# A Decomposition of tensors into irreps (Georgi's book)

Tags:
1. Jul 22, 2015

### PineApple2

Hi. In Georgi's book page 143, eqn. (10.29) he gives an example of decomposing a tensor product into irreps:
$$u^iv_k^j=\frac{1}{2} \left( u^iv_k^j+u^jv_k^i-\frac{1}{4}\delta_k^iu^\ell v_\ell^j-\frac{1}{4}\delta_k^ju^\ell v_\ell^i \right)\\ +\frac{1}{4} \varepsilon^{ij\ell} \left( \varepsilon_{\ell m n}u^m v_k^n + \varepsilon_{kmn}u^m v_\ell ^n \right)\\ +\frac{1}{8} \left( 3 \delta_k^i u^\ell v_\ell^j - \delta_k^j u^\ell v_\ell^i \right)$$
1. In the first line, he symmetrizes and removes the trace. Looking at the second line, it can be written as $\frac{1}{2} (u^i v_k^j-u^jv_k^i) - \frac{1}{4} \delta_k^i u^\ell v_\ell ^j + \frac{1}{4} \delta_k^j u^\ell v_\ell ^i$ which corresponds to also removing the trace from the anti-symmetric part. Why do we need to also remove the trace from the anti-symmetric part? aren't the irreducible representations - traceless symmetric, anti-symmetric, and trace?
2. Why is the trace $v_j^j$ zero?
3. How do we get the dimensions of the irreps. For example, how do we figure out that the first line corresponds to $\mathbf{\bar{15}}$?

Thanks.

2. Jul 22, 2015

### fzero

Georgi's upper indices correspond to $\mathbf{3}$s, while the lower indices are $\mathbf{\bar{3}}$s. The invariant tensor $\delta^i_j$ contracts a $\mathbf{3}$ index with a $\mathbf{\bar{3}}$ index. So the irreducible representations are symmetric, antisymmetric and with the traces removed from pairing upper with lower indices. But you don't get a symmetric, traceless representation for $SU(3)$. For instance, the two index symmetric product of $\mathbf{3}$s is the $\mathbf{6}$ given by (10.22). There's no invariant tensor available to remove the trace.

Perhaps you are confused with orthogonal groups, which have real representations and so the invariant tensor has indices of the same type (both down or both up depending on convention).

So in the first line of the expression you quote, the traces to be removed are those between the upper indices $i$ and $j$ and the lower index $k$. Note that there is no particular symmetry between the upper and lower indices.

$v^k_j$ is assumed to be an irreducible representation (actually the adjoint in particular), so it must be traceless (since it has an upper and lower index).

If we symmetrize in $ij$, we count $3\cdot 4/2 = 6$ independent components with respect to those indices. There are $6\cdot 3 = 18$ ways to pair this with the lower index $k$, but we have to subtract the (symmetrized) trace between the upper and lower indices, which removes a total of $3$ terms. So we are left with $18-3=15$ independent components for this combination.

