Generators of a semi-simple lie algebra are traceless

1. Oct 16, 2015

CAF123

1. The problem statement, all variables and given/known data
Consider a semi simple lie algebra. Show that if $T_a$ are the generators of a semi simple Lie algebra then $\text{Tr}T_a=0$.

2. Relevant equations

The commutation relations of the generators and the cyclic properties of the trace.

3. The attempt at a solution
$$[T_a, T_b] = ic^c_{ab}T_c \Rightarrow \text{Tr}[T_a, T_b] = ic^c_{ab}\text{Tr}(T_c) = 0.$$

I also know that $g_{ab} = -c^c_{ad}c^{d}_{bc}\,\,(1)$. So since the l.h.s of the equation above is zero, I could multiply by another structure constant, thereby introducing the metric and then since the lie algebra is semi simple, multiply by the inverse of the metric and thus just have $\text{Tr}(T_a)$ on the l.h.s meaning it has to vanish.

That is my argument but I can't write it down because i don't see a way to introduce another structure constant because in the definition $(1)$, $d$ is contracted but in my expression $d$ is also contracted with the generator. So the placements of the indices mean I can't just multiply by another structure constant with the required indices without violating the summation convention.

Any hints on how to make progress? Thanks!

2. Oct 16, 2015

fzero

You can sum this against $c^{cab}$ and use the total antisymmetry of $c_{abc}$ to relate the expression to that for the metric.

Edit: I meant $c^{dab}$, so that you're not repeating the index that was already summed over.

Last edited: Oct 17, 2015
3. Oct 17, 2015

CAF123

Hi fzero,
Ok, so I have $$c^{dab}c^c_{ab} \text{Tr}T_c = 0,$$ but I want to put the indices in the right places so that I can use the fact that $g_{ab} = -c^{c}_{ad}c^{d}_{bc}$ but I can't see how to do that without introducing more metrics which violate the summation convention. Or perhaps there is some properties of the structure constants that I am not aware of, the ones I know are that $c^{c}_{ab}=-c^{c}_{ba}$ and $c_{abc}=c_{bca}=c_{cab}$, with all indices lowered in the last result. Thanks!

4. Oct 17, 2015

samalkhaiat

You have $$C_{ab}{}^{c}\mbox{Tr}(T_{c}) = C_{abc}\mbox{Tr}(g^{cd}T_{d}) = C_{abc} \mbox{Tr}(T^{c}) = 0.$$ Since the structure constant is totally antisymmetric, you get $$C_{cab} \mbox{Tr}(T^{c}) = 0. \ \ \ \ (1)$$
Now, make all $\mbox{Tr}(T^{a})=0$ except for one, say $T^{A}$: you can think of $\mbox{Tr}(T^{a})$’s being the components of a vector, so you can rotate it to a new vector with only one non-zero component. In this case, (1) gives you $$C_{Aab}\mbox{Tr}(T^{A}) = 0, \ \Rightarrow \ C_{Aab}=0.$$ Thus $$[T_{A},T_{a}]=0, \ \forall a.$$ This contradicts the semi-simple nature of the algebra. Thus $\mbox{Tr}(T^{a})=0, \ \ \forall a.$

5. Oct 17, 2015

fzero

You can do manipulations like
$$g_{ab} = -c^{c}_{ad}c^{d}_{bc} \rightarrow g^{ab} = - g_{ce}g_{df} c^{cad} c^{fbe} = g_{ce}g_{df} c^{acd} c^{bef} = c^{acd} c^b_{cd}.$$

Then you'll use the fact that the metric is nondegenerate for a semisimple Lie algebra.

6. Oct 18, 2015

CAF123

Hi samalkhaiat, thanks for this alternative argument, I have a few questions:
I see what are you doing here but I am trying to see how this is consistent with the fact that the trace of a matrix is invariant under a change of basis?

Contradicts in the sense that for a semi simple group, it must not admit any abelian invariant subalgebras? So the statement $[T_A, T_{a}]=0$ contradicts this, since it implies that the lie group is abelian and so the lie algebra of any subgroup will also be abelian?

