Bruce Bartlett Frobenius algebra stuff

This represents some previous work with Jamie Vicary.

Frobenius algebra stuff

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

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

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

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

Note: in the following lemma, I need to use a different definition of ** then the one on page 3 of your paper. See bottom of page. What is going on? (Bruce: I think this is old)

Proposition. Suppose AA is a Frobenius algebra with pairing (,)(\cdot, \cdot ) and also a Hilbert space with inner product ,\langle \cdot, \cdot \rangle. Define the map *:AA* : A \rightarrow A by setting a *a^* to be the unique element such that a *,b=(a,b)\langle a^*, b \rangle = (a, b). Then the following are equivalent: * AA is a \dagger-Frobenius algebra. * (ab) *=b *a *(ab)^* = b^* a^* and (a *) *=a(a^*)^* = a. (It will follow automatically that 1 *=11^* = 1). * L a =L a *L_a^\dagger = L_{a^*}.

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

(1)(ab) *,c =(ab) *1,c =1,(ab)c =1,a(bc) =a *,bc =b *a *,c. \begin{aligned} \langle (ab)^*, c \rangle &= \langle (ab)^* 1, c \rangle \\ &=\langle 1, (ab) c \rangle \\ &=\langle 1, a (bc) \rangle \\ &= \langle a^*, bc \rangle \\ &= \langle b^* a^*, c \rangle. \end{aligned}

To show that 1 *=11^* = 1, we compute:

(2)1 *,a =1 *1,a =1,1a =1,a. \begin{aligned} \langle 1^*, a \rangle &= \langle 1^* 1, a \rangle \\ &= \langle 1, 1a \rangle \\ &= \langle 1, a \rangle. \end{aligned}

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

(3)(a *) *,b =(a *) *1,b =1,a *b =a *b,1¯ =b,a¯ =a,b. \begin{aligned} \langle (a^*)^*, b \rangle &= \langle (a^*)^* 1, b \rangle \\ &= \langle 1, a^* b \rangle \\ &= \overline{\langle a^* b, 1 \rangle } \\ &= \overline{ \langle b, a \rangle } &= \langle a, b \rangle. \end{aligned}

(2) \Rightarrow (3):

(4)L a b,c =b,L ac =b,ac =(b *) *,ac =(b *,ac) =(b *a,c) =(b *a) *,c =a *b,c \begin{aligned} \langle L_a^\dagger b, c \rangle &= \langle b, L_a c \rangle \\ &= \langle b, ac \rangle \\ &= \langle (b^*)^*, ac \rangle \\ &= (b^*, ac) \\ &= (b^*a, c) \\ &= \langle (b^*a)^*, c \rangle \\ &= \langle a^*b, c \rangle \end{aligned}

which is what we needed to show.

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

We calculate:

(5)(a,b) =1,(a,b) =1,u (ab) =u(1),ab A =1,ab =a *,b. \begin{aligned} (a,b) &= \langle 1, (a,b) \rangle_{\mathbb{C}} \\ &= \langle 1, u^\dagger (ab) \rangle_{\mathbb{C}} \\ &= \langle u(1), ab\rangle_A \\ &= \langle 1, ab \rangle &= \langle a^*, b \rangle. \end{aligned}

Since ** is an isomorphism, this shows that the pairing (a,b)(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 dd is actually m m^\dagger.

\square

Here is another thing I am puzzled about. By my calculations, your formula for α\alpha' on page 3 computes as (if you allow me to change right multiplication into left multiplication): The element a *a^* is the unique element such that a *,b=1,ab\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 a *,b=(a,b)\langle a^*, b \rangle = (a, b). Here’s a quick lemma though:

Lemma. If we define a *a^* via a *,b=(a,b)\langle a^*, b \rangle = (a,b), and if we assume that 1 *=11^* = 1, then your (Jamie’s) definition automaticaly follows. Proof. We compute:

(6)a *,b =(a,b) =(1 *,ab) =(1,ab). \begin}{aligned} \langle a^*, b \rangle &= (a,b) \\ &= (1^*, ab) \\ &= (1, ab). \\ \end{aligned}

I am unable to go in the reverse direction!

Revised on May 12, 2010 at 13:58:51 by Bruce Bartlett