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Let $A$ be a graded vector space, and suppose that two commuting differentials $d_1$ and $d_2$ of degree +1 act on $A$, such that $A$ equipped with either is a chain complex.

We now construct $C(A)$, the "chiral ring" of $A$, defined as follows: Let $A^\prime$ be the intersection of $\textrm{Ker}(d_1)$ and $\textrm{Ker}(d_2)$ inside $A$. That is, it is the space of vectors that are annihilated by both differentials. $C(A)$ is then a particular quotient of $A^\prime$. Then, we take its quotient by $\textrm{Im}(d_1)$ and $\textrm{Im}(d_2)$. Because space may not be a subspace of $A^\prime$, the quotient is taken by the intersection of ($\textrm{Im}(d_1)\cup \textrm{Im}(d_1)$) with $A^\prime$. In fact, the chiral ring $C(A)$ is NOT the cohomology of $A$, with respect to any differential.

If $A$ is endowed with a ring structure (commutative differential graded algebra), $A^\prime$ is obviously a subalgebra. So the product of two chiral operators is chiral. It remains to check whether or not the quotient is by an ideal in $A^\prime$. Since the kernel of the quotient $A^\prime \to C(A)$ is an ideal, $C(A)$ inherits a multiplication from $A$. The chiral ring is a ring.

Question: Is this construction well-known in the mathematical literature, and does it have a name?

In physics, the chiral ring is defined in this way (see the paper for a nice review). In supersymmetric gauge theories, chiral operators are particularly interesting because a correlation function of chiral operators is independent of positions of the operators. I wonder if mathematicians study properties of a chiral ring.

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  • $\begingroup$ Thanks to David Ben-Zvi's comment, I have included the condition that the two differentials commute. $\endgroup$ Commented Nov 9, 2015 at 19:46

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[Edit: I assumed you were talking about the case of commuting differentials. Are the examples in physics you're interested in not of this kind? If not, the commutator of the supercharges is a translation - are you not restricting to modes invariant by that translation (effectively making them commute in the reduced theory)? if not I'll delete this answer. Of course if they do commute there's the full theory of spectral sequences eg addressing the situation, of which the mixed complexes I mention are a special case, so maybe should delete anyway.]

I don't have an answer but a few comments:

It's clear that there's more structure in local operators of SUSY QFT than has been fully exploited in the math literature. In making twisted field theories one usually fixes a supercharge $Q^2=0$ and considers its cohomology. One then often asks how this depends on the choice of $Q$, e.g. in the work of Kapustin-Witten on Geometric Langlands, resulting in an interesting family of [in their case topological] field theories. But as you say these chiral rings are different objects, they are ways of extracting more structure out of the collection of available supercharges.

I'm not sure about the comment about independence of positions of operators - I thought that had to do with specific properties of the Q's rather than the abstract construction of chiral rings: if a given translation operator is Q-exact then correlation functions in Q-cohomology are independent of that translation, and I assume you're saying something similar for this bi-Q-cohomology, but that feels like an independent feature that one finds.

The closest structure to working with many Q's at once I've seen studied in math I learned about in a talk by Vera Serganova at MSRI last year. Look in the attached notes on p.4, where one considers (following Serganova and Duflo) the cone C of all square nilpotents in a Lie superalgebra, and for any module M we localize it over C by considering the various Q-cohomologies. The support of this is a version of the theory of support varieties, a key tool in many contexts in representation theory. In any case I assume one can describe the multi-Q-cohomology similarly over a space of commuting pairs. This also has a flavor common in representation theory, where "higher characters" are functions of k-tuples of commuting elements in a group.

Finally I should mention the theory of mixed complexes, namely complexes with two commuting differentials, which is the standard algebraic setting for the theory of cyclic homology. In this case we have a preserved charge (ie a graded vector space) and the two differentials have opposite charges (cohomological and homological differentials), which won't always be the case in physics. These model complexes with an action of the circle: the cohomological remnant of a circle action on a space is an action of the homology of $S^1$, which (over the reals) is the enveloping algebra of the Lie superalgebra ${\mathbb R}^{0|1}$, i.e. a square zero differential, which is indeed how cyclic homology arises in physics. e.g. we can start with a SUSY field theory (with fixed supercharge $Q$) in two dimensions and compactify on the circle (leading to a circle-rotation action) -- this is how de Rham cohomology arises out of the B-model say. So your setting with more commuting $Q$'s is an algebraic version of a torus action, and I think the cohomological invariants like chiral rings should have a natural interpretation in this setting.

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  • $\begingroup$ Thank you very much for your comments. Yes, you are right. In physics, we consider two commuting differentials $d_1$ and $d_2$. I think that I should have included this condition in the definition. $\endgroup$ Commented Nov 9, 2015 at 19:27
  • $\begingroup$ However, I don't think the theory of spectral sequence apply for chiral rings. $\endgroup$ Commented Nov 9, 2015 at 19:37
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    $\begingroup$ An example makes this easy to see. Let $A$ be of total rank three, generated by $x$ in deg 0 and $e_1$ and $e_2$ in deg 1. Let $d_1(x) = e_1$ and $d_2(x) = e_2$; everything else is zero for degree reasons. In this case, $A$ is in fact a bicomplex, and we can consider various cohomology theories: in order of decreasing size, we have $H(A,d_1)$, $H(A,d_2)$ with respect to single differentials; iterated cohomologies like $H(H(A,d_1),d_2)$; and finally $H(A,d_1+d_2)$ with respect to the total differential. For our example, all of these agree and have one generator in deg 1. $\endgroup$ Commented Nov 9, 2015 at 19:41
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    $\begingroup$ However, $C(A)$ has rank zero! $A^\prime$ is spanned by $e_1$ and $e_2$, which are both in the ideal $\textrm{Im}(d_1)+\textrm{Im}(d_2)$. The chiral ring is empty. Since rk $C(A)$ is not equal (mod 2) to rk $A$, there is no differential d such that $C(A) = H(A,d)$. $\endgroup$ Commented Nov 9, 2015 at 19:41
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    $\begingroup$ Point well taken. By the way is it really natural to ask for the two differentials to have degree specifically 1? in a general theory we just know they're odd. Giving a Z-grading corresponds to an invariant $U(1)$ charge in the theory, and depending on the charge the differentials could eg have degrees $\pm 1$ as in the cyclic homology setting, no? e.g. you could have a $U(1)$ acting nontrivially on the "twistor line" of linear combinations of $d_1,d_2$ (and maybe this even happens in geometric Langlands, ie $\mathcal N=4$ super Yang Mills in GL twist?) $\endgroup$ Commented Nov 9, 2015 at 21:34

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