I have a question about the tensor decomposition of $\mathrm{SU(3)}$. According to Georgi (page 142 and 143), a tensor $T^i{}_j$ decomposes as: \begin{equation} \mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1} \end{equation} where the $\mathbf{1}$ represents the trace. However, I do not understand why we cannot further decompose the traceless part into a symmetric and an antisymmetric part.

In order to understand my logic: A general tensor $\varphi^i$ transforms as: \begin{equation} \varphi^i \rightarrow U^i{}_j \varphi^j \end{equation} whereas $\varphi_i$ transforms as: \begin{equation} \varphi_i \rightarrow (U^*)_i{}^j \varphi_j \end{equation} where $U \in \mathrm{SU(3)}$ is a $3 \times 3$ matrix. Now, I will let $S^i{}_j$ denote the traceless part of $T^i{}_j$ (i.e. $S^i{}_j$ has dimensions $\mathbf{8}$) and we can decompose this in the "symmetric" and "antisymmetric" part as usual: \begin{equation} S^i{}_j = \frac{1}{2}(S^i{}_j + S_j{}^i) + \frac{1}{2}(S^i{}_j - S_j{}^i) \end{equation} Then under an $\mathrm{SU(3)}$ transformation: \begin{equation} S^i{}_j + S_j{}^i \rightarrow U^i{}_k (U^*)_j{}^l S^k{}_l + U^i{}_k (U^*)_j{}^l S^k{}_l = U^i{}_k (U^*)_j{}^l (S^i{}_j + S_j{}^i) \end{equation} and: \begin{equation} S^i{}_j - S_j{}^i \rightarrow U^i{}_k (U^*)_j{}^l S^k{}_l - U^i{}_k (U^*)_j{}^l S^k{}_l = U^i{}_k (U^*)_j{}^l (S^i{}_j - S_j{}^i) \end{equation} Therefore, the symmetric part keeps its symmetry and the antisymmetric part keeps its antisymmetry. Thus two invariant subspaces are created and the representation is reducible? To sum up, I would think we decompose $T^i{}_j$ as: \begin{equation} \mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{3} \oplus \mathbf{5} \oplus \mathbf{1} \end{equation} where $\mathbf{3}$ denotes the dimensions of the antisymmetric part and $\mathbf{5}$ denotes the dimensions of the symmetric part. Where am I going wrong?

Edit: I got my convention from "Invariances in Physics and Group Theory" by Jean-Bernard Zuber:

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    $\begingroup$ How would the tensor $S_j{}^i$ be related to $S^i{}_j$? $\endgroup$
    – Olof
    Dec 6, 2013 at 8:56
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    $\begingroup$ I think there is a problem in the transformations of the symmetric and antisymmetric parts... I don't agree with these equations, let me check it! $\endgroup$ Dec 6, 2013 at 9:28
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    $\begingroup$ @AstoundingJB: That's my point. It looks like Hunter tries to raise and lower the indices in order to construct the "symmetric" and "anti-symmetric" tensors, but there is no way to do this in $SU(3)$. So the only sensible interpretation is $S_j{}^i - S^i{}_j = 0$, which of course gives an irreducible representation, but not a very interesting one :) $\endgroup$
    – Olof
    Dec 6, 2013 at 10:57
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    $\begingroup$ Good one @Olof ! +1 I didn't get your suggestion at first! :) Probably, this is the key point for one can't decompose further $\boldsymbol{8}$ into... something! $\endgroup$ Dec 6, 2013 at 11:08
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    $\begingroup$ Uhm.. wait! I'm wandering if you're confusing the meaning of a transformation matrix, say those you named $U_i^{\phantom{i}j}$, and a base state for this representation, say $S^i_j$ or $T^i_j$... They are very different objects, indeed Georgi write these bases like $\big| ^i_j\big\rangle$, while the $U$s are matrices... The point is that the transpose of a base tensor like $S^i_j=\big| ^i_j\big\rangle$ doesn't actually have much sense... It has no sense mixing the two indices, one fro $\boldsymbol{3}$ and one from $\bar{\boldsymbol{3}}$... $\endgroup$ Dec 6, 2013 at 13:49

