Brian Bi

## Section 2.16. Problems on Lie algebras

Problem 2.16.1 Following the Hint, we proceed by induction in dimension.

If $$\mathfrak{g}$$ is one-dimensional, and $$a \in \mathfrak{g}$$ is nonzero, then let $$v \in V$$ be an eigenvector of $$\rho(a)$$; the span of $$v$$ alone, a one-dimensional subspace of $$V$$, is a one-dimensional subrepresentation of $$V$$; so if $$V$$ is irreducible then it must be one-dimensional.

If $$d \geq 2$$, assume that the Lie theorem holds for all Lie algebras of dimension strictly less than $$d$$. Let $$\mathfrak{g}$$ be a $$d$$-dimensional solvable Lie algebra. If $$K(\mathfrak{g})$$ is trivial, then $$\mathfrak{g}$$ is abelian and the only irreducible representations of $$\mathfrak{g}$$ are one-dimensional, as desired. If $$K(\mathfrak{g})$$ is nontrivial, then the following hold:

1. $$K(\mathfrak{g})$$ is a solvable Lie algebra of dimension strictly less than $$\mathfrak{g}$$;
2. $$Q = \mathfrak{g}/K(\mathfrak{g})$$ is an abelian Lie algebra.

Applying the inductive hypothesis to $$V$$ considered as a representation of $$K(\mathfrak{g})$$, we see that contained within $$V$$ is an irreducible subrepresentation of $$K(\mathfrak{g})$$ of dimension one. Let $$v$$ be a nonzero vector belonging to this subrepresentation. If we define $$\chi: K(\mathfrak{g}) \to \mathbb{C}$$ such that $$av = \chi(a)v$$ for each $$a$$, then $$\chi$$ is a linear function.

Suppose $$x \in \mathfrak{g}$$ and $$a \in K(\mathfrak{g})$$. Consider the smallest vector space $$U$$ containing $$v$$ and invariant under $$x$$, that is, $$\span \{v, xv, x^2v, \ldots\}$$. Since $$U$$ is finite-dimensional, a basis for $$U$$ is given by $$\{v, xv, \ldots, x^{\dim U - 1}v\}$$. If $$n \in \mathbb{N}$$, then $$a x^n v = x^n a v + [a, x^n]v = \chi(a) x^n v + \chi([a, x^n])v \label{eqn:2.16.1}$$ $$(\ref{eqn:2.16.1})$$ implies that $$a x^n v \in U$$. Since this holds for any $$a \in K(\mathfrak{g})$$, $$U$$ is invariant under $$K(\mathfrak{g})$$. Furthermore, if $$1 \leq n < \dim U$$, then the coefficient of $$x^n v$$ in $$a x^n v$$ is $$\chi(a)$$, while if $$n = 0$$, then the coefficient is again $$\chi(a)$$ since $$[a, x^n] = [a, \Id] = 0$$. Therefore $$a$$ acts with trace $$(\dim U) \chi(a)$$ on $$U$$. In particular, $$\tr([x, a]) = \tr(xa) - \tr(ax) = 0$$, implying $$0 = \dim(U) \chi([x, a])$$, so \begin{equation*} \chi([x, a]) = 0 \end{equation*}

Using this result, we find that $$0 = [x, a]v = xav - axv = \chi(a)xv - axv$$, so if $$v$$ is a common eigenvector of all $$a \in K(\mathfrak{g})$$, then each $$xv$$ is also a common eigenvector of $$K(\mathfrak{g})$$. Although $$xv$$ may not necessarily lie in $$\span(v)$$, we can consider the subspace $$W \subseteq V$$ such that $$aw = \chi(a)w$$ for all $$w \in W$$; then $$W$$ is closed under the action of $$\mathfrak{g}$$. Now $$Q$$ acts linearly on $$W$$ and is abelian, so $$W$$ contains some common eigenvector of $$Q$$, which is therefore a common eigenvector of $$\mathfrak{g}$$. Therefore $$V$$ can be irreducible only if it is one-dimensional.

(Note from Brian): It's not clear to me whether this solution is the one intended by the text. The Hint leaves off after show that $$\mathfrak{g}$$ preserves common eigenspaces of $$K(\mathfrak{g})$$. Taking the quotient $$\mathfrak{g}/K(\mathfrak{g})$$ seems to be the cleanest way to proceed, but is not so obvious. A more common technique involves applying the inductive hypothesis to an ideal of codimension one, so there is only one outside element for which an eigenvector needs to be found.

Problem 2.16.2 In the case of characteristic zero, Lie's theorem applies, and every finite-dimensional irreducible representation is one-dimensional. It's easy to see that $$Y$$ must act as the zero operator, while $$X$$ is any scalar multiple of $$\Id$$. There is therefore one irreducible representation for each element of the ground field.

We now focus on the case of characteristic $$p > 0$$. Use $$L$$ to denote the Lie algebra described in the problem. Let $$V$$ be a finite-dimensional irrep of $$L$$ and let $$v$$ be an eigenvector of $$x$$, so that $$xv = \lambda v$$. Observe that $$x(yv) = yxv + [x, y]v = (\lambda + 1)yv$$, that is, $$yv$$ is either zero or an eigenvector with eigenvalue $$\lambda + 1$$. If the sequence $$v, yv, y^2 v, \ldots$$ terminates in zero, then we have found a simultaneous eigenvector of $$x$$ and $$y$$, and hence a one-dimensional subrepresentation. So $$V$$ itself must be one-dimensional, and consist only of $$v$$, so that $$yv = 0$$. So we have found one class of irreducible representations $$V_\lambda$$ of the form $$Y = 0, X = \lambda \Id$$ for some $$\lambda \in k$$.

