Compatibility of two definitions of Koszul dual

Question

Let $k$ be a field and $A$ a nonnegatively graded ring over $k$. Assume $A_0 = k.$ We have a bigrading on $\operatorname{Ext}(k,k)$ (one corresponding to homological degree, one corresponding to the grading on $A$). We assume $A$ is Koszul, i.e., that $\operatorname{Ext}^{ij}(k,k)=0$ whenever $i\neq j$.

Following Beilinson-Ginzburg-Soergel, "Koszul duality patterns in representation theory", we can define the algebra $E(A)=\oplus\operatorname{Ext}^{ii}(k,k).$ They also define the quadratic dual, $A^{!}$. Their Theorem 2.10.1 then states that $E(A)=A^{!\operatorname{opp}}$.

On the other hand, the start of chapter $2$ of "Quadratic Algebras", Polishchuk&Positselski states instead that $E(A)=A^{!}$. Why is there this disparity? This is probably something stupid, but I can't figure it out.

(On a related note: I'm also confused as to how BGS is using the functor $\operatorname{RHom}(k,-)$ to go from $D(A-\operatorname{mof})$ to $D(A^{!}-\operatorname{mof});$ I would have assumed that applying that functor would give you a module over $\operatorname{RHom}(k,k)$, which is formal and thus should just correspond to $E(A)$, not $A^{!}$. Could someone clarify this as well?)

Answer 1

Let $A$ be a quadratic graded algebra over a field $k$ with finite-dimensional components $A_n$ and $A_0=k$. Then the construction of the quadratic dual algebra $A^!$ involves setting $A^!_1$ to be the dual $k$-vector space to $A_1$ and the subspace of quadratic relations in $A^!$ to be the orthogonal complement to the subspace of quadratic relations $R\subset A_1\otimes A_1$. The orthogonal complement is taken with respect to a natural nondegenerate pairing between the vector spaces $A_1\otimes A_1$ and $A_1^*\otimes A_1^*$.

Accordingly, the definition of the algebra $A^!$ depends on the choice of this pairing, or in other words, on the choice of a natural isomorphism $(A_1\otimes A_1)^*\simeq A_1^*\otimes A_1^*$. There are, basically, two options: one can set $\langle f\otimes g, u\otimes v\rangle=\langle f,u\rangle\langle g,v\rangle$ or $\langle f\otimes g, u\otimes v\rangle=\langle f,v\rangle\langle g,u\rangle$. Switching between these two choices replaces the algebra $A^!$ with the opposite algebra.

In our book, the convention is to use the first of these two pairings, which leads to $E(A)=A^!$. This looks like the natural choice for as long as you work with quadratic algebras over a field. However, when you replace a field with a noncommutative ring (e.g., a noncommutative semisimple algebra) in the role of $A_0$, the choice-of-pairing issue becomes the one of constructing a natural isomorphism $$Hom_{A_0}(A_1\otimes_{A_0}A_1,A_0)\simeq Hom_{A_0}(A_1,A_0)\otimes_{A_0} Hom_{A_0}(A_1,A_0),$$ where $Hom_{A_0}$ denotes the homomorphisms of (say) left $A_0$-modules. Then it turns out that the only such isomorphism is a particular case of the natural isomorphism $$Hom_{A_0}(U\otimes_{A_0}V,A_0)\simeq Hom_{A_0}(V,A_0)\otimes_{A_0} Hom_{A_0}(U,A_0)$$ for $A_0$-$A_0$-bimodules $U$ and $V$ that are finitely generated and projective as left $A_0$-modules. Notice that $U$ and $V$ switch places from the left-hand side to the right-hand side of the latter formula, which basically means that the only one of the above two commutative pairings that admits a noncommutative generalization is the second one.

A simple way to distinguish between the two opposite versions of the quadratic dual algebra in the case of a noncommutative ring $A_0$ is to look on the degree-zero components. Notice that $Ext^0_A(A_0,A_0) = Hom_{A_0}(A_0,A_0)=A_0^{opp}$ is the opposite ring to $A_0$ (if $Ext_A$ denotes the $Ext$ of left $A$-modules), while the algebra $A^!$ would be probably defined in such a way that $A^!_0=A_0$. Then the algebras $E(A)$ and $A^!$ cannot possibly be isomorphic, but only anti-isomorphic.

One of the consequences of these complications is that the functors $A\longmapsto A^!$ and $A\longmapsto A^{opp}$ do not commute with each other when the ring $A_0$ is noncommutative. Furthermore, if the functor $A\longmapsto A^!$ is defined in such a way that $A^!_0=A_0$ rather than $A_0^{opp}$, then this functor is not self-inverse when $A_0$ is noncommutative. Instead, one has $(A^{!opp})^{!opp}=A$. One can also say that the functor $A\longmapsto A^!$ comes in two versions, the left and the right one, $A\longmapsto A^!$ and $A\longmapsto {}^!\!A$, with $(A^!)_0=A_0=({}^!\!A)_0$ and ${}^!\!A=((A^{opp})^!)^{opp}$, which are inverse to each other, ${}^!(A^!)=A=({}^!\!A)^!$.