We know that a supermanifold $M$ is a locally ringed space $(M,O_M)$ which is locally isomorphic to $(U,C^\infty(U) \otimes \wedge W^\ast)$, where $U$ is an open subset of $\mathbb{R}^n$, $W$ is a finite dimensional real vector space and the above isomorphism defined in the category of $\mathbb{Z}_2$ graded algebra i.e. the parity $\bigoplus_{k \geq 0}C^\infty(U) \otimes \wedge^k W^\ast \rightarrow \mathbb{Z}_2$ defined by $f \otimes x \rightarrow f \otimes x:=x=k \mod 2$. I would like to know how can we geometrically think of this. For example we know that the algebra of differential forms $\Omega(M)$ on a manifold $M$ which is locally isomorphic to $C^\infty(U) \otimes T^\ast_x M$ for some $x \in U$, therefore the sheaf of differential forms on a manifold corresponds to a supermanifold. How can we geometrically visualize this? Moreover what is the significance of defining a supermanifold structure for the sheaf of differential forms for a manifold $M$.
1 Answer
The Wikipedia page on supermanifolds does a decent job of presenting different definitions and their relations. See in particular the statement of Batchelor's theorem.
Forgetting about the "super" part, already an ordinary manifold can be described as a ringed space $(M,O_M)$, locally isomorphic to $(U,C^\infty(U))$. When $U = \mathbb{R}^n$, with standard coordinates $x^i$, the coordinates $x^i$ play the role of commuting generators (up to taking limits of polynomial expressions) of $C^\infty(\mathbb{R}^n)$. The point of this description is that in the pair $(M,O_M)$, the sheaf of algebras $O_M$ should be interpreted as the sheaf of algebras of functions on open subsets of $M$. But since you are looking into ringed spaces at all, I suspect that you have already seen this interpretation.
Supergeometry allows the sheaf $O_M$ to be a supercommutative algebra (not just a commutative one). The heuristic is that the supermanifold $M$ should be covered by charts of the form $\mathbb{R}^n_{even} \times \mathbb{R}^m_{odd}$ with corresponding algebras of functions $C^\infty(\mathbb{R}^n_{even} \times \mathbb{R}^m_{odd}) := C^\infty(\mathbb{R}^n_{even}) \otimes \bigwedge^\bullet (\mathbb{R}^m_{odd})^*$, where the second tensor factor is the algebra of functions generated by the odd (hence supercommuting) "coordinates" $\theta^j$ on $\mathbb{R}^m_{odd}$. Now, since $\mathbb{R}^n_{even} \times \mathbb{R}^m_{odd}$ does not exist as a topological space, we simply consider the algebra $C^\infty(\mathbb{R}^n_{even} \times \mathbb{R}^m_{odd})$ as assigned to the $\mathbb{R}^n_{even}$ factor, which does exist as a topological space.
This is exactly analogous to how the sheaf of functions on a fibered manifold $N \to M$, locally modeled on charts of of the form $V\times U \to U$, gives rise to the ringed space $(M,O_M)$ where which locally looks like $(U, C^\infty(V\times U))$. Batchelor's theorem makes this precise, by identifying any supermanifold $(M,O_M)$ with the total space of a fibered supermanifold $M \to M_0$, where $M_0$ is an ordinary manifold and the fibers of $M \to M_0$ are purely odd.
In this sense, the example that you gave, $(M,\Omega^\bullet)$, corresponds to the total space of the fibered supermanifold $(\Pi T)M \to M$, where $\Pi$ denotes the parity shift (from even to odd) of the cotangent fibers. There might be a common name for this supermanifold, but I can't recall it at the moment.

$\begingroup$ Is 'parity shirt' supposed to be 'parity shift'? $\endgroup$– LSpiceDec 23, 2019 at 1:39

1$\begingroup$ @LSpice thanks for spotting that! Typo fixed. $\endgroup$ Dec 23, 2019 at 8:17