3 Algebra and co-algebras

Recall that an algebra \(\mathcal{A}\) is given by a multiplication map \(m\colon \mathcal{A}\otimes \mathcal{A}\to \mathcal{A}\) and a unit map \(\eta \colon \mathbb {C}\to \mathcal{A}\), with associativity \(m(m\otimes \operatorname {id})=m(\operatorname {id}\otimes \, m)\) and the property \(m(\eta \otimes \operatorname {id})=\operatorname {id}=m(\operatorname {id}\otimes \, \eta )\). We say \((\mathcal{A},m,\eta )\) is an algebra when those properties are satisfied.

Also recall that we say \(\mathcal{A}\) is a co-algebra when it has a co-multiplication map \(\mu \colon \mathcal{A}\to \mathcal{A}\otimes \mathcal{A}\) and a co-unit map \(\varpi \colon \mathcal{A}\to \mathbb {C}\) such that co-associativity is satisfied, i.e., \((\mu \otimes \operatorname {id})\mu =(\operatorname {id}\otimes \, \mu )\mu \), and the property \((\varpi \otimes \operatorname {id})\mu =\operatorname {id}=(\operatorname {id}\otimes \, \varpi )\mu \) is satisfied. We say \((\mathcal{A},\mu ,\varpi )\) is a co-algebra when those properties are satisfied.

Theorem 3.0.1 the Frobenius equations

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Given an algebra and co-algebra \((\mathcal{A},m,\eta ,\mu ,\varpi )\), then if

\[ (\operatorname {id}\otimes \, m)(\mu \otimes \operatorname {id})=(m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu ), \]

then we get the following equations (the Frobenius equations),

\[ (\operatorname {id}\otimes \, m)(\mu \otimes \operatorname {id})=\mu m=(m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu ). \]

Proof ▶

\begin{align*} (m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu ) & = (m\otimes \operatorname {id})((\varpi \otimes \operatorname {id})\mu \otimes \mu ) & \text{by (counit.id)}\\ & = (m\otimes \operatorname {id})(\varpi \otimes \operatorname {id}^{\otimes 3})\mu ^{\otimes 2}\\ & = (m(\varpi \otimes \operatorname {id}^{\otimes 2})\otimes \operatorname {id})\mu ^{\otimes 2}\\ & = ((\operatorname {id}_\mathbb {C}\otimes \, m)(\varpi \otimes \operatorname {id}^{\otimes 2})\otimes \operatorname {id})\mu ^{\otimes 2}\\ & = ((\varpi \otimes \operatorname {id})(\operatorname {id}\otimes \, m)\otimes \operatorname {id})\mu ^{\otimes 2}\\ & = (\varpi \otimes \operatorname {id}^{\otimes 2})(\operatorname {id}\otimes \, m\otimes \operatorname {id})(\mu \otimes \operatorname {id}^{\otimes 2})(\operatorname {id}\otimes \, \mu )\\ & = (\varpi \otimes \operatorname {id}^{\otimes 2})((\operatorname {id}\otimes \, m)(\mu \otimes \operatorname {id})\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu )\\ & = (\varpi \otimes \operatorname {id}^{\otimes 2})((m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu )\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu ) & \text{by hyp}\\ & = (\varpi \otimes \operatorname {id}^{\otimes 2})(m\otimes \operatorname {id}^{\otimes 2})(\operatorname {id}\otimes \, (\mu \otimes \operatorname {id})\mu )\\ & = (\varpi m\otimes \operatorname {id}^{\otimes 2})(\operatorname {id}\otimes \, (\operatorname {id}\otimes \mu )\mu ) & \text{by (comul.assoc)}\\ & = (\varpi \otimes \operatorname {id}^{\otimes 2})(m\otimes \operatorname {id}^{\otimes 2})(\operatorname {id}^{\otimes 2}\otimes \mu )(\operatorname {id}\otimes \, \mu )\\ & = (\varpi \otimes \operatorname {id}^{\otimes 2})(\operatorname {id}\otimes \, \mu )(m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu )\\ & = (\varpi \otimes \operatorname {id}^{\otimes 2})(\operatorname {id}\otimes \, \mu )(\operatorname {id}\otimes \, m)(\mu \otimes \operatorname {id}) & \text{by hyp}\\ & = (\varpi \otimes \operatorname {id}^{\otimes 2})(\operatorname {id}\otimes \, \mu m)(\mu \otimes \operatorname {id})\\ & = (\operatorname {id}_\mathbb {C}\otimes \, \mu m)(\varpi \otimes \operatorname {id}^{\otimes 2})(\mu \otimes \operatorname {id})\\ & = \mu m((\varpi \otimes \operatorname {id})\mu \otimes \operatorname {id})\\ & = \mu m(\operatorname {id}\otimes \, \operatorname {id}) & \text{by (counit.id)}\\ & = \mu m. \end{align*}

Definition 3.0.2

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We say an algebra and co-algebra \((\mathcal{A},m,\eta ,\mu ,\varpi )\) is a Frobenius algebra when it satisfies the Frobenius equation condition,

\[ (m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu )=(\operatorname {id}\otimes \, m)(\mu \otimes \operatorname {id}). \]

Corollary 3.0.3

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Given a Frobenius algebra \((\mathcal{A},m,\eta ,\mu ,\varpi )\), we get,

\[ (m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu \eta )=\mu . \]

Proof ▶

By Theorem 3.0.1 and definition.

Corollary 3.0.4

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Given a Frobenius algebra \((\mathcal{A},m,\eta ,\mu ,\varpi )\), we get,

\[ (\operatorname {id}\otimes \, m)(\mu {\eta }\otimes \operatorname {id})=\mu . \]

Proof ▶

By Theorem 3.0.1 and definition.

