2. Preliminaries

We will use without further reference well-known basic material on stable module categories of symmetric algebras, see e.g. [Reference Linckelmann22, § 2.13]. We briefly review the main properties of shift functors on stable module categories for symmetric algebra, mainly to adopt some notational abuse for simplicity of exposition later on.

Let R be a commutative Noetherian ring with unit element. Let A be an R-algebra. An A-module U is called relatively R-projective if the canonical surjection of A-modules $A\otimes_R U \to U$ sending $a\otimes u$ to au is split, and U is called relatively R-injective if the canonical injection of A-modules $U\to \mathrm{Hom}_R(A,U)$, $u\mapsto (a\mapsto au)$ is split (where $u\in U$ and $a\in A$), see [Reference Linckelmann22, § 2.6] for details.

Assume now that A is a symmetric R-algebra. Then the two classes of relatively R-projective and relatively R-injective modules coincide (cf. [Reference Linckelmann22, Theorem 2.15.1]). We denote by $\underline{\mathrm{mod}}(A)$ the relatively R-stable category of finitely generated A-modules. The objects of $\underline{\mathrm{mod}}(A)$ are the finitely generated A-modules, and morphisms in $\underline{\mathrm{mod}}(A)$ are classes of A-homomorphisms $\underline{\mathrm{Hom}}_A(U,V)=$ $\mathrm{Hom}_A(U,V)/\mathrm{Hom}^\mathrm{pr}_A(U,V)$, where U, V are finitely generated A-modules and $\mathrm{Hom}^\mathrm{pr}_A(U,V)$ is the R-module of A-homomorphisms from U to V which factor through a finitely generated relatively R-projective A-module. Composition in $\underline{\mathrm{mod}}(A)$ is induced by that in the category of finitely generated A-modules $\mathrm{mod}(A)$. Since R is Noetherian, the category $\mathrm{mod}(A)$ is a full abelian subcategory of the category $\mathrm{Mod}(A)$ of all A-modules. The category $\underline{\mathrm{mod}}(A)$ is no longer abelian but triangulated (cf. [Reference Linckelmann23, § A.3]),with shift functor $\Sigma_A$ which sends a finitely generated A-module U to the cokernel of U → I for some relatively R-injective envelope I of U. As an A-module, $\Sigma_A(U)$ depends on the choice of U → I but this assignment is unique up to unique isomorphism in $\underline{\mathrm{mod}}(A)$, hence does indeed induce a functor on $\underline{\mathrm{mod}}(A)$ (cf. [Reference Linckelmann22, Theorem 2.14.4]), still denoted $\Sigma_A$, which is unique up to unique isomorphism. Moreover, since A is symmetric, the functor $\Sigma_A$ is an equivalence on $\underline{\mathrm{mod}}(A)$ through which $\underline{\mathrm{mod}}(A)$ becomes a triangulated category (cf. [Reference Linckelmann23, Theorem A.3.2]). An inverse, denoted $\Sigma_A^{-1}$, of $\Sigma_A$ is induced by the assignment sending a finitely generated A-module U to the kernel of a relatively R-projective cover P → U of U, which as before depends, as an A-module, on the choice of P → U but is unique up to unique isomorphism in $\underline{\mathrm{mod}}(A)$. For any non-negative integer n we define $\Sigma^n_A(U)$ as the nth cokernel of a relatively R-injective resolution of U, and $\Sigma_A^{-n}(U)$ as the nth kernel of a relatively R-projective resolution of U. As before, these A-modules depend on the choices of resolutions, but their images in $\underline{\mathrm{mod}}(A)$ are unique up to unique isomorphism. As functors on $\underline{\mathrm{mod}}(A)$, we have canonical isomorphisms $\Sigma^n_A \cong (\Sigma_A)^n$ and $\Sigma_A^{-n}\cong$ $(\Sigma_A^{-1})^n$. We adopt the convention that $\Sigma_A^0$ is the identity functor on $\underline{\mathrm{mod}}(A)$. For any two integers n, m, we have canonical identifications of functors $\Sigma_A^n\circ \Sigma_A^m \cong \Sigma_A^{n+m}$ on $\underline{\mathrm{mod}}(A)$. For any integer n, we set

(2.1)\begin{equation} \widehat{\mathrm{Ext}}^n_A(U,V)=\underline{\mathrm{Hom}}_A(U,\Sigma_A^n(V)). \end{equation}

Note that if U is finitely generated projective as an R-module, then a projective cover (respectively resolution) of U is also a relatively R-projective cover (respectively resolution). If U is finitely generated projective as an R-module, then a relatively R-injective envelope (respectively resolution) can be constructed by taking the R-dual of a projective cover (respectively resolution) of the R-dual of U. If U, V are two A-modules which are finitely generated projective as R-modules, then $\mathrm{Hom}_A^\mathrm{pr}(U,V)$ consists of the space of A-homomorphisms from U to V which factor through a finitely generated projective A-module, and for n > 0 we have $\widehat{\mathrm{Ext}}^n_A(U,V)=$ $\mathrm{Ext}^n_A(U,V)$. In general, U need not have an injective resolution which consists of finitely generated injective A-modules.

The functor $\Sigma_A$ on $\underline{\mathrm{mod}}(A)$ lifts to an exact functor (albeit not an equivalence in general) on $\mathrm{mod}(A)$. To see this, set $A^e=$ $A\otimes_R A^\mathrm{op}$, and consider A as an A^e-module. As an A-A-bimodule, $\Sigma^{-1}_{A^e}(A)$ can be chosen to be the kernel of the multiplication map $A\otimes_R A \to A$, $a\otimes b\mapsto ab$, because this is a projective cover (not necessary minimal) of A as an A^e-module. This choice ensures that $\Sigma^{-1}_{A^e}(A)$ is finitely generated projective as a left and right A-module. The R-dual of the multiplication map together with the symmetry of A yields a relatively R-injective envelope $A\to A\otimes_R A$. Again with this choice, $\Sigma_{A^e}(A)$ is an A^e-module which is finitely generated projective as a left and right A-module, and the functor $\Sigma_{A^e}(A)\otimes_A -$ on $\mathrm{mod}(A)$ is exact and induces the equivalence $\Sigma_A$ on $\underline{\mathrm{mod}}(A)$. We emphasise that $\Sigma_{A^e}(A)\otimes_A-$ regarded as a functor on $\underline{\mathrm{mod}}(A)$ does not depend on this choice, but the exact lift to a functor on $\mathrm{mod}(A)$ does.

It follows from the above that choosing a symmetrising form of A uniquely determines in $\underline{\mathrm{mod}}(A^e)$ an isomorphism

\begin{equation*}\Sigma_{A^e}(A) \cong (\Sigma_{A^e}^{-1}(A))^\vee\end{equation*}

and hence more generally, uniquely determines isomorphisms

\begin{equation*}\Sigma_{A^e}^n(A) \cong (\Sigma_{A^e}^{-n}(A))^\vee\end{equation*}

in $\underline{\mathrm{mod}}(A^e)$, for all integers n. The Tate analogue $\widehat{H\!H}^*(A)$ of the Hochschild cohomology $H\!H^*(A)$ of A is

(2.2)\begin{equation} \widehat{H\!H}^n(A) = \widehat{\mathrm{Ext}}_{A^e}^n(A, A) = \underline{\mathrm{Hom}}_{A^e}(A, \Sigma^n(A)). \end{equation}

Since A is symmetric, hence finitely generated projective as an R-module, it follows as before that for n > 0 we have $\widehat{H\!H}^n(A)=$ $H\!H^n(A)$.

Let A, B, C be symmetric R-algebras. Then the R-algebras $A^\mathrm{op}$, $A\otimes_R B$, and $A\otimes_R B^\mathrm{op}$ are symmetric. An A-B-bimodule, or equivalently, an $A\otimes_R B^\mathrm{op}$-module, is called perfect if it is finitely generated projective as a left A-module and as a right B-module. We denote by $\mathrm{perf}(A,B)$ the category of perfect A-B-bimodules. Note that all modules in $\mathrm{perf}(A,B)$ are finitely generated projective as R-modules. The category $\mathrm{perf}(A,B)$ is a full R-linear subcategory of $\mathrm{mod}(A\otimes_R B^\mathrm{op})$ which is closed under taking direct summands. We denote by $\underline{\mathrm{perf}}(A, B)$ the image of $\mathrm{perf}(A, B)$ in $\underline{\mathrm{mod}}(A\otimes_R B^\mathrm{op})$; this is a thick subcategory of the triangulated category $\underline{\mathrm{mod}}(A\otimes_R B^\mathrm{op})$. If M is a perfect A-B-bimodule and N a perfect B-C-bimodule, then $M\otimes_B N$ is a perfect A-C-bimodule. In particular, we may choose $\Sigma^n_{A^e}(A)$ in $\mathrm{perf}(A,A)$ and $\Sigma^n_{B^e}(B)$ in $\mathrm{perf}(B,B)$. With such a choice, the exact functors $\Sigma^n_{A^e}(A)\otimes_A -$ and $-\otimes_B \Sigma^n_{B^e}(B)$ on $\mathrm{mod}(A\otimes_R B^\mathrm{op})$ restrict to exact functors on $\mathrm{perf}(A, B)$, and they induce functors on $\underline{\mathrm{mod}}(A\otimes_R B^\mathrm{op})$ which are both canonically isomorphic to the functor $\Sigma^n_{A\otimes_R B^\mathrm{op}}$ on $\underline{\mathrm{mod}}(A\otimes_R B^\mathrm{op})$. Note that $\mathrm{perf}(A,A)$ is closed under the tensor product over A, and hence $\underline{\mathrm{perf}}(A,A)$ is a tensor triangulated category, with tensor product $-\otimes_A -$.

