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The separability projector realizes the relative tensor product

Statement

Let $C$ and the special Frobenius algebra data be as in MTC-valued Frobenius algebra and defect data for oriented rational CFT. Let $A, B$ be such algebras, $X$ an $A$ - $B$ -bimodule, and $M$ a left $B$ -module. Put $e_{B} = Δ_{B} \circ η_{B} : 1 \to B \otimes B$ . With canonical associators and unitors inserted, define $P_{X, M} = (ρ_{X}^{r} \otimes ρ_{M}) \circ ({id}_{X} \otimes e_{B} \otimes {id}_{M}) : X \otimes M \to X \otimes M .$ Then $P_{X, M}$ is an idempotent, and its split image is the coequalizer $X \otimes_{B} M$ of the right $B$ -action on $X$ and left $B$ -action on $M$ . The left $A$ -action on $X$ descends to this image.

Statement references

MTC-valued Frobenius algebra and defect data for oriented rational CFT

Facts & Assumptions

Given: The finite semisimple complex ribbon category, the special Frobenius algebra $B$ , the $A$ - $B$ -bimodule $X$ , and the left $B$ -module $M$ with the action and associator conventions of the linked definition.

[F1]

MTC-valued Frobenius algebra and defect data for oriented rational CFT: $m_{B} \circ Δ_{B} = {id}_{B}$ , the Frobenius identities hold, and the left and right actions on $X$ commute. The relative tensor product is the coequalizer of the two stated action maps.

Proof

1.1givenF1algebra

Define $e_{B} = Δ_{B} η_{B}$ . The Frobenius identities and the two unit identities give the centrality equality $(m_{B} \otimes {id}_{B}) \circ ({id}_{B} \otimes e_{B}) = Δ_{B} = ({id}_{B} \otimes m_{B}) \circ (e_{B} \otimes {id}_{B}),$ as maps $B \to B \otimes B$ . Specialness gives $m_{B} \circ e_{B} = η_{B}$ . These are equalities of morphisms; in Sweedler shorthand they say $b e_{B}^{(1)} \otimes e_{B}^{(2)} = e_{B}^{(1)} \otimes e_{B}^{(2)} b$ and $e_{B}^{(1)} e_{B}^{(2)} = 1_{B}$ .

2.1givenF1step 1.1algebra

In the same shorthand, interpreted as composition with the morphism $e_{B}$ and the indicated module actions rather than as elements of the objects, $P_{X, M} (x \otimes m) = x e_{B}^{(1)} \otimes e_{B}^{(2)} m$ . Applying $P$ twice gives $x e_{B}^{(1)} e_{B}^{'}^{(1)} \otimes e_{B}^{'}^{(2)} e_{B}^{(2)} m$ , where the primed copy denotes the second occurrence of $e_{B}$ . Apply centrality to the primed copairing with the input slot $e_{B}^{(2)}$ : $e_{B}^{'}^{(1)} \otimes e_{B}^{'}^{(2)} e_{B}^{(2)} = e_{B}^{(2)} e_{B}^{'}^{(1)} \otimes e_{B}^{'}^{(2)}$ . The double application therefore equals $x e_{B}^{(1)} e_{B}^{(2)} e_{B}^{'}^{(1)} \otimes e_{B}^{'}^{(2)} m$ . Now $m_{B} e_{B} = η_{B}$ and the unital right action on $X$ reduce it to $x e_{B}^{'}^{(1)} \otimes e_{B}^{'}^{(2)} m = P_{X, M} (x \otimes m)$ . This substitution composes the centrality morphism with the second output of the unprimed copairing; no interchange of the two outputs or braiding is used. Thus $P^{2} = P$ as endomorphisms of $X \otimes M$ .

3.1givenF1step 1.1step 2.1algebra

Centrality also gives $P \circ (ρ_{X}^{r} \otimes {id}_{M}) = P \circ ({id}_{X} \otimes ρ_{M})$ . If $q : X \otimes M \to Z$ equalizes those action maps, then $q P (x \otimes m) = q (x e_{B}^{(1)} \otimes e_{B}^{(2)} m) = q (x \otimes e_{B}^{(1)} e_{B}^{(2)} m) = q (x \otimes m),$ so $q P = q$ . Conversely, if $q P = q$ , the displayed equality for $P$ implies that $q$ equalizes the two action maps. Hence the coequalizing maps are exactly the maps fixed by precomposition with $P$ .

4.1givenF1step 2.1step 3.1algebra∎

Finite semisimplicity splits $P$ as $X \otimes_{B} M \overset{i}{\to} X \otimes M \overset{r}{\to} X \otimes_{B} M$ with $i r = P$ and $r i = id$ . For every coequalizer $q$ , step 3.1 gives $q P = q$ , so $q = (q i) r$ ; if $q^{'} = q^{''} r$ , then $q^{''} = q^{'} i$ by $r i = id$ . Thus $r$ has the coequalizer universal property. Since the left $A$ -action on $X$ commutes with its right $B$ -action, its tensor with ${id}_{M}$ commutes with $P$ and preserves the image; restricting along $i, r$ defines the descended left $A$ -action.

Depends on

MTC-valued Frobenius algebra and defect data for oriented rational CFT