# $\mathbb Z$-graded manifold is isomorphic to the structure sheaf of supermanifold locally

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?

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, 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.

• I think it is a typo: the second local model has infinite real dimensional fibers over every point of $M$, while the first doesn't. Maybe the vector space in the second example is odd? May 11 '20 at 12:08
• @BenMcKay I'm sorry, I missed the condition that $V$ in the second definition is a graded vector space. May 11 '20 at 12:17
• If you set $V_{\bar{0}} = 0$ and $V_{\bar{1}} = V$, then a supermanifold is straight forwardly a graded manifold, as basically you've already shown. A graded manifold with bosonic directions is not a supermanifold, so there cannot be a functor going in the other direction. What do you think is missing? May 11 '20 at 19:01
• The odd components are odd. May 12 '20 at 3:22
• In the updated question you write "The only way to eliminate this term is to require $V_{\bar{0}} = 0$ [...]" That is correct. A supermanifold is a graded manifold. But not every graded manifold is a supermanifold, as you've noticed. The Remark is talking about an isomorphism between the structure sheaves of any one supermanifold and a corresponding graded manifold. It cannot be referring to an isomorphism between the two categories, because such an isomorphism does not exist. May 12 '20 at 9:57

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 $$V_{\bar{0}} = 0$$.