Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  Theo Johnson-Freyd Sep 13 '10 at 6:13
    
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 $G-{\rm rep}_{\mathbb K}$. –  Yemon Choi Sep 13 '10 at 6:36
    
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. –  Victor Protsak Sep 13 '10 at 7:18
1  
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. –  Victor Protsak Sep 13 '10 at 7:28
8  
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. –  S. Carnahan Sep 13 '10 at 7:56
show 2 more comments

1 Answer

up vote 6 down vote accepted

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.

share|improve this answer
1  
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. –  S. Carnahan Sep 13 '10 at 9:32
    
I'm slightly ashamed of myself for not thinking of this when I first read the question. –  Ben Webster Sep 13 '10 at 22:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.