3
$\begingroup$

I have a number of questions relating to the construction of $SU(n)$ tensors.

  1. It is unclear to me how to use Young tableaux to construct tensors, as there seem to be a number of different conventions around. For example, consider the adjoint $(1, 1)$ of $SU(3)$ with the tableau shown below. One convention (see e.g. Georgi eq. 12.6) is to symmetrise over $i$ and $j$ first, and then antisymmetrise with $k$, yielding $$ \psi_{ij;k} = \frac{1}{2}\left(\psi_{[i|j|k]}+\psi_{[j|i|k]} \right)= \frac{1}{4}\left( \psi_{ijk}-\psi_{kji}+\psi_{jik}-\psi_{jki}\right). \tag{1} $$ However, I have also seen people (see e.g. page 5 of "Note 5: Tensor Method in SU(n)" here) antisymmetrise over $k$ first, and then symmetrise over $ij$. This gives an entirely different tensor $$ \phi_{ij;k} = \frac{1}{2}\left(\phi_{(ij)k}-\phi_{k(ij)}\right)=\frac{1}{4}\left(\phi_{ijk}+\phi_{jik}-\phi_{kij}-\phi_{kji}\right). \tag{2} $$ Which of these is correct? Alternatively, do they both give the correct answer?

enter image description here

  1. If I understand correctly, one purpose of Young diagrams is to provide an easy way to write down tensors which are (anti)symmetrised in particular indices. However, it appears to me that neither the tensors (1) nor (2) are both symmetric with respect to the exchange of $i$ and $j$, and antisymmetric with respect to the exchange of $i$ and $k$. That is, exchanging $i$ and $k$ for $\psi_{ij;k}$ gives $$ \psi_{ij;k} \overset{i\leftrightarrow k}{\longrightarrow} \psi_{kj;i} = \frac{1}{4}\left( \psi_{kji}-\psi_{ijk}+\psi_{jki}-\psi_{jik}\right) = -\psi_{ij;k} $$ but $$ \psi_{ij;k} \overset{i\leftrightarrow j}{\longrightarrow} \psi_{ji;k} = \frac{1}{4}\left( \psi_{jik}-\psi_{kij}+\psi_{ijk}-\psi_{ikj}\right) \neq \psi_{ij;k}. $$ So $\psi_{ij;k}$ appears to be antisymmetric w.r.t $i\leftrightarrow k$, but is not symmetric w.r.t $i\leftrightarrow j$. On the other hand for $\phi_{ij;k}$ we get $$ \phi_{ij;k} \overset{i\leftrightarrow k}{\longrightarrow} \phi_{kj;i} = \frac{1}{4}\left(\phi_{kji}+\phi_{jki}-\phi_{ikj}-\phi_{ijk}\right) \neq \phi_{ij;k} $$ but $$ \phi_{ij;k} \overset{i\leftrightarrow j}{\longrightarrow} \phi_{ji;k} = \frac{1}{4}\left(\phi_{jik}+\phi_{ijk}-\phi_{kji}-\phi_{kji}\right) = \phi_{ij;k} $$ So $\psi_{ij;k}$ is not antisymmetric w.r.t $i\leftrightarrow k$, but it is symmetric w.r.t $i\leftrightarrow j$. What is going on here?

  2. I do not understand how to generalise this to higher dimensions. For example, how would one explicitly construct the $SU(5)$ tensor for the Young diagram below?

enter image description here

Thank you!

$\endgroup$
1
$\begingroup$

With mixed symmetry tensors (i.e. tensors that are not just one row or just one column) it is impossible to have symmetry under exchanges in the same row and antisymmetry under exchanges in the same column at the same time. One of them has to give, as you correctly noted.

The difference between the two tensors is that in case (1) you do the symmetrization fist and the antisymmetrization last, so the manifest symmetry is lost. Whereas in case (2) you antisymmetrize first and symmetrize last, so manifest antisimmetrization is lost.

Ultimately you are free to do either of those, as long as you are consistent. It is the same as choosing to compose permutations left to right or vice versa.

The general case can be seen from e.g. Wikipedia where the author picked a convention equivalent to your case (2). First you list all symmetries of the rows, $$ P_{\lambda }=\{g\in S_{n}:g{\text{ preserves each row of }}\lambda \}\,, $$ then all the symmetries of the columns $$ Q_{\lambda }=\{g\in S_{n}:g{\text{ preserves each column of }}\lambda \}\,. $$ Then you build the symmetrizer and the antisymmetrizer $$ a_{\lambda }=\sum _{{g\in P_{\lambda }}}e_{g} \,,\qquad b_\lambda = \sum _{{g\in Q_{\lambda }}}\mathrm{sgn}(g) e_g\,. $$ $a_\lambda$ and $b_\lambda$ are elements of the group algebra, namely the vector space generated by the group elements with formal linear combinations as vector space operations and group multiplication as binary operation.

Finally you multiply them $$ c_{\lambda }:=a_{\lambda }b_{\lambda }=\sum _{g\in P_{\lambda },h\in Q_{\lambda }}\operatorname {sgn}(h)e_{gh} \,. $$ The order $a_{\lambda }b_{\lambda }$ vs. $b_{\lambda }a_{\lambda }$ is the ambiguity we were discussing about. Finally you act on your tensor with the distributive property, e.g. $$ (e_{\mathbf{id}} + e_{(12)} + e_{(123)})\,\phi_{ijk} = \phi_{ijk} + \phi_{jik} + \phi_{jki}\,. $$

$\endgroup$
2
  • $\begingroup$ That clears quite a lot up, thank you! In the formulae you list, does it matter which standard Young tableau $\lambda$ is taken? $\endgroup$ – Arthur Morris Jan 11 at 15:38
  • 1
    $\begingroup$ It shouldn't because choosing a different tableau is the same as conjugating all elements by some constant element $g \to g_0^{-1} g g_0$. (I think) $\endgroup$ – MannyC Jan 11 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.