Brian Bi
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## Section 2.15. Representations of $$\mathfrak{sl}(2)$$

Problem 2.15.1

1. Let $$v \in V, \lambda \in \mathbb{C}$$. Then, using the commutation relations for $$E$$ and $$H$$, \begin{equation*} (H - (\lambda + 2)I)(Ev) = E(H - (\lambda + 2)I)v + 2Ev = E(H - \lambda I)v \end{equation*} It follows that \begin{equation*} (H - (\lambda + 2)I)^n (Ev) = E(H - \lambda I)^n v \end{equation*} so if $$v$$ is a generalized eigenvector of $$H$$ with eigenvalue $$\lambda$$, then $$Ev$$ is either a generalized eigenvector of $$H$$ with eigenvalue $$\lambda + 2$$, or zero. If $$\lambda$$ is the eigenvalue with maximum real part, then $$Ev$$ must be zero.
2. Lemma 1: If $$k \geq 1$$ is an integer, then \begin{equation} [H, F^k] = -2k F^k \label{eqn:2.15.1b1} \end{equation}

Proof: Use the formula $$[A, BC] = B[A, C] + [A, B]C$$ with $$A = H, B = F^{k-1}, C = F$$ together with induction on $$k$$.

Lemma 2: If $$k \geq 1$$ is an integer, then \begin{equation} [E, F^k] = kF^{k-1}H - k(k-1) F^{k-1} \label{eqn:2.15.1b2} \end{equation}

Proof: By induction. The base case $$k = 1$$ is easily verified. For $$k \geq 2$$, \begin{align*} [E, F^k] &= F[E, F^{k-1}] + [E, F]F^{k-1} \\ &= F((k-1) F^{k-2} H - (k-1)(k-2)F^{k-2}) + HF^{k-1} \\ &= (k-1)F^{k-1}H - (k-1)(k-2) F^{k-1} + F^{k-1} H + [H, F^{k-1}] \\ &= k F^{k-1}H - (k-1)(k-2) F^{k-1} - 2(k-1) F^{k-1} \\ &= k F^{k-1}H - k(k-1) F^{k-1} \end{align*}

Lemma 3: If $$k \geq 0$$ is an integer, then \begin{equation*} EH^k w = 0 \end{equation*}

Proof: Use the fact that $$EH^k = (HE + [E,H])H^{k-1}$$ together with induction on $$k$$.

It immediately follows that $$EP(H)w = 0$$ where $$P$$ is any polynomial.

Now we are ready to prove the main result. Let $$P$$ be a polynomial. Then \begin{align*} E^k F^k P(H) w &= E^{k-1} (E F^k) P(H) w \\ &= E^{k-1} (F^k E + [E, F^k]) P(H) w \\ &= E^{k-1} (k F^{k-1}H - k(k-1) F^{k-1}) P(H) w \\ &= E^{k-1} F^{k-1} k(H - (k-1)) P(H) w \end{align*} Therefore \begin{equation*} E^k F^k w = k! H (H - 1) (H - 2) \ldots (H - (k-1)) w \end{equation*}

3. This is very similar to part (a), but $$F$$ acts to lower the real part of the eigenvalue by 2, in contrast to $$E$$ which raises it. Since $$V$$ is finite-dimensional, it can only have finitely many eigenvalues, so the sequence $$v, Fv, F^2v, \ldots$$ must end in zero. Indeed, we can make the stronger statement that $$F^{\dim V}v = 0$$.
4. Lemma: If $$v$$ is a generalized eigenvector of $$H$$ with eigenvalue $$\lambda$$, and $$a_1, \ldots a_n \in \mathbb{C}$$, and $$(H-a_1)(H-a_2) \ldots (H-a_n)v = 0$$, then there is some $$i$$ such that $$a_i = \lambda$$.

Proof: By induction. For $$n = 1$$, $$v$$ is just an ordinary eigenvector with eigenvalue $$a_1$$. If $$n \geq 2$$, then either $$(H - a_n)v = 0$$, in which case $$\lambda = a_n$$, or $$(H - a_n)v$$ is a generalized eigenvector with eigenvalue $$\lambda$$, and by the inductive hypothesis, there is some $$i \leq n - 1$$ with $$a_i = \lambda$$.