If the algebra under consideration is clear from the context, we will simply write Σ for the shift functor on the stable module category, and sometimes use the same letter Σ for some exact lift to the category of finitely generated modules. This is to keep notation under control, but requires some care when it comes to establishing that all constructions are well-defined.

3. Adjunction for symmetric algebras

We briefly review without proofs some formalities on bimodules over symmetric algebras; broader expositions can be found in many sources such as [Reference Broué4], [Reference Broué5], [Reference Linckelmann19, § 6 Appendix], [Reference Linckelmann21, § 3], [Reference Linckelmann22, § 2.12]. Let R be a commutative Noetherian ring (with unit element), and let A, B be symmetric R-algebras, with symmetrising forms s and t, respectively. The functor $\mathrm{Hom}_R(-,R)$ is contravariant, and for U an R-module, we write $U^\vee=\mathrm{Hom}_A(U,R)$. Let M be a perfect A-B-bimodule. Since A, B are symmetric, the R-dual $M^\vee$ is a perfect B-A-bimodule. We have a B-A-bimodule isomorphism

(3.1)\begin{equation} \mathrm{Hom}_A(M,A)\cong M^\vee \end{equation}

sending $\alpha\in$ $\mathrm{Hom}_A(M,A)$ to $s\circ \alpha$, and we have a B-A-bimodule isomorphism

(3.2)\begin{equation} \mathrm{Hom}_{B^\mathrm{op}}(M,B)\cong M^\vee \end{equation}

sending $\beta\in $ $\mathrm{Hom}_{B^\mathrm{op}}(M,B)$ to $t\circ\beta$. For any A-module U, we have natural isomorphisms

(3.3)\begin{equation} M^\vee\otimes_A U \cong \mathrm{Hom}_A(M,A)\otimes_A U \cong \mathrm{Hom}_A(M,U) \end{equation}

where the first map is induced by the isomorphism from Equation (3.1) and the second map sends $\lambda\otimes u $ to the map $m\mapsto \lambda(m)u$, for $u\in U$, $m\in M$, $\lambda \in \mathrm{Hom}_A(M,A)$. Using that M is finitely generated projective as an A-module one sees that this is indeed an isomorphism. Combining this with the tensor-Hom adjunction shows that the functors $M\otimes_B-$ and $M^\vee\otimes_A-$ between the categories $\mathrm{mod}(A)$ and $\mathrm{mod}(B)$ of finitely generated modules over A and B, respectively, are left and right adjoint to each other. More precisely, the choices of symmetrising forms s, t determine adjunction isomorphisms

(3.4)\begin{equation} \mathrm{Hom}_A(M\otimes_B V,U) \cong \mathrm{Hom}_B(V,M^\vee\otimes_A U) \end{equation}

where U is a finitely generated A-module and V a finitely generated B-module. This isomorphism has the property that it sends a map of the form $\lambda_{\gamma,u}$ to the map $v\mapsto$ $s\circ\gamma_v\otimes u$, where $\gamma\in$ $\mathrm{Hom}_A(M\otimes_B V,A)$, $u\in$ U, where $\lambda_{\gamma,u}\in$ $\mathrm{Hom}_A(M\otimes_B V,U)$ is defined by $\lambda_{\gamma,u}(m\otimes v)=$ $\gamma(m\otimes v)u$, and where $\gamma_v\in$ $\mathrm{Hom}_A(M,A)$ is defined by $\gamma_v(m)=$ $\gamma(m\otimes v)$, for all $m\in$ M, $v\in$ V. Maps of the form $\lambda_{\gamma,u}$ are precisely the maps which factor through A, hence span the subspace $\mathrm{Hom}_A^\mathrm{pr}(M\otimes_B V,U)$ of $\mathrm{Hom}_A(M\otimes_B V,U)$. The unit and counit of the adjunction (3.4) are represented by bimodule homomorphisms

(3.5)\begin{align} & \epsilon_M : B \longrightarrow M^\vee\otimes_A M\ ,\ \ 1_B \mapsto \sum_{i\in I}\ (s\circ \alpha_i) \otimes m_i\ ,\\ & \eta_M : M\otimes_B M^\vee\longrightarrow A\ ,\ \ m\otimes (s\circ\alpha)\mapsto \alpha(m)\ ,\nonumber \end{align}

where I is a finite indexing set, $\alpha_i\in$ $\mathrm{Hom}_A(M,A)$ and $m_i\in$ M such that $\sum_{i\in I}\ \alpha_i(m^{\prime})m_i=$ mʹ for all $m^{\prime}\in$ M. Similarly, we have a natural isomorphism

(3.6)\begin{equation} \mathrm{Hom}_B(M^\vee\otimes_A U,V) \cong \mathrm{Hom}_A(U,M\otimes_B V) \end{equation}

obtained from Equation (3.4) by exchanging the roles of A and B and using $M^\vee$ instead of M together with the canonical double duality $M^{\vee\vee}\cong$ M. The adjunction unit and counit of this adjunction are represented by bimodule homomorphisms

(3.7)\begin{align} & \epsilon_{M^\vee} : A \longrightarrow M\otimes_B M^\vee\ ,\ \ 1_A \mapsto \sum_{j\in J}\ m_j\otimes (t\circ \beta_j)\ ,\\ & \eta_{M^\vee} : M^\vee\otimes_A M\longrightarrow B\ ,\ \ (t\circ\beta)\otimes m \mapsto \beta(m)\ , \nonumber \end{align}

where J is a finite indexing set, $\beta_j\in$ $\mathrm{Hom}_{B^\mathrm{op}}(M,B)$, $m_j\in$ M, such that $\sum_{j\in J}\ m_j\beta_j(m^{\prime})=$ mʹ for all $m^{\prime}\in$ M, where $m\in$ M and $\beta\in$ $\mathrm{Hom}_{B^\mathrm{op}}(M,B)$. Note that $\eta_M\circ \epsilon_{M^\vee}$ is an A-A-bimodule endomorphism of A, hence given by left or right multiplication with an element in Z(A). Similarly, $\eta_{M^\vee}\circ \epsilon_{M}$ is a B-B-bimodule endomorphism of B, hence given by left or right multiplication with an element in Z(B). Following [Reference Linckelmann19, Definition 3.1], we set

(3.8)\begin{align} \pi_M & = (\eta_M\circ \epsilon_{M^\vee})(1_A)\\ \pi_{M^\vee} &= (\eta_{M^\vee}\circ \epsilon_M)(1_B). \nonumber \end{align}

We call π_M the relatively M-projective element of Z(A). Similarly, $\pi_{M^\vee}$ is called the relatively $M^\vee$-projective element of Z(B). These elements depend on the choices of the symmetrising forms of A and B, see [Reference Linckelmann19, Remark 3.2] for details.

Remark 3.10. The adjunction isomorphism (3.4) is additive in M. Thus, the adjunction unit and counit in Equation (3.5) are additive in M in the following sense: given two perfect A-B-bimodules M, N, the adjunction unit

\begin{equation*}\epsilon_{M\oplus N} : B \to (M\oplus N)^\vee \otimes_A (M\oplus N)\end{equation*}

is equal to the composition of

\begin{equation*}\epsilon_M + \epsilon_N : B\to M^\vee\otimes_A M \oplus N^\vee \otimes_A N\end{equation*}

followed by the canonical inclusion of the right side into $(M\oplus N)^\vee \otimes_A (M\oplus N)$. Similarly, the adjunction counit

\begin{equation*}\eta_{M\oplus N} : (M\oplus N)\otimes_B (M\oplus N)^\vee \to A\end{equation*}

is equal to the map

\begin{equation*}\eta_M+\eta_N : M\otimes_A M^\vee \oplus N\otimes_A N^\vee \to A\end{equation*}

extended by zero on the mixed summands $M\otimes_A N^\vee$ and $N\otimes_A M^\vee$. The analogous statements hold for the adjunction isomorphism (3.6) and the corresponding adjunction unit and counit in Equation (3.7).

