Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 27 September 2014

Lecture 36: Revision: [t]-modules

(1) Let A= ( 123 456 987 ) . A acts on 3={(a1a2a3)|a1,a2,a3}. 3 has basis {e1,e2,e3} where e1=(100), e2=(010), e3=(001). Define an action of [t] on 3 by, for example, (3+4t+5t2) (312)= (3+4A+5A2) (312). This is the [t]-module defined by A.

(2) Let [t]3= { (g1g2g3) |g1,g2,g3[t] } so that, for example, ( 3t+7 t3-2t+1 5t2+t6 ) [t]3 and(312) [t]3. [t] acts on [t]3 by, for example, (3+4t+5t2) (312)= ( 9+12t+15t2 3+4t+5t2 6+8t+10t2 ) . [t]3 has basis: { (100), (010), (001), (t00), (0t0), (00t), (t200), (0t20), (00t2), } . [t]3 is infinite dimensional. [t]3 is the free [t]-module generated by { (100), (010), (001) } . A [t]-module is a vector space V over with a function [t]×V V (g,v) gv such that

(a) If g1,g2[t] and vV then (g1g2)v=g1(g2v),
(b) If vV then 1·v=v,
(c) If c1,v2, g1,g2[t] and vV then (c1g1+c2g2)v= c1g1v+c2g2v,
(d) If c1,c2, g[t] and v1,v2V then g(c1v1+c2v2)= c1gv1+c2gv2.

A [t]-module homomorphism from V to W is a linear transformation f:VW such that if g[t] and vV then f(gv)=gf(v).

The kernel of f is kerf={vV|f(v)=0} and the image of f is imf={f(v)|vV}.

(3) The map Φ: [t]3 3 (g1g2g3) g1(A)(100)+ g2(A)(010)+ g3(A)(001) is a [t]-module homomorphism.


More specifically, t-A = ( t-1-2-3 -4t-5-6 -9-8t-7 ) and kerΦ = { ( t-1-2-3 -4t-5-6 -9-8t-7 ) ( g1(t) g2(t) g3(t) ) | g1(t), g2(t), g3(t) } .

Consequence (1st homomorphism theorem): 3 [t]3 (t-A)[t]3 as [t]-modules.

POINT: A acting in 3 is very similar to a "clock" 12.

Use row reduction to find invertible matrices L and R (with entries in [t]) such that L(t-A)R is diagonal.

A=(3-715). Then (t-A)= ( t-37 -1t-5 ) and ( t-37 -1t-5 ) ( -1t-5 t-37 ) ( 1-(t-5) t-37 ) ( 1t-5 07+(t-5)(t-3) ) ( 1t-5 0t2-8t+22 ) ( 10 0t2-8t+22 ) . t2-8t+22=(t-λ)(t-μ) where {μ,λ}= {8±64-442}= {8±202}= {4±5} and L = (1-(t-3)01) (-1001) (0110)= (-(t-3)-110), R = (1-(t-5)01). So L(t-A)R= L(t-37-1t-5)R= (100t2-8t+22) and [t]2 (t-A)[t]2 L[t]2 L(t-A)[t]2 = [t]2 L(t-A)R[t]2 (g1g2) L(g1g2) and [t]2 (100t2-8t+22)[t]2 = [t][t] [t](t2-8t+22)[t]= [t](t2-8t+22)[t]. [t](t2-8t+22)[t] has basis {1,t} with t-action on this basis given by the matrix (0-2218).

Notes and References

These are a typed copy of Lecture 36 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 28, 2011.

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