2 Factorisable finite tensor categories

In this section we introduce notation and recall some definitions and results that will be used later, such as the different characterisations of a factorisable finite tensor category. The reader is invited to just skim through this section and return to it when necessary.

For ${\mathcal{A}}$ an essentially small abelian category we write (see e.g. [Reference Etingof, Gelaki, Nikshych and OstrikEGNO15] for more details):

• ∙ $\operatorname{Irr}({\mathcal{A}})$: a choice of representatives for the isomorphism classes of simple objects of ${\mathcal{A}}$, i.e. every simple object is isomorphic to one and only one element of $\operatorname{Irr}({\mathcal{A}})$;

• ∙ ${\mathcal{P}}\text{roj}({\mathcal{A}})$: the full subcategory of ${\mathcal{A}}$ consisting of all projective objects;

• ∙ $\operatorname{Gr}({\mathcal{A}})$: the Grothendieck group of ${\mathcal{A}}$. We write $[X]\in \operatorname{Gr}({\mathcal{A}})$ for the class of $X\in {\mathcal{A}}$ in the Grothendieck group;

• ∙ $\operatorname{Gr}_{k}({\mathcal{A}})$: if ${\mathcal{A}}$ is $k$-linear we set $\operatorname{Gr}_{k}({\mathcal{A}}):=k\otimes _{\mathbb{Z}}\operatorname{Gr}({\mathcal{A}})$. Note that the canonical ring homomorphism $\operatorname{Gr}({\mathcal{A}})\rightarrow \operatorname{Gr}_{k}({\mathcal{A}})$ is injective if and only if $k$ has characteristic zero.

A $k$-linear abelian category ${\mathcal{A}}$ is called locally finite if all Hom-spaces are finite-dimensional over $k$ and every object of ${\mathcal{A}}$ has finite length. A locally finite abelian category is called finite if $\operatorname{Irr}({\mathcal{A}})$ is finite and every simple object has a projective cover. For a finite abelian category ${\mathcal{A}}$ we set

• ∙ $P_{U}$: a choice of projective cover $\unicode[STIX]{x1D70B}_{U}:P_{U}\rightarrow U$ for each simple $U\in \operatorname{Irr}({\mathcal{A}})$.

One can show that a $k$-linear abelian category is finite if and only if it is equivalent as a $k$-linear category to the category of finite-dimensional modules over a finite-dimensional $k$-algebra.

For a finite abelian category ${\mathcal{A}}$ we have $\operatorname{Gr}({\mathcal{A}})=\bigoplus _{U\in \operatorname{Irr}({\mathcal{A}})}\mathbb{Z}\,[U]$. (On the other hand, if infinite direct sums are allowed in ${\mathcal{A}}$ we have $\operatorname{Gr}({\mathcal{A}})=0$ because of the short exact sequence $X\rightarrow \bigoplus _{\mathbb{N}}X\rightarrow \bigoplus _{\mathbb{N}}X$.)

A monoidal category ${\mathcal{M}}$ is called rigid if for each $X\in {\mathcal{M}}$ there is $X^{\ast }\in {\mathcal{M}}$ (the left dual) and $\text{}^{\ast }\!X$ (the right dual), together with morphisms

(2.1)$$\begin{eqnarray}\displaystyle \operatorname{ev}_{X}:X^{\ast }\otimes X\rightarrow \mathbf{1}, & \quad & \displaystyle \operatorname{coev}_{X}:\mathbf{1}\rightarrow X\otimes X^{\ast },\nonumber\\ \displaystyle \widetilde{\text{ev}}_{\!X}:X\otimes \text{}^{\ast }\!X\rightarrow \mathbf{1}, & \quad & \displaystyle \widetilde{\operatorname{coev}}_{X}:\mathbf{1}\rightarrow \text{}^{\ast }\!X\otimes X,\end{eqnarray}$$

where $\mathbf{1}\in {\mathcal{M}}$ is the tensor unit. These morphisms have to satisfy the zig-zag identities, see e.g. [Reference Etingof, Gelaki, Nikshych and OstrikEGNO15] for more details. We will later use string diagram notation. Our diagrams will be read bottom to top and the above duality morphisms will be written as

(2.2)

A monoidal category ${\mathcal{M}}$ with left duals is pivotal if it is equipped with a natural monoidal isomorphism

(2.3)$$\begin{eqnarray}\unicode[STIX]{x1D6FF}:\text{Id}_{{\mathcal{M}}}\rightarrow (-)^{\ast \ast }.\end{eqnarray}$$

It is then automatically rigid via $\text{}^{\ast }\!X:=X^{\ast }$ and (we omit the tensor product between objects for better readability)

(2.4)$$\begin{eqnarray}\displaystyle \widetilde{\operatorname{ev}}_{X} & = & \displaystyle [XX^{\ast }\xrightarrow[{}]{\unicode[STIX]{x1D6FF}_{X}\,\otimes \,\text{id}}X^{\ast \ast }X^{\ast }\xrightarrow[{}]{\operatorname{ev}_{X^{\ast }}}\mathbf{1}],\nonumber\\ \displaystyle \widetilde{\operatorname{coev}}_{X} & = & \displaystyle [\mathbf{1}\xrightarrow[{}]{\operatorname{coev}_{X^{\ast }}}X^{\ast }X^{\ast \ast }\xrightarrow[{}]{\text{id}\,\otimes \,\unicode[STIX]{x1D6FF}_{X}^{-1}}X^{\ast }X].\end{eqnarray}$$

Let ${\mathcal{M}}$ be a monoidal category with right duals. Below we will make use of the functor $\text{}^{\ast }(-)\otimes (-):{\mathcal{M}}^{\text{op}}\times {\mathcal{M}}\rightarrow {\mathcal{M}}$, $(X,Y)\mapsto \text{}^{\ast }\!X\otimes Y$, and of dinatural transformations between this functor and the constant functor with value $X$ for some $X\in {\mathcal{M}}$. We refer to [Reference Mac LaneMac98] for dinatural transformations in general and to [Reference Fuchs and SchweigertFS10, §4] or [Reference Farsad, Gainutdinov and RunkelFGR19, §3] for these specific functors.

The following lemma is a slight generalisation of [Reference Geer, Kujawa and Patureau-MirandGKP11, Lemma 2.5.1] which in turn follows [Reference DeligneDel07]. It is proven in the same way as in [Reference Geer, Kujawa and Patureau-MirandGKP11] and we relegate the proof to Appendix A.

Lemma 2.1. Let ${\mathcal{M}}$ be an abelian monoidal category with right duals and biexact tensor product functor. Suppose that there is a projective object $P\in {\mathcal{M}}$ and a surjection $p:P\rightarrow \mathbf{1}$. Let $X\in {\mathcal{M}}$ and let $\unicode[STIX]{x1D702}$ be a dinatural transformation from $\text{}^{\ast }(-)\otimes (-)$ to $X$. Consider the following commuting diagram in ${\mathcal{M}}$ with exact rows.

(2.5)

In this case, the following equality of morphisms $\mathbf{1}\rightarrow X$ holds for any choice of endomorphisms $a$, $b$, and $c$ that make (

2.5

) commute:

(2.6)$$\begin{eqnarray}\unicode[STIX]{x1D702}_{B}\circ (\operatorname{id}_{\text{}^{\ast }\!B}\,\otimes \,b)\circ \widetilde{\operatorname{coev}}_{B}=\unicode[STIX]{x1D702}_{A}\circ (\operatorname{id}_{\text{}^{\ast }\!A}\,\otimes \,a)\circ \widetilde{\operatorname{coev}}_{A}+\unicode[STIX]{x1D702}_{C}\circ (\operatorname{id}_{\text{}^{\ast }\!C}\,\otimes \,c)\circ \widetilde{\operatorname{coev}}_{C}.\end{eqnarray}$$

Remark 2.2. (1) A corresponding statement as that in Lemma 2.1 holds for a dinatural transformation $\unicode[STIX]{x1D709}$ from $(-)\otimes (-)^{\ast }$ to $X$. In this case, for $a,b,c$ as in (2.5) we have

(2.7)$$\begin{eqnarray}\unicode[STIX]{x1D709}_{B}\circ (b\otimes \operatorname{id}_{B^{\ast }})\circ \operatorname{coev}_{B}=\unicode[STIX]{x1D709}_{A}\circ (a\otimes \operatorname{id}_{A^{\ast }})\circ \operatorname{coev}_{A}\,+\,\unicode[STIX]{x1D709}_{C}\circ (c\otimes \operatorname{id}_{C^{\ast }})\circ \operatorname{coev}_{C}.\end{eqnarray}$$

(2) As a consequence of Lemma 2.1, the morphism $\unicode[STIX]{x1D702}_{A}\circ \widetilde{\operatorname{coev}}_{A}\in {\mathcal{M}}(\mathbf{1},X)$ only depends on the class $[A]\in \operatorname{Gr}({\mathcal{M}})$. Indeed, if one chooses $a,b,c$ in (2.5) to be identities, then (2.6) states that given a short exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ we have the identity

(2.8)$$\begin{eqnarray}\unicode[STIX]{x1D702}_{B}\circ \widetilde{\operatorname{coev}}_{B}=\unicode[STIX]{x1D702}_{A}\circ \widetilde{\operatorname{coev}}_{A}+\unicode[STIX]{x1D702}_{C}\circ \widetilde{\operatorname{coev}}_{C}\end{eqnarray}$$

of morphisms in ${\mathcal{M}}(\mathbf{1},X)$. For example, if ${\mathcal{M}}$ is in addition pivotal, one can take $X=\mathbf{1}$ and the dinatural transformation $\unicode[STIX]{x1D702}_{A}=\operatorname{ev}_{A}$. One obtains the statement that the quantum dimension only depends on the class in $\operatorname{Gr}({\mathcal{M}})$.

A finite tensor category ${\mathcal{C}}$ is a category which (see [Reference Etingof and OstrikEO04]):

• ∙ is a $k$-linear finite abelian category;

• ∙ is a rigid monoidal category with $k$-bilinear tensor product functor;

• ∙ has a simple tensor unit.

The tensor product functor of an abelian rigid monoidal category is automatically biadditive and biexact [Reference Etingof, Gelaki, Nikshych and OstrikEGNO15, Proposition 4.2.1]. A finite tensor category is called unimodular if $(P_{\mathbf{1}})^{\ast }\cong P_{\mathbf{1}}$, see [Reference Etingof, Nikshych and OstrikENO04, Reference Etingof, Gelaki, Nikshych and OstrikEGNO15] for more details.

The following technical corollary to Lemma 2.1 generalises [Reference Geer, Kujawa and Patureau-MirandGKP11, Corollary 2.5.2] and will be needed later. The proof is given in Appendix A.

For the rest of this section, we fix

• ∙ ${\mathcal{C}}$: a braided and pivotal finite tensor category over $k$.

Our notation for the coherence and braiding isomorphisms in ${\mathcal{C}}$ is, for $U,V,W\in {\mathcal{C}}$,

(2.10)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FC}_{U,V,W} & : & \displaystyle U\otimes (V\otimes W)\longrightarrow (U\otimes V)\otimes W\hphantom{\text{(unit isoi,}}\qquad \text{(associator)},\nonumber\\ \displaystyle \unicode[STIX]{x1D706}_{U} & : & \displaystyle \mathbf{1}\otimes U\longrightarrow U,\quad \unicode[STIX]{x1D70C}_{U}:U\otimes \mathbf{1}\longrightarrow U\qquad \text{(unit isomorphisms)},\nonumber\\ \displaystyle c_{U,V} & : & \displaystyle U\otimes V\longrightarrow V\otimes U\qquad \hphantom{\text{(unit isomorphisms)};}\qquad \text{(braiding)}.\end{eqnarray}$$

We write $\operatorname{End}(\text{Id}_{{\mathcal{C}}})$ for the natural endomorphisms of the identity functor on ${\mathcal{C}}$.

Given $V\in {\mathcal{C}}$, consider the natural transformation $\unicode[STIX]{x1D70E}(V)\in \operatorname{End}(\text{Id}_{{\mathcal{C}}})$: for all $X\in {\mathcal{C}}$,

(2.11)

where in $(\ast )$ we used properties of the braiding to deform the string diagram, as well as pivotality to exchange left and right duality morphisms. This natural transformation appears in [Reference TuraevTur94, § I.1.5] and also in [Reference Costantino, Geer and Patureau-MirandCGP15, Reference Creutzig and GannonCG17] where it is called ‘general Hopf link’ and ‘open Hopf link operator’, respectively (see [Reference Gainutdinov and RunkelGR19, Remark 3.10(2)] for the conventions used here).

