2.1 Auxiliary lemmas

Let $\mathcal {C}$ be as in the last section. By convention, we suppress all associators and unitors. Let $X\in \mathcal {C}$ be non-negligible, that is, there is an indecomposable $Y\in \mathcal {C}$ such that is a direct summand of $Y\otimes X\cong X\otimes Y$ . We denote X by an upwards oriented strand and Y by a downwards oriented strand. We begin with a well-known observation.

Proof. By rescaling, we may assume $z=\operatorname {\mathrm {id}}_{X}$ . Then is a subdiagram of $z^2=z=\operatorname {\mathrm {id}}_X$ , which implies $z'\neq 0$ . Since $z^{\prime 2}=z'$ , $z'=\operatorname {\mathrm {id}}_Y$ as Y is indecomposable. Using the braiding,

Thus X is rigid with ${}^*X\cong Y\cong X^*$ .

For $W,Z\in \mathcal {C}$ , a morphism $r: W\to Z$ is called a retraction or a split surjection if it admits a splitting $s: Z\to W$ such that $rs = \operatorname {\mathrm {id}}_Z$ . Clearly, the existence of a retraction $r:W\to Z$ is equivalent to Z being a direct summand of W. Thus there exists a retraction . Next, we carefully choose one via the following lemma.

Let $R(X)\subset \operatorname {\mathrm {End}}(X)$ be the nilradical, and choose a retraction and splitting satisfying Lemma 2.2 for $Z=X\otimes Y$ and $N:=R(X)\otimes \operatorname {\mathrm {id}}_Y\subset \operatorname {\mathrm {End}}(X\otimes Y)$ . We prove the contrapositive of Theorem 1.1, that is, if X is not rigid, then $\dim \operatorname {\mathrm {End}}(X^{\otimes n})\ge n!$ for all $n\in \Bbb N$ . Using the braiding, we get a retraction with a splitting:

Observe that if any of the following morphisms is invertible, then after redefining , X is rigid with ${}^*X\cong Y\cong X^*$ by Lemma 2.1.

(2.1)

Recall that since X is indecomposable, the finite-dimensional algebra $\operatorname {\mathrm {End}}(X)$ is local, hence every element of $\operatorname {\mathrm {End}}(X)$ is either invertible or nilpotent [Reference Assem, Simson and SkowrońskiASS06, Cor. 4.8(b)]. We may thus assume that the morphisms in (2.1) are all nilpotent, that is, lie in $R(X)$ .

Now consider the n-strand braid group

Given an element of $B_n$ , we number its strands from left to right on the bottom; for example, $\sigma _i$ swaps the i-th and $i+1$ -th strands. The standard set embedding of $S_n$ into $B_n$ is given by taking a reduced word $w\in S_n$ and writing the same word in $B_n$ . In braid diagrams, these are the elements that can be written in the following form:

• • any two distinct strands cross at most once, and

• • if the i-th and j-th strands cross with $i<j$ , then the i-th strand passes over the j-th strand.

Every element of $S_n\subset B_n$ can be written uniquely as $w_{n-1}w_{n-2}\cdots w_2w_1$ where for each index $1\leq j\leq n-1$ , denoting multiple strands by thick colored strands,

(2.2)

We identify $B_n,S_n$ with their respective images in $B_{n+1},S_{n+1}$ under adding a strand to the right.

Proof. If $s^{-1}t\in B_{n-1}$ , we are finished. Otherwise, since the n-th strands of $s,t$ pass behind all other strands, there are $s',t'\in S_{n-1}$ such that $s= \sigma _i \sigma _{i+1}\cdots \sigma _{n-1}s'$ and $t= \sigma _j \sigma _{j+1}\cdots \sigma _{n-1}t'$ with $i\neq j$ . Without loss of generality, we may assume $s=\sigma _i \sigma _{i+1}\cdots \sigma _{n-1}$ and $t=\sigma _j \sigma _{j+1}\cdots \sigma _{n-1}$ with $i<j$ . The result now follows by inspection; representing multiple strands by thick lines, we see

2.2 Proof of Theorem 1.1

As above, we assume the morphisms in (2.1) are nilpotent, that is, lie in the nilradical $R(X)\subset \operatorname {\mathrm {End}}(X)$ . For $t\in S_n$ , let $f_t\in \operatorname {\mathrm {End}}(X^{\otimes n})$ denote the corresponding morphism. We claim that the set is linearly independent, which implies the result.

