## 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:

- \([x^i y^j, x] = j x^i y^{j-1}\)
- \([y, x^i y^j] = i x^{i-1} y^j\)

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

*Proof:*

- By induction on \(j\).
- By induction on \(i\).

- 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.
- 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 \(A\).
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 the90 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).

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.

- 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.
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.