7. Oct 18, 2015

samalkhaiat

But, we did not change basis! If the dimension of the algebra is $n$, then we have $n$ numbers given by $\mbox{Tr}(T^{a})$. We group these numbers together to define the $n$-vector: $V^{a}\equiv \mbox{Tr}(T^{a})$. Now, we can rotate this vector by $O(n)$ matrix: $\bar{V}^{a}= R^{ab} V^{b}$, and choose the rotation matrix $R$ so that only one component $V^{A}\equiv \mbox{Tr}(T^{A}) \neq 0$.
Correct.
Correct.
No it does not mean that the group/algebra is Abelian, $[T^{A},T^{a}]=0$ means that the generator $T^{A}$, (whose trace we assumed non-zero), generates an invariant abelian subgroup of the original group. This is not possible because the original group is semi-simple. Therefore our assumption $\mbox{Tr}(T^{A}) \neq 0$ cannot be true, i.e., we must have $\mbox{Tr}(T^{A}) = 0$. Since we made the traces vanish for the other $(n-1)$ generators, thus $\mbox{Tr}(T^{a}) = 0$ for all $a$.

8. Oct 19, 2015

CAF123

Ok many thanks I see. Can I ask another related question? Show that there are no non abelian lie groups of dimension 2.
Attempt:
Lie algebra of dim 2 means the lie algebra is the set $\left\{T_a, T_b \right\}$. Suppose there is a non abelian group, then $[T_a, T_b] = ic^c_{ab}T_c$. But by definition, c can only be a or b. Hence $[T_a, T_b] = ic^{a}_{ab}T_a + ic^{b}_{ab}T_b$. Then lower the indices on the c's and use the anisymmetry $c^c_{ab}=-c^{c}_{ba}$. This means the r.h.s is zero. This contradicts the original assumption that the group was non abelian. Hence it is must be abelian. Is it ok?

9. Oct 19, 2015

samalkhaiat

Yeah, it is okay, but you don't need to over do things:
The structure constant carries 3 indices and is totally antisymmetric. Therefore, it is necessarily zero in 2 dimensions.
Here is a question for you. Show that the Cartan metric is invariant under inner group automorphism (i.e., the adjoint map). Then, conclude that the structure constant of any Lie group is totally antisymmetric.

10. Oct 20, 2015

CAF123

Ok thanks, I'll keep this exercise in mind. Apparantly, the question I posed before was not correct and it should have been 'Show that there are no semi simple lie groups of dimension 2' . I thought the 'proof' I gave answered the original question but it seems there is an abelian two dim lie group: the affine group. So I am trying to reconcile what I did and the fact there does indeed exist a lie group of dim 2.

Is it maybe because the statement $[T_a,T_b]=0$ that I derived means that if $T_a$ or $T_b$ generate a subalgebra, then the commutator with the other generator generating the lie algebra of the group gives zero so $T_a$ or $T_b$ generates an abelian invariant subalgebra thus there are no semi simple lie groups of dim 2.
While the statement $[T_a,T_b]=0$ implies that the algebra is abelian, it doesn't necessarily mean the group is abelian?
Thanks

11. Oct 20, 2015

samalkhaiat

2-dimensional Lie groups/algebras are necessarily abelian.
All abelian Lie groups/algebras admit 1-dimensional representation.

I have no idea what you are talking about. If you only have two commuting generators, then their algebra as well as the Lie group they generate must be abelian.

12. Oct 20, 2015

CAF123

13. Oct 21, 2015

samalkhaiat

14. Oct 21, 2015

CAF123

Ok I see, I'll write out the new claim and my attempt at the proof:

Show that there are no semi simple lie groups of dimension 2.

Attempt:
Lie algebra of dim 2 means the lie algebra is the set $\left\{T_a, T_b \right\}$. So, $[T_a, T_b] = ic^c_{ab}T_c$. But by definition, c can only be a or b. Hence $[T_a, T_b] = ic^{a}_{ab}T_a + ic^{b}_{ab}T_b$. Now suppose that there did exist a semi simple lie group of dim 2. Then can lower the indices on the c's and use the anisymmetry $c^c_{ab}=-c^{c}_{ba}$. This means the r.h.s is zero. So, it means either $T_a$ or $T_b$ generate an abelian invariant subalgebra. Contradiction. So no semi simple lie groups of dim 2. Is that ok?