# Explicit integral representation theory

The representation theory of finite groups over the complex numbers is classical und it is usually quite easy to compute the set of isomorphism classes of irreducible representations, at least for small examples. Now, sometimes one is not content with the representation theory over the complex numbers or even over any field, but one wants to consider representations over $\mathbb{Z}$ or $\mathbb{Z}_{(p)}$. At least for the non-expert it is not so easy to obtain a complete list of isomorphism classes of indecomposable representations in these cases even for small examples. So, what I want to ask is the following:

1) Can one formulate a "guide" how to obtain such a list?

2) Is there a place in the literature where a list of indecomposables/irreducibles is given for some small examples as $S_3$ (the symmetric group on three elements)? In the last example I am especially interested.

• The complex irreducible representations of S_n are all defined over Z. I don't know whether that constitutes a list of the indecomposable representations, though. – Qiaochu Yuan Jul 17 '10 at 21:57
• The keyword is "modular representation theory" (and more narrowly, "integral representation theory"). Specifically for symmetric groups, there has been quite a bit of work, some of which is surveyed in James and Kerber's book. – Victor Protsak Jul 17 '10 at 23:53

• irreducible Z[G] modules,
• Z-forms of irreducible Q[G] modules,
• indecomposable Z[G] modules, or
• indecomposable Z[G] modules that are finitely generated and free as Z-modules.

I'll assume the last is the main concern.

The irreducible modules of ZS3 are all finite and have an elementary abelian p-group as their additive group. For p=2,3 there are 2 each, and for p>3, there are 3 each.

The irreducible CS3 modules are all realizable over Q. Every such module may be realized over Z, but the two-dimensional representation has two distinct Z-forms, giving four total "irreducible" Z-free ZS3 modules, that is, four total Z-forms of irreducible QS3 modules.

Indecomposable ZS3 modules up to isomorphism are more complicated than the human mind can possibly comprehend. Indeed, even those in which S_3 acts as the identity are much too complex. Luckily they divide up into several types: annihilated by a prime p (then classified by modular representation theory), torsion (more complicated, but basically now p-adic integral reps), Gorenstein projective (Z-free, so covered in the next bullet point), or madness (that is, the rest).

The indecomposable ZS3 modules that are free as Z-modules are classified in:

Lee, Myrna Pike. "Integral representations of dihedral groups of order 2p." Trans. Amer. Math. Soc. 110 (1964) 213–231. MR 156896 doi:10.2307/1993702

There are 10 of them, and the Krull-Schmidt theorem fails for them. Not only are indecomposables not completely reducible, the decomposition of a finitely generated Z-free module into indecomposable summands is not unique. In other words, integral representations of even very small groups are quite complicated.

• Integral representation theory of all flavors is definitely a hard subject, even for groups of very small order. Irving Reiner did extensive work, as have others. For textbook coverage, the two-volume book Methods of Representation Theory Reiner wrote with C.W. Curtis (Wiley-Interscience, 1981/1987) has big chapters on this subject along with some related algebraic K-theory. (These books are unfortunately hard to find except in some libraries.) As with representation theory over fields, small examples alone probably won't give much insight without added theory. – Jim Humphreys Jul 18 '10 at 14:03
• Regarding Curtis-Reiner, surprisingly their other book (from 1962) is available from the AMS: ams.org/bookstore-getitem/item=CHEL-356-H. This book also has some useful sections on integral representation theory. – fherzig Jan 17 '11 at 21:45
• For instance, it discusses the Jordan-Zassenhaus theorem which says that for any finite group $G$, a f.d. $\mathbf QG$-module contains only finitely many $\mathbf ZG$-lattices up to isomorphism. In particular, there are only finitely many indecomposable $\mathbf ZG$-modules (that are $\mathbf Z$-free) of bounded rank. – fherzig Jan 17 '11 at 21:50
• Is there a brief argument for why the two dimensional representation over $\mathbb{C}$ has two distinct $\mathbb{Z}$-forms? Garnir straightening with the basis of standard young tableaux gives one particular form. How can I find the matrices associated with the other form? In general, If I begin with some representation over $\mathbb{C}$, how can one predict what the various $\mathbb{Z}$-forms associated with this representation is? I suspect that this has to do with change of basis that is possible in $\mathbb{C}$ but not in $\mathbb{\Z}$.. but it is still fuzzy for me.Thank you in advance! – Maithreya Sitaraman Oct 29 '19 at 4:44