Let $G$ be a finite group and $\mathbb{C}G$ its group ring.

Now taking the approach of orangeskid, consider the space $\mathbb{C}G$ as a Hilbert space with orthonormal basis $\delta^g$. $G$ acts on it by left multiplication which are isometries. $$L_g \colon \delta^h \mapsto \delta^{g h}.$$ Hence the adjoint of the map $L_g$ is $L_{g}^{-1} = L_{g^{-1}}$.

Now $\mathbb{C}G$ acts on $\mathbb{C}G$ by left multiplication so we can look at $\mathbb{C}G$ as an algebra of linear operators on the Hilberts spaced $\mathbb{C}G$. This algebra is closed under $*$, since $$\left(\sum a_g g\right)^* = \sum \bar a_g g^{-1}$$ so it's a $*$-subalgebra of $End(\mathbb{C}G)$, hence a $\mathrm{C}^*$-algebra.

I am interested in the states of $\mathbb{C}G$. Now the states of $\mathbb{C}G$ lie in the second dual of $F(G)$ in other words they lie in $F(G)$.

$\mathbb{C}G$ is unital with unit $1_{\mathbb{C}G}=\delta^e$ and we have (e.g. Murphy, Corollary 3.3.4):

If $\tau$ is a bounded linear functional on a unital $\mathrm{C}^*$-algebra, then $\tau$ is positive if and only if $\tau(1)=\|\tau\|$.

Now... I have no idea what the $\mathrm{C}^*$-norm on $\mathbb{C}G$ actually is... this is something that I just get very confused about (any information here also greatly appreciated).

If I knew what the norm on $\mathbb{C}G$ was I could use this... so instead what I want to do is find the condition on linear functional on $\mathbb{C}G$ being positive and from there I can say that, well, we must have $\|\tau\|=\tau(\delta^e)$ and this means that for $$\tau=\sum_{g\in G}\alpha_g\delta_g$$ to be a state we must have $\alpha_e=1$.

For a state to be positive we must have $$\tau(v)\geq 0$$ for all $v\in \mathbb{C}G^+$. Now an element of $\mathbb{C}G$ is positive if it is of the form $\phi^*\star \phi$... that is for $\varphi$ in to be positive there must exist complex numbers $\{\beta_g:g\in G\}$ such that $$\varphi=\sum_g\left(\sum_{t\in G}\overline{\beta_t}\beta_{tg}\right)\delta^g,$$

and if $\tau=\sum_{s\in G}\alpha_s\delta_s$ a linear functional is to be positive we must have

$$\sum_{g\in G}\alpha_g\left(\sum_{t\in G}\overline{\beta_t}\beta_{tg}\right)\geq 0,$$

for all possible selections of $\{\beta_g:g\in G\}\subset \mathbb{C}$.

Are things as (computationally) awkward as this or is there another way to characterise/recognise the states?

Context: I want to study states on $\mathbb{C}G$ that have the property that

$$\lim_{k\rightarrow \infty} \sum_{g\in G}a_g\nu(\delta^g)^k=a_e;$$ for $\nu$ a state and $\{a_g\}\subset\mathbb{C}$.

These are states on the (algebra of functions on the) quantum group $\mathbb{C}G$ such that their convolution powers converges to the Haar state.

  • $\begingroup$ Can you consider $\Bbb C G$ as the algebra of operators acting on $\Bbb C(G)$ the free $\Bbb{C}$ vector space with generators from $G$, and give it an operator norm that way? $\endgroup$ – snulty Jun 24 '16 at 15:04
  • $\begingroup$ @snulty ... have we agreement with what $\mathbb{C} G$ is... $\endgroup$ – JP McCarthy Jun 25 '16 at 12:28
  • $\begingroup$ I believe that as sets $\Bbb{C} G$ the group ring, and $\Bbb{C}(G)$ which I'm denoting as the free vector space over the field $\Bbb C$ with generators in the group $G$ are the same. What I'm calling $C(G)$ is only a vector space where there is a representation of $G$ acting on it. Your $\Bbb{C}G$ the group ring has more structure in that it's an algebra over $\Bbb C$. You can take elements $z_1 g_1*z_2 g_2=(z_1*z_2)g_1g_2$ and this product is defined. In what I was mentioning I only have basis vectors $e_g$ and no vector product defined. $\endgroup$ – snulty Jun 25 '16 at 13:24
  • $\begingroup$ @snulty I'll get back to you Monday. $\endgroup$ – JP McCarthy Jun 25 '16 at 15:39
  • $\begingroup$ I probably don't know enough about this, I was just curious if the operator norm would work as a $*$-norm, in this case. $\endgroup$ – snulty Jun 25 '16 at 15:41

The states on $\mathbb{C}G$ are given by positive definite functions in $F(G)$ such that $u(e)=1$.

Therefore, states on $\mathbb{C} G$ are given by unitary representations $\rho:G\rightarrow \hom(V)$ and unit vectors $\xi\in V$ according to

$$u(g)=\langle \rho(g)\xi,\xi\rangle.$$


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