Proof. Let $X\in {\mathcal{C}}$ be arbitrary and consider the dinatural transformation from $(-)^{\ast }\otimes (-)$ to $X\otimes X^{\ast }$ given by (see also [Reference Farsad, Gainutdinov and RunkelFGR19, §4.4])

(2.12)

Note that $(\unicode[STIX]{x1D70E}(V)_{X}\otimes \operatorname{id}_{X^{\ast }})\circ \operatorname{coev}_{X}=\unicode[STIX]{x1D702}_{V}\circ \widetilde{\operatorname{coev}}_{V}$. By Lemma 2.1 (or rather Remark 2.2(2)), the right-hand side factors through $\operatorname{Gr}({\mathcal{C}})$, hence so does $\unicode[STIX]{x1D70E}$. That the resulting map is a ring map follows from the identity $\unicode[STIX]{x1D70E}(V\otimes W)_{X}=\unicode[STIX]{x1D70E}(V)_{X}\circ \unicode[STIX]{x1D70E}(W)_{X}$, which one can verify by a straightforward calculation [Reference TuraevTur94, §4.5] (see also [Reference Creutzig and GannonCG17, §3.1]).◻

By abuse of notation, we also use the symbol $\unicode[STIX]{x1D70E}$ for the resulting ring homomorphism

(2.13)

Let ${\mathcal{L}}$ be the coend (see e.g. [Reference Kerler and LyubashenkoKL01] or the review in [Reference Fuchs and SchweigertFS10, §4] or in [Reference Farsad, Gainutdinov and RunkelFGR19, §3])

(2.14)$$\begin{eqnarray}{\mathcal{L}}=\int ^{X\in {\mathcal{C}}}X^{\ast }\otimes X\end{eqnarray}$$

and denote by $\unicode[STIX]{x1D704}_{X}:X^{\ast }\otimes X\rightarrow {\mathcal{L}}$, $X\in {\mathcal{C}}$, the corresponding dinatural transformation. The coend ${\mathcal{L}}$ exists since ${\mathcal{C}}$ is a finite tensor category (see e.g. [Reference Kerler and LyubashenkoKL01, Corollary 5.1.8]). It carries the structure of a Hopf algebra in ${\mathcal{C}}$ and is equipped with a Hopf pairing $\unicode[STIX]{x1D714}:{\mathcal{L}}\otimes {\mathcal{L}}\rightarrow \mathbf{1}$ [Reference LyubashenkoLyu95a]. We denote the product, coproduct, unit, counit and antipode of the Hopf algebra ${\mathcal{L}}$ by $\unicode[STIX]{x1D707}_{{\mathcal{L}}}$, $\unicode[STIX]{x1D6E5}_{{\mathcal{L}}}$, $\unicode[STIX]{x1D702}_{{\mathcal{L}}}$, $\unicode[STIX]{x1D700}_{{\mathcal{L}}}$, $S_{{\mathcal{L}}}$, respectively (we use the conventions in [Reference Farsad, Gainutdinov and RunkelFGR19, §3]).

If ${\mathcal{C}}$ is unimodular, the coend ${\mathcal{L}}$ admits a two-sided integral $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}:\mathbf{1}\rightarrow {\mathcal{L}}$ [Reference ShimizuShi17a, Theorem 6.8].Footnote ² The integral is unique up to a scalar factor. Dually, ${\mathcal{L}}$ admits a two-sided cointegral $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}:{\mathcal{L}}\rightarrow \mathbf{1}$ which can be normalised such that $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D6EC}_{{\mathcal{L}}}=\operatorname{id}_{\mathbf{1}}$ (see e.g. [Reference Kerler and LyubashenkoKL01, §4.2.3]).

We will need the linear maps

(2.15)

which are defined as follows.

• ∙ For $\unicode[STIX]{x1D6FC}\in \operatorname{End}(\text{Id}_{{\mathcal{C}}})$, the value $\unicode[STIX]{x1D713}(\unicode[STIX]{x1D6FC})$ is defined uniquely by the universal property of the coend $({\mathcal{L}},\unicode[STIX]{x1D704})$ via, for all $X\in {\mathcal{C}}$,

  (2.16)$$\begin{eqnarray}\unicode[STIX]{x1D713}(\unicode[STIX]{x1D6FC})\circ \unicode[STIX]{x1D704}_{X}=\operatorname{ev}_{X}\circ (\text{id}\,\otimes \,\unicode[STIX]{x1D6FC}_{X}).\end{eqnarray}$$

• ∙ The linear map $\unicode[STIX]{x1D6FA}$ is defined via the Hopf pairing of ${\mathcal{L}}$ as, for $f:\mathbf{1}\rightarrow {\mathcal{L}}$,

  (2.17)$$\begin{eqnarray}\unicode[STIX]{x1D6FA}(f)=[{\mathcal{L}}\xrightarrow[{}]{{\sim}}\mathbf{1}{\mathcal{L}}\xrightarrow[{}]{f\,\otimes \,\text{id}}{\mathcal{L}}{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D714}}\mathbf{1}].\end{eqnarray}$$

• ∙ The definition of $\unicode[STIX]{x1D70C}$ requires ${\mathcal{C}}$ to be unimodular and the choice of a non-zero integral $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}:1\rightarrow {\mathcal{L}}$. In this case we set, for $g:{\mathcal{L}}\rightarrow \mathbf{1}$,

  (2.18)$$\begin{eqnarray}\unicode[STIX]{x1D70C}(g)=[\mathbf{1}\xrightarrow[{}]{\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}}{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D6E5}_{{\mathcal{L}}}}{\mathcal{L}}{\mathcal{L}}\xrightarrow[{}]{g\,\otimes \,\text{id}}\mathbf{1}{\mathcal{L}}\xrightarrow[{}]{{\sim}}{\mathcal{L}}].\end{eqnarray}$$

The bialgebra structure on ${\mathcal{L}}$ allows one to endow ${\mathcal{C}}(\mathbf{1},{\mathcal{L}})$ and ${\mathcal{C}}({\mathcal{L}},\mathbf{1})$ with the structure of a $k$-algebra. $\operatorname{End}(\text{Id}_{{\mathcal{C}}})$ is equally a $k$-algebra. We have the following.

For the proof of this lemma, see e.g. [Reference LyubashenkoLyu95a], [Reference Kerler and LyubashenkoKL01, Proposition 5.2.5], [Reference Kerler and LyubashenkoKL01, Corollary 4.2.13] and [Reference ShimizuShi19, §3.1] (and also [Reference Gainutdinov and RunkelGR19, §2]). One verifies that the inverses of $\unicode[STIX]{x1D713}$ and $\unicode[STIX]{x1D70C}$ are, for $X\in {\mathcal{C}}$, $f:{\mathcal{L}}\rightarrow \mathbf{1}$, $g:\mathbf{1}\rightarrow {\mathcal{L}}$,

(2.19)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D713}^{-1}(f)_{X} & = & \displaystyle [X\xrightarrow[{}]{{\sim}}\mathbf{1}X\xrightarrow[{}]{\text{coev}_{X}\,\otimes \,\text{id}}(XX^{\ast })X\xrightarrow[{}]{{\sim}}X(X^{\ast }X)\xrightarrow[{}]{\text{id}\,\otimes \,\unicode[STIX]{x1D704}_{X}}X{\mathcal{L}}\xrightarrow[{}]{\text{id}\,\otimes \,f}X\mathbf{1}\xrightarrow[{}]{{\sim}}X],\nonumber\\ \displaystyle \unicode[STIX]{x1D70C}^{-1}(g) & = & \displaystyle [{\mathcal{L}}\xrightarrow[{}]{{\sim}}\mathbf{1}{\mathcal{L}}\xrightarrow[{}]{g\,\otimes \,\text{id}}{\mathcal{L}}{\mathcal{L}}\xrightarrow[{}]{S_{{\mathcal{L}}}\,\otimes \,\text{id}}{\mathcal{L}}{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D707}_{{\mathcal{L}}}}{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}}\mathbf{1}].\end{eqnarray}$$

The internal character of $V\in {\mathcal{C}}$ is the element $\unicode[STIX]{x1D712}_{V}\in {\mathcal{C}}(\mathbf{1},{\mathcal{L}})$ given by [Reference Fuchs and SchweigertFS10, Reference ShimizuShi17b] (we use the convention in [Reference Gainutdinov and RunkelGR19])

(2.20)$$\begin{eqnarray}\unicode[STIX]{x1D712}_{V}=[\mathbf{1}\xrightarrow[{}]{\widetilde{\operatorname{coev}}_{V}}V^{\ast }\otimes V\xrightarrow[{}]{\unicode[STIX]{x1D704}_{V}}{\mathcal{L}}].\end{eqnarray}$$

By Lemma 2.1, this map factors through the Grothendieck ring, i.e. we obtain a map

(2.21)$$\begin{eqnarray}\unicode[STIX]{x1D712}:\operatorname{Gr}({\mathcal{C}})\rightarrow {\mathcal{C}}(\mathbf{1},{\mathcal{L}}),\quad [V]\mapsto \unicode[STIX]{x1D712}_{V}.\end{eqnarray}$$

By abuse of notation, we will denote the induced map $\operatorname{Gr}_{k}({\mathcal{C}})\rightarrow {\mathcal{C}}(\mathbf{1},{\mathcal{L}})$, $1\otimes _{\mathbb{Z}}[V]\mapsto \unicode[STIX]{x1D712}_{V}$ by $\unicode[STIX]{x1D712}$ as well.

This theorem remains true for ${\mathcal{C}}$ not braided (but still a pivotal finite tensor category over an algebraically closed field).

Define, for ${\mathcal{C}}$ unimodular and $M\in {\mathcal{C}}$,

(2.22)$$\begin{eqnarray}\unicode[STIX]{x1D719}_{M}:=\unicode[STIX]{x1D713}^{-1}(\unicode[STIX]{x1D70C}^{-1}(\unicode[STIX]{x1D712}_{M}))\in \operatorname{End}(\text{Id}_{{\mathcal{C}}}).\end{eqnarray}$$

After substituting the definitions, one can check that, for all $X\in {\mathcal{C}}$,

(2.23)

As $\unicode[STIX]{x1D712}_{M}$ only depends on $[M]\in \operatorname{Gr}({\mathcal{C}})$, so does $\unicode[STIX]{x1D719}_{M}$. Since $\unicode[STIX]{x1D713}$ and $\unicode[STIX]{x1D70C}$ are isomorphisms, by Theorem 2.6 the set $\{\unicode[STIX]{x1D719}_{U}\mid U\in \operatorname{Irr}({\mathcal{C}})\}$ is linearly independent in $\operatorname{End}(\text{Id}_{{\mathcal{C}}})$. We note that since $\unicode[STIX]{x1D70C}$ is not necessarily an algebra map, neither is the linear map $\operatorname{Gr}_{k}({\mathcal{C}})\rightarrow \operatorname{End}(\text{Id}_{{\mathcal{C}}})$, $[M]\mapsto \unicode[STIX]{x1D719}_{M}$.

We define the linear map $\mathscr{S}_{{\mathcal{C}}}:\operatorname{End}(\text{Id}_{{\mathcal{C}}})\rightarrow \operatorname{End}(\text{Id}_{{\mathcal{C}}})$, the modular $S$-transformation, as

(2.24)$$\begin{eqnarray}\mathscr{S}_{{\mathcal{C}}}=[\operatorname{End}(\text{Id}_{{\mathcal{C}}})\xrightarrow[{}]{\,\unicode[STIX]{x1D713}\,}{\mathcal{C}}({\mathcal{L}},\mathbf{1})\xrightarrow[{}]{~\unicode[STIX]{x1D70C}~}{\mathcal{C}}(\mathbf{1},{\mathcal{L}})\xrightarrow[{}]{~\unicode[STIX]{x1D6FA}~}{\mathcal{C}}({\mathcal{L}},\mathbf{1})\xrightarrow[{}]{~\unicode[STIX]{x1D713}^{-1}~}\operatorname{End}(\text{Id}_{{\mathcal{C}}})].\end{eqnarray}$$

We have seen that $\unicode[STIX]{x1D713}$ and $\unicode[STIX]{x1D6FA}$ are algebra maps, while $\unicode[STIX]{x1D70C}$ is in general not. Thus in general $\mathscr{S}_{{\mathcal{C}}}$ is not an algebra map, either. Recall the definition of $\unicode[STIX]{x1D70E}$ in (2.13). A straightforward calculation shows that $\unicode[STIX]{x1D70E}([M])=\unicode[STIX]{x1D713}^{-1}\circ \unicode[STIX]{x1D6FA}(\unicode[STIX]{x1D712}_{M})$. Combining this with (2.22) and (2.24) gives, for all $M\in {\mathcal{C}}$,

(2.25)$$\begin{eqnarray}\unicode[STIX]{x1D70E}([M])=\mathscr{S}_{{\mathcal{C}}}(\unicode[STIX]{x1D719}_{M}),\end{eqnarray}$$

cf. [Reference Gainutdinov and RunkelGR19, Remark 3.10(2)]. In particular, by Lemma 2.4, the combination $[M]\mapsto \mathscr{S}_{{\mathcal{C}}}(\unicode[STIX]{x1D719}_{M})$ is an algebra homomorphism from $\operatorname{Gr}_{k}({\mathcal{C}})$ to $\operatorname{End}(\text{Id}_{{\mathcal{C}}})$.

