Recall that a finite dimensional associative algebra $A$ over a field $k$ is called a symmetric Frobenius algebra (sometimes called "closed" Frobenius algebra) if its equipped with a symmetric non degenerate bilinear form $< , >: A \otimes A \to k$ satisfying $<ab,c>=<ca,b>$.

We can obtain a coproduct $v: A \to A \otimes A$ through the following composition of maps:

$A \cong A^* \to (A \otimes A)^* \cong A^* \otimes A^* \cong A \otimes A$

where the first map is the isomorphism of $A$-bimodules between $A$ and its dual $A^*$ defined by the pairing, the second is the dual of the product in $A$, the third follows from the finite dimensionality of $A$, and the last is again obtained from isomorphism induced by the pairing. One can check that that the compatibility of the product in $A$ with $<, >$ implies that $v$ is a map of $A$-bimodules.

This yields the notion of a "Frobenius bialgebra" (sometimes called an "open" Frobenius algebra), which is essentially an associative algebra $A$ equipped with a coassociative coproduct which is a map of $A$-bimodules. A closed Frobenius algebra is equivalent to an open Frobenius algebra with unit and counit.

There is a homotopy notion of a closed Frobenius algebra - I think originally due to Kontsevich- known as a cyclic $A_{\infty}$-algebra, which is by definition a finite dimensional $A_{\infty}$-algebra $(A, m_k:A^{\otimes k} \to A)$ equipped a non degenerate bilinear form $< , >: A \otimes A \to k$ cyclically compatible with the $m_k$'s in the sense that $<m_k(a_1,...,a_k),a_{k+1}>= \pm <m_k(a_{k+1},a_1,...,a_{k-1}),a_k>$. In particular, this cyclic compatibility condition implies that the isomorphism $A \to A^*$ is a map of $A_{\infty}$-bimodules over $A$.

My question is the following: Is there an explicit notion of a Frobenius $A_{\infty}$-bialgebra? In particular, given an a cyclic $A_{\infty}$-algebra $A$ we can obtain an $A_{\infty}$-coalgebra structure on $A$ by dualizing the structure maps and using the isomorphism $A \cong A^*$ induced by the pairing as we did above for the strict case. What are the explicit compatibilities between the $A_{\infty}$-algebra and $A_{\infty}$-coalgebra structure maps characterizing such structure?


Certainly one can ask for the data of a (unital or nonunital) $A_\infty$-algebra $A$ equipped with a "comultiplication" map $A \to A \otimes A$ of $A_\infty$ bimodules. Of course, this is infinitely much data --- all the higher coherences explaining how the comultiplication intertwines with the left and right actions. I will call these higher coherences "(higher) Frobeniators".

Such data will not, in and of itself, give the notion you are after, because it does not include the coassociativity data. It also clearly does not suffice to ask that the underlying map of vector spaces $A \to A \otimes A$ be part of a co-$A_\infty$ structure, as then you may not have good coherence between the (higher) coassociators and the (higher) Frobeniators. Recall that the coassociator is a homotopy between two different maps $A \to A \otimes A \otimes A$. What I expect is that you want to ask that it be a homotopy of maps of $A_\infty$-bimodules.

Almost surely that will give you a good notion of "open $A_\infty$ Frobenius algebra", although I don't know if/where it's been studied in the literature. If it has, almost surely it's in the context of Poincare duality.

Here's another option. Following Gan, call by the name dioperad a collection of chain complexes $\{P(m,n)\}_{m,n\in \mathbb N}$ with symmetric group actions and associative ``composition'' maps indexed by directed, but not necessarily rooted, trees. Dioperads are one possibly many-to-many generalization of operads which, unlike cyclic operads, admit a straightforward theory of Koszul duality. The open and coopen Frobenius dioperad is the one for which $P(m,n)$ is empty if $mn=0$ and otherwise is the free bimodule for the symmetric groups on $m$ and $n$ letters. It is generated by a class in $P(1,2)$ and a class in $P(2,1)$ subject to "associativity", "coassociativity", and "Frobenius" relations.

