Explicit integral representation theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T12:06:44Z http://mathoverflow.net/feeds/question/32311 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32311/explicit-integral-representation-theory Explicit integral representation theory Lennart Meier 2010-07-17T20:24:19Z 2010-07-17T22:40:06Z <p>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:</p> <p>1) Can one formulate a "guide" how to obtain such a list?</p> <p>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. </p> http://mathoverflow.net/questions/32311/explicit-integral-representation-theory/32321#32321 Answer by Jack Schmidt for Explicit integral representation theory Jack Schmidt 2010-07-17T22:23:59Z 2010-07-17T22:40:06Z <p>You might be asking about four separate types of modules: </p> <ul> <li>irreducible Z[G] modules,</li> <li>Z-forms of irreducible Q[G] modules,</li> <li>indecomposable Z[G] modules, or</li> <li>indecomposable Z[G] modules that are finitely generated and free as Z-modules.</li> </ul> <p>I'll assume the last is the main concern.</p> <p>The irreducible modules of ZS<sub>3</sub> 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.</p> <p>The irreducible CS<sub>3</sub> 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 ZS<sub>3</sub> modules, that is, four total Z-forms of irreducible QS<sub>3</sub> modules.</p> <p>Indecomposable ZS<sub>3</sub> 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).</p> <p>The indecomposable ZS<sub>3</sub> modules that are free as Z-modules are classified in:</p> <p>Lee, Myrna Pike. "Integral representations of dihedral groups of order 2p." Trans. Amer. Math. Soc. 110 (1964) 213–231. <a href="http://www.ams.org/mathscinet-getitem?mr=156896" rel="nofollow">MR 156896</a> <a href="http://dx.doi.org/10.2307/1993702" rel="nofollow">doi:10.2307/1993702</a></p> <p>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.</p>