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The tensor product and contragredient dual of a finite-group module

Example

Let $C_{2} = {1, s}$ with $s^{2} = 1$ , let $H = k [C_{2}]$ , and give $V = k e_{0} \oplus k e_{1}$ the representation $s e_{0} = e_{1}$ , $s e_{1} = e_{0}$ . With $Δ (g) = g \otimes g$ , $ε (g) = 1$ , and $S (g) = g^{- 1}$ on group elements, the tensor product action is diagonal. The left-dual action in this page's convention is the contragredient action $(g \cdot f) (v) = f (g^{- 1} v)$ .

Facts & Assumptions

Given: A field $k$ , the group $C_{2} = {1, s}$ with $s^{2} = 1$ , $H = k [C_{2}]$ , the displayed two-dimensional representation $V$ , and its basis dual $e^{0}, e^{1}$ .

[F1]

Group and abelian group: a group has an associative multiplication and an identity and each element has an inverse.

[F2]

The module category of a bialgebra and its tensor action: the tensor action is $h (v \otimes w) = \sum h_{(1)} v \otimes h_{(2)} w$ and the unit acts by the counit.

[F3]

The left dual of a finite-dimensional Hopf module: the dual action is $(h \cdot f) (v) = f (S (h) v)$ and the coevaluation tensor is $\sum_{i} e_{i} \otimes e^{i}$ .

[F4]

Evaluation, coevaluation, and triangle identities for left duals: for a finite-dimensional module these evaluation and coevaluation maps are $H$ -linear and satisfy both triangle identities.

Verification

Given: The data in Facts & Assumptions.

1.1F1givenalgebra

The group algebra $H$ has basis $1, s$ and relation $s^{2} = 1$ . Define $Δ (1) = 1 \otimes 1$ , $Δ (s) = s \otimes s$ , $ε (1) = ε (s) = 1$ , and $S (1) = 1$ , $S (s) = s$ , extending linearly. For $a, b, c, d \in k$ , both $Δ ((a + b s) (c + d s))$ and $Δ (a + b s) Δ (c + d s)$ equal $(a c + b d) 1 \otimes 1 + (a d + b c) s \otimes s$ ; both $ε ((a + b s) (c + d s))$ and $ε (a + b s) ε (c + d s)$ equal $(a + b) (c + d)$ . These maps preserve units. Coassociativity and the two counit identities hold on $1, s$ by direct substitution, hence on $H$ by linearity. Also $S (s) s = s S (s) = 1$ and the same holds for $1$ , so both antipode identities hold on the basis and then on all of $H$ . Thus these formulas give the required Hopf structure.

2.1F2givenstep 1.1

The operator $s$ on $V$ squares to the identity, so the stated action defines a unital $H$ -module. The diagonal tensor action sends $e_{i} \otimes e_{j}$ to $e_{1 - i} \otimes e_{1 - j}$ because $Δ (s) = s \otimes s$ ; applying it twice fixes every basis tensor. The unit module has $s$ acting by $1$ .

2.2F3step 1.1algebra

Since $S (s) = s^{- 1} = s$ , $(s \cdot e^{0}) (e_{0}) = e^{0} (e_{1}) = 0$ and $(s \cdot e^{0}) (e_{1}) = e^{0} (e_{0}) = 1$ , hence $s \cdot e^{0} = e^{1}$ . Similarly $s \cdot e^{1} = e^{0}$ . This is exactly the contragredient action, and it squares to the identity, so it is a left $H$ -module.

3.1F2F3givenstep 2.1step 2.2

The evaluation pairing is invariant because $ev (s \cdot (e^{i} \otimes e_{j})) = ev (e^{1 - i} \otimes e_{1 - j}) = δ_{1 - i, 1 - j} = δ_{i, j} = ε (s) ev (e^{i} \otimes e_{j})$ . The coevaluation tensor $e_{0} \otimes e^{0} + e_{1} \otimes e^{1}$ is fixed by $s$ , since $s$ interchanges its two summands; it is therefore $H$ -linear. Since $1, s$ span $H$ , these checks prove linearity for all $h \in H$ .

4.1F3F4step 3.1algebra∎

The two triangle composites send $e_{j}$ to $\sum_{i} e_{i} e^{i} (e_{j}) = e_{j}$ and send $e^{j}$ to $\sum_{i} e^{j} (e_{i}) e^{i} = e^{j}$ . The basis vectors span $V$ and $V^{*}$ , so both composites are identities; this also verifies the finite-group instance of [F4].

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