Now use the Hint from the problem. Take $$N = \dim V$$. From (b), we have \begin{equation} \label{eqn:2.15.1.d} 0 = E^N F^N v = N! H(H-1)(H-2)\ldots(H-N+1) v \end{equation} By the Lemma, $$\lambda$$ is an integer between 0 and $$N-1$$. Since polynomials of $$H$$ commute with each other, we can rewrite $$(\ref{eqn:2.15.1.d})$$ as \begin{equation*} 0 = \left(\prod_{0 \leq i < N, i \neq \lambda} H-i\right) (H-\lambda)v \end{equation*} Let $$v' = (H-\lambda)v$$. By the contrapositive of the Lemma, $$v'$$ cannot be a generalized eigenvector of $$H$$ with eigenvalue $$\lambda$$, since none of the $$i$$'s can equal $$\lambda$$. Therefore $$v' = 0$$, and $$v$$ is an ordinary eigenvector. Since this holds for all $$v \in \overline{V}(\lambda)$$, it follows that $$H$$ is diagonalizable on $$\overline{V}(\lambda)$$.

5. This follows immediately from the Lemma proven in (d).
6. Suppose $$V$$ is an irreducible finite-dimensional representation of $$\mathfrak{sl}(2)$$. Let $$\lambda$$ be as in (a) and $$v \in V$$ be an eigenvector of $$H$$ with eigenvalue $$\lambda$$. From (e), we have that $$F^{\lambda+1}v = 0$$ and $$F^\lambda v \neq 0$$. Therefore $$v, Fv, \ldots, F^\lambda v$$ are eigenvectors of $$H$$ with eigenvalues $$\lambda, \lambda-2, \ldots, 2, 0, -2, \ldots, -\lambda$$, respectively, which implies that they are all linearly independent. Furthermore, $$\span\{v, Fv, \ldots, F^\lambda v\}$$ is obviously invariant under $$F$$ and $$H$$, and is also invariant under $$E$$ since, by $$(\ref{eqn:2.15.1b2})$$, for any positive integer $$k$$, \begin{equation*} EF^k v = F^k Ev + kF^{k-1}Hv -k(k-1)F^{k-1}v = k(\lambda-k+1) F^{k-1}v \end{equation*} Since $$V$$ is irreducible, this span must equal $$V$$.

With respect to the basis $$\{v, Fv, \ldots, F^\lambda v\}$$, the action of $$H$$ on $$V'$$ takes the matrix form $$\diag(N-1, N-3, \ldots, -(N-3), -(N-1))$$, while $$F$$ obviously has ones on the subdiagonal and zeroes everywhere else. The raising property of $$E$$ implies that it is nonzero only along the superdiagonal, and the equation $$[E, F] = H$$ then fixes $$E$$ as well, so that the superdiagonal entries are easily seen to be $$1(N-1), 2(N-2), \ldots, (N-1)(N-(N-1)) = N-1$$. Call this irreducible representation $$V_\lambda$$. It has dimension $$N = \lambda + 1$$; thus, the value of $$\lambda$$ fixes the dimension of the irreducible representation and also fixes the representation itself up to isomorphism.

We have not proven that $$V_\lambda$$ is actually irreducible. Let us do so now. Suppose $$w \in V_\lambda$$ is nonzero. Write $$w = \sum_i c_i F^i v$$, where $$Hv = \lambda v$$ as above. Let $$m$$ be the smallest integer such that $$c_m$$ is nonzero. Then $$\frac{1}{c_m} F^{\lambda - m} w = F^\lambda v$$ and by applying $$E$$ to $$F^\lambda v$$ we can regenerate the basis. We conclude that there is exactly one irreducible representation of each positive finite dimension up to isomorphism, namely $$V_\lambda$$ where $$\lambda$$ is one less than the dimension.