4 Answers 4


SECTION A : What remains invariant for a complex $\:3\times 3\:$ tensor depends upon its transformation law under $\:U \in SU(3)\:$

CASE 1 : $\:\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}=\boldsymbol{6}\boldsymbol{\oplus}\overline{\boldsymbol{3}}\:$

The transformation law for the complex $\:3\times 3\:$ tensor $\:\mathrm{X}\:$ in this case is \begin{equation} \mathrm{X }^{\prime}=U\mathrm{X}U^{\mathsf{T}}\quad \tag{A-01} \end{equation} Here the symmetry (+) or antisymmetry (-) is invariant since \begin{equation} \mathrm{X}^{\mathsf{T}}=\pm\:\mathrm{X} \Longrightarrow {(\mathrm{X }^{\prime})}^{\mathsf{T}}=(U\mathrm{X}U^{\mathsf{T}})^{\mathsf{T}}= {(U^{\mathsf{T}})}^{\mathsf{T}}\mathrm{X}^{\mathsf{T}}U^{\mathsf{T}}=U(\pm\:\mathrm{X})U^{\mathsf{T}}=\pm\:\mathrm{X }^{\prime} \tag{A-02} \end{equation}

In this case it makes sense to split the tensor in its symmetric and anti-symmetric parts \begin{equation} \mathrm{\Psi}=\dfrac{1}{2} \left(\mathrm{X}+\mathrm{X}^{\mathsf{T}}\right)\:, \quad\mathrm{\Omega}=\dfrac{1}{2} \left(\mathrm{X}-\mathrm{X}^{\mathsf{T}}\right) \tag{A-03} \end{equation} The symmetric part $\:\mathrm{\Psi}\:$ depends on 6 parameters, so is identical to a complex $\:6$-vector $\:\boldsymbol{\psi}\:$ which belongs to a complex 6-dimensional invariant subspace and is transformed under a special unitary transformation $\:W \in SU(6)\:$

\begin{equation} \boldsymbol{\psi}^{\prime}=W\boldsymbol{\psi}\:, \quad W \in SU(6) \tag{A-04} \end{equation}

while, on the other hand, the anti-symmetric part $\:\mathrm{\Omega}\:$ depends on 3 parameters, so is identical to a complex $\:3$-vector $\:\boldsymbol{\omega}\:$ which belongs to a complex 3-dimensional invariant subspace and is transformed under the special unitary transformation $\:\overline{U} \in SU(3)\:$

\begin{equation} \boldsymbol{\omega}^{\prime}=\overline{U}\boldsymbol{\omega}\:, \quad \overline{U} \in SU(3) \tag{A-05} \end{equation} That's why the symmetric and anti-symmetric parts give rise to the terms $\:\boldsymbol{6}\:$ and $\:\overline{\boldsymbol{3}}\:$ in the right hand of equation $\:\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}=\boldsymbol{6}\boldsymbol{\oplus}\overline{\boldsymbol{3}}\:$ respectively.

CASE 2 : $\:\boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}=\boldsymbol{8}\boldsymbol{\oplus}\boldsymbol{1}\:$

The transformation law for the complex $\:3\times 3\:$ tensor $\:\mathrm{X}\:$ in this case is \begin{equation} \mathrm{X }^{\prime}=U\mathrm{X}U^{\boldsymbol{*}}=U\mathrm{X}U^{-1} \tag{A-06} \end{equation} For those interested, this is proved in SECTION B, motivated by the adventure of explaining structure of mesons under quark theory.
Here the symmetry (+) or antisymmetry (-) is NOT invariant \begin{equation} \mathrm{X}^{\mathsf{T}}=\pm\:\mathrm{X} \Longrightarrow {(\mathrm{X }^{\prime})}^{\mathsf{T}}=(U\mathrm{X}U^{\boldsymbol{*}})^{\mathsf{T}}= {(U^{\boldsymbol{*}})}^{\mathsf{T}}\mathrm{X}^{\mathsf{T}}U^{\mathsf{T}}=\overline{U}(\pm\:\mathrm{X})U^{\mathsf{T}} \ne\pm\:\mathrm{X }^{\prime} \tag{A-07} \end{equation}

