# An irreducible decomposition in SU(3)

1. Jul 1, 2013

### Wonchu

Hi,
I'v got a problem on Chapter 10 of Georgi's "Lie Algebras in Particle Physics".
I want to prove

$u^i v^j_k = \frac{1}{2}A_k^{ij} + \frac{1}{4}B_k^{ij} + \frac{1}{8}C_k^{ij} \ \ \ \ \ \ \ (10.29) \\$
i.e.
$3 \bigotimes 8 = 15 \bigoplus \bar{6} \bigoplus 3 \\$
where
$A_k^{ij} \equiv u^i v^j_k - u^j v^i_k - \frac{1}{4}\delta^i_k u^l v^j_l - \frac{1}{4}\delta^j_k u^l v^i_l \\ B_k^{ij} \equiv \epsilon^{ijl} ( \epsilon_{lmn}u^m v^n_k - \epsilon_{kmn}u^m v^n_l ) \\ C_k^{ij} \equiv 3\delta^i_k u^l v^j_l - \delta^j_k u^l v^i_l \\$

I persist in explaining my procedure in detail to check my calculation.
First, noting that
$u^i v^j_k = \frac{1}{2}(u^i v^j_k + u^j v^i_k) + \frac{1}{2}(u^i v^j_k - u^j v^i_k)$
In order to have a symmetric(with respect to i and j) and traceless tensor from $(u^i v^j_k + u^j v^i_k)$ ,I define $A^{ij}_k$ ,
$A^{ij}_k \equiv u^i v^j_k + u^j v^i_k -\frac{1}{4}Y^{ij}_k$
where
\begin{equation*}
\begin{split}
Y^{ij}_k & \equiv \delta^{i}_k (u^l v^j_l + u^j v^l_l) + (i \leftrightarrow j) \\
& = \delta^{i}_k u^l v^j_l + \delta^{j}_k u^l v^i_l \\
( ∴Y^{ij}_i & = 4 u^l v^j_l ) \\
\end{split}
\end{equation*}

Using $A^{ij}_k$,I'v got

$u^i v^j_k = \frac{1}{2}A^{ij}_k + \frac{1}{2} (u^i v^j_k - u^j v^i_k + \frac{1}{4}\delta^i_k u^l v^j_l + \frac{1}{4}\delta^j_k u^l v^i_l ) \\$

Here,
\begin{equation*}
\begin{split}
\frac{1}{2}(u^i v^j_k - u^j v^i_k) & = \frac{1}{2} \epsilon^{ijl}\epsilon_{lmn}u^m v^n_k \\
& \equiv \frac{1}{2}\epsilon^{ijl}X_{lk} \\
\end{split}
\end{equation*}

Noting that

$X_{lk} = \frac{1}{2}(X_{lk}+X_{kl}) - \frac{1}{2}(X_{lk}-X_{kl}) \\$
I'v got
\begin{equation*}
\begin{split}
\frac{1}{2}(u^i v^j_k - u^j v^i_k)
& = \frac{1}{4}\epsilon^{ijl}(X_{lk}+X_{kl}) + \frac{1}{4}\epsilon^{ijl}(X_{lk}-X_{kl}) \\
& = \frac{1}{4}\epsilon^{ijl}(\epsilon_{lmn}u^m v^n_k + \epsilon_{kmn}u^m v^n_l) +
\frac{1}{4}\epsilon^{ijl}(\epsilon_{lmn}u^m v^n_k - \epsilon_{kmn} u^m v^n_l) \\
& \equiv \frac{1}{4}B^{ij}_k + \frac{1}{4} \cdot \frac{1}{2} \tilde{D}^{ij}_k \\
\end{split}
\end{equation*}
where
$\tilde{D}^{ij}_k=2\epsilon^{ijl}(\epsilon_{lmn}u^m v^n_k - \epsilon_{kmn} u^m v^n_l) \\$

\begin{equation*}
\begin{split}
∴ u^i v^j_k
& = \frac{1}{2}A^{ij}_k + \frac{1}{4}B^{ij}_k + \frac{1}{8}(\tilde{D}^{ij}_k + \delta^i_k u^l v^j_l + \delta^j_k u^l v^i_l ) \\
& \equiv \frac{1}{2}A^{ij}_k + \frac{1}{4}B^{ij}_k + \frac{1}{8}D^{ij}_k \\
\end{split}
\end{equation*}
where
$D_k^{ij} \equiv 2 \epsilon^{ijl} ( \epsilon_{lmn}u^m v^n_k - \epsilon_{kmn}u^m v^n_l ) + \delta^i_k u^l v^j_l + \delta^j_k u^l v^i_l$

Then the following equation must hold to get (10.29) in the textbook,
$C_k^{ij} = D_k^{ij}$
i.e.
$\epsilon^{ijl}\epsilon_{kmn}u^m v^n_l = u^i v^j_k - u^j v^i_k - \delta^i_k u^l v^j_l + \delta^j_k u^l v^i_l \\$

Why is this ?
Or have I mistaken anywhere ?
Besides,
$C^{ij}_j =0 , C^{ij}_i \neq 0 \\$
What implication does exist in the background ?