Next we introduce four equivalent non-degeneracy requirements for the braiding. (This does not require a pivotal structure, hence we use the letter ${\mathcal{D}}$ and reserve ${\mathcal{C}}$ for the pivotal case as declared above.)

In [Reference Etingof, Nikshych and OstrikENO04, Proposition 4.5] the following result is proved (using formulation 2 of factorisability in Theorem 2.8).

In particular, for factorisable ${\mathcal{C}}$ the Hopf algebra given by the coend ${\mathcal{L}}$ admits integrals and cointegrals. Moreover, in this case we can normalise the integral such that

(2.26)$$\begin{eqnarray}\unicode[STIX]{x1D714}\circ (\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}\otimes \unicode[STIX]{x1D6EC}_{{\mathcal{L}}})=\operatorname{id}_{\mathbf{1}},\end{eqnarray}$$

see e.g. [Reference Kerler and LyubashenkoKL01, §5.2.3]. Since the space of integrals is one-dimensional, this determines $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}$ up to a sign.

Using Lemma 2.4 and (2.22), (2.25), we get the following corollary to Theorem 2.6.

The following important theorem is proved in [Reference LyubashenkoLyu95a], see also [Reference Farsad, Gainutdinov and RunkelFGR19, §5.1] for a summary in the notation used here.

Theorem 2.13. Let ${\mathcal{C}}$ be factorisable and ribbon with ribbon twist $\unicode[STIX]{x1D703}\in \operatorname{End}(\text{Id}_{{\mathcal{C}}})$. Then there is a projective representation of $\text{SL}(2,\mathbb{Z})$ on $\operatorname{End}(\text{Id}_{{\mathcal{C}}})$ for which the $S$ and $T$ generators act as

(2.27)$$\begin{eqnarray}\big(\begin{smallmatrix}0 & -1\\ 1 & 0\end{smallmatrix}\big)\longmapsto \mathscr{S}_{{\mathcal{C}}},\quad \big(\begin{smallmatrix}1 & 1\\ 0 & 1\end{smallmatrix}\big)\longmapsto \unicode[STIX]{x1D703}\circ (-).\end{eqnarray}$$

3 Existence of a simple projective object

Throughout the rest of this paper, the following technical condition will play an important role.

Two important classes of categories which satisfy Condition P are described in the next two lemmas.

Proof. Since by assumption all objects have finite length, we have $\operatorname{Gr}({\mathcal{C}})=\bigoplus _{U\in \operatorname{Irr}({\mathcal{C}})}\mathbb{Z}[U]$. Thus for $X\in {\mathcal{C}}$ it follows that $X=0$ if and only if $[X]=0$.

The map $\operatorname{ev}_{X}:X^{\ast }\otimes X\rightarrow \mathbf{1}$ is non-zero, hence $X^{\ast }\otimes X\neq 0$, and so $[X^{\ast }][X]\neq 0$. Iterating this argument shows $(X^{\ast }\otimes X)^{\ast }\otimes (X^{\ast }\otimes X)\neq 0$, i.e. $([X^{\ast }][X])^{2}\neq 0$, etc. We used here Drinfeld’s canonical isomorphism $u_{X}:X\rightarrow X^{\ast \ast }$ given by

(3.1)

Using commutativity of $\operatorname{Gr}({\mathcal{C}})$, after $m$ steps we find $[X^{\ast }]^{m}[X]^{m}\neq 0$.

As a finite ${\mathcal{C}}$ has enough projectives, it contains a non-zero projective object $P$. Since in characteristic zero, the canonical ring homomorphism $\operatorname{Gr}({\mathcal{C}})\rightarrow \operatorname{Gr}_{k}({\mathcal{C}})$ is injective, $[P]^{m}\neq 0$ for all $m>0$ also holds in $\operatorname{Gr}_{k}({\mathcal{C}})$.◻

We can now state our first main result.

The proof relies on a series of lemmas and will be given at the end of this section. The idea is very simple: we show that in the absence of simple projectives, the injective algebra homomorphism $\unicode[STIX]{x1D70E}$ from Corollary 2.12 would map the class $[P]$ of a projective object to a nilpotent natural endomorphism, which is a contradiction to Condition P.

We now turn to the proof of Theorem 3.4. We first gather the tools to see that $\unicode[STIX]{x1D70E}([P])$ is nilpotent for projective $P$ unless there is a simple projective object.

Proof. We use induction on the length of an object. Given an object $B$ of finite length greater than $1$, choose a short exact sequence $0\rightarrow A\xrightarrow[{}]{f}B\xrightarrow[{}]{g}C\rightarrow 0$ such that $A,C$ have smaller lengths. Then

(3.2)

commutes. By induction hypothesis, $\unicode[STIX]{x1D702}_{A}$ and $\unicode[STIX]{x1D702}_{C}$ are nilpotent. Hence there is $m>0$ such that

(3.3)

commutes. Then $g\circ (\unicode[STIX]{x1D702}_{B})^{m}=0$ and $(\unicode[STIX]{x1D702}_{B})^{m}\circ f=0$, and since the image of $f$ is the kernel of $g$ we have the inclusions $\text{im}(\unicode[STIX]{x1D702}_{B})^{m}\subset \text{im}f\subset \text{ker}(\unicode[STIX]{x1D702}_{B})^{m}$. This shows that $(\unicode[STIX]{x1D702}_{B})^{2m}=0$.◻

Combining the above results, the proof of Theorem 3.4 is now a straightforward consequence of Corollary 2.12.

4 Non-degenerate trace on the projective ideal

In this section we will review some aspects of the theory of modified traces from [Reference Geer, Kujawa and Patureau-MirandGKP11, Reference Geer, Patureau-Mirand and VirelizierGPV13, Reference Geer, Kujawa and Patureau-MirandGKP13]. In particular, we will see that the existence of a simple projective object established in Theorem 3.4 guarantees the existence of a unique-up-to-scalars non-zero modified trace on the projective ideal, and that this trace induces non-degenerate pairings on Hom-spaces.

Let ${\mathcal{C}}$ be a pivotal finite tensor category over $k$. The full subcategory ${\mathcal{P}}\text{roj}({\mathcal{C}})$ of all projective objects in ${\mathcal{C}}$ is a tensor ideal. Following [Reference Geer, Patureau-Mirand and VirelizierGPV13], a modified right trace on ${\mathcal{P}}\text{roj}({\mathcal{C}})$ is a family of $k$-linear functions $\{t_{P}:\operatorname{End}(P)\rightarrow k\mid P\in {\mathcal{P}}\text{roj}({\mathcal{C}})\}$ such that

1. (1) (partial trace) For all $P\in {\mathcal{P}}\text{roj}({\mathcal{C}})$, $X\in {\mathcal{C}}$ and $f\in \operatorname{End}(P\otimes X)$ we have

   (4.1)$$\begin{eqnarray}t_{P\otimes X}(f)=t_{P}(\text{tr}_{X}^{r}(f)),\end{eqnarray}$$
   where $\text{tr}_{X}^{r}$ denotes the partial right trace
   (4.2)

   

2. (2) (cyclicity) For all $P,P^{\prime }\in {\mathcal{P}}\text{roj}({\mathcal{C}})$, $f:P\rightarrow P^{\prime }$, $g:P^{\prime }\rightarrow P$ we have

   (4.3)$$\begin{eqnarray}t_{P^{\prime }}(f\circ g)=t_{P}(g\circ f).\end{eqnarray}$$

Given a modified right trace, one can define a family of pairings by setting, for all $M\in {\mathcal{C}}$, $P\in {\mathcal{P}}\text{roj}({\mathcal{C}})$,

(4.4)$$\begin{eqnarray}{\mathcal{C}}(M,P)\times {\mathcal{C}}(P,M)\rightarrow k,\quad (f,g)\mapsto t_{P}(f\circ g).\end{eqnarray}$$

Note that the cyclicity property (4.3) only applies when also $M$ is projective.

In this paper we will only use modified right traces, and we will refer to these just as ‘modified traces’.

The first part of the proposition is proved in [Reference Geer, Kujawa and Patureau-MirandGKP13, Corollary 3.2.1].Footnote ⁴ We will review the argument below as we hope it to be helpful to have the more general construction of [Reference Geer, Patureau-Mirand and VirelizierGPV13, Reference Geer, Kujawa and Patureau-MirandGKP13] specialised to the present setting, and since we will need the explicit construction of the modified trace below anyway. The second part of the proposition is proved in [Reference Costantino, Geer and Patureau-MirandCGP15, Proposition 6.6] for the Hopf algebra $\overline{U}_{q}^{H}\text{sl}(2)$, but the same proof works in general as we also review below.

Before turning to the proof of Proposition 4.2, we need some preparation. For the rest of this section, we fix

• ∙ ${\mathcal{C}}$: a unimodular pivotal finite tensor category over $k$;

• ∙ $Q$: a simple projective object in ${\mathcal{C}}$ (in particular we assume that such a $Q$ exists).

Lemma 4.3 [Reference Geer, Kujawa and Patureau-MirandGKP13, Theorem 3.1.3 and §3.2].

For all $g\in \operatorname{End}(Q\otimes Q^{\ast })$ we have

(4.5)

Proof. Write $Q\otimes Q^{\ast }=\bigoplus _{U\in \operatorname{Irr}({\mathcal{C}})}P_{U}^{\oplus m_{U}}$ and let $i_{U}:P_{U}^{\oplus m_{U}}\rightarrow Q\otimes Q^{\ast }$ and $p_{U}:Q\otimes Q^{\ast }\rightarrow P_{U}^{\oplus m_{U}}$ be the corresponding embedding and projection maps. The resulting idempotent will be denoted by $e_{U}=i_{U}\circ p_{U}\in \operatorname{End}(Q\otimes Q^{\ast })$. Since ${\mathcal{C}}(Q\otimes Q^{\ast },\mathbf{1})\cong {\mathcal{C}}(Q,Q)$ is one-dimensional ($Q$ being simple and $k$ algebraically closed), we have $m_{\mathbf{1}}=1$.

Since $P_{\mathbf{1}}$ is indecomposable, there is an $\unicode[STIX]{x1D6FC}\in k$ and a nilpotent element $n\in \operatorname{End}(P_{\mathbf{1}})$ such that (using again algebraically closedness of $k$)Footnote ⁵

(4.6)$$\begin{eqnarray}p_{\mathbf{1}}\circ g\circ i_{\mathbf{1}}=\unicode[STIX]{x1D6FC}\cdot \text{id}_{P_{\mathbf{1}}}+n.\end{eqnarray}$$

For all $u:\mathbf{1}\rightarrow P_{\mathbf{1}}$ and $v:P_{\mathbf{1}}\rightarrow \mathbf{1}$ we have

(4.7)$$\begin{eqnarray}n\circ u=0\quad \text{and}\quad v\circ n=0.\end{eqnarray}$$

To see $n\circ u=0$, first note that, by unimodularity (i.e. since $(P_{\mathbf{1}})^{\ast }\cong P_{\mathbf{1}}$), ${\mathcal{C}}(\mathbf{1},P_{\mathbf{1}})$ is one-dimensional. Suppose $u\neq 0$, so that $u$ forms a basis of ${\mathcal{C}}(\mathbf{1},P_{\mathbf{1}})$. Then there is $\unicode[STIX]{x1D706}\in k$ such that $n\circ u=\unicode[STIX]{x1D706}\cdot u$. But $n$ is nilpotent, say $n^{m}=0$, and so applying $n$ on both sides of the latter equality $m$ times we get $\unicode[STIX]{x1D706}^{m}=0$ and therefore $\unicode[STIX]{x1D706}=0$. For $v$ the argument is analogous.

Next insert the identity $g=\sum _{U,V}e_{U}\circ g\circ e_{V}$ into (4.5). Since for $U\ncong \mathbf{1}$, ${\mathcal{C}}(P_{U},\mathbf{1})=0$ and ${\mathcal{C}}(\mathbf{1},P_{U})=0$ (using unimodularity), only the term with $U=V=\mathbf{1}$ will contribute on either side of (4.5).

