Brian Bi
\[ \DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref[1]{(\ref{#1})} \]

Section 2.3. Representations

Parenthetical to example 2.3.3 We are asked to explain why a representation of \(A = k\langle x_1, \ldots, x_n\rangle\) is just a vector space \(V\) over \(k\) with a collection of arbitrary linear operators \(\rho(x_1), \ldots, \rho(x_n): V \to V\).

This follows from the universal property of the free algebra: for every assignment of operators \(\rho_1, \ldots, \rho_n\) to the generators, there is a unique algebra homomorphism \(\rho : A \to \End V\) such that \(\rho(x_i) = \rho_i\) for each \(i\). But this is precisely a representation of \(A\).

Parenthetical to example 2.3.6 We are asked to check that if \(\phi : V_1 \to V_2\) is an isomorphism of representations, then \(\phi^{-1} : V_2 \to V_1\) is also.

Since \(\phi\) is an isomorphism of vector spaces, so is \(\phi^{-1}\). We need only show that \(\phi^{-1}\) is a homomorphism of representations. Let \(a \in A, v \in V_2\) where \(A\) is the algebra being represented. Then, \(\phi(\phi^{-1}(av)) = av = a\phi(\phi^{-1}(v)) = \phi(a\phi^{-1}(v))\), therefore, since \(\phi\) is an isomorphism, \(\phi^{-1}(av) = a\phi^{-1}(v)\).

Problem 2.3.15 For finite dimensional representations, we perform induction on the dimension. For dimension 1, \(V\) itself is irreducible. For dimension \(n\), either \(V\) is irreducible itself, or it has a nonzero proper subrepresentation, \(V'\). Since \(\dim V' < n\), by the inductive hypothesis, \(V'\) has an irreducible subrepresentation, which is also an irreducible subrepresentation of \(V\).

For the infinite dimensional case, let \(A = k[x]\), the polynomial algebra with one generator. Let \(V = \mathbb{R}^\infty\), and let \(x\) act on \(V\) by a right shift, that is, \(x(v_1, v_2, \ldots) = (0, v_1, v_2, \ldots)\). We claim that this algebra has no irreducible subrepresentations. To see this, let \(V'\) be a nonzero subrepresentation of \(V\). Then \(V'\) contains some nonzero vector, say, \(v\), and therefore necessarily contains as a subrepresentation \(Av\) (which may be \(V'\) itself). Consider now \(Axv\), which is a nonzero subrepresentation of \(Av\). If the first nonzero entry of \(v\) is located at index \(i\), then \(xv\)'s first nonzero entry is located at index \(i+1\), and therefore \(v \notin Axv\). Therefore \(Axv\) is a nonzero proper subrepresentation of \(Av\) and of \(V'\), so \(V'\) is not irreducible.

Remark: Note that for a representation to have no irreducible subrepresentation, it must have no nonzero finite-dimensional subrepresentations, otherwise the first part of the problem would apply.

Problem 2.3.16
  1. The operator \(\rho(z)\) is automatically an intertwining operator since \((az)v = (za)v\). By the corollary to Schur's lemma for algebraically closed fields, \(\rho(z)\) is a scalar operator. Since \(\chi_V : Z(A) \to k\) is just the map \(\lambda \, \mathrm{Id} \mapsto \lambda\), it is obviously a homomorphism.
  2. Fix \(z\). Then \(V\) has a basis of generalized eigenvectors and is a direct sum of generalized eigenspaces of \(z\). Let \(\lambda\) be an eigenvalue, and suppose \((z - \lambda \, \mathrm{Id})^m v = 0\). Then \((z - \lambda \, \mathrm{Id})^m av = a(z - \lambda \, \mathrm{Id})^m v = 0\), since \((z - \lambda \, \mathrm{Id})^m\) is central. That is, the generalized eigenspaces are subrepresentations of \(V\). Since \(V\) is indecomposable, there can only be one of these, and hence a single eigenvalue of \(z\). That this is the scalar by which \(z\) acts on an irreducible subrepresentation follows from problem 2.3.15 and part (a) of this problem.
  3. No, for example, for \(A = k[x]\), \(\rho(x)\) can be any Jordan block, as in example 2.3.14.

Problem 2.3.17 Let \(a^*\) denote the element in \(A^{\mathrm{op}}\) corresponding to \(a \in A\). Let \(\phi \in \End_A{A}\). Then assign to \(\phi\) the element \((\phi(1))^*\), which belongs to the opposite algebra. Likewise, to an element of the opposite algebra \(a^*\), assign the endomorphism \(x \mapsto xa\). One can then easily verify that this establishes an algebra isomorphism between \(A^{\mathrm{op}}\) and \(\End_A{A}\).

Problem 2.3.18 By the usual Schur's lemma, each \(\phi \in D\) is an isomorphism, therefore it has an inverse in the algebra \(D\). Since \(D\) is a division ring, \(V\) is a free \(D\)-module. This implies that \(\dim V \geq \dim D\) and in particular \(D\) is at most countably dimensional. If \(\phi\) is not a scalar, it is transcendental in \(\mathbb{C}\) since the latter is algebraically closed. The set of all elements of \(\mathbb{C}(\phi)\) of the form \(\frac{1}{\phi - k}\) with \(k \in \mathbb{C}\) are then linearly independent, so \(\mathbb{C}(\phi)\) is uncountably dimensional, and so is \(D \supseteq \mathbb{C}(\phi)\). We have arrived at a contradiction, so our assumption that some \(\phi \in D\) exists which is not a scalar must be false.