So it makes NO SENSE to split the tensor in its symmetric and anti-symmetric parts.

On the contrary :

(1) if $\:\mathrm{X}\:$ is a constant tensor, that is a scalar multiple of the identity, $\:\mathrm{X}=z\mathrm{I}\:$ ($\:z \in \mathbb{C}\:$) , then is invariant $\:\mathrm{X }^{\prime}= U\mathrm{X}U^{-1}=U\left(z\mathrm{I}\right)U^{-1}=z\mathrm{I}=\mathrm{X}\:$


(2) since the transformation (A-06) is a similarity transformation, it preserves the Trace (=sum of the elements on the main diagonal) of $\:\mathrm{X}\:$, that is $\:Tr \left(\mathrm{X}^{\prime}\right)=Tr\left(\mathrm{X}\right)\:$. So a traceless tensor remains traceless.

It would sound not very well, but in this case the invariants are the "tracelessness" and the "scalarness".

In this case it makes sense to split the tensor in a traceless and in a scalar part : \begin{equation} \mathrm{\Phi}=\mathrm{X}-\left[\dfrac{1}{3}Tr\left(\mathrm{X}\right)\right]\cdot\mathrm{I}\:, \quad \mathrm{\Upsilon}=\left[\dfrac{1}{3}Tr\left(\mathrm{X}\right)\right]\cdot\mathrm{I} \tag{A-08} \end{equation} The traceless part $\:\mathrm{\Phi}\:$ depends on 8 (=3x3-1) parameters, so is identical to a complex $\:8$-vector $\:\boldsymbol{\phi}\:$ which belongs to a complex 8-dimensional invariant subspace NOT FURTHER REDUCED TO INVARIANTS SUBSPACES and is transformed under a special unitary transformation $\:V \in SU(8)\:$ \begin{equation} \boldsymbol{\phi}^{\prime}=V\boldsymbol{\phi}\:, \quad V \in SU(8) \tag{A-09} \end{equation} while, on the other hand, the scalar part $\:\mathrm{\Upsilon}\:$ depends on 1 parameter, so is identical to a complex $\:1$-vector $\:\boldsymbol{\upsilon}\:$ which belongs to a complex 1-dimensional invariant subspace (identical to the set of complex numbers $\:\mathbb{C}\:$) and is transformed under the special unitary transformation $\:\mathrm{I} \in SU(1)\:$
(identical to the identity) \begin{equation} \boldsymbol{\upsilon}^{\prime}=\mathrm{I}\boldsymbol{\upsilon}=\boldsymbol{\upsilon} \tag{A-10} \end{equation} Note that $\:SU(1)\equiv \{\:\mathrm{I}\:\}\:$, that is the group $\:SU(1)\:$ has only one element, the identity $\:\mathrm{I}\:$, while $\:U(1)\equiv\{\:U\::\:U=e^{i\theta}\mathrm{I}\:, \quad \theta \in \mathbb{R} \}\:$, that is mathematically identical to the unit circle in $\:\mathbb{C}\:$.