Consider the bilinear form given by

For $s,t\in S_n$ , we claim that $ \Phi (f_s^{-1},f_t) = \delta _{s,t}$ , which immediately implies that is linearly independent:

$$\begin{align*}0 = \sum_{t\in S_n} \lambda_t f_t \qquad\Longrightarrow\qquad 0 = \Phi\left( f_s^{-1}, \sum_{t\in S_n} \lambda_t f_t \right ) = \sum_{t\in S_n} \lambda_t \delta_{s,t} = \lambda_s \qquad \forall s\in S_n. \end{align*}$$

Clearly $\Phi (f_s^{-1}, f_s) = 1$ by construction. If $s\neq t$ , then writing $s=v_{n-1}\cdots v_{1}$ and $t=w_{n-1}\cdots w_{1}$ as in (2.2), pick k maximal such that $w_k\neq v_k$ . Then $s^{-1}t$ lies in the image of $S_k$ included into $S_n$ , but the k-th lower boundary point of $s^{-1}t$ does not connect to the k-th upper boundary point. By Lemma 2.3, there are $u,v\in B_{k-1}\subset B_n$ such that $s^{-1}t=u\sigma _{k-1}^{\pm 1}v$ . Then has $n-k$ closed loops which can be removed, but the k-th strand performs a self-crossing of the form

which lie in $R(X)$ by assumption. By naturality of the braiding, there is an such that

2.3 Non-negligible rigid objects

Let $\mathbb {k}$ be algebraically closed. It is well known that when $\mathcal {C}$ is braided and $X\in \mathcal {C}$ is rigid, we have a distinguished quantum trace using the Drinfeld isomorphism $u_X: X\to X^{**}$ :

(2.3)

It is straightforward to verify that $\operatorname {\mathrm {Tr}}_X(fg)=\operatorname {\mathrm {Tr}}_X(gf)$ for all $f,g\in \operatorname {\mathrm {End}}(X)$ .

Proof. (i) If such an f exists, we can take $Y=X^*$ , $r:= \operatorname {\mathrm {ev}}_{X^{*}} (u_X f \otimes \operatorname {\mathrm {id}}_{X^*})$ and $s:= \operatorname {\mathrm {Tr}}_X(f)^{-1} \operatorname {\mathrm {coev}}_X$ so that $rs=1$ , and thus X is non-negligible. Conversely, if is a retraction with splitting , then set

(2.4)

(ii) If is a retraction with splitting , then the f from (2.4) above is invertible as $\operatorname {\mathrm {Tr}}_X(f)=1$ , $\ker (\operatorname {\mathrm {Tr}}_X)=R(X)$ , and X is indecomposable. Dualizing, we see that $f^*: X^*\to X^*$ is an isomorphism. But $f^*$ can be written as $g\circ h$ , where

This implies that g is a retraction with splitting $h\circ (f^*)^{-1}$ , so h identifies $X^*$ with a direct summand of Y. Since Y is indecomposable, it follows that h is an isomorphism.

2.4 Proof of Corollary 1.5

When X is rigid and non-negligible, then $\operatorname {\mathrm {Tr}}_X: \operatorname {\mathrm {End}}(X)\to \mathbb {k}$ is nonzero by Lemma 2.5. Since $\dim \operatorname {\mathrm {End}}(X)<\infty $ and $R(X)$ is a maximal ideal as X is indecomposable, it suffices to prove that when X is rigid, non-negligible, and $\dim \operatorname {\mathrm {End}}(X^{\otimes n})<n!$ for some n, then $R(X)\subseteq \ker (\operatorname {\mathrm {Tr}}_X)$ . To do so, we show that if X is rigid and non-negligible and there is a nilpotent $f\in R(X)$ with $\operatorname {\mathrm {Tr}}_X(f)\neq 0$ , then $\dim \operatorname {\mathrm {End}}(X^{\otimes n})\ge n!$ for all n.

First, we mimic the proof of Lemma 2.2. Set $Y=X^*$ , $r:= \operatorname {\mathrm {ev}}_{X^{*}}(u_X\otimes \operatorname {\mathrm {id}}_{X^*})$ and $N:=R(X)\otimes \operatorname {\mathrm {id}}_{X^*}$ . Let $k\in \mathbb {N}$ be minimal such that $rN^ka= 0$ for all . Since $\operatorname {\mathrm {Tr}}_X(f)\neq 0$ , taking $a=\operatorname {\mathrm {coev}}_X$ shows that $k\geq 2$ . Pick $g\in R(X)^{k-1}$ and so that $r(g\otimes \operatorname {\mathrm {id}}_{X^*})s_0=1$ . Setting $r_0:= r(g\otimes \operatorname {\mathrm {id}}_{X^*})$ , we have $r_0s_0=1$ and $r_0Ns_0=0$ . Observe that

both lie in $R(X)$ as they contain $g\in R(X)^{k-1}\subset R(X)$ . Now by mimicking the proof of Theorem 1.1, we can use $r_0,s_0$ in place of the cap/cup to see that $\dim \operatorname {\mathrm {End}}(X^{\otimes n})\ge n!$ for all n.

