Decomposition of direct product into symmetric/antisymmetric parts

[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.

 

[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?

 

[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?

 

[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

 

[arivero]

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

 

[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 $V\in 10$.

 

[samalkhaiat]

I thought the general pattern was

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

Why did you think that?
The general rule is

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

This is nothing but the tensor identity

$$x_{a} \otimes y_{b} \equiv T_{ab} = T_{(ab)} + T_{[ab]}$$

For n = 3, a pair of antisymmetrized lower indices [ab] is equivalent to an upper index "vector", i.e., to the conjugate representation;

$$T_{[ab]} \equiv \epsilon_{abc} Z^{c} \in [\overline{3}]$$

For n = 4, the lower pair [ab] is equivalent to an upper (conjugate) pair;

$$T_{[ab]} \equiv \epsilon_{abcd} A^{[cd]}$$

So for SU(4), $[6] \equiv [\overline{6}]$ and $4 \otimes 4 = 10 \oplus 6$.

For n = 5, the cojugate rep. does not show up because (as you showed)) contracting $T_{[ab]}$ with $\epsilon^{abcde}$ produces a higher rank tensor.

regards

sam

 

[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?

 

[jdstokes]

Why did you think that?

A lecturer told me.

The general rule is

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

This is nothing but the tensor identity

$$x_{a} \otimes y_{b} \equiv T_{ab} = T_{(ab)} + T_{[ab]}$$

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.

For n = 3, a pair of antisymmetrized lower indices [ab] is equivalent to an upper index "vector", i.e., to the conjugate representation;

$$T_{[ab]} \equiv \epsilon_{abc} Z^{c} \in [\overline{3}]$$

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.

For n = 4, the lower pair [ab] is equivalent to an upper (conjugate) pair;

$$T_{[ab]} \equiv \epsilon_{abcd} A^{[cd]}$$

So for SU(4), $[6] \equiv [\overline{6}]$ and $4 \otimes 4 = 10 \oplus 6$.

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 $\mathbf{6} \cong \overline{\mathbf{6}}$.

For n = 5, the cojugate rep. does not show up because (as you showed)) contracting $T_{[ab]}$ with $\epsilon^{abcde}$ produces a higher rank tensor.

regards

sam

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.

 

[samalkhaiat]

A lecturer told me.

OK, for n = 3, that happened to be the case because, and only because, the invariant tensor $\epsilon$ has three indices in SU(3).

I've had trouble reconciling this with physicists' notation for SU(3) where the anti-symmetric part is represented by $\overline{3}$.

As you might know, the decomposition of n X n

$$X^{i}Y^{j} = X^{(i}Y^{j)} + X^{[i}Y^{j]}$$

is invariant under the action of SU(3), i.e., $T^{(ij)} = X^{(i}Y^{j)}$ (or linear combination of it) and $T^{[ij]} = X^{[i}Y^{j]}$ (or linear combination of it) do not mix under the SU(3) transformation. Since $T^{(ij)}$ and $T^{[ij]}$ cannot be decomposed any further, they (or linear combinations of each) thus span irreducible subspaces of dimension n(n+1)/2 and n(n-1)/2 respectively. The story so far is trure for any SU(n). Now, for SU(3), you can easily show that the linear combination

$$\epsilon_{ijk}T^{[ij]}$$

transforms exactly like the $Z_{k}$ of $\overline{3}$. This shows that the 3-dimensional space spanned by the $T^{[ij]}$ is nothing but the $\overline{3}$.

So when physicists say that upper antisymmetrized pair of indices [ij] is equivalent to a single lower indix k, they mean that in the same sense as the equivalence between the angular momentum tensor

$$J_{ij} = \frac{1}{2}(x_{i}p_{j} - x_{j}p_{i})$$

and the angular momentum vector

$$J_{k} = (\vec{x} \times \vec{p})_{k} = \epsilon_{ijk}x_{i}p_{j} = \epsilon_{ijk}J_{ij}$$

Mathematically, $J_{ij}$ and $J_{k}$ span one and the same 3-dimensional vector space; the Lie algebra of SO(3). However, in contrast to the case of SU(3), we do not distinguish between upper and lower indices in SO(3) = SU(2). Do you know why?

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.

Think of $T^{ab}$ as basis of representation space of dimension D. This means that T has D independent components. If you run out of the tricks that invariantly divide T into "sub-tensors", then the representation space is irreducible and T enters into the Clebsch-Gordan series.
[note that for Lorentz group, $T^{(\mu\nu)}$ and $T^{[\mu\nu]}$ are reducible.Do you know why?]
The task of finding irreducible tensors of an arbitrary rank involves forming a complete set of permutation operations on their indices. And the problem of finding the irreducible representation of the permutation group has a complete solution in terms of the Young tableaux.

regards

sam