SECTION B : Mesons from three quarks

Suppose we know the existence of three quarks only : $\boldsymbol{u}$, $\boldsymbol{d}$ and $\boldsymbol{s}$. Under full symmetry (the same mass) these are the basic states, let
\begin{equation} \boldsymbol{u}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \qquad \boldsymbol{d}= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \qquad \boldsymbol{s}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \tag{B-01} \end{equation} of a 3-dimensional complex Hilbert space of quarks, say $\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}$. A quark $\boldsymbol{\xi} \in \mathbf{Q}$ is expressed in terms of these basic states as \begin{equation} \boldsymbol{\xi}=\xi_1\boldsymbol{u}+\xi_2\boldsymbol{d}+\xi_3\boldsymbol{s}= \begin{bmatrix} \xi_1\\ \xi_2\\ \xi_3 \end{bmatrix} \qquad \xi_1,\xi_2,\xi_3 \in \mathbb{C} \tag{B-02} \end{equation} For a quark $\boldsymbol{\eta} \in \mathbf{Q}$ \begin{equation} \boldsymbol{\eta}=\eta_1\boldsymbol{u}+\eta_2\boldsymbol{d}+\eta_3\boldsymbol{s}= \begin{bmatrix} \eta_1\\ \eta_2\\ \eta_3 \end{bmatrix} \tag{B-03} \end{equation}

the respective antiquark $\overline{\boldsymbol{\eta}}$ is expressed by the complex conjugates of the coordinates

\begin{equation} \overline{\boldsymbol{\eta}}=\overline{\eta}_1 \overline{\boldsymbol{u}}+\overline{\eta}_2\overline{\boldsymbol{d}}+\overline{\eta}_3\overline{\boldsymbol{s}}= \begin{bmatrix} \overline{\eta}_1\\ \overline{\eta}_2\\ \overline{\eta}_3 \end{bmatrix} \tag{B-04} \end{equation} with respect to the basic states
\begin{equation} \overline{\boldsymbol{u}}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \qquad \overline{\boldsymbol{d}}= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \qquad \overline{\boldsymbol{s}}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \tag{B-05} \end{equation} the antiquarks of $\boldsymbol{u},\boldsymbol{d}$ and $\boldsymbol{s}$ respectively. The antiquarks belong to a different space, the space of antiquarks $\overline{\mathbf{Q}}\equiv \mathbb{C}^{\boldsymbol{3}}$.

Since a meson is a quark-antiquark pair, we'll try to find the product space \begin{equation} \mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\: \left(\equiv \mathbb{C}^{\boldsymbol{9}}\right) \tag{B-06} \end{equation}

Using the expressions (B-02) and (B-04) of the quark $\boldsymbol{\xi} \in \mathbf{Q}$ and the antiquark $\overline{\boldsymbol{\eta}} \in \overline{\mathbf{Q}}$ respectively, we have for the product meson state $ \mathrm{X} \in \mathbf{M}$ \begin{equation} \begin{split} \mathrm{X}=\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\eta}}=&\xi_1\overline{\eta}_1 \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_1\overline{\eta}_2 \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_1\overline{\eta}_3 \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+ \\ &\xi_2\overline{\eta}_1 \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_2\overline{\eta}_2 \left( \boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_2\overline{\eta}_3 \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+\\ &\xi_3\overline{\eta}_1 \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_3\overline{\eta}_2 \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_3\overline{\eta}_3 \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right) \end{split} \tag{B-07} \end{equation}

