Brian Bi
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## 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.