I'm trying to solve the following exercise:

Construct the physical states of the $\mathcal{N} = 4$ massless graviton supermultiplet, starting from a Clifford vacuum of helicity $λ_0 = 0$.

Decompose the $\mathcal{N} = 4$ massless graviton supermultiplet into a direct sum of $\mathcal{N} = 1$ massless supermultiplets: how many $\mathcal{N} = 1$ graviton, gravitino, vector and chiral multiplets does the $\mathcal{N} = 4$ graviton multiplet contain?

My main problem lies in the second question (the decomposition) but I will answer the first question as to make myself better understood.

I know that the helicity values and its CPT values for $\mathcal{N}=4$ follows:

$$\left(\lambda_0 \hspace{1mm}; \hspace{1mm} 4 \times \left(\lambda_0+\frac{1}{2} \right) \hspace{1mm}; \hspace{1mm} 6 \times \left(\lambda_0+1 \right) \hspace{1mm};4 \times \left(\lambda_0+\frac{3}{2} \right) \hspace{1mm} ; \lambda_0 +2 \right)\tag{1}$$

I know that a graviton multiple means that the helicity must be $\lambda_0 =-2$ but I am told to start from $\lambda_0 =0$ so I obtained the following helicity content:

$$\lambda_0 =0 \hspace{1mm};\hspace{1mm} \left( 0;4 \times \frac{1}{2} ; 6 \times 1 ; 4 \times \frac{3}{2}; 2 \right) \tag{2}$$

$$\lambda_0 =-\frac{1}{2}\hspace{1mm};\hspace{1mm} \left(-\frac{1}{2} ;4 \times 0 ; 6 \times \frac{1}{2} ; 4 \times 1; \frac{3}{2} \right)\tag{3}$$

$$\lambda_0 =-1\hspace{1mm};\hspace{1mm} \left( -1;4 \times \frac{-1}{2} ; 6 \times 0 ; 4 \times \frac{1}{2}; 1\right)\tag{3}$$

$$\lambda_0 =-\frac{3}{2}\hspace{1mm};\hspace{1mm} \left( \frac{3}{2};4 \times (-1) ; 6 \times \frac{-1}{2} ; 4 \times 0; \frac{1}{2}\right) \tag{4}$$

$$\lambda_0 =-2\hspace{1mm};\hspace{1mm} \left( -2;4 \times \frac{-3}{2} ; 6 \times (-1) ; 4 \times \frac{-1}{2}; 0\right)\tag{5}$$

Because I know that the graviton multiple means $\lambda_0=-2$, I believe (but I am not entirely certain) I must only consider the helicity content on $(5)$. But, if that is the case, I don't see any possible way in which I can decompose $(5)$ into a direct sum of $\mathcal{N}=1$ supermultiplets.

How can overcome this issue ?

Must I instead consider all the helicity content from $(2)$ to $(5)$? If so, why?

These are the helicity contents for the $\mathcal{N}=1$ massless supermultiplets:

$$\lambda_0 = 0 \hspace{1mm}; \hspace{1mm} \left( \frac{-1}{2} , 2 \times 0 , \frac{1}{2}\right) \tag{6}$$ $$\lambda_0 = \frac{1}{2} \hspace{1mm}; \hspace{1mm}\left( -1, \frac{-1}{2}, \frac{1}{2}, 1\right)\tag{7} $$ $$\lambda_0 = 1 \hspace{1mm}; \hspace{1mm} \left( \frac{-3}{2}, -1, 1, \frac{3}{2} \right)\tag{8}$$ $$\lambda_0 = \frac{3}{2} \hspace{1mm}; \hspace{1mm} \left( -2, \frac{-3}{2}, \frac{3}{2}, 2\right)\tag{9}$$


As for the massless $\mathcal{N}=1 $ supermultiplet, you have only two supercharges, and one of the can be put to zero, so the shortest possible mulptiplet is formed by $\left(\lambda_0, \lambda_0 + \frac{1}{2} \right)$. Therefore you have to look, in what whay can you decompose $\mathcal{N}=4$ into such small multiplets. There is one state with the lowest helicity of $0$ and one with the highest $2$. So one can see a chiral multiplet with $(0, 1/2)$ states, and a graviton multiplet $(3/2, 2)$. At the moment we have left $(3 \times 1/2 , 6 \times 1, 3 \times 3/2)$. It is easy to see 3 gauge $ \ \mathcal{N}=1 $ multiplets $(1/2, 1)$, and the 3 gravitino $(1, 3/2)$ multiplets.

  • $\begingroup$ But in the questions it has asked me to decompose it in terms of N=1 massless supermultiplets, I thought that meant that I had to consider the helicity content that contained the CPT symmetry But if it is as you say and I don't need to consider the CPT symmetry than the summations make sense knowing if I can simply use $(\lambda_0 , \lambda_0 + \frac{1}{2})$. Thank you for your help. $\endgroup$ – user7077252 Apr 26 '20 at 17:06

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