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

## Section 2.7. Examples of algebras

Parenthetical to definition 2.7.3 We are asked to show that if $$\char k = 0$$, then $$k[t]$$ is a faithful representation of the Weyl algebra, where $$x$$ acts by multiplication by $$t$$ and $$y$$ acts by differentiation with respect to $$t$$. For this we just need the kernel of the homomorphism $$\rho$$ from the Weyl algebra $$A$$ to $$\End k[t]$$ to be trivial. Let $$a \in A$$ be nonzero. Then $$a$$ can be written uniquely as $$\sum_{j} P_j(x) y^j$$ where $$P_j(x)$$ is a polynomial in $$x$$ and at least one $$P_j$$ is nonzero. If $$J$$ is the minimum $$j$$ with nonzero $$P_j$$, then all $$y^j$$ with $$j > J$$ annihilate $$t^J$$. Therefore $$at^J = P_J(x) y^J t^J = J! P_J(x) \neq 0$$. The desired result follows.

If $$x, y$$ are the generators of the Weyl algebra as defined in the text, and $$[\cdot, \cdot]$$ denotes the commutator, then:

1. $$[x^i y^j, x] = j x^i y^{j-1}$$
2. $$[y, x^i y^j] = i x^{i-1} y^j$$

Therefore, $$[\cdot, x]$$ acts as formal differentiation with respect to $$y$$, and $$[y, \cdot]$$ acts as formal differentiation with respect to $$x$$.

Proof:

1. By induction on $$j$$.
2. By induction on $$i$$.
Problem 2.7.4
1. In an $$n$$-dimensional representation of $$A$$, we have $$n = \tr(I_n) = \tr(\rho(yx - xy)) = \tr(\rho(y)\rho(x)) - \tr(\rho(x)\rho(y)) = 0$$, so only the trivial representation is finite-dimensional. A nonzero two-sided ideal $$I \subseteq A$$ is closed under commutators with all elements of $$A$$, and contains some nonzero $$a$$, which can be uniquely written as $$a = \sum_j P_j(x) y^j$$ where each $$P_j(x)$$ is a polynomial in $$x$$ and at least one $$P_j$$ is nonzero. Let $$J$$ be the greatest $$j$$ with nonzero $$P_j$$. Then, by taking the commutator with $$x$$ iterated $$J$$ times, and using the Lemma, we obtain that $$I$$ contains $$P_J(x) J!$$ and therefore $$P_J(x)$$, a nonzero polynomial. By taking the commutator with $$y$$ iterated $$\deg P_J$$ times, and using the Lemma again, we obtain that $$I$$ contains a nonzero scalar, so $$I = A$$. We conclude that the Weyl algebra is simple.
2. That the elements $$x^p$$ and $$y^p$$ commute with $$x$$ and $$y$$ follows immediately from the Lemma. Since they commute with the generators, they commute with all elements of $$A$$. The centre of $$A$$ certainly contains $$k[x^p, y^p]$$. If we have some $$a \in A \setminus k[x^p, y^p]$$, then write $$a$$ with respect to the basis $$\{x^i y^j\}$$ and let $$b$$ consist of $$a$$ with all terms of the form $$c_{ij} x^i y^j$$ removed where $$p \mid i, j$$. Such $$b$$ is nonzero. If it contains some $$c_{ij} x^i y^j$$ with $$p \not\mid i$$, then write it in the form $$b = \sum_i x^i P_i(y)$$ where each $$P_i(y)$$ is a polynomial in $$y$$; then $$[y, b] = \sum_i i x^{i-1} P_i(y)$$ by the Lemma, and is nonzero. Otherwise, it must contain some $$c_{ij} x^i y^j$$ with $$p \not\mid j$$, in which case write it as $$b = \sum_j P_j(x) y^j$$, and the Lemma gives $$[b, x] = \sum_j P_j(x) j y^{j-1}$$, which is nonzero. In either case, $$b$$ is not central. We conclude that $$k[x^p, y^p]$$ is the entire centre of $$A$$.
3. Following the Hint, let $$V$$ be an irreducible finite-dimensional representation of $$A$$, and let $$v$$ be an eigenvector of $$y$$. Since $$V$$ is irreducible, $$v$$ is cyclic, implying that $$\{x^i y^j v \mid x, y \in \mathbb{N}\}$$ spans $$V$$. But $$x^i y^j v = \lambda^j x^i v$$ where $$\lambda$$ is the eigenvalue corresponding to $$v$$, so $$\{x^i v \mid x \in \mathbb{N}\}$$ spans $$V$$. Furthermore, since $$x^p$$ is central, it acts as a scalar in $$V$$ (Problem 2.3.16), so $$\{v, xv, \ldots, x^{p-1}v\}$$ spans $$V$$. To see that this set is linearly independent (and therefore a basis), observe that if $$av = 0$$, then $$[y, a]v = 0$$ since $$v$$ is an eigenvector of $$y$$. If $$\sum_i c_i x^i v = 0$$, then $$(\sum_i c_i x^i)v = 0$$ and taking the commutator with $$y$$ iterated $$p-1$$ times immediately gives that $$(p-1)! c_{p-1} v = 0$$, so $$c_{p-1} = 0$$; repeating this process for each $$i$$ in descending order then shows that all $$c_i$$ vanish, as required. Thus all irreducible finite-dimensional representations of $$A$$, if any, are $$p$$-dimensional. Note that by the trace argument used in part (a), any $$p$$-dimensional representation of $$A$$ is also automatically irreducible.