In order to simplify the expressions, the product symbol $"\boldsymbol{\otimes}"$ is omitted and so \begin{equation} \begin{split} \mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\eta}}=&\xi_1\overline{\eta}_1 \left(\boldsymbol{u}\overline{\boldsymbol{u}}\right)+\xi_1\overline{\eta}_2 \left(\boldsymbol{u}\overline{\boldsymbol{d}}\right)+\xi_1\overline{\eta}_3 \left(\boldsymbol{u}\overline{\boldsymbol{s}}\right)+ \\ &\xi_2\overline{\eta}_1 \left(\boldsymbol{d}\overline{\boldsymbol{u}}\right)+\xi_2\overline{\eta}_2 \left( \boldsymbol{d}\overline{\boldsymbol{d}}\right)+\xi_2\overline{\eta}_3 \left(\boldsymbol{d}\overline{\boldsymbol{s}}\right)+\\ &\xi_3\overline{\eta}_1 \left(\boldsymbol{s}\overline{\boldsymbol{u}}\right)+\xi_3\overline{\eta}_2 \left(\boldsymbol{s}\overline{\boldsymbol{d}}\right)+\xi_3\overline{\eta}_3 \left(\boldsymbol{s}\overline{\boldsymbol{s}}\right) \end{split} \tag{B-08} \end{equation} Due to the fact that $\mathbf{Q}$ and $\overline{\mathbf{Q}}$ are of the same dimension, it's convenient to represent the meson states in the product 9-dimensional complex space $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ by square $3 \times 3$ matrices instead of row or column vectors

\begin{equation} \mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\eta}}= \begin{bmatrix} \xi_1\overline{\eta}_1 & \xi_1\overline{\eta}_2 & \xi_1\overline{\eta}_3\\ \xi_2\overline{\eta}_1 & \xi_2\overline{\eta}_2 & \xi_2\overline{\eta}_3\\ \xi_3\overline{\eta}_1 & \xi_3\overline{\eta}_2 & \xi_s\overline{\eta}_3 \end{bmatrix}= \begin{bmatrix} \xi_1\\ \xi_2\\ \xi_3 \end{bmatrix} \begin{bmatrix} \overline{\eta}_1 \\ \overline{\eta}_2 \\ \overline{\eta}_3 \end{bmatrix}^{\mathsf{T}} = \begin{bmatrix} \xi_1\\ \xi_2\\ \xi_3 \end{bmatrix} \begin{bmatrix} \overline{\eta}_1 & \overline{\eta}_2 & \overline{\eta}_3 \end{bmatrix} \tag{B-09} \end{equation}

Now, under a unitary transformation $\;U \in SU(3)\;$ in the 3-dimensional space of quarks $\;\mathbf{Q}\;$, we have \begin{equation} \boldsymbol{\xi}^{\prime}= U\boldsymbol{\xi} \tag{B-10} \end{equation} so in the space of antiquarks $\overline{\mathbf{Q}}\;$, since $\;\boldsymbol{\eta}^{\prime}=U\boldsymbol{\eta}\;$ \begin{equation} \overline{\boldsymbol{\eta}^{\prime}}= \overline{U}\;\overline{\boldsymbol{\eta}} \tag{B-11} \end{equation} and for the meson state \begin{equation} \mathrm{X}^{\prime}=\boldsymbol{\xi}^{\prime}\boldsymbol{\otimes}\overline{\boldsymbol{\eta}^{\prime}}=\left(U\boldsymbol{\xi}\right)\left(\overline{U}\overline{\boldsymbol{\eta}}\right) = \Biggl(U\begin{bmatrix} \xi_1\\ \xi_2\\ \xi_3 \end{bmatrix}\Biggr) \Biggl(\overline{U}\begin{bmatrix} \overline{\eta}_1\\ \overline{\eta}_2\\ \overline{\eta}_3 \end{bmatrix}\Biggr)^{\mathsf{T}} \\= U\Biggl(\begin{bmatrix} \xi_1\\ \xi_2\\ \xi_3 \end{bmatrix} \begin{bmatrix} \overline{\eta}_1 & \overline{\eta}_2 & \overline{\eta}_3 \end{bmatrix}\Biggr)\overline{U}^{\mathsf{T}} = U\left(\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\eta}}\right)U^{*}=U\;\mathrm{X}\;U^{*} \tag{B-12} \end{equation} so proving the transformation law (A-06).

