Let $A^{k}(U,{\mathbb{R}})$ be the space of smooth real differential forms on an open subset $U$ of $N_{\mathbb{R}}$, then a superform of bidegree $(p,q)$ on $U$ is an element of

$$A^{p,q}(U):=A^{p}(U,{\mathbb{R}})\otimes_{C^{\infty}(U)}A^{q}(U,{\mathbb{R}})=C^{\infty}(U)\otimes_{\mathbb{Z}}\Lambda^{p}M\otimes_{\mathbb{Z}}\Lambda^{q}M.$$

Formally, such a superform $\alpha$ may be written as

$$\alpha=\sum_{|I|=p,|J|=q}\alpha_{IJ}d^{\prime}x_{I}\wedge{d^{\prime\prime}x_{J}}$$

where $I$ (resp. $J$) consists of $i_{1}<\dots<i_{p}$ (resp. $j_{1}<\dots<j_{q}$), $\alpha_{IJ}\in C^{\infty}(U)$ and

$$d^{\prime}x_{I}\wedge{d^{\prime\prime}x_{J}}:=(dx_{i_{1}}\wedge\dots\wedge dx_{i_{p}})\otimes(dx_{j_{1}}\wedge\dots\wedge dx_{j_{q}}).$$

The wedge product is defined in the usual way on the space of superforms $A(U):=\oplus_{p,q\leq n}A^{p,q}(U)$. There is a canonical $C^{\infty}(U)$-linear isomorphism $J^{p,q}:A^{p,q}(U)\rightarrow A^{q,p}(U)$ obtained by switching factors in the tensor product. The inverse of $J^{p,q}$ is $J^{q,p}$. We call $\alpha\in A^{p,p}(U)$ symmetric if $J^{p,p}\alpha=\alpha$.