Our objective now is to construct all $$p$$-dimensional representations. We know that $$y^p$$ is a scalar, say, $$c \, \mathrm{Id}$$. This implies that $$y$$ has only one eigenvalue, $$\lambda = \sqrt[p]{c}$$, since $$p$$th roots are unique in characteristic $$p$$. Furthermore, we claim that there is only one linearly independent eigenvector of $$y$$. To see this, say $$yv_1 = \lambda v_1$$ and $$yv_2 = \lambda v_2$$, then let $$P(x)$$ be the polynomial of degree at most $$p-1$$ such that $$v_2 = P(x)v_1$$; the argument given in the previous paragraph implies that such $$P$$ exists and is unique. Now $$P'(x)v_1 = [y, P(x)]v_1 = yP(x)v_1 - P(x)yv_1 = \lambda v_2 - \lambda v_2 = 0$$, so $$P' = 0$$, therefore $$v_2$$ is a constant multiple of $$v_1$$. Thus, the Jordan form of $$y$$ has $$\lambda$$ for each diagonal entry and 1 for each superdiagonal entry. Call this matrix $$Y$$. We always have one representation given by \begin{equation*} X_{ij} = \begin{cases} j & \text{if $i = j+1$} \\ 0 & \text{otherwise} \end{cases} \end{equation*} where $$X = \rho(x)$$. For example, for $$p = 5$$, this gives the following representation: $X = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 \end{pmatrix} \qquad Y = \begin{pmatrix} \lambda & 1 & 0 & 0 & 0 \\ 0 & \lambda & 1 & 0 & 0 \\ 0 & 0 & \lambda & 1 & 0 \\ 0 & 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & 0 & \lambda \end{pmatrix}$ Once $$\lambda \in k$$ is fixed, the objective is to find all $$X$$ such that $$[Y, X] = I_p$$; we then set $$\rho(x) = X, \rho(y) = Y$$ to obtain a representation. To find all such $$X$$, notice that $$[Y, X] = I_p$$ is a linear equation in $$X$$, so the general solution is obtained as the sum of the particular solution previously given, which we shall call $$X_p$$, and the solution to the homogeneous equation $$[Y, X] = 0$$. It is easy to verify that $$Y - \lambda \, \mathrm{Id}$$ (and therefore $$Y$$) commutes with $$X$$ precisely when $$X$$ is upper triangular and constant along each diagonal. All such representations are pairwise nonisomorphic, since either they have different Jordan forms $$Y$$ for $$\rho(y)$$, or, when they have the same $$Y$$, they have different $$X$$. For the case $$p = 5$$ this general form is illustrated below: $X = \begin{pmatrix} a & b & c & d & e \\ 1 & a & b & c & d \\ 0 & 2 & a & b & c \\ 0 & 0 & 3 & a & b \\ 0 & 0 & 0 & 4 & a \end{pmatrix}$ Remark: If we regard $$V$$ as $$k[t]/\langle t^p\rangle$$, with the $$i$$th entry being the coefficient of $$t^{p-i-1}$$, then the result we have obtained can be interpreted as that all the representations over $$V$$, up to isomorphism, are given by \begin{align*} y &= t + \lambda \\ x &= -\frac{\partial}{\partial t} + P(t) \end{align*} with $$\lambda$$ drawn from $$k$$ and $$P(t)$$ drawn from $$k[t]/\langle t^p\rangle$$. We can cast this into a more familiar form by noting that given a representation $$\rho$$ of $$A$$, we can always construct the 90 degrees rotated representation $$\rho'$$ given by $$\rho'(x) = \rho(y); \rho'(y) = -\rho(x)$$. This gives \begin{align*} x &= t + \lambda \\ y &= \frac{\partial}{\partial t} - P(t) \end{align*} It should also be apparent that this is the inspiration for the ansatz for $$X_p$$.

