# Scalar field transformation behaviour of N=4 SYM after a topological twist

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Why do scalars transform as 2-forms after a topological twist?

I'm currently looking at this paper https://arxiv.org/abs/1403.2530. There they analyze a topological twist on N=4 SYM theory on Kähler 2-fold.

After doing a topological twist with the $\mathcal{R}$-symmetry as well as an additional $U(1)$ symmetry in a $N=4$ SYM, the 6 real scalars of the multiplet are reorganized under the twisted holonomy group: $$G = SU(2)_L \times U(1)_{J'}$$ with $U(1)_{J'} \subset SU(2)_R$. The important part is now that the 6 scalars can be reorganized in a way including two complex fields: $$(\boldsymbol{1})_2 \oplus (\boldsymbol{1})_{-2}$$ It is now stated that these two fields can be identified with a (2,0) and a (0,2) form. My question is, how can this identification be reached exactly ?

It is of course obvious that after twisting and obtaining charge under the $U(1)_{J'} \subset U(2)$ of the holonomy group, these fields cannot transform simply as scalars anymore, but how can i conclude from their $U(1)$ charge on the exact transformation behaviour ?

This is probably connected to another question of mine, in another paper on this topic it is stated that the spinors in these theories could be understood as sections of $\boldsymbol{S}^{\pm} \otimes K^{1/2}$ with $\boldsymbol{S}^{\pm}$ denoting the spin bundle and $K^{1/2}$ the square root of the canonical bundle of the base manifold and that by twisting with the $\mathcal{R}$-symmetry we just untwist these twisted differential forms, but I don't get why spinors are not just sections of the spin bundle solely in the first place.

An answer for the first question only would be totally enough, just wanted to illuminate my confusion here a bit further.

This post imported from StackExchange Physics at 2016-12-25 09:32 (UTC), posted by SE-user Moe

edited Dec 31, 2016

Spinors are globally defined on spin manifolds. In order to get spinors globally defined you need a $Spin$ structure on the manifold. Not all Kahler manifolds are $Spin$. Despite that, all Kahler manifolds are $Spin_c$ which requires twisting the Spin bundle with the "square-root" of a line bundle such that sign ambiguities are washed up. This should answer the second part of your question.
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