If the sequence does not terminate in zero, note that $$y^p$$ is central (the proof is left as an exercise for the reader), and if $$V$$ is irreducible, then $$y^p$$ acts as a scalar in $$V$$, so $$\span\{v, yv, \ldots, y^{p-1}v\}$$ is a subrepresentation of $$V$$. Since $$V$$ is irreducible, $$V = \span\{v, yv, \ldots, y^{p-1}v\}$$. Furthermore, since $$y^k v$$ is an eigenvector of $$x$$ with eigenvalue $$\lambda + k$$, the $$y^k v$$ ($$0 \leq k \leq p-1$$) have different eigenvalues and are linearly independent. So they form a basis of $$V$$. With respect to this basis, $$X$$ takes the form \begin{equation*} X_\lambda = \begin{pmatrix} \lambda & 0 & \cdots & 0 \\ 0 & \lambda + 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda + p - 1 \end{pmatrix} \end{equation*} If $$y^p = c \Id$$, and $$\mu$$ is the unique $$p$$th root of $$c$$, then we can always normalize the eigenvectors of $$X$$ so that $$Y$$ takes the form \begin{equation*} Y_\mu = \begin{pmatrix} 0 & 0 & \cdots & 0 & \mu \\ \mu & 0 & \cdots & 0 & 0 \\ 0 & \mu & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & \mu & 0 \end{pmatrix} \end{equation*} Clearly different $$c$$, and hence $$\mu$$, yield nonisomorphic representations. The only constraint is that $$\mu \neq 0$$ in order for the representation to be irreducible. Two $$\lambda$$'s will yield isomorphic representations if they differ by $$1 + 1 + \ldots$$, otherwise the spectra of $$X$$ will be different so the representations will be nonisomorphic.

So all finite-dimensional irreps of $$L$$ are given by the $$\{V_\lambda \mid \lambda \in k\}$$ previously defined, and the $$\{W_{\lambda,\mu} \mid \lambda \in k/C_p, \mu \in k \setminus \{0\}\}$$ given by $$\{X = X_\lambda, Y = Y_\mu\}$$. (The notation $$k/C_p$$ means the quotient with the cyclic group of order $$p$$.) Evidently, the Lie theorem is not true in positive characteristic.

Problem 2.16.3

1. One useful fact is that if $$x$$ and $$y$$ generate a Lie algebra $$\mathfrak{g}$$, and $$[x, z] = [y, z] = 0$$ for some $$z \in \mathfrak{g}$$, then $$[\mathfrak{g}, z] = \{0\}$$. This is because the Jacobi identity allows us to flatten a Lie bracket of the form $$[[a, b], c]$$ as $$[a, [b, c]] - [b, [a, c]]$$. By iterating this process, whatever the form of $$a \in \mathfrak{g}$$, we can flatten out $$[a, z]$$ into a linear combination of terms of the form $$\ad(a_1) \ad(a_2) \ldots \ad(a_n) z$$ in which each $$a_i$$ is either $$x$$ or $$y$$, so the result is zero. (We can formalize this using proof by induction.)

Hence, the Lie algebra $$\mathfrak{g}_1$$ can be no more than 3-dimensional since $$[x, [x, y]] = [y, [x, y]] = 0$$. To see that it is exactly 3-dimensional, we consider it as the Lie algebra spanned by $$e_1, e_2, e_3$$ with $$[e_1, e_2] = e_3$$ and all other Lie brackets zero. This satisfies the Jacobi identity since all $$[a, [b, c]]$$ vanish, so it is a valid Lie algebra.

In $$\mathfrak{g}_2$$, $$[y, [y, [x, y]]] = 0$$ by definition while the Jacobi identity implies that $$[x, [y, [x, y]]] = [y, [x, [x, y]]] + [[x, y], [x, y]] = 0$$ so $$\mathfrak{g}_2$$ can be no more than 4-dimensional. To show that it is exactly 4-dimensional, we use the same technique as with $$\mathfrak{g}_1$$. Identifying $$e_1, e_2, e_3, e_4$$ with $$x, y, [x, y], [y, [x, y]]$$, respectively, we get the Lie brackets \begin{equation*} [e_1, e_2] = e_3; \qquad [e_2, e_3] = e_4 \end{equation*} with all other Lie brackets zero. We only need to check the Jacobi identity for $$[e_2, [e_1, e_2]]$$ since this is the only nonzero double Lie bracket. Indeed $$[e_2, [e_1, e_2]] + [e_1, [e_2, e_2]] + [e_2, [e_2, e_1]] = 0$$ so $$\span\{e_1, e_2, e_3, e_4\}$$ with these brackets forms a valid Lie algebra.

In $$\mathfrak{g}_3$$, we note that $$[x, [y, [x, y]]]$$ must still be zero since the proof of that only used the fact that $$\ad^2(x)(y) = 0$$, which holds also in $$\mathfrak{g}_3$$. To continue to explore $$\mathfrak{g}_3$$, it suffices to keep on iterating the application of $$\ad(x)$$ or $$\ad(y)$$. Now $$[y, [y, [x, y]]]$$ is plausibly nonzero. From there we can continue to $$[y, [y, [y, [x, y]]]]$$, which is zero by definition, and $$[x, [y, [y, [x, y]]]]$$, which is also plausibly nonzero. Using the Jacobi identity at the outermost level, we can rewrite this in the form: \begin{equation*} [x, [y, [y, [x, y]]]] = [y, [x, [y, [x, y]]]] + [[x, y], [y, [x, y]]] = [[x, y], [y, [x, y]]] \end{equation*} since $$[x, [y, [x, y]]]$$ vanishes. Beyond that point, a further application of $$\ad(x)$$ yields zero, which we can see by again applying the Jacobi identity at the outermost level: \begin{align*} [x, [[x, y], [y, [x, y]]]] = [[x, y], [x, [y, [x, y]]]] + [[x, [x, y]], [y, [x, y]]] = 0 \end{align*} since $$[x, [y, [x, y]]]$$ and $$[x, [x, y]]$$ both vanish. Finally, the application of $$\ad(y)$$ one final time also yields zero, which we can see by repeated application of the Jacobi identity: \begin{align*} [y, [[x, y], [y, [x, y]]] &= [[y, [x, y]], [y, [x, y]]] + [[x, y], [y, [y, [x, y]]]] \\ &= [x, [y, [y, [y, [x, y]]]]] - [y, [x, [y, [y, [x, y]]]] \\ &= -[y, [y, [x, [y, [x, y]]]]] - [y, [[x, y], [y, [x, y]]]] \end{align*} (where, at each step, we use the fact that the first term on each RHS is zero). Thus, in characteristic other than 2, $$[y, [[x, y], [y, [x, y]]]] = 0$$. (In characteristic 2 it appears this problem has a different answer, but I assume the ground field is implied to be $$\mathbb{C}$$.)

