Taken from Franz & Gohm.

Let return to the map $b:M\times G\rightarrow M$ considered in the beginning of the previous section. If $G$ is a group, then $b:M\times G\rightarrow M$ is called a left action of $G$ on $M$, if it satisfies the following axioms expressing associativity and unit ($\checkmark0$),

$b(b(x,g),h)=b(x,hg)$, and $b(x,e)=x$

for all $x\in M$$g,h\in G$; and where $e\in G$ is the identity. As before we have the unital *-homomorphisms $\alpha_g:F(M)\rightarrow F(M)$. Actually, in order to get a representation of $G$ on $F(M)$, i.e. $\alpha_g\alpha_h=\alpha_{gh}$ for all $g,\,h\in G$ we modify  the definition and use $\alpha_g(f)(x):=f(b(x,g^{-1}))$. (Otherwise we get an anti-representation. But this is a minor point at this stage). In the associated coaction $\beta:F(M)\rightarrow F(M)\otimes F(G)$ the axioms above are turned into the coassociativity and counit properties. These make perfect sense not only for groups but also for quantum groups and we state them at once in this more general setting. We are rewarded with a particular interesting class of quantum Markov chains associated to quantum groups which we call random walks and are the subject of this lecture.

A bialgebra is a unital associative algebra $A$ equipped with two unital algebra homomorphisms $\varepsilon:A\rightarrow \mathbb{C}$ and $\Delta:A\rightarrow A\otimes A$ such that

$(I_A\otimes \Delta)\circ\Delta=(\Delta\otimes I_A)\circ \Delta$

$(I_A\otimes \varepsilon)\circ \Delta=I_A=(\varepsilon\otimes I_A)\circ \Delta$.

To see where these are motivated from (i.e. quantising $F(G)$), we will examine these relations for, once again, $\mathbf{1}_{\{g\}}$. The naive view at this point would suggest that the comultiplication and counit are of the form:

$\varepsilon(f)=f(e)$,

$\Delta(\mathbf{1}_{\{g\}})=\sum_{h\in G}\mathbf{1}_{\{gh^{-1}\}}\otimes\mathbf{1}_{\{h\}}$.

In fact this is the correct form of the comultiplication. Now for the first condition above:

$(I_A\otimes \Delta)\circ \Delta(g)=\sum_{s\in G}\mathbf{1}_{\{gs^{-1}\}}\otimes\Delta(\mathbf{1}_{\{s\}})$.

Having done a few calculations, I think that this equals:

$\sum_{s,t\in G}\mathbf{1}_{\{gs^{-1}\}}\otimes\mathbf{1}_{\{st^{-1}\}}\otimes\mathbf{1}_{\{t\}}$.

Similarly,

$(\Delta\otimes I_{F(G)})\circ\Delta(\mathbf{1}_{\{g\}})=\sum_{s,t\in G}\mathbf{1}_{\{gs^{-1}t^{-1}\}}\otimes\mathbf{1}_{\{t\}}\otimes\mathbf{1}_{\{s\}}$.

I’m sure this is supposed to encode the associativity law $g_1(g_2g_3)=(g_1g_2)g_3$, but I can’t really see how exactly. In fact seeing where the ‘group encodings’ are coming from exactly is the biggest problem. I can see how they’re supposed to encode associativity, the identity and inverses but Thomas Cooney defines $\Delta:F(G)\rightarrow F(G)\otimes F(G)=F(G\times G)$ by $\Delta(f)(g_1,g_2)=f(g_1g_2)$ and I’m failing to get to these relations via this definition.

Next the condition on the counit. First the calculations:

$(I_{F(G)}\otimes\varepsilon)\circ\Delta(\mathbf{1}_{\{g\}})=\sum_{t\in G}\mathbf{1}_{\{gt^{-1}\}}\otimes\varepsilon(\mathbf{1}_{\{t\}})=\mathbf{1}_{\{g\}}$.

Similarly, $(\varepsilon\otimes I_{F(G)})\circ\Delta=I_{F(G)}$. It is a little easier to see the encoding here of $eg=g=ge$.