This dioperad is known to be Koszul (which follows e.g. from the Koszulity of the associative operad along with some results of Vallette), and its homotopy algebras are a reasonable notion of "homotopy Frobenius algebra". I believe, but have not checked, that they are precisely the same as the "open $A_\infty$ Frobenius algebras" mentioned above.

Dioperads only involve trees, and so everything in, say, "homotopy Frobenius algebra" when defined in terms of dioperads only involves trees. So let me actually call such things "tree-level homotopy Frobenius algebras". There is another reasonable many-to-many generalization of operads known as PROPs, which involve graphs. In particular, there is a PROP who describes Frobenius algebras. PROPs do not admit as good a Koszul duality theory, making it hard to work out "homotopy algebra", but pretty much every PROP in nature is the universal enveloping PROP of a properad, a notion due to Vallette, and moreover Vallette has shown that the universal enveloping functor from properads to PROPs is exact. In particular, there is a properad of Frobenius algebras (generated by the same presentation as the dioperadic version), and PROPic and properadic homotopy algebras thereof are equivalent. I will call homotopy algebras for the properad/PROP of Frobenius algebras "graph-level homotopy Frobenius algebras", since PROPs and properads involve non-tree graphs.

An aside: It is very hard to prove that the properad of Frobenius algebras is Koszul, and I don't think a proof is in the literature. Its commutative cousin is known to be Koszul by arXiv:1402.4048, but that paper was a long time coming and required some new techniques. Without Koszulity, you can still work out what a large presentation of the notion of "homotopy ...", but not a small one.

Now for the warning: the universal enveloping functor from dioperads to properads is NOT EXACT. So you should not expect the notions of graph- and tree-level homotopy Frobenius algebra to be equivalent.

I studied some related questions in arXiv:1412.4664 and arXiv:1307.5812, where I showed that tree-level homotopy "shifted commutative" Frobenius algebras can fail to extend to graph-level ones, or if they do extend then the extension can fail to be canonical. Indeed, the extension from tree- to graph-level is a "perturbative quantization" problem closely related to Feynman diagrams in quantum field theory.

The punchline is just that you should pay attention to which version of "homotopy Frobenius algebra" you want — graph- or tree-level — and be careful not to assume you have one when you've built the other.

| cite | improve this answer | |
  • $\begingroup$ Hi Theo, thanks for your answer! Are you claiming in your first comments that if we are given a cyclic $A_{\infty}$ algebra $(A, m_k: A^{\otimes k} \to A, <,>)$ and we dualize the $m_k$'s and use the isomorphism $A \cong A^*$ induced by $<,>$ we will obtain maps $u_k: A \to A^{\otimes k}$ defining an $A_{\infty}$-coalgebra structure and moreover each $u_k$ will be a map of $A_{\infty}$-bimodules over $A$ where the $A_{\infty}$-bimoldue structure on $A^{\otimes k}$ is given by the the left and right $A_{\infty}$-algebra strucutre of the first and last factors? $\endgroup$ – Manuel Rivera Jul 29 '15 at 16:06
  • $\begingroup$ Ah, maybe I want more data, since as you suggest $u_3$ should interact with the higher coherences that explain how $u_2: A \to A \otimes A$ interwines with the $A_{\infty}$-algebra actions on the left and right. $\endgroup$ – Manuel Rivera Jul 29 '15 at 16:31
  • $\begingroup$ @ManuelRivera I haven't thought enough about cyclic algebras enough to confidently make any claims. Well, let me say that I would not immediately expect $u_k$ to be itself a map of bimodules (of some cohomological degree), but instead the first Taylor coefficient of a homotopy between maps of bimodules, which I think is different. Then somehow you're supposed to extract the remaining coefficients by considering signed combinations of compositions of $u_k$s and $m_j$s. At least, that's my expectation. $\endgroup$ – Theo Johnson-Freyd Jul 29 '15 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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