Timeline for What is the correct formulation of the CDE triangle?
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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 |