Representations of \(\mathfrak{sl}(2,\mathbb{C})\)

The Classification

We have built, for each integer \(m \geq 0\), an explicit polynomial representation \(\Pi_m\) of \(SU(2)\) of dimension \(m + 1\), and we proved each one irreducible. Each \(\Pi_m\) differentiates to a representation \(\pi_m\) of \(\mathfrak{su}(2)\), which extends complex-linearly to a representation of \(\mathfrak{sl}(2,\mathbb{C})\), still written \(\pi_m\). The hypothesis of that extension, \(\mathfrak{su}(2) \cap i\,\mathfrak{su}(2) = \{0\}\), was verified when \(\pi_m\) was constructed. Because \(SU(2)\) is connected (sending \(U\) to its first column identifies it with the unit sphere in \(\mathbb{C}^2\)), the group-algebra correspondence makes \(\pi_m\) irreducible as a representation of \(\mathfrak{su}(2)\), and the extension is then irreducible as a representation of \(\mathfrak{sl}(2,\mathbb{C})\). That was the constructive half of the story, and it produced one irreducible representation of \(\mathfrak{sl}(2,\mathbb{C})\) in each dimension. The present section completes it. We will show that these are, up to isomorphism, the only finite-dimensional irreducible complex representations of \(\mathfrak{sl}(2,\mathbb{C})\), and that nothing else exists. The classification is exact and complete.

The computation is worth doing carefully for three reasons. First, \(\mathfrak{sl}(2,\mathbb{C})\) is the complexification of \(\mathfrak{su}(2)\), which is in turn isomorphic to \(\mathfrak{so}(3)\). The representations of \(\mathfrak{so}(3)\) are the mathematics of angular momentum in quantum mechanics, and the calculation we are about to perform is the one found in physics texts under that heading, in the normalization where \(H\) plays the role of \(2J_z\). Second, the method is a template. It shows how the commutation relations, applied to a finite-dimensional irreducible representation, force its entire structure. Third, the representation theory of larger semisimple Lie algebras is built by locating copies of \(\mathfrak{sl}(2,\mathbb{C})\) inside them and applying precisely this result.

The Working Basis

We use the basis of \(\mathfrak{sl}(2,\mathbb{C})\) introduced when we formed the complexification of \(\mathfrak{su}(2)\), \[ H = \begin{pmatrix} 1 & 0 \\\\ 0 & -1 \end{pmatrix}, \quad E = \begin{pmatrix} 0 & 1 \\\\ 0 & 0 \end{pmatrix}, \quad F = \begin{pmatrix} 0 & 0 \\\\ 1 & 0 \