# Image of Comultiplication on Finite Quantum Groups/Hopf Algebras

Let $$A=:F(G)$$ be the algebra of functions on a finite quantum groups aka a finite dimensional C*-Hopf Algebra.

Suppose that $$F(G)$$ is neither commutative nor cocommutative.

In their 1966 paper Kac and Paljutkin, show that when we write $$F(G)$$ as a multimatrix algebra, one of which must be one-dimensional (to account for the counit), that we can decompose into one-dimensional factors and $$n_j>1$$ dimensional factors: $$F(G)=\left(\bigoplus_{g\in G_1}\mathbb{C}\delta_g\right)\oplus\left( \bigoplus_{j=1}^mM_{n_j}(\mathbb{C})\right):=A_1\oplus B,\qquad (\star)$$

where $$G_1$$ is a finite group.

Kac and Paljutkin show that:

$$\Delta(A_1)\subseteq A_1\otimes A_1+B\otimes B.$$

Assumption 1: Kac and Paljutkin then assume that $$m=1$$ so that $$B=M_{n_1}(\mathbb{C})$$ contains a single summand.

They show that: $$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1+B\otimes B.$$

Let $$f\in B$$ and write:

$$\Delta(f)=\underbrace{T(f)}_{\in A_1\otimes B+B\otimes A_1}+\underbrace{K(f)}_{\in B\otimes B}.$$

Assumption 2: Then they assume further that $$|G_1|=n_1^2$$.

From there they show that $$K(f)=0$$ so that in fact

$$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1.$$

I am not sure if the assumptions were made to write down a quicker/easier proof for the purposes of illustration... or if there is a finite quantum group such that this inclusion does not hold.

I am wondering in the 50 odd years hence has anyone working in Finite Quantum Groups (finite dimensional Hopf *-algebras $$H$$ with $$f^*f\neq 0$$ for all $$f\in H$$) showed that with respect to the decomposition $$(\star)$$ that $$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1?$$

Question 1: Does this inclusion hold? If not, what is a counterexample?

Question 2: Is it possible that $$\Delta(M_{n_j}(\mathbb{C}))$$ never falls into $$M_{n_j}(\mathbb{C})\otimes M_{n_j}(\mathbb{C})$$? It might be easier to show this more specific result (that must also be useful for me).

Some Efforts: Let $$E^j_{mn}$$ be a matrix unit in $$M_{n_j}(\mathbb{C})$$. We want to show that $$(I_{n_j}\otimes I_{n_j})\Delta(E^j_{mn})=0.$$ This would be sufficient for my needs. I have the following assumption about how the antipode relates to the multi-matrix structure but I cannot find any suggestion that it is true. The only thing that seems remotely consistent with it is that $$S^2=I_{F(G)}$$. The assumption is that on the $$m$$ non-commutative factors, the antipode is the transpose. This allows a partial result to be proved using the antipodal property.

Take a matrix unit. We have that $$\varepsilon(E_{mn}^j)=0$$ and this implies further that $$m\circ (S\otimes I_{F(G)})\circ\Delta(E^j_{mn})=0=m\circ (I_{F(G)}\otimes S)\circ \Delta(E^j_{mn}).$$

Think about the $$A_1\otimes B$$ and $$B\otimes A_1$$ parts of $$\Delta(E^j_{mn})$$ (there is no $$A_1\otimes A_1$$ part by Kac and Paljutkin). We know that $$S(A_1)\subset A_1$$, $$S(B)\subset B$$, and moreover $$A_1B=0$$, and so when we multiply $$m\circ (A_1\otimes B)$$ we get zero.

Now choose lots of structure constants (for a fixed $$j$$) $$\alpha^{mn}_{xy,zw}\in\mathbb{C}$$: $$(I_{n_j}\otimes I_{n_j})\Delta(E^j_{mn})=\sum_{x,y,z,w=1}^{n^j}\alpha_{xy,zw}^{mn}(E^{j}_{xy}\otimes E^j_{zw}).$$ There are probably lots of conditions on the $$\alpha$$, but under the assumption that the antipode is the transpose, and the linear independence of the matrix units, it is the case that for any $$x,y,w=1,\dots,n_j$$, $$\alpha_{xy,wy}=0$$ and $$\alpha_{xy,xw}=0$$.

I am hoping that the homomorphism property of $$\Delta$$ might 'finish off' the other structure constants.

I guess what you are looking for is the inclusion matrix for the unital inclusion of finite dimensional $${\rm C}^{\star}$$-algebras $$\Delta(A) \subset A \otimes A$$. It is given by the fusion rules for $$Rep(A)$$, see Proposition 7.4 in my preprint arXiv:1704.00745v5.

