# Irrep decomposition of direct product of stress tensors

I have stress tensors direct product of the form $T^{ab}(x)T^{cd}(y)$. I want to write this in terms of a tensor $I^{abcd}$ in the form. $T^{ab}(x)T^{cd}(y)= I^{abcd}$. This is like decomposing the direct product of the Rank 2 symmetric tensor into its irreps. How can I do a decomposition like this for $T\otimes T\otimes T...$???

• Note that the tensors I am considering are for the conformal field theory. Hence the group structure is $SO(d,2)$ but it will also be valid for the $SO(N)$ group as well. – user50183 Jun 9 '14 at 11:26
• You mean you want to write a generic tensor $T^{ab}(x)$ in the form of $T_1^{a}(x)T_2^{b}(x)?$ If so, I don't think it is possible. – Antonio Ragagnin Jun 9 '14 at 11:58

Assuming that the tensors are those of $SO(N)$, which in particular means they are traceless, we can compute the tensor product of $(2,0,...)$ and $(2,0,...)$ where the notation refers to the Young diagrams ($(s_1,s_2,...)$ denote a Young diagram that has $s_1$ cells in the first row, $s_2$ in the second and so on), as follows $$(2,0)\otimes (2,0)= (4,0)\oplus (3,1)\oplus (2,0)\oplus (1,1) \oplus (2,2) \oplus (0,0)$$ It seems that you need a symmetric part of the tensor product since the operators are identical, which misses some of the components, namely the symemtric part is $$(2,0)\otimes (2,0)|_S= (4,0)\oplus (2,0) \oplus (2,2) \oplus (0,0)$$ Now the notation: $(4,0)$ is a rank four symmetric; $(2,2)$ is rank-four but has the symmetry of the Riemann tensor; $(3,1)$ rank four, symmetric in three indices, has certain extra conditiond; $(2,0)$ rank-two symmetric, appears as the trace; $(1,1)$ rank-two anti-symmetric; $(0,0)$ scalar $T_{ab}T^{ab}$, appears as a double trace.
Now it seems that you would like to say these are the structures for the stress-tensor two-point function. The two-point function is unique and is fixed by $so(d,2)$ symmetry, while we made use only of the Lorentz symmetry $so(d-1,1)$, so we cannot fix it completely. Even if we use $so(d,2)$ we still have to impose conservation as an independent condition, which is not about tensor products at all.
Now, with rank three you are in big trouble as this requires good knowledge of tensor products, here is the symmetric part of it $$(2,0)\otimes(2,0)\otimes(2,0)|_S=(6,0)\oplus (4,2)\oplus(4,0)\oplus(2,2)\oplus(2,0)\oplus(0,0)$$ and again we know there are three conformal! structures in three-point function, but this we cannot see without taking into account conservation and conformal symmetry