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In a spherical category the left and right traces agree

Statement

Let $C$ be a spherical tensor category over an algebraically closed field $k$ , with spherical structure $a$ . For every object $X$ and every endomorphism $x : X \to X$ ,

${Tr}_{L} (a_{X} x) = {Tr}_{R} (x a_{X}^{- 1}) .$

Facts & Assumptions

Given: A spherical tensor category $(C, a)$ over an algebraically closed field $k$ , an object $X$ , and an endomorphism $x : X \to X$ .

[L1]

Spherical means $\dim_{a} (U) = \dim_{a} (U^{\lor})$ for every object $U$ (Spherical structure). The pivotal comparison is a monoidal natural isomorphism (Pivotal structure).

[L2]

Traces are the composites of The categorical trace of a morphism into the double dual. They are additive on direct sums and satisfy duality and cyclicity, including the different-object mate calculation in its proof (Basic properties of the categorical trace).

[L3]

A tensor category is locally finite, $k$ -linear abelian with bilinear tensor product and $End (1) = k$ ; in particular objects have finite length and endomorphism spaces have finite dimension (Tensor and multitensor categories).

Proof

technique · direct

1.1L1L2L3givenalgebra

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

2.1L3step 1.1induction

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 $n = i p$ through its image $W$ , with $p : U ↠ W$ and $i : W ↪ U$ . Since $n$ is nilpotent and $U \neq 0$ , 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 $p i$ of $W$ is nilpotent, because $(p i)^{m + 1} = p n^{m} i = 0$ for a sufficiently large $m$ . Step 1.1 gives $t_{U} (n) = t_{W} (p i) = 0$ and $r_{U} (n) = r_{W} (p i) = 0$ by induction.

3.1L2L3step 1.1step 2.1algebra∎

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

Remarks

The algebraically closed ground-field hypothesis is essential for the dimension-based definition of spherical structure used here. Over $R$ , take finite-dimensional real representations of the cyclic group of order three and let the pivotal comparison be the canonical double-dual map followed by the central generator $g$ . Every real representation has the same character at $g$ and $g^{- 1}$ , so its two pivotal dimensions agree. On the two-dimensional rotation representation, however, $x = g$ has left pivotal trace $tr (g^{2}) = - 1$ and right pivotal trace $tr (1) = 2$ . Definitions of spherical category that require equality of traces of all endomorphisms over arbitrary fields use a stronger defining condition.

Depends on

Used by

FALSE: the left and right traces always agree False statement
RT handlebody pairings and surgery gluing Lemma
The complete Kirby-color band-slide identity with open colored graph coupons Lemma

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories, Theorem 4.7.15 (standard reference, not scraped)