The claim of the lemma now follows from the observation that taking $u=p_{\mathbf{1}}\circ \operatorname{coev}_{Q}$ and $v=\widetilde{\operatorname{ev}}_{Q}\circ i_{\mathbf{1}}$ we have by (4.6) and (4.7)

(4.8)$$\begin{eqnarray}\widetilde{\operatorname{ev}}_{Q}\circ e_{\mathbf{1}}\circ g\circ e_{\mathbf{1}}=\unicode[STIX]{x1D6FC}\cdot \widetilde{\operatorname{ev}}_{Q},\quad e_{\mathbf{1}}\circ g\circ e_{\mathbf{1}}\circ \operatorname{coev}_{Q}=\unicode[STIX]{x1D6FC}\cdot \operatorname{coev}_{Q},\end{eqnarray}$$

where we used that $\widetilde{\operatorname{ev}}_{Q}\circ e_{\mathbf{1}}=\widetilde{\operatorname{ev}}_{Q}$ and $e_{\mathbf{1}}\circ \operatorname{coev}_{Q}=\operatorname{coev}_{Q}$ (because $(P_{\mathbf{1}})^{\ast }\cong P_{\mathbf{1}}$). Both the left- and right-hand sides of (4.5) are therefore equal to $\unicode[STIX]{x1D6FC}\operatorname{id}_{Q}$.◻

The above lemma will be used in the following form: for all $f\in \operatorname{End}(Q^{\ast }\otimes Q)$,

(4.9)

This is a consequence of (4.5) upon substituting

(4.10)

Let us fix a non-zero linear function for the given simple projective $Q$,

(4.11)$$\begin{eqnarray}t_{Q}:\operatorname{End}(Q)\rightarrow k.\end{eqnarray}$$

Since $\operatorname{End}(Q)=k\operatorname{id}_{Q}$, such a function is unique up to a constant, and it is uniquely determined by its value $t_{Q}(\operatorname{id}_{Q})\in k^{\times }$.

Let $P\in {\mathcal{P}}\text{roj}({\mathcal{C}})$ and choose an object $X\in {\mathcal{C}}$ such that there is a surjection $p:Q\otimes X\rightarrow P$. Such an $X$ always exists, for example $X=Q^{\ast }\otimes P$. As $P$ is projective, there is a morphism $i:P\rightarrow Q\otimes X$ such that $p\circ i=\operatorname{id}_{P}$. That is, $i,p$ realise $P$ as a direct summand of $Q\otimes X$. Consider the function

(4.12)$$\begin{eqnarray}t_{P}:\operatorname{End}(P)\rightarrow k,\quad h\mapsto t_{Q}(\text{tr}_{X}^{r}(i\circ h\circ p)).\end{eqnarray}$$

The next lemma shows in particular that this notation is indeed consistent with (4.11) when setting $P=Q$ (in the lemma, take $X=\mathbf{1}$ and $i,p$ (inverse) unit isomorphisms).

Proof. This proof is taken from [Reference Geer, Patureau-Mirand and VirelizierGPV13, §4.5]. The argument is more general than needed for the statement of the lemma, but we can reuse it later to show cyclicity.

Let $P^{\prime }\in {\mathcal{P}}\text{roj}({\mathcal{C}})$ and pick $X^{\prime }$, $p^{\prime }:Q\otimes X^{\prime }\rightarrow P^{\prime }$, $i^{\prime }:P^{\prime }\rightarrow Q\otimes X^{\prime }$. For $u:P\rightarrow P^{\prime }$ and $v:P^{\prime }\rightarrow P$ arbitrary, consider the morphism $f\in \operatorname{End}(Q^{\ast }\otimes Q)$ given by

(4.13)

It is easy to check that inserting $f$ into (4.9) results in the identity

(4.14)$$\begin{eqnarray}\text{tr}_{X^{\prime }}^{r}(i^{\prime }\circ u\circ v\circ p^{\prime })=\text{tr}_{X}^{r}(i\circ v\circ u\circ p),\end{eqnarray}$$

where for the right-hand side we used cyclicity of the categorical trace, the zig-zag axiom and that $p^{\prime }\circ i^{\prime }=\operatorname{id}_{P^{\prime }}$. For $P=P^{\prime }$ and $u=h$, $v=\operatorname{id}_{P}$, this shows the statement of the lemma.◻

Proof of Proposition 4.2.

Part (1) (following [Reference Geer, Patureau-Mirand and VirelizierGPV13, §4.5]).

Existence. By Lemma 4.4 we have a family of functions $(t_{P}:\operatorname{End}(P)\rightarrow k)_{P\in {\mathcal{P}}\text{roj}({\mathcal{C}})}$. This family is not identically zero since by the definition in (4.11), $t_{Q}$ is not zero. By (4.14) in the proof of Lemma 4.4, the $t_{P}$ satisfy the cyclicity requirement (4.3).

For the partial trace property (4.1) choose $Y\in {\mathcal{C}}$ such that there are $p:Q\otimes Y\rightarrow P$, $i:P\rightarrow Q\otimes Y$, $p\circ i=\operatorname{id}_{P}$, in order to compute $t_{P}$ from the definition in (4.12) with $X$ replaced by $Y$. Then fix

(4.15)$$\begin{eqnarray}p^{\prime }=[Q(YX)\xrightarrow[{}]{{\sim}}(QY)X\xrightarrow[{}]{p\,\otimes \,\text{id}_{X}}PX],\quad i^{\prime }=[PX\xrightarrow[{}]{i\,\otimes \,\text{id}_{X}}(QY)X\xrightarrow[{}]{{\sim}}Q(YX)]\end{eqnarray}$$

to compute $t_{P\otimes X}$ from (4.12). A short calculation shows (4.1).

Uniqueness. First note that for the modified trace $t_{P}$ in (4.12) (or in fact for any modified trace) we have, for $f\in \operatorname{End}(Q\otimes X)$,

(4.16)$$\begin{eqnarray}\text{tr}_{X}^{r}(f)=\frac{t_{Q\otimes X}(f)}{t_{Q}(\operatorname{id}_{Q})}\cdot \text{id}_{Q}.\end{eqnarray}$$

To see this, apply $t_{Q}$ to both sides. Let now $(t_{P}^{\prime }:\operatorname{End}(P)\rightarrow k)_{P\in {\mathcal{P}}\text{roj}({\mathcal{C}})}$ be a modified trace. Let $P\in {\mathcal{P}}\text{roj}({\mathcal{C}})$ and $h\in \operatorname{End}(P)$ be given. Pick $X$, $p:Q\otimes X\rightarrow P$, $i:P\rightarrow Q\otimes X$ such that $p\circ i=\operatorname{id}_{P}$ and compute

(4.17)

Part (2) (following the proof of [Reference Costantino, Geer and Patureau-MirandCGP15, Proposition 6.6]).

We will show non-degeneracy in the first argument of the pairing (4.4). Non-degeneracy in the second argument can be seen analogously.

Let $M\in {\mathcal{C}}$, $P\in {\mathcal{P}}\text{roj}({\mathcal{C}})$ and $f\in {\mathcal{C}}(M,P)$, $f\neq 0$, be given. We need to show that there is a $g\in {\mathcal{C}}(P,M)$ such that $t_{P}(f\circ g)\neq 0$. Pick $X$, $p:Q\otimes X\rightarrow P$, $i:P\rightarrow Q\otimes X$ such that $p\circ i=\operatorname{id}_{P}$. Then also $i\circ f\neq 0$. Define

(4.18)$$\begin{eqnarray}u:=[MX^{\ast }\xrightarrow[{}]{(i\circ f)\,\otimes \,\text{id}}(QX)X^{\ast }\xrightarrow[{}]{{\sim}}Q(XX^{\ast })\xrightarrow[{}]{\text{id}\,\otimes \,\widetilde{\text{ev}}_{X}}Q\mathbf{1}\xrightarrow[{}]{{\sim}}Q].\end{eqnarray}$$

Using the zig-zag identity, we can recover $i\circ f$ from $u$ as

(4.19)$$\begin{eqnarray}i\circ f=[M\xrightarrow[{}]{{\sim}}M\mathbf{1}\xrightarrow[{}]{\text{id}\,\otimes \,\widetilde{\text{coev}}_{X}}M(X^{\ast }X)\xrightarrow[{}]{{\sim}}(MX^{\ast })X\xrightarrow[{}]{u\,\otimes \,\text{id}_{X}}QX].\end{eqnarray}$$

In particular, since $i\circ f\neq 0$, we must have $u\neq 0$. Since $Q$ is simple, $u$ is surjective. Since $Q$ is projective, there is $v:Q\rightarrow M\otimes X^{\ast }$ such that $u\circ v=\operatorname{id}_{Q}$. Define

(4.20)$$\begin{eqnarray}g:=[P\xrightarrow[{}]{i}QX\xrightarrow[{}]{v\,\otimes \,\text{id}}(MX^{\ast })X\xrightarrow[{}]{{\sim}}M(X^{\ast }X)\xrightarrow[{}]{\text{id}\,\otimes \,\text{ev}_{X}}M\mathbf{1}\xrightarrow[{}]{{\sim}}M].\end{eqnarray}$$

Then

(4.21)

where for the first equality we used the partial trace property (4.1) on the right-hand side and then the pivotal structure in (2.4). ◻

Lemma 4.5. Suppose ${\mathcal{C}}$ is in addition braided. Let $t$ be a modified trace on ${\mathcal{C}}$ and let $R,S\in {\mathcal{P}}\text{roj}({\mathcal{C}})$ and $f\in \operatorname{End}(R)$, $g\in \operatorname{End}(S)$. Then

(4.22)

Proof. One computes

(4.23)

where

(4.24)

Step $(\ast )$ follows from cyclicity of the modified trace, as well as naturality of the braiding. In step $(\ast \ast )$ we used that the partial trace over $S\otimes R^{\ast }$ amounts to closing the double string, see (4.2). In the last equality, to straighten the line coloured by $R$ we also used (2.4) and the property of the pivotal structure that $\unicode[STIX]{x1D6FF}_{R}^{\ast }=(\unicode[STIX]{x1D6FF}_{R^{\ast }})^{-1}$.◻

Remark 4.6. (1) The result of Proposition 4.2 is remarkable, because the categorical trace defined on all of ${\mathcal{C}}$ via the pivotal structure vanishes on ${\mathcal{P}}\text{roj}({\mathcal{C}})$ unless ${\mathcal{C}}$ is semisimple. Indeed, ${\mathcal{C}}$ is semisimple if and only if $P_{\mathbf{1}}=\mathbf{1}$, which in turn is equivalent to the existence of $f,g$ such that $\mathbf{1}\xrightarrow[{}]{f}P_{\mathbf{1}}\xrightarrow[{}]{g}\mathbf{1}$ is non-zero. But for any projective object $P$, the categorical trace

(4.25)$$\begin{eqnarray}\mathbf{1}\xrightarrow[{}]{\operatorname{coev}_{P}}P\otimes P^{\ast }\xrightarrow[{}]{f\,\otimes \,\text{id}}P\otimes P^{\ast }\xrightarrow[{}]{\widetilde{\operatorname{ev}}_{P}}\mathbf{1},\end{eqnarray}$$

factors through $P_{\mathbf{1}}$ in this way.

(2) For each $P\in {\mathcal{P}}\text{roj}({\mathcal{C}})$, $t_{P}$ turns the $k$-algebra $\operatorname{End}(P)$ into a symmetric Frobenius algebra, or a symmetric algebra for short. We will review some aspects of symmetric algebras and their categories of modules in § 5.

Since a factorisable finite tensor category is automatically unimodular [Reference Etingof, Nikshych and OstrikENO04, Proposition 4.5], we state the following consequence of Theorem 3.4 and Proposition 4.2 for later use.

This result generalises the existence of a modified trace for factorisable Hopf algebras (over an algebraically closed field of characteristic zero) proven in [Reference Geer, Kujawa and Patureau-MirandGKP13, Theorem 4.3.1].

5 Symmetric algebras and ideals in $\operatorname{End}(\text{Id}_{{\mathcal{C}}})$

In this section we will collect some facts about symmetric algebras following [Reference BrouéBro09, Reference Cohen and WestreichCW08] and translate them into categorical statements. In the end we give implications for categories with modified trace as in the previous section.

For a $k$-algebra $A$, denote by

(5.1)$$\begin{eqnarray}C(A)=\{\unicode[STIX]{x1D711}:A\rightarrow k\mid \unicode[STIX]{x1D711}(ab)=\unicode[STIX]{x1D711}(ba)\text{ for all }a,b\in A\}\end{eqnarray}$$

the space of central forms on $A$. A symmetric algebra $A$ over $k$ (also known as a symmetric Frobenius algebra) is a finite-dimensional $k$-algebra together with a central form $\unicode[STIX]{x1D700}$ such that the induced pairing $(a,b)\mapsto \unicode[STIX]{x1D700}(ab)$ is non-degenerate. In this case the map

(5.2)$$\begin{eqnarray}\unicode[STIX]{x1D701}:Z(A)\rightarrow C(A),\quad z\mapsto \unicode[STIX]{x1D700}(z\cdot (-))\end{eqnarray}$$

is an isomorphism, and the invertible elements of $Z(A)$ are precisely mapped to those elements of $C(A)$ that induce a non-degenerate pairing (see e.g. [Reference BrouéBro09, Lemma 2.5]).

