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, 2020 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, 2020 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, 2020 at 19:01
  • 1
    $\begingroup$ The odd components are odd. $\endgroup$ May 12, 2020 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, 2020 at 9:57

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.