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

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.

## 1 Answer

So, the answer gets obscured in some sense by the explicit dimensionality d=4 of this setting. The scalars transform by definition as a singlet of the U(2) holonomy before the twist, i.e. the scalars are elements of $\Omega^0(M)$ and are also uncharged under $U(1)_J$, the $U(1)$ part of the holonomy group.

But the six scalars in the $N=4$ SYM theory are rotated into each other by the $SU(4) \simeq SO(6)$ and form a vector of SO(6). Now, for the topological twist we split $$SO(6) \rightarrow SU(2)_A \times SU(2)_B \times U(1)_\mathcal{R}$$ under which the scalar fields decompose into four real scalar fields transforming as a $(\boldsymbol{2}, \boldsymbol{2})$ under $SU(2)_A \times SU(2)_B$ and a pair of complex scalars charged under $U(1)_\mathcal{R}$. These are just the scalars mentioned in my question, $$(\boldsymbol{1})_{0,1} \oplus (\boldsymbol{1})_{0,-1} \text{ under } SU(2) \times U(1)_J \times U(1)_\mathcal{R}.$$ The above $SU(2)$ is from the $U(2)$ holonomy. Of course, they both are still "scalars" from the spacetime perspective.

Now, we twist the $U(1)_J$ of the holonomy group via $$J'= J + 2\mathcal{R}$$ and the previously uncharged (under the holonomy group) states transform now as $$(\boldsymbol{1})_{2} \oplus (\boldsymbol{1})_{-2}$$ under $SU(2) \times U(1)_{J'}$. But this representation is just the same representation of a 2-form on a Kähler 2-fold! It is a singlet under the $SU(2)$ since it is a top form on the 2-fold, i.e. an element of $\Omega^{2,0}$ or $\Omega^{0,2}$ but the charge of 2 under the U(1) indicate that it is an element of the 2nd exterior power $\Lambda^2 T^*M$ and not just a function.

It will be more obvious if you look at a spinor after the twist. Before the twist a right handed spinor transforms as a $(\boldsymbol{2,1})$ under the $Spin(4) \simeq SU(2) \times SU(2)_J$ which is certainly not a 1-form. After the twist they decompose and we get a spinor in the representation $$(\boldsymbol{2})_{1} \text{ of } SU(2) \times U(1)_{J'} \subset U(2)$$ which is exactly the decomposition of an element in the vector rep. of $U(2)$ in the given branching, therefore the spinor is a one-form or an element of $\Lambda^1 T^*M$ after the twist.