Remark 3.11. If U, V have in addition right C-module structures for some further R-algebra C, then the isomorphisms in Equations (3.1) and (3.3) are isomorphisms of right C-modules. Thus the isomorphism (3.4) induces an isomorphism

\begin{equation*}\mathrm{Hom}_{A\otimes_R C^\mathrm{op}}(M\otimes_B V,U) \cong \mathrm{Hom}_{B\otimes_R C^\mathrm{op}}(V,M^\vee\otimes_A U) .\end{equation*}

4. Transfer for symmetric algebras

Let R be a commutative Noetherian ring (with unit element), and let A, B be symmetric R-algebras, with symmetrising forms s, t, respectively. Let M be a perfect A-B-bimodule. Following [Reference Broué4], for finitely generated A-modules U, V, we have a transfer map

(4.1)\begin{equation} \mathrm{tr}_M = \mathrm{tr}_M(U,V) : \mathrm{Hom}_B(M^\vee\otimes_A U, M^\vee \otimes_A V) \to \mathrm{Hom}_A(U,V) \end{equation}

sending a B-homomorphism $\beta : M^\vee\otimes_A U\to M^\vee\otimes_A V$ to the A-homomorphism

(4.2)\begin{equation} \mathrm{tr}_M(\beta) = (\eta_M\otimes\mathrm{Id}_V) \circ (\mathrm{Id}_M\otimes \beta) \circ (\epsilon_{M^\vee} \otimes \mathrm{Id}_U). \end{equation}

More explicitly, $\mathrm{tr}_M(\beta)$ is the composition of A-homomorphisms

with the standard identifications $A\otimes_A U=U$ and $A\otimes_A V=V$. The functors $M\otimes_B-$ and $M^\vee\otimes_A-$ are exact and preserve finitely generated projective modules. Therefore, if β factorises through a projective B-module, then $\mathrm{tr}_M(\beta)$ factorises through a projective A-module, and hence $\mathrm{tr}_M$ induces a well-defined map, still denoted

(4.3)\begin{equation} \mathrm{tr}_M : \underline{\mathrm{Hom}}_B(M^\vee\otimes_A U, M^\vee \otimes_A V) \to \underline{\mathrm{Hom}}_A(U,V). \end{equation}

It also follows that in the stable module category $\underline{\mathrm{mod}}(A)$, for any integer n, we have unique isomorphisms

\begin{equation*}\Sigma_A^n(M\otimes_A M^\vee \otimes_A U) = M \otimes_A \Sigma_B^n(M^\vee\otimes_A U) = M\otimes_B M^\vee \otimes_A \Sigma_A^n(U)\end{equation*}

and through these identifications and their analogues, we have an equality of morphisms in the stable category $\underline{\mathrm{mod}}(A)$

(4.4)\begin{equation} \Sigma^n_A(\mathrm{tr}_M(\beta)) = \mathrm{tr}_M(\Sigma_B^n(\beta)) : \Sigma_A^n(U) \to \Sigma_A^n(V). \end{equation}

By [Reference Linckelmann20, § 7.1], there are graded versions of these transfer maps for Tate and Tate–Hochschild cohomology. An element in $\widehat{\mathrm{Ext}}^n_B(M^\vee\otimes_A U,M^\vee\otimes_A V)$ is represented by a B-homomorphism $\beta : M^\vee\otimes_A U\to$ $M^\vee\otimes_A \Sigma^n(V)$, where we identify $\Sigma^n(M^\vee\otimes_A U)=$ $M^\vee\otimes_A \Sigma^n(V)$ and where we use the same letter Σ for either $\Sigma_A$ or $\Sigma_B$. The transfer map $\mathrm{tr}_{M}$ sends β to the element $\mathrm{tr}_M(\beta)$ in $\mathrm{Ext}_A^n(U, V)$ represented by the A-homomorphism, abusively also denoted $\mathrm{tr}_M(\beta)$, given by

(4.5)\begin{equation} \mathrm{tr}_M(\beta) = (\eta_M\otimes \mathrm{Id}_{\Sigma^n(V)})\circ (\mathrm{Id}_M\otimes \beta) \circ (\epsilon_{M^\vee} \otimes \mathrm{Id}_U) \end{equation}

with the standard identifications $A\otimes_A U=U$ and $A\otimes_A \Sigma^n(V)=\Sigma^n(V)$. More explicitly, $\mathrm{tr}_M(\beta)$ is obtained as the composition

A variation of the same principle applied to bimodules yields in particular a transfer for Tate–Hochschild cohomology. We use again simply Σ instead of $\Sigma_{A\otimes_k A^{\mathrm{op}}}$ or $\Sigma_{B\otimes_k B^{\mathrm{op}}}$. An element $\zeta\in$ $\widehat{H\!H}^n(B)$ is represented by a B-B-bimodule homomorphism, abusively denoted by the same letter, $\zeta : B\to$ $\Sigma^n(B)$. We denote by $\mathrm{tr}_M(\zeta)$ the element in $\widehat{H\!H}^n(A)$ represented by the A-A-bimodule homomorphism

precomposed with the adjunction unit $\epsilon_{M^\vee}: A\to$ $M\otimes_B M^\vee$ and composed with the ‘shifted’ adjunction counit $\Sigma^n(\eta_M) : \Sigma^n(M\otimes_B M^\vee)\to$ $\Sigma^n(A)$. The identification $M\otimes_B \Sigma^n(B)\otimes_B M^\vee=$ $\Sigma^n(M\otimes_B M^\vee)$ is to be understood as the canonical isomorphism in $\underline{\mathrm{mod}}(A\otimes_k A^{\mathrm{op}})$, using the fact that the functor $M\otimes_B - \otimes_B M^\vee$ sends a projective resolution of the B-B-bimodule B to a projective resolution of the A-A-bimodule $M\otimes_B M^\vee$. Modulo this identification, we thus have graded k-linear map

(4.7)\begin{equation} \mathrm{tr}_M : \widehat{H\!H}^*(B) \longrightarrow \widehat{H\!H}^*(A) \end{equation}

defined by

(4.8)\begin{equation} \mathrm{tr}_M(\zeta) = \Sigma^n(\eta_M)\circ (\mathrm{Id}_M\otimes\zeta\otimes\mathrm{Id}_{M^\vee})\circ \epsilon_{M^\vee}. \end{equation}

Note that $\mathrm{tr}_M$ is not necessarily a multiplicative map from $\widehat{H\!H}^*(B)$ to $\widehat{H\!H}^*(A)$. In all the cases above, we have analogous transfer maps $\mathrm{tr}_{M^\vee}$ obtained from exchanging the roles of A and B. Using $\mathrm{Ext}$ instead of $\widehat{\mathrm{Ext}}$ yields the transfer maps introduced in [Reference Linckelmann19]. The two are well-known to coincide for n > 0.

Suppose that A is R-free. Let X be an R-basis of A and $X^\vee$ the dual basis with respect to the symmetrising form s on A; that is, we have a bijection $x\mapsto x^\vee$ from X to $X^\vee$ such that $s(xx^\vee)=1$ for $x\in X$ and $s(xy^\vee)=0$ for x, $y \in X$ such that x ≠ y. The element

(4.9)\begin{equation} z_A = \sum_{x\in X} x x^\vee \end{equation}

is called the relative projective element with respect to s . One easily checks that this is an element in Z(A) which does not depend on the choice of the basis X, but which does depend on the choice of s. If sʹ is another symmetrising form, then there is a unique element $z\in Z(A)^\times$ such that $s^{\prime}(a) = s(za)$ for all $a\in A$. If $X^\vee$ is as before the dual basis of X with respect to s, then $z^{-1}X^\vee$ is the dual basis of X with respect to sʹ, and hence the relative projective element with respect to sʹ is equal to $z^{\prime}_A = z^{-1}z_A$.

Tate duality for Tate–Hochschild cohomology involves bimodules, and hence we will need the following well-known description of relative projective elements for tensor products of symmetric algebras as well as their compatibility with the passage to blocks.

Proof. A trivial verification shows that s, $s\otimes t$, and s + t are symmetrising forms of $A^\mathrm{op}$, $A\otimes_R B$, and A × B, respectively. Let X be an R-basis of A, with dual basis Xʹ and corresponding bijection $x\mapsto x^{\prime}$ from X to Xʹ as in Equation (4.9). Then X and Xʹ are also dual to each other with respect to s as a symmetrising form of $A^\mathrm{op}$. This implies $z_{A^\mathrm{op}}=z_A$, whence (i). Let Y be an R-basis of B with dual basis Yʹ and corresponding bijection $y\mapsto y^{\prime}$ for $y\in Y$. Then the image in $A\otimes_R B$ of $X\otimes Y$ is an R-basis, and its dual basis with respect to $s\otimes t$ is $X^{\prime}\otimes Y^{\prime}$, with the bijection from $X\otimes Y$ to $X^{\prime}\otimes Y^{\prime}$ mapping $x\otimes y$ to $x^{\prime}\otimes y^{\prime}$, for $x\in X$ and $y\in Y$. It follows that

\begin{equation*} z_{A\otimes_R B} = \sum_{x\in X, y\in Y}\ xx^{\prime} \otimes yy^{\prime} = (\sum_{x\in X} xx^{\prime}) \otimes (\sum_{y\in Y} yy^{\prime}) = z_A \otimes z_B \end{equation*}

as stated in (ii). The union $(X \times \{0\}) \cup (\{0\}\times Y)$ is an R-basis of A × B with dual basis $(X^{\prime}\times \{0\}) \cup (\{0\}\times Y^{\prime})$. Statement (iii) follows.

