The Voronov Product of Operads

#math #operads #swiss-cheese

My first real post in a while! It turns out that writing an actual paper (cf. previous blog post) takes a lot of time and effort. Who knew?

The Voronov product of operads is an operation introduced by Voronov in his paper The Swiss-cheese operad (he just called it “the product”). It combines an operad and a multiplicative operad to yield a new colored operad; the main example I know is the homology of the Swiss-cheese operad. This is a construction that I use in my preprint Swiss-Cheese operad and Drinfeld center, where as far as I know I coined the name “Voronov product” – I haven’t seen this operation at all outside of Voronov’s paper. I wanted to advertise it a bit because I find it quite interesting and I’m eager to see what people can do with it.

Voronov products

The setting is as follows. Consider two symmetric one-colored operads, \(\mathtt{P}\) and \(\mathtt{Q}\), in some monoidal category. Suppose that you’re also given a morphism of operads \(\mathtt{Com} \to \mathtt{P}\), where \(\mathtt{Com}\) is the operad of commutative algebras. Then Voronov builds a new, bicolored operad \(\mathtt{P} \otimes \mathtt{Q}\).

This operad has two colors, \(\mathfrak{c}\) and \(\mathfrak{o}\), that you can think of as “open” and “closed” colors. The operations with closed output are exactly given by \(\mathtt{P}\), that is:

$$(\mathtt{P} \otimes \mathtt{Q})(\mathfrak{c}, \dots, \mathfrak{c}; \mathfrak{c}) = \mathtt{P}(n),$$

whereas if any of the colors \(x_i\) is not \(\mathfrak{c}\),

$$(\mathtt{P} \otimes \mathtt{Q})(x_1, \dots, x_n; \mathfrak{c}) = \varnothing.$$

This is an example of a relative operad (over \(\mathtt{P}\)), also known as a Swiss-cheese type operad. This type of operad can equivalently be seen as an operad in the category of right modules over \(\mathtt{P}\).

Composition of such operations is given by the composition of \(\mathtt{P}\). The operations of \(\mathtt{P} \otimes \mathtt{Q}\) with \(n\) open inputs, \(m\) closed inputs, and an open output, are given by:

$$(\mathtt{P} \otimes \mathtt{Q})(n,m) = \mathtt{P}(m) \otimes \mathtt{Q}(n).$$

There are two kinds of composition to define. To insert an operation with closed output in an operation with open output, one must define:

$$\circ_{i}^{\mathfrak c} : \bigl( \mathtt{P}(m) \otimes \mathtt{Q}(n) \bigr) \otimes \mathtt{P}(m') \to \mathtt{P}(m+m'-1) \otimes \mathtt{Q}(n).$$

This composition doesn’t touch the \(\mathtt{Q}(n)\) factor, and uses the composition of \(\mathtt{P}\) on the rest. To insert an operation with open output, one must also define:

$$\circ_{i}^{\mathfrak c} : \bigl( \mathtt{P}(m) \otimes \mathtt{Q}(n) \bigr) \otimes \bigl( \mathtt{P}(m') \otimes \mathtt{Q}(n') \bigr) \to \mathtt{P}(m+m') \otimes \mathtt{Q}(n+n'-1).$$

On the \(\mathtt{Q}\) factors, this is simply given by the composition of \(\mathtt{Q}\). On the \(\mathtt{P}\) factors, recall that we are given a morphism of operads \(\mathtt{Com} \to \mathtt{P}\); we thus have some multiplication \(\mu \in \mathtt{P}(2)\), and we can use it to define:

$$\begin{align} \mathtt{P}(m) \otimes \mathtt{P}(m') & \to \mathtt{P}(m+m') \\ p \otimes p' & \mapsto \mu(p, p'). \end{align}$$

