Discuss representations of \(\ZZ\), the infinite cyclic group. What is the group algebra of \(\ZZ\)?
Solution
Fix a field \(k\) and consider \(k\)-linear representations.
1. A representation \(\rho:\ZZ\to\operatorname{GL}(V)\) is determined by the single invertible operator\[ T=\rho(1)\in\operatorname{GL}(V), \]and every invertible operator arises this way.
Since \(\ZZ\) is generated by \(1\), every \(n\in\ZZ\) satisfies \[ \rho(n)=\rho(1)^n=T^n. \] Conversely, any \(T\in\operatorname{GL}(V)\) defines a representation by \(\rho(n)=T^n\).
2. The group algebra of \(\ZZ\) over \(k\) is the Laurent polynomial ring\[ k[\ZZ]\cong k[t,t^{-1}]. \]
Let \(e_n\) denote the basis element of \(k[\ZZ]\) corresponding to \(n\in\ZZ\), so that \[ e_m e_n=e_{m+n}. \] The assignment \(e_n\mapsto t^n\) extends \(k\)-linearly to an algebra homomorphism \(k[\ZZ]\to k[t,t^{-1}]\), because \(t^mt^n=t^{m+n}\). It carries the \(k\)-basis \(\{e_n\}_{n\in\ZZ}\) bijectively onto the \(k\)-basis \(\{t^n\}_{n\in\ZZ}\), so it is an isomorphism.
3. Representations of \(\ZZ\) over \(k\) are exactly modules over \(k[t,t^{-1}]\).
1.These constructions are inverse to each other.
4. Two finite-dimensional representations are isomorphic exactly when their operators are similar.Hence they are classified by the rational canonical forms of invertible matrices, and over an algebraically closed field by Jordan normal forms with nonzero eigenvalues.
A linear isomorphism \(f:V\to W\) intertwines the representations determined by \(T\) and \(U\) exactly when \(fT=Uf\), that is, when \(U=fTf^{-1}\). Similarity classes of matrices are classified by rational canonical form, and over an algebraically closed field by Jordan normal form. A matrix is invertible exactly when \(0\) is not an eigenvalue.
5. If \(k\) is algebraically closed, every finite-dimensional irreducible representation of \(\ZZ\) is one-dimensional, and these are indexed by \(\lambda\in k^\times\), with \(1\in\ZZ\) acting as multiplication by \(\lambda\).
Let \(V\neq0\) be finite-dimensional and irreducible, with operator \(T\). Since \(k\) is algebraically closed, \(T\) has an eigenvector \(v\neq0\) with \(Tv=\lambda v\), and \(\lambda\neq0\) because \(T\) is invertible. The line \(kv\) is invariant under \(T\) and \(T^{-1}\), hence under \(\ZZ\), so irreducibility forces \(V=kv\). A one-dimensional representation is determined by the scalar \(T=\lambda\in k^\times\), and distinct scalars give non-isomorphic representations by 4.