So it appears that a basis of $$\mathfrak{g}_3$$ is given by $$x, y, [x, y], [y, [x, y]], [y, [y, [x, y]]], [[x, y], [y, [x, y]]]$$. To prove this, we again use the same technique as with $$\mathfrak{g}_1$$ and $$\mathfrak{g}_2$$. Identifying $$e_1, e_2, e_3, e_4, e_5, e_6$$ with the purported basis vectors in the order previously given, we can compute that the only nonzero Lie brackets are: \begin{equation*} [e_1, e_2] = e_3; \qquad [e_1, e_5] = e_6; \qquad [e_2, e_3] = e_4; \qquad [e_2, e_4] = e_5; \qquad [e_3, e_4] = e_6; \end{equation*} (Note that the Lie brackets [16], [26], [34], [35], [36], [45], [46], and [56] vanish because they are too deep: by using the Jacobi identity to flatten out a nested Lie bracket expression with six or more generators, we can always see that it vanishes based on the work above.) We have to verify that this is a valid Lie algebra by verifying the Jacobi identity for all double Lie brackets. The only nonzero ones are $$[1[24]], [2[12]], [2[23]], [3[23]]$$. Antisymmetry automatically implies the Jacobi identity if any two of the three operands are the same. For the case [1[24]], we can easily check that \begin{equation*} [e_1, [e_2, e_4]] + [e_2, [e_4, e_1]] + [e_4, [e_1, e_2]] = [e_1, e_5] + 0 + [e_4, e_3] = e_6 - e_6 = 0 \end{equation*} so we are done; the dimension of $$\mathfrak{g}_3$$ is 6.

2. See here for the write-up.

Problem 2.16.4 We will use the letter $$K$$ to denote the ground field. Let $$V$$ be an irreducible finite-dimensional representation of $$\mathfrak{sl}_2(K)$$. (Etingof has clarified that the problem is intended to be about the finite-dimensional irreps only, but adds that infinite-dimensional irreps do not exist here.)

Recall a number of useful facts:

1. As we know from Problem 2.15.1, if $$v \in V$$ is an eigenvector of $$H$$ with eigenvalue $$\lambda$$, then $$Ev$$ (if nonzero) is an eigenvector of $$H$$ as well, with eigenvalue $$\lambda + 2$$, and $$Fv$$ (if nonzero) is an eigenvector of $$H$$ with eigenvalue $$\lambda -2$$.
2. We also have some results from parts (a) and (b) of Problem 2.15.1. We collect them here together with some similar results that can be proven by induction in a similar way: \begin{gather} [H, E^k] = 2k E^k \label{eqn:2.16.4.1} \\ [E^k, F] = k E^{k-1} H + k(k-1) E^{k-1} \label{eqn:2.16.4.2} \\ [H, F^k] = -2k F^k \tag{\ref{eqn:2.15.1b1}} \\ [E, F^k] = kF^{k-1} H - k(k-1) F^{k-1} \tag{\ref{eqn:2.15.1b2}} \end{gather}
3. From the above, we immediately see that in characteristic $$p$$, the elements $$E^p$$ and $$F^p$$ are central.
4. In part (g), we showed that $$C = EF + FE + H^2/2$$ is central. Therefore, it acts as a scalar in $$V$$, say, $$C = c \Id$$. If $$v$$ is an eigenvector of $$H$$, then using $$Cv = cv$$ together with $$(EF - FE)v = Hv = \lambda v$$, we obtain that $$EFv$$ and $$FEv$$ are scalar multiples of $$v$$.

Since $$V$$ is irreducible, we know that $$E^p$$ and $$F^p$$ both act as scalars. Note that $$E^p$$ being nonzero automatically implies that the dimension of $$V$$ is at least $$p$$. To see this, observe that $$E^p$$ being nonzero would imply that $$E$$ is nonsingular; and if $$v \in V$$ were an eigenvector of $$H$$, the sequence $$v, Ev, \ldots, E^{p-1}v$$ would then be a sequence of eigenvectors of $$H$$ with different eigenvalues, so they would be linearly independent, proving that the dimension is at least $$p$$. A similar argument shows that $$F^p$$ being nonzero would imply that $$\dim V \geq p$$.

This allows us to easily classify the irreducible representations of dimension strictly less than $$p$$. For these, both $$E^p$$ and $$F^p$$ act as the zero operator on $$V$$. Pick some eigenvector of $$H$$. By iterating the action of $$E$$ on this eigenvector, we obtain a sequence of eigenvectors of $$H$$ that terminates in zero. So there exist $$v \in V \setminus \{0\}, \lambda \in K$$ with $$Hv = \lambda v, Ev = 0$$. Let $$n$$ be the smallest integer such that $$F^n v = 0$$. Obviously, $$n \leq \dim V < p$$. The results of 2.15.1(b)–(f) apply, and the vectors $$v, Fv, \ldots, F^{n-1}v$$ form the basis of a $$n$$-dimensional irreducible representation of $$\mathfrak{sl}_2(K)$$, where $$\lambda = n -1$$, which we may again denote by $$V_\lambda$$; and the $$V_\lambda$$ exhaust the irreps of dimension less than $$p$$, up to isomorphism.

