Tensor and Fusion Categories

Two cyclic linear functionals. Put $t_U(z)={\mathrm{Tr}}_L(a_{U}z)$ and $r_U(z)={\mathrm{Tr}}_R(z{a}_U^{-1})$. Bilinearity of tensor product makes these $k$-linear. If $f:U\to V$ and $g:V\to U$, generalized cyclicity in [L2] and $a_{V}f=f^{\vee \vee }{a}_U$ give $t_U(gf)=t_V(fg)$; the mirrored mate calculation gives $r_U(gf)=r_V(fg)$. Both functionals are additive under finite direct sums. Naturality of $a$ at evaluation, together with monoidality, says that $a_{U^{\vee }}$ is the inverse of $a_U^{\vee }$: after inserting either into evaluation the same pairing results, and uniqueness of mates identifies them. The dual-trace identity in [L2], applied on $U^{\vee }$, therefore gives $r_U(1_U)={\dim }_a(U^{\vee })={\dim }_a(U)=t_U(1_U)$.

Nilpotents have zero trace. We prove simultaneously that $t_U(n)=r_U(n)=0$ for any nilpotent endomorphism $n$ by induction on the finite length of $U$. The zero object and $n=0$ give zero by linearity. Otherwise factor $n=ip$ through its image $W$, with $p:U\twoheadrightarrow W$ and $i:W\hookrightarrow U$. Since $n$ is nilpotent and $U\mathrm{\ne }0$, its image $W$ is a proper subobject of $U$: otherwise $n$ would be epic and a power would be both zero and epic. Thus $W$ has smaller length than $U$. The endomorphism $pi$ of $W$ is nilpotent, because $(pi{)}^{m+1}=p{n}^{m}i=0$ for a sufficiently large $m$. Step 1.1 gives $t_U(n)=t_W(pi)=0$ and $r_U(n)=r_W(pi)=0$ by induction.

Primary summands. Finite-dimensionality of $\mathrm{End}(X)$ gives a nonzero polynomial annihilating $x$. Since $k$ is algebraically closed, it factors into powers of the distinct linear factors $T-\lambda$. The Chinese remainder identity for these pairwise coprime factors supplies mutually orthogonal polynomial idempotents $e_{\lambda }(x)$ summing to $1_X$ and with $(x-\lambda {)}^{m_{\lambda }}{e}_{\lambda }(x)=0$. Idempotents split in an abelian category: $X={⨁}_{\lambda }\mathrm{im}{e}_{\lambda }(x)$, since the sum of their inclusion/projection pairs is the identity. On the summand $X_{\lambda }$, write $x=\lambda 1+n_{\lambda }$ with $n_{\lambda }$ nilpotent. Additivity, linearity and step 2.1 give $t_X(x)={\sum }_{\lambda }\lambda t_{X_{\lambda }}(1)$ and $r_X(x)={\sum }_{\lambda }\lambda r_{X_{\lambda }}(1)$. Step 1.1 equates each pair of summands, proving the displayed identity.

By [F1] choose left and right duals of $X$. By [L1], each of $X\otimes -$ and $-\otimes X$ is both a left and a right adjoint (using the appropriate dual).

Thus each tensoring functor preserves kernels and cokernels by [L2]. It is additive because the tensor product is $k$-bilinear.

By [L3] both functors are exact. Since $X$ was arbitrary, tensoring is exact in either variable.

Rigidity makes the contravariant functor of [L1] a duality: its quasi-inverse is the chosen right-dual functor. Thus it is an equivalence ${\mathcal{C}}^{\mathrm{o}\mathrm{p}}\simeq \mathcal{C}$.

Both source and target are abelian, so [L2] makes this equivalence exact. Equivalently, a short exact sequence is carried, with arrows reversed, to a short exact sequence.

Factor $f$ and $g$ as an epimorphism followed by a monomorphism through their images, as specified by [F2].

