# How to formulate supercommutativity in a characteristic free way?

I would never dare posting this here, but the question https://math.stackexchange.com/q/3019853/214353 on math.SE did not receive any feedback (except for 13 views and one upvote) since November 30, so...

I believe I've seen an answer to this somewhere (some Deligne notes?) but cannot recover it.

In the $$\mathbb Z/2$$-graded formulation, supercommutativity means that two odd degree elements anticommute while all other kinds of homogeneous elements commute.

If 2 is invertible, to name a $$\mathbb Z/2$$-grading is the same as to name an involution: for $$\tau:A\to A$$ with $$\tau^2$$ equal to identity, even elements are those with $$\tau e=e$$ and the odd ones those with $$\tau o=-o$$. Then every $$a$$ is uniquely the sum of the even element $$(a+\tau a)/2$$ and the odd element $$(a-\tau a)/2$$.

Now suppose 2 is not invertible, and we still want to formulate supercommutativity in terms of $$\tau$$ alone, in such a way that it becomes precisely supercommutativity as soon as 2 becomes invertible. How to do it?

My attempts produced ugly and inconclusive things like $$2yx=xy+x\tau(y)+\tau(x)y-\tau(x)\tau(y)$$ or $$yx-\tau(y)x-y\tau(x)+\tau(y)\tau(x)=-xy+x\tau(y)+\tau(x)y-\tau(x)\tau(y).$$

There must be something much better, what is it?

• Why do you expect this to be possible? What if the characteristic is $2$? Note that speaking just in terms of gradings, over any (connected?) base a $\mathbb{Z}/2$-grading is the same as an action of the group scheme $\mu_2$. In characteristic $2$ this group scheme is non-reduced and in particular badly fails to be isomorphic to the constant group scheme $\mathbb{Z}/2$. – Qiaochu Yuan Jan 4 at 8:16
• @QiaochuYuan Well I don't even see any rigorous impossibility statement that could be formulated. – მამუკა ჯიბლაძე Jan 4 at 9:55
• Could you explain more about $\mu_2$-actions? Do not they just give $\mathbb Z/2$-graded plainly commutative algebras? Or you mean $\mu_2$-equivariant vector bundles over $\mu_2$-schemes? – მამუკა ჯიბლაძე Jan 4 at 10:05
• Btw what are tensor powers of $\mu_2$? – მამუკა ჯიბლაძე Jan 4 at 10:06
• @HarryGindi Actually I now also remember Deligne working with $\mu_2$ there... – მამუკა ჯიბლაძე Jan 4 at 10:13