Let $(A,\unicode[STIX]{x1D700})$ be a symmetric algebra and denote by $\unicode[STIX]{x1D6FE}\in A\otimes A$, $\unicode[STIX]{x1D6FE}=\sum _{(\unicode[STIX]{x1D6FE})}\unicode[STIX]{x1D6FE}^{\prime }\otimes \unicode[STIX]{x1D6FE}^{\prime \prime }$, the copairing, that is, for all $a\in A$ we have $a=\sum _{(\unicode[STIX]{x1D6FE})}\unicode[STIX]{x1D700}(a\unicode[STIX]{x1D6FE}^{\prime })\unicode[STIX]{x1D6FE}^{\prime \prime }=\sum _{(\unicode[STIX]{x1D6FE})}\unicode[STIX]{x1D6FE}^{\prime }\unicode[STIX]{x1D700}(\unicode[STIX]{x1D6FE}^{\prime \prime }a)$. We recall that the copairing is unique and that, since $\unicode[STIX]{x1D700}$ is central, the copairing is symmetric. Define the map

(5.3)$$\begin{eqnarray}\unicode[STIX]{x1D70F}:A\rightarrow Z(A),\quad a\mapsto \mathop{\sum }_{(\unicode[STIX]{x1D6FE})}\unicode[STIX]{x1D6FE}^{\prime }a\unicode[STIX]{x1D6FE}^{\prime \prime }.\end{eqnarray}$$

It is easy to check that the image of $\unicode[STIX]{x1D70F}$ does indeed lie in the centre of $A$ (see e.g. [Reference BrouéBro09, §3.A]). The argument uses that the copairing $\unicode[STIX]{x1D6FE}$ of a non-degenerate invariant pairing satisfies $\sum _{(\unicode[STIX]{x1D6FE})}(a\unicode[STIX]{x1D6FE}^{\prime })\otimes \unicode[STIX]{x1D6FE}^{\prime \prime }=\sum _{(\unicode[STIX]{x1D6FE})}\unicode[STIX]{x1D6FE}^{\prime }\otimes (\unicode[STIX]{x1D6FE}^{\prime \prime }a)$ for all $a\in A$.

We will need the following chains of inclusions in $C(A)$ and $Z(A)$, compatible with the map $\unicode[STIX]{x1D701}$ from (5.2) [Reference Cohen and WestreichCW08].

(5.4)

The individual subsets are defined as follows.

• ∙ $\text{Rey}(A)=\text{ann}_{Z(A)}(\text{Jac}(A))$ is the Reynolds ideal. (Symmetry of $A$ is not required for this definition.) Here, $\text{Jac}(A)$ is the Jacobson radical of $A$. $\text{Rey}(A)$ is an ideal in $Z(A)$.

• ∙ $\text{Hig}(A)=\text{im}(\unicode[STIX]{x1D70F})$ is the Higman ideal or projective centre.Footnote ⁶ It is an ideal in $Z(A)$ as $z\unicode[STIX]{x1D70F}(a)=\unicode[STIX]{x1D70F}(za)$ for $z\in Z(A)$, and by [Reference Héthelyi, Horváth, Külshammer and MurrayHHKM05, Lemma 4.1]Footnote ⁷ it is contained in $\text{Rey}(A)$. Note that even though $\unicode[STIX]{x1D70F}$ depends on the choice of a non-degenerate central form $\unicode[STIX]{x1D700}$, $\text{im}(\unicode[STIX]{x1D70F})$ does not. Indeed, any two $\unicode[STIX]{x1D700}$, $\unicode[STIX]{x1D700}^{\prime }$ would be related by an invertible $z\in Z(A)$ as $\unicode[STIX]{x1D700}^{\prime }(a)=\unicode[STIX]{x1D700}(za)$, and one checks that $\unicode[STIX]{x1D70F}^{\prime }(a)=\unicode[STIX]{x1D70F}(z^{-1}a)$.

• ∙ $R(A)=\text{span}_{k}\{\unicode[STIX]{x1D712}_{M}\mid M\text{ right }A\text{-module}\}$, where $\unicode[STIX]{x1D712}_{M}(a)=\text{tr}_{M}(a)$, the trace in $M$ over the linear map given by acting with $a\in A$. That $\unicode[STIX]{x1D701}(\text{Rey}(A))=R(A)$ is shown in [Reference LorenzLor97, Theorem 1.6] (for finite-dimensional not necessarily symmetric algebras).

• ∙ $I(A)=\text{span}_{k}\{\unicode[STIX]{x1D712}_{P}\mid P\text{ projective right }A\text{-module}\}$. In [Reference Cohen and WestreichCW08, Proposition 2.1] it is shown that $\unicode[STIX]{x1D701}(\text{Hig}(A))=I(A)$.

The Cartan matrix of a finite-dimensional algebra $A$ over some field is the $\operatorname{Irr}(\text{mod}\text{-}A)\times \operatorname{Irr}(\text{mod}\text{-}A)$ matrix $\mathsf{C}(A)$ with entries

(5.5)$$\begin{eqnarray}\mathsf{C}(A)_{U,V}=[P_{U}:V],\end{eqnarray}$$

that is, the multiplicity of the simple $A$-module $V$ in the composition series of the projective cover $P_{U}$ of $U$.

Let $e_{U}$, $U\in \operatorname{Irr}(\text{mod}\text{-}A)$ be a choice of primitive idempotents such that $e_{U}A\cong P_{U}$ as right $A$-modules. If the underlying field is given by $k$, which we assumed to be algebraically closed in the outset of the paper, the endomorphism spaces of simple $A$-modules are one-dimensional and we can write

(5.6)$$\begin{eqnarray}\mathsf{C}(A)_{U,V}=\dim _{k}\operatorname{Hom}_{A}(P_{V},P_{U})=\dim _{k}(e_{U}Ae_{V}).\end{eqnarray}$$

The last equality follows by noting that every right $A$-module map $f:e_{V}A\rightarrow e_{U}A$ is given by left multiplication with an element $e_{U}ae_{V}$, $a\in A$. Namely, in one direction we set $f\mapsto f(e_{V})e_{V}\in e_{U}Ae_{V}$, while in the other direction $e_{U}ae_{V}\mapsto e_{U}ae_{V}\cdot (-)$ which is in $\operatorname{Hom}_{A}(e_{V}A,e_{U}A)$, and one checks that both the maps are inverses of each other.

Note that if $k$ has characteristic $0$, then $\text{rank}(\tilde{\mathsf{C}})$ is just the rank of $\mathsf{C}(A)$ over $\mathbb{Q}$.

We now give a categorical formulation of the inclusions $Z(A)\supset \text{Rey}(A)\supset \text{Hig}(A)$ in (5.4), which will be applicable in particular in the presence of a modified trace as in § 4. For the rest of this section, let us fix

• ∙ ${\mathcal{A}}$: a finite abelian category over $k$;

• ∙ $G$: a projective generator of ${\mathcal{A}}$;

• ∙ $E=\operatorname{End}(G)$ the $k$-algebra of endomorphisms of $G$.

Recall the Hom-tensor adjoint equivalence

(5.8)

of $k$-linear categories between ${\mathcal{A}}$ and the category $\text{mod}\text{-}E$ of finite-dimensional right $E$-modules (see e.g. [Reference Etingof, Gelaki, Nikshych and OstrikEGNO15, §1.8]).

It is known that being symmetric is a Morita-invariant property of an algebra, as is the chain of ideals $Z(E)\supset \text{Rey}(E)\supset \text{Hig}(E)$ (see e.g. [Reference Héthelyi, Horváth, Külshammer and MurrayHHKM05, Reference BrouéBro09, Reference Liu, Zhou and ZimmermannLZZ12] for related results). We can therefore use the equivalence in (5.8) and the ideals in (5.4) for the choice $A=E$ and for a given projective generator $G$ to define a chain of ideals in $\operatorname{End}(\text{Id}_{{\mathcal{A}}})$ as follows.

(5.9)

Rather than showing directly that the above definition is independent of the choice of $G$, we will now give a different definition of $\text{Rey}({\mathcal{A}})$ and $\text{Hig}({\mathcal{A}})$ which does not require a choice of a projective generator. We show in Propositions 5.3 and 5.8 that this definition agrees with the one above.

$\text{Rey}({\mathcal{A}})$

For each $P\in {\mathcal{P}}\text{roj}({\mathcal{A}})$ define the subspace $J_{P}\subset \operatorname{End}(P)$ as follows:

(5.10)$$\begin{eqnarray}J_{P}=\{f:P\rightarrow P\mid \forall \,U\in {\mathcal{A}}\text{simple},u\in {\mathcal{A}}(P,U):u\circ f=0\}.\end{eqnarray}$$

By definition, the Jacobson radical $\text{Jac}(R)$ of a ring $R$ consists of all $r\in R$ which act as zero on all simple $R$-modules. Since $G\in {\mathcal{A}}$ is a projective generator, ${\mathcal{A}}(G,U)$, where $U$ runs over all simple $U\in {\mathcal{A}}$, gives all simple right $E$-modules up to isomorphism (by the equivalence (5.8)). From (5.10) we therefore get

(5.11)$$\begin{eqnarray}J_{G}=\text{Jac}(E).\end{eqnarray}$$

In the case ${\mathcal{A}}=\text{mod}\text{-}A$ we can take $G=A_{A}$, so that $\operatorname{End}(G)=A$ and hence $J_{A}=\text{Jac}(A)$.

Using the subspaces (5.10), we define

(5.12)$$\begin{eqnarray}\text{Rey}({\mathcal{A}})=\{\unicode[STIX]{x1D702}\in \operatorname{End}(\text{Id}_{{\mathcal{A}}})\mid \forall \,P\in {\mathcal{P}}\text{roj}({\mathcal{A}}),f\in J_{P}:\unicode[STIX]{x1D702}_{P}\circ f=0\}.\end{eqnarray}$$

From this definition it is immediate that $\text{Rey}({\mathcal{A}})$ is an ideal.

Recall that for a projective generator $G\in {\mathcal{A}}$, the map

(5.13)$$\begin{eqnarray}\unicode[STIX]{x1D709}:\operatorname{End}(\text{Id}_{{\mathcal{A}}})\rightarrow Z(E),\quad \unicode[STIX]{x1D702}\mapsto \unicode[STIX]{x1D702}_{G}\end{eqnarray}$$

is an algebra isomorphism (that $\unicode[STIX]{x1D702}_{G}$ is central in $E$ follows from the naturality of $\unicode[STIX]{x1D702}$; that (5.13) is a bijection is clear in the case ${\mathcal{A}}=\text{mod}\text{-}E$, and for general ${\mathcal{A}}$ use the equivalence (5.8)). We can use this to relate $\text{Rey}({\mathcal{A}})$ to the previous definition in terms of algebras.

Proposition 5.3. Let $A$ be a finite-dimensional $k$-algebra. Then $\unicode[STIX]{x1D709}$ in (5.13) restricts to an isomorphism of algebras $\unicode[STIX]{x1D709}:\text{Rey}(\text{mod}\text{-}A)\rightarrow \text{Rey}(A)$.

Proof. We set ${\mathcal{A}}=\text{mod}\text{-}A$ and choose the projective generator $G=A_{A}$ and thus $E=A$. We have to show that $\unicode[STIX]{x1D709}(\text{Rey}({\mathcal{A}}))=\text{Rey}(E)$.

$\subset$:

Let $\unicode[STIX]{x1D702}\in \text{Rey}({\mathcal{A}})$. Then $\unicode[STIX]{x1D709}(\unicode[STIX]{x1D702})=\unicode[STIX]{x1D702}_{G}$. By definition of $\text{Rey}({\mathcal{A}})$ we have in particular that for all $f\in J_{G}=\text{Jac}(E)$ (recall (5.11)), $\unicode[STIX]{x1D702}_{G}\circ f=0$. Thus $\unicode[STIX]{x1D702}_{G}$ annihilates $\text{Jac}(E)$, i.e. $\unicode[STIX]{x1D702}_{G}\in \text{Rey}(E)$.