If $A$ has an involution $*:A\rightarrow A$ such that $\varepsilon$ and $\Delta$ are *-homomorphisms, then we call $A$ a *-bialgebra.

Let $a\in A$ and suppose $\Delta(a)=\sum_{i=1}^na_i\otimes b_i$. If there exists furthermore a linear map $S:A\rightarrow A$ (called the antipode)  satisfying

$\sum_{i=1}^na_iS(b_i)=\varepsilon(a_1)A=\sum_{i=1}^nS(a_i)b_i$,

then we call $A$ a *-Hopf algebra. If we define a multiplication map $\nabla:A\otimes A\rightarrow A$$\nabla(a_1\otimes a_2)=a_1a_2$, then this can be rewritten as

$\nabla(I_A\otimes S)\Delta(a)=\varepsilon(a)1_A=\nabla(S\otimes I_A)\Delta(a).$

## Definition A.1

A finite quantum group is a finite dimensional C*-Hopf algebra.

Note that finite dimensional C*-algebras are very concrete objects, namely they are multi-matrix algebras $\bigoplus_{n=1}^NM_{k_n}(\mathbb{C})$ (see Theorem 6.3.8). Not every multi-matrix algebra carries a Hopf algebra structure. The simplest example is that of the group algebra of a finite group. For example, the direct sum must contain a one-dimensional summand to make possible the existence of a counit.

Proof : The only homomorphisms from $M_n(\mathbb{C})\rightarrow \mathbf{C}$ occur for $n=1$ (Second Example) $\bullet$

Let $A$ be a finite quantum group with comultiplication $\Delta$ and counit $\varepsilon$.  A C*-algebra $B$ is an $A$-comodule algebra if there exists a unital *-algebra homomorphism $\beta:B\rightarrow B\otimes A$ such that

$(\beta\otimes I_A)\circ \beta=(I_B\otimes \Delta)\circ\beta$,$(I_B\otimes\varepsilon)\circ\beta=I_B$.

Such a map $\beta$ is a coaction.

To see that this is the ‘encoding’ of an action we examine this for the algebra $F(G)$ of a finite group. The coaction is going to be related to the action of $G$ on itself — morryah the coaction of $F(G)$ onto $F(G)\otimes F(G)$ (?). Let $\beta(\mathbf{1}_{\{g\}})=\sum_{i=1}^n a_i\otimes b_i$. The calculations show, for $f=\mathbf{1}_{\{g\}}$:

$(\beta\otimes I_{F(G)})\circ\beta(f)=\sum_{i=1}^n \beta(a_i)\otimes b_i$.

$(I_{F(G)}\otimes \Delta)\circ\beta(f)=\sum_{i=1}^na_i\otimes\Delta(b_i)$.

This is supposed to encode the fact, that for actions, $b:M\times G\rightarrow M$,

$b(b(x,g),h)=b(x,hg)$,

but once again I can’t see it exactly.

Now for the other condition:

$(I_{F(G)}\otimes\varepsilon)\circ\beta(\mathbf{1}_{\{g\}})=\sum_{i=1}^na_i\varepsilon(a_i)\overset{!}{=}\mathbf{1}_{\{g\}}$.

This we might be able to do something with. $F(G)\otimes F(G)$ has basis $E=\{\mathbf{1}_{\{g_i\}}\otimes\mathbf{1}_{\{g_j\}}:g_i,\,g_j\in G\}$, we can write, for complex numbers $\{\alpha_{ij}\}_{i,j=1}^{|G|}$:

$\beta(\mathbf{1}_{\{g\}})=\sum_{g_i,g_j\in G}\alpha_{ij}(\mathbf{1}_{\{g_i\}}\otimes\mathbf{1}_{\{g_j\}})$

$\Rightarrow (I_{F(G)}\otimes\varepsilon)\circ\beta(\mathbf{1}_{\{g\}})=\sum_{g_i,\,g_j\in G}\alpha_{ij}(\mathbf{1}_{\{g_i\}}\mathbf{1}_{\{g_j\}}(e))$

$=\sum_{t\in G}\alpha_t\mathbf{1}_{\{t\}}$.

