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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.

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    $\begingroup$ 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! $\endgroup$ – Sebastien Palcoux Nov 20 '19 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. $\endgroup$ – JP McCarthy Nov 20 '19 at 9:54
  • $\begingroup$ Do you consider that the elements $\{p_i:i=0,\dots,d-1\}$ span $F(G)$ ? $\endgroup$ – Konstantinos Kanakoglou Jan 20 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. $\endgroup$ – JP McCarthy Jan 20 at 7:17
  • $\begingroup$ The $\dim G$s of course should be $\dim F(G)$s. $\endgroup$ – JP McCarthy Jan 20 at 12:24
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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.

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  • $\begingroup$ 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. $\endgroup$ – JP McCarthy Jan 23 at 9:13
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    $\begingroup$ @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. $\endgroup$ – Konstantinos Kanakoglou Jan 23 at 16:34
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    $\begingroup$ 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). $\endgroup$ – Konstantinos Kanakoglou Jan 23 at 17:31
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    $\begingroup$ 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 $F(G)=\big(k(N\times H)\big)^*\cong(kH)^*\otimes(kN)^*$. $\endgroup$ – Konstantinos Kanakoglou Jan 23 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). $\endgroup$ – JP McCarthy Jan 27 at 18:48
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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.

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