$\supset$:

Conversely, let $r\in \text{Rey}(E)$ and let $\unicode[STIX]{x1D702}=\unicode[STIX]{x1D709}^{-1}(r)\in \operatorname{End}(\text{Id}_{{\mathcal{C}}})$. Let $P\in {\mathcal{P}}\text{roj}({\mathcal{C}})$, and $f\in J_{P}$. We need to show $\unicode[STIX]{x1D702}_{P}\circ f=0$. Pick a surjection $p:G^{\oplus n}\rightarrow P$ and a right-inverse $i:P\rightarrow G^{\oplus n}$. We have $i\circ f\circ p\in J_{G^{\oplus n}}$ since for all $u:G^{\oplus n}\rightarrow U$, where $U$ is simple, $u\circ i\circ f\circ p=0$ since $u\circ i:P\rightarrow U$ and $f\in J_{P}$. Note then that $r^{\oplus n}:G^{\oplus n}\rightarrow G^{\oplus n}$ and $r^{\oplus n}\in \text{Rey}(\operatorname{End}(G^{\oplus n}))$, therefore $r^{\oplus n}\circ i\circ f\circ p=0$. Since $r^{\oplus n}=\unicode[STIX]{x1D702}_{G^{\oplus n}}$ we get

(5.14)$$\begin{eqnarray}0=\unicode[STIX]{x1D702}_{G^{\oplus n}}\circ i\circ f\circ p=i\circ \unicode[STIX]{x1D702}_{P}\circ f\circ p.\end{eqnarray}$$

But $i$ is injective and $p$ surjective, and so $\unicode[STIX]{x1D702}_{P}\circ f=0$, as required.◻

$\text{Hig}({\mathcal{A}})$

The condition replacing that the algebra is symmetric will be that ${\mathcal{P}}\text{roj}({\mathcal{A}})$ is Calabi–Yau (see e.g. [Reference HesseHes17] for a recent treatment of the relation between symmetric Frobenius algebras and Calabi–Yau categories in the semisimple case).

Definition 5.4. A $k$-linear additive category ${\mathcal{K}}$ with finite-dimensional Hom-spaces is called Calabi–Yau if it is equipped with a family of $k$-linear maps (the trace maps)

(5.15)$$\begin{eqnarray}(\mathsf{t}_{M}:\operatorname{End}(M)\rightarrow k)_{M\in {\mathcal{K}}}\end{eqnarray}$$

such that for all $M,N\in {\mathcal{K}}$ we have the following.

1. (1) (cyclicity) For all $f:M\rightarrow N$, $g:N\rightarrow M$ we have $\mathsf{t}_{M}(g\circ f)=\mathsf{t}_{N}(f\circ g)$.

2. (2) (non-degeneracy) The pairing $(-,-):{\mathcal{K}}(M,N)\times {\mathcal{K}}(N,M)\rightarrow k$, $(f,g)=\mathsf{t}_{N}(f\circ g)$ is non-degenerate.

We denote by $\text{CY}({\mathcal{K}})$ the set of all families of trace maps $(\mathsf{t}_{M})_{M\in {\mathcal{K}}}$ which turn ${\mathcal{K}}$ into a Calabi–Yau category.

A Calabi–Yau category with one object is the same as a symmetric Frobenius algebra. The generalisation of the isomorphism (5.2) to the present setting is as follows.

Lemma 5.5. Let ${\mathcal{K}}$ be as in Definition 5.4 and let $\operatorname{End}(\text{Id}_{{\mathcal{K}}})^{\times }$ be the subset of invertible endotransformations. Then for each $(\mathsf{t}_{M})_{M\in {\mathcal{K}}}\in \text{CY}({\mathcal{K}})$ the map

(5.16)$$\begin{eqnarray}\operatorname{End}(\text{Id}_{{\mathcal{K}}})^{\times }\longrightarrow \text{CY}({\mathcal{K}}),\quad \unicode[STIX]{x1D702}\longmapsto (f\mapsto \mathsf{t}_{M}(\unicode[STIX]{x1D702}_{M}\circ f))\end{eqnarray}$$

is a bijection.

Proof. Clearly, $(\mathsf{t}_{M}(\unicode[STIX]{x1D702}_{M}\circ -))_{M\in {\mathcal{K}}}$ defines a family of cyclic and non-degenerate traces (for non-degeneracy, one notes that $\unicode[STIX]{x1D702}_{M}\circ g$ is non-zero for any non-zero $g$ because $\unicode[STIX]{x1D702}_{M}$ is invertible). Conversely, given $(\tilde{\mathsf{t}}_{M})_{M\in {\mathcal{K}}}\in \text{CY}({\mathcal{K}})$, by non-degeneracy for each $M\in {\mathcal{K}}$ there exists a unique $\unicode[STIX]{x1D702}_{M}\in \operatorname{End}(M)^{\times }$ such that $\tilde{\mathsf{t}}_{M}(-)=\mathsf{t}_{M}(\unicode[STIX]{x1D702}_{M}\circ -)$. Rewriting the cyclicity condition $\tilde{\mathsf{t}}_{M}(g\circ f)=\tilde{\mathsf{t}}_{N}(f\circ g)$ for all $f:M\rightarrow N$ and $g:N\rightarrow M$ in terms of the family $\mathsf{t}_{M}$ and using its cyclicity property shows $\mathsf{t}_{M}(g\circ (\unicode[STIX]{x1D702}_{N}\circ f-f\circ \unicode[STIX]{x1D702}_{M}))=0$ for any $g$, and thus by non-degeneracy of $\mathsf{t}_{M}$ we have $\unicode[STIX]{x1D702}_{N}\circ f=f\circ \unicode[STIX]{x1D702}_{M}$ for all $M$, $N$, and $f$.◻

Suppose now that the finite abelian category ${\mathcal{A}}$ is such that ${\mathcal{P}}\text{roj}({\mathcal{A}})$ is Calabi–Yau with trace maps $(\mathsf{t}_{P})_{P\in {\mathcal{P}}\text{roj}({\mathcal{A}})}$. For the pairings of morphisms between the projective objects $P,R\in {\mathcal{P}}\text{roj}({\mathcal{A}})$ we write

(5.17)$$\begin{eqnarray}(-,-)_{PR}:{\mathcal{A}}(P,R)\times {\mathcal{A}}(R,P)\rightarrow k,\quad (f,g)_{PR}=\mathsf{t}_{R}(f\circ g).\end{eqnarray}$$

Denote by $\unicode[STIX]{x1D6FE}_{PR}\in {\mathcal{A}}(R,P)\otimes {\mathcal{A}}(P,R)$ the corresponding (unique) copairing, that is $\unicode[STIX]{x1D6FE}_{PR}=\sum _{(\unicode[STIX]{x1D6FE}_{PR})}\unicode[STIX]{x1D6FE}_{PR}^{\prime }\otimes \unicode[STIX]{x1D6FE}_{PR}^{\prime \prime }$ and for all $x:P\rightarrow R$, $y:R\rightarrow P$,

(5.18)$$\begin{eqnarray}\mathop{\sum }_{(\unicode[STIX]{x1D6FE}_{PR})}(x,\unicode[STIX]{x1D6FE}_{PR}^{\prime })_{PR}\unicode[STIX]{x1D6FE}_{PR}^{\prime \prime }=x,\quad \mathop{\sum }_{(\unicode[STIX]{x1D6FE}_{PR})}\unicode[STIX]{x1D6FE}_{PR}^{\prime }(\unicode[STIX]{x1D6FE}_{PR}^{\prime \prime },y)_{PR}=y.\end{eqnarray}$$

We need the following lemma, whose proof is given in Appendix B.

Lemma 5.6. For each $R\in {\mathcal{P}}\text{roj}({\mathcal{A}})$ and $y\in \operatorname{End}(R)$ there exists a unique $\unicode[STIX]{x1D702}\in \operatorname{End}(\text{Id}_{{\mathcal{A}}})$ such that $\unicode[STIX]{x1D702}_{P}=\sum _{(\unicode[STIX]{x1D6FE}_{PR})}\unicode[STIX]{x1D6FE}_{PR}^{\prime }\circ y\circ \unicode[STIX]{x1D6FE}_{PR}^{\prime \prime }$ for all $P\in {\mathcal{P}}\text{roj}({\mathcal{A}})$.

This lemma allows us to define, for each $R\in {\mathcal{P}}\text{roj}({\mathcal{A}})$, a $k$-linear map $\unicode[STIX]{x1D70F}_{R}:\operatorname{End}(R)\rightarrow \operatorname{End}(\text{Id}_{{\mathcal{A}}})$ by

(5.19)$$\begin{eqnarray}\unicode[STIX]{x1D70F}_{R}(y)=\unicode[STIX]{x1D702}\quad \text{where for all }P\in {\mathcal{P}}\text{roj}({\mathcal{A}})\quad \unicode[STIX]{x1D702}_{P}=\mathop{\sum }_{(\unicode[STIX]{x1D6FE}_{PR})}\unicode[STIX]{x1D6FE}_{PR}^{\prime }\circ y\circ \unicode[STIX]{x1D6FE}_{PR}^{\prime \prime }.\end{eqnarray}$$

These maps are the analogue of the map $\unicode[STIX]{x1D70F}$ from (5.3). Indeed, $\text{Hig}({\mathcal{A}})$ is now defined as the joint image of all the maps $\unicode[STIX]{x1D70F}_{R}$:

(5.20)$$\begin{eqnarray}\displaystyle \text{Hig}({\mathcal{A}})=\{\unicode[STIX]{x1D702}\in \operatorname{End}(\text{Id}_{{\mathcal{A}}})\mid \exists \,R\in {\mathcal{P}}\text{roj}({\mathcal{A}}),y\in \operatorname{End}(R):\unicode[STIX]{x1D702}=\unicode[STIX]{x1D70F}_{R}(y)\}. & & \displaystyle\end{eqnarray}$$

The point of the above definition is to avoid the choice of a projective generator. To actually compute $\text{Hig}({\mathcal{A}})$, choosing a projective generator $G$ may nonetheless be useful. For example, we show in Corollary B.3 that $\text{Hig}({\mathcal{A}})=\text{im}(\unicode[STIX]{x1D70F}_{G})$.

As was the case for the Higman ideal of a symmetric algebra, the next proposition shows that also $\text{Hig}({\mathcal{A}})$ is independent of the choice of traces.

Proposition 5.7. Let ${\mathcal{A}}$ be a finite abelian category over $k$ such that ${\mathcal{P}}\text{roj}({\mathcal{A}})$ is Calabi–Yau. Then $\text{Hig}({\mathcal{A}})$ is independent of the choice of the family of traces $(\mathsf{t}_{P})_{P\in {\mathcal{P}}\text{roj}({\mathcal{A}})}$.

Proof. Let $\mathsf{t},\tilde{\mathsf{t}}\in \text{CY}({\mathcal{P}}\text{roj}({\mathcal{A}}))$. By Lemma 5.5 there is $\unicode[STIX]{x1D702}\in \operatorname{End}(\text{Id}_{{\mathcal{P}}\text{roj}({\mathcal{A}})})^{\times }$ such that $\tilde{\mathsf{t}}_{P}(f)=\mathsf{t}_{P}(\unicode[STIX]{x1D702}_{P}\circ f)$ for all $P\in {\mathcal{P}}\text{roj}({\mathcal{A}})$ and $f\in \operatorname{End}(P)$. Let $\unicode[STIX]{x1D6FE}_{PR}$ and $\tilde{\unicode[STIX]{x1D6FE}}_{PR}$ be the copairings arising from $\mathsf{t}$ and $\tilde{\mathsf{t}}$ as in (5.18). If we write $\unicode[STIX]{x1D6FE}_{PR}=\sum _{(\unicode[STIX]{x1D6FE}_{PR})}\unicode[STIX]{x1D6FE}_{PR}^{\prime }\otimes \unicode[STIX]{x1D6FE}_{PR}^{\prime \prime }$, it is easy to see that

(5.21)$$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6FE}}_{PR}=\mathop{\sum }_{(\unicode[STIX]{x1D6FE}_{PR})}(\unicode[STIX]{x1D6FE}_{PR}^{\prime }\circ \unicode[STIX]{x1D702}_{R}^{-1})\otimes \unicode[STIX]{x1D6FE}_{PR}^{\prime \prime }=\mathop{\sum }_{(\unicode[STIX]{x1D6FE}_{PR})}\unicode[STIX]{x1D6FE}_{PR}^{\prime }\otimes (\unicode[STIX]{x1D702}_{R}^{-1}\circ \unicode[STIX]{x1D6FE}_{PR}^{\prime \prime }).\end{eqnarray}$$

Write $\tilde{\unicode[STIX]{x1D70F}}_{R}$ for the map (5.19) computed from $\tilde{\unicode[STIX]{x1D6FE}}$. It is immediate from the above equality that for $y\in \operatorname{End}(R)$,

(5.22)$$\begin{eqnarray}\tilde{\unicode[STIX]{x1D70F}}_{R}(y)=\unicode[STIX]{x1D70F}_{R}(\unicode[STIX]{x1D702}_{R}^{-1}\circ y).\end{eqnarray}$$

Thus indeed the joint image of the $\unicode[STIX]{x1D70F}_{R}$ is independent of the choice of traces $\mathsf{t}_{P}$.◻

The next proposition makes the connection between the definition (5.20) of $\text{Hig}({\mathcal{A}})$ and $\text{Hig}(A)$ for a symmetric algebra $A$ when taking ${\mathcal{A}}=\text{mod}\text{-}A$. The proof has been relegated to Appendix B.