Well this means that if $g_1=e$, then the ‘$\alpha_{i1}$‘ coaction of $\beta$ acts on $\mathbf{1}_{\{g\}}$ by the vector $\delta^g\in M_p(G)$ — in other words it does nothing really, consolidating $b(x,e)=x$

If we start with such a coaction $\beta$ then we can look at the quatum Markov chain constructed in the previous section in a different way. Define for $n\geq 1$:

$k_n:A\rightarrow B\otimes\hat{A}$,

$a\mapsto 1_B\otimes1\otimes\cdots\otimes 1\otimes a\otimes1\otimes\cdots$

where $a$ is inserted at the $n$-th copy of $A$. We can interpret the $k_n$ as non-commutative random variables. Note that the $k_n$ are identically distributed (Cf. the construction of a random walk on a finite group [Equation 1.5, page 15]). Further the sequence $j_0,\,k_1,\,k_2,\dots$ is a sequence of tensor independent random variables (their ranges commute and the state acts as a product state on them). The convolution $j_0\star k_1$ is another random variable, defined by

$j_o\star k_1(b):=\sum_{i=1}^nj_0(b_i)k_1(a_i)=\sum_{i=1}b_ik_1(a_i)$.

where $\beta(b)=\sum_{i=1}^nb_i\otimes a_i$. By my calculations, this is just $j_0\star k_1=\beta$?!! This would give the homomorphism property though.

In a similar way we  can form the convolution of the $k_n$ among each other. By induction we can prove the following formulas for the random variables $j_n$ of the chain.

## Proposition 3.1

$j_n=(\beta\otimes I_A\otimes\cdots I_A)\cdots(\beta\otimes I_A\otimes I_A)(\beta\otimes I_A)\beta$

$=(I_A\otimes I_A\otimes\cdots \otimes \Delta)(\cdots)(I_A\otimes I_A\otimes\Delta)(I_A\otimes \Delta)\beta$

$=j_0\star k_1\star\cdots\star k_n$.

Proof: That the second formula is true is quickly shown using a telescoping argument. Let $P(n)$ be the proposition that the third formula is true. Clearly $P(0)$ and $P(1)$ is true. Assume that $P(m)$ is true:

$j_m=j_0\star k_1\star\cdots\star k_m$.

What about $P(m+1)$? Again using $\beta(b)=\sum_{i=1}^nb_i\otimes a_i$:

$j_m\star k_{m+1}=\sum_{i=1}^nj_m(b_i)k_{m+1}(a_i)$

$=\sum_{i=1}^nj_m(b_i)(1\otimes\cdots\otimes a_i\otimes\cdots)=j_{m+1}(b)$,

where the $a_i$ term is in the $(m+1)$st position $\bullet$

Note that by the properties of coactions  and comultiplications the convolution is associative and we do not need to insert brackets. The statement $j_n=j_0\star k_1\star \cdots\star k_n$ can be put into words by saying that the Markov chain associated to a coaction is a chain with (tensor-) independent and stationary increments (???). Using the convolution of states we can write the distribution of $j_n=j_0\star k_1\star\cdots\star k_n$ as $\psi\star \phi^{\star n}$. For all $b\in B$ and $n\geq 1$ the Transition operator satisfies

$\psi(T^n\phi(b))=\Psi(j_n(b))=\psi\star\phi^{\star n}(b)$,

and thus from this we can verify that

$T_\phi^n=(I_B\otimes \phi^{\star n})\circ\beta$.