$===================\text{end of answer}=======================$

The quark structure of $\:\boldsymbol{\eta}^{\prime}\:$,$\:\boldsymbol{\eta}\:$ and $\:\boldsymbol{\pi}^{0}\:$  mesons

FIGURE : The quark structure of $\:\boldsymbol{\eta}^{\prime}\:$,$\:\boldsymbol{\eta}\:$ and $\:\boldsymbol{\pi}^{0}\:$ mesons

(Note: Meson symbols $\:\boldsymbol{\eta}^{\prime}\:$ and $\:\boldsymbol{\eta}\:$ must not be confused with the complex 3-vectors in the text)

(1) Meson $\:\boldsymbol{\eta}^{\prime}\:$ is a singlet, representative of $\:\boldsymbol{1}\:$ in $\:\boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}=\boldsymbol{8}\boldsymbol{\oplus}\boldsymbol{1}\:$
(2) Mesons $\:\boldsymbol{\eta}\:$ and $\:\boldsymbol{\pi}^{0}\:$ are members of the octet $\;\boldsymbol{\lbrace}\boldsymbol{\pi}^{+},\boldsymbol{\pi}^{-},\boldsymbol{\pi}^{0},\mathbf{K}^{+},\mathbf{K}^{-},\mathbf{K}^{0},\overline{\mathbf{K}}^{0},\boldsymbol{\eta}\boldsymbol{\rbrace}\;$, basic meson states of $\boldsymbol{8}\:$ in $\:\boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}=\boldsymbol{8}\boldsymbol{\oplus}\boldsymbol{1}\:$ where

$\:\boldsymbol{\pi}^{+}\equiv\boldsymbol{u}\overline{\boldsymbol{d}}\:$ , $\:\boldsymbol{\pi}^{-}\equiv\boldsymbol{d}\overline{\boldsymbol{u}}\:$ , $\:\mathbf{K}^{+}\equiv\boldsymbol{u}\overline{\boldsymbol{s}}\:$ , $\:\mathbf{K}^{-}\equiv\boldsymbol{s}\overline{\boldsymbol{u}}\:$ , $\:\mathbf{K}^{0}\equiv\boldsymbol{d}\overline{\boldsymbol{s}}\:$ , $\:\overline{\mathbf{K}}^{0}\equiv\boldsymbol{s}\overline{\boldsymbol{d}}\:$ .



Ok, I think there is a mistake here:

A general tensor $\varphi^i$ transforms as: $$\varphi^i\rightarrow U^i_{\phantom{1}j}\varphi^j$$ whereas $\varphi_i$ transforms as: $$\varphi_i\rightarrow (U^\boldsymbol{\ast})_i^{\phantom{1}j}\varphi_j$$

Where did you find these equations? The unitary matrix element in the second line should not be a complex conjugate. I don't remember Giorgi's conventions but the customary notation I'm used to is this one: $$U_i^{\phantom{i}j}=U_{ij},\quad \varphi_i\rightarrow U_i^{\phantom{1}j}\varphi_j\\ U^i_{\phantom{i}j}=U^\ast_{ij},\quad \varphi^\ast_i\equiv \varphi^i\rightarrow U^i_{\phantom{1}j}\varphi^j\equiv(U_i^{\phantom{i}j}\varphi_j)^\ast.$$ Hence, in your equations I'd understand: $$(U^\ast)_i^{\phantom{i}j}\equiv U^\ast_{ij}=U^i_{\phantom{i}j}$$ and it doesn't provide the right transformation law for $\varphi_i$.