Now we come to the $$p$$-dimensional or higher irreps (we will see that the dimension cannot be greater than $$p$$). We begin with the case where $$\rho(E^p) = a \Id$$ with $$a \neq 0$$. Let $$v_0$$ be an eigenvector of $$H$$ with eigenvalue $$\lambda$$ and let $$v_i = E^i v_0$$ for $$i = 1, 2, \ldots, p-1$$; then the $$v_i$$'s are all linearly independent, with $$Hv_i = (\lambda + 2i)v_i$$. Obviously, $$Ev_{p-1} = av_0$$. Since $$FEv$$ is a scalar multiple of $$v$$ for each $$v \in V$$, it follows that $$Fv_i = FEv_{i-1}$$ is a scalar multiple of $$v_{i-1}$$ for $$i = 1, 2, \ldots, p-1$$ and similarly $$Fv_0 = a^{-1} FEv_{p-1}$$ is a scalar multiple of $$v_{p-1}$$. So the $$v_i$$'s form a basis for an invariant subspace of $$V$$, which must be $$V$$ itself if the latter is to be irreducible. Now suppose $$Fv_0 = bv_{p-1}$$. Then we can work out $$Fv_1 = FEv_0 = EFv_0 - Hv_0 = (ab - \lambda) v_0$$. Continuing, we obtain $$Fv_2 = FEv_1 = EFv_1 - Hv_1 = (ab - 2\lambda - 2)v_1$$ and so on. We must check that the commutation relations are also satisfied at the end, that is, that $$(EF - FE)v_{p-1} = Hv_{p-1}$$. Luckily, this turns out to be the case, so we have obtained a valid representation, in which: \begin{align} Hv_i &= (\lambda + 2i) v_i \label{eqn:RH} \\ Ev_i &= \begin{cases} v_{i+1} & i < p-1 \\ av_0 & i = p-1 \end{cases} \label{eqn:RE} \\ Fv_i &= \begin{cases} bv_{p-1} & i = 0 \\ (ab - i\lambda - i(i-1)) v_{i-1} & i > 0 \end{cases} \label{eqn:RF} \\ C &= \left[\frac{\lambda^2}{2} - \lambda + 2ab\right]\Id \end{align} Denote this representation by $$R(a, b, \lambda)$$. If $$R'$$ is a proper subrepresentation of $$R(a, b, \lambda)$$, then $$E^p$$ must act as the zero operator on $$R'$$ since $$R'$$ is less than $$p$$-dimensional. But since $$E^p$$ acts as a nonzero scalar on $$R(a, b, \lambda)$$, $$R'$$ must be the zero representation. Therefore all $$R(a, b, \lambda)$$ are irreducible.

Obviously any two $$R(a, b, \lambda)$$ with different $$a$$ are nonisomorphic. The representations $$R(a, b_1, \lambda_1), R(a, b_2, \lambda_2)$$ where $$\lambda_1 - \lambda_2 \notin \mathbb{F}_p$$ are also nonisomorphic as the spectra of $$H$$ differ. However, the representations $$R(a, b, \lambda)$$ and $$R(a, b - a^{-1}\lambda, \lambda + 2)$$ are isomorphic. If $$\{v_0, v_1, \ldots, v_{p-1}\}$$ is a basis for $$R(a, b, \lambda)$$ satisfying the above relations, and $$\{w_0, w_1, \ldots, w_{p-1}\}$$ is a basis for $$R(a, b - a^{-1}\lambda, \lambda')$$ with $$Hw_0 = \lambda' w_0$$ and so on ($$\lambda' = \lambda + 2$$), then an isomorphism is given by \begin{equation*} \phi(v_i) = \begin{cases} w_{i-1} & i > 0 \\ a^{-1} w_{p-1} & i = 0 \end{cases} \end{equation*} as can be easily verified. Moreover, in a pair of isomorphic representations, the Casimir operators must act as the same scalar. This scalar is linear in $$b$$ so it follows that $$R(a, b, \lambda)$$ cannot be isomorphic to $$R(a, b', \lambda + 2)$$ for any $$b'$$ other than $$b - a^{-1}\lambda$$. Consequently, up to isomorphism, there is exactly one irrep of the form $$R(a, b, \lambda)$$ for each $$a \in K \setminus \{0\}, b \in K$$, and $$\lambda$$ the representative of some element of the quotient group $$K/\mathbb{F}_p$$, and all other representations of the form $$R(a, b, \lambda)$$ with $$a \neq 0$$ are isomorphic to one of these.

By considering $$\rho(F^p) = a \Id$$ with $$a \neq 0$$ instead, we obtain an analogous set of $$p$$-dimensional irreducible representations $$S(a, b, \lambda)$$ for which \begin{align} Hv_i &= (\lambda - 2i) v_i \label{eqn:SH} \\ Ev_i &= \begin{cases} bv_{p-1} & i = 0 \\ (ab + i\lambda - i(i-1)) v_{i-1} & i > 0 \end{cases} \label{eqn:SE} \\ Fv_i &= \begin{cases} v_{i+1} & i < p-1 \\ av_0 & i = p - 1 \end{cases} \label{eqn:SF} \\ C &= \left[\frac{\lambda^2}{2} + \lambda + 2ab\right]\Id \end{align} The isomorphism situation is analogous to that of the $$R$$'s: by considering Casimir invariants we see that all possible isomorphisms between different $$S$$'s are generated by the isomorphism between $$S(a, b, \lambda)$$ and $$S(a, b + a^{-1}\lambda, \lambda - 2)$$.

Note that if in a representation $$S(a, b, \lambda)$$ the scalar operator $$\rho(E^p)$$ is nonzero then $$S(a, b, \lambda)$$ must be isomorphic to some $$R(a', b', \lambda')$$ since the $$R$$'s comprise all irreducible $$p$$-dimensional representations where $$\rho(E^p)$$ is nonzero. Therefore the only $$S$$'s that are not isomorphic to any $$R$$ are those for which the product $$b(ab+\lambda)(ab+2\lambda-2) \ldots (ab-\lambda-2)$$ is zero, namely the representations $$S_i(a, \lambda) = S(a, -ia^{-1}(\lambda-i+1), \lambda)$$ where $$a \in K \setminus \{0\}, \lambda \in K, i \in \{0, 1, \ldots, p-1\}$$.