Proof of the first equality:  First off a few calculations of powers of the transition operator:

$T_\phi(b)=\sum_{i=1}^m\phi(a_i)b_i$

$\Rightarrow T^2_\phi(b)=(I_B\otimes\phi)\sum_{i=1}^m\beta(\phi(a_i)b_i)=(I_B\otimes \phi)\sum_{i_1=1}^{m_1}\phi(a_i)\beta(b_i)$

$=(I_B\otimes \phi)\sum_{i_1=1}^{m_1}\phi(a_i)\left(\sum_{i_2=1}^{m_2}b_{i_2}\otimes a_{i_2}\right)$

$=\sum_{i_1=1}^{m_1}\phi(a_i)\left(\sum_{i_2=1}^{m_2}\phi(a_{i_2})b_{i_2}\right)$

Inductively,

$T^n_\phi(b)=\sum_{i_1=1}^{m_1}\phi(a_{i_1})\left(\sum_{i_2=1}^{m_2}\phi(a_{i_2})\left(\cdots\left(\sum_{i_n=1}^{m_n}\phi(a_{i_n})b_{i_n}\right)\cdots\right)\right)$

Now hitting this with $\psi$ we hit the $b_{i_n}$ term. We must get an expression for $j_n(b)$ therefore.  We have that $j_o(b)=b$$j_1(b)=\beta(b)=\sum_{i=1}^mb_i\otimes a_i$. Now

$j_2(b)=(j_1\otimes I_A)\circ\beta=(j_1\otimes I_A)\left(\sum_{i=1}^mb_i\otimes a_i\right)$

$=\sum_{i=1}^mj_1(b)\otimes a_i=\sum_{i=1}^m\beta(b)\otimes a_i$

$=\sum_{i_1=1}^{m_1}\left(\sum_{i_2=1}^{m_2}b_{i_2}\otimes a_{i_2}\right)\otimes a_{i_1}$.

Hence, inductively:

$j_n(b)=\sum_{i_1=1}^{m_1}\left(\cdots\left(\sum_{i_{n-1}=1}^{m_{n-1}}\left(\sum_{i_n=1}^{m_n}b_{i_n}\otimes a_{i_n}\right)\otimes a_{i_{n-1}}\right)\cdots\right)\otimes a_{i_1}$.

Hitting this with $\Psi$ yields $\psi(T_\phi^n(b))=\Psi(j_n(b))$ we have that $\psi\circ T_\phi^n=\Psi\circ j_n$ $\bullet$

i.e. given $\beta$, the semigroup of transition operators $\{T_\phi^n\}$ and the semigroup $\{\phi^{\star n}\}$ of convolution powers of the transition state are essentially the same thing.

A quantum Markov chain associated to such a coaction is called a random walk on the $A$-comodule algebra $B$. We have seem that in the commutative case this construction describes the action of a group on a set and the random walk derived from it. Because of this background, some authors call a coaction the action of a quantum group.

## Theorem A.2

Let $A$ be a finite quantum group. Then there exists a unique state $\eta$ on $A$ such that

$(I_A\otimes\eta)\circ\Delta(a)=\eta(a)1_A$for all $a\in A$.

The state $\eta$ is called the Haar State of $A$. The defining property (above) is called left-invariance. On finite (and more generally on compact) quantum groups, left invariance is equivalent to right invariance; i.e. the Haar state satisfies also

$(\eta\otimes I_A)\circ\Delta(a)=\eta(a)1_A$.

One can show that it is a faithful trace, i.e. $\eta(a^*a)=0$ implies $a=0$ and

$\eta(ab)=\eta(ba)$, for $a,\,b\in A$.

Using the unique Haar state, we also get the distinguished inner product on $A$, namely for $a,\,b\in A$:

$\langle a,b\rangle\eta(a^*b)$.

This theorem is stated in the paper without proof. A proof is found here (including the faithfulness property). This is the approach we follow.

Proof : Let $A$ be a finite quantum group and let $A^*$ be it’s dual. We know that the functionals on $A^*$$A^{**}$ are of the form:

$\hat{a}(\varphi)=\varphi(a)$,

for $a\in A$ and $\varphi\in A^*$.

### Claim 1

There exists a non-zero element $h\in A$ such that $ah=\varepsilon(a)h$.

Proof : Let $\{e_i\}$ be a basis for $A$ and let $\{f_i\}$ be the dual basis (of $A^*$). Suppose that $\Delta(e_i)=\sum_{j=1}^{n_i}a_{ij}\otimes b_{ij}$ and let $J_k=\{1,2,\dots,n_k\}$ for $i=1,2,\dots,n$.  For any $c\in A$ define $h\in A$ by:

$h=\sum_{i=1;\,j\in J_i}^{i=|A|}\langle f_i|S^2(b_{ij})c\rangle a_{ij}$.