3. Jul 23, 2015

### samalkhaiat

If all the components of a tensor $T$ are independent of each other, we say that the tensor is irreducible, i.e., if such tensor has $d$ components then we have a d-vector in (an irreducible) d-dimensional vector space. We write this irreducible representation space as $[d]$. This means that the group in question acts on (transforms) such vector by $d \times d$ matrix representation of the group elements. However, in general an arbitrary tensor is reducible, i.e., not all components are independent of each other. Reducible tensor can be decomposed (in a group invariant way) into irreducible tensors by the processes of symmetrization, antisymmetrization and contraction between upper and lower indices, i.e., subtracting all possible traces.
Okay, let us talk about $SU(3)$. Consider the antisymmetric tensor $T^{[ij]}_{k} = T^{ij}_{k} - T^{ji}_{k}$. It has $3(\frac{1}{2}(2)(3)) = 9$ components but it is not irreducible. To reduce it, we impose the 3 trace relations $T^{[ij]}_{j}=0$. So, we end up with $9-3=6$ independent components. But 6 is exactly the number of independent components in a symmetric tensor $S_{(kl)}$. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. This is exactly what you have done in the second line of your equation. In practise, we actually never need to subtract a trace from the antisymmetric part because by subtracting the trace from the symmetric part we also have to balance our equation, i.e., we have to add what we have subtracted and this added piece will make the antisymmetric part traceless (see below).
Now, let us count the number of independent components in the traceless symmetric tensor $T^{(ij)}_{k}$ (this is the same tensor in the first line of your equation). Again, symmetry implies that for each $k$, we have $\frac{1}{2}(3)(4) = 6$ components, i.e., a total of $3 \times 6 = 18$ components in the symmetric mixed tensor. Then, the 3 trace relations $T^{(ij)}_{i}=0$ reduces the number to $18-3=15$ independent components. If you associate upper indices with the conjugate representation, then $$T^{(ij)}_{k} \equiv \frac{1}{2} u^{(i}v^{j)}_{k} - \frac{1}{8} \delta^{(i}_{k}v^{j)}_{n}u^{n} \in [\bar{15}] .$$
Now, the number of components of the totally symmetric (irreducible) $SU(3)$ tensor $T^{(k_{1} k_{2} \cdots k_{m})}$ is $\frac{1}{2}(m+1)(m+2)$. Therefore, the tensor $T^{(k_{1} k_{2} \cdot k_{m})}_{(j_{1} j_{2} \cdots j_{n})}$ has $\frac{1}{4}(m+1)(m+2)(n+1)(n+2)$ components. The number of components of the corresponding (traceless) irreducible tensor $\hat{T}^{(\cdots)}_{(\cdots)}$ will be less than that of $T^{(\cdots)}_{(\cdots)}$ by the number of independent components in the reduced-rank tensor $T^{(k_{1} \cdots k_{m-1})}_{(j_{1} \cdots j_{n-1})}$, i.e., the dimension of the space of traces:
$$\mbox{dim}\left( \hat{T}^{(k_{1} \cdots k_{m})}_{(j_{1} \cdots j_{n})} \right) = \mbox{dim}\left( T^{(k_{1} \cdots k_{m})}_{(j_{1} \cdots j_{n})} \right) - \mbox{dim}\left( T^{(k_{1} \cdots k_{m-1})}_{(j_{1} \cdots j_{n-1})} \right) ,$$ or
$$\mbox{dim}(m,n) = \frac{1}{4}(m+1)(m+2)(n+1)(n+2) - \frac{1}{4}m(m+1)n(n+1) = \frac{1}{2}(m+1)(n+1)(m+n+2) .$$
As for why $v^{i}_{j}$ is traceless: well it is (1,1) irreducible tensor. So, by the above relation of the dimension you see that $v^{i}_{j}$ belongs to the adjoint representation $[8]$ which is traceless by construction. Do you recall the relation $[3] \otimes [\bar{3}] = [8] \oplus [1]$? In tensor language, we take the tensor product of the fundamental representation $\phi^{i}$ with its conjugate representation $\phi_{j}$ and subtract the invariant trace: $$\phi^{i} \otimes \phi_{j} = \left( \phi^{i} \ \phi_{j} - \frac{1}{3} \ \delta^{i}_{j} \ \phi^{n}\phi_{n}\right) + \frac{1}{3} \ \delta^{i}_{j} \ \phi^{n}\phi_{n} .$$ So, your $v^{i}_{j}$ is just $$v^{i}_{j} = \phi^{i} \ \phi_{j} - \frac{1}{3} \ \delta^{i}_{j} \ \phi^{n}\phi_{n} ,$$ which is traceless and has 8 independent components, i.e., it belongs to the adjoint representation $[8]$.
Finally, let me do what I should have done first. Notice first that $u^{i}v^{j}_{k} \in [3] \otimes [8]$, i.e. it is a reducible tensor. So, let us decompose it into irreducible parts. First, we split the tensor into symmetric and antisymmetric tensors:
$$u^{i}v^{j}_{k} = \frac{1}{2} u^{(i}v^{j)}_{k} + \frac{1}{2} u^{[i}v^{j]}_{k} .$$ To make the symmetric part traceless, we subtract (and add) the symmetric combinations of traces
$$u^{i}v^{j}_{k} = T^{(ij)}_{k} + \frac{1}{2} u^{[i}v^{j]}_{k} + \frac{1}{8}\delta^{(i}_{k}v^{j)}_{n}u^{n} , \ \ (1)$$ where $T$ is the traceless symmetric tensor
$$T^{(ij)}_{k} = \frac{1}{2} u^{(i}v^{j)}_{k} - \frac{1}{8}\delta^{(i}_{k}v^{j)}_{n}u^{n} \in [15]. \ \ \ \ (2)$$
Now, in Eq(1) we use the identity (which you can verify easily)
$$\delta^{(i}_{k}v^{j)}_{n} = -2 \delta^{[i}_{k}v^{j]}_{n} + 3 \delta^{i}_{k}v^{j}_{n} - \delta^{j}_{k}v^{i}_{n} ,$$ and arrange the terms:
$$u^{i}v^{j}_{k} = T^{(ij)}_{k} + T^{[ij]}_{k} + \frac{3}{8}\delta^{i}_{k}v^{j}_{n}u^{n} - \frac{1}{8}\delta^{j}_{k}v^{i}_{n}u^{n} ,$$ where we have defined the following traceless antisymmetric tensor
$$T^{[ij]}_{k} = \frac{1}{2} u^{[i}v^{j]}_{k} - \frac{1}{4}\delta^{[i}_{k}v^{j]}_{n}u^{n} .$$ This can be rewritten as
$$T^{[ij]}_{k} = \frac{1}{4} \epsilon^{ijl}S_{(kl)} \in [\bar{6}] , \ \ \ (3)$$
where
$$S_{(kl)} = \epsilon_{kmn}u^{m}v^{n}_{l} + \epsilon_{lmn}u^{m}v^{n}_{k} . \ \ \ (4)$$ Now, if you put (2), (3), and (4) in (1), you obtain the equation you wrote.
After all that, we finally proved
$$3 \times 8 = 15 + \bar{6} + 3 .$$

