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Let $G$ be a (discrete, say) group and $\mathbb K$ a field. The regular representation $G^{\mathbb K}$ is the vector space of all functions $G \to \mathbb K$. It is a (left, say) $G$-module: given $g\in G$ and $f: G \to \mathbb K$, the action is $g\cdot f: x \mapsto f(xg)$. Then $G^{\mathbb K}$ is a commutative algebra object in $G\text{-rep}_{\mathbb K}$, the symmetric monoidal category of $\mathbb K$-valued $G$-representations, under pointwise multiplication $f_1f_2: x \to f_1(x)f_2(x)$.

But the pointwise product is not necessarily the only commutative algebra (in $G\text{-rep}$) structure that can be put on $G^{\mathbb K}$. For example, when $\mathbb K = \mathbb R$ and $G = \mathbb Z/2$, as an algebra $G^{\mathbb K} \cong \mathbb R[\epsilon]/(\epsilon^2 = 1)$, with the $G$-action corresponding to conjugation $\epsilon \mapsto -\epsilon$. The same $G$-module supports the algebra structure $\mathbb R[\epsilon]/(\epsilon^2 = -1) = \mathbb C$, which is patently a different algebra.

My question is whether there are any examples with $\mathbb K = \mathbb C$? I.e.:

Does there exist a group $G$ so that there is a commutative algebra object in $G\text{-rep}_{\mathbb C}$ that is isomorphic to $G^{\mathbb C}$ as a representation but not as an algebra?

I believe that any such group must be rather large: in particular, I'm sure that it cannot be finite.

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  • $\begingroup$ I will mention an idea in the comments, since I'm very unsure of its correctness. Take G=PGL(2,C) and forget its algebraic structure: think of it just as a discrete group. It acts on the field C(x) as automorphisms, and the invariant subspace of C(x) is just C. So G^C and C(x) have the same dimension and the same invariant subspace, and that's almost a proof that they're the same representation. Unfortunately, math isn't that easy, and I don't see how to really get my hands on either representation. $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 13, 2010 at 6:13
  • $\begingroup$ Silly/stupid question: when you say ``${\mathbb K}$-valued representation of $G$'', do you mean a repn of G as endomorphisms of some ${\mathbb K}$-vector space? I also don't follow what it means for $G^{\mathbb K}$ to be an algebra object in the category $G-{\rm rep}_{\mathbb K}$. $\endgroup$
    – Yemon Choi
    Commented Sep 13, 2010 at 6:36
  • $\begingroup$ Yemon, yes to the first question; "algebra object etc" simply means that the multiplication map on functions is $G$-equivariant (well, the same for scalar multiplication, but that is automatic). You can also call it a "commutative $G$-algebra", which is fairly standard. However, the correct notion of isomorphism should take both structures into account (in particular, $A$ and $B$ may two $G$-algebras that are isomorphic as algebras and as $G$-modules, but not as $G$-algebras); this question is about a stronger property of being non-isomorphic as algebras only. $\endgroup$
    – Victor Protsak
    Commented Sep 13, 2010 at 7:18
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    $\begingroup$ Aren't $A=\mathbb{C}[x]/(x^n)$ and $B=\mathbb{C}[x]/(x^n-1)\cong\mathbb{C}^G$ two commutative $G$-algebras isomorphic to the regular representation of $G=\mathbb{Z}_n$ but non-isomorphic as algebras (e.g. because $A$ is not semisimple)? In both cases, the standard generator of $G$ acts on $x$ as the multiplication by the fixed primitive $n$th root of unity. $\endgroup$
    – Victor Protsak
    Commented Sep 13, 2010 at 7:28
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    $\begingroup$ I think your exponential notation is at odds with standard convention: $G^{\mathbb{K}}$ usually means maps from $\mathbb{K}$ to $G$, and you want the opposite. $\endgroup$
    – S. Carnahan ♦
    Commented Sep 13, 2010 at 7:56

1 Answer 1

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Summary Yes, there are many such examples, even for finite groups.

Construction Let $W<GL(V)$ be a complex reflection group, $A=\mathbb{C}[V]$ be the algebra of polynomial functions on $V$ and $A^W$ be the subalgebra of $W$-invariants. Then by the Chevalley–Shephard–Todd theorem, $A^W$ is a polynomial algebra and $A$ is a free $A^W$-module. This may be viewed as a deformation of $\mathbb{C}^W$ as follows. For any $z\in\text{Spec}A^W,$ consider the fiber $A_z=A/zA.$ By the freeness property, each $A_z$ carries the regular representation of $W.$ For a regular $z,$ corresponding to a $W$-orbit of a regular point in $V,$ the algebra $A_z$ may be identified with the algebra of functions on the orbit, which consists of $|W|$ points; in particular, $A_z\cong \mathbb{C}^W$ as a $W$-algebra. But for values of $z$ corresponding to singular orbits, algebras $A_z$ are not reduced. The most singular fiber is $A_0=\mathbb{C}[V]/(\mathbb{C}[V]^W_{+})$ and is a graded nilpotent Frobenius algebra, with one-dimensional socle and radical, called the covariant algebra of $W.$

Example Let $W=\mathbb{Z}_n$ acting on the one-dimensional vector space with coordinate $x$ by the $n$th roots of unity. Then regular fibers $\mathbb{C}[x]/(x^n-a)$ with $a\ne 0$ are semisimple and isomorphic to $\mathbb{C}^{\mathbb{Z}_n}$ (explicitly, $x$ is mapped to $b\sum_k \zeta^{-k}\delta_k,$ where $b^n=a$), whereas the singular fiber $\mathbb{C}[x]/(x^n)$ is a graded nilpotent algebra.

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    $\begingroup$ The case $n=2$ is by itself rather instructive. One only has the choice of specifying multiplication on the sign representation summand, and asking whether the map $sign \otimes sign \to triv$ is the zero map or an isomorphism. $\endgroup$
    – S. Carnahan ♦
    Commented Sep 13, 2010 at 9:32
  • $\begingroup$ I'm slightly ashamed of myself for not thinking of this when I first read the question. $\endgroup$
    – Ben Webster ♦
    Commented Sep 13, 2010 at 22:50

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