# Tag Info

3

The idea behind the notation is that the operator $F$ is supposed to count the number of fermions in an expression, i.e. $$[F,A_n]= n A_n$$ if the operator $A_n$ contains $n$ fermions, what that means. Then $$[f(F),A_n]= f(n) A_n$$ for a sufficiently well-behaved function $f:\mathbb{C}\to \mathbb{C}$. In particular, for $f(x)=(-1)^x$, one has ...

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Found the issue. The scalars in the N=8 gravity multiplet are real scalars, not complex. Same for the vector multiplet. Thus the N=8 gravity multiplet can be decomposed as 1 copy of the N=4 gravity multiplet, 4 copies of the N=4 gravitino multiplet and 6 copies of the N=4 vector multiplet.

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You can try to transform $F \epsilon$ under Lorentz transformations. You will notice that it does not transform like a Lorenz invariant bosonic quantity, i.e. vector, scalar, rank-2 tensor,... You have to have $\bar\epsilon F$ to correspond to a bosonic quantity.

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I think what they mean is FI-term is not gauge invariant under the full gauge symmetry of the theory, but under this remaining gauge freedom after WZ gauge, which is $U(1)$.

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The space of semi-infinite forms is basically the name used by mathematicians for the fermionic Fock space please see for example: Friedrich Wagemann lecture, page 8. Given an infinite dimensional vector space with spanned by: $\{ e_i, i\in \mathbb{Z}\}$, let its dual space be spanned by $\{ f_i, i\in \mathbb{Z}\} (\langle f_j, e_i \rangle = \delta_{ij}$) ...

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They shouldn't be thought of as operators i.e. $q$-numbers; instead, they should be thought of as $c$-numbers. They're mutually anticommuting but otherwise they play exactly the same role as $\Delta x^\mu$ for translations or angles $\varphi$ for rotations. They're spinor variables which means that under a Lorentz transformation $\Lambda\in SO(3,1)$, they ...

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There is a long and formal way, and also an easy and dirty way. I will tell you the easy option. The algebra tells you that $[\delta_Q (\epsilon_1), \delta_Q (\epsilon_2)] = \delta_{P}(\xi^\mu_3)$ where $\epsilon$ is your SUSY parameter and $\xi^\mu_3 = \bar\epsilon_1 \gamma^\mu \epsilon_2$ is your translation parameter. Now, the only Lorentz vector that ...

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