Algebras over Voronov products

Algebras over \(\mathtt{P} \otimes \mathtt{Q}\) have a particularly nice description. Such an algebra is a pair \((A,B)\) where \(A\) is an algebra over \(\mathtt{P}\) and \(B\) is an algebra over \(\mathtt{Q}\). Since we are given a fixed morphism \(\mathtt{Com} \to \mathtt{P}\), it follows that \(A\) is endowed with a commutative algebra. There is finally an action of \(A\) on \(B\):

$$\nu : A \otimes B \to B,$$

corresponding to \(\operatorname{id} \otimes \operatorname{id} \in \mathtt{P}(1) \otimes \mathtt{Q}(1)\). This action has to satisfy the following condition, for all \(q \in \mathtt{Q}(n)\):

$$q(a_1 \cdot b_1, \dots, a_n \cdot b_n) = (a_1 \dots a_n) \cdot q(b_1, \dots, b_n).$$

Example: the homology of the Swiss-cheese operad

The main example of a Voronov product I know is the homology of the Swiss-cheese operad \(\mathtt{SC}\). Morally speaking, the Swiss-cheese operad is a combination of the little disks operad and the little intervals operad. It makes sense that its homology is given by a combination of their respective homologies.

This is indeed the case. The homology of the little disks operad \(\mathtt{Ger}\), the operad encoding Gerstenhaber algebras, and the homology of the little intervals operad is \(\mathtt{Ass}\), the operad encoding associative algebras. If we consider Voronov’s original version of the Swiss-cheese operad, which forbids operations with an open output and no closed input, then the homology is given by the Voronov product \(\mathtt{Ger} \otimes \mathtt{Ass}\)! That’s as good as can be expected. An algebra over this homology is a pair \((A,B)\) where \(A\) is a Gerstenhaber algebra and \(B\) is an associative algebra which is also a module over the underlying commutative algebra of \(A\), satisfying:

$$(a \cdot b) \cdot (a' \cdot b') = aa' \cdot bb', \; \forall a,a' \in A, b,b' \in B.$$

(Here we see the Eckmann–Hilton argument appearing in the background…)

If we now allow operations with an open output and only closed inputs, things get a bit more complicated. The description of the homology of this new operad can be found in the paper “Open-closed homotopy algebras and strong homotopy Leibniz pairs through Koszul operad theory” by Hoefel and Livernet. Just like before, an algebra over this operad is given by a pair consisting of a Gerstenhaber algebra \(A\) and an associative algebra \(B\). Instead of an action of \(A\) on \(B\), there is a morphism of commutative algebras from \(A\) to the center of the algebra \(B\). If \(B\) is a unital algebra, this is exactly the same thing as before, with \(f(b) = b \cdot 1_A\) (and the Eckmann–Hilton argument shows that this lands in the center of \(A\)).

This new operad can almost be described as the Voronov product of two operads. The remark about unital algebras tips us off. Instead of \(\mathtt{Ger}\) and \(\mathtt{Ass}\), consider instead \(\mathtt{Ger}_+\) and \(\mathtt{Ass}_+\), the operads encoding unital Gerstenhaber algebras and unital associative algebras. There is still a morphism \(\mathtt{Com} \to \mathtt{Ger}_+\), so we can build the Voronov product \(\mathtt{Ger}_+ \otimes \mathtt{Ass}_+\).

This is not quite right: this encodes a pair consisting of a unital Gerstenhaber algebra, a unital associative algebra, and a central morphism from the former to the latter. To recover the homology of the variant of Swiss-cheese, one simply removes the operations with zero inputs, something I denote \(\mathtt{Ger}_+ \otimes_0 \mathtt{Ass}_+\) in my paper (section 4). When we remove these operations we don’t have units anymore in our algebras, but we keep a central morphism \(A \to Z(B)\) and an action \(A \otimes B \to B\), related by:

$$a \cdot b = f(a) \cdot b$$

where the first dot is the action of \(A\) on \(B\) and the second one the multiplication in \(B\).

The main motivation for my paper was to try and “lift” this splitting of the homology of Swiss-cheese to the topological level. Due to the non-formality of the Swiss-cheese operad (cf. Livernet, Non-formality of the Swiss-cheese operad), it is not actually possible to do; nevertheless I think I succeeded in showing that the Swiss-cheese operad splits as a “shuffled” Voronov product, a notion that I’d like to formalize someday – read my paper for more details ;).