2.3. Representations 13 every subspace of V is a subrepresentation. But V is irreducible, so 0 and V are the only subspaces of V . This means that dim V = 1 (since V = 0). Example 2.3.14. 1. A = k. Since representations of A are simply vector spaces, V = A is the only irreducible and the only indecom- posable representation. 2. A = k[x]. Since this algebra is commutative, the irreducible representations of A are its 1-dimensional representations. As we discussed above, they are defined by a single operator ρ(x). In the 1- dimensional case, this is just a number from k. So all the irreducible representations of A are Vλ = k, λ ∈ k, in which the action of A is defined by ρ(x) = λ. Clearly, these representations are pairwise nonisomorphic. The classification of indecomposable representations of k[x] is more interesting. To obtain it, recall that any linear operator on a finite dimensional vector space V can be brought to Jordan nor- mal form. More specifically, recall that the Jordan block Jλ,n is the operator on kn which in the standard basis is given by the formulas Jλ,nei = λei + ei−1 for i 1 and Jλ,ne1 = λe1. Then for any linear operator B : V → V there exists a basis of V such that the matrix of B in this basis is a direct sum of Jordan blocks. This implies that all the indecomposable representations of A are Vλ,n = kn, λ ∈ k, with ρ(x) = Jλ,n. The fact that these representations are indecom- posable and pairwise nonisomorphic follows from the Jordan normal form theorem (which in particular says that the Jordan normal form of an operator is unique up to permutation of blocks). This example shows that an indecomposable representation of an algebra need not be irreducible. 3. The group algebra A = k[G], where G is a group. A represen- tation of A is the same thing as a representation of G, i.e., a vector space V together with a group homomorphism ρ : G → Aut(V ), where Aut(V ) = GL(V ) denotes the group of invertible linear maps from the space V to itself (the general linear group of V ).

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