Take $a\in A$ (this first line makes no sense to me actually: I thought the counit $\varepsilon:A\rightarrow \mathbb{C}$ was a homomorphism so that $\varepsilon(\alpha a)=\alpha \varepsilon(a)$. I therefore cannot see how the counit gets inside the bracket to take advantage of the $\sum a_iS(b_i)=\varepsilon(a)1_A$ relation?? Therefore I might just ‘take his word’ as I can’t follow an argument I don’t understand.$\bullet$

This proves the existence of left Haar functional on $A^*$. By duality we also get the existence of a left Haar measure on $A$ (I presume this means via the map $\varphi:A\rightarrow A^{*}$$h=\sum_{i=1}^{|A|}\alpha_i e_i\mapsto\sum_{i=1}^{|A|}\alpha_if_i$ ??).

Uniqueness will follow easily from the next result.

### Claim 2

Let $h$ be a non-zero element of  $A$ such that $ah=\varepsilon(a)h$ for all $a\in H$. Then the map $A^*\rightarrow A$$\rho\mapsto (I_A\otimes \rho)\Delta(h)$ is bijective.

Proof : It is enough to prove that this map is injective. We can show that the map is surjective if for any complex numbers $\{\beta_{k}:k=1,\dots,|A|\}$$\{\alpha_{ij}:i,=1,2,\dots,<,\,j=1,\dots,|A|\}$ (where $M\leq|A|^2$), $\{\gamma_k:k=1,\dots,|A|\}$ we can solve the $M|A|$ equations for the $|A|$ unknowns $\{x_j\}$:

$\gamma_k=\beta_k\left(\sum_{j=1}^{|A|}\alpha_{ij}x_j\right)$

where $x_j=\rho(e_j)$…(!)

So assume $\rho\in A^*$ and $(I_A\otimes\rho)\circ\Delta(h)=0$ (the map is linear so one-to-one at zero implies injective)… actually I’m totally confused looking at this so I’m just going to assume the existence of the Haar measure at this point $\bullet$

Concerning stationarity we get:

## Proposition 3.2

For a state $\psi$ on $B$ the following assertions are equivalent:

1. $(\psi\otimes I_A)\circ\beta(b)=\psi(b)1_A$
2. $(\psi\otimes\phi)\circ\beta=\psi$ for all states $\phi$ on $A$
3. $(\psi\otimes\eta)\circ\beta=\psi$, where $\eta$ is the Haar state on $A$.
Proof : That (a) $\Rightarrow$ (b) can be seen by applying $\phi$ ti both sides of (a). I’m not so sure of how to go the other way although it is probably pretty clear. (b) $\Rightarrow$ (c) is patently obvious.
Assuming (c) and using the invariance properties of $\eta$ we get for all states $\phi$ on $A$:
$\psi=(\psi\otimes\eta)\beta=$… I can’t make head nor tail of this! $\bullet$

## The Eight-Dimensional Kac-Paljutkin Quantum Group

In this section we give the defining relations and the main structure of an eight-dimensional quantum group introduced by Kac and Paljutkin. This is actually the smallest finite quantum groups that is not a group algebra (e.g. $F(\mathbb{Z}_5)$). In other words, it is the non-commutative C*-Hopf algebra of smallest dimension (Franz & Gohm talk about both the group algebra and the algebra of functions on the group: what is the difference? He also talks about commutative and cocommutative. What is the difference?).

Consider the multi-matrix algebra $A=\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\mathbb{C}\oplus M_2(\mathbb{C})$, with the usual multiplication and involution. We shall use the basis $e_1=(1,0,0,0,0)$ (with $e_2,\,e_3,\,e_4$ defined in the same way) and

$a_{11}=0+ 0+ 0+ 0+\left(\begin{array}{cc}1&0\&0\end{array}\right)$,

with the other $a_{ij}$ defined in the same way. The algebra $A$ is an eight-dimensional C*-algebra. Its unit is of course $1_A=e_1+e_2+e_3+e_4+a_{11}+a_{22}$. We shall need the trace $\text{Tr}$ on $A$.