Using the fact that $$S(a, b, \lambda)$$ is isomorphic to $$S(a, b + a^{-1}\lambda, \lambda - 2)$$ we obtain that $$S_i(a, \lambda)$$ is isomorphic to $$S_{i-1}(a, \lambda - 2)$$. It follows that all the $$S_i$$'s are isomorphic to some $$S_0$$. So we can restrict our attention to the representations of the form $$S_0(a, \lambda)$$. For a given $$S_0(a, \lambda)$$, all isomorphic $$S$$'s are given by $$\{S_i(a, \lambda+2i) \mid i \in \{1, 2, \ldots, p-1\}\}$$ since these are the representations generated by the isomorphism previously noted between $$S(a, b, \lambda)$$ and $$S(a, b + a^{-1}\lambda, \lambda-2)$$. Therefore an isomorphism exists between $$S(a, 0, \lambda)$$ and $$S(a, 0, \lambda + 2i)$$ if and only if $$S(a, 0, \lambda + 2i) = S_i(a, \lambda + 2i)$$, that is, if $$0 = -ia^{-1}(\lambda + 2i - i + 1)$$, which is solved to obtain $$i = 0$$ or $$\lambda = -i-1$$. Therefore for each $$a \in K \setminus \{0\}$$, the only isomorphisms between different $$S(a, 0, \lambda)$$s are those between $$S(a, 0, -i-1)$$ and $$S(a, 0, i-1)$$ for each $$i \in \mathbb{F}_p$$. These partition the $$S(a, 0, i)$$s into the isomorphism classes $$\{\{p-1\}, \{0, p-2\}, \{1, p-3\}, \ldots, \{(p-3)/2, (p-1)/2\}\}$$ with respect to $$i$$.

Finally, it falls to us to consider any remaining representations (dimension $$p$$ or greater) for which both $$E^p$$ and $$F^p$$ act as the zero operator. In such a representation, we can always find a common eigenvector of $$H$$ and $$E$$, that is, some $$v_0 \neq 0, \lambda$$ with $$Ev_0 = 0, Hv_0 = \lambda v_0$$. Writing $$v_i = F^i v_0$$, and applying reasoning similar to that which we previously considered when $$F^p$$ was nonzero, we obtain that $$\{v_0, v_1, \ldots, v_{p-1}\}$$ form a basis for $$V$$ and that $$Ev_i = (i\lambda - i(i-1))v_{i-1}$$ for each $$i = 1, 2, \ldots, p-1$$. Thus: \begin{align*} Hv_i &= (\lambda - 2i)v_i \\ Ev_i &= \begin{cases} 0 & i = 0 \\ (i\lambda-i(i-1))v_{i-1} & i > 0 \end{cases} \\ Fv_i &= \begin{cases} v_{i+1} & i < p - 1 \\ 0 & i = p - 1 \end{cases} \end{align*} Note that this can be regarded as $$S(0, 0, \lambda)$$. No two such representations can be isomorphic to each other since in $$S(0, 0, \lambda)$$, we can uniquely determine $$\lambda$$ using the fact that the unique eigenvector of $$F$$ has the eigenvalue $$\lambda + 2$$ with respect to $$H$$. Note however that not all $$S(0, 0, \lambda)$$'s are irreducible. For $$\lambda \notin \mathbb{F}_p$$ every nonzero vector is cyclic since we can iterate $$F$$ until we are left with a multiple of $$v_{p-1}$$ and then iterate $$E$$ repeatedly to generate a basis, but for $$\lambda \in \{0, 1, \ldots, p-2\}$$ the subspace spanned by $$\{v_{\lambda+1}, v_{\lambda+2}, \ldots, v_{\lambda+p-1}\}$$ is invariant. For $$\lambda = p-1$$, the representation $$S(0, 0, \lambda)$$ is isomorphic to $$V_{p-1}$$.

In conclusion, all nonisomorphic irreducible representations of $$\mathfrak{sl}_2(K)$$ are given by:

• The $$n$$-dimensional representation $$V_{n - 1}$$ for each $$n = 1, 2, \ldots, p$$, as defined in Problem 2.15.1,
• The $$p$$-dimensional representation $$R(a, b, \lambda)$$ for each $$a \in K \setminus \{0\}$$, $$b \in K$$, $$\lambda$$ the representative of some equivalence class of $$K$$ modulo $$\mathbb{F}_p$$, as defined by $$(\ref{eqn:RH})$$ through $$(\ref{eqn:RF})$$,
• The $$p$$-dimensional representation $$S(a, 0, \lambda)$$ for each $$a \in K \setminus \{0\}$$, $$\lambda \in (K \setminus \mathbb{F}_p) \cup \{-1, 0, 1, \ldots, (p-3)/2\}$$, as defined by $$(\ref{eqn:SH})$$ through $$(\ref{eqn:SF})$$,
• The $$p$$-dimensional representation $$S(0, 0, \lambda)$$ for each $$\lambda \in K$$.

There are no irreducible representations of dimension greater than $$p$$.

Problem 2.16.5 Etingof has clarified that this problem was intended to only be about the finite-dimensional irreps (but adds that there are no infinite-dimensional irreps). This problem turns out to be very similar to Problem 2.16.4, so some similar techniques will be abbreviated in this solution.

First consider the case with $$q$$ not a root of unity. Suppose $$V$$ is an irreducible finite-dimensional representation of $$\mathcal{U}_q(\mathfrak{sl}(2))$$. Let $$v$$ be an eigenvector of $$\rho(K)$$ be an eigenvector of $$K$$ with eigenvalue $$\lambda$$. Then $$K(ev) = \lambda Ke (K^{-1}v) = \lambda q^2 ev$$, therefore $$ev$$, if nonzero, is another eigenvector of $$\rho(K)$$ with eigenvalue $$q^2\lambda$$. Likewise, $$fv$$ is either zero or an eigenvector of $$\rho(K)$$ with eigenvalue $$q^{-2}\lambda$$. By starting with some eigenvector of $$\rho(K)$$ and iterating the action of $$e$$, we must always eventually reach zero (we cannot get an infinite number of eigenvectors of $$\rho(K)$$ all with different eigenvalues).