2.5 Proof of Lemma 1.6

Suppose

$$ \begin{align*}Z=\bigoplus_{i=1}^r V_i\otimes X_i, \end{align*} $$

where $X_i$ are pairwise nonisomorphic indecomposable objects of $\mathcal {C}$ and $V_i$ are finite-dimensional vector spaces. Then the nilradical $R(Z)$ of $\operatorname {\mathrm {End}}(Z)$ has the form

$$ \begin{align*}R(Z)=\bigoplus_{i=1}^r \operatorname{\mathrm{End}}(V_i)\otimes R(X_i) \oplus \bigoplus_{i\ne j}\operatorname{\mathrm{Hom}}(V_i,V_j)\otimes \operatorname{\mathrm{Hom}}(X_i,X_j). \end{align*} $$

So $\operatorname {\mathrm {Tr}}_Z$ vanishes on $R(Z)$ , and on the quotient

$$ \begin{align*}\operatorname{\mathrm{End}}(Z)/R(Z)\cong \bigoplus_{i=1}^r \operatorname{\mathrm{End}}(V_i) \end{align*} $$

induces the linear functional $T(f_1,\cdots ,f_r):=\sum _{i=1}^r c_i\operatorname {\mathrm {Tr}}_{\operatorname {\mathrm {End}}(V_i)}(f_i)$ , where $c_i:=\operatorname {\mathrm {Tr}}_{X_i}(\mathrm {id}_{X_i})$ . If $f: Z\to Z$ is a nilpotent endomorphism, then its image $(f_1,\dots , f_r)\in \operatorname {\mathrm {End}}(Z)/R(Z)$ satisfies that $f_i\in \operatorname {\mathrm {End}}(V_i)$ is nilpotent for each i, and thus $\operatorname {\mathrm {Tr}}_{\operatorname {\mathrm {End}}(V_i)}(f_i)=0$ for each i. We conclude that $\operatorname {\mathrm {Tr}}_Z(f)=T(f_1,\cdots ,f_r)=0$ .

3.1 The linearized category of crystals

The assumption that $\mathcal {C}$ is braided in Theorem 1.1 cannot be removed. A counterexample is the $\Bbb C$ -linearization $\mathrm {Crys}(G)$ of the category of crystals for the quantum group $G_q$ attached to a simply connected simple complex group G [Reference Henriques and KamnitzerHK06], which is a semisimple category of moderate growth but is not braided, nor rigid (only weakly rigid). This category is a limit when $q\to 0$ of the braided, rigid categories $\mathsf {Rep}(G_q)$ with the same Grothendieck ring, but the braiding and rigidity are destroyed in the limit; in fact, the only rigid objects in $\mathrm {Crys}(G)$ are multiples of . Indeed, in $\mathsf {Rep}(G_q)$ , if $V_\lambda $ is the irreducible representation with highest weight $\lambda $ and $f: V_\lambda \to V_\lambda ^{**}$ is an isomorphism, then $\operatorname {\mathrm {Tr}}(f)\operatorname {\mathrm {Tr}}((f^*)^{-1})=(\dim _q V_\lambda )^2$ , which goes to infinity as $q\to 0$ for any $\lambda \ne 0$ , as do those eigenvalues of the squared braiding on $V_\lambda \otimes V_\mu $ which are negative powers of q. The category $\mathrm {Crys}(G)$ is, however, a coboundary category [Reference Henriques and KamnitzerHK06]: it carries a symmetric commutor (a natural isomorphism $\sigma _{X,Y}:X\otimes Y\to Y\otimes X$ ) satisfying the Reidemeister move R2, but not R3. This shows that the move R3 plays a crucial role in the proof of Theorem 1.1.

To make things concrete, let us restrict to the case $G=SL(2)$ . In this case, $\mathrm {Crys}(G)$ can be realized as the asymptotic Temperley-Lieb-Jones category $TLJ(\infty )$ . Namely, recall that the usual Temperley-Lieb-Jones category $TLJ(\delta )$ is defined so that the circle evaluates to $\delta $ , while the zig-zag evaluates to $1$ (namely, $TLJ(\delta )=\mathsf {Rep}(SL(2)_q)$ where $\delta =-q-q^{-1}$ ). By renormalizing the diagrams, we may arrange that the circle evaluates to $1$ , while the zig-zag evaluates to $\delta ^{-1}$ . This allows us to specialize $TLJ(\delta )$ at $\delta =\infty $ (i.e., $\delta ^{-1}=0$ ), which yields a semisimple category $TLJ(\infty )$ with Grothendieck ring the representation ring of $SL(2)$ where the circle evaluates to $1$ while the zig-zag evaluates to $0$ (see [Reference VirkVir, §3]), which is not braided, nor rigid (although is weakly rigid). Namely, $\operatorname {\mathrm {Hom}}([n],[m])$ in $TLJ(\infty )$ has the usual basis of Temperley-Lieb-Jones diagrams with n inputs and m outputs and composition given by concatenation of diagrams and removing the circles, but when the concatenation contains a zig-zag, instead of straightening it, the composition is declared to be zero.

