2
$\begingroup$

Let $F(G)$ be the algebra of functions on a finite quantum group $G$ (so that $F(G)$ is a finite dimensional $\mathrm{C}^*$-Hopf algebra).

Suppose that $\{p_i:i=0,\dots,d-1\}\subset F(G)$ is a partition of unity, in other words $p^2_i=p_i^*=p_i$, $$p_ip_j=p_jp_i=\delta_{i,j}\,p_i,$$ and

$$\sum_{i=0}^{d-1}p_i=\mathbf{1}_G:=1_{F(G)},$$ the unit of $F(G)$.

Edit: The following condition was added after Konstantinos' answer:

Suppose that a state $\nu\in M_p(G):=\mathcal{S}(F(G))$ has the property that for all projections $q\in F(G)$, there exists $k_q$ such that $\nu^{\star k_q}(q)\neq 0$, where $$\nu\star \nu=(\nu\otimes \nu)\circ \Delta.$$

Suppose furthermore that $\nu\in M_p(G)$ has the property that:

$$\nu(p_i)=\begin{cases}1 & \text{ if }i=1\\ 0 & \text{else}\end{cases},$$

and we also have that, where $\varepsilon\in M_p(G)$ is the counit:

$$\varepsilon(p_i)=\begin{cases}1 & \text{ if }i=0\\ 0 & \text{else}\end{cases}.$$

Furthermore,

$$(\nu\otimes I_{F(G)})\circ \Delta(p_i)=:T_\nu(p_i)=p_{i-1},$$ with $T_\nu(p_0)=p_{d-1}$.

Note that $\Delta$ is a *-homomorphism, and, where $\int_G:=h\in M_p(G)$ is the Haar state of $F(G)$, we can show that:

$$\int_Gp_i=\frac{1}{d}.$$

Is it the case that

$$\Delta(p_i)=\sum_{k=0}^{d-1}p_{i-k}\otimes p_k?$$

If $F(G)$ is commutative, this condition holds.

$\endgroup$
8
  • 1
    $\begingroup$ If I am not mistaken, what you expect "should imply" that $p_{i-k} * p_k = p_i$ (up to a multiplicative constant) which "should imply" each $p_i$ to be rank one. Now the set of projections $p_i$ also provides a partition of the minimal central projections, but what you expect seems so strong that the minimal central projection "should" also be of rank one. Then we are reduced to a finite group $G = \{g_0, \dots , g_{d-1} \}$ with $g_{i-k}g_{k} = g_i$ which "should imply" that $G = C_d$ the cyclic group of order $d$ with $g_k = e^{2ik\pi/d}$. I did know think that much so I could be wrong! $\endgroup$ Nov 20, 2019 at 7:55
  • $\begingroup$ @SebastienPalcoux I believe that it might be possible to show that $S(p_i)=p_{d-i}$ in which case what I expect implies $p_i\star p_j=\frac{1}{d}p_{i+j}$... but the $p_i$ need not be rank one. If $F(G)$ is commutative, then the $p_i$ are indicator functions on the cosets of a normal subgroup $N\rhd G$ such that $G/N\cong \mathbb{Z}_d$. I am trying to show something similar here but I am not sure if I have a first fundamental theorem. I think it might be possible that $p_0$ is only a group like projection (that is normal in a sense) but not necessarily a quantum subgroup. $\endgroup$ Nov 20, 2019 at 9:54
  • $\begingroup$ Do you consider that the elements $\{p_i:i=0,\dots,d-1\}$ span $F(G)$ ? $\endgroup$ Jan 20, 2020 at 3:07
  • $\begingroup$ @Konstantinos for the purposes of what I am looking at usually $d$ is less than $\dim G$. I would be interested if there is a $d=\dim G$ counterexample. $\endgroup$ Jan 20, 2020 at 7:17
  • $\begingroup$ The $\dim G$s of course should be $\dim F(G)$s. $\endgroup$ Jan 20, 2020 at 12:24

2 Answers 2

2
+200
$\begingroup$

This answer was given before the edit

(with the understanding that under the stated assumptions the comultiplication described in the OP is cocommutative for the $d$ idempotents $p_i$ and $k$ is an algebraically closed field of char zero).

  • An example where "it is the case":
    Consider the finite group $N$ and the cyclic group $C_d$ of order $d$. Then $k(N\times C_d)=kN\otimes kC_d$ is the group hopf algebra of the group $N\times C_d$. If we take its dual let us set: $$ F(G):=\big(k(N\times C_d)\big)^*\cong\big(kN\otimes kC_d\big)^*\cong(kC_d)^*\otimes(kN)^*\cong kC_d\otimes (kN)^* $$ because $(kC_d)^*\cong kC_d$ as hopf algebras, for any finite abelian group. Inside $(kC_d)^*$ the multiplication and the comultiplication are exactly as in the OP.
    (Actually, any finite abelian group $H$ of order $d$, in place of $C_d$ would do the job).