So there is some $$v \in V$$ that satisfies $$Hv = \lambda v, ev = 0, v \neq 0$$. The sequence $$v, fv, f^2 v, \ldots$$ must likewise terminate in zero. Suppose $$f^{n-1} v \neq 0$$ and $$f^n v = 0$$. Then we claim that $$\{v, fv, \ldots, f^{n-1}v\}$$ span an invariant subspace (and hence form a basis of $$V$$). To do so it suffices to show that $$ef^i v$$ is a scalar multiple of $$f^{i-1} v$$ for each $$i = 1, 2, \ldots, n-1$$. This can be done by induction:

1. For $$i = 1$$, $$ef^i v = efv = fev + [e, f]v = [e, f]v$$ which is a scalar multiple of $$v$$ since $$v$$ is an eigenvector of $$\rho(K)$$.
2. For $$i > 1$$, $$ef^i v = ef(f^{i-1}v) = fe(f^{i-1}v) + [e, f](f^{i-1}v) = f(af^{i-2}v) + bf^{i-1}v = (a + b)f^{i-1}v$$ where the scalar $$a$$ is guaranteed to exist by the inductive hypothesis and $$b$$ by the fact that $$f^{i-1}v$$ is an eigenvector of $$\rho(K)$$.

Therefore, if $$N = \dim V$$ and $$v_i$$ denotes $$f^i v$$ then \begin{align*} Kv_i &= q^{-2i}\lambda v_i \\ fv_i &= \begin{cases} v_{i+1} & i < N-1 \\ 0 & i = N-1 \end{cases} \\ ev_i &= \begin{cases} 0 & i = 0 \\ \frac{v_{i-1}}{q-q^{-1}} \sum_{j=0}^{i-1} (q^{-2j}\lambda - q^{2j} \lambda^{-1}) & i > 0 \end{cases} \end{align*} where the expression for $$ev_i$$ has been obtained by induction using the commutation relation for $$e, f$$. Taking this one step further and setting $$0 = efv_{N-1} = fev_{N-1} + [e,f]v_{N-1}$$ yields the relation \begin{equation*} 0 = \sum_{j=0}^N (q^{-2j}\lambda - q^{2j}\lambda^{-1}) \end{equation*} which can be rearranged to yield $$\lambda^2 = q^{2N-2}$$, therefore $$\lambda = \pm q^{N-1}$$. Using this value of $$\lambda$$ with the above relations and doing some algebra, we obtain \begin{align} Kv_i &= \pm q^{N-2i-1} v_i \label{eqn:Kq1} \\ fv_i &= \begin{cases} v_{i+1} & i < N-1 \\ 0 & i = N-1 \end{cases} \label{eqn:fq1} \\ ev_i &= \begin{cases} 0 & i = 0 \\ \pm \frac{v_{i-1}}{(q-q^{-1})^2} [q^N + q^{-N} - q^{N-2i} - q^{-N+2i}] & i > 0 \label{eqn:eq1} \end{cases} \end{align} Call the representations obtained by the two choices of sign $$V^+_{q,N}$$ and $$V^-_{q,N}$$. It is straightforward but tedious to verify that they are valid representations. To see that they are irreducible, note that we can start with any nonzero vector, iterate $$f$$ repeatedly until we get a multiple of $$v_{N-1}$$, then iterate $$e$$ repeatedly to regenerate the basis. This latter procedure must work because $$q^N + q^{-N} - q^{N-2i} - q^{-N+2i}$$ can't be zero unless $$q^{2i} = 1$$ or $$q^{2(N-i)} = 1$$, neither of which is possible for any $$0 < i < N$$ unless $$q$$ is a root of unity. Since $$v_0$$ is the only eigenvector of $$\rho(K)$$ that is annihilated by $$e$$, and this eigenvector has two different eigenvalues in $$V^+_{q,N}$$ and $$V^-_{q,N}$$, these two irreps are nonisomorphic. So for $$q$$ not a root of unity, there is a countably infinite series of finite-dimensional irreducible representations $$V^\pm_{q,N}$$, two of each positive dimension $$N$$.

Consider now the case where $$q$$ is a root of unity. Let $$M$$ be the smallest positive integer such that $$q^{2M} = 1$$. Suppose $$1 \leq N < M$$ and let $$V$$ be an irrep of dimension $$N$$. Suppose $$v \in V$$ is an eigenvector of $$\rho(K)$$. It must be the case that $$e^N v = 0$$. For if this were not the case, then the vectors $$v, ev, \ldots, e^N v$$ would have $$N+1$$ distinct eigenvalues, which is a contradiction. Thus, we can always find some nonzero $$v_0 \in V$$ with $$Kv_0 = \lambda v_0, ev_0 = 0$$. As before, if $$v_i = f^i v$$ with $$v_n = 0, v_{n-1} \neq 0$$, then $$\{v_0, \ldots, v_{n-1}\}$$ form a basis of an invariant subspace of $$V$$, hence a basis of $$V$$. So $$n = N$$ and the basis is $$\{v_0, \ldots, v_{N-1}\}$$ and the action of the generators is given by $$(\ref{eqn:Kq1})$$ through $$(\ref{eqn:eq1})$$, in other words, we have the two nonisomorphic irreps $$V^\pm_{q,N}$$ where, as before, $$\lambda = \pm q^{N-1}$$.

Now consider $$N \geq M$$. Observe that $$Ke^M K^{-1} = (KeK^{-1})^M = (q^2 e)^M = q^{2M} e^M = e^M$$. Likewise $$Kf^M K^{-1} = f^M$$. Furthermore $$K$$ commutes with $$fe$$ and hence shares a common eigenvector with it. We will also make use of the following results:

Lemma 1: For each positive integer $$i$$, $$Ke^i = q^{2i}e^i K$$ and $$e^i K^{-1} = q^{2i} K^{-1} e^i$$.

Proof: Note that the defining relations imply $$Ke = q^2 eK$$ and $$eK^{-1} = q^2 K^{-1}e$$. The desired result immediately follows.

