In this [paper](https://arxiv.org/abs/1905.08047), definition 4.4.1 about supermanifold and definition 4.6.1 about graded manifold:



> 
> **Definition 4.4.1:** An supermanifold $\mathcal{M}$ is a locally ringed space $(M,\mathcal O\_M)$ which is locally isomorphic to $(U, C^\infty(U)\otimes \wedge^\bullet V^\*)$ where $U\subset \mathbb R^n$ is open and $V$ is some finite-dimensional real vector space.
> 
> 
> **Definition 4.6.1:** A graded manifold $\mathcal M$ is a manifold $M$ which locally looks like $(U, C^\infty(U)\otimes \text{Sym} (V^\*))$, where $U ⊂ \mathbb R^n$ is open and $V$ is a graded vector space.
> 
> 
> 


Then it claims without proof in Remark 4.6.1 that:



> 
> **Remark 4.6.1:**
> One can construct an isomorphism between the the structure sheaf of a supermanifold and the local model of a graded manifold, which will be in the category of $\mathbb Z$-graded algebras.
> 
> 
> 


My question:


Using definition 4.4.1 for supermanifold and definition 4.6.1 for graded manifold above, how can we construct such isomorphism?




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My trial:
In the second definition, write graded vector space $V = \bigoplus\_{i \in \mathbb Z} V\_{i}$.


Let $V\_{\bar{0}} = \bigoplus\_{k \in 2 \mathbb Z} V\_k$ and $V\_{\bar{1}} = \bigoplus\_{k \in 2 \mathbb Z+1} V\_k$, then $V = V\_{\bar{0}}\oplus V\_{\bar{1}}$ and $V^{\*} = V\_{\bar{0}}^{\*} \oplus V\_{\bar{1}}^{\*}$.


$$\text{Sym}^{n}(V^{\*}) = \text{Sym}^{n}(V\_{\bar{0}}^{\*} \oplus V\_{\bar{1}}^{\*}) = \bigoplus\_{0 \le k \le n}(\text{Sym}^{k}V\_{\bar{0}}^{\*}\otimes\bigwedge^{n-k}V\_{\bar{1}}^{\*}).$$


$$\text{Sym}(V^{\*}) 
= \bigoplus\_{n} \text{Sym}^{n}(V\_{\bar{0}}^{\*} \oplus V\_{\bar{1}}^{\*}) \\
= \bigoplus\_{n} \bigoplus\_{0 \le k \le n}(\text{Sym}^{k}V\_{\bar{0}}^{\*}\otimes\bigwedge^{n-k}V\_{\bar{1}}^{\*})
= \text{Sym} V\_{\bar{0}}^{\*} \otimes \bigwedge^{\bullet} V\_{\bar{1}}.$$


Therefore, locally a graded manifold $\mathcal M$ looks like $(U, C^{\infty}(U) \otimes \text{Sym} V\_{\bar{0}}^{\*} \otimes \bigwedge^{\bullet} V\_{\bar{1}})$.


Compared to $(U, C^\infty(U)\otimes \wedge^\bullet V^\*)$ in definition 4.4.1, I got an extra term $\text{Sym} V\_{\bar{0}}^{\*}$.


The only way to eliminate this term is to require $V\_{\bar{0}} = 0$, i.e. $V = V\_{\bar{1}} = \bigoplus\_{k \in 2 \mathbb Z+1} V\_k$ is made of odd components.


However, in definition of graded manifold, there is no such requirement on the graded vector space $V$, so Definition 4.6.1 and Remark 4.6.1 are not consistent, and this is where I got puzzled.


A way to fix this might be like this, use a slightly different definition of $\mathbb Z$-supermanifold:


For example, in Mnev's [paper](https://arxiv.org/abs/1707.08096), definition 4.22 and the following, it requires the open set $U$ belongs to a graded vector space $W$(the target of **even** characters, as the $V\_\bar{1}$ we defined here), and $V$ the odd fiber.


Thanks for your time and effort.



$\mathbb Z$-graded manifold is isomorphic to the structure sheaf of supermanifold locally
A supermanifold is a graded manifold with only odd components, more precisely the equivalence is seen by setting $V\_{\bar{0}}=0$ and $V\_{\bar{1}} = V$ in your notation. In the opposite direction, a graded manifold is a supermanifold **only** if it has no even components, that is $V\_{\bar{0}} = 0$.


