In this paper, 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?

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.

  • 1
    $\begingroup$ 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? $\endgroup$
    – Ben McKay
    May 11 '20 at 12:08
  • $\begingroup$ @BenMcKay I'm sorry, I missed the condition that $V$ in the second definition is a graded vector space. $\endgroup$
    – Andrews
    May 11 '20 at 12:17
  • $\begingroup$ 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? $\endgroup$ May 11 '20 at 19:01
  • 1
    $\begingroup$ The odd components are odd. $\endgroup$ May 12 '20 at 3:22
  • 1
    $\begingroup$ 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. $\endgroup$ 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 $V_{\bar{0}} = 0$.


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