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I am working out the number of wick contraction of a number $n$ of stress-energy tensor in 4D CFT.

The strategy is as follows:

  1. For 1 stress energy tensor $T_{\alpha\beta}$, you have only one possibility that is the spin 2 $T_{\alpha\beta}$.
  2. For 2 stress tensor $T_{\alpha\beta}T_{\mu\nu}$, one can get 1 spin 4, the fully uncontracted, one can get 1 spin 2 which is two indices contracted and two not contracted and a spin 0 which is the fully contracted result. Since the stress tensor is fully symmetric and we cannot distinguish the stress tensor, the way we do the contraction does not matter.
  3. For 3 stress tensor, we got 1 spin 6, 1 spin 4, 2 spin 2 and 1 spin 0
  4. And so on...

So in principle, we can write something like \begin{align} 2 =& 1\\ 2 \otimes2=& 1 \oplus1\oplus1\\ 2\otimes2\otimes2=& 1\oplus 1\oplus 2 \oplus1\\ 2\otimes2\otimes2\otimes2=& 1\oplus 1\oplus 2 \oplus1\\ 2\otimes 2 \otimes 2\otimes2\otimes2=& 1\oplus 1\oplus 3 \oplus3\oplus 2\\ 2\otimes 2\otimes 2\otimes 2\otimes2\otimes2=& 1\oplus 1\oplus 3 \oplus3\oplus5\oplus3\\ %2\otimes 2\otimes 2\otimes 2\otimes2\otimes2\otimes2=& 1\oplus 1\oplus3\oplus5\oplus6\oplus5\oplus5 \end{align} and so the total number is 1, 3, 5, 10, 16,... I have good reasons (oeis.org) to believe that this can be classified using the number of plane partitions of n with at most two rows. I have been searching a "irrep" way of obtaining those numbers but failed. Is there an easy solution that I missed ?

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  • $\begingroup$ Can you define precisely what you mean by wick contractions? $\endgroup$ – Peter Kravchuk Jun 12 at 15:49
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Not sure where OP wants to take this, but it is in principle possible to decompose tensor powers of the symmetric SEM tensor $$T~=~10~=~9~\oplus~ 1~=~(1,1)~\oplus~ (0,0)$$ in a plethora of irreducible representations of the 3+1D Lorentz group: $$\begin{align} T^{\otimes 2}&=~[(1,1)~\oplus~ (0,0)]^{\otimes 2} ~=~(2\oplus 1\oplus 0,2\oplus 1\oplus 0)~\oplus~ 2~(1,1)~\oplus~ (0,0)~=~\ldots\cr T^{\otimes 3}&=~[(1,1)~\oplus~ (0,0)]^{\otimes 3}~=~\ldots\cr \vdots & \end{align}$$ and so forth.

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