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

Problem 2.15.1

- 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.
*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*}

- 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 \(F^{\dim V}v = 0\).
*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 \(\overline{V}(\lambda)\).

- This follows immediately from the Lemma proven in (d).
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.

- 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.
- 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.
- 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 \(n \ge 2\).
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 \(V\).

- 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.
- 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 \(E = A\).
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*}

- 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 \(J_{M+N-1}(0) \oplus J_{M+N-3}(0) \oplus \ldots \oplus J_{M-N+1}(0)\).