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I might have thought about this problem a little longer but am quite confused so said I would put this question to the good people here...

Consider a finite group $G$ or rather the algebra of functions on $G$, $\mathbb{C}^G=F(G)$ with multiplication in $F(G)$ defined pointwise. We have an involution on $F(G)$:

$$f^*(g)=\overline{f(g)}.$$

If we take the supremum norm on $F(G)$:

$$\|f\|_\infty=\max_{g\in G}|f(g)|,$$

we have a C*-algebra. The positive elements of $F(G)$ are those of the form $f^*f$ and so those that have positive coefficients with respect to the basis $\{\delta_s:s\in G\}$ are the positive functions.

Denote by $\mathbb{C}G$ the linear functionals on $F(G)$. This set is spanned by elements dual to the $\{\delta^s:s\in G\}$:

$$\delta^s(\delta_t)=\delta_{t,s},$$ where the latter $\delta$ is the Kronecker delta. It is not difficult to show that the positive linear functionals of $\mathbb{C}(G)$ are those with positive coefficients with respect to the $\{\delta^s:s\in G\}$ basis.

So for example, if $G=\{\mathbb{Z}_3,\,\oplus_3\}$ then $\displaystyle \nu=\frac{\delta^1+\delta^2}{2}\in\mathbb{C}G$ is a positive linear functional.

However one can also give $\mathbb{C}G$ the structure of a C*-algebra. Extend by linearity the convolution product $$(\delta^s,\delta^t)\mapsto \delta^{st};\qquad\delta^s\star\delta^t=\delta^{st},$$ to the whole of $\mathbb{C}G$. My understanding is that there are C*-norms on $\mathbb{C}G$ when equipped with the involution:

$$\nu^*(\delta_s)=\overline{\nu(\delta_{s^{-1}})}.$$

Now positivity in this C*-algebra is not the same as positivity as a linear functional. In particular it is not difficult to show that there is no $\mu\in\mathbb{C}\mathbb{Z}_3$ such that

$$\mu^*\star \mu=\frac{\delta^1+\delta^2}{2}=:\nu.$$

Therefore, as an element of the C*-algebra $\mathbb{C}\mathbb{Z}_3$, $\nu$ is not positive.

Is there an easy way of recognising if an element of the C*-algebra $\mathbb{C}G$ is positive?

Random thoughts... is it necessary for $\nu(\delta_e)\neq0$, symmetry plays a role? $\nu(\delta_s)=\nu(\delta_{s^{-1}})$... have we any positive elements of the algebra which are not positive functionals. Alternatively I need to look at self-adjoint --- $\nu(\delta_s)=\overline{\nu(\delta_{s^{-1}})}$ --- and positive spectrum... can I find the spectrum of an element of $\mathbb{C}G$?

Thank you.

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Consider the space $\mathbb{C}G$ as a Hilbert space with orthonormal basis $e_g$. $G$ acts on it by left multiplication which are isometries. $$L_g \colon h \mapsto g h$$ 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 $C^{*}$ algebra. Now, the positivity is preserved under the imbeddings of $C^{*}$ algebras. Now, a basic fact about $C^{*}$ algebras is that an element of $\mathbb{C}G$ is positive, if and only if its image in $End(\mathbb{C}G)$ is positive. So you consider the corresponding $|G|$ dimensional matrix check whether that one is positive.

In general, an element $\sum a_g g$ will have the corresponding matrix \begin{eqnarray} \sum a_g g &\mapsto& (x_{gh})_{g, h \in G}\\ x_{gh} &=& a_{g h^{-1}} \end{eqnarray} a so called group matrix.

Consider an element of the group algebra $\mathbb{C} ( \mathbb{Z}/3)$ $$\alpha = a_0 e_0 + a_1 e_1 + a_2 e_2$$ with corresponding matrix $$ \left( \begin{array}{ccc} a_0 & a_2 & a_1 \\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0 \end{array} \right ) $$ The element $\alpha$ and corresponding matrix are selfadjoint if and only if $a_2 = \bar a_1$ and $a_0 \in \mathbb{R}$. So take $\alpha = 1 \cdot e_1 + 1 \cdot e_2$. The corresponding matrix $$ \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1& 0 \end{array} \right ) $$ is not positive definite so neither is $\alpha$.

One can do this in general for other groups.

The idea: Map your $C^{*}$ algebra into a matrix algebra and test positivity there.

You can also map $\mathbb{C}G$ isomorphically to a product of matrix algebras ( representation of finite groups ) and test positivity on each component. Eg: for $G = \mathbb{Z}/n$, a self-adjoint matrix $(a_{m-n})$ is positive if and only if all the sums $\sum a_n \omega^{n}$ are positive, where $\omega$ is an $n-th$ root of $1$ ( these are the eigenvalues of this circular matrix).

$\bf{Added:}$

Let's mention that every injective morphism of $C^{*}$ algebras preserves the norms. The way it goes with $\mathbb{C} G \subset End (\mathbb{C}G)$ that's clear Now let's assume that some $\alpha$ from $\mathbb{C} G$ has a square root in $End (\mathbb{C}G)$. Then that square root is the limit of a sequence of form $r_{n+1} = \frac{1}{2} ( \alpha + \frac{r_n}{\alpha})$, $r_0 = 1$, so that will also be in $\mathbb{C}G$.

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  • $\begingroup$ Thank you very much...have you a reference for "an element of $A$ is positive if and only if its image in $\text{End}(A)$" is positive...also I have elements of $\mathbb{C}\mathbb{Z}_3$ that are self-adjoint but not positive such as can be seen in math.stackexchange.com/questions/1327302/… $\endgroup$ – JP McCarthy Jun 17 '15 at 6:28
  • $\begingroup$ @Jp McCarthy: No worries... Added some details. For $C^{*}$ algebras, perhaps Dixmier. $\endgroup$ – Orest Bucicovschi Jun 17 '15 at 7:00
  • $\begingroup$ Thank you... what about $\nu=(1/3,\frac16 (2+\sqrt{3}i),\frac{1}{6}(2-\sqrt{3}i))$. According to my calculations, and the answer to the linked answer, there is no $\mu\in\mathbb{C}\mathbb{Z}_3$ such that $\nu=\mu^*\mu$... does this not mean that $\nu$ is not positive (but self-adjoint yes)? $\endgroup$ – JP McCarthy Jun 17 '15 at 7:31
  • $\begingroup$ @Jp McCarthy: No, it's not positive, one of the sums involving the $3$-rd roots of $1$ is $-\frac{1}{2}$ ( the other are $1$ and $-\frac{1}{2}$). All three of them have to be positive. The nice thing is that the conditions for positivity in the case of abelian groups are linear in the coefficients. $\endgroup$ – Orest Bucicovschi Jun 17 '15 at 7:45
  • $\begingroup$ Lovely thank you. This gives me enough to go on to give you the bounty anyway. $\endgroup$ – JP McCarthy Jun 17 '15 at 8:00

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