The endomorphism algebra of the object $[n]$ in $TLJ(\infty )$ is thus the asymptotic Temperley-Lieb-Jones algebra $TLJ_n(\infty )$ spanned by diagrams with n inputs and n outputs (no longer generated by the usual generators $e_i$ if $n\ge 3$ , however). We obtain a quick proof that $TLJ_n(\infty )$ and hence the category $TLJ(\infty )$ are semisimple by defining a filtrationFootnote ¹³ on $TLJ_n(\infty )$ in which the degree of a diagram u is its number of cups.

We can even identify $TLJ_n(\infty )$ as a multimatrix algebra by observing that the Jones-Wenzl idempotents exist and are nonzero for all $n\in \mathbb {N}$ . Indeed, the usual Jones-Wenzl recurrence relation [Reference WenzlWen87] simplifies greatly as zig-zags are zero:

$$\begin{align*}JW_{n+1} = JW_n\otimes 1 - (JW_n\otimes 1) e_n (JW_n\otimes 1). \end{align*}$$

One can also check via this recurrence relation that $JW_n$ is the linear combination of all projections p with no nested cups/caps where the coefficient of p is $(-1)^{\#\text { caps in }p}$ . Matrix units for the summands of $TLJ_n(\infty )$ are then obtained by cabling the ‘waists’ of basis elements by the appropriate Jones-Wenzl idempotents. It then follows that $TLJ_n(\infty )$ is a multimatrix algebra, as diagrams whose through strings are cabled by different Jones-Wenzl idempotents are orthogonal.Footnote ¹⁴

All these counterexamples have infinitely many simple objects, however. This gives rise to the following question, which is open even for weakly rigid categorifications of based rings (with duality).

3.2 Rigidity for small growth dimension in nonbraided categories

For a positive element Z of a based ring A as in [Reference Etingof, Gelaki, Nikshych and OstrikEGNO15, Def. 3.1.3], define the growth dimension $\operatorname {\mathrm {gd}}(Z)$ to be

$$ \begin{align*}\operatorname{\mathrm{gd}}(Z):=\lim_{n\to \infty}\mathrm{length}(Z^n)^{\frac{1}{n}} \end{align*} $$

(see [Reference Coulembier, Etingof and OstrikCEO23, Def. 4.1]), where the length of a positive element is the sum of its coefficients.Footnote ¹⁵ For example, for A the character ring of a complex reductive group G, $\operatorname {\mathrm {gd}}(Z)$ is the usual dimension of the representation Z, and if A is finite, then $\operatorname {\mathrm {gd}}(Z)$ coincides with the Frobenius-Perron dimension $\operatorname {\mathrm {FPdim}}(Z)$ .

Proof. Fix a retraction (cap) and splitting (cup) such that $r_+s_+=1$ and similarly and such that $r_-s_-=1$ . We therefore have the zig-zags

$$ \begin{align*}z_1=(r_+\otimes 1)(1\otimes s_-),\ z_2=(1\otimes r_+)(s_-\otimes 1),\ z_3=(r_-\otimes 1)(1\otimes s_+),\ z_4=(1\otimes r_-)(s_+\otimes 1)\in \mathbb{k}. \end{align*} $$

Moreover, $z_1^2=z_1z_2=z_2^2$ , so $z_1=z_2=:z$ , and similarly $z_3=z_4=:z_*$ .Footnote ¹⁶

For a noncrossing matching u on the word $(YX)^n$ , let $D(u)$ be the number of $YX$ -matchings in u (so the number of $XY$ -matchings is $n-D(u)$ ). Consider the pairing

given by $\Phi (f,g)=f\circ g$ . For every non-crossing matching u of $[1,2n]$ we have elements defined using $r_+,r_-$ and defined using $s_+,s_-$ . It is easy to see that for any two noncrossing matchings $u,v$ , we have

$$ \begin{align*}\Phi(f_u,g_v)=z^{a(u,v)}z_*^{b(u,v)}, \end{align*} $$

where $a(u,v)-b(u,v)=D(u)-D(v)$ , $a(u,v)+b(u,v)>0$ if $u\ne v$ , and $\Phi (f_u,g_u)=1$ . Thus if $z=0$ or $z_*=0$ , then the matrix $[\Phi (f_u,g_v)]$ is upper or lower triangular in any ordering compatible with the D-grading, with $1$ on the diagonal, hence nondegenerate. It then follows that $\lbrace g_u\rbrace $ are linearly independent. Since the number of noncrossing matchings is $C_n$ (the n-th Catalan number), this implies that

Since $\lim _{n\to \infty }C_n^{\frac {1}{n}}=4$ , it follows that $\operatorname {\mathrm {gd}}(Y\otimes X)\ge 4$ . Thus if $\operatorname {\mathrm {gd}}(Y\otimes X)<4$ then $z,z_*\ne 0$ , implying that X is rigid and $Y\cong X^*$ .

