Algebraic K-Theory by E. M. Friedlander, M. R. Stein

By E. M. Friedlander, M. R. Stein

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Proof. Write xi = r λir vr and yj = Now expand both sides using (4). s µjs ws for some scalars λir and µjs . 23. Let M, N be CG-modules with bases m1 , . . and n1 , . . respectively. Then defining g · (mi ⊗ nj ) = (g · mi ) ⊗ (g · nj ) and extending bilinearly gives M ⊗ N the structure of an CG-module. Extending bilinearly means that if x = nj ∈ M ⊗ N then x · t is defined to be g λg g ∈ CG and t = i,j λij mi ⊗ λg λij g · (mi ⊗ nj ). g,i,j When we do this the two linearity conditions in the definition of a module are guaranteed to hold, and we only need to check that acting gh is the same as acting h then g.

Then for any g ∈ G we have χΩ (g) = | fixΩ (g)|. Version of Sunday 29th March, 2015 at 20:00 40 Proof. Consider the matrix for the action of g with respect to the basis Ω of CΩ. The column of this matrix corresponding to ω ∈ Ω has a 1 at position g · ω and zeros elsewhere, so it contributes 1 to the trace if g · ω = ω and zero otherwise. Thus the trace of this matrix is the number of elements of Ω fixed by g. 4. The character χ of the regular CG-module satisfies χ(g) = 0 if g = e and χ(e) = |G|.

So x = s ∈ SI ∩ V = {0}. This finishes the proof. 2. With the notation of the previous lemma, let f : S1 ⊕ · · · ⊕ Sn → N be a module homomorphism. Then im f is isomorphic to a direct sum of some of the Si . Proof. ker f is a submodule of M , so it has a complement C isomorphic to the direct sum of some of the Si by the previous lemma. Because C is a complement to the kernel of f , the restriction f |C : C → im f defined by f |C (c) = f (c) is an isomorphism of modules. So im f ∼ = C which is isomorphic to a direct sum of some of the Si .

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