Remark 4.13. The transfer map in Tate–Hochschild cohomology (4.7) and (4.8) is not strictly speaking a special case of the transfer maps $\mathrm{tr}_M(U,V)$, but the two are related via a generalisation of $\mathrm{tr}_M(U,V)$. Let A, B, C be symmetric R-algebras, and let U,V be finitely generated $A\otimes_R C^\mathrm{op}$-modules. The transfer map $\mathrm{tr}_M=$ $\mathrm{tr}_M(U,V)$ from Equation (4.1) induces a map, yet again denoted

\begin{equation*}\mathrm{tr}_M = \mathrm{tr}_M(U,V) : \mathrm{Hom}_{B\otimes_R C^\mathrm{op}}(M^\vee\otimes_A U, M^\vee \otimes_A V) \to \mathrm{Hom}_{A\otimes_R C^\mathrm{op}}(U,V)\end{equation*}

sending a $B\otimes_R C^\mathrm{op}$-homomorphism $\beta : M^\vee\otimes_A U\to M^\vee\otimes_A V$ to the $A\otimes_R C^\mathrm{op}$-homomorphism

\begin{equation*}\mathrm{tr}_M(\beta) = (\eta_M\otimes\mathrm{Id}_V) \circ (\mathrm{Id}_M\otimes \beta) \circ (\epsilon_{M^\vee} \otimes \mathrm{Id}_U).\end{equation*}

The functor $M\otimes_B-$ sends a projective $B\otimes_R C^\mathrm{op}$-module to a projective $A\otimes_R C^\mathrm{op}$-module, and hence if β factors through a projective $B\otimes_R C^\mathrm{op}$-module, then $\mathrm{Id}_M\otimes\beta$ factors through a projective $A\otimes_R C^\mathrm{op}$-module. Thus $\mathrm{tr}_M$ induces a well-defined map

\begin{equation*}\mathrm{tr}_M : \underline{\mathrm{Hom}}_{B\otimes_R C^\mathrm{op}}(M^\vee\otimes_A U, M^\vee \otimes_A V) \to \underline{\mathrm{Hom}}_{A\otimes_R C^\mathrm{op}}(U,V).\end{equation*}

Applied with C = A, U = A, $V=\Sigma_{A^e}^n(A)$, this yields a map

(4.14)\begin{equation} \mathrm{tr}_M : \underline{\mathrm{Hom}}_{B\otimes_R A^\mathrm{op}}(M^\vee, \Sigma^n(M^\vee)) \to \underline{\mathrm{Hom}}_{A^e}(A,\Sigma^n(A)) \end{equation}

where we have made use of the standard identifications $\Sigma^n(B)\otimes_B M^\vee\cong$ $\Sigma^n(M^\vee)\cong$ $M^\vee\otimes_A \Sigma^n(A)$ in the stable category $\underline{\mathrm{mod}}(B\otimes_R A^\mathrm{op})$. The functor $-\otimes_B M^\vee$ induces a graded algebra homomorphism $\widehat{H\!H}^*(B) =$ $\widehat{\mathrm{Ext}}^*_{B^e}(B,B) \to$ $\widehat{\mathrm{Ext}}^*_{B\otimes_R A^\mathrm{op}}(M^\vee, M^\vee)$, and composing this with the map $\mathrm{tr}_M$ from Equation (4.14) yields the transfer map in Hochschild cohomology $\widehat{H\!H}^*(B) \to$ $\widehat{H\!H}^*(A)$ from Equations (4.7) and (4.8).

5. Adjunction maps for matrix algebras

We need to identify the adjunction maps and the transfer maps reviewed in the previous section in the case that A, B are matrix algebras. This is elementary linear algebra, so we just give some pointers towards verifications. Let R be a commutative ring. Let U, V be free R-modules of finite ranks over R. Set $A=\mathrm{End}_R(U)$ and $B=\mathrm{End}_R(V)$. Then A and B are symmetric R-algebras with symmetrising forms the trace maps $\mathrm{trace}_U$, $\mathrm{trace}_V$, sending a R-linear endomorphism of U, V to its trace, respectively. Any other symmetrising form of A, B is of the form $\rho\cdot\mathrm{trace}_U$, $\rho\cdot\mathrm{trace}_V$ for some $\rho \in R^\times$, respectively. Set $M=U\otimes_R V^\vee$. Tensoring with M and its dual is the simplest instance of a Morita equivalence; all we need to make sure in this section is that the standard maps in this context are indeed the adjunction maps with respect to the trace maps as symmetrising forms. These verifications make use of the following well-known Lemma which links traces to adjunction maps.

Lemma 5.1. We have an isomorphism $\mathrm{Hom}_A(U,A)\cong U^\vee$ sending $\lambda\in$ $\mathrm{Hom}_A(U,A)$ to $\mathrm{trace}_U\circ \lambda$. We have a commutative diagram of A-A-bimodule homomorphisms

where α sends $u\otimes \mu$ to the endomorphism $u^{\prime}\mapsto \mu(u^{\prime})\otimes u$, σ sends $u\otimes \lambda$ to $u\otimes (\mathrm{trace}_U\circ\lambda)$, ρ sends $u\otimes \lambda$ to $\lambda(u)$, and τ sends $u\otimes \mu$ to $\mu(u)$, for all u, $u^{\prime}\in U$, $\lambda\in\mathrm{Hom}_A(U,A)$, and $\mu\in U^\vee$.

We identify $M^\vee$ with $V\otimes_R U^\vee$ via the obvious isomorphisms

\begin{equation*}M^\vee= (U\otimes_R V^\vee)^\vee= V^{\vee\vee}\otimes_R U^\vee= V\otimes_R U^\vee. \end{equation*}

This leads to identifications

\begin{equation*}M\otimes_B M^\vee = U\otimes_R V^\vee \otimes_B V \otimes_R U^\vee = U\otimes_R U^\vee\end{equation*}

where we identify $V^\vee\otimes_B V = R$ via the map $\nu \otimes v \to \nu(v)$, for $v\in V$ and $\nu \in V^\vee$. Similarly, we identify $M^\vee\otimes_A M = V\otimes_R V^\vee$. Let ${\mathcal{B}}$ be an R-basis of U, with dual basis in $U^\vee$ denoted ${\mathcal{B}}^\vee$. For $u\in {\mathcal{B}}$, we denote by $u^\vee$ the unique element in ${\mathcal{B}}^\vee$ satisfying $u^\vee(u)=1$ and $u^\vee(u^{\prime})=0$ for $u^{\prime}\in$ ${\mathcal{B}}$, $u^{\prime}\neq u$. Similarly, let ${\mathcal{C}}$ be an R-basis of V, with dual basis in $V^\vee$ denoted analogously ${\mathcal{C}}^\vee$. The adjunction units and counits from the preceding section in this case (with the choice of symmetrising forms $\mathrm{trace}_U$, $\mathrm{trace}_V$) are all isomorphisms, and their precise descriptions are as follows:

(5.2)\begin{align} & \epsilon_M : B \longrightarrow M^\vee\otimes_A M = V\otimes_R V^\vee \ ,\ \ 1_B \mapsto \sum_{v\in{\mathcal{C}}}\ v \otimes v^\vee,\\ & \eta_M : U\otimes_R U^\vee = M\otimes_B M^\vee\longrightarrow A\ ,\ \ u \otimes \mu\mapsto (y \mapsto \mu(y)u), \nonumber \end{align}

where u, $y\in U$, $\mu\in U^\vee$.

(5.3)\begin{align} \epsilon_{M^\vee} : A \longrightarrow M\otimes_B M^\vee = U\otimes_R U^\vee\ ,\ \ 1_A \mapsto \sum_{u\in {\mathcal{B}}}\ u \otimes u^\vee\ ,\\ \eta_{M^\vee} : V\otimes_R V^\vee = M^\vee\otimes_A M\longrightarrow B\ ,\ \ v\otimes \nu \mapsto (w \mapsto \nu(w)v)\ , \nonumber \end{align}

where v, $w\in V$ and $\nu \in V^\vee$. We further note that

(5.4)\begin{equation} \epsilon_{M^\vee} = (\eta_M)^{-1},\ \ \ \ \ \eta_{M^\vee} = (\epsilon_M)^{-1}. \end{equation}

An easy verification shows that the relative projective elements in Z(A) and Z(B) with respect to the symmetrising forms $\mathrm{trace}_U$ and $\mathrm{trace}_V$, respectively, are equal to

(5.5)\begin{equation} z_A = \mathrm{rk}_R(U)\cdot 1_R, \ \ \ \ \ \ \ \ z_B = \mathrm{rk}_R(V)\cdot 1_R. \end{equation}