3.3 Categories of nonmoderate growth

The assumption that X has moderate growth in Theorem 1.1 cannot be removed, even for symmetric categories. A counterexample is the symmetric oriented Brauer category $OB(\infty )$ , which is the limit as $t\to \infty $ of the oriented Brauer categories $OB(t)$ , also known as the Deligne categories $\mathsf {Rep}(GL_t)$ , $t\in \Bbb C$ [Reference Etingof, Gelaki, Nikshych and OstrikEGNO15, §9.12]. Namely, recall that $OB(t)$ has objects $[n,m]$ for $n,m\in \mathbb {N}$ , and $\operatorname {\mathrm {Hom}}([n_1,m_1],[n_2,m_2])$ is spanned by appropriate walled Brauer diagrams, with composition given by concatenation of diagrams, so that the circle evaluates to t. Similar to the construction of $TLJ(\infty )$ in §3.1, we may renormalize these diagrams by suitable powers of t so that the circle evaluates to $1$ , but the zig-zags are $t^{-1}$ . This allows us to evaluate at $t=\infty $ (i.e., $t^{-1}=0$ ) and, upon Cauchy completion, obtain a symmetric category $OB(\infty )$ . In this category, diagrams are composed by concatenation and removing circles, but if the concatenated diagram contains a zig-zag, the composition is declared to be zero.

For example, $\operatorname {\mathrm {End}}([n,m])$ is the asymptotic walled Brauer algebra $W_{n,m}(\infty )$ obtained by the above limiting procedure from the usual walled Brauer algebra $W_{n,m}(t)$ . This algebra has the usual basis of walled Brauer diagrams with n inputs and n outputs to the left of the wall and m inputs and m outputs to the right of the wall. As in §3.1, we can define a filtration on $W_{n,m}(\infty )$ in which the degree of a diagram u is its number of cups, and the associated graded algebra $\operatorname {\mathrm {gr}}(W_{n,m}(\infty ))=\bigoplus _k I_k/I_{k+1}$ is manifestly semisimple with k-th summand

$$\begin{align*}I_k/I_{k+1} \cong \mathbb{C}[S_{n-k}]\otimes \mathbb{C}[S_{m-k}]\otimes \operatorname{Mat}\left(\frac{n!m!}{(n-k)!(m-k)!k!}\right). \end{align*}$$

This implies that $W_{n,m}(\infty )$ and thus the category $OB(\infty )$ are semisimple, with simples $X_{\lambda ,\mu }$ labeled by pairs of partitions, and the same fusion rules as in $OB(t)$ . However, the only rigid objects in this category are multiples of : the dimension $\dim V_{\lambda ,\mu }(t)$ in $OB(t)$ is a nonconstant polynomial of t, hence goes to infinity as $t\to \infty $ . Unsurprisingly, these are also the only objects in this category that have moderate growth.

One can also perform the same limiting procedure to the unoriented Brauer category $UB(t)=\mathsf {Rep}(O_t)$ ([Reference Etingof, Gelaki, Nikshych and OstrikEGNO15, §9.12]), getting a semisimple category $UB(\infty )$ which is likewise a counterexample to Theorem 1.1 without moderate growth.

3.4 Categories with nilpotent endomorphisms of nonzero trace

The moderate growth assumption in Corollary 1.5 cannot be removed, even for symmetric categories, as was first observed by Deligne [Reference DeligneDel07, §5.8]. Namely, there exist symmetric categories (of nonmoderate growth) in which the trace of a nilpotent endomorphism of an object can be nonzero. A nice example is the oriented Brauer category $OB(t)$ over a field $\mathbb {k}$ of characteristic $2$ , where $t\in \mathbb {k}$ is not equal to $0$ or $1$ . Then we can take $V:=[1,0]$ , $X=V^{\otimes 2}=[2,0]$ and $z=1-s\in \operatorname {\mathrm {End}}(X)$ where s is the swap $V^{\otimes 2}\to V^{\otimes 2}$ . Then $z^2=0$ but $\operatorname {\mathrm {Tr}}(z)=\operatorname {\mathrm {Tr}}(1)-\operatorname {\mathrm {Tr}}(s)=t^2-t\ne 0$ . Many other examples of categories containing nilpotent endomorphisms with nonzero trace appear in [Reference Khovanov, Ostrik and KononovKOK22]. Of course, such a symmetric category cannot have monoidal functors to abelian symmetric tensor categories, since in the latter the trace of any nilpotent endomorphism z must be zero (as z is strictly upper triangular in the filtration by kernels of its powers).

4.1 2-out-of-3 rigidity

Let $\mathcal {C}$ be a braided monoidal category. Recall that if $X\in \mathcal {C}$ is rigid then so is $X^*\cong {}^*X$ .

Assume in addition that $\mathcal {C}$ is abelian and artinian with right exact tensor product, so that $\operatorname {\mathrm {Tor}}(X,Y)$ is well defined for every $X,Y\in \mathcal {C}$ , possibly as a pro-object.Footnote ¹⁷ An object $X\in \mathcal {C}$ is called flat if the functor $X\otimes -$ is exact.

