Bruce Bartlett Frobenius algebra stuff

Frobenius algebra stuff

We define a $\dagger$-Frobenius algebra (following your paper ‘Categorical formulation of quantum algebras’) as an algebra $(A, m, u)$ equipped with an inner product $\langle , \rangle$ which has the property that $(A, m, u, m^\dagger, u^\dagger)$ is a Frobenius algebra.

The following lemma shows that simultaneously demanding that the comultiplication is $m^\dagger$ and that the Frobenius form is $u^\dagger$ is redundant; i.e. if one is true then the other is automatically true.

Lemma. Suppose $(A, m, u)$ is an algebra equipped with an inner product $\langle, \rangle$. Then the following are equivalent:

• $A$ is a $\dagger$-Frobenius algebra.
• The form $u^\dagger \colon A \rightarrow \mathbb{C}$ gives $A$ the structure of a Frobenius algebra, and in this Frobenius algebra, the comultiplication turns out to be $m^\dagger$.
• The comultiplication map $m^\dagger \colon A \rightarrow A \otimes A$ gives $A$ the structure of a Frobenius algebra, and in this Frobenius algebra, the Frobenius form turns out to be $u^\dagger$.

Proof I computed it in a basis… would prefer a more elegant proof… \square

Proposition. Suppose $A$ is a Frobenius algebra with pairing $(\cdot, \cdot )$ and also a Hilbert space with inner product $\langle \cdot, \cdot \rangle$. Define the map $* : A \rightarrow A$ by setting $a^*$ to be the unique element such that $\langle a^*, b \rangle = (a, b)$. Then the following are equivalent: * $A$ is a $\dagger$-Frobenius algebra. * $(ab)^* = b^* a^*$ and $(a^*)^* = a$. (It will follow automatically that $1^* = 1$). * $L_a^\dagger = L_{a^*}$.

Proof. (3) $\Rightarrow$ (2). To show that $(ab)^* = b^*a^*$, we compute:

To show that $1^* = 1$, we compute:

To show that $(a^*)^* = a$, we need to use the symmetry of the inner product:

(2) $\Rightarrow$ (3):

which is what we needed to show.

(2) and (3) $\Rightarrow$ (1): We use the definition of a Frobenius algebra as “an algebra $A$ equipped with a form $\theta \colon A \rightarrow \mathbb{C}$ such that the pairing defined by $(a,b) = \theta(ab)$ is nondegenerate”. We want to show that $u^\dagger$ is a Frobenius form on $A$; it will automatically follow that the comultiplication map must be $m^\dagger$.

We calculate:

Since $*$ is an isomorphism, this shows that the pairing $(a,b)$ is nondegenerate (as the inner product is nondegenerate).

Now we should check that in this Frobenius algebra, we really do have that the comultiplication $d$ is actually $m^\dagger$.

$\square$

Here is another thing I am puzzled about. By my calculations, your formula for $\alpha'$ on page 3 computes as (if you allow me to change right multiplication into left multiplication): The element $a^*$ is the unique element such that $\langle a^*, b \rangle = \langle 1, ab \rangle$. But I have been unable to use this for the above lemma! Instead, I was forced to make the definition that $\langle a^*, b \rangle = (a, b)$. Here’s a quick lemma though:

Lemma. If we define $a^*$ via $\langle a^*, b \rangle = (a,b)$, and if we assume that $1^* = 1$, then your (Jamie’s) definition automaticaly follows. Proof. We compute:

I am unable to go in the reverse direction!