Remark 5.6. Let s, t be symmetrising forms of A, B, respectively. Then $s = \lambda \cdot\mathrm{trace}_U$ and $t = \mu \cdot \mathrm{trace}_V$ for some λ, $\mu \in R^\times$. Denoting by $\epsilon^{\prime}_M$, $\eta^{\prime}_M$, $\epsilon^{\prime}_{M^\vee}$, $\eta^{\prime}_{M^\vee}$ the adjunction maps from Equations (5.2) and (5.3) with respect to s, t, it follows that

\begin{equation*}\epsilon^{\prime}_M = \lambda\epsilon_M,\ \ \ \ \ \eta^{\prime}_M = \lambda^{-1} \eta_M, \ \ \ \ \ \epsilon_{M^\vee} = \mu \epsilon_{M^\vee}, \ \ \ \ \ \eta^{\prime}_{M^\vee} = \mu^{-1} \eta_{M^\vee}.\end{equation*}

Thus the trace map $\mathrm{tr}^{\prime}_M : \mathrm{End}_B(M^\vee\otimes_A U) \to \mathrm{End}_A(U)$ with respect to s and t satisfies

\begin{equation*}\mathrm{tr}^{\prime}_M = \lambda^{-1} \mu \mathrm{tr}_M.\end{equation*}

The relative projective central elements $z^{\prime}_A$ and $z^{\prime}_B$ with respect to s, t are

\begin{equation*}z^{\prime}_A = \lambda^{-1} z_A = \lambda^{-1}\mathrm{rk}_R(U), \ \ \ \ \ z^{\prime}_B = \mu^{-1} z_B= \mu^{-1} \mathrm{rk}_R(V).\end{equation*}

6. Transfer for matrix algebras

Let K be a field of characteristic zero, and let U, V be finite-dimensional K-vector spaces. We set $A=\mathrm{End}_K(U)$ and $B=\mathrm{End}_K(V)$, regarded as symmetric algebras with symmetrising forms $\mathrm{trace}_U$ and $\mathrm{trace}_V$, respectively. We set $M = U\otimes_K V^\vee$. We note that since $A\otimes_K A^\mathrm{op}$ and $A\otimes_K B^\mathrm{op}$ are simple algebras, it follows that every finitely generated A-A-bimodule is projective and isomorphic to a finite direct sum of copies of A, and every finitely generated A-B-bimodule is projective and isomorphic to a finite direct sum of copies of the simple A-B-bimodule M. For finitely generated A-modules Uʹ, Uʹʹ we denote by

\begin{equation*}\varphi_A : \mathrm{Hom}_A(U^{\prime},U^{\prime\prime}) \times \mathrm{Hom}_A(U^{\prime\prime},U^{\prime}) \to K\end{equation*}

the bilinear map sending $(\alpha, \beta)$ to $z_A^{-1} \mathrm{trace}_{U^{\prime}}(\beta\circ \alpha)$. We use the analogous notation φ_B for finitely generated B-modules. We keep the above notation throughout this section.

Proposition 6.1. Let Uʹ, Uʹʹ be finitely generated A-modules. Then the bilinear map

\begin{equation*}\varphi_A : \mathrm{Hom}_A(U^{\prime},U^{\prime\prime}) \times \mathrm{Hom}_A(U^{\prime\prime},U^{\prime}) \to K\end{equation*}

is non-degenerate.

The following result is needed for the proof of Theorem 1.1.

Proposition 6.2. Let Uʹ, Uʹʹ be finite-dimensional A-modules. Let $\alpha : U^{\prime}\to U^{\prime\prime}$ be an A-homomorphism and let $\beta : M^\vee\otimes_A U^{\prime} \to M^\vee\otimes_A U^{\prime\prime}$ be a B-homomorphism. For any choice of symmetrising forms on A and B, we have

\begin{equation*}\varphi_A(\alpha, \mathrm{tr}_M(\beta)) = \varphi_B(\mathrm{Id}_{M^\vee}\otimes \alpha, \beta). \end{equation*}

Proof. Assume first that the symmetrising forms are $\mathrm{trace}_U$ and $\mathrm{trace}_V$. Note that Uʹ, Uʹʹ are isomorphic to finite direct sums of copies of U. Since φ_A, φ_B are additive in both components, we may assume that $U^{\prime}=U^{\prime\prime}=U$. Then α is a K-linear multiple of $\mathrm{Id}_U$, and since both sides are bilinear, we may assume that $\alpha=\mathrm{Id}_U$. Thus stated equation is equivalent to

\begin{equation*}z_A^{-1}\mathrm{trace}_U(\mathrm{tr}_M(\beta)) = z_B^{-1} \mathrm{trace}_{M^\vee\otimes_A U}(\beta).\end{equation*}

Note that $M^\vee\otimes_A U\cong V$, and hence $\beta=\lambda \mathrm{Id}_{M^\vee\otimes_A U}$ for some $\lambda \in K$. In particular, we have

\begin{equation*}\mathrm{trace}_{M^\vee\otimes_A U}(\beta) = \lambda \mathrm{dim}_K(V).\end{equation*}

By Equation (4.2), we have $\mathrm{tr}_M(\beta) = $ $\eta_M \circ (\mathrm{Id}_M\otimes \beta) \circ \epsilon_{M^\vee} =$ $\lambda \eta_M\circ \epsilon_{M^\vee}=$ $\lambda \mathrm{Id}_U$, and hence

\begin{equation*}\mathrm{trace}_U(\mathrm{tr}_M(\beta)) = \lambda \mathrm{dim}_K(U) .\end{equation*}

Using $z_A=\mathrm{dim}_K(U)\cdot 1_K$ and $z_B=\mathrm{dim}_K(V)\cdot 1_K$, the result follows for the chosen symmetrising forms $\mathrm{trace}_U$ and $\mathrm{trace}_V$. The result for arbitrary symmetrising forms follows easily from the Remark 5.6.

The next result, which is a variation of the previous Proposition, will be needed in the proof of Theorem 1.3. Set $A^e = A\otimes_K A^\mathrm{op}$. By Lemma 4.11, we have $z_{A^e} =$ $z_A\otimes z_A$. Thus, the inverse of this element acts on an A-A-bimodule by simultaneously multiplying by $z_A^{-1}$ on the left and on the right. In particular, this element acts on A by multiplication by $z_A^{-2}$.

Proposition 6.3. Let X be a finitely generated A-A-bimodule and let Y be a finitely generated B-B-bimodule. Let $\zeta : A\to X$ be an A-A-bimodule homomorphism, let $\xi : X\otimes_A M \to M\otimes_B Y$ be an A-B-bimodule homomorphism, and let $\sigma : Y \to B$ be a B-B-bimodule homomorphism. For any choice of symmetrising forms on A and on B, the trace on A of the map

\begin{equation*}z_A^{-2}\cdot \eta_M\circ (\mathrm{Id}_M\otimes\sigma\otimes\mathrm{Id}_{M^\vee}) \circ (\xi \otimes\mathrm{Id}_{M^\vee}) \circ (\mathrm{Id}_X \otimes \epsilon_{M^\vee}) \circ \zeta\end{equation*}

is equal to the trace on B of the map

\begin{equation*}z_B^{-2}\cdot \sigma \circ \eta_{M^\vee} \circ (\mathrm{Id}_{M^\vee}\otimes\xi) \circ (\mathrm{Id}_{M^\vee}\otimes\zeta \otimes\mathrm{Id}_M) \circ \epsilon_M.\end{equation*}

Proof. Both maps in the statement are additive in X. Since A is up to isomorphism the unique indecomposable A-A-bimodule, it follows that X is isomorphic to a finite direct sum of copies of A, and hence we may assume that X = A. For the same reason, we may assume that Y = B. Then, ξ becomes an A-B-bimodule endomorphism of the simple A-B-bimodule $M=U\otimes_K V^\vee$, hence is equal to multiplication by a scalar, which we will denote abusively again by ξ. Similarly, σ becomes a bimodule endomorphism of B, so is given by multiplication with a scalar, again denoted by σ, and ζ becomes a bimodule endomorphism of A, given by multiplication with a scalar, again denoted by ζ. Thus, the two maps in the statement take the form

(6.4)\begin{align} z_A^{-2}\sigma\xi\zeta \cdot (\eta_M \circ \epsilon_{M^\vee}),\qquad \qquad z_B^{-2}\sigma\xi\zeta \cdot (\eta_{M^\vee}\circ\epsilon_M). \end{align}

Let s, t be symmetrising forms of A, B. Then $s = \lambda \cdot\mathrm{trace}_U$ and $t = \mu \cdot\mathrm{trace}_V$ for some λ, $\mu\in K^\times$. It follows from the Remark 5.6 and 5.4 that then the adjunction units and counits with respect to these symmetrising forms satisfy

\begin{align*} \eta_M\circ \epsilon_{M^\vee} &= \lambda^{-1}\mu\mathrm{Id}_A, \\ \eta_{M^\vee}\circ\epsilon_M &= \lambda\mu^{-1}\mathrm{Id}_B. \end{align*}

Also by the Remark 5.6, the relative projective elements are $z_A = \lambda^{-1} \mathrm{dim}_K(U)$ and $z_B = \mu^{-1}\mathrm{dim}_K(V)$. Thus $z_A^{-2}=$ $\lambda^2\mathrm{dim}_K(U)^{-2}$ and $z_B^{-2}=$ $\mu^2\mathrm{dim}_K(V)^{-2}$. Therefore, the two maps in Equation (6.4) are equal to the two maps

\begin{align*} & \mathrm{dim}_K(U)^{-2}\lambda\mu\sigma\xi\zeta \cdot \mathrm{Id}_A,\\ &\mathrm{dim}_K(V)^{-2}\lambda\mu\sigma\xi\zeta\cdot \mathrm{Id}_B. \end{align*}

Since $\mathrm{dim}_K(U)^2=\mathrm{dim}_K(A)$ and $\mathrm{dim}_K(V)^2=\mathrm{dim}_K(B)$, it follows that both maps have the same trace, equal to $\lambda\mu\sigma\xi\zeta$.