Proof. (i) Since $X,Y$ are rigid, by Lemma 4.1 they are flat. Using the long exact sequence of Tor groups, we see that Z is flat as well. Also Z represents an element

Let $Z^\vee $ be the extension of $Y^*$ by $X^*$ defined by the element $-\alpha $ , so it is also flat. Since $X,Y,Z,X^*,Y^*,Z^\vee $ are flat, we see that the object $Z\otimes Z^\vee $ has a three-step filtration with successive quotients $Y\otimes X^*,X\otimes X^*\oplus Y\otimes Y^*,X\otimes Y^*$ , meaning that we have subobjects

$$ \begin{align*}F_1=Y\otimes X^* \subset F_2= \ker( Z\otimes Z^\vee \to X\otimes Y^* ) \subset F_3= Z\otimes Z^\vee \end{align*} $$

with $F_2/F_1\cong X\otimes X^*\oplus Y\otimes Y^*$ and $F_3/F_2\cong X\otimes Y^*$ . By construction of $Z^\vee $ , the morphism lifts to a morphism , hence to a morphism , which we denote by $\operatorname {\mathrm {coev}}_Z$ . To see this, one looks at the exact sequence

$$ \begin{align*}0\to F_1=Y\otimes X^*\to F_2 \to F_2/F_1 \cong X\otimes X^*\oplus Y\otimes Y^*\to 0 \end{align*} $$

and computes that the pullback in of its class in $\operatorname {\mathrm {Ext}}^1(X\otimes X^*\oplus Y\otimes Y^*, Y\otimes X^*)$ is trivial since we used $-\alpha $ to define $Z^\vee $ , whence our lift exists. We remark that writing $i^\vee : Z^\vee \to Y^*$ and $p^\vee :X^*\to Z^\vee $ , the following diagram commutes.

This can be seen by analyzing the other faces of the following larger diagram.

By a dual argument, we can analyze $Z^\vee \otimes Z$ by a three-step filtration with successive quotients $X^*\otimes Y, X^*\otimes X\oplus Y^*\otimes Y,Y^*\otimes X$ given by

$$ \begin{align*}G_1= X^*\otimes Y \subset G_2 = \ker(Z^\vee\otimes Z \to Y^*\otimes X) \subset G_3 = Z^\vee \otimes Z \end{align*} $$

to construct an evaluation map . One looks at the exact sequence

$$ \begin{align*}0 \to X^*\otimes X\oplus Y^*\otimes Y \cong G_2/G_1 \to \operatorname{coker}(X^*\otimes Y\to Z^\vee\otimes Z) \cong G_3/G_1 \to Y^*\otimes X \to 0. \end{align*} $$

By reasoning dually as before, the morphism extends to a morphism , hence to a morphism , which we denote by $\operatorname {\mathrm {ev}}_Z$ . By similar reasoning, $\operatorname {\mathrm {ev}}_Z$ makes the following diagram commute.

Now one verifies that both $\operatorname {\mathrm {ev}}_Z,\operatorname {\mathrm {coev}}_Z$ make the following diagram commute.

Looking at the outside rectangle above and the zig-zag $z:Z\to Z$ in the middle, since the zig-zags for $X,Y$ are both identities, our short exact sequence fits in the following commutative diagram.

Since $\mathcal {C}$ is abelian, z is an isomorphism by the Short Five Lemma. We may thus renormalize one of $\operatorname {\mathrm {ev}}_Z,\operatorname {\mathrm {coev}}_Z$ so that the zig-zag axiom holds on the nose. One then uses a similar diagram to check that the zig-zag for $Z^\vee $ , which is an idempotent,Footnote ¹⁸ is also invertible, and is thus the identity.

(ii) We have an epimorphism $p: Z\to X$ whose kernel is Y. We have

and we write $p^*:X^*\to Z^*$ for the corresponding morphism.Footnote ¹⁹ Denote by $Y^\vee $ the cokernel of $p^*$ . We have a morphism , where $\pi : Z^*\to Y^\vee $ is the projection.

Since X is rigid, it is flat by Lemma 4.1, so $\mathrm {Tor}^1(X,Y^\vee )=0$ . Hence we have a short exact sequence

$$ \begin{align*}0\to Y\otimes Y^\vee\to Z\otimes Y^\vee\to X\otimes Y^\vee\to 0 \end{align*} $$

(cf. [Reference Kazhdan and LusztigKL94, Lem. 4.6]). The morphism $(p\otimes \operatorname {\mathrm {id}}_{Y^\vee })\circ (\operatorname {\mathrm {id}}_Z\otimes \pi )\circ \operatorname {\mathrm {coev}}_Z$ is the morphism corresponding to the composition $X^*\to Z^*\to Y^\vee $ , which is zero by definition of $Y^\vee $ . Thus the morphism $(\operatorname {\mathrm {id}}_Z\otimes \pi )\circ \operatorname {\mathrm {coev}}_Z$ lands in $Y\otimes Y^\vee $ , and we denote this morphism by $\operatorname {\mathrm {coev}}_Y$ .

