four dimensions (classically) has SL$(2,\fieldR)$ symmetry, due to the presence of additional tensor multiplets which transform nontrivially [@BGHL]. Setting the scalars in these multiplets to nonvanishing constants results in a breakdown of the symmetry and leaves only the above transformations as residual invariances.

The equations of motion for the Lagrange multipliers $\lambda_a$ imply that the $H^a$ are closed. Writing $H^a={\mathrm{d}}B^a$, one obtains the double-tensor multiplet. Integrating out the tensors instead gives the duality relation $$\label{dual-rel}
  {\mathrm{d}}\lambda_a = - M_{ab} *\! H^b\ .$$ Substituting this back yields the action for the universal hypermultiplet , upon identifying $$\label{mult1}
  \lambda_1 = \varphi\ ,\quad \lambda_2 = \sigma\ .$$ The dual formulation in terms of the double-tensor multiplet is not unique[^3]. We can start with , but write everywhere $\varphi$ instead of $\chi$. Dualizing the tensors and identifying $$\label{mult2}
  \lambda_1 = - \chi\ ,\quad \lambda_2 = \sigma + \varphi\, \chi$$ yields the same hypermultiplet action, as one can easily check.

In addition, the dualization procedure yields a boundary term which has to be added to the hypermultiplet action, $$\label{surf-term}
  {\mathcal{L}}_\mathrm{bnd} = (-)^D\, {\mathrm{d}}\big[ \lambda_a\, (M^{-1})^{ab} *{\mathrm{d}}\lambda_b\big]\ ,$$ where we used that, when acting on a $p$-form in Minkowski space, $**=
-(-)^{(D-1)p}$. The different choices corresponding 