Sam

Last edited: Jul 23, 2015
4. Jul 23, 2015

### PineApple2

I see, thanks. So is it correct to say that the irreducible representations need to be traceless with respect to upper and lower indices, because otherwise the invariant tensor $\delta^i_j$ will contract them into a smaller dimensional representation, which would mean that they were not irreducible in the first place? and is that also the reason that $v^j_j$ is zero, otherwise it would be contracted into a singlet?

5. Jul 23, 2015

### fzero

Exactly. If we can use $\delta^i_j$ to pull a singlet out of an expression, then the expression was in a reducible representation.

6. Jul 23, 2015

### PineApple2

Thanks!

@samalkhaiat: Thank you very much for the detailed post, that is helpful. However, I don't understand a few things: how do you know in eqn. (2) that the coefficient is $-\frac{1}{8}$? Also, how do you know that you should use the relation $\delta_k^{(i}v_n^{j)}=-2\delta_k^{[i}v_n^{j]}+3\delta_k^iv_n^j-\delta_k^jv_n^i$ and not some different relation between $\delta_k^{(i}v_n^{j)}$ and $\delta_k^{[i}v_n^{j]}$? I guess the answer is that you are looking (both in this case and in the case of the $1/8$) for such linear combinations that would eventually give traceless combinations. But how do you see these combinations at the stage where you have written it?

7. Jul 23, 2015

### samalkhaiat

1) Write $$T^{(ij)}_{k} = \frac{1}{2} u^{(i}v^{j)}_{k} + a \ \delta^{(i}_{k}v^{j)}_{n}u^{n} .$$ Now, the demand that this symmetric tensor is traceless uniquely determine the value of the constant $a$: $$T^{(kj)}_{k} = 0 = \frac{1}{2} u^{k}v^{j}_{k} + 4 \ a \ v^{j}_{n}u^{n} , \ \ \Rightarrow \ \ 8 \ a = -1 .$$
2) Write $$\delta^{(i}_{k}v^{j)}_{n} = a \ \delta^{[i}_{k}v^{j]}_{n} + b \ \delta^{i}_{k}v^{j}_{n} + c \ \delta^{j}_{k}v^{i}_{n} . \ \ (2)$$ Now, if you contract $(i)$ with $(n)$, you get $a + b = 1$. And contracting $(j)$ with $(n)$ gives $c - a = 1$. And that is all, you can not get more information from (2). However, as in (1), the demand $T^{[kj]}_{k} = 0$ gives you $a = -2$ which in turn leads to $b = 3$ and $c = -1$
Okay, here is another exercise for you. Decompose the tensor $$u^{i}v^{jk} \in [3] \otimes [6] ,$$ into $SU(3)$-irreducible parts.