Proposition 5.8. Let $A$ be a finite-dimensional $k$-algebra.

1. (1) $A$ admits a central form turning it into a symmetric algebra if and only if ${\mathcal{P}}\text{roj}(\text{mod}\text{-}A)$ admits a family of traces turning it into a Calabi–Yau category.

2. (2) If $A$ is symmetric, the map $\unicode[STIX]{x1D709}$ from (5.13) for $G=A_{A}$ restricts to an isomorphism of algebras $\unicode[STIX]{x1D709}:\text{Hig}(\text{mod}\text{-}A)\rightarrow \text{Hig}(A)$.

Combining this proposition and Proposition 5.3 with the equivalence (5.8) and the inclusions (5.4) we get that we indeed have the inclusions (5.9).

As for algebras, the Cartan matrix of ${\mathcal{A}}$ is the $\operatorname{Irr}({\mathcal{A}})\times \operatorname{Irr}({\mathcal{A}})$-matrix $\mathsf{C}({\mathcal{A}})$ with entries $\mathsf{C}({\mathcal{A}})_{UV}=[P_{U}:V]$, i.e. in the Grothendieck group $\operatorname{Gr}({\mathcal{A}})$ we have

(5.23)$$\begin{eqnarray}[P_{U}]=\mathop{\sum }_{V\in \operatorname{Irr}({\mathcal{A}})}\mathsf{C}({\mathcal{A}})_{UV}\,[V].\end{eqnarray}$$

Since $k$ is algebraically closed, we can write

(5.24)$$\begin{eqnarray}\mathsf{C}({\mathcal{A}})_{UV}=\dim _{k}{\mathcal{A}}(P_{V},P_{U}).\end{eqnarray}$$

In view of Proposition 5.8 and the equivalence (5.8), we have the following corollary to Proposition 5.2.

Corollary 5.9. Let ${\mathcal{A}}$ be a finite abelian category over $k$ such that ${\mathcal{P}}\text{roj}({\mathcal{A}})$ is Calabi–Yau. Then we have the following.

1. (1) The Cartan matrix of ${\mathcal{A}}$ is symmetric, $\mathsf{C}({\mathcal{A}})_{U,V}=\mathsf{C}({\mathcal{A}})_{V,U}$.

2. (2) Let $n=|\text{Irr}({\mathcal{A}})|$ and let $\widehat{\mathsf{C}}\in \text{Mat}_{n}(k)$ be the image of $\mathsf{C}({\mathcal{A}})$ under the canonical homomorphism $\text{Mat}_{n}(\mathbb{Z})\rightarrow \text{Mat}_{n}(k)$. We have

   (5.25)$$\begin{eqnarray}\text{rank}(\widehat{\mathsf{C}})=\dim _{k}\text{Hig}({\mathcal{A}}).\end{eqnarray}$$

In the context of this paper we are particularly interested in the case of factorisable and pivotal finite tensor categories ${\mathcal{C}}$. In this case we will later give a more direct description of $\text{Rey}({\mathcal{C}})$ and $\text{Hig}({\mathcal{C}})$ in terms of the $\unicode[STIX]{x1D719}_{M}$ defined in (2.22), see Proposition 7.1 below. The next proposition shows in particular that Corollary 5.9 is applicable to such ${\mathcal{C}}$.

Proposition 5.10. Let ${\mathcal{C}}$ be a factorisable and pivotal finite tensor category over $k$ which satisfies Condition P. Then,

1. (1) the tensor ideal ${\mathcal{P}}\text{roj}({\mathcal{C}})$ admits a family of traces turning it into a Calabi–Yau category;

2. (2) the category ${\mathcal{C}}$ is semisimple if and only if any one of the inclusions $\operatorname{End}(\text{Id}_{{\mathcal{C}}})\supset \text{Rey}({\mathcal{C}})\supset \text{Hig}({\mathcal{C}})$ is an equality.

Proof. Part (1) is immediate from Corollary 4.7 by setting $\mathsf{t}_{P}:=t_{P}$ for any $P\in {\mathcal{P}}\text{roj}({\mathcal{C}})$, where $t$ is the modified trace on ${\mathcal{P}}\text{roj}({\mathcal{C}})$. For part (2), the equivalence of semisimplicity to $\operatorname{End}(\text{Id}_{{\mathcal{C}}})=\text{Rey}({\mathcal{C}})$ follows from Proposition 5.3 and Lemma 5.1.

To show that $\text{Rey}({\mathcal{C}})=\text{Hig}({\mathcal{C}})$ is equivalent to semisimplicity of ${\mathcal{C}}$, we pick a projective generator $G$ and write $E=\operatorname{End}(G)$. Via the equivalence (5.8) to $\text{mod}\text{-}E$, and by Propositions 5.3 and 5.8, the equality $\text{Rey}({\mathcal{C}})=\text{Hig}({\mathcal{C}})$ is equivalent to $\text{Rey}(E)=\text{Hig}(E)$, and by the isomorphisms in (5.4) in turn to $R(E)=I(E)$.

Let $\widehat{\mathsf{C}}$ be the image of $\mathsf{C}({\mathcal{C}})$ in $\text{Mat}_{n}(k)$ as in Corollary 5.9. We now show the following.

Claim. The equality $R(E)=I(E)$ holds if and only if  $\widehat{\mathsf{C}}$ is non-degenerate.

This claim completes the proof of part (2), since, by [Reference Etingof, Gelaki, Nikshych and OstrikEGNO15, Theorem 6.6.1], a pivotal finite tensor category over $k$ is semisimple if and only if $\widehat{\mathsf{C}}$ is non-degenerate.

Proof of Claim.

Recall from (5.4) that the characters $\unicode[STIX]{x1D712}_{M}$ of $E$-modules $M$ span $R(E)$ (over $k$) and those of projective $E$-modules span $I(E)$. Using again the equivalence (5.8), we have $\unicode[STIX]{x1D712}_{\tilde{P}_{U}}=\sum _{V\in \operatorname{Irr}({\mathcal{C}})}\widehat{\mathsf{C}}_{UV}\,\unicode[STIX]{x1D712}_{\tilde{V}}$, where $\tilde{X}:={\mathcal{C}}(G,X)$ is the image of $X$ under the equivalence (5.8). In particular, the image of the $k$-linear map described by the matrix $\widehat{\mathsf{C}}$ in the basis $\{{\unicode[STIX]{x1D712}_{\tilde{V}}\}}_{V\in \operatorname{Irr}({\mathcal{C}})}$ of $R(E)$, is precisely $I(E)$. We conclude that $R(E)=I(E)$ is equivalent to non-degeneracy of $\widehat{\mathsf{C}}$, proving the claim.◻

Part (2) of the proposition also follows from [Reference ShimizuShi17b, Corollary 4.2 and Theorem 5.12] together with Proposition 7.1 below.Footnote ⁸

6 Characters and the modified trace

In this section we fix

• ∙ ${\mathcal{C}}$: a factorisable and pivotal finite tensor category over $k$ which satisfies Condition P given in the beginning of § 3;

• ∙ $t$: a choice of non-zero modified trace on ${\mathcal{P}}\text{roj}(C)$ via Corollary 4.7;

• ∙ $G$: a projective generator for ${\mathcal{C}}$;

• ∙ $E:=\operatorname{End}(G)$, a symmetric $k$-algebra via $t_{G}$.

Recall from (2.22) the natural endomorphism $\unicode[STIX]{x1D719}_{M}\in \operatorname{End}(\text{Id}_{{\mathcal{C}}})$ assigned to each $M\in {\mathcal{C}}$. Note that $\unicode[STIX]{x1D719}_{M}$ is uniquely determined by $(\unicode[STIX]{x1D719}_{M})_{G}\in E$, and that in fact $(\unicode[STIX]{x1D719}_{M})_{G}\in Z(E)$.

For a given $M\in {\mathcal{C}}$ we can now define two central forms on $E$. Let $x\in E$. The first central form uses the Hom-tensor equivalence (5.8) and is given by $x\mapsto \text{tr}_{{\mathcal{C}}(G,M)}(x)$, the trace in the right $E$-module ${\mathcal{C}}(G,M)$ over the right action of $x$. The second form is $x\mapsto t_{G}((\unicode[STIX]{x1D719}_{M})_{G}\circ x)$. The aim of this section is to prove our second main result.

Theorem 6.1. The modified trace $t_{P_{\mathbf{1}}}((\unicode[STIX]{x1D719}_{\mathbf{1}})_{P_{\mathbf{1}}})$ is non-zero, and for all $M\in {\mathcal{C}}$ the following equality of central forms on $E$ holds:

(6.1)$$\begin{eqnarray}\text{tr}_{{\mathcal{C}}(G,M)}(-)=\frac{t_{G}((\unicode[STIX]{x1D719}_{M})_{G}\circ -)}{t_{P_{\mathbf{1}}}((\unicode[STIX]{x1D719}_{\mathbf{1}})_{P_{\mathbf{1}}})}.\end{eqnarray}$$

The proof relies on several intermediate results, which we present now.

Lemma 6.2. Let $f:A\rightarrow B$ and suppose that for all $X\in {\mathcal{C}}$ we have

(6.2)$$\begin{eqnarray}(f\otimes \operatorname{id}_{X})\circ c_{X,A}\circ c_{A,X}=f\otimes \text{id}_{X}.\end{eqnarray}$$

Then there is $m>0$ and $a:A\rightarrow \mathbf{1}^{\oplus m}$, $b:\mathbf{1}^{\oplus m}\rightarrow B$ such that $f=b\circ a$.

Proof. Write $f=[A\xrightarrow[{}]{a}M\xrightarrow[{}]{b}B]$, where $a$ is epi and $b$ is mono. Rewrite (6.2) as

(6.3)$$\begin{eqnarray}(b\otimes \operatorname{id}_{X})\circ c_{X,M}\circ c_{M,X}\circ (a\otimes \operatorname{id}_{X})=(b\otimes \operatorname{id}_{X})\circ (a\otimes \operatorname{id}_{X}).\end{eqnarray}$$

By exactness of $\otimes$, also $a\,\otimes \,\text{id}_{X}$ is epi and $b\,\otimes \,\text{id}_{X}$ is mono. We conclude that $c_{X,M}\circ c_{M,X}=\operatorname{id}_{M\otimes X}$ for all $M$. From characterisation (1) of factorisability in Theorem 2.8 it follows that $M\cong \mathbf{1}^{\oplus m}$ for some $m$.◻

Recall the coend ${\mathcal{L}}$ with its family $\unicode[STIX]{x1D704}_{X}$ of dinatural transformations and its non-zero cointegral $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}$ from § 2.

Lemma 6.3. For each $X\in {\mathcal{C}}$ there exist $m>0$, $a:X\rightarrow \mathbf{1}^{\oplus m}$, $b:\mathbf{1}^{\oplus m}\rightarrow X$ (possibly zero) such that

(6.4)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{X} & = & \displaystyle \operatorname{ev}_{X}\circ (\operatorname{id}_{X^{\ast }}\otimes (b\circ a))\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}=1}^{m}[X^{\ast }X\xrightarrow[{}]{(b_{\unicode[STIX]{x1D6FC}})^{\ast }\,\otimes \,a_{\unicode[STIX]{x1D6FC}}}\mathbf{1}^{\ast }\mathbf{1}\xrightarrow[{}]{\operatorname{ev}_{\mathbf{1}}}\mathbf{1}],\end{eqnarray}$$

where $a_{\unicode[STIX]{x1D6FC}}:X\rightarrow \mathbf{1}$, $b_{\unicode[STIX]{x1D6FC}}:\mathbf{1}\rightarrow X$ are the components of $a,b$.

