# This weaker version of CR-structure: is it studied somewhere

When I study 5-dimensional $\mathcal{N} = 1$ supersymmetry, I came across such structure as follows.

$(R, \kappa, \Phi, M)$ is an almost contact 5-manifold, such that $$\kappa \left( R \right) = 1,\;\;\;\;\Phi R = \kappa \circ \Phi = 0,\;\;\;\;{\Phi ^2} = - 1 + \kappa \otimes R.$$ Supersymmetry requires these quantities satisfy an extra condition: $${N_\Phi }\left( {X,Y} \right) + d\kappa \left( {\Phi X,\Phi Y} \right) = 0,\;\;\;\;\text{for} \; \kappa(X) = \kappa(Y) = 0\;\;\;\;\;\;\;\;\;\;\;(*)$$ where ${N_\Phi }\left( {X,Y} \right) \equiv {\Phi ^2}\left[ {X,Y} \right] + \left[ {\Phi X,\Phi Y} \right] - \Phi \left[ {\Phi X,Y} \right] - \Phi \left[ {X,\Phi Y} \right]$.

1) If $d\kappa \left( {\Phi X,\Phi Y} \right) = d\kappa \left( {X,Y} \right) \Leftrightarrow \kappa \left( {\left[ {X,\Phi Y} \right] + \left[ {\Phi X,Y} \right]} \right) = 0$, then the structure is Cauchy-Riemann.

2) $(*)$ implies $$\left[ {{X^{1,0}},{Y^{1,0}}} \right] = {Z^{1,0}} - d\kappa \left( {{X^{1,0}},{Y^{1,0}}} \right)R,$$ which further implies ${\left( {d{\omega ^{0,1}}} \right)^{2,0}} = 0$, and therefore I naively think we have usual decomposition $d = d_V + \partial + \bar\partial$.

So my question is whether the almost contact structure with $(*)$ is studied somewhere? Reference would be great!

========Update:========

Let me ask a enhanced version of question. Suppose the (1,0)+(0,1) decomposition induced by the almost contact structure satisfies the conditions (for some $Z^{1,0}$ and $Z'^{1,0}$ ) $$\left[ {{X^{1,0}},{Y^{1,0}}} \right] = {Z^{1,0}} - d\kappa \left( {{X^{1,0}},{Y^{1,0}}} \right)R,$$ $$\left[ {{X^{1,0}},R} \right] = {Z'^{1,0}} - \left( {...} \right)R,$$

does it mean that one can find special local coordinates $(\theta, z_{1,2}, \bar z_{1,2})$, such that the transformations between the coordinates take the form $$\begin{array}{l} z' = f\left( z \right)\\ \theta ' = \theta + g\left( {z,\bar z} \right) \end{array}$$

• Notation: $R$ is a vector field, $\kappa$ a 1-form and $\Phi$ an endomorphism of the tangent bundle? – Ben McKay Sep 8 '14 at 15:34
• @BenMcKay:Yes they are. – Lelouch Sep 8 '14 at 16:23