7. Proof of Theorem 1.1

Let ${\mathcal{O}}$ be a complete discrete valuation ring with field of fractions K of characteristic zero. Let A, B be symmetric ${\mathcal{O}}$-algebras such that $K\otimes_{\mathcal{O}} A$ and $K\otimes_{\mathcal{O}} B$ are semisimple. Note that then $K\otimes_{\mathcal{O}} A$ and $K\otimes_{\mathcal{O}} B$ are separable since $\mathrm{char}(K)=0$. Fix symmetrising forms s, t of A, B, respectively. Let M be an A-B-bimodule which is finitely generated projective as a left A-module and as a right B-module. Let U, V be finitely generated ${\mathcal{O}}$-free A-modules. We write KA instead of $K\otimes_{\mathcal{O}} A$ and KU instead of $K\otimes_{\mathcal{O}} U$; similarly for B and V. We identify $\mathrm{Hom}_{KA}(KU, KV)=$ $K\mathrm{Hom}_A(U,V)$ whenever convenient, and we identify $\mathrm{Hom}_A(U,V)$ with its image in this space.

In degree zero, we have $\widehat{\mathrm{Ext}}^0_A(U,V)=$ $\underline{\mathrm{Hom}}_A(U,V)$, and Tate duality takes the following form. By [Reference Eisele, Geline, Kessar and Linckelmann9, Proposition 2.1], we have a non-degenerate bilinear form

(7.1)\begin{equation} \varphi_{KA}(-,-) : K\mathrm{Hom}_A(U,V) \times K\mathrm{Hom}_A(V,U) \to K \end{equation}

which sends $(\alpha, \beta)\in$ $K\mathrm{Hom}_A(U,V)\times K\mathrm{Hom}_A(V,U)$ to the trace on KU of the KA-endomorphism $z_A^{-1}\beta\circ\alpha$ of KU. This restricts to an ${\mathcal{O}}$-bilinear form

(7.2)\begin{equation} \varphi_A : \mathrm{Hom}_A(U,V) \times \mathrm{Hom}_A(V,U) \to K. \end{equation}

By [Reference Eisele, Geline, Kessar and Linckelmann9, Theorem 1.3] and its proof in [Reference Eisele, Geline, Kessar and Linckelmann9, § 2], this form sends $\mathrm{Hom}_A^\mathrm{pr}(U,V)\times\mathrm{Hom}_A(V,U)$ and $\mathrm{Hom}_A(U,V)\times \mathrm{Hom}^\mathrm{pr}_A(V,U)$ to ${\mathcal{O}}$, and the induced bilinear form

(7.3)\begin{equation} \langle - , - \rangle_A : \underline{\mathrm{Hom}}_A(U,V) \times \underline{\mathrm{Hom}}_A(V,U) \to K/{\mathcal{O}} \end{equation}

is non-degenerate. We note that for any $\alpha\in\mathrm{Hom}_A(U,V)$ and $\beta\in\mathrm{Hom}_A(V,U)$ we have

(7.4)\begin{equation} \begin{aligned} \varphi_{KA}(\alpha, \beta) & = \varphi_{KA}(\beta,\alpha),\\ \langle \alpha, \beta \rangle_A & = \langle \beta , \alpha \rangle_A. \end{aligned} \end{equation}

To see this, observe that left multiplication by $z_A^{-1}$ commutes with all KA-homomorphisms. Thus, the KA-endomorphism $z_A^{-1} (\beta \circ \alpha)$ of KU is equal to $(z_A^{-1}\beta) \circ \alpha$, hence has the same trace on KU as the endomorphism $\alpha\circ (z_A^{-1}\beta)$ on KV. The latter is equal to $z_A^{-1}(\alpha\circ \beta)$.

Proof of Theorem 1.1

We start by proving Theorem 1.1 in degree zero. Tate duality in degree zero takes the form as reviewed in Equations (7.1), (7.2) and (7.3). Since the functors $M\otimes_B-$ and $M^\vee\otimes_A-$ preserve finitely generated projective modules over A and B, it follows that if $\beta\in\mathrm{Hom}_A^\mathrm{pr}(M^\vee\otimes_A U, M^\vee\otimes_A V)$, then $\mathrm{tr}_M(\beta) \in \mathrm{Hom}_A^\mathrm{pr}(U,V)$. Similarly, if $\alpha\in\mathrm{Hom}_A^\mathrm{pr}(U,V)$, then $\mathrm{Id}_{M^\vee} \otimes \alpha\in$ $\mathrm{Hom}_B^\mathrm{pr}(M^\vee\otimes_A U, M^\vee\otimes_A V)$. That is, it suffices to show the equality

\begin{equation*}\varphi_{KA}(\alpha, \mathrm{tr}_M(\beta)) = \varphi_{KB}(\mathrm{Id}_{KM^\vee}\otimes\alpha, \beta)\end{equation*}

where $\alpha\in K\mathrm{Hom}_A(U,V)$ and $\beta \in$ $ K\mathrm{Hom}_B(M^\vee\otimes_A U, M^\vee\otimes_A V)$. This equation holds if and only if it holds for field extensions of K, so we may assume that KA, KB are split semisimple. That is, KA, KB are direct products of matrix algebras. Since both sides are additive, we may in fact assume that KA, KB are matrix algebras. In that case, the equation follows from Proposition 6.2. Together with Equation (7.4), this proves Theorem 1.1 for n = 0.

To prove Theorem 1.1 in an arbitrary degree n, we need to show that the above is compatible with the shift functors $\Sigma_A$ and $\Sigma_B$ on the relatively ${\mathcal{O}}$-stable categories $\underline{\mathrm{mod}}(A)$ and $\underline{\mathrm{mod}}(B)$. For simplicity, we denote both shift functors by Σ. We have $\widehat{\mathrm{Ext}}^n_A(U,V) = \underline{\mathrm{Hom}}_A(U,\Sigma^n(V))$, and $\widehat{\mathrm{Ext}}^{-n}_A(V,U)=\underline{\mathrm{Hom}}_A(V,\Sigma^{-n}(U))\cong \underline{\mathrm{Hom}}_A(\Sigma^n(V),U)$, where the second isomorphism is obtained from applying the functor $\Sigma^n$. The Tate duality

\begin{equation*}\langle - , - \rangle_A : \widehat{\mathrm{Ext}}_A^{n}(U,V)\times \widehat{\mathrm{Ext}}_A^{-n}(V,U) \to K/{\mathcal{O}}\end{equation*}

is induced by the map sending $(\alpha, \gamma)\in$ $\mathrm{Hom}_A(U,\Sigma^n(V))\times \mathrm{Hom}_A(V, \Sigma^{-n}(U))$ to the trace on KU of the endomorphism

\begin{equation*}z_A^{-1} \Sigma^n(\gamma) \circ \alpha;\end{equation*}

in other words, this is induced by the degree zero duality applied to U and $\Sigma^n(V)$, combined with the shift functor $\Sigma^n$. Applying the degree zero case to U and $\Sigma^n(V)$ yields the equation

\begin{equation*}\langle \alpha, \mathrm{tr}_M(\Sigma^n(\beta))\rangle_A = \langle \mathrm{Id}_{M^\vee}\otimes\alpha, \Sigma^n(\beta) \rangle_B\end{equation*}

in $K/{\mathcal{O}}$. It remains to show that the left side is equal to $\langle \alpha, \Sigma^n(\mathrm{tr}_M(\beta)) \rangle_A$. This expression depends only on the images of the morphisms α, β in their respective stable categories. By Equation (4.4), the images in the stable category $\underline{\mathrm{mod}}(A)$ of $\Sigma^n(\mathrm{tr}_M(\beta)$ and $\mathrm{tr}_M(\Sigma^n(\beta))$ are equal. Again using Equation (7.4), the result follows.