$\text{Tr}\left(x_1+x_2+x_3+x_4+\left(\begin{array}{cc}c_{11} & c_{12}\\ c_{21}&c_{22}\end{array}\right)\right)=x_1+x_2+x_3+x_4+c_{11}+c_{22}$.

Note that Tr is normalised to be equal to one on minimal projections ($\checkmark$).

The following defines a coproduct on $A$,

$\Delta(e_1)=e_1\otimes e_1+e_2\otimes e_2+e_3\otimes e_3+e_4\otimes e_4+$

$\frac{1}{2}a_{11}\otimes a_{11}+\frac{1}{2}a_{12}\otimes a_{12}+\frac{1}{2}a_{21}\otimes a_{21}+\frac{1}{2}a_{22}\otimes a_{22}$,

$\Delta(e_2)=e_1\otimes e_2+e_2\otimes e_1+e_3\otimes e_4+e_4\otimes e_3+$

$\frac{1}{2}a_{11}\otimes a_{22}+\frac{1}{2}a_{22}\otimes a_{11}+\frac{i}{2}a_{21}\otimes a_{12}-\frac{i}{2}a_{12}\otimes a_{21}$,

$\Delta(e_3)=e_1\otimes e_3+e_3\otimes e_1+e_2\otimes e_4+e_4\otimes e_2+$

$\frac{1}{2}a_{11}\otimes a_{11}+\frac{1}{2}a_{22}\otimes a_{11}-\frac{i}{2}a_{21}\otimes a_{12}+\frac{i}{2}a_{12}\otimes a_{21}$,

$\Delta(e_4)=e_1\otimes e_4+e_4\otimes e_1+e_2\otimes e_3+e_3\otimes e_2+$

$\frac{1}{2}a_{11}\otimes a_{11}+\frac{1}{2}a_{22}\otimes a_{22}-\frac{1}{2}a_{12}\otimes a_{21}-\frac{1}{2}a_{21}\otimes a_{21}$,

$\Delta(a_{11})=e_1\otimes a_{11}+a_{11}\otimes e_1+e_2\otimes a_{22}+a_{22}\otimes e_2+$

$e_3\otimes a_{22}+a_{22}\otimes e_3+e_4\otimes a_{11}+a_{11}\otimes e_4$,

$\Delta(a_{12})=e_1\otimes a_{12}+a_{12}\otimes e_1+ie_2\otimes a_{21}-ia_{12}\otimes e_2+$

$-ie_3\otimes a_{21}+ia_{21}\otimes e_3-e_4\otimes a_{12}-a_{12}\otimes e_4$,

$\Delta(a_{21})=e_1\otimes a_{21}+a_{21}\otimes e_1-ie_2\otimes a_{12}+ia_{12}\otimes e_2+$

$ie_3\otimes a_{12}-ia_{12}\otimes e_3-e_4\otimes a_{21}-a_{21}\otimes e_4$,

$\Delta(a_{22})=e_1\otimes a_{22}+a_{22}\otimes e_1+e_2\otimes a_{11}+a_{11}\otimes e_2+$

$e_3\otimes a_{11}+a_{11}\otimes e_3+e_{4}\otimes a_{22}+a_{22}\otimes e_4$.

The counit is given by (looking at this we can see the relationship between the counit and the comultiplication with respect to the unit and multiplication in a commutative, group algebra, case. To encode $ge=g=eg$ any time there is a term of the form $x\otimes e_1$, there must be a term of the form $e_1\otimes x$ — to capture the left and right symmetry $R_\varepsilon=(I_A\otimes\varepsilon)\circ\Delta=(\varepsilon\otimes I_A)\circ \Delta=L_\varepsilon$. This also shows that, in this case, $\Delta(x)$ must contain $x\otimes e_1$ and $e_1\otimes x$ to encode $L_\varepsilon=I_A=R\varepsilon$):

$\varepsilon\left(x_1+x_2+x_3+x_4+\left(\begin{array}{cc}c_{11} &c_{12}\\ c_{21}&c_{22}\end{array}\right)\right)=x_1$.