8. Jul 24, 2015

### PineApple2

Here is my trial. Please let me know if there is a better method to use.
In this case, we have only upper indices, which we should symmetrize / anti-symmetrize, and not require being traceless. So,
$$T^{ijk}=u^iv^{jk}=T^{(ijk)}+T^{[ijk]}$$
now
$$T^{(ijk)} = \frac{1}{6}(u^iv^{jk}+u^iv^{kj}+u^jv^{ik}+u^jv^{ki}+u^kv^{ij}+u^kv^{ji})$$
now we assume that $v^{ij}=v^{ji}$ since this is the $\mathbf{6}$, which must be symmetric (otherwise it would have 9 deg. of freedom), using that we have
$$T^{(ijk)}=\frac{1}{3}(u^iv^{jk}+u^jv^{ik}+u^kv^{ij})$$
this has dimension $\frac{5!}{3! 2!}=10$. Now consider the anti-symmetric part. anti-symmetrize each time with respect to a different pair of indices.
$$T^{[ijk]}=\frac{1}{3} (\varepsilon^{ij\ell}\varepsilon_{\ell mn}u^mv^{nk} + \varepsilon^{ik\ell}\varepsilon_{\ell mn}u^mv^{jn} + \varepsilon^{jk\ell}\varepsilon_{\ell mn}u^iv^{mn})$$
now the last term vanishes because of the symmetry of $v^{mn}$. I am not sure how to calculate the dimension of the first two terms. Could I say that each of them is an anti-symmetric tensor in two of the indices, which is 3 deg. of freedom, and multiplied by the 3 deg. of the remaining index gives 3x3=9. I should get 8, but the tensor with only upper indices should not be traceless.

Last edited: Jul 24, 2015
9. Jul 24, 2015

### samalkhaiat

You have made few fundamental mistakes:
For a general rank-3 tensor $T^{ijk}$, we have
$$T^{(ijk)} + T^{[ijk]} = \frac{2}{3!} (T^{ijk} + T^{jki} + T^{kij}) \neq T^{ijk}.$$
In our case, since $T^{ijk} = T^{ikj}$, then $T^{[ijk]} = 0$. So, we should try
$$T^{ijk} = T^{(ijk)} + a \ T^{[ij]k} + b \ T^{[ik]j} . \ \ \ (1)$$ Then, $$T^{ijk} = T^{ikj} \ \ \ \Rightarrow \ a = b.$$ So, the last two tensors in (1) are just symmetric combination of the same irreducible tensor. In order to find the constant $a$, we write (1) explicitly in terms of $T^{ijk}=u^{i}v^{jk}$:
$$T^{ijk}= (\frac{1}{3} + 2a) T^{ijk} + (\frac{1}{3} - a) T^{jki} + (\frac{1}{3} - a) T^{kij} .$$ This is true if and only if $a = 1/3$.
Now, since $T^{(ijk)}$ is irreducible, it belongs to the representation space $(3,0)$, which has dimension: $$\mbox{dim}(3,0) = (1/2)(3+1)(3+2)=10.$$ But the original tensor $T^{ijk}=u^{i}v^{jk}$ has $18$ components. Therefore, $(T^{[ij]k} + T^{[ik]j})$ must belong to the (traceless) adjoint representation $[8]$. To see this, we write
$$T^{[ij]k} = u^{i}v^{jk}- u^{j}v^{ik} = ( \delta^{i}_{m}\delta^{j}_{n} - \delta^{j}_{m}\delta^{i}_{n}) u^{m}v^{nk},$$ or
$$T^{[ij]k} = \epsilon^{ijl} \epsilon_{lmn}u^{m}v^{nk} \equiv \epsilon^{ijl} \Gamma^{k}_{l} .$$ Similarly $$T^{[ik]j} = \epsilon^{ikl} \Gamma^{j}_{l} .$$ Notice that $$\mbox{Tr}(\Gamma) = \Gamma^{k}_{k} = \epsilon_{mnk}u^{m}v^{nk} = 0 .$$ Thus $\Gamma$ is an irreducible tensor in the adjoint representation $(1,1) = [8]$:
$$\mbox{dim}(1,1) = \frac{1}{2} (1+1)(1+1)(1+1+2) = 8.$$ So you finally have
$$u^{i}v^{jk} = T^{(ijk)} + \frac{1}{3} \left( \epsilon^{ijl}\ \Gamma^{k}_{l} + \epsilon^{ikl} \ \Gamma^{j}_{l} \right) .$$ That is the long way to prove $$3 \times 6 = 10 + 8 .$$
Good luck

Sam

10. Jul 26, 2015

### PineApple2

Hi. Thank you for the detailed answer, that helped me understand. I have a few further questions please. You found the dimension of the $\mathbf{8}$ as the dimension of the traceless antisymmetric tensor $\Gamma^{j}_{l}$. and I understand that the tensor $\varepsilon^{ikl}$ doesn't matter for that calculation since it's invariant. Do you have a way of figuring out the dimension directly from the expression $\epsilon^{ijl} \epsilon_{lmn}u^{m}v^{nk}$ which has only upper indices?
Also, is there a general way to write any irreducible tensor only in terms of epsilon and delta tensors? how would you write, for example, $T^{(ijk)}$ using that? or in a different case $T^{[ijkl]}$?

