## 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.

- 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.
- 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.
- 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.