# Invariant tensors of SU(3)

1. Nov 12, 2015

### CAF123

'Using the following normalization in the su(3) algebra $[\lambda_i, \lambda_j] = 2if_{ijk}\lambda_k$, we see that $g_{ij} = 4f_{ikl}f_{jkl} = 12 \delta_{ij}$ and, by expanding the anticommutator in invariant tensors, we have further that $$\left\{\lambda_i, \lambda_j\right\} = \frac{4}{3}\delta_{ij} + 2d_{ijk}\lambda_k.$$
The first statement about $g_{ij}$ I understand but how did the one about the anticommutator come about?
I can reexpress $\left\{\lambda_i, \lambda_j\right\} = [\lambda_i, \lambda_j] + 2 \lambda_j \lambda_i = 2if_{ijk}\lambda_k + 2 \lambda_j \lambda_i$. Now, $\lambda_j \lambda_i$ is a second rank tensor so can be written as $a \delta_{ij}$, for some a. I was thinking I could then consider a single case to determine a (i.e i=j=1) but this didn't work.

Any tips would be great!

Thanks!

2. Nov 12, 2015

You are incorrect to say that $\lambda_j \lambda_i$ can be written as $a \delta_{i j}$, which explains why your approach didn't work. Unfortunately, I don't know how to prove your identity. What is $d_{i j k}$?

3. Nov 12, 2015

### CAF123

Yup, I saw that I was incorrect in that after I posted my thread. $\delta_{ij}$ and $d_{ijk}$ are supposed to be the two invariant tensors for SU(3). In my notes it also says that $d_{ijk} = \frac{1}{4}\text{Tr} \lambda_i \left\{\lambda_j, \lambda_k\right\}$, so I guess I can use this fact. So then $$d_{ijk} = \frac{1}{4}\text{Tr} \lambda_i ([\lambda_j, \lambda_k] + 2 \lambda_k \lambda_{\ell}) = \frac{1}{4} \text{Tr} \lambda_i (2if_{jk \ell }\lambda_{\ell} + 2 \lambda_k \lambda_{j}) = \frac{i}{2} f_{jk\ell}\text{Tr} \lambda_i \lambda_{\ell} + \frac{1}{2}\text{Tr} (\lambda_i \lambda_k \lambda_j)$$ using the given normalisation of the algebra. Any ideas how to continue?

Thanks!

4. Nov 12, 2015

### samalkhaiat

Any $n \times n$ hermitian matrix $M$ can be expanded in terms of the $n \times n$ hermitian traceless matrices $\lambda^{a}$ , $a = 1,2, \cdots , n^{2}-1$, and the $n \times n$ identity matrix $I_{n}$ as follow
$$M = \frac{1}{n} \mbox{Tr}(M) \ I_{n} + \frac{1}{2} \sum_{c=1}^{n^{2}-1} \mbox{Tr}(M \lambda^{c}) \ \lambda^{c} . \ \ \ (1)$$
Now, take $M = \{ \lambda^{a} \ , \lambda^{b} \}$ and define the symmetric invariant tensor
$$d^{abc} \equiv \frac{1}{4}\mbox{Tr}\left( \{ \lambda^{a} \ , \lambda^{b} \} \lambda^{c} \right) .$$ With the $\lambda$’s normalized according to $\mbox{Tr}(\lambda^{a}\lambda^{b}) = 2 \delta^{ab}$, equation (1) becomes
$$\{\lambda^{a} , \lambda^{b} \} = \frac{4}{n} \delta^{ab} \ I_{n} + 2 d^{abc} \lambda^{c} .$$
Adding this to the algebra $[\lambda^{a} , \lambda^{b}] = 2 i f^{abc} \lambda^{c}$, and multiplying by another $\lambda$, you get
$$\lambda^{a} \lambda^{b} \lambda^{e} = \frac{2}{n} \delta^{ab} \lambda^{e} + (i f^{abc} + d^{abc} ) \ \lambda^{c} \lambda^{e} .$$ Taking the trace, you get
$$\frac{1}{2} \mbox{Tr} (\lambda^{a} \lambda^{b} \lambda^{c}) = i f^{abc} + d^{abc} .$$

Last edited: Nov 13, 2015
5. Nov 13, 2015

### CAF123

Hi samalkhaiat,
I see, thanks. The only thing I didn't understand was this statement above^^. In general for any irreducible representation we have that $\text{Tr}T_a T_b = C(R) \delta_{ab}$ where $C(R)$ is the Casimir of the representation. In this case, the $\lambda_a$ constitute the fundamental representation for SU(3) so are indeed irreducible. I am just not sure how to get $\text{Tr}\lambda_i \lambda_j = 2\delta_{ij}$, i.e showing the casimir of the defining rep of SU(3) is 2. Any ideas on this?

I think I could also obtain the result by writing $\left\{\lambda_i, \lambda_j\right\} = a\delta_{ij} + b d_{ijk}\lambda_k$, which computing some traces gives me a and b.
Thanks.

6. Nov 13, 2015

### samalkhaiat

Don’t confuse normalization (i.e., orthogonal transformation plus scalling) of generators with the Dynkin’s index of irreducible representations.
The quadratic Casmir of Lie algebra is given by
$$C_{2}(r) = g^{ab} \ t^{(r)}_{a} \ t^{(r)}_{b} = d_{(r)} \ I_{r}$$ where $d_{r}$ is a representation-dependent number called the Dynkin’s index, and $I_{r}$ is the identity matrix in the irreducible representation, i.e., $\mbox{Tr}(I_{r})$ is the dimension of the representation space $\mbox{dim}(r)$. Taking the trace, you get $$g^{ab} \ \mbox{Tr}(t^{(r)}_{a} \ t^{(r)}_{b}) = d_{(r)} \ \mbox{dim}(r) .$$ In the Adjoint representation, you can normalize the generators $A_{a}$ so that $$\mbox{Tr}(A_{a} \ A_{b}) = g_{ab}.$$ So, for any Lie algebra you have $$g^{ab} \ g_{ab} = d_{A} \ \mbox{dim}(A) , \ \ \Rightarrow \ \ d_{A} = 1.$$ This is because $g^{ab}g_{ab} = \delta^{c}_{c} = \mbox{dim}(A)$.

For $SU(n)$ and $SO(n)$ you can always make the following normalization convention $\mbox{Tr}(T_{i} \ T_{j}) = \lambda \ \delta_{ij}$, because $\mbox{Tr}(T_{i} \ T_{j})$ is a real symmetric matrix and can be diagonalized by taking an appropriate real linear combination of the generators, with diagonal coefficients set to a constant $\lambda$. With this basis of the algebra, the structure constants are given by $$C^{k}_{mn} = - \frac{i}{\lambda} \ \mbox{Tr}(T_{k}[T_{m},T_{n}]) ,$$ which implies that $C^{k}_{mn}$ is totally antisymmetric in all three indices.
For example $SU(2)$, where $g^{ab}= \frac{1}{2}\delta^{ab}$, we choose, for the Fundamental representation, $$\mbox{Tr}(T_{a}^{(F)} \ T_{b}^{(F)}) = \mbox{Tr}( \frac{\sigma_{a}}{2} \ \frac{\sigma_{b}}{2} )= \frac{1}{2} \ \delta_{ab} ,$$ So, $$d_{(F)} \ \mbox{dim}(F) = \frac{1}{2} \ \delta^{ab} \ \frac{1}{2} \ \delta_{ab} = \frac{3}{4} .$$ From this we find the Dynkin’s index $d_{(F)} = \frac{3}{8}$, because $\mbox{dim}(F) = 2$.