Example: consider the finite dimensional Hopf $${\rm C}^{\star}$$-algebra $$A=\mathbb{C}S_3$$ with the symmetric group $$S_3$$. Then as an algebra $$A \simeq \mathbb{C} \oplus \mathbb{C} \oplus M_2(\mathbb{C}).$$ Now, $$Rep(A)=Rep(S_3)$$ and the fusion rules $$(n_{ij}^k)$$ are given by the following matrices:
$$\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right), \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right), \left(\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{matrix} \right)$$

This provides a counter-example for your questions because $$n_{33}^3 = 1 \neq 0$$.
[If you want, you can find the explicit formulas for the comultiplication computed here.]

You can get many counter-examples from some non-abelian finite groups.
See the following fusion rules for $$Rep(A_5)$$:

$$\left(\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1 \end{smallmatrix} \right), \left(\begin{smallmatrix}0&1&0&0&0\\1&1&0&0&1\\0&0&0&1&1\\0&0&1&1&1\\0&1&1&1&1 \end{smallmatrix} \right), \left(\begin{smallmatrix}0&0&1&0&0\\0&0&0&1&1\\1&0&1&0&1\\0&1&0&1&1\\0&1&1&1&1 \end{smallmatrix} \right), \left(\begin{smallmatrix}0&0&0&1&0\\0&0&1&1&1\\0&1&0&1&1\\1&1&1&1&1\\0&1&1&1&2 \end{smallmatrix} \right), \left(\begin{smallmatrix}0&0&0&0&1\\0&1&1&1&1\\0&1&1&1&1\\0&1&1&1&2\\1&1&1&2&2 \end{smallmatrix} \right)$$

Counter-examples (for your questions) which are neither commutative nor cocommutative
Let $$\mathcal{C}$$ be the class of non-abelian finite groups $$G$$ having an irreducible complex representation $$V$$ with $$\dim(V)>1$$ and which is an irreducible component of $$V \otimes V$$ (the groups $$S_3$$ and $$A_5$$ are of class $$\mathcal{C}$$, as shown above). Let $$G$$ be of class $$\mathcal{C}$$, and let $$H$$ be any non-abelian finite group. Let $$A,B$$ be the finite dimensional Hopf $${\rm C}^{\star}$$-algebras $$\mathbb{C}G$$ and $$\mathbb{C}H$$ (respectively), then the usual tensor product of Hopf algebras $$A \otimes B^*$$ is a non-commutative and non-cocommutative counter-example of your questions.
In addition, for any twist $$J$$ of $$A$$, $$A_J$$ has the same algebra structure than $$A$$. Moreover, $$J$$ is an invertible element of $$A \otimes A$$ and $$\Delta_J = J^{-1} \Delta J$$, so the inclusions $$\Delta(A) \subset A \otimes A$$ and $$\Delta_J(A_J) \subset A_J \otimes A_J$$ have the same inclusion matrices. Thus, $$Rep(A_J)$$ and $$Rep(A)$$ have the same fusion rules. Take a group $$G$$ of class $$\mathcal{C}$$, which admits a non-trivial twisting $$(\mathbb{C}G)_J$$ as for $$G=A_5$$ (by a result of Nikshych). It follows that $$(\mathbb{C}G)_J$$ is also a non-commutative and non-cocommutative counter-example of your questions.

• Thank you for this answer but I am assuming that the algebra of functions is neither commutative nor cocommutative. – JP McCarthy Jul 14 '19 at 21:00
• @JPMcCarthy: Take $A$ as above, then $A \otimes A^*$ is neither commutative nor cocommutative and should still be a counter-example for your questions. – Sebastien Palcoux Jul 14 '19 at 21:29
• @JPMcCarthy: I just edited my answer with additional non-commutative and non-cocommutative counter-examples. – Sebastien Palcoux Jul 15 '19 at 6:10
• Concretely: consider $\mathbb{C}S_3\otimes F(S_3)$. This has algebra $(\mathbb{C}^2\oplus M_2(\mathbb{C}))\otimes \mathbb{C}^6$. Unless I am being an idiot, this is isomorphic to twelve copies of $\mathbb{C}$, and six copies of $M_2(\mathbb{C})$. In my head the six copies of $M_2(\mathbb{C})$ are indexed by $S_3$, so, for example, $a_{12}\otimes \delta_{(12)}\in M_{2,(12)}(\mathbb{C})$. $\Delta(a_{11}\otimes\delta_e)$ includes $(2a_{22}\otimes\delta_e)\otimes(a_{22}\otimes\delta_e)$: from $M_{2,e}(\mathbb{C})$ to $M_{2,e}(\mathbb{C})\otimes M_{2,e}(\mathbb{C})$. – JP McCarthy Sep 11 '19 at 16:39