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]

[QUOTE]

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]

[QUOTE=jdstokes;1860376]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