3
$\begingroup$

I'm concerned with equation 2.24 of http://arxiv.org/abs/1601.00482

The superconformal hypermultiplets in this paper have a conic hyperkahler target manifold and the authors want to gauge some isometries of this manifold. Letting the isometry group be $G$ and to have an associated Lie algebra $\mathfrak{g}$ generated by Killing vectors $k_I$, we can express this as $\mathcal{L}_{k_I} g=0$ where $g$ is the metric on the conic hyperkahler manifold.

Then, in order to not break SUSY, the $k_I$ must commute with the SUSY generators. Apparently this is equivalent to the Killing vectors being triholomorphic $\mathcal{L}_{K_I} J_\alpha=0$ where $J_\alpha$ are the triplet of complex structures. Does anyone know why this is the case?

Secondly, they say in 2.24 of this paper that the moment maps associated to these symmetries must satisfy the "equivariance condition". Unfortunately they don't offer any explanation of what this is or where it comes from. There is some discussion in other literature along the lines of "we can also derive the equivariance condition...." but they never say what it is or explain how they found it. The best I've found is in the Freedman/van Proeyen Supergravity book where in eqn (13.61), they seem to say it comes from requiring the moment maps to transform in the adjoint:

$$(k_I^\alpha \partial_\alpha + k_I^{\bar{\alpha}} \partial_{\bar{\alpha}} ) \mu_J = f_{IJ}^K \mu_K$$

They then use some identities to write this as (13.62):

$$k_I^\alpha g_{\alpha \bar{\beta}} k_J^{\bar{\beta}} - k_J^\alpha g_{\alpha \bar{\beta}} k_I^{\bar{\beta}} = i f_{IJ}^K \mu_K$$

Although I don't see how this looks anything like (2.24) of the attached paper.

If anyone can offer any help or thoughts on either of these issues I'd greatly appreciate it!

$\endgroup$
1
  • $\begingroup$ Equivariance is usually used in the context of group actions---symmetry groups, in this case, I suppose. $\endgroup$
    – Danu
    May 4, 2016 at 19:16

1 Answer 1

0
$\begingroup$

For your first question:

The requirement ${\cal L}_k J=0$ can be understood from the commutator of supersymmetry and the gauge transformations. E.g. in ${\cal N}=1$ for the chiral multiplet: calculate the commutator of the supersymmetry transformation (which contains only $\bar \epsilon P_L$) with a gauge transformation $\delta Z^\alpha =\theta ^A k_A^\alpha (Z,\bar Z)$. There is one term that will contain $\bar \epsilon P_R$, namely in $$\delta _\epsilon \delta _\theta Z^\alpha = \theta ^A (\frac{\partial }{\bar Z^{\bar \beta }} k_A{}^\alpha (Z,\bar Z))\bar \epsilon P_R\chi ^{\bar \beta }\,. $$ Since this term cannot be in the commutator, we find that $k_A{}^\alpha $ should be holomorphic. On the other hand $$ \left( {\cal L}_{k_A} {J}\right){} _a{}^b\equiv k_A{}^c \partial_c {J}_a{}^b -\partial_c k_A{}^b {J}_ a{}^c+\partial_a k_A{}^c {J}_c{}^b $$ where $a,b,c$ are either holomorphic indices $\alpha $ or antiholomorphic $\bar \alpha $. In this basis $J^\alpha {}_\beta = i\,\delta ^\alpha {}_\beta$, and its complex conjugate: $J^{\bar\alpha} {}_{\bar\beta} = -i\,\delta ^{\bar\alpha}{}_{\bar\beta}$. With this form of the complex structure, the expression for $\left( {\cal L}_{k_A} {J}\right){} _a{}^b$ is trivially zero if $a$ and $b$ are both holomorphic (or antiholomorphic). The nontrivial part is for $a=\alpha $ and $b=\bar \beta $: $$ \left( {\cal L}_{k_A} {J}\right){} _a{}^{\bar \beta }= -2i\partial_\alpha k_A{}^{\bar b} $$ and this is thus the holomorphicity condition.

For rigid supersymmetry in ${\cal N}=2$ the same is obtained when you consider the commutator of supersymmetry and gauge transformations on the hyperscalars. It needs more formalism to explain, but the principle is the same. However, for supergravity there is a modification, see below.

On your second question, the equivariance condition. Indeed, one can choose the constants in the moment maps for non-Abelian theories such that they transfer in the adjoint representation, and this is the equivariance condition. The moment maps appear in supersymmetric theories in the transformation of the gauginos and in the action. Then this equivariance condition is necessary to prove supersymmetry. For ${\cal N}=2$ in my book with Dan Freedman, the equation is first written in (20.51). But for supergravity it becomes (20.175), which is the same as in the paper you refer to after translating symbols, $\mu ^n\rightarrow \frac{1}{2}\vec P$, $\omega ^n\rightarrow \vec{J}$ and taking $\nu =-\kappa ^2=-1$. The extra term with $\kappa ^2\vec{P}_I\times \vec{P}_J$ can be understood in the superconformal framework from the fact that the scalars of the compensating multiplets have to be included in the sum over indices, and they also transform with an SU(2) transformation. Maybe it is sufficiently explained in the book, but more details can be found in Sec. 4.6.3 of the book with E. Lauria, also available as e-Print 2004.11433. Notice that due to the same mechanism, the equation ${\cal L}_k J=0$ is now also modified as can be understood from the contributions of the compensating multiplets. See e.g. (5.4) in hep-th/0411209 [hep-th]. The right-hand side contains $\vec{r}_I$, which appears then also in the non-zero commutator of gauge transformations and supersymmetries (to make contact with the start of this text).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.