Category of a Quantum Groupoid

A873C024-16E2-408D-8521-AC452457B0C4

For a quantum groupoid H let Rep(H) be the category of representations of H, whose objects are finite-dimensional left H -modules and whose morphisms are H -linear homomorphisms. We shall show that Rep(H) has a natural structure of a monoidal category with duality.

For objects V, W of Rep(H) set

V ⊗ W = x ∈ V ⊗k W|x = ∆(1) · x ⊂ V ⊗k W —– (1)

with the obvious action of H via the comultiplication ∆ (here ⊗k denotes the usual tensor product of vector spaces). Note that ∆(1) is an idempotent and therefore V ⊗ W = ∆(1) × (V ⊗k W). The tensor product of morphisms is the restriction of usual tensor product of homomorphisms. The standard associativity isomorphisms (U ⊗ V ) ⊗ W → U ⊗ (V ⊗ W ) are functorial and satisfy the pentagon condition, since ∆ is coassociative. We will suppress these isomorphisms and write simply U ⊗ V ⊗ W.

The target counital subalgebra Ht ⊂ H has an H-module structure given by h · z = εt(hz),where h ∈ H, z ∈ Ht.

Ht is the unit object of Rep(H).

Define a k-linear homomorphism lV : Ht ⊗ V → V by lV(1(1) · z ⊗ 1(2) · v) = z · v, z ∈ Ht, v ∈ V.

This map is H-linear, since

lV h · (1(1) · z ⊗ 1(2) · v) = lV(h(1) · z ⊗ h(2) · v) = εt(h(1)z)h(2) · v = hz · v = h · lV (1(1) · z ⊗ 1(2) · v),

∀ h ∈ H. The inverse map l−1V: → Ht ⊗ V is given by V

l−1V(v) = S(1(1)) ⊗ (1(2) · v) = (1(1) · 1) ⊗ (1(2) · v)

The collection {lV}V gives a natural equivalence between the functor Ht ⊗ (·) and the identity functor. Indeed, for any H -linear homomorphism f : V → U we have:

lU ◦ (id ⊗ f)(1(1) · z ⊗ 1(2) · v) = lU 1(1) · z ⊗ 1(2) · f(v) = z · f(v) = f(z·v) = f ◦ lV(1(1) · z ⊗ 1(2) · v)

Similarly, the k-linear homomorphism rV : V ⊗ Ht → V defined by rV(1(1) · v ⊗ 1(2) · z) = S(z) · v, z ∈ Ht, v ∈ V, has the inverse r−1V(v) = 1(1) · v ⊗ 1(2) and satisfies the necessary properties.

Finally, we can check the triangle axiom idV ⊗ lW = rV ⊗ idW : V ⊗ Ht ⊗ W → V ⊗ W ∀ objects V, W of Rep(H). For v ∈ V, w ∈ W we have

(idV ⊗ lW)(1(1) · v ⊗ 1′(1)1(2) · z ⊗ 1′(2) · w)

= 1(1) · v ⊗ 1′(2)z · w) = 1(1)S(z) · v ⊗ 1(2) · w

=(rV ⊗ idW) (1′(1) · v ⊗ 1′(2) 1(1) · z ⊗ 1(2) · w),

therefore, idV ⊗ lW = rV ⊗ idW

Using the antipode S of H, we can provide Rep(H) with a duality. For any object V of Rep(H), define the action of H on V = Homk(V, k) by

(h · φ)(v) = φ S(h) · v —– (2)

where h ∈ H , v ∈ V , φ ∈ V. For any morphism f : V → W , let f: W → V be the morphism dual to f. For any V in Rep(H), we define the duality morphisms dV : V ⊗ V → Ht, bV : Ht → V ⊗ V∗ as follows. For ∑j φj ⊗ vj ∈ V* ⊗ V, set

dV(∑j φj ⊗ vj)  = ∑j φj (1(1) · vj) 1(2) —– (3)

Let {fi}i and {ξi}i be bases of V and V, respectively, dual to each other. The element ∑i fi ⊗ ξi does not depend on choice of these bases; moreover, ∀ v ∈ V, φ ∈ V one has φ = ∑i φ(fi) ξi and v = ∑i fi ξi (v). Set

bV(z) = z · (∑i fi ⊗ ξi) —– (4)

The category Rep(H) is a monoidal category with duality. We know already that Rep(H) is monoidal, it remains to prove that dV and bV are H-linear and satisfy the identities

(idV ⊗ dV)(bV ⊗ idV) = idV, (dV ⊗ idV)(idV ⊗ bV) = idV.

Take ∑j φj ⊗ vj ∈ V ⊗ V, z ∈ Ht, h ∈ H. Using the axioms of a quantum groupoid, we have

h · dV(∑j φj ⊗ vj) = ((∑j φj (1(1) · vj) εt(h1(2))

= (∑j φj ⊗ εs(1(1)h) · vj 1(2)j φj S(h(1))1(1)h(2) · vj 1(2)

= (∑j h(1) · φj )(1(1) · (h(2) · vj))1(2)

= (∑j dV(h(1) · φj  ⊗ h(2) · vj) = dV(h · ∑j φj ⊗ vj)

therefore, dV is H-linear. To check the H-linearity of bV we have to show that h · bV(z) =

bV (h · z), i.e., that

i h(1)z · fi ⊗ h(2) · ξi = ∑i 1(1) εt(hz) · fi ⊗ 1(2) · ξi

Since both sides of the above equality are elements of V ⊗k V, evaluating the second factor on v ∈ V, we get the equivalent condition

h(1)zS(h(2)) · v = 1(1)εt (hz)S(1(2)) · v, which is easy to check. Thus, bV is H-linear.

Using the isomorphisms lV and rV identifying Ht ⊗ V, V ⊗ Ht, and V, ∀ v ∈ V and φ ∈ V we have:

(idV ⊗ dV)(bV ⊗ idV)(v)

=(idV ⊗ dV)bV(1(1) · 1) ⊗ 1(2) · v

= (idV ⊗ dV)bV(1(2)) ⊗ S−1(1(1)) · v

= ∑i (idV ⊗ dV) 1(2) · fi ⊗ 1(3) · ξi ⊗ S−1 (1(1)) · v

= ∑1(2) · fi ⊗ 1(3) · ξi (1′(1)S-1 (1(1)) · v) 1′(2)

= 1(2) S(1(3)) 1′(1) S-1 (1(1)) · v ⊗ 1′(2) = v

(dV ⊗ idV)(idV ⊗ bV)(φ)

= (dV ⊗ idV) 1(1) · φ ⊗ bV(1(2))

= ∑i (dV ⊗ idV)(1(1) · φ ⊗ 1(2) · 1(2) · 1(3) · ξi )

= ∑i (1(1) · φ (1′(1)1(2) · fi)1′(2) ⊗ 1(3) · ξi

= 1′(2) ⊗ 1(3)1(1) S(1′(1)1(2)) · φ = φ,

QED.

 

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s