# Decomposition of direct product into symmetric/antisymmetric parts

1. Sep 2, 2008

### jdstokes

Can anyone explain to me why

the 3-rep of SU(3) gives

$3\otimes 3 = \overline{3}\oplus 6$

whereas for the 5 of SU(5)

$5\otimes 5 = 10\oplus 15$?

I thought the general pattern was

$N \otimes N = \overline{\frac{1}{2}N(N-1)}\oplus \frac{1}{2}N(N+1)$

but this second example seems to contradict that.

Edit: I wrote 15* instead of 15.

Last edited: Sep 2, 2008
2. Sep 2, 2008

### arivero

Slansly report is available in the net, scanned by KEK
http://www.slac.stanford.edu/spires/find/hep/www?j=PRPLC,79,1

SU(3) is here
http://ccdb4fs.kek.jp/cgi-bin/img/reduced_gif?198102061+194+265
and SU(5) here
http://ccdb4fs.kek.jp/cgi-bin/img/reduced_gif?198102061+199+265

Someone has different conventions somewhere.

A related thing that intrigues me, btw. The authors of this
http://www.slac.stanford.edu/spires/find/hep/www?j=PHLTA,B188,58
are doing, classically, $5\otimes 5$ or $5\otimes \bar 5$ or $(5 \oplus \bar 5) \otimes (5 \oplus \bar 5)$ but the point is that they get SO(32). How is it? Does SO(32) and SU(5)^2 share representations, or dimensionality of representations?

3. Sep 2, 2008

### jdstokes

I don't think it's a difference in convention.

Slansky gets $5\otimes 5 = 10 \oplus 15$ for SU(5) and $\overline{3}\otimes \overline{3} = \overline{6}\oplus 3$ for SU(3) which is equivalent to $3\otimes 3 = 6\oplus\overline{3}$.

I understand how to derive 3 x 3 = 6 + 3* by contracting with the permutation symbol.

Do you know how to derive 5 x 5 = 10 + 15?

4. Sep 2, 2008

### Haelfix

Probably the easiest way is to do a Young Tableux calculation. Its not a very long calculation (about 4 - 5 lines) but completely impossible to show in this sort of forum

You can check yourself with the other possibility
5 * 5bar = 1 + 24

5. Sep 2, 2008

### arivero

but note the subscript _s in Slansky to mark the symmetric. It differs from the bar.

6. Sep 2, 2008

### jdstokes

Here's my proposed proof. Work under the assumption that a tensor is irreducible if it a tensor of lower rank can not be formed by contraction with the isotropic tensors.

Consider an arbitrary element of the tensor product space $T \in 5\otimes 5$.

Contracting with the lower index permutation symbol gives

$U_{klm} = \varepsilon_{ijklm}T^{ij}$.

Contracting again with the upper index permutation symbol gives

$V^{no} = \varepsilon^{klmno}U_{klm}$.

V is irreducible, has two upper anti-symmetric indices. Therefore $V\in 10$.

7. Sep 3, 2008

8. Sep 3, 2008

### arivero

So, the original question is based in confusion between "conjugate" and "anti-symmetric", isn't it? Is this confusion a notational or a conceptual one?

9. Sep 5, 2008

### jdstokes

A lecturer told me.

Yes, I believe that $T^2(V) = S^2(V)\oplus \Lambda^2(V)$. I've had trouble reconciling this with physicists' notation for SU(3) where the anti-symmetric part is represented by $\overline{3}$. For consistency let's use upper indices to denote tensors which transform according to the fundamental rep from now on.

I think you're saying that there's an isomorphism between the group actions on $\Lambda^2(\mathbb{C}^3)$ (represented by $T^{ij}=T^{[ij]}$, say) and the representation on the dual space $(\mathbb{C}^{3})^\ast$, which is defined (in component notation) by $\mathrm{SU}(3) : v_i \mapsto {U_{i}}^j v_j$ where ${U_{i}}^j \equiv {U^{i}}_j^\ast$ and the action on the fundamental representation is $\mathrm{SU}(3) : v^i \mapsto {U^{i}}_j v^j$.

I suppose the isomorphism is $Z_i \mapsto \varepsilon^{ijk}Z_k$ but I don't see why this is an isomorphism. Ie it's not obvious why they have the same transformation properties.

Once again, it's not obvious to me why for SU(4), $T_{[ij]}$ transforms in the same way as $T^{[ij]}$, and thus why $\mathbf{6} \cong \overline{\mathbf{6}}$.

I understand the concept of irreducibility by the non-existence of proper non-trivial subspaces which transform only amongst themselves. What I fail to see is why this is equivalent to the inability for form lower rank tensors by contraction with the isotropic tensors.

10. Sep 8, 2008

### samalkhaiat

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