Problem 2.7.5 We should first of all observe that $$x^{-1}y = qyx^{-1}$$ and $$xy^{-1} = qy^{-1}x$$ (proof left as exercise for reader).

1. It is easy to see that: \begin{align*} [x^i y^j, x] &= (q^j - 1)x^{i+1} y^j \\ [x^i y^j, x^{-1}] &= (q^{-j} - 1)x^{i-1} y^j \\ [y, x^i y^j] &= (q^i - 1) x^i y^{j+1} \\ [y^{-1}, x^i y^j] &= (q^{-i} - 1) x^i y^{j-1} \end{align*} Since $$\{x^i y^j \mid i, j \in \mathbb{Z}\}$$ is a basis, an element $$a \in A_q$$ commutes with $$x$$ and $$x^{-1}$$ if and only if all $$x^i y^j$$ with nonzero coefficients in $$a$$ satisfy $$q^j = q^{-j} = 1$$, and with $$y$$ and $$y^{-1}$$ if and only if all $$x^i y^j$$ with nonzero coefficients in $$a$$ satisfy $$q^i = q^{-i} = 1$$. So for $$q$$ not a root of unity, the centre is trivial, while for $$q$$ a primitive $$n$$th root of unity, the centre is $$k[x^n, y^n, x^{-n}, y^{-n}]$$; elements belonging to this set will commute with the generators and hence all of $$A_q$$.

For $$q$$ not a root of unity, suppose we have a nonzero ideal containing the nonzero element $$a = \sum_{i \in \mathbb{Z}} Q_i(x) y^i$$, where each $$Q_i \in \langle x, x^{-1}\rangle$$ and therefore commutes with $$x$$. By right-multiplying by $$y$$ or $$y^i$$, without loss of generality, we can take $$Q_i = 0$$ for $$i < 0$$ and $$Q_0 \neq 0$$. If $$Q_i = 0$$ for $$i > 0$$, then we have $$a \in k[x, x^{-1}]$$. Otherwise, take the commutator with $$x$$: \begin{equation*} [a, x] = \sum_{i \in \mathbb{N}} (q^i - 1)x Q_i(x) y^i \\ \end{equation*} This eliminates the $$i = 0$$ term and since $$q$$ is not a root of unity, it does not eliminate $$i \neq 0$$ terms. By left-multiplying by $$x^{-1}$$ and repeating this process, we can obtain that the ideal contains some $$b \in k[x, x^{-1}]$$. By taking commutators with $$y$$, we can analogously eliminate terms one by one, eventually finding that $$1$$ belongs to the ideal. Thus, $$A_q$$ is simple whenever $$q$$ is not a root of unity.

