This went “under my radar” for a while, I am not sure why, but anyway…

**Q-manifolds**

Recall that a Q-manifold is a supermanifold \(M\) equipped with a homological vector field, that is an odd vector field \(Q\) such that \([Q,Q]=2 Q^{2}=0\).

The algebra of functions \(C^{\infty}(M)\) is thus a supercommutative differential algebra. The product of two functions is again a function and importantly this is supercommutative:

\(f g = (-1)^{\widetilde{f} \widetilde{g} } g f\),

where I denote the Grassmann parity of a homogeneous function by the “tilde”. Just as on a classical manifold, the product of two functions is associative:

\(f(gh) = (fg)h\),

for all functions f,g,h. Also important is that the product itself does not carry any Grassmann parity, thus

\(\widetilde{fg} = \widetilde{f} + \widetilde{g}\).

The prime example here is to take \(M = \Pi TN\) where \(N\) is a classical manifold. The functions \(C^{\infty}(\Pi TM)\) are identified with (pseudo)differential forms and the homological vector field is just the de Rham differential.

**The derived product**

**Definition** Let \((M,Q)\) be a Q-manifold. Then the derived product on \(C^{\infty}(M)\) is defined as

\(f \ast g = (-1)^{\widetilde{f}+1} Q(f)g\).

Some comments are in order.

- The derived product is Grassmann odd, \(\widetilde{f\ast g} = \widetilde{f} + \widetilde{g}+1\)
- The derived product is clearly not supercommutative

**Remark** One can consider a derived product on any differential algebra, one does not need the supercommutivity, but as I am interested in supergeometry the definition here suits well.

Right away the derived product seems a strange beast, it carries Grassmann parity itself. Being noncommutative is also interesting, but we are used to noncommutative forms of multiplication.

**Theorem** * The derived product is associative.*

**Proof** Explicitly

\(f \ast (g \ast h) = (-1)^{\widetilde{f} + \widetilde{g}} Q(f)Q(g)h\).

Now consider

\((f \ast g)\ast h = (-1)^{\widetilde{f} + \widetilde{g}}Q ((-1)^{\widetilde{f}+1}Q(f)g )h\).

Using the fact that \(Q\) is homological, that is odd and squares to zero we get

\((f \ast g)\ast h =(-1)^{\widetilde{f} + \widetilde{g}} Q(f)Q(g) h\).

*QED*

**Theorem** \(f \ast g = (-1)^{(\widetilde{f}+1)(\widetilde{g}+1)} g \ast f + (-1)^{\widetilde{f}+1}Q(fg)\).

**Proof** Left as an exercise for the readers.

The above theorem shows very explicitly that the derived product is not supercommutative. It is however antisymmetric if and only if \(Q(fg)=0\).

**Where has this come from?**

As far as I am aware, the notion of a derived product can be found in the work of Loday;

J.L. Loday. Dialgebras. In *Dialgebras and related operads*, 7-66,* Lecture Notes in Math.*, **1763**, Springer, Berlin 2001.

*Jean-Louis Loday,*

(12 January 1946 – 6 June 2012)

**Concluding remarks**

We have a kind of noncommutative deformation of the algebra of functions on a Q-manifold via the derived product. This “new product” carries Grassman parity and is associative, it is closer to the standard notion of a product than say a Lie algebra.

I am sure there is lots more to say about derived products, but this is something I am only just beginning to explore. Watch this space…