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Feb 17, 2010 at 20:38 comment added Bruce Westbury The intuition for the question is to take say the Specht modules which are defined over $\mathbb{Z}[q,q^{-1}]$. Then I can change base to $\mathbb{Q}(\omega)$ or to $\mathbb{F}$. Then I can (in principle) find multiplicities of simples. Alternatively I can change base to $\mathbb{Z}(\omega)$. Then starting from there I hope to have a CDE square and Ben has explained what is involved there. Then I would expect two of these decomposition matrices to multiply to give the third. Also no-one has mentioned the relation that $D$ and $E$ are transposes.
Feb 17, 2010 at 20:03 comment added Ben Webster Torsten- I'm very confused by what "the whole theory only sees simple modules" means. Are you saying the Cartan matrix of R_k won't change? that sounds pretty hard to believe.
Feb 17, 2010 at 17:06 comment added Torsten Ekedahl One comment about a CDE square vs a triangle. The whole theory only sees simple modules so one may start by dividing out by the radical of $R_K$ (and its intersection with $R$) making $R_K$ semi-simple.
Feb 17, 2010 at 17:01 comment added Torsten Ekedahl There are two things here. To define $C$ one needs for $A$ to be local Henselian only (a minor point). To have $D$ defined it is necessary to have that the kernel of $K_0(R_A) \to K_0(R_K)$ ($K_0$ being the Grothendieck group of f.g. modules) map to zero under the reduction map $K_0(R_A) \to K_0(R_k)$ (and for that map to be defined one needs $R$ to be regular). The traditional case is that $A$ is a DVR, I am unsure about the right generality. One could however assume only that $(A,(q))$ is a Henselian couple ($(q)$ need not be maximal) and $k=A/(q)$ and $K=A[q^{-1}]$.
Feb 17, 2010 at 16:55 comment added Bruce Westbury P.S. Why doesn't jsMath work in a comment?
Feb 17, 2010 at 16:53 comment added Bruce Westbury That's a nice account of what I meant in the first question. If I understand this then we have three CDE squares with $(K,k)$ given by $(\mathbb{Q}(q),\mathbb{Q}(\omega))$, $(\mathbb{Q}(\omega),\mathbb{F})$ and $(\mathbb{Q}(q),\mathbb{F})$. Then I want to say that the first two CDE squares glue together to give the third. Experts say that if you know the decomposition numbers for $q$ a root of unity then you know the classical decomposition numbers using the Steinberg representation. Does this mean they know the second CDE square? and are then combining them?
Feb 17, 2010 at 16:10 history answered Ben Webster CC BY-SA 2.5