I've been looking into extended Supersymmetry in higher dimensions recently. What I keep wondering about are some components of the gravity multiplet, that seem to appear from the construction of Supergravity from superconformal methods.

The appearance of the graviphoton is clear from the N=2 structure. But what puzzles me is the anti-selfdual two-tensor that I usually find as a part of the gravity multiplet.

How does the anti-selfdual tensor get introduced into the gravity multiplet? I would be most grateful for an answer that can at least shed some light on this without referring explicitly to superconformal tensor calculus.

  • $\begingroup$ Do you ask about the origin of the $2$-form $B$, in the $N=2, D=6$ supergravity, such as the $3$-form $H= dB$ is anti-self-dual ? (for instance, see list page $9$, of this paper) $\endgroup$ – Trimok Nov 27 '13 at 20:52
  • $\begingroup$ @Trimok Yes, I wonder about the $B_{\mu \nu}^-$. $\endgroup$ – Neuneck Nov 27 '13 at 21:12

While keeping the array page $9$ in ref1, already given, in mind, we add a new ref2, especially fig $1$ page $7$, paragraph $2.2.3$. $D = 6$, page $11$, table $5$ page $13$, and discussion page $12$

From fig $1$, page $7$, we see, that in $D=6$, the $N=2$ supersymmetry corresponds to a $(N_+, N_-) = (1,0)$ supersymmetry

Looking at the discussion page $12$, about the table $5$, page $13$, the idea is to begin by small multiplet (hypermultiplet), and to tensor with helicities representations to obtain other multiplets.

The representations are about $SU_+(2)\otimes SU_-(2)\otimes USP(2N_+) \otimes USP(2N_-)$ (which is the 6D massless little group), so here, it is simply $SU_+(2)\otimes SU_-(2)\otimes USP(2)$, keeping in mind that $USP(2) \sim SU(2)$

For instance, beginning with the hyper multiplet fermionic part (we can drop the $N_-$ part, as we chose $(N_+, N_-) = (1, 0)$) $ (2,1; 1)$, and tensor product by the $ (2,1; 1)$ representation, we get $ (3,1; 1) + (1,1; 1)$, while taking the hyper multiplet bosonic part $ (1,1; 2)$ and tensor product by the same $ (2,1; 1)$ representation, we get $ (2,1; 2)$. So, we see, that taking the hypermultiplet and tensor product by $ (2,1; 1)$, we get the tensor multiplet.

If we take the hyper multiplet and tensor product with the $ (1,2; 1)$ representation, we get the vector multiplet.

If we take the hyper multiplet and tensor product with the $ (2,3; 1)$ representation, we get the supergravity multiplet.

Now, looking, in the bosonic part of the supergravity part, we have the representation $(1,3; 1)$, which corresponds to the strengh of $B_{\mu\nu}^-$ in ref1. Now, $(1,3)$ is the same representation, thinking about electromagnetic field in $4D$, that $F_{\mu\nu}^- \sim (E-iB)$ (while $F_{\mu\nu}^+ \sim (E+iB)$) (not sure about the sign of $B$, but this is the idea). In $4D$, the total electromagnetic field representation is $(1,3) \oplus (3,1) = F_{\mu\nu}^- \oplus F_{\mu\nu}^+$

Of course, we notice, that we are in $6D$, so the strengh of $B_{\mu\nu}^-$ is a $3$-form, so there is a possibility for self-duality and anti-self-duality

  • $\begingroup$ Well-explained! Thank you. I did not get it, though, until I worked through a hard to obtain work by Stathdee. $\endgroup$ – Neuneck Nov 30 '13 at 9:05
  • $\begingroup$ What still keeps puzzling me is that the rep's given above refer to the little group and not to the 6D Poincaré group. Is there any reason, why e.g. $(2,2;1)$ (a 6-vector) should correspond to the 4D Lorentz rep $(2, 2)$, which is a 4-vector? $\endgroup$ – Neuneck Nov 30 '13 at 11:00
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    $\begingroup$ @Neuneck : Yes, Stathdee is cited like the reference (I did not sucess to find a free version.). Yes, reps refer to the little group, which is, for massless representations in $D=6$, $SO(4)$. There are $4$ on-shell freedom degrees for the vector representation, so it is natural to have a $(2,2)$ representation of $SU_+(2)\otimes SU_-(2)$. It is not exactly a vector Lorentz representation, because Lorentz is $SO(3,1)$ and not $SO(4)$. Moreover, the $(2,2)$ vector Lorentz representation is rarely used in $D=4$, for instance, the electromagnetic fied $A_\mu$ does not transform following this rep. $\endgroup$ – Trimok Nov 30 '13 at 19:01

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