EDIT: well, provided the previous comments, let me clarify some issues with the notation, that may led to confuse the meaning of these transformation laws. Let us choose the convention to denote $SU(N)$ transformations, that is $N\times N$ unitary matrices with unit determinant, with uppercase letters, like $U$, and base states (scalars, vectors and tensors) with lowercase Greek letters, $\psi\in \mathbb{C}^N$. For example vector states transform as: $$\psi\to U\psi,\quad \psi_i\to U_{ij}\psi_j\equiv U_i^{\ j}\psi_j$$ Note that here I followed the convention of writing base states of the fundamental or vector representation with lower indices, as Georgi does and as you can find here. This is the convention I'm used to, but nothing stops you to do the contrary, choosing upper indices! Note also that $U\psi$ represents the ordinary product of an $N\times N$ matrix by a vector $\psi=(\psi_1,\ldots,\psi_N)^T$, and produce a vector of the same type. In the notation $U_{ij}$ the index $i$ represents the rows whereas the second index $j$ represents the columns. It's customary to write it like $U_i^{\ j}$ to distinguish rows and columns. $\psi_i$ is a column vector and $i$ counts its rows. You can define the conjugate representation by means of the conjugate vectors $\psi_i^\ast$, whose transformation law is $$\psi^\ast\to (U\psi)^\ast=\psi^\ast U^\ast,\quad \psi_i^\ast\to (U^\ast)_{ij}\psi_j^\ast=\psi_j^\ast(U^\dagger)_{ji}$$ Since these conjugate vectors transform in a different way with respect to $\psi_i$, it's useful to introduce upper indices to distinguish them: $$\psi^i\equiv \psi_i^\ast \to U^\ast_{ij}\psi_j^\ast\equiv U^i_{\ \ j} \psi^j.$$ As you can see, now indices are "summed on the bottom-right". The extension to any arbitrary $(p,q)$-tensor is trivial, their transformation law are those of the direct (diagonal) product of $p$ type $\psi^i$ vectors and $q$ type $\psi_i$ vectors: $$\psi^{i_1\ldots i_p}_{j_1\ldots j_q}\to \big(U_{j_1}^{\ \ j'_1}\cdot\ldots\cdot U_{j_q}^{\ \ j'_q}\big)\big(U^{i_1}_{\ \ i'_1}\cdot\ldots\cdot U^{i_p}_{\ \ i'_p}\big)\psi^{i'_1\ldots i'_p}_{j'_1\ldots j'_q}.$$ Since upper and lower indices represents different objects it has no sense mixing them.

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    $\begingroup$ Ok, I just took a look at Giorgi's book. The conventions I listed are consistent with Giorgi's eq. (10.6-8)! Also, take a look at this tutorial, you could find it very usefull: phys.nthu.edu.tw/~class/group_theory2012fall/doc/tensor.pdf $\endgroup$ Dec 6, 2013 at 10:27
  • $\begingroup$ Thanks for you reply. I have made an edit so you can see where I got my conventions from. Maybe using all these different conventions are confusing me, and I need to stick to the one you are suggesting. $\endgroup$
    – Hunter
    Dec 6, 2013 at 13:31
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    $\begingroup$ Here it is the edit! I had to wait for the coffee break! ;) $\endgroup$ Dec 6, 2013 at 16:28
  • $\begingroup$ Great answer! Thank you; really clarifies a lot of my confusion. $\endgroup$
    – Hunter
    Dec 6, 2013 at 16:42

First, if you take the fundamental representation (representation $N$) of $SU(N)$ made of $N$ objects $\varphi^i$, the transformation law is :

$\varphi^i \to U^i{}_j \varphi^j$.

By taking the complex conjugate, you get : $\varphi^{*i} \to (U^*)^i{}_j \varphi^{*j}= (U^\dagger)^j{}_i \varphi^{*j}$.

Now, looking at the last expression with $U^\dagger$, one sees that it is more practical to define objects $\varphi_i$, wich transform like $\varphi^{*i}$ :

$\varphi_{i} \to (U^\dagger)^j{}_i \varphi_{j}$,

This is the representation $\bar N$

Now clearly, when you make the product of the two representations $N$ and $\bar N$, you have a representation $T^i_j$ which transforms as $\varphi^i\varphi_j$ :

$T^i_j \to (U)^i_k (U^\dagger)^l_j T^k_l$

Secondly, you cannot symmetrize or anti-symmetrize the representation $N \otimes \bar N$, that is $T^i_j$, because the indices $i$ and $j$ have a different nature, and correspond to different representations.