The antipode is the transpose map, i.e.

$S(e_i)=e_i$, and $S(a_{jk})=a_{ji}$.

Which I’ve checked satisfies $\nabla(I_A\otimes S)\Delta(a)=\varepsilon(a)1_A=\nabla(S\otimes I_A)\delta(a)$ for $a=e_1$ and $e_2$.

### The Haar State

Finite quantum groups have unique Haar elements $h$ satisfying $h^*=h=h^2$$\varepsilon(h)=1$ and

$ah=\varepsilon(a)h=ha$, for all $a\in A$.

For the Kac-Paljutkin quantum group, it is easy to see that $h=e_1$. An invariant functional is given by $\phi(a)=\text{Tr}(aK^{-1})$ with $K=(\text{Tr}\otimes I_A)\Delta(h)-e_1+e_2+e_3+e_4+\frac{1}{2}(a_{11}+a_{22})$ and $K^{-1}=e_1+e_2+e_3+e_4+2(a_{11}+a_{22})$. On an arbitrary element of $A$ the action of $\phi$ is given by

$\phi\left(x_1+x_2+x_3+x_4+\left(\begin{array}{cc}c_{11}&c_{12}\\ c_{21}&c_{22}\end{array}\right)\right)=x_1+x_2+x_3+x_4+2c_{11}+2c_{22}$.

Normalising $\eta$ so that $\eta(1)=1$, we get the Haar state $\eta=\frac{1}{8}\phi$.

### The dual of $A$

The dual $A^*$ of a finite quantum group is again a finite quantum group. Its morphisms are the duals of the morphisms (do these look right???) of $A$, e.g.

$\nabla_{A^*}:\Delta^*_A:A^*\otimes A^*\cong (A\otimes A)^*\rightarrow A^*$$\nabla_{A^*}(\phi_1\otimes \phi_2)=(\phi_1\otimes\phi_2)\circ\Delta$

and

$\Delta_{A^*}=\nabla^*_{A}:A^*\rightarrow A^*\otimes A^*\cong(A\otimes A)^*$$\Delta_{A_{A^*}}\phi=\phi\circ m_A$.

The involution of $A^*$ is given by $\phi^*(a)=\overline{\phi(S(a)^*)}$ for $\phi\in A^*$$a\in A$ (this is done by showing that the routine calculation $a=S(S(a)^*)^*$ holds for all $a\in A.$).

To show that $A^*$ is indeed a C*-algebra, once can show that the dual regular action of $A^*$ on $A$ defined by $T_\phi a=\sum_{i=1}^n\phi(b_i)a_i$ for $\phi\in A^*$$a\in A$, is a faithful *-representation of $A^*$ with respect to the inner product on $A$ defined by

$\langle a,b\rangle=\eta(a^*b)$

for $a,\,b\in A$.

For the Kac-Paljutkin quantum group $A$ the dual $A^*$ actually turns out to be isomorphic to $A$ itself.

Denote by $\{\eta_1,\eta_2,\eta_3,\eta_4,\alpha_{11},\alpha_{12},\alpha_{21},\alpha_{22}\}$ the basis of $A^*$ that is dual to $\{e_1,e_2,e_3,e_4,a_{11},a_{12},a_{21},a_{22}\}$, i.e. the functionals on $A$ defined by

$\eta_i(e_j)=\delta_{ij}$$\eta_{i}(a_{rs})=0$,

$\alpha_{kl}(e_j)=0$$\alpha_{kl}(\alpha_{rs})=\delta_{kr}\delta_{ls}$,

for $i,j=1,2,3,4$$k,l,r,s=1,2$,

We leave the following as an exercise (yeah; what is a minimal projection? I have a little proposition (16.5 (a)) but I can’t quite seem to make sense of how it applies to $A^*$):

