# Is every graded manifold affine, and is this definition of graded manifold the right one?

The following definition is from:

• Dmitry Roytenberg, "AKSZ-BV formalism and Courant algebroid-induced topological field theories", Letters in Mathematical Physics, 2007 vol. 79 (2) pp. 143-159, MR2301393.

A graded manifold $M$ over base $M_0$ is a sheaf of $\mathbb Z$-graded commutative algebras ${\rm C}(M)$ over a smooth manifold $M_0$ locally isomorphic to an algebra of the form ${\rm C}^\infty(U) \otimes {\rm S}(V)$ where $U \subseteq M$ is an open set, $V$ is a graded vector space whose degree-zero component $V_0$ vanishes, and ${\rm S}(V)$ is the free graded-commutative algebra on $V$.

Presumably there is an equivalent definition that builds in the axioms for $M_0$ to be a smooth classical manifold — we could start with the topological space with a sheaf of Frechet algebras locally isomorphic to something.

Recall, from, for example the MO question Algebraic description of compact smooth manifolds?, that all classical manifolds are affine in the sense that, although presented as sheaves, a complete invariant is the algebra of global functions.

Question: Is it true that the algebra of global sections of the sheaf ${\rm C}(M)$ is a complete invariant? I.e. can I recover a graded manifold from its algebra of global smooth functions?

The follow-up question would be to describe, a la the above-linked MO question, which algebras are algebras of smooth functions on a graded manifold. The orthogonal follow-up question is the same one for "dg manifolds", which is a graded manifold with a square-zero degree $-1$ vector field, thought of as a derivation of the sheaf algebras of functions: does knowing the derivation on global sections determine it on all local sections?

Since no one has yet posted an answer, let me generalize the question to include:

Question: Is the above definition the right one?

Now let me motivate this generalization. First, the answer to my original question above is trivially "yes" provided that the underlying manifold $M_0$ can be read from the algebra, as then I would have access to the right partitions of unity to really get my hands on the whole sheaf. This certainly happens when the "fiber" (whose linear functions are) $V$ has only positive gradings: then the manifold is recovered from the degree-zero subalgebra. But if $V$ has both positive and negative degrees, then the degree-zero subalgebra is too big. Instead, the base $M_0$ should be recovered as the maximal degree-zero quotient algebra, and I have less intuition for whether such a thing should exist.

But then it's clear that the sheaf in the above definition is not, generally, a sheaf of Frechet algebras. For example, consider the vector space $\mathbb R^{1t^2 + 1t^{-2}}$ with one dimension in degree $2$ and one in degree $-2$, and call the coordinate functions $x$ and $y$. Then there is a quadratic map $\mathbb R^{1t^2 + 1t^{-2}} \to \mathbb R^1$ corresponding to the subalgebra of polynomials in $xy$, and this degree-zero subalgebra is not Frechet-complete. Rather, to exhibit the map as a map of smooth rather than algebraic spaces requires that ${\rm C^\infty}(\mathbb R^{1t^2 + 1t^{-2}})$ include the Frechet algebra ${\rm C^\infty}(\mathbb R^1)$ as a subalgebra.

So what's going on? If I add the words "Frechet completion" in the correct spot in the above definition, is it the answer to the first question "yes"?

There is another, older meaning of "affine", namely "properly embeds into (finite-dimensional) affine space". Before adding this extra content, the title of this question included a parenthetical "and in what sense". So at the risk of making this question long enough that the correct answer is a good reference:

Question: For the correct definition of "graded manifold", is it true that every graded manifold embeds smoothly into some finite-dimensional graded vector space"?

• Define "correct definition". (This is a serious request, not a snarky comment.) – Kevin H. Lin Jul 30 '10 at 7:41
• @Kevin Lin: I totally agree that "correct definition" needs a definition, but I don't have one. At some level, the correct definition should allow the examples people consider, behave a lot like classical manifolds, and be presentable without too much cruft. – Theo Johnson-Freyd Jul 30 '10 at 12:09
• Well, what examples do people consider? – Kevin H. Lin Jul 30 '10 at 16:49

## 1 Answer

First of all, you're completely right in that I didn't need to stipulate that $M_0$ be a manifold in the definition: it follows from a graded manifold being a locally ringed space with a particular local model (it was a physically motivated survey article, so I wasn't much concerned with foundational issues). It also follows that a map of graded manifolds must be polynomial in coordinates of nonzero degree, so things like $e^{xy}$ on $\mathbb{R}[-2]\times\mathbb{R}$ are not allowed. Furthermore, the body of $\mathbb{R}[-2]\times\mathbb{R}$ is a single point, so there are no non-constant maps from $\mathbb{R}[-2]\times\mathbb{R}$ to $\mathbb{R}$; in particular, your example is not a map of graded manifolds. In general, the body $M_0$ of a graded manifold $M$ can be recovered as the spec of $C^0(M)/(I\cap C^0(M))$, where $I$ is the ideal generated by coordinates of non-zero degree -- much like it is done for supermanifolds. Of course, in the non-negatively graded case, $M_0$ can also be recovered as $0\cdot M$.

Now, manifolds as well as supermanifolds and graded (super)manifolds are "affine" in the sense you asked. More precisely, taking global sections of the structure sheaf defines a fully faithful (contravariant) functor to $\mathbb{R}$-algebras (notice that you don't need a compactness assumption on the body, nor any topology on the algebra, which is a blessing for those who, like me, are put off by functional analysis). For manifolds this is classical (see eg. Cor 35.10, p. 301 of Kolar-Michor-Slovak

http://www.emis.de/monographs/KSM/kmsbookh.pdf ), the super/graded case follows easily. The differentials are okay as well, since they're nothing but vector fields.

Finally, whether my definition is the "right" one depends on the application one has in mind, I guess. For me, coordinates of positive degree generate the symmetries, while those of negative degrees serve to cut out the zero locus of the homological vector field (the solution set of Maurer-Cartan or Euler-Lagrange equations) and resolve its singularities. The relevant completions are then pro/ind completions, not something as awful as Frechet. Of course, that doesn't rule out applications for which the latter would be relevant.

ADDENDUM. Oh yes, graded manifolds are affine in the other sense as well, at least the non-negatively graded ones are (I haven't thought through the general case). It follows from the fact that they are isomorphic to total spaces of graded vector bundles over manifolds, by a result analogous to Batchelor's theorem for supermanifolds.

• Thanks for the great answer. I remember some talks by Peter Teichner in which he defined supermanifolds as commutative Frechet algebras in supervect with some other properties, so I was assuming that Frechet would be the right word for the graded case (I try to avoid analysis, so have no intuition of my own). I continue to be bothered that there is a polynomial map that is not a smooth map. But as my interest is in trying to put together precisely one paper on physics and Lie algebroids (hence reading your paper), my question is just idle curiosity: I don't have any particular applications. – Theo Johnson-Freyd Aug 6 '10 at 3:01
• Sure, no prob. I guess Teichner was giving a characterization of the essential image of the global sections functor into algebras -- an issue, in some sense, orthogonal to the answer I gave. In the graded case, the characterization will look different unless you complete, which may be a sensible thing to do for some purposes. Cheers! – Dmitry Roytenberg Aug 6 '10 at 17:47