Similarly, consider the morphism . It pulls back to a morphism . Moreover, since the pullback of this morphism to $X^*\otimes Y$ is zero (as it corresponds to the composition $Y\to Z\to X$ ) and since we have an exact sequence $X^*\otimes Y\to Z^*\otimes Y\to Y^\vee \otimes Y\to 0$ , the morphism $\operatorname {\mathrm {ev}}_Z\circ (\operatorname {\mathrm {id}}_{Z^*}\otimes i)$ is the pullback of a unique morphism , which we denote $\operatorname {\mathrm {ev}}_Y$ . Now diagram chasing shows that the morphisms $\operatorname {\mathrm {ev}}_Y,\operatorname {\mathrm {coev}}_Y$ equip $Y^\vee $ with the structure of the left dual $Y^*$ of Y, as desired.

(iii) We have a monomorphism $i: Y\to Z$ which by assumption gives rise to an epimorphism $i^*: Z^*\to Y^*$ , and we may define $X^\vee $ to be the kernel of $i^*$ , so that we have a short exact sequence

$$ \begin{align*}0\to X^\vee\to Z^*\to Y^*\to 0.\end{align*} $$

Since $Y^*$ is flat, $\mathrm {Tor}^1(X,Y^*)=0$ , so we obtain a short exact sequence

$$ \begin{align*}0\to X\otimes X^\vee\to X\otimes Z^*\to X\otimes Y^*\to 0. \end{align*} $$

We have a morphism where $p: Z\to X$ is the projection. We have $(\operatorname {\mathrm {id}}_X\otimes i^*)\circ (p\otimes \operatorname {\mathrm {id}}_{Z^*})\circ \operatorname {\mathrm {coev}}_Z=0$ , so the morphism $(p\otimes \operatorname {\mathrm {id}}_{Z^*})\circ \operatorname {\mathrm {coev}}_Z$ lands in the kernel of the morphism $\operatorname {\mathrm {id}}_X\otimes i^*: X\otimes Z^*\to X\otimes Y^*$ . Thus $(p\otimes \operatorname {\mathrm {id}}_{Z^*})\circ \operatorname {\mathrm {coev}}_Z$ lands in $X\otimes X^\vee $ , and we denote this morphism by $\operatorname {\mathrm {coev}}_X$ .

Similarly, we have a morphism , where $\iota : X^\vee \hookrightarrow Z^*$ is the inclusion, and its pullback to a morphism corresponds to the morphism $X^\vee \to Y^*$ , which is zero by definition of $X^\vee $ . But we have an exact sequence

$$ \begin{align*}X^\vee\otimes Y\to X^\vee \otimes Z\to X^\vee\otimes X\to 0, \end{align*} $$

so $\operatorname {\mathrm {ev}}_Z\circ (\iota \otimes \operatorname {\mathrm {id}}_Z)$ is the pullback of the unique morphism , which we denote by $\operatorname {\mathrm {ev}}_X$ . Now diagram chasing shows that the morphisms $\operatorname {\mathrm {ev}}_X,\operatorname {\mathrm {coev}}_X$ equip $X^\vee $ with the structure of the left dual $X^*$ of X, as desired.

The following corollary follows immediately from Lemma 4.2.

4.2 Recursive filtrations

Let $\mathbb {k}$ be an algebraically closed field.

Note that in such a category, so

$$ \begin{align*}\dim \operatorname{\mathrm{Hom}}(X,Y)\le \mathrm{length}(X\otimes Y^\vee).\end{align*} $$

Hence $\operatorname {\mathrm {Hom}}(X,Y)$ is finite dimensional, so $\mathcal {C}$ is artinian and .

Proof. It suffices to show that all simple objects in $\mathcal {C}$ are rigid. To this end, it suffices to show that for all n, all $X\in \operatorname {\mathrm {Irr}}_n(\mathcal {C})$ are rigid. We induct on n. The base case is trivial. To pass from n to $n+1$ , let $X\in \operatorname {\mathrm {Irr}}_{n+1}(\mathcal {C})$ and let $X_1,\cdots ,X_m\in \mathcal {C}$ be objects with composition factors in S such that the composition series of $X_1\otimes \cdots \otimes X_m$ consists of a single copy of X and objects in $\operatorname {\mathrm {Irr}}_n(\mathcal {C})$ . So there is a three-step filtration of $X_1\otimes \cdots \otimes X_m$ with terms $Z,X,Y$ , where all composition factors of Z and Y are in $\operatorname {\mathrm {Irr}}_n(\mathcal {C})$ . Let T be the kernel of the morphism $X_1\otimes \cdots \otimes X_m\to Y$ , so we have a short exact sequence

$$ \begin{align*}0\to T\to X_1\otimes\cdots\otimes X_m\to Y\to 0. \end{align*} $$

Since $X_i$ and (by the induction assumption) $Y,Z$ have rigid composition factors, they are rigid by the 2-out-of-3 property. Hence $X_1\otimes \cdots \otimes X_m$ is rigid. Thus T is rigid by the 2-out-of-3 property. But we also have a short exact sequence

$$ \begin{align*}0\to Z\to T\to X\to 0. \end{align*} $$

Since $Z,T$ are rigid, we conclude by the 2-out-of-3 property that X is rigid, as needed.

