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Tensor products of modules over a bialgebra without an antipode

Example

Let $H = k [t]$ over a field $k$ , with

$Δ (t) = t \otimes t, ε (t) = 1,$

extended linearly by $Δ (\sum_{n} a_{n} t^{n}) = \sum_{n} a_{n} t^{n} \otimes t^{n}$ and $ε (\sum_{n} a_{n} t^{n}) = \sum_{n} a_{n}$ . This is a bialgebra but not a Hopf algebra. The one-dimensional modules $V = W = k$ on which $t$ acts by zero have a tensor product module with the diagonal action; hence the tensor product construction does not require an antipode.

The statement that this bialgebra has no antipode means no antipode exists for the specified coproduct and counit. It does not say that no module structure can ever be put on an underlying dual vector space; in this example the $t = 0$ module has a one-dimensional dual with $t$ acting by zero.

Facts & Assumptions

Given: A field $k$ , the polynomial ring $k [t]$ with coefficientwise addition and convolution multiplication, the displayed maps $Δ, ε$ , and the modules $V = W = k$ where $t$ acts by zero.

[F1]

The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution: $k [t]$ consists of finite coefficient sums $\sum_{n} a_{n} t^{n}$ , with convolution multiplication and unit the constant polynomial $1$ .

[F2]

Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring: these operations make $k [t]$ a unital commutative ring.

[F3]

The module category of a bialgebra and its tensor action: for modules $V, W$ , $h (v \otimes w) = \sum h_{(1)} v \otimes h_{(2)} w$ is the proposed tensor action.

[F4]

Bialgebra module tensor actions and their coherence: when $H$ is a bialgebra, that formula makes $V \otimes W$ a left $H$ -module.

[F5]

Field: a field satisfies $0_{k} \neq 1_{k}$ .

[F6]

Linear functionals and the algebraic dual $V^{*} = L (V, F)$ : $V^{*}$ is the vector space of all linear functionals $V \to k$ .

Verification

Given: The data and conventions in Facts & Assumptions.

1.1givenF1F2algebra

The maps $Δ$ and $ε$ are well-defined linear maps because every polynomial has a finite coefficient expansion [F1]. On monomials, $Δ (t^{m} t^{n}) = t^{m + n} \otimes t^{m + n} = Δ (t^{m}) Δ (t^{n})$ and $ε (t^{m} t^{n}) = 1 = ε (t^{m}) ε (t^{n})$ ; the formulas also preserve $1$ . Bilinearity extends multiplicativity to all polynomials.

1.2givenF1F5algebra

If an antipode $S$ existed, its left convolution identity at $t$ would give $S (t) t = ε (t) 1 = 1$ . Every polynomial multiplied by $t$ has constant coefficient zero, while $1$ has nonzero constant coefficient by [F5]. This is impossible, so the bialgebra is not Hopf.

2.1givenF1step 1.1algebra

For each monomial $t^{n}$ , $(Δ \otimes i d) Δ (t^{n}) = t^{n} \otimes t^{n} \otimes t^{n} = (i d \otimes Δ) Δ (t^{n})$ and $(ε \otimes i d) Δ (t^{n}) = t^{n} = (i d \otimes ε) Δ (t^{n})$ . Since monomials span $k [t]$ , these identities hold for all polynomials; together with step 1.1 they prove that $H$ is a bialgebra.

3.1F1F3F4step 1.1step 2.1

The actions $p (t) \cdot c : = p (0) c$ make $V = W = k$ unital left $H$ -modules because evaluation at zero is a unital algebra map. On $1 \otimes 1 \in V \otimes W$ , $t$ acts by $Δ (t) = t \otimes t$ , hence by $0 \otimes 0 = 0$ . The theorem [F4] makes the diagonal formula a module action on all of $V \otimes W$ ; it uses only the bialgebra structure proved in steps 1.1 and 2.1.

4.1F1F3F5F6step 1.1step 3.1algebra∎

Thus the tensor module exists without an antipode. To check the example's claim about the underlying dual vector space, take $e = 1 \in V = k$ and $e^{*} (c) = c$ , which spans $V^{*}$ by [F6]. Define $p (t) \cdot e^{*} = p (0) e^{*}$ . For $p (t) = \sum_{n} a_{n} t^{n}$ and $q (t) = \sum_{n} b_{n} t^{n}$ , the constant coefficient of $p q$ is $a_{0} b_{0} = p (0) q (0)$ , and $1 (0) = 1$ ; hence this is a unital $H$ -module action. It is not a left-dual object in $H - M o d$ : $t$ acts by zero on $e^{*} \otimes e$ under the diagonal action, so evaluation sends $t \cdot (e^{*} \otimes e)$ to $0$ , while $t \cdot ev (e^{*} \otimes e) = t \cdot 1 = ε (t) 1 = 1$ . Since $0 \neq 1$ by [F5], evaluation is not $H$ -linear.

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