Proof. By definition of the cointegral, we have

(6.5)$$\begin{eqnarray}[{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D6E5}_{{\mathcal{L}}}}{\mathcal{L}}{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\,\otimes \,\text{id}_{{\mathcal{L}}}}\mathbf{1}{\mathcal{L}}\xrightarrow[{}]{{\sim}}{\mathcal{L}}]=[{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}}\mathbf{1}\xrightarrow[{}]{\unicode[STIX]{x1D702}_{{\mathcal{L}}}}{\mathcal{L}}].\end{eqnarray}$$

Combining this with the pairing $\unicode[STIX]{x1D714}_{{\mathcal{L}}}$ we obtain the following equality:

(6.6)$$\begin{eqnarray}[{\mathcal{L}}{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D6E5}_{{\mathcal{L}}}\,\otimes \,\text{id}_{{\mathcal{L}}}}({\mathcal{L}}{\mathcal{L}}){\mathcal{L}}\xrightarrow[{}]{{\sim}}{\mathcal{L}}({\mathcal{L}}{\mathcal{L}})\xrightarrow[{}]{\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\,\otimes \,\unicode[STIX]{x1D714}_{{\mathcal{L}}}}\mathbf{1}\mathbf{1}]=[{\mathcal{L}}{\mathcal{L}}\xrightarrow[{}]{\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\,\otimes \,\unicode[STIX]{x1D700}_{{\mathcal{L}}}}\mathbf{1}\mathbf{1}],\end{eqnarray}$$

where we used the property of the Hopf pairing that $\unicode[STIX]{x1D714}_{{\mathcal{L}}}\circ (\unicode[STIX]{x1D702}_{{\mathcal{L}}}\,\otimes \,\text{id}_{{\mathcal{L}}})\circ \unicode[STIX]{x1D706}_{{\mathcal{L}}}^{-1}=\unicode[STIX]{x1D700}_{{\mathcal{L}}}$. After pre-composing with $\unicode[STIX]{x1D704}_{X}\otimes \unicode[STIX]{x1D704}_{Y}$ and substituting the defining equations, a short calculation shows

(6.7)$$\begin{eqnarray}\displaystyle & & \displaystyle [\,(X^{\ast }X)(Y^{\ast }Y)\xrightarrow[{}]{{\sim}}X^{\ast }((XY^{\ast })Y)\xrightarrow[{}]{\text{id}\otimes (c_{Y^{\ast },X}\circ c_{X,Y^{\ast }})\,\otimes \,\text{id}}X^{\ast }((XY^{\ast })Y)\nonumber\\ \displaystyle & & \displaystyle \qquad \xrightarrow[{}]{{\sim}}(X^{\ast }X)(Y^{\ast }Y)\xrightarrow[{}]{(\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{X})\,\otimes \,\text{ev}_{Y}}\mathbf{1}\mathbf{1}]\nonumber\\ \displaystyle & & \displaystyle \quad =[(X^{\ast }X)(Y^{\ast }Y)\xrightarrow[{}]{(\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{X})\,\otimes \,\text{ev}_{Y}}\mathbf{1}\mathbf{1}].\end{eqnarray}$$

Using the isomorphism ${\mathcal{C}}(X^{\ast }\otimes M,N)\xrightarrow[{}]{{\sim}}{\mathcal{C}}(M,X\otimes N)$,

(6.8)$$\begin{eqnarray}h\longmapsto [M\xrightarrow[{}]{{\sim}}\mathbf{1}M\xrightarrow[{}]{\text{coev}_{X}\,\otimes \,\text{id}}(XX^{\ast })M\xrightarrow[{}]{{\sim}}X(X^{\ast }M)\xrightarrow[{}]{\text{id}\otimes h}XN],\end{eqnarray}$$

and similarly for ${\mathcal{C}}(M\otimes Y,N)\xrightarrow[{}]{{\sim}}{\mathcal{C}}(M,N\otimes Y^{\ast })$, the latter equality can be brought to the form $(f\otimes \operatorname{id}_{Y^{\ast }})\circ c_{Y^{\ast },X}\circ c_{X,Y^{\ast }}=f\otimes \operatorname{id}_{Y^{\ast }}$ with

(6.9)$$\begin{eqnarray}f=[X\xrightarrow[{}]{{\sim}}\mathbf{1}X\xrightarrow[{}]{\text{coev}_{X}\,\otimes \,\text{id}}(XX^{\ast })X\xrightarrow[{}]{{\sim}}X(X^{\ast }X)\xrightarrow[{}]{\text{id}\otimes (\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{X})}X\mathbf{1}\xrightarrow[{}]{{\sim}}X].\end{eqnarray}$$

Lemma 6.2 now implies the claim. ◻

Corollary 6.4. For $U\in {\mathcal{C}}$ simple, $U\ncong \mathbf{1}$, we have $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{P_{U}}=0$. Furthermore, there exists a unique $\unicode[STIX]{x1D708}_{\mathbf{1}}:\mathbf{1}\rightarrow P_{\mathbf{1}}$ such that

(6.10)$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{P_{\mathbf{1}}}=\operatorname{ev}_{\mathbf{1}}\circ (\unicode[STIX]{x1D708}_{\mathbf{1}}^{\ast }\otimes \unicode[STIX]{x1D70B}_{\boldsymbol{ 1}}),\end{eqnarray}$$

where $\unicode[STIX]{x1D70B}_{\mathbf{1}}:P_{\mathbf{1}}\rightarrow \mathbf{1}$ is the canonical projection of the projective cover.

Recall the natural endomorphism $\unicode[STIX]{x1D719}_{M}$ in (2.23). The next proposition computes its values on projective objects and also shows that $\unicode[STIX]{x1D708}_{\mathbf{1}}$ (and all $\unicode[STIX]{x1D708}_{U}$ defined there) are non-zero.

Proposition 6.5. Let $U,V\in \operatorname{Irr}({\mathcal{C}})$. Then $(\unicode[STIX]{x1D719}_{U})_{P_{U}}\neq 0$ and there exists a unique non-zero $\unicode[STIX]{x1D708}_{U}:U\rightarrow P_{U}$ such that

(6.11)$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{U})_{P_{V}}=\left\{\begin{array}{@{}ll@{}}0,\quad & V\neq U,\\ \unicode[STIX]{x1D708}_{U}\circ \unicode[STIX]{x1D70B}_{U},\quad & V=U,\end{array}\right.\end{eqnarray}$$

where $\unicode[STIX]{x1D70B}_{U}:P_{U}\rightarrow U$ is the canonical projection of the projective cover, and where $\unicode[STIX]{x1D708}_{\mathbf{1}}$ is the same as in Corollary 6.4. Moreover, there exists a choice of $a:P_{\mathbf{1}}\rightarrow P_{U}\otimes U^{\ast }$, $b:P_{U}\otimes U^{\ast }\rightarrow P_{\mathbf{1}}$ which realise $P_{\mathbf{1}}$ as a direct summand of $P_{U}\otimes U^{\ast }$ (i.e. $b\circ a=\operatorname{id}_{P_{\mathbf{1}}}$), such that

(6.12)

Proof. In the expression (2.23) for $(\unicode[STIX]{x1D719}_{M})_{X}$ set $M=U$ and $X=P_{V}$. Write $P_{V}\otimes U^{\ast }\cong \bigoplus _{W\in \operatorname{Irr}({\mathcal{C}})}P_{W}^{\oplus n_{W}}$ and pick $a_{W,\unicode[STIX]{x1D6FC}}\in {\mathcal{C}}(P_{W},P_{V}\otimes U^{\ast })$ and $b_{W,\unicode[STIX]{x1D6FC}}\in {\mathcal{C}}(P_{V}\otimes U^{\ast },P_{W})$, $\unicode[STIX]{x1D6FC}=1,\ldots ,n_{W}$, which realise the direct sum decomposition:

(6.13)$$\begin{eqnarray}b_{W,\unicode[STIX]{x1D6FC}}\circ a_{W^{\prime },\unicode[STIX]{x1D6FC}^{\prime }}=\unicode[STIX]{x1D6FF}_{W,W^{\prime }}\,\unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FC}^{\prime }}\text{id}_{P_{W}},\quad \mathop{\sum }_{W\in \operatorname{Irr}({\mathcal{C}})}\mathop{\sum }_{\unicode[STIX]{x1D6FC}=1}^{n_{W}}a_{W,\unicode[STIX]{x1D6FC}}\circ b_{W,\unicode[STIX]{x1D6FC}}=\operatorname{id}_{P_{V}\,\otimes \,U^{\ast }}\!.\end{eqnarray}$$

For the dinatural transformation $\unicode[STIX]{x1D704}_{P_{V}\,\otimes \,U^{\ast }}$ in (2.23) consider the equalities

(6.14)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D704}_{P_{V}\,\otimes \,U^{\ast }} & = & \displaystyle \mathop{\sum }_{W\in \operatorname{Irr}({\mathcal{C}})}\mathop{\sum }_{\unicode[STIX]{x1D6FC}=1}^{n_{W}}\unicode[STIX]{x1D704}_{P_{V}\,\otimes \,U^{\ast }}\circ (\operatorname{id}_{(P_{V}\,\otimes \,U^{\ast })^{\ast }}\otimes (a_{W,\unicode[STIX]{x1D6FC}}\circ b_{W,\unicode[STIX]{x1D6FC}}))\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{W\in \operatorname{Irr}({\mathcal{C}})}\mathop{\sum }_{\unicode[STIX]{x1D6FC}=1}^{n_{W}}\unicode[STIX]{x1D704}_{P_{W}}\circ (a_{W,\unicode[STIX]{x1D6FC}}^{\ast }\otimes b_{W,\unicode[STIX]{x1D6FC}}).\end{eqnarray}$$

From Corollary 6.4, $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{P_{W}}=0$ for $W\ncong \mathbf{1}$, and so in the expression for $\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{P_{V}\otimes U^{\ast }}$ the sum over $W$ in (6.14) reduces to $W=\mathbf{1}$. Since $\dim {\mathcal{C}}(P_{W},R)=\unicode[STIX]{x1D6FF}_{W,R}$ for $W,R\in \operatorname{Irr}({\mathcal{C}})$, the multi-plicities $n_{W}$ are given by $n_{W}=\dim {\mathcal{C}}(P_{V}\otimes U^{\ast },W)$. In particular,

(6.15)$$\begin{eqnarray}n_{\mathbf{1}}=\dim {\mathcal{C}}(P_{V}\otimes U^{\ast },\mathbf{1})=\dim {\mathcal{C}}(P_{V},U)=\unicode[STIX]{x1D6FF}_{U,V}.\end{eqnarray}$$

So in (6.14) the sum over $\unicode[STIX]{x1D6FC}$ is zero for $U\neq V$ and reduces to $\unicode[STIX]{x1D6FC}=1$ for $U=V$. Together with (6.10) we obtain

(6.16)$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{{\mathcal{L}}}^{\text{co}}\circ \unicode[STIX]{x1D704}_{P_{V}\,\otimes \,U^{\ast }}=\unicode[STIX]{x1D6FF}_{U,V}\,\operatorname{ev}_{\mathbf{1}}\circ ((a_{\mathbf{1},1}\circ \unicode[STIX]{x1D708}_{\mathbf{1}})^{\ast }\otimes (\unicode[STIX]{x1D70B}_{\boldsymbol{ 1}}\circ b_{\mathbf{1},1})).\end{eqnarray}$$

Substituting this into (2.23) yields $(\unicode[STIX]{x1D719}_{U})_{P_{V}}=\unicode[STIX]{x1D6FF}_{U,V}\,\tilde{\unicode[STIX]{x1D708}}\circ \tilde{\unicode[STIX]{x1D70B}}$ where

(6.17)

It remains to show that we can achieve $\tilde{\unicode[STIX]{x1D70B}}=\unicode[STIX]{x1D70B}_{U}$, the chosen projection $P_{U}\rightarrow U$, and that for $U=\mathbf{1}$, $\tilde{\unicode[STIX]{x1D708}}$ agrees with $\unicode[STIX]{x1D708}_{\mathbf{1}}$ in (6.10).

By Theorem 2.6, in particular $\unicode[STIX]{x1D712}_{U}\neq 0$. Since $\unicode[STIX]{x1D713}$ and $\unicode[STIX]{x1D70C}$ are isomorphisms, also $\unicode[STIX]{x1D719}_{U}\neq 0$, and thus $\tilde{\unicode[STIX]{x1D708}},\tilde{\unicode[STIX]{x1D70B}}\neq 0$ (or otherwise $\unicode[STIX]{x1D719}_{U}$ would vanish on all indecomposable projectives and hence be zero). Since ${\mathcal{C}}(P_{U},U)$ is one-dimensional, by rescaling $b_{\mathbf{1},1}$ if necessary (and thus rescaling $a_{\mathbf{1},1}$ accordingly by the inverse factor) one can achieve $\tilde{\unicode[STIX]{x1D70B}}=\unicode[STIX]{x1D70B}_{U}$. This finally gives (6.12).

For $U=V=\mathbf{1}$, one can alternatively compute $(\unicode[STIX]{x1D719}_{\mathbf{1}})_{P_{\mathbf{1}}}$ directly from (2.23) and (6.10). A short calculation confirms that $\unicode[STIX]{x1D708}_{\mathbf{1}}$ from (6.10) agrees with $\tilde{\unicode[STIX]{x1D708}}$.◻