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Basic properties of the categorical trace

Statement

Let $a : X \to X^{\lor \lor}$ and $b : Y \to Y^{\lor \lor}$ .

${Tr}_{L} (a) = {Tr}_{R} (a^{\lor})$ .
If the category is additive, then ${Tr}_{L} (a \oplus b) = {Tr}_{L} (a) + {Tr}_{L} (b)$ .
${Tr}_{L} (a \otimes b) = {Tr}_{L} (a) {Tr}_{L} (b)$ .
For every endomorphism $c : X \to X$ , ${Tr}_{L} (a c) = {Tr}_{L} (c^{\lor \lor} a)$ .

The corresponding right-trace statements hold by the same formulas with left and right exchanged.

Facts & Assumptions

Given: A rigid monoidal category, morphisms $a : X \to X^{\lor \lor}$ and $b : Y \to Y^{\lor \lor}$ , and when needed an additive structure.

[F1]

EGNO Proposition 4.7.3 proves exactly the four displayed properties, with the additive clause explicitly restricted to additive categories.

[L1]

The duality functor is contravariant and antimonoidal (Left duality is a contravariant antimonoidal functor).

[L2]

The traces are the ones from The categorical trace of a morphism into the double dual.

Proof

technique · direct

1.1

The formula in [L2] for ${Tr}_{L} (a)$ is obtained by inserting $a$ between one coevaluation and one evaluation. Dualizing that composite and using the contravariant antimonoidality from [L1] reverses the order and turns it into the defining formula for ${Tr}_{R} (a^{\lor})$ , which is the first clause recorded in [F1].

givenF1L1L2

1.2

The same proposition [F1] states that direct sums split the trace additively in additive categories, tensor products split it multiplicatively, and composing with an endomorphism may be cycled across the trace at the cost of a double dual: ${Tr}_{L} (a c) = {Tr}_{L} (c^{\lor \lor} a) .$

2.1

Therefore all four displayed identities hold, and the right-trace versions follow by applying the same argument to the mirrored formulas in [L2].

step 1.1step 1.2∎

Depends on

Used by

Exact-sequence additivity of trace and its missing hypotheses Remark

Dependency tree · two levels

11 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, Proposition 4.7.3 (standard reference, not scraped)