Combining Proposition 4.10 with our main result Theorem 1.1, we thus obtain:

4.3 Application: rigidity of the Kazhdan-Lusztig category at negative rational levels

An example of application of Theorem 4.11 is a simplification of the proof of rigidity of the category $\mathcal O_\kappa $ in [Reference Kazhdan and LusztigKL94]. Namely, let $\kappa \in \Bbb Q_{<0}$ , $\mathfrak {g}$ be a simply laced simple finite-dimensional complex Lie algebra, and $\mathcal O_\kappa $ be the Kazhdan-Lusztig category of $\mathfrak {g}[[t]]$ -integrable finitely generated representations of the affine Lie algebra $\widehat {\mathfrak {g}}$ at level $\kappa -h^\vee $ , where $h^\vee $ is the dual Coxeter number of $\mathfrak {g}$ . In [Reference Kazhdan and LusztigKL94, Prop 31.2, p. 410] it is proved that $\mathcal O_\kappa $ is a braided, semirigid category, with $\operatorname {\mathrm {Irr}}(\mathcal {C})=P_+$ , the set of dominant integral weights, and it is easy to see that it has moderate growth. Let $S:=\lbrace \omega _1,\cdots ,\omega _r\rbrace $ be the fundamental weights for $\mathfrak {g}$ , let $\rho =\sum _i \omega _i$ , and let $\operatorname {\mathrm {Irr}}_n(\mathcal {C})$ be the set of weights $\lambda \in P_+$ with $(\lambda ,\rho )\le n$ . Assume that $\kappa $ is such that the Weyl module $V_\lambda ^\kappa $ over $\widehat {\mathfrak {g}}$ is irreducible, and its quantum dimension $d_\lambda (q):=\prod _{\alpha>0} \frac {[(\omega _i+\rho ,\alpha )]_q}{[(\rho ,\alpha _i)]_q}$ is nonzero at $q=e^{\pi i/\kappa }$ (this rules out an explicit finite set of numerators for $\kappa $ , cf. [Reference Kazhdan and LusztigKL94, p. 413]). Then, as shown in [Reference Kazhdan and LusztigKL94] (see p. 413), $V_{\omega _i}^\kappa \otimes V_{\omega _{i^*}}^\kappa $ contains as a direct summand, so the $V_{\omega _i}^\kappa $ are non-negligible. If $\lambda =\sum _i n_i\omega _i$ , $n=\sum _i n_i$ , and $L_\lambda ^\kappa \in \mathcal O_\kappa $ is the simple module with highest weight $\lambda $ , then the composition series of $\bigotimes _i (V_{\omega _i}^\kappa )^{\otimes n_i}$ consists of $L_\lambda ^\kappa $ and elements of $\operatorname {\mathrm {Irr}}_{n-1}(\mathcal {C})$ , so $\mathcal {C}$ admits a recursive filtration relative to S. Thus applying Theorem 4.11, we get that $\mathcal O_\kappa $ is rigid, which is [Reference Kazhdan and LusztigKL94, Thm. 32.1, p. 419]. This allows us to avoid quite a few technicalities in the proof of this theorem.

Remark 4.13. We thank Robert McRae for an email exchange providing the following additional example showing the assumption that $i^*$ is an epimorphism in Lemma 4.2(iii) cannot be dropped. His example comes from Grothendieck-Verdier categories which are not r-categories, and is artinian, whereas the example in Remark 4.3 above is not.

The Weyl modules $V_n=V^{p/q}_{n\omega _1}$ for affine $\mathfrak {sl}_2$ obtained by parabolic induction from the $(n+1)$ -dimensional simple $\mathfrak {sl}_2$ -module, at level $k = -2 + p/q$ where $p,q$ are relatively prime positive integers, give an exact sequence

$$ \begin{align*}0 \to V_{p-2} \to V_1\otimes V_{p-1}\to V_p \to 0. \end{align*} $$

Here, $V_{p-1}$ is the analogue of the Steinberg representation for quantum $\mathfrak {sl}_2$ , or for $sl_2$ in positive characteristic. In this case, $V_{p-2}$ and $V_1\otimes V_{p-1}$ are rigid, but $V_p$ is not. Although the Grothendieck-Verdier dual of the inclusion $V_{p-2} \to V_1\otimes V_{p-1}$ is surjective, the rigid dual of $V_{p-2}$ is not the same as its Grothendieck-Verdier dual, so the rigid dual of the inclusion $V_{p-2} \to V_1\otimes V_{p-1}$ cannot be surjective.