- #1

- 96

- 1

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

Any ideas?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, the conversation discusses the commutation relations of the Lie algebra of SO(3) and how they lead to the understanding that the algebra has a rank of 1. The concept of the rank is linked to the number of mutually commuting generators and the construction of Casimir operators. The conversation also touches on the rank of the Lie group and the dimension of the Cartan subalgebra, and how this applies to different groups such as SU(3) and SL(2,C). Finally, it considers the case of SU(2)xSU(2) and how it relates to the rank of 1.

- #1

- 96

- 1

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

Any ideas?

Physics news on Phys.org

- #2

Gold Member

- 3,381

- 463

- #3

- 96

- 1

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

Gold Member

- 3,381

- 463

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

- #5

- 96

- 1

Why the reference to SU(3) ?

- #6

Gold Member

- 3,381

- 463

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

- #7

- 96

- 1

- #8

Gold Member

- 3,381

- 463

gentsagree said:Is it correct to think that the Cartan is of dimension 1 since each generator commutes only with itself?

- #9

- 96

- 1

I see. Thanks for the help anyway.ChrisVer said:ΟΚ then, it seems I misunderstood what you meant with this:

- #10

- 96

- 1

Here the commutators read

[tex]

\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}

[/tex]

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 [itex]J_{1}[/itex], commutes with itself.

- #11

Gold Member

- 3,381

- 463

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:

[itex] M = a_0 I_2 + \sum_{i=1}^3 a_i \sigma_i[/itex]

Obviously you can simultaneously diagonalize the [itex]\sigma_3[/itex] and [itex]I_2[/itex], so it's rank 2...

Now in the case of the su(2)x su(2)

I am pretty sure that taking [itex]J_3,K_3[/itex] as the diagonalizable matrices.

So the [itex]\{J_3,K_3\}[/itex] (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 [itex]K^{\pm}=c( K_1 \pm i K_2 )[/itex], [itex]J^{\pm}=c(J_1 \pm iJ_2)[/itex] and [itex]K_3,J_3[/itex]?

[itex] M = a_0 I_2 + \sum_{i=1}^3 a_i \sigma_i[/itex]

Obviously you can simultaneously diagonalize the [itex]\sigma_3[/itex] and [itex]I_2[/itex], so it's rank 2...

Now in the case of the su(2)x su(2)

I am pretty sure that taking [itex]J_3,K_3[/itex] as the diagonalizable matrices.

So the [itex]\{J_3,K_3\}[/itex] (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 [itex]K^{\pm}=c( K_1 \pm i K_2 )[/itex], [itex]J^{\pm}=c(J_1 \pm iJ_2)[/itex] and [itex]K_3,J_3[/itex]?

Last edited:

Share:

- Replies
- 5

- Views
- 1K

- Replies
- 1

- Views
- 700

- Replies
- 17

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Replies
- 8

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 4

- Views
- 1K

- Replies
- 2

- Views
- 1K