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By Lam T.Y., Magid A.R. (eds.)

ISBN-10: 0821810871

ISBN-13: 9780821810873

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Extra info for Algebra, K-Theory, Groups, and Education: On the Occasion of Hyman Bass's 65th Birthday

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If ϕ1 and ϕ2 are non-isomorphic, then χ1 , χ2 = 1 χ1 (g)χ2 (g) = 0. |G| g∈G If ϕ1 and ϕ2 are isomorphic, then χ1 , χ2 = χ1 , χ1 = 1. In other words, the set of irreducible representations of G forms an orthonormal set in the space of class functions. Proof. Let dim V = m and dim W = n, and define eij to be the n × m matrix with a 1 in the (i, j)th position and 0’s everywhere else. First, suppose that V and W are non-isomorphic. 2, the matrix fij = |G| g∈G ϕ1 (g)eij ϕ2 (g ) corresponds to an intertwiner V → W .

Now, suppose that V is a module over a group algebra k[G]. Then, V is a vector space over k ⊂ k[G] (since modules over fields are vector spaces), and each element g of G acts on the vector space V by a k-linear map. Denote this k-linear map by ϕ(g) : V → V . Then, for V to be a module, we must have ϕ(g)ϕ(h) = ϕ(gh) for all g and h in G. e. the group of invertible linear maps from V to itself. Notice that the module structure of V is completely characterized by ϕ and makes no reference to group algebras.

Example. Let G = C∞ , and let V be a 2-dimensional vector space over a field k. Pick a generator g of G and a basis {v1 , v2 } of V . Then, to define a representation of G, we just have to define a homomorphism G → GL(V ). It suffices to pick where we send g in this map; so, we define a representation by sending g → ( 10 11 ). Then, we have gv1 = v1 , so that kv1 ⊆ V is a k[G]-submodule of V which is fixed by the action of G, hence it is a subrepresentation. However, its complement in the vector space is kv2 , which is not fixed by G.

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Algebra, K-Theory, Groups, and Education: On the Occasion of Hyman Bass's 65th Birthday by Lam T.Y., Magid A.R. (eds.)


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