Exactness in [F1] preserves those epimorphisms and monomorphisms after tensoring, first in one variable and then in the other. Hence $f\otimes g$ factors as an epimorphism onto $\mathrm{im}(f)\otimes \mathrm{im}(g)$ followed by a monomorphism.

In an abelian category this epi--mono factorization identifies its middle object with the image. Therefore the displayed canonical map is an isomorphism.

The left dual gives $\mathrm{Hom}(P\otimes X,-)\cong \mathrm{Hom}(P,-\otimes X^{\vee }),$ while the mirrored adjunction for the right dual gives $\mathrm{Hom}(X\otimes P,-)\cong \mathrm{Hom}(P,{}^{\vee }X\otimes -)\mathrm{.}$

An epimorphism remains an epimorphism after applying either exact tensor functor on the right sides of step 1.1. Then [F2] lifts every map out of $P$, and transport through the corresponding adjunction proves the lifting property for both stated tensor products.

If $\mathbf{1}$ is projective, every $X\cong \mathbf{1}\otimes X$ is projective by [F1]. In this finite-length abelian setting, all objects projective implies that all short exact sequences split, hence every object is a direct sum of its simple factors and is semisimple in the sense of [F2].

Conversely, in a semisimple abelian category every epimorphism splits after decomposing its codomain into simples; thus every object, in particular $\mathbf{1}$, is projective.

The Eckmann--Hilton argument makes $E=\mathrm{End}(\mathbf{1})$ a finite-dimensional commutative $k$-algebra. If $a^2=0$, put $J=\mathrm{im}a$ and $K=\ker a$. By [F3], $J\otimes J=0$ and $K\otimes J=0$. Tensoring $0\to K\to \mathbf{1}\to J\to 0$ by $J$ and using [F2] then gives $J=0$, hence $a=0$. A nonzero nilpotent has a nonzero square-zero power, so $E$ is reduced. Thus the commutative Artinian algebra $E$ is a finite product of fields. Its primitive idempotents split $\mathbf{1}$ as a finite direct sum of indecomposable component units ${\mathbf{1}}_i$, each with $\mathrm{End}({\mathbf{1}}_i)$ a field (not necessarily $k$).

Fix a component and a simple subobject $S\subseteq {\mathbf{1}}_i$, which exists by [F1]. Dualizing $0\to S\to {\mathbf{1}}_i\to Q\to 0$ and then tensoring on the left by $S$ gives an exact sequence $0\to S\otimes Q^{\vee }\to S\to S\otimes S^{\vee }\to 0$. The last object is nonzero by the coevaluation zig-zag, so simplicity of $S$ makes $S\to S\otimes S^{\vee }$ an isomorphism.

The coevaluation followed by the inverse of the isomorphism in step 2.1 is a nonzero epimorphism $p:{\mathbf{1}}_i\twoheadrightarrow S$. If $j:S\hookrightarrow {\mathbf{1}}_i$ is the inclusion, then $jp$ is a nonzero element of the field $\mathrm{End}({\mathbf{1}}_i)$, hence an isomorphism. Therefore $j$ is also epic and thus an isomorphism. So every component unit is simple, and step 1.1 together with [F4] makes $\mathbf{1}$ semisimple.

By [F1], write $\mathbf{1}$ as a finite direct sum of simple objects. A decomposition with at least two nonzero summands supplies a nontrivial idempotent projection in $\mathrm{End}(\mathbf{1})$.

But [F2] identifies this endomorphism algebra with the field $k$, whose only idempotents are $0$ and $1$. Thus there is one nonzero summand, and $\mathbf{1}$ is simple.

The zig-zag identity shows that evaluation is nonzero: otherwise its composite giving $1_X$ would vanish. Its image is therefore a nonzero subobject of the simple object $\mathbf{1}$, hence all of $\mathbf{1}$ by [F1]. Thus evaluation is epic.

The other zig-zag identity shows directly that coevaluation is nonzero: if it vanished, its composite giving $1_X$ would vanish. Since its source $\mathbf{1}$ is simple by [F1], its kernel is either $0$ or $\mathbf{1}$; the nonzero map excludes the latter. Thus coevaluation is monic.

