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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.

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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.

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  • $\begingroup$ Thank you for this answer but I am assuming that the algebra of functions is neither commutative nor cocommutative. $\endgroup$ – JP McCarthy yesterday
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    $\begingroup$ @JPMcCarthy: Take $A$ as above, then $A \otimes A^*$ is neither commutative nor cocommutative and should still be a counter-example for your questions. $\endgroup$ – Sebastien Palcoux yesterday
  • $\begingroup$ @JPMcCarthy: I just edited my answer with additional non-commutative and non-cocommutative counter-examples. $\endgroup$ – Sebastien Palcoux 20 hours ago

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