5
$\begingroup$

Consider a discrete group $G$ (not necessarily amenable), and let $\mathbb{C} G$ be its group algebra (with the usual $\star$-algebra structure). Is every $\star$-subalgebra of $\mathbb{C} G$ isomorphic (as $\star$-algebras) to $\mathbb{C}H$ for some subgroup $H$ of $G$?

This happens to be the case for all examples I considered (some simple cases of subalgebras where $G$ is a free group), but I can't find a proof in general.

In case the above is true, what can one say about ideals? It appears somewhat intuitive that an ideal cannot just be the group algebra of some subgroup (but my intuition is by no means well formed here). This is the principal reason I believe for now that the initial statement might actually be untrue.

My main motivation here is to (eventually) study group $C^\star$ algebras in this light, if possible...

I also noted that perhaps the field being $\mathbb{C}$ is of crucial importance here: a couple of related questions on Math Stack Exchange and Math Overflow seem to imply that the characteristic of the field is a deciding factor.

$\endgroup$
0
14
$\begingroup$

Given any element of $\alpha \in\mathbb C[G]$ the image of $\mathbb C[x]$ sending $ax^n\mapsto a\alpha^n$ is a ring homomorphism.

If $\alpha^*=\alpha,$ then the image $\mathbb C[\alpha]$ is a *-subalgebra.

Now, if $G$ is a non-trivial finite group of odd order, take $\alpha =\sum_{g\in G} g$ then $\alpha^2=|G|\alpha,$ and $\alpha^*=\alpha,$ so the *-subalgebra $\mathbb C[\alpha]$ is $2$-dimensional as a vector space. But $G$ has no subgroup of order $2.$

An example is $G=(\mathbb Z/3\mathbb Z,+).$ Then $$\begin{align}\mathbb C[G]&\cong \mathbb C[x]/\langle x^3-1\rangle\\ &\cong \mathbb C[x]/\langle x-1\rangle \times \mathbb C[x]/\langle x-\omega\rangle\times\mathbb C[x]/\langle x-\omega^2\rangle\end{align}$$ where $\omega=\frac{-1+\sqrt{-3}}2.$

Then $\alpha=1+x+x^2$ corresponds to the element $(3,0,0)$ in the product ring. And $\mathbb C[\alpha]$ consists of elements in the product ring of the form $(a,b,b),$ for any $a,b\in\mathbb C.$ It is isomorphic to $\mathbb C\times \mathbb C.$

In these cases, $\mathbb C[\alpha]$ is isomorphic to a group ring, namely for $H\cong C_2.$ But $H$ is not isomorphic to a subgroup of $G.$


Another Example

A “large” sub algebra when $G$ is commutative can be generate by elements of the form $g+g^{-1}.$

Again, $(g+g^{-1})^*=g+g^{-1}.$

Also, given $g,h\in G$ we can get $$(g+g^{-1})(h+h^{-1})=(gh+(gh)^{-1})+(gh^{-1}+(gh^{-1})^{-1}).$$

So this sub-algebra has the $g+g^{-1}$ as a linear basis.

If $G$ is finite, this algebra has dimension $\frac{|G|+k}2$ where $k$ is number of elements $g\in G$ such that $g=g^{-1}.$ Since $k\geq 1,$ this number is only a divisor of $|G|$ if $k=|G|.$

So if there is some element $g\in G$ such that $g\neq g^{-1}$ then this sub-algebra is not isomorphic to the group ring of a subgroup.


Even if $G$ is not commutative, if we have $h\in G$ of odd order $d,$ then $\mathbb C[h+h^{-1}]$ has dimension $\frac{d+1}2,$ which means you’d need $\frac{d+1}{2}\mid |G|.$ For example, if $|G|$ is a multiple of $5$ but not $3,$ then an element $h$ of order $5$ gives a sub-algebra $\mathbb C[h+h^{-1}]$ which is not isomorphic to the group ring of a subgroup.


Infinite case

If $G=\mathbb Z$ then $$\mathbb C[G]\cong\mathbb C[x,x^{-1}].$$

But show that $$\mathbb C[x+x^{-1}]\cong \mathbb C[y]$$ and the units of $\mathbb C[y]$ are just the non-zero elements of $\mathbb C.$ But there are units other than $a\cdot 1$ in $\mathbb C[H]$ when $H$ is a non-trivial group group. So this sub-algebra is not a group ring at all.

$\endgroup$
1
$\begingroup$

Since you are interested in $C^*$-algebras, let me give a rather goofy example in the commutative $C^*$-case.

Let $G = \bigoplus_n \mathbb{Z}_2$. Then by the Pontryagin duality, $C^*(G) \cong C(\Delta)$, the algebra of continuous functions on $\Delta$, where $\Delta = \prod_n \mathbb{Z}_2$ is topologically the Cantor set. Now, fix a convergent sequence (together with the limit) $K\subset \Delta$ that is not eventually constant. There is a continuos map $\varphi\colon \Delta\to K$, so $C(K)$ is a sub-C*-algebra of $C(\Delta)$ via the embedding $f\mapsto f\circ \varphi$ ($f\in C(K)$).

It is not a group $C^*$-algebra because $K$ is not even homeomorphic to a topological group being topologically non-homogenous. (Abelian group $C^*$-algebras are of the form $C(\Gamma)$ for some compact Abelian group $\Gamma$.)

$\endgroup$

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.