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Equivariance Relation - Superconformal Hypermultiplets

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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!

This post imported from StackExchange Physics at 2016-06-26 09:50 (UTC), posted by SE-user user11128
asked May 4, 2016 in Theoretical Physics by user11128 (90 points) [ no revision ]
Equivariance is usually used in the context of group actions---symmetry groups, in this case, I suppose.

This post imported from StackExchange Physics at 2016-06-26 09:50 (UTC), posted by SE-user Danu

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