# Decomposing the $\mathcal{N} = 4$ massless graviton supermultiplet

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.
• 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. Apr 26, 2020 at 17:06