2. In an $$n$$-dimensional representation of $$A_q$$, we have $$\det\rho(y)\det\rho(x) = \det(\rho(yx)) = \det(\rho(qxy)) = q^n\det\rho(x)\det\rho(y)$$. Neither $$\rho(x)$$ nor $$\rho(y)$$ can be singular, since $$\rho(x^{-1}) = \rho(x)^{-1}$$ and $$\rho(y^{-1}) = \rho(y)^{-1}$$. So we get finite-dimensional representations only when $$q$$ is a root of unity. If $$q^n = 1$$, then we can get $$n$$-dimensional representations.
3. Say $$q$$ is a primitive $$n$$th root of unity. Let $$v \in A_q$$ be an eigenvector of $$y$$ in a finite-dimensional representation $$V$$, so that $$yv = \lambda v$$. Since $$y$$ has an inverse, $$\lambda \neq 0$$, and $$y^{-1}v = \lambda^{-1}v$$. If this representation is irreducible, then it is cyclic, and $$\{x^i y^j v \mid i, j \in \mathbb{Z}\}$$ spans $$V$$. But since $$y^j v = \lambda^j v$$, it must be that $$\{x^i v \mid i \in \mathbb{Z}\}$$ spans $$V$$. But $$x^n$$ and $$x^{-n}$$ act as scalars in $$V$$ (Problem 2.3.16), so $$\{v, xv, \ldots, x^{n-1}v\}$$ spans $$V$$. Note that $$yxv = qxyv = q\lambda xv$$, so $$xv$$ is an eigenvector of $$y$$ with eigenvalue $$q\lambda$$. We immediately see that all of $$\{v, xv, \ldots, x^{n-1}v\}$$ are eigenvectors of $$y$$ with distinct eigenvalues $$\lambda, q\lambda, \ldots, q^{n-1}\lambda$$, so they are linearly independent (and hence a basis). So $$\rho(y)$$ has a Jordan form like the following (taking $$n = 4$$): \begin{equation*} \rho(y) = \lambda \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & q & 0 & 0 \\ 0 & 0 & q^2 & 0 \\ 0 & 0 & 0 & q^3 \end{pmatrix} \end{equation*} Since $$\rho(x)$$ promotes an eigenvector to an eigenvector with $$q$$ times the eigenvalue, the most general form it takes on is as follows: \begin{equation*} \rho(x) = \begin{pmatrix} 0 & 0 & 0 & \mu_0 \\ \mu_1 & 0 & 0 & 0 \\ 0 & \mu_2 & 0 & 0 \\ 0 & 0 & \mu_3 & 0 \end{pmatrix} \end{equation*} where $$\mu_i \neq 0$$, so that $$\rho(x)$$ has an inverse given by \begin{equation*} \rho(x^{-1}) = \begin{pmatrix} 0 & \mu_1^{-1} & 0 & 0 \\ 0 & 0 & \mu_2^{-1} & 0 \\ 0 & 0 & 0 & \mu_3^{-1} \\ \mu_0^{-1} & 0 & 0 & 0 \end{pmatrix} \end{equation*} We should choose $$\lambda$$ with argument in the range $$[0, 2\pi/n)$$, to avoid getting the same eigenvalues for $$y$$ with different values of $$\lambda$$. With this restriction, clearly all representations with different $$\lambda$$ are nonisomorphic, and for the same $$\lambda$$, by choosing $$\mu_i$$ differently, we get different forms for $$\rho(x)$$, so all representations obtained in this way are pairwise nonisomorphic.

Remark: In the special case that $$q = 1$$, we obtain only one-dimensional irreps. This is consistent with Corollary 2.3.12.