# Classification of finite complex reflection groups

Background:

Let $K$ be a field and let $V$ be a finite-dimensional $K$-vector space. A pseudoreflection (or usually imprecisely just reflection) in $V$ is an element $1 \neq s \in \mathrm{GL}(V)$ fixing a hyperplane. A reflection representation of a group $W$ over $K$ is a $K$-linear representation $\rho:W \rightarrow \mathrm{GL}(V)$, such that $\rho(W)$ is generated by reflections. A group $W$ is called a reflection group over $K$ if it admits a reflection representation over $K$.

Shephard-Todd classified (see below) the finite irreducible reflection groups over $\mathbb{C}$ (i.e. those finite groups admitting an irreducible reflection representation over $\mathbb{C}$).

Question:

Is there also a classification of the finite irreducible reflection representations over $\mathbb{C}$?

Edit: This question is very imprecise as indicated in the comments below. I should say what "classification of representations" means, and I have to admit: I don't know. A few ideas in this direction are:

• determine the isomorphism classes of finite irreducible reflection representations over $\mathbb{C}$, where an isomorphism between two reflection representations $\rho:W \rightarrow \mathrm{GL}(V)$, $\rho':W' \rightarrow \mathrm{GL}(V')$ is a vector space isomorphism $f:V \rightarrow V'$ such that $f \rho(G) f^{-1} = \rho'(G)$. (I think the Shephard-Todd classification is a classification relative to this notion!?)

• the same as above but an isomorphism is a vector space isomorphism $f:V \rightarrow V'$ and a group isomorphism $\varphi:W \rightarrow W'$ such that $f \rho(g) f^{-1} = \rho'( \varphi(g) )$ for all $g \in W$.

• consider pairs $(W,T)$ consisting of a finite irreducible reflection group over $\mathbb{C}$ and a subset $T$ which are generating reflections of some irreducible reflection representation of $W$ and then determine isomorphism classes of such pairs.

• [Insert your idea here].

My motivation for this question is something like this: A Cherednik-Algebra is defined for any finite irreducible reflection representation over $\mathbb{C}$. In what sense does the algebra depend on the group alone and not on the choice of a particular reflection representation?

• Your definition of isomorphism of representations is somewhat different from usual no? Normally one would require an elementwise compatibility $f \rho(g) f^{-1}=\rho'(g)$ for all $g \in G$. – GS Jun 22 '10 at 9:04
• Good question. I don't know. I came across this definition in "Reflection groups and invariant theory" by Richard Kane (p. 156). But he's not really using this concept... – user717 Jun 22 '10 at 9:07
• I don't have Kane's book to hand, but it seems likely to me that he introduces this definition in order to discuss the classification of subgroups of GL(V) up to conjugacy---which is really a different from classifying representations of a particular group up to isomorphism! The latter is a finer (fewer elements in each equivalence class) classification, and more interesting/difficult. – GS Jun 22 '10 at 9:13
• ...though Shephard-Todd probably gets more credit for completing the classification than is really deserved. Most of the work had been done, I believe, by the time that paper was written---especially for the classification of the primitive groups. – GS Jun 22 '10 at 10:24
• I just want to point out that the point of Shephard-Todd was not classification, rather it was the Chevalley-Shephard-Todd theorem. Since their proof is case-by-case, they listed complex reflection groups and so their paper became a convenient reference. – Victor Protsak Jun 22 '10 at 11:10

## 2 Answers

This was done by Shephard-Todd. A recent book on this is:

MR2542964 (Review) Lehrer, Gustav I. ; Taylor, Donald E.
Unitary reflection groups. Australian Mathematical Society Lecture Series, 20.
Cambridge University Press, Cambridge, 2009. viii+294 pp. ISBN: 978-0-521-74989-3

• Of course I did not look at their original work... :( Let's see. – user717 Jun 22 '10 at 8:57
• Dear Bruce, I'm still somewhat confused by the intent of the original question, but it matters whether one is classifying reflection representations of a group up to isomorphism, or just reflection subgroups of GL(V) up to conjugacy. The former was definitely not done by Shephard-Todd. Best, Stephen – GS Jun 22 '10 at 9:47

Answering the first question, if the field has characteristic zero then the classification will be reduced to Shephard-Todd. By this I mean that every finite reflection group will be on Shephard-Todd list. In the opposite direction, finite Weyl groups will appear over every field and the rest of them will need some algebraic integers to be present in your field...

In positive characteristic, the life gets tough as pseudoreflections can be unipotent. I do not know whether classification is known but I can say that the list will get much longer...

• The question was about $\CC$... but you're right, the list gets longer. Here is one reference: MR0603578 (82i:20060) Zalesskiĭ, A. E.; Serežkin, V. N. Finite linear groups generated by reflections. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 44 (1980), no. 6, 1279--1307, 38. – GS Jun 22 '10 at 10:07
• ...and here is a paper which might be easier to get: springerlink.com.ezproxy.webfeat.lib.ed.ac.uk/content/… – GS Jun 22 '10 at 10:19
• Your link brings me to UofEdinburgh site that asks me for some stupid password and my command of whiterussian is somewhat rusty but thanks for info, anyway! – Bugs Bunny Jun 23 '10 at 7:11
• BTW, the question is about $K$, not $\CC$, unless I missed something – Bugs Bunny Jun 23 '10 at 7:13
• Wassup Bugs, Sorry about the broken tex/link! The question (see above) is "Is there also a classification of the finite irreducible reflection representations over $\mathbb{C}$?" The OP began by defining refl. groups over a field K, but then switched the char. 0. The paper (which I have not found available free online) was "Determination of the finite primitive reflection groups over an arbitrary field of characteristic not 2". I. Geom. Dedicata 9 (1980), no. 2, 239--253 by A. Wagner. It's part of a three part series. – GS Jun 23 '10 at 10:33