7. Using the identity $$[A, BC] = B[A, C] + [A, B]C$$: \begin{align*} [E, C] &= [E, EF + FE + H^2/2] \\ &= [E, EF] + [E, FE] + [E, H^2/2] \\ &= [E, E]F + E[E, F] + F[E, E] + [E, F]E + \frac{1}{2}H[E, H] + \frac{1}{2}[E, H]H \\ &= EH + HE - HE - EH \\ &= 0 \\ [F, C] &= [F, EF + FE + H^2/2] \\ &= [F, EF] + [F, FE] + [F, H^2/2] \\ &= E[F, F] + [F, E]F + F[F, E] + [F, F]E + \frac{1}{2}H[F, H] + \frac{1}{2}[F, H]H \\ &= -HF -FH + HF + FH \\ &= 0 \\ [H, C] &= [H, EF + FE + H^2/2] \\ &= [H, EF] + [H, FE] + [H, H^2/2] \\ &= E[H, F] + [H, E]F + F[H, E] + [H, F]E \\ &= -2EF + 2EF + 2FE - 2FE \\ &= 0 \end{align*} Since $$C$$ commutes with all the generators, it is central, and by the result of Problem 2.3.16(a), it therefore acts as a scalar on $$V_\lambda$$. If we take $$v$$ such that $$Ev = 0$$, we can readily compute $$Cv = \frac{\lambda(\lambda+2)}{2}v$$ using the result of the previous part.
8. Since $$V$$ is assumed indecomposable, we can use the result of Problem 2.3.16(b) to conclude that $$C$$ must act in $$V$$ as an operator with only one eigenvalue, namely, the scalar by which it acts on some irreducible subrepresentation. If that irreducible subrepresentation is $$V_\lambda$$, then that single eigenvalue is $$\lambda(\lambda+2)/2$$ by the result of the previous part.
9. Since $$W$$ is reducible and finite-dimensional, it has to have some irreducible proper subrepresentation, and if $$C$$'s single eigenvalue is $$\lambda(\lambda+2)/2$$, then that subrepresentation is isomorphic to $$V_\lambda$$. The quotient representation $$V/W$$ is also a nonzero representation of $$\mathfrak{sl}(2)$$ but is of lower dimension than $$V$$. Either $$V/W$$ is irreducible, in which case it must also be isomorphic to $$V_\lambda$$ since $$C$$ has only one eigenvalue, so that $$n = 1$$; or else, if $$V/W$$ is reducible, then since $$V$$ was assumed to be the smallest reducible representation which is not a direct sum of irreps, then $$V/W$$ must be a direct sum of irreps, and again, since $$C$$ has only the single eigenvalue, each of those irreps must be isomorphic to $$V_\lambda$$. In this case $$n \ge 2$$.
10. If the eigenspace $$V(\lambda)$$ of $$H$$ is $$m$$-dimensional, then, since the eigenspace $$W(\lambda)$$ is one-dimensional, it follows that the eigenspace $$(V/W)(\lambda)$$ is $$(m-1)$$-dimensional. If $$V/W = nV_\lambda$$, then $$m - 1 = n$$, so $$V(\lambda)$$ is $$(n+1)$$-dimensional.

Let $$\{v_1, \ldots, v_{n+1}\}$$ be a basis for $$V(\lambda)$$. The result of part (e) guarantees that for each $$j$$ in $$\{0, \ldots, \lambda\}$$, the vectors in the set $$S_j = \{F^j v_1, \ldots, F^j v_{n+1}\}$$ are all nonzero. We also know that the elements of $$S_j$$ are all eigenvectors of $$H$$ with eigenvalue $$\lambda - 2j$$. The result of part (b) implies that each of $$E^j F^j v_i$$ is a nonzero scalar multiple of $$v_i$$, so that $$\span\{E^j S_j\}$$ is $$(n+1)$$-dimensional, which implies that $$\span S_j$$ is likewise $$(n+1)$$-dimensional; so $$S_j$$ is linearly independent. Since eigenvectors with different eigenvalues are always linearly independent, the union $$S = \cup_j S_j$$ is linearly independent. Since $$|S| = (n+1)(\lambda+1) = \dim V$$, $$S$$ is a basis of $$V$$.

