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 = 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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