Lemma 2: For each positive integer $$i$$, \begin{equation*} [e^i, f] = \frac{1 - q^{2i}}{(1-q^2)(q-q^{-1})} (e^{i-1}K - K^{-1}e^{i-1}) \end{equation*}

Proof: By induction. For $$i = 1$$ this follows immediately from the defining relation for $$[e,f]$$. For $$i > 1$$, write $$[e^i, f] = e[e^{i-1}f] + [e,f]e^{i-1}$$ and apply the inductive hypothesis.

Corollary: $$[e^M, f] = 0$$. Together with the fact that $$Ke^M K^{-1} = e^M$$, this implies that $$e^M$$ is central.

Since $$e^M$$ is central, it follows that in the irrep $$V$$, it acts as a scalar. First assume this scalar is a nonzero value $$a$$. Therefore if $$v_0$$ is some common eigenvector of $$\rho(K)$$ and $$\rho(fe)$$, and $$v_i = e^i v_0$$, then the vectors $$v_0, v_1, \ldots, v_{M-1}$$ have distinct eigenvalues with respect to $$K$$, and therefore form a basis of a subspace invariant under $$e$$ and $$K$$. We claim that it is invariant under $$f$$ as well, which is proven by induction:

1. $$fv_1 = fev_0$$ is a scalar multiple of $$v_0$$ since $$v_0$$ was assumed an eigenvector of $$\rho(fe)$$.
2. For $$2 \leq i < M$$, we have that $$fv_i = fev_{i-1} = efv_{i-1} - [e,f]v_{i-1}$$ which is a scalar multiple of $$v_{i-1}$$ by the inductive hypothesis and the relation for $$[e,f]$$.
3. Furthermore, $$fv_0 = a^{-1}fev_{M-1} = a^{-1}(efv_{M-1} - [e,f]v_{M-1})$$ which is similarly a scalar multiple of $$v_{M-1}$$.

Therefore $$V$$ is precisely $$M$$-dimensional with basis $$\{v_0, v_1, \ldots, v_{M-1}\}$$. Let $$b \in k$$ be such that $$fv_0 = bv_{M-1}$$ and let $$\lambda$$ be the eigenvalue of $$v_0$$ with respect to $$K$$. Then, by using Lemma 2 to compute $$fe^i v_0$$, we obtain: \begin{align} Kv_i &= q^{2i} \lambda v_i \label{eqn:KqR} \\ ev_i &= \begin{cases} v_{i+1} & i < M - 1 \\ av_0 & i = M - 1 \end{cases} \label{eqn:eqR} \\ fv_i &= \begin{cases} bv_{M-1} & i = 0 \\ \left[ab - \frac{(1-q^{2i})(\lambda - q^{-2i+2}\lambda^{-1})} {(1-q^2)(q-q^{-1})}\right] v_{i-1} & i > 0 \end{cases} \label{eqn:fqR} \end{align} This turns out to be a valid representation for any $$\lambda \neq 0$$ (we omit some algebra). We call this representation $$R(a,b,\lambda)$$. If $$R'$$ is a proper subrepresentation of $$R(a,b,\lambda)$$, then $$e^M$$ must act as the zero operator on $$R'$$ since $$R'$$ is less than $$M$$-dimensional. But since $$e^M$$ acts as a nonzero scalar on $$R(a,b,\lambda)$$, $$R'$$ must be the zero representation. Therefore all $$R(a,b,\lambda)$$ are irreducible.

To investigate isomorphisms we will use the quadratic Casimir element for $$\mathcal{U}_q(\mathfrak{sl}(2))$$ [Kerler, 1995]: $$C = ef + \frac{qK^{-1} + q^{-1}K}{(q-q^{-1})^2}$$ The element $$C$$ is central, so it acts as a scalar in an irrep. For $$R(a,b,\lambda)$$ that scalar is given by $$c = ab + (q\lambda^{-1} + q^{-1}\lambda)/(q-q^{-1})^2$$, which is linear in $$b$$. Now two $$R$$'s can only be isomorphic when they have equal $$a$$ and the ratio of their $$\lambda$$'s is a power of $$q^2$$ (to ensure that the spectra are equal). Because the scalar $$c$$ is linear in $$b$$, it's not possible for $$R(a,b,\lambda)$$ to be isomorphic to $$R(a,b',\lambda)$$ unless $$b = b'$$. Indeed, by equating $$c$$ between $$R(a,b,\lambda)$$ and $$R(a,b', q^2\lambda)$$ we see that the only possible isomorphism must be obtained by setting $$b' = b - a^{-1}(\lambda - \lambda^{-1})/(q - q^{-1})$$. Explicitly, the isomorphism between $$R(a, b, \lambda)$$ and $$R(a, b', q^2\lambda)$$ is given by \begin{equation*} \phi(v_i) = \begin{cases} a^{-1} w_{M-1} & i = 0 \\ w_{i-1} & i > 0 \end{cases} \end{equation*} where $$\{v_0, \ldots, v_{N-1}\}$$ and $$\{w_0, \ldots, w_{N-1}\}$$ are respectively bases of $$R(a,b,\lambda)$$ and $$R(a, b', q^2\lambda)$$ satisfying Eqns $$(\ref{eqn:KqR})$$ through $$(\ref{eqn:fqR})$$. It is straightforward but tedious to verify that $$\phi$$ is indeed an isomorphism. So all nonisomorphic $$R$$'s are given by $$\{R(a, b, \lambda) \mid a \in k \setminus \{0\}, b \in k, \lambda \in k/\langle q^2\rangle\}$$.