11. Jul 29, 2015

### samalkhaiat

Dimension of what? The 9-component tensor $T^{[ij]k} = \epsilon^{ijl} (\epsilon_{lmn} u^{n}v^{kn})$ is still reducible. The dimension of the irreducible subspace is given by the traseless part $(\epsilon_{lmn} u^{n}v^{kn})$.
I don’t know what you mean. You can not obtain a lower rank tensor from irreducible tensor. But you can write $$T^{(ijk)} = \frac{1}{6} (\delta^{i}_{l}\delta^{j}_{m}\delta^{k}_{n} + \cdots ) T^{lmn} ,$$ which means that you are isolating $10$ independent components from the $27$ components. Symbolically; $$T^{(10)} \sim (\delta \delta \delta) T^{(27)} .$$ Similarly, in 3-dimension, you can use the determinant identity for $\epsilon^{ijk}\epsilon_{lmn}$ to write $$T^{[ijk]} = \frac{1}{6} \epsilon^{ijk}( \epsilon_{lmn} T^{lmn}) = \epsilon^{ijk} [1],$$ which means that a one component tensor (i.e., scalar) is obtained from the $27$-component tensor $T^{ijk}$: $T^{(1)} \sim (\epsilon) T^{(27)}$.

In 3-dimension $T^{[ijkl]}=0$, because the indices run from 1 to 3.
Look, you can understand this stuff by doing more examples and relating them to the representation matrices of the group algebra. Another nice and important decomposition is $$[3] \otimes [3] \otimes [3] = [10] \oplus [1] \oplus [8] \oplus [8] .$$ So, have a go: write $T^{ijk}=q^{i}q^{j}q^{k}$, and show that
$$q^{i}q^{j}q^{k} = T^{(ijk)} + \frac{1}{6} \epsilon^{ijk}(\epsilon_{lmn}q^{l}q^{m}q^{n}) + \frac{1}{3} \epsilon^{ijl}\Lambda^{k}_{l} + \frac{1}{3}\epsilon^{ikl}\Gamma^{j}_{l} ,$$
where,
$$\Lambda^{k}_{l} = \epsilon_{lmn}( T^{mnk} + T^{knm}) \in [8] ,$$
$$\Gamma^{j}_{l} = \epsilon_{lmn} ( T^{mjn} + T^{jmn}) \in [8] .$$

12. May 13, 2016

### PineApple2

I get this decomposition:
$$T^{ijk}=T^{(ijk)}+T^{[ijk]}+\frac{4}{3}T^{(i[j)k]}+\frac{4}{3}T^{[i(j]k)}$$
is that right? because the last two terms don't match to your last two terms. For example
$$\frac{4}{3}T^{(i[j)k]}=\frac{4}{3}\cdot\frac{1}{4}(T^{ijk} -T^{ikj}+T^{jik}-T^{jki})=\frac{1}{3}(\epsilon^{jk\ell}\epsilon_{\ell mn}T^{imn}+\epsilon^{ik\ell}\epsilon_{\ell mn}T^{jmn})$$
which is not equal to either your third or your fourth term (even though close).

13. May 14, 2016

### samalkhaiat

Eleven months is a long time to spend on that problem, don’t you think?
$$T^{(abc)} + T^{[abc]} = \frac{1}{3} \left( T^{abc} + T^{cab} + T^{bca} \right) .$$
$$T^{abc} = T^{(abc)} + T^{[abc]} + \frac{1}{3} \left(T^{abc} - T^{bca}\right) + \frac{1}{3} \left(T^{abc} - T^{cab}\right) .$$
Now, add $\frac{1}{3}(T^{bac} - T^{cba})$ to the third term (to make it antisymmetric in $(ac)$), and subtract it from the fourth term to make it antisymmetric in $(ab)$. Okay, I do the third term for you and you do the fourth.
\begin{align*} \mbox{third term} &= \frac{1}{3} \left( \delta^{a}_{m} \delta^{c}_{n} - \delta^{a}_{n} \delta^{c}_{m} \right) \left( T^{mbn} + T^{bmn} \right) \\ &= \frac{1}{3} \epsilon^{acl}\epsilon_{lmn} \left( T^{mbn} + T^{bmn} \right) \\ &= \frac{1}{3} \epsilon^{acl} \Gamma_{l}^{b} . \end{align*}
Notice that $\Gamma_{b}^{b} = 0$. Thus, $\Gamma^{a}_{c} \sim [8]$.