Now, if you consider the representation $N \otimes N$, that is some representation $S^{ij}$, then here you may separe in a symmetric and anti-symmetric part, for instance, you have :

$3 \otimes 3 = 6 \oplus \bar 3 $

The $6$ is the symmetric part, while the $\bar 3$ is dual (equivalent) to the anti-symmetric part, thanks to the Levi-Civita tensor : $\varphi_i = \epsilon_{ijk} \varphi^{jk}$


Due to OP comments, some precisions :

You have $U^\dagger = (U^*)^T$, where $T$ means transposed operation. Transposition means exchange of the row and columns of the matrix, that is exchange of the $i$ and $j$ indices. If you put the row indice as an upper indice and the column indice as a lower indice, then the exchange necessarily will put the row indice as a lower indice, and the column indice as a upper indice. Your notation $(U^*)^i{}_j = (U^\dagger)_j{}^i$ is a not-too-good equivalent notation, I say not-too-good, because you loose the orginal meaning that I describe above .About the representations, this is a different thing (these are not the same $i$ and $j$...), the upper indice transforms as a $N$ representation, and the lower indice transforms as a $\bar N$ representation, so it is like apples and bananas, you can only symmetrize or anti-symmetrize equivalent quantities which transform in the same manner ($2$ apples or $2$ bananas), but not $1$ banana + $1$ apple

  • $\begingroup$ Thank you for your reply. I have two questions, I was hoping you could elaborate on? i) The second equation you write in your message implies $(U^*)^i{}_j = (U^\dagger)^j{}_i$ but I don't understand this as I have learned to write this as: $(U^*)^i{}_j = (U^\dagger)_j{}^i$. Could you explain me your reasoning? $\endgroup$
    – Hunter
    Dec 6, 2013 at 13:38
  • $\begingroup$ ii) What do you mean with: "the indices $i$ and $j$ have a different nature" (and thus cannot be symetrized)? $\endgroup$
    – Hunter
    Dec 6, 2013 at 13:40
  • $\begingroup$ Any help is much appreciated! $\endgroup$
    – Hunter
    Dec 6, 2013 at 13:40
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    $\begingroup$ @Hunter: I updated the answer $\endgroup$
    – Trimok
    Dec 6, 2013 at 20:25
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    $\begingroup$ Good answer! Should've gotten more points. Although one should perhaps choose a different label in e.g. $\varphi^i\varphi_j$ i.e. something like $\varphi^i\eta_j$? $\endgroup$ Sep 28, 2014 at 23:08

The problem is: you are not allowed to (anti)symmetrize upper indices with lower indices. In other words $$ S^{i}_{\;j} = S^{\;i}_{j}\,. $$ Therefore the decomposition into symmetric + antisymmetric is just the decomposition whole vector space + zero vector space, which is trivial.

Why aren't we allowed to mix upper and lower indices?

  1. Because strictly speaking they act on different spaces, so there is no natural action of the permutation group over it. A tensor of type $(p,q)$ is a map $$ T(p,q) : V^{\otimes p} \otimes \tilde{V}{}^{\otimes q} \to \mathbb{R}\,, $$ where $V$ is a vector space and $\tilde{V}$ its dual (space of functionals $f:V\to\mathbb{R}$).

  2. Because you can actually get rid of the upper indices and work only with lower ones by doing $$ \epsilon_{ijk}\,v^{i}_{\ldots} = v_{jk\ldots}\,. $$ then there is not meaning into permuting a single index with a pair of antisymmetrized indices.

  3. The irreducible representation dimensions that one can get are given by the Weyl dimension formula $$ \mathrm{dim}[a,b] = \tfrac12 [(a+1)(b+1)(a+b+2)]\,, \quad a \geq 0, \,\,b\geq 0\,,\, a,b\in \mathbb{N} $$ and you can check that there is no way to get $5$ out of it.

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