8. Proof of Theorem 1.3

As in the previous section, let ${\mathcal{O}}$ be a complete discrete valuation ring with field of fractions K of characteristic zero. Let A, B be symmetric ${\mathcal{O}}$-algebras such that $KA=K\otimes_{\mathcal{O}} A$ and $KB=K\otimes_{\mathcal{O}} B$ are semisimple. Fix symmetrising forms s, t of A, B, respectively. Let M be an A-B-bimodule which is finitely generated projective as a left A-module and as a right B-module. As before, we set $A^e=A\otimes_{\mathcal{O}} A^\mathrm{op}$ and $B^e=B\otimes_{\mathcal{O}} B^\mathrm{op}$. Let n be an integer. The Tate duality

\begin{equation*}\langle - , - \rangle_{A^e} : \widehat{H\!H}^n(A) \times \widehat{H\!H}^{-n}(A) \to K/{\mathcal{O}}\end{equation*}

is induced by a map

\begin{equation*}\varphi_{A^e} : \mathrm{Hom}_{A^e}(A,\Sigma^n(A)) \times \mathrm{Hom}_{A^e}(A,\Sigma^{-n}(A)) \to K\end{equation*}

sending $(\alpha,\beta)\in$ $\mathrm{Hom}_{A^e}(A,\Sigma^n(A)) \times \mathrm{Hom}_{A^e}(A,\Sigma^{-n}(A))$ to the trace on KA of the A^e-endomorphism

\begin{equation*}z_A^{-2} \cdot \Sigma^n(\beta)\circ\alpha\end{equation*}

where we use that the projective element $z_{A^e}=z_A\otimes z_A$ acts as multiplication by $z_A^2$ on the A^e-module A. Note that φ_A depends on the choices of $\Sigma^n(A)$ and $\Sigma^n(\beta)$, but the induced map to $K/{\mathcal{O}}$ does not. We have the analogous description for B instead of A.

As described in § 2, the functor $\Sigma^n$ on $\underline{\mathrm{mod}}(A^e)$ preserves the full subcategory $\underline{\mathrm{perf}}(A, A)$ of perfect A-A-bimodules. Slightly more generally, the functor $\Sigma^n$ on $\underline{\mathrm{mod}}(A\otimes_{\mathcal{O}} B^\mathrm{op})$ preserves $\underline{\mathrm{perf}}(A,B)$. We may choose $X=$ $\Sigma^n(A)$ to be in $\mathrm{perf}(A,A)$. Then, the functor Σ on $\underline{\mathrm{mod}}(A^e)$ restricted to $\underline{\mathrm{perf}}(A,A)$ is canonically isomorphic to the functor induced by any of the two exact functors $X\otimes_A -$ and $-\otimes_A X$ on $\mathrm{perf}(A, A)$, and the functor $\Sigma^n$ on $\underline{\mathrm{mod}}(A\otimes_{\mathcal{O}} B^\mathrm{op})$ is canonically isomorphic to the functor induced by the exact functor $X\otimes_A-$ on $\mathrm{perf}(A, B)$. Similarly, choosing $Y=\Sigma^n_{B^e}(B)$ in $\mathrm{perf}(B,B)$, the functor $\Sigma^n$ restricted to $\underline{\mathrm{perf}}(B,B)$ is canonically isomorphic to any of the two functors induced by the exact functors $Y\otimes_B-$ and $-\otimes_B Y$ on $\mathrm{perf}(B,B)$, and the functor $\Sigma^n$ on $\underline{\mathrm{mod}}(A\otimes_{\mathcal{O}} B^\mathrm{op})$ is canonically isomorphic to the functor induced by the exact functor $-\otimes_B Y$ on $\mathrm{perf}(A,B)$. With this notation, we will need the identification

(8.1)\begin{equation} X\otimes_A M= \Sigma^n_{A\otimes_{\mathcal{O}} B^\mathrm{op}}(M) = M\otimes_B Y, \end{equation}

in $\underline{\mathrm{perf}}(A,B)$. Denote by

(8.2)\begin{equation} \xi : X \otimes_A M \longrightarrow M \otimes_B Y \end{equation}

and A-B-bimodule homomorphism which induces the identification in Equation (8.1). Since ξ induces an isomorphism in $\underline{\mathrm{perf}}(A,B)$, the kernel and cokernel of ξ are projective $A\otimes_{\mathcal{O}} B^\mathrm{op}$-modules.

Proof of Theorem 1.3

Let $\zeta\in\widehat{H\!H}^n(A)$, and $\tau\in \widehat{H\!H}^{-n}(B)$. Represent these classes by bimodule homomorphisms, abusively denoted by the same letters,

\begin{equation*}\zeta : A \to \Sigma^n(A) = X,\ \ \ \ \ \tau : B \to \Sigma^{-n}(B).\end{equation*}

Then $\Sigma^n(\tau)$ is represented by a morphism, again denoted by the same letter,

\begin{equation*}\Sigma^n(\tau) : \Sigma^n(B) = Y \to B\end{equation*}

where we have used the identification $\Sigma^n(\Sigma^{-n}(B))=B$ in $\underline{\mathrm{perf}}(B,B)$. We need to show that the trace on KA of

\begin{equation*}z_A^{-2} \cdot (\Sigma^n(\mathrm{tr}_M(\tau))\circ \zeta)\end{equation*}

is equal to the trace on KB of

\begin{equation*}z_B^{-2} \cdot (\Sigma^n(\tau)\circ \mathrm{tr}_{M^\vee}(\zeta)).\end{equation*}

For simplicity, all identity homomorphisms on any of the bimodules M, $M^\vee$, X, Y are denoted $\mathrm{Id}$. The A^e-homomorphism $\mathrm{tr}_M(\tau)$ is equal to the composition

where we have identified $M\otimes_B \Sigma^n(B) \otimes_B M^\vee$ and $\Sigma^n(M\otimes_B M^\vee)$ along a bimodule homomorphism inducing the canonical isomorphism in the stable module category $\underline{\mathrm{mod}}(A^e)$. Thus, $\Sigma^n(\mathrm{tr}_M(\tau))\circ \zeta$ is the composition

where in the third term we use the identification $\Sigma^n(M\otimes_B M^\vee)=$ $M\otimes_B \Sigma^n(B) \otimes_B M^\vee$ in $\underline{\mathrm{mod}}(A^e)$, and in the fourth term we identify $M\otimes_B M^\vee=$ $M\otimes_B B\otimes_B M^\vee$. In terms of the bimodules X and Y, as well as replacing $\Sigma^n$ by $X\otimes_A-$ as appropriate, this shows that $\Sigma^n(\mathrm{tr}_M(\tau))\circ \zeta$ is represented by the composition of morphisms in $\mathrm{perf}(A)$

(8.3)

Similarly, the map $\Sigma^n(\tau)\circ \mathrm{tr}_{M^\vee}(\zeta)$ is represented by the composition

(8.4)

Since we need to compare traces of endomorphisms of KA, KB, we may extend coefficients to any field extension of K. Relative projective elements are compatible with these coefficient extensions, and hence we may assume that ${\mathcal{O}}=K$ is a splitting field of A and B. Thus, we may assume that A, B are finite direct products of matrix algebras over K. What we need to show is that the traces of the two maps in 8.3 and 8.4 multiplied by $z_A^{-1}$ and $z_B^{-2}$, respectively, are equal. Since traces are additive and all maps above (in particular, the adjunction maps, hence relative projective elements) are compatible with the block decompositions of A, B, we may assume that $A=$ $\mathrm{End}_K(U)$ and $B=\mathrm{End}_K(V)$ for some finite-dimensional K-vector spaces U, V. The adjunction maps are also additive in M, so we may assume that $M=U\otimes_K V^\vee$ (this is, up to isomorphism, the unique finite-dimensional indecomposable A-B-bimodule). The result follows from Proposition 6.3 with $\Sigma^n(\tau)$ instead of σ.

9. Remarks on transfer for Calabi–Yau triangulated categories

There are two ways to associate transfer maps to a triangle functor ${\mathcal{F}} : {\mathcal{C}}\to {\mathcal{D}}$ of Calabi–Yau categories: either by making use of a biadjoint functor ${\mathcal{G}}$ (if there is such a functor, as described in many sources such as [Reference Chouinard7], [Reference Linckelmann20]), or by making use of Serre duality. Theorem 1.1 and [Reference Linckelmann21, Theorem 1.1] state that for stable categories of symmetric algebras over fields or complete discrete valuation rings these two constructions coincide.

To be more precise, let ${\mathcal{C}}$, ${\mathcal{D}}$ be triangulated categories over a field k. We use the same letter Σ for the shift functors in ${\mathcal{C}}$ and ${\mathcal{D}}$. Suppose that homomorphism spaces between objects in either category are finite-dimensional and that ${\mathcal{C}}$, ${\mathcal{D}}$ admit Serre functors ${\mathbb{S}}$, ${\mathbb{T}}$, respectively. Let ${\mathcal{F}} : {\mathcal{C}}\to {\mathcal{D}}$ be a k-linear functor. Let X, Y be objects in ${\mathcal{C}}$. Dualising the map

\begin{equation*}\mathrm{Hom}_{\mathcal{C}}(Y,X) \to \mathrm{Hom}_{\mathcal{D}}({\mathcal{F}}(Y), {\mathcal{F}}(X))\end{equation*}

induced by ${\mathcal{F}}$, and making use of the defining property of a Serre functor, yields a map

\begin{equation*}\mathrm{tr}_{{\mathcal{C}}, {\mathcal{D}}} : \mathrm{Hom}_{\mathcal{D}}({\mathcal{F}}(X), {\mathbb{T}}({\mathcal{F}}(Y))) \to \mathrm{Hom}_{\mathcal{C}}(X, {\mathbb{S}}(Y)) .\end{equation*}

which makes the following diagram commutative.