If $X\mathrm{\ne }0$, [F2] and exactness imply that $F(\mathbf{1})\to F(X\otimes X^{\vee })$ is monic. Strong monoidality identifies its source with the nonzero unit of $\mathcal{D}$, so its target is nonzero. Since $F(X\otimes X^{\vee })\cong F(X)\otimes F(X^{\vee })$, this forces $F(X)\mathrm{\ne }0$.

If $F(f)=0$, exactness gives $F(\mathrm{im}f)=0$. Step 1.1 therefore implies $\mathrm{im}f=0$, hence $f=0$ in the abelian source category. Thus $F$ is faithful.

If $0\to A\to B\to C\to 0$ is exact, then [F1] makes each of its tensors with $X$ exact. Thus $[B\otimes X]=[A\otimes X]+[C\otimes X]$, and similarly in the other variable.

Hence the bilinear rule on generators descends through the relations of [F2]. The associator and unitors identify $(X\otimes Y)\otimes Z$ with $X\otimes (Y\otimes Z)$ and $\mathbf{1}\otimes X$ with $X$, so the descended product is associative with unit $[\mathbf{1}]$.

By [F1], dualization takes each short exact relation in [F3] to a short exact relation, so $[X]\mapsto [X^{\vee }]$ is a well-defined additive map.

By [F2], it reverses products: $([X][Y]{)}^{\ast }=[Y{]}^{\ast }[X{]}^{\ast }$. Chosen right duals give the inverse map on isomorphism classes and hence on $K_0$. Therefore this is an anti-isomorphism; no identification of $X$ with $X^{\vee \vee }$ is asserted.

By semisimplicity, decompose $X$ into simple summands. For a simple summand $V$ and a simple object $Z$, dual transposition from [F2] gives $\mathrm{Hom}(\mathbf{1},V\otimes Z)\cong \mathrm{Hom}(V^{\vee },Z),\mathrm{Hom}(V\otimes Z,\mathbf{1})\cong \mathrm{Hom}(Z,{}^{\vee }V)\mathrm{.}$ Rigidity makes left and right dualization quasi-inverse contravariant equivalences, and exactness in [F3] therefore preserves simple objects. Hence the first space is nonzero exactly when $Z\cong V^{\vee }$, and the second exactly when $Z\cong {}^{\vee }V$.

This comparison does not require $k$ to be algebraically closed. Write the semisimple objects as $\mathbf{1}\cong {⨁}_i{S}_i^{\oplus a_i}$ and $V\otimes Z\cong {⨁}_i{S}_i^{\oplus b_i}$ with pairwise nonisomorphic simples $S_i$. Then both ${\dim }_k\mathrm{Hom}(\mathbf{1},V\otimes Z)$ and ${\dim }_k\mathrm{Hom}(V\otimes Z,\mathbf{1})$ equal ${\sum }_i{a}_i{b}_i{\dim }_k\mathrm{End}(S_i)$. Hence the two unique simples detected in step 1.1 agree: $V^{\vee }\cong {}^{\vee }V$. Taking finite direct sums over the simple summands gives the result for $X$.

Let $A=k[\epsilon ]\mathrm{/}({\epsilon }^2)$ and let $\mathcal{C}=A\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$ be the category of finite-dimensional left $A$-modules. It is finite: $A$ is finite-dimensional, it has the single simple module $S=A\mathrm{/}(\epsilon )$, and the quotient $A\twoheadrightarrow S$ is its projective cover.

The sequence $0\to S\stackrel{1\mapsto \epsilon }{\to }A\to S\to 0$ does not split. Indeed, every vector killed by $\epsilon$ in $A$ is a multiple of $\epsilon$ and maps to zero in the quotient, so no section exists. Thus $A$ is not a direct sum of simples, and [F2] says that $\mathcal{C}$ is not semisimple.

