices and the identity matrix... In fact any complex matrix 2x2 can be written as a linear combination of those four:
$M = a_0 I_2 + \sum_{i=1}^3 a_i \sigma_i$
Obviously you can simultaneously diagonalize the $\sigma_3$ and $I_2$, so it's rank 2...

Now in the case of the su(2)x su(2)
I am pretty sure that taking $J_3,K_3$ as the diagonalizable matrices.
So the $\{J_3,K_3\}$ (not anticommutation) is a set of mutually commuting/simultaneously diagonalizable generators???
In fact this is like have 2 SU(2)s, so the rank you'd expect is 1 ( either because SU(2) is pretty much like SO(3) or by checking out the commutations each has 1 diagonalizable generator per time which we choose by convention the 3rd J3 ) from each, sum=2.
Would it be easier for you if you tried to define $K^{\pm}=c( K_1 \pm i K_2 )$, $J^{\pm}=c(J_1 \pm iJ_2)$ and $K_3,J_3$?

Last edited: Apr 18, 2016