As in Problem 2.16.4, we can likewise investigate $$M$$-dimensional irreps where $$\rho(f^M) = a \Id$$ with $$a \neq 0$$. We omit a lot of analogous calculations here, observing only that it suffices to swap $$e$$ and $$f$$ and replace $$q$$ by its inverse in the calculations above. This gives irreducible representations with basis $$\{v_0, \ldots, v_{M-1}\}$$ parametrized by $$b, \lambda \in k$$ where \begin{align} Kv_i &= q^{-2i} \lambda v_i \label{eqn:KqS} \\ fv_i &= \begin{cases} v_{i+1} & i < M-1 \\ av_0 & i = M-1 \end{cases} \label{eqn:fqS} \\ ev_i &= \begin{cases} bv_{M-1} & i = 0 \\ \left[ab + \frac{(1 - q^{-2i})(\lambda - q^{2i-2}\lambda^{-1})} {(1 - q^{-2})(q - q^{-1})}\right] v_{i-1} & i > 0 \end{cases} \label{eqn:eqS} \\ Cv_i &= \left[ab + \frac{q\lambda + q^{-1}\lambda^{-1}}{(q-q^{-1})^2} \right] v_i \end{align} We shall denote this representation by $$S(a, b, \lambda)$$; it can be verified that $$S(a, b, \lambda) \cong S(a, b + a^{-1}(\lambda - \lambda^{-1})/(q-q^{-1}), q^{-2}\lambda) \label{eqn:Siso}$$ where the $$\cong$$ symbol denotes isomorphism; and that (since the Casimir invariant is linear in $$b$$) this isomorphism generates all isomorphisms between different $$S$$'s.

Now, any $$S(a, b, \lambda)$$ in which $$\rho(e^M)$$ is nonzero must be isomorphic to one of the $$R$$'s, so we consider only those where $$\rho(e^M) = 0$$. For these, one of the coefficients in $$(\ref{eqn:eqS})$$ must vanish, therefore these representations are given by \begin{equation*} S_i(a, \lambda) = S\left(a, -\frac{a^{-1}(1-q^{-2i})(\lambda - q^{2i-2} \lambda^{-1})}{(1-q^{-2})(q-q^{-1})}, \lambda\right) \end{equation*} for some $$i \in \{0, 1, \ldots, M-1\}$$. With a bit of algebra, we can show that $$(\ref{eqn:Siso})$$ is equivalent to the statement that $$S_i(a, \lambda)$$ is isomorphic to $$S_{i-1}(a, q^{-2}\lambda)$$ for all $$i \in \{0, 1, \ldots, M-1\}$$. Therefore all $$S_i$$'s can be put into isomorphism with some $$S_0(a, \lambda) = S(a, 0, \lambda)$$, and for a given $$S_0(a, \lambda)$$, all isomorphic $$S$$'s are given by $$\{S_i(a, q^{2i}\lambda) \mid i \in \{1, 2, \ldots, M-1\}\}$$. An isomorphism between $$S(a, 0, \lambda)$$ and $$S(a, 0, q^{2i} \lambda)$$ with $$i \in \{1, 2, \ldots, M-1\}$$ will therefore occur if and only if $$S(a, 0, q^{2i} \lambda) = S_i(a, q^{2i}\lambda)$$, that is, when \begin{equation*} 0 = -\frac{a^{-1} (1 - q^{-2i})(q^{2i}\lambda - q^{2i-2} q^{-2i} \lambda^{-1})}{(1-q^{-2})(q-q^{-1})} \end{equation*} which implies $$q^{2i} = 1$$ or $$\lambda = \pm q^{-i-1}$$. By definition of $$M$$, the former cannot happen. So all the isomorphisms between different $$S_0$$'s are those between $$S(a, 0, \pm q^{-i-1})$$ and $$S(a, 0, \pm q^{i-1})$$ for $$i \in \mathbb{Z}$$. For example, if $$M$$ is odd and $$q$$ is a primitive $$M$$th root of unity, then all pairs of isomorphic $$S_0$$'s are given by $$\{S(a, 0, q^0), S(a, 0, q^{M-2})\}, \{S(a, 0, q^1), S(a, 0, q^{M-3})\}, \ldots, \{S(a, 0, q^{(M-3)/2}), S(a, 0, q^{(M-1)/2})\}$$, and similarly with each $$q^i$$ replaced by $$-q^i$$. We will skip the explicit construction of the isomorphism classes for the other cases ($$M$$ odd with $$q$$ a primitive $$2M$$th root of unity, or $$M$$ even).

We now turn our attention to $$M$$-dimensional representations in which both $$e^M$$ and $$f^M$$ act as the zero operator. We proceed in a similar manner as in Problem 2.16.4, starting with $$v_0$$ a common eigenvector of $$\rho(K)$$ and $$\rho(e)$$ with eigenvalues $$\lambda$$ and 0, respectively, and writing $$v_i = f^i v_0$$. The resulting representation is then $$S(0, 0, \lambda)$$ for some $$\lambda \neq 0$$. No two such representations can be isomorphic since in $$S(0, 0, \lambda)$$ we can uniquely determine $$\lambda$$ using the fact that the unique eigenvector of $$\rho(f)$$ has the eigenvalue $$q^2\lambda$$ with respect to $$K$$. However, some of these representations are not irreducible, namely precisely those in which the coefficient in $$(\ref{eqn:eqS})$$ vanishes for some $$1 \leq i \leq M-1$$. Using some algebra we see that this happens when $$\lambda = \pm q^{i-1}$$.

In conclusion, the nonisomorphic irreducible representations are given by:

• The $$N$$-dimensional representations $$V^\pm_{q,N}$$, where $$1 \leq N < M$$, as defined by $$(\ref{eqn:Kq1})$$ through $$(\ref{eqn:eq1})$$;
• The $$M$$-dimensional representation $$R(a,b,\lambda)$$ for each $$a \in k^\times, b \in k, \lambda \in k^\times$$, as defined by $$(\ref{eqn:KqR})$$ through $$(\ref{eqn:fqR})$$;
• The $$M$$-dimensional representation $$S(a, 0, \lambda)$$ for each $$a \in k^\times, \lambda \in k^\times$$, as defined by $$(\ref{eqn:KqS})$$ through $$(\ref{eqn:eqS})$$, subject to the fact that for each $$i \in \mathbb{Z}$$, the representations $$S(a, 0, q^{-i-1})$$ and $$S(a, 0, q^{i-1})$$ are isomorphic, and the representations $$S(a, 0, -q^{-i-1})$$ and $$S(a, 0, -q^{i-1})$$ are isomorphic;
• The $$M$$-dimensional representations $$S(0, 0, \lambda)$$ for each $$\lambda \in k^\times \setminus \{\pm q^i \mid i \in \{0, 1, \ldots, M-2\}\}$$.