  • A counterexample (where "it is not the case"):
    Consider the finite group $N$ and the finite non-abelian group $H$ of order $d$. Then $k(N\times H)=kN\otimes kH$ is the group hopf algebra of the group $N\times H$. If we take its dual let us set: $$ F(G):=\big(k(N\times H)\big)^*\cong\big(kN\otimes kH\big)^*\cong(kH)^*\otimes(kN)^* $$ Inside $(kH)^*$ the multiplication is isomorphic to the one described in the OP (i.e. the orthogonal idempotents providing a partition of unity) but the comultiplication cannot be the one suggested in the OP. The reason is that since $kH$ is non-commutative then its dual hopf algebra $(kH)^*$ cannot be cocommutative.

$\endgroup$
9
  • $\begingroup$ I don't know why you think the comultiplication is cocommutative? If it is cocommutative on these $d$ elements I am not sure why it follows more generally (but this is by-the-by). I must construct a small example and see how the counterexample works. Thank you. $\endgroup$ Jan 23, 2020 at 9:13
  • 1
    $\begingroup$ @JP McCarthy, i am not claiming that the comultiplication is cocommutative on $F(G)$. (I guess my introductory comment was misleading). I was just refering to the set of the $d$ orthogonal idempotents: what i had in mind, was that if your comultiplication is valid then the $d$ idempotents span a hopf subalgebra. This is the cocommutative one. $\endgroup$ Jan 23, 2020 at 16:34
  • 1
    $\begingroup$ and since it is also commutative, then it will be isomorphic to a (self-dual) group hopf algebra for a finite abelian group. so it will be $kC_d$ (or $kH$ fore some finite abelian group). this corresponds to my first example (notice that $F(G)=\big(k(N\times C_d)\big)^*\cong kC_d\otimes (kN)^*$ is generally not cocommutative if $N$ is not abelian). $\endgroup$ Jan 23, 2020 at 17:31
  • 1
    $\begingroup$ On the other hand, if we consider both $N$, $H$ to be non-abelian then none of their dual hopf algebras $(kN)^*$, $(kH)^*$ (which are both hopf subalgebras of $F(G)$) will be cocommutative; this is the situation in the counterexample $F(G)=\big(k(N\times H)\big)^*\cong(kH)^*\otimes(kN)^*$. $\endgroup$ Jan 23, 2020 at 17:35
  • $\begingroup$ I don't think a $F(G)$-commutative counterexample is possible. I reckon in this case $p_0$ is the indicator function on a normal subgroup $H$, and the $p_i$ the indicator function on the cosets and $G/H\cong C_d$ and the formulae for $\Delta(p_i)$ follows. I think I can show this with the homomorphism $\operatorname{supp} p_i\rightarrow i\in C_d$; will get back to it Wednesday. I am possibly missing the condition that for all projections $q\in F(G)$, there exists $k_q$ such that $\nu^{\star k_q}(q)\neq 0$. That is an assumption I am missing (but can possibly do without). $\endgroup$ Jan 27, 2020 at 18:48
0
$\begingroup$

These are additions to Konstantinos' answer and was given before the edit

A Counterexample

This is inspired from here and here.

Let $G=S_3\times C_2$ and $\nu=\frac{1}{2}(\delta^{(e,1)}+\delta^{((12),1)})$.

Now define $S_0=\{(e,0),((12),0),((13),1),((23),1),((123),1),((132),1)\}$ and $S_1=G\backslash S_0$.

Consider $p_0=\mathbf{1}_{S_0}$ and $p_1=\mathbf{1}_{S_1}$.

These projections have all the properties given above.

Consider $$p_1=\delta_{((123),0)}+ \cdots.$$ Note $$\begin{align*} \Delta(p_1)&=\Delta(\delta_{((123),0)}+\cdots) \\&=\underbrace{\delta_{((132),0)}\otimes \delta_{((132),0)}}_{\in p_1F(G)\otimes p_1F(G)}+\cdots,\end{align*}$$ and so $$\Delta(p_1)\neq p_0\otimes p_1+p_1\otimes p_0.$$

A Missing Condition

I am missing the following condition. We suppose in addition that for all projections $q\in F(G)$, there exists $k\in \mathbb{N}$ such that $\nu^{\star k}(q)\neq 0$, where $$\nu\star \nu=(\nu\otimes\nu)\circ \Delta.$$ This will be added to the question.

In the classical case, where $F(G)$ is commutative, it can be shown that $p_0=\mathbf{1}_N$, where $N\lhd G$, $p_1=\mathbb{1}_{Ng}$, and $p_m=\mathbb{1}_{Ng^m}$, i.e. the $p_i$ are indicator functions on cosets of the normal subgroup $N\lhd G$. Furthermore, $G/N\cong C_d$, and so with the missing condition, in the classical case, the projections satisfy:

$$\Delta(p_i)=\sum_{k=0}^{d-1}p_{i-k}\otimes p_k.$$

I will edit the original question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.