(9.1)

where ${\mathcal{F}}^\vee$ is the dual of the map induces by ${\mathcal{F}}$ on morphisms, and where the vertical isomorphism are Serre duality isomorphisms.

Assume now that ${\mathcal{C}}$ and ${\mathcal{D}}$ are d-Calabi–Yau triangulated categories for some integer d. That is, the Serre functors ${\mathbb{S}}$, ${\mathbb{T}}$ are isomorphic to $\Sigma^d$ on ${\mathcal{C}}$, ${\mathcal{D}}$, respectively (see Kontsevich [Reference Kontsevich, Bellaiche, Dat, Marin and Racinet18] or also Keller [Reference Keller16] for background material and a long list of references on Calabi–Yau triangulated categories – what we call Calabi–Yau would be called weakly Calabi–Yau in many sources). Then, the previous map $\mathrm{tr}_{{\mathcal{C}},{\mathcal{D}}}$ takes the form

(9.2)\begin{align} &\mathrm{tr}_{{\mathcal{F}}^\vee} : \mathrm{Ext}^d_{\mathcal{D}}({\mathcal{F}}(X), {\mathcal{F}}(Y))=\mathrm{Hom}_{\mathcal{D}}({\mathcal{F}}(X), \Sigma^d({\mathcal{F}}(Y))) \to \mathrm{Hom}_{\mathcal{C}}(X, \Sigma^d(Y))\nonumber\\ &\qquad\quad= \mathrm{Ext}^d_{\mathcal{C}}(X,Y). \end{align}

If ${\mathcal{F}}$ is a functor of triangulated categories, then the functors $\Sigma\circ{\mathcal{F}}$ and ${\mathcal{F}}\circ\Sigma$ are isomorphic, so upon replacing Y by $\Sigma^{n-d}(Y)$, where n is an integer, we get in particular a map

(9.3)\begin{align} &\mathrm{tr}_{{\mathcal{F}}^\vee} : \mathrm{Ext}^n_{\mathcal{D}}({\mathcal{F}}(X), {\mathcal{F}}(Y))=\mathrm{Hom}_{\mathcal{D}}({\mathcal{F}}(X), \Sigma^n({\mathcal{F}}(Y))) \to \mathrm{Hom}_{\mathcal{C}}(X, \Sigma^n(Y))\nonumber\\ &\qquad\qquad= \mathrm{Ext}^n_{\mathcal{C}}(X,Y). \end{align}

For n = 0 this yields a map

(9.4)\begin{equation} \mathrm{tr}_{{\mathcal{F}}^\vee} : \mathrm{Hom}_{\mathcal{D}}({\mathcal{F}}(X), {\mathcal{F}}(Y)) \to \mathrm{Hom}_{\mathcal{C}}(X,Y). \end{equation}

Combining the above, the map $\mathrm{tr}_{{\mathcal{F}}^\vee}$ makes the following diagram commutative:

(9.5)

Here the vertical isomorphisms are given by a choice of Calabi–Yau duality together with a choice of an isomorphism $\Sigma\circ{\mathcal{F}}\cong$ ${\mathcal{F}}\circ\Sigma$, and the bottom horizontal is the dual of the map induced by ${\mathcal{F}}$. (Note the dependence on choices.)

If ${\mathcal{F}}$ has a biadjoint functor of triangulated categories ${\mathcal{G}} : {\mathcal{D}}\to {\mathcal{C}}$, then (following e.g. [Reference Chouinard7] or [Reference Linckelmann20, § 4]) we also have a transfer map

(9.6)\begin{equation} \mathrm{tr}_{\mathcal{G}} : \mathrm{Hom}_{\mathcal{D}}({\mathcal{F}}(X),{\mathcal{F}}(Y)) \longrightarrow \mathrm{Hom}_{\mathcal{C}}(X,Y) \end{equation}

sending a morphism $\psi : {\mathcal{F}}(X)\to {\mathcal{F}}(Y)$ to the composition of morphisms

where the first map is the adjunction unit of ${\mathcal{F}}$ being left adjoint to ${\mathcal{G}}$, and the last map is the adjunction counit of ${\mathcal{F}}$ being right adjoint to ${\mathcal{G}}$. The map $\mathrm{tr}_{\mathcal{G}}$ depends on the choice of adjunction isomorphisms.

One can rephrase [Reference Linckelmann21, Theorem 1.2] as stating that $\mathrm{tr}_{{\mathcal{F}}^\vee}=\mathrm{tr}_{\mathcal{G}}$. The proof amounts to showing that the choices made for adjunction isomorphisms and Tate duality determined by the choices of symmetrising forms are compatible. Indeed, if A is a symmetric k-algebra, then Tate duality turns $\underline{\mathrm{mod}}(A)$ into a $(-1)$-Calabi–Yau triangulated category. Given two symmetric k-algebras A, B and an A-B-bimodule M in $\mathrm{perf}(A,B)$, Theorem [Reference Linckelmann21, Theorem 1.2] can be rephrased as stating that the transfer maps $\mathrm{tr}_M(U,V)$ are special cases of the construction of $\mathrm{tr}_{{\mathcal{F}}^\vee}$ in the above diagram 9.5; in other words, the maps in Equations (9.4) and (9.6) coincide.

Let now A be a symmetric algebra over a complete discrete valuation ring ${\mathcal{O}}$ with a field of fractions K of characteristic 0 such that $K\otimes_{\mathcal{O}} A$ is semisimple. Extend the notion of Calabi–Yau triangulated categories to ${\mathcal{O}}$-linear triangulated categories by using duality with respect to the Matlis module $K/{\mathcal{O}}$. Note that $K/{\mathcal{O}}$ is the degree 1-term of the injective resolution $K\to K/{\mathcal{O}}$ of ${\mathcal{O}}$. Note further that there are no non-zero ${\mathcal{O}}$-linear maps from the torsion ${\mathcal{O}}$-modules $\underline{\mathrm{Hom}}_A(U,V)$ to the torsion free ${\mathcal{O}}$-module K (where U, V are A-lattices), and hence Matlis duality coincides with $\mathrm{RHom}(-, {\mathcal{O}})$ on the morphism spaces in the stable module category, except for a degree shift.

Regardless of dimension considerations, Theorem 1.1 shows that the transfer maps $\mathrm{tr}_M(U,V)$ in Theorem 1.1 are special cases of the construction given by the diagram 9.5 with k-duality replaced by $K/{\mathcal{O}}$-duality. As pointed out earlier, there are choices to be made: Tate duality depends on the choices of symmetrising forms, and showing that the maps from Equations (9.4) and (9.6) are equal in Theorem 1.1 boils down to being able to make compatible choices.

10. On products in negative degrees of Tate cohomology

Non-zero products in negative degree in Tate cohomology have implications for the depth of the non-negative part. This phenomenon was first observed in [Reference Benson and Carlson2] in finite group cohomology, and then generalised in [Reference Bergh, Jorgensen and Oppermann3], [Reference Linckelmann21, § 8]. This arises over complete discrete valuation rings as well, with essentially the same arguments used in [Reference Benson and Carlson2].

Let A be a symmetric algebra over a complete discrete valuation ring ${\mathcal{O}}$ with a field of fractions K of characteristic zero. Let U, V, W be finitely generated A-modules. Let m, n be integers. Let $\alpha\in\widehat{\mathrm{Ext}}^m_A(U,V)$, $\beta\in\widehat{\mathrm{Ext}}^n_A(V,W)$, and $\gamma\in$ $\widehat{\mathrm{Ext}}^{-m-n}_A(W,U)$. We denote by $\beta\alpha\in$ $\widehat{\mathrm{Ext}}^{m+n}_A(U,W)$ the Yoneda product; that is, βα is represented by $\Sigma^m(\beta)\circ\alpha$, where we use the same letters α, β for representatives of their classes in $\mathrm{Hom}_A(U,\Sigma^m(V))$, $\mathrm{Hom}_A(V,\Sigma^n(W))$. We have

(10.1)\begin{equation} \langle \beta\alpha, \gamma\rangle_A = \langle \alpha, \gamma\beta\rangle_A \end{equation}

because both sides are equal to the image in $K/{\mathcal{O}}$ of the trace on KU of the endomorphism $z_A^{-1}\cdot(\Sigma^{m+n}(\gamma)\circ \Sigma^m(\beta)\circ\alpha)$ of KU, whereas before we use abusively the same letters α, β, γ for representatives in $\mathrm{Hom}_A(U,\Sigma^m(V))$, $\mathrm{Hom}_A(V,\Sigma^n(W))$, $\mathrm{Hom}_A(W, \Sigma^{-m-n}(U))$ of their classes. Applied with U = W and $\gamma=\mathrm{Id}_U$ this yields

(10.2)\begin{equation} \langle \beta\alpha, \mathrm{Id}_U\rangle_A = \langle \alpha, \beta\rangle_A. \end{equation}