# Tag Info

5

Only the $su(4)$ generators appear on the right hand side of the $u(2,2|4)$ commutation relations, so superconformal invariance does not prevent an anomaly in the $u(1)$ reducing the symmetry to $su(2,2|4)$. In $\mathcal{N}=4$ SYM the central charge is furthermore zero, so the actual symmetry is $psu(2,2|4)$. The breaking of the generator with non-zero ...

5

It's a scenario that has heavy scalars and relatively light gauginos, so it's one example of a class of "split SUSY" or "mini-split SUSY" scenarios that have survived most of the constraints. In this kind of scenario, collider bounds put the lightest superpartners, namely the winos, above about 270 GeV. Gluinos are constrained to be somewhere north of a TeV, ...

3

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 $12$, ...

2

Say we have a supercharge $Q$ in $\mathbb{R}^{10}$. To turn this into a supercharge on the $\mathbb{R}^4$ effective theory obtained by compactifying on $X$, we need to contract $Q$ with a covariantly constant spinor on $X$. The reason why we want it to be covariantly constant is because we want to take the size of $X$ to zero. Covariant constant spinors are ...

2

It is not an answer, but maybe some information which could be useful : In an other post, it has been also noticed that the commutators of the $R$-symmetry generators with supercharge generators are: $[R^a_b,Q^c_{\alpha}]=\delta^c_bQ^a_{\alpha}-\frac{1}{4}\delta^a_bQ^c_{\alpha}$ So, taking the trace (on $a,b$), with $\mathcal N=4$, gives a null ...

1

Let me elaborate on Ryan's correct comments. The flat background makes all components of the spinors covariantly constant; so the geometry is compatible with all of SUSY. A generic curved 6-real-dimensional manifold has an $O(6)$ holonomy or $SO(6)\sim SU(4)$ if it is orientable. The $SU(3)$ subgroup preserves 1/4 of the original supercharges – it is the ...

1

In the first action the $A_{\mu}$ are Hermitian. In the second action the $A_{\mu}$ are anti-Hermitian since we let $A_{\mu}\to\frac{i}{g}A_{\mu}$. The commutator of anti-Hermitian matrices are also anti-Hermitian. If we have $\text{Tr}(M^{2})$ , with $M$ being anti-Hermitian, then we can write it as $\text{Tr}(M^{2})=-\text{Tr}((iM)^{2})$ , with $iM$ ...

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