11. As $$i$$ ranges from 1 to $$n+1$$ and $$j$$ ranges from 0 to $$\lambda$$, the $$F^j v_i$$ range through all distinct basis vectors of $$V$$ as found in the previous part. By partitioning these basis vectors into the subsets $$T_i = \{F^j v_i \mid 0 \leq j \leq \lambda\}$$ for each $$i$$, we obtain the subspaces $$W_i = \span T_i$$ whose direct sum equals $$V$$. By the argument given in the solution to part (f), each of these subspaces is invariant under $$E$$, $$F$$, and $$H$$, and is therefore a subrepresentation of $$V$$. So we have derived a contradiction with the assumption that $$V$$ is not the direct sum of subrepresentations.
12. The Jordan form of $$E$$ in the irrep $$V_\lambda$$, as found in (f), is a single Jordan block with eigenvalue zero (and size $$\lambda + 1$$, of course). Therefore, the Jordan form of $$E$$ in the general representation $$V = \large\oplus_i V_{\lambda_i}$$ is the direct sum of Jordan blocks with eigenvalue zero and sizes $$\lambda_i + 1$$; since all finite-dimensional representations of $$\mathfrak{sl}(2)$$ are of this form, we can conclude that for every direct sum of Jordan blocks with eigenvalue zero, there is exactly one representation of $$\mathfrak{sl}(2)$$ in which $$E$$ takes that Jordan normal form. Given a nilpotent operator $$A : V \to V$$, its Jordan normal form is a direct sum of Jordan blocks with eigenvalue zero, so there is exactly one representation, up to isomorphism, in which $$E = A$$.
13. Following the Hint, we define the character of a finite-dimensional representation $$V$$ as $$\chi_V(x) = \tr(e^{xH})$$. If $$V$$ and $$W$$ are two representations, then \begin{align*} \chi_{V \oplus W}(x) &= \tr \exp(x\rho_{V\oplus W}(H)) \\ &= \tr \exp(x\rho_V(H) \oplus x\rho_W(H)) \\ &= \tr(\exp(x\rho_V(H)) \oplus \exp(x\rho_W(H))) \\ &= \tr \exp(x\rho_V(H)) + \tr \exp(x\rho_W(H)) \\ &= \chi_V(x) + \chi_W(x) \end{align*} Also, \begin{align*} \chi_{V \otimes W}(x) &= \tr \exp(x(\rho_V(H) \otimes \Id_W + \Id_V \otimes \rho_W(H))) \\ &= \tr[\exp(x\rho_V(H) \otimes \Id_W) \exp(x\Id_V \otimes \rho_W(H))] \\ &= \tr[(\exp(x \rho_V(H)) \otimes \Id_W ) (\Id_V \otimes \exp(x \rho_W(H)))] \\ &= \tr[\exp(x \rho_V(H)) \otimes \exp(x \rho_W(H))] \\ &= \tr \exp(x \rho_V(H)) \tr \exp(x \rho_W(H)) \\ &= \chi_V(x) \chi_W(x) \end{align*} where between the first and second lines we have used the fact that $$\rho_V(H) \otimes \Id_W$$ and $$\Id_V \otimes \rho_W(H)$$ commute; and between the second and third lines we have used the identity that $$\exp(A \otimes \Id) = \exp(A) \otimes \Id$$, which follows from the power series expansion of the operator exponential.

The character of $$V_\lambda$$ is easily seen to be $$e^{\lambda x} + e^{(\lambda - 2)x} + \ldots + e^{-\lambda x}$$ from the result of part (f). Using the formula derived in the previous paragraph, the character of $$V_\lambda \otimes V_\mu$$ is $$\chi_{V_\lambda}\chi_{V_\mu}$$. If we assume without loss of generality that $$\lambda \geq \mu$$, then \begin{equation*} (e^{\lambda x} + \ldots + e^{-\lambda x})(e^{\mu x} + \ldots + e^{-\mu x}) = e^{(\lambda + \mu)x} + 2e^{(\lambda + \mu - 2)x} + \ldots + \mu e^{(\lambda - \mu)x} + \mu e^{(\lambda - \mu - 2)x} + \ldots + \mu e^{-(\lambda - \mu)x} + (\mu - 1) e^{-(\lambda - \mu + 2)x} + \ldots + e^{-(\lambda + \mu)x} \end{equation*} To write this as the sum of characters of irreps is easy because if $$V_k$$ is the highest-dimensional irrep that appears in the sum, then $$e^{kx}$$ will be the highest term that appears in the character; so we can successively peel off $$\chi_{V_k}(x)$$ from $$\chi_{V_\lambda \otimes V_\mu}(x)$$ where at each stage $$k$$ is taken from the highest exponential remaining. The result is \begin{equation*} \chi_{V_\lambda \otimes V_\mu}(x) = \chi_{V_\lambda + V_\mu} + \chi_{V_{\lambda + \mu - 2}} + \ldots + \chi_{V_{\lambda - \mu}} \end{equation*} so the desired decomposition is \begin{equation*} V_\lambda \otimes V_\mu \simeq V_{\lambda + \mu} \oplus V_{\lambda + \mu - 2} \oplus \ldots \oplus V_{\lambda - \mu} \end{equation*}

14. Using (l), there exists a representation in which $$E = A = J_M(0) \otimes \Id_N + \Id_M \otimes J_N(0)$$. But this is just the tensor product of representations in which $$E = J_M(0)$$ and $$E = J_N(0)$$, which are the irreps $$V_{M-1}$$ and $$V_{N-1}$$. This representation therefore decomposes as the direct sum $$V_{M+N-2} \oplus V_{M+N-4} \oplus \ldots \oplus V_{M-N}$$, where, without loss of generality, we have assumed that $$M \geq N$$, so $$E$$'s Jordan normal form, and hence that of $$A$$, must be $$J_{M+N-1}(0) \oplus J_{M+N-3}(0) \oplus \ldots \oplus J_{M-N+1}(0)$$.