Corollary 3.0.5

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Given a Frobenius algebra \((\mathcal{A},m,\eta ,\mu ,\varpi )\), we get,

\[ (\operatorname {id}\otimes \, \varpi m)(\mu \otimes \operatorname {id})=m. \]

Proof ▶

Use Corollary 3.0.4.

Corollary 3.0.6

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Given a Frobenius algebra \((\mathcal{A},m,\eta ,\mu ,\varpi )\), we get,

\[ (\varpi m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu )=m. \]

Proof ▶

Use Corollary 3.0.3.

The following two results are known as the “snake equations”.

Corollary 3.0.7

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Given a Frobenius algebra \((\mathcal{A},m,\eta ,\mu ,\varpi )\), we get,

\[ (\varpi m\otimes \operatorname {id})(\operatorname {id}\otimes \, \mu \eta )=\operatorname {id}. \]

Proof ▶

Use Corollary 3.0.6.

Corollary 3.0.8

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Given a Frobenius algebra \((\mathcal{A},m,\eta ,\mu ,\varpi )\), we get,

\[ (\operatorname {id}\otimes \, \varpi m\otimes )(\mu \eta \otimes \operatorname {id})=\operatorname {id}. \]

Proof ▶

Use Corollary 3.0.5.

3.1 Finite-dimensional algebras

Proposition 3.1.1

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Let \((\mathcal{A},m,\eta )\) be a finite-dimensional algebra and Hilbert space. Then we can form a co-algebra by letting \(m^*\) be the co-multiplication and \(\eta ^*\) be the co-unit.

Proof ▶

comul_assoc:

We have the following equivalences,

\begin{align*} (m^*\otimes \operatorname {id})m^*=(\operatorname {id}\otimes \, m^*)m^* & \Leftrightarrow {(m(m\otimes \operatorname {id}))}^*={(m(\operatorname {id}\otimes \, m)}^*\\ & \Leftrightarrow m(m\otimes \operatorname {id})=m(\operatorname {id}\otimes \, m), \end{align*}

which is true since \(\mathcal{A}\) is an algebra.

counit_comul_id:

Similarly, taking adjoints, we get

\begin{align*} & \, (\eta ^*\otimes \operatorname {id})m^*=\operatorname {id}=(\operatorname {id}\otimes \, \eta ^*)m^* \, \Leftrightarrow \, m(\eta \otimes \operatorname {id})=\operatorname {id}=m(\operatorname {id}\otimes \, \eta ), \end{align*}

which is true since \(\mathcal{A}\) is an algebra.

Lemma 3.1.2

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Let \((\mathcal{A},m,\eta )\) be a finite-dimensional algebra and Hilbert space. Then the counit is exactly \(\left\langle 1\right\rvert \).

Proof ▶

By definition and Lemma 1.1.6.

Proposition 3.1.3

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Let \((\mathcal{A}_1,m_1,\eta _1),(\mathcal{A}_2,m_2,\eta _2)\) be finite-dimensional algebras and Hilbert spaces, and let \(f\colon \mathcal{A}_1\to \mathcal{A}_2\) be a linear map. Then \(f\) is an algebra homomorphism if and only if \(f^*\) is a co-algebra homomorphism.

Proof ▶

\begin{align*} f^*\text{ is a co-algebra hom} & \Leftrightarrow (f^*\otimes f^*)\circ m_2^*=m_1^*\circ f^* \text{ and }\eta _1^*\circ f^*=\eta _2^*\\ & \Leftrightarrow m_2\circ (f\otimes f)=f\circ m_1\text{ and }f\circ \eta _1=\eta _2\\ & \Leftrightarrow f\text{ is an algebra hom}. \end{align*}

Proposition 3.1.4

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For any finite-dimensional algebra \((B,m,\eta )\) that is also a Hilbert space, if \(\left\langle xy \mid z\right\rangle =\left\langle y \mid x^*z\right\rangle \) for all \(x,y,z\in B\), then \((B,m,\eta )\) is a Frobenius algebra.

Proof ▶

We already know that our algebra is also a co-algebra using Proposition 3.1.1 with co-multiplication \(m^*\) and co-unit \(\eta ^*\). So then it suffices to show \((m\otimes \, \operatorname {id})(\operatorname {id}\otimes \, m^*)=m^*m\), as we get the other equality by taking adjoints.

Let \(x,y,z,w\in {B}\). Let \(m^*(y)=\sum _i\alpha _i\otimes \beta _i\) for some tuples \((\alpha _i),(\beta _i)\) in \(B\). Then we compute,

\begin{align*} \left\langle (m\otimes \operatorname {id})(\operatorname {id}\otimes \, m^*)(x\otimes y) \mid z\otimes w\right\rangle & = \sum _i\left\langle (m\otimes \operatorname {id})(x\otimes \alpha _i\otimes \beta _i) \mid z\otimes w\right\rangle \\ & = \sum _i\left\langle x\alpha _i\otimes \beta _i \mid z\otimes w\right\rangle = \sum _i\left\langle x\alpha _i \mid z\right\rangle \left\langle \beta _i \mid w\right\rangle \\ = \sum _i\left\langle \alpha _i \mid x^*z\right\rangle \left\langle \beta _i \mid w\right\rangle \\ & = \sum _i\left\langle \alpha _i\otimes \beta _i \mid x^*z\otimes w\right\rangle =\left\langle m^*(y) \mid x^*z\otimes w\right\rangle \\ & = \left\langle y \mid x^*zw\right\rangle =\left\langle xy \mid zw\right\rangle \\ & = \left\langle m^*m(x\otimes y) \mid z\otimes w\right\rangle . \end{align*}

Thus \((m\otimes \operatorname {id})(\operatorname {id}\otimes \, m^*)=m^*m\).