The functionals

$f_1=\frac{1}{8}(\eta_1+\eta_2+\eta_3+\eta_4+2\alpha_{11}+2\alpha_{22})$,

$f_2=\frac{1}{8}(\eta_1-\eta_2-\eta_3+\eta_4-2\alpha_{11}+2\alpha_{22})$,

$f_3=\frac{1}{8}(\eta_1-\eta_2-\eta_3+\eta_4+2\alpha_{11}-2\alpha_{22})$,

$f_4=\frac{1}{8}(\eta_1+\eta_2+\eta_3+\eta_4-2\alpha_{11}-2\alpha_{22})$,

are minimal projections in $A^*$, Furthermore

$b_{aa}=\frac{1}{4}(\eta_1+\eta_2-\eta_3-\eta_4)$,

$b_{12}=\frac{1-i}{2\sqrt{2}}(\alpha_{12}+i\alpha_{21})$,

$b_{21}=\frac{1+i}{2\sqrt{2}}(\alpha_{12}-i\alpha_{21})$,

$b_{22}=\frac{1}{4}(\eta_1-\eta_2+\eta_3-\eta_4)$,

are matrix units; i.e. satisfy the relations

$b_{ij}b_{kl}=\delta_{jk}b_{il}$ and $b_{ij}^*=b_{ji}$,

and the mixed products vanish,

$f_ib_{jk}=0=b_{jk}f_i$$i=1,2,3,4$$j,k=1,2$.

Therefore $A^*\cong\mathbb{C}^4\oplus M_2(\mathbb{C})\cong \mathbb{A}$ as an algebra. But actually, $e_i\mapsto f_i$ and $a_{ij}\mapsto b_{ij}$ even defines a C*-Hopf algebra isomorphism from $A$ to $A^*$.

### The States on $\mathcal{A}$

On $\mathbb{C}$ there exists only one state, the identity map. States on $M_2(\mathbb{C})$ are given by density matrices, i.e. positive semi-definite matrices with trace one. More precisely, for any state $\rho$ on $M_2(\mathbb{C})$ there exists a unique density matrix $\theta\in M_2(\mathbb{C})$ such that

$\rho(A)=\text{Tr }(\theta A)$,

for all $A\in M_2(\mathbb{C})$. The $2\times 2$ density matrices can be parameterised by the unit ball $B_1=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2\leq1\}$,

$\displaystyle \theta(x,y,z)=\frac{1}{2}\left(\begin{array}{cc}1+z&x+iy\\x-iy&1-z\end{array}\right)$.

A state on $\mathcal{A}$ is a convex combination of states on the four copies of $\mathbb{C}$ and a state on $M_2(\mathbb{C})$. All states on $\mathcal{A}$ can therefore be parameterised by $B_1$ and the set

$\{(\mu_1,\mu_2,\mu_3,\mu_4,\mu_5):\sum_{i}\mu_i=1,\,\mu_i\geq 0\}$.

### Example 3.3

Consider the commutative subalgebra $\mathcal{B}=\mathbb{C}^4$ of the Kac-Paljutkin Quantum Group $\mathcal{A}$ with standard basis $v_1,\dots,v_4$ and component-wise multiplication. Franz & Gohh on page 9 define an $\mathcal{A}$-coaction $\beta$ which makes $\mathbb{C}$ into an $\mathcal{A}$-comodule algebra.

Let $\rho$ be an arbitrary state on $\mathcal{A}$. It can be parameterised by the $\mu_i$ and $x,\,y,\,z$. Then the transition operator $T_\rho=(I_{\mathcal{A}}\otimes\rho)\circ\beta$ on $\mathbb{C}^4$ has a matrix representation, with respect to the basis $\{v_i\}_{i=1,\dots,4}$, given on p.10 of Franz & Gohm.

The state $\pi:\mathcal{B}\rightarrow\mathbb{C}$ defined by $\pi(v_i)=1/4$ for $i=1,\dots,4$ is invariant, i.e. we have

$\pi\star\rho=(\pi\otimes\rho)\circ\beta=\pi$

for any state $\rho$ on $\mathcal{A}$.