Take $\mathcal{D}={\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k^{\mathrm{f}\mathrm{d}}\times {\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k^{\mathrm{f}\mathrm{d}}$ with componentwise tensor product. It is $k$-linear and abelian, and $(V,W)$ has dual $(V^{\vee },W^{\vee })$, so it is rigid.

Its unit is $(k,k)$, whose endomorphism algebra is $k\times k$, not $k$. Thus $\mathcal{D}$ fails the scalar-unit condition in [F2] and is not a tensor category.

Over $k=\mathbb{C}$, send a finite-dimensional vector space $V$ to its conjugate vector space $\bar{V}$ and a linear map to the same underlying additive map. The canonical maps $\bar{V}\otimes \bar{W}\cong \overline{V\otimes W}$ and $\mathbb{C}\cong \bar{\mathbb{C}}$ make this a strong monoidal endofunctor of ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_{\mathbb{C}}^{\mathrm{f}\mathrm{d}}$.

It is not $\mathbb{C}$-linear on hom-spaces: for a nonreal scalar $\lambda$, the images of $\lambda f$ and of $\lambda F(f)$ differ. Hence it fails the $k$-linearity required by [F2], so a strong monoidal functor need not be a tensor functor under this convention.

Let $G=S_3$ and let ${\mathbf{V}\mathbf{e}\mathbf{c}}_G$ be the category of finite-dimensional $G$-graded vector spaces. Its simple objects ${\delta }_g$ are indexed by $g\in G$, and ${\delta }_g\otimes {\delta }_h\cong {\delta }_{gh}$; it is a tensor category.

Thus $K_0({\mathbf{V}\mathbf{e}\mathbf{c}}_G)\cong \mathbb{Z}[G]$. Since $S_3$ is nonabelian, for example $(12)(23)\mathrm{\ne }(23)(12)$, this ring is not commutative. The order reversal in [F1] is consistent with, but does not remove, this counterexample.

[F1] produces isomorphisms separately for objects.

By [F2], these need not be natural in morphisms or multiplicative under tensor product. They therefore need not form a pivotal structure.

Let $\mathcal{C}$ be the tensor category of finite-dimensional $\mathbb{Z}$-graded vector spaces over a field $k$, and fix $q\in k^{\times }$ with $q\mathrm{\ne }\pm 1$. The usual double-dual map is a pivotal structure, and multiplying its component on degree-$n$ vectors by $q^n$ gives another pivotal structure $a^{\mathrm{\prime }}$.

Let $X=k[1]$ be the one-dimensional object concentrated in degree $1$. Then $X^{\vee }=k[-1]$, so ${\dim }_{a^{\mathrm{\prime }}}(X)=q,{\dim }_{a^{\mathrm{\prime }}}(X^{\vee })=q^{-1}\mathrm{.}$ Since $q\mathrm{\ne }{q}^{-1}$, the pivotal structure $a^{\mathrm{\prime }}$ is not spherical.

For the identity endomorphism $1_X$, the two trace expressions are exactly those two dimensions: ${\mathrm{Tr}}_L(a_X^{\mathrm{\prime }}{1}_X)=q,{\mathrm{Tr}}_R(1_X(a_X^{\mathrm{\prime }}{)}^{-1})=q^{-1}\mathrm{.}$ They are unequal, so left and right traces do not always agree. This also supplies a witness over $k=\mathbb{C}$, within the ground-field convention of In a spherical category the left and right traces agree; the pivotal structure is deliberately non-spherical.

By [L1], the standard pivotal structure on graded vector spaces can be multiplied by the monoidal automorphism acting as $q^n$ in degree $n$. This gives the modified pivotal structure $a^{\mathrm{\prime }}$ used in [L3].

The standard pivotal structure gives $\dim (X)=1_k$ for the degree-$1$ line $X=k[1]$, while [L3] gives ${\dim }_{a^{\mathrm{\prime }}}(X)=q\mathrm{\ne }{1}_k$. These are exactly the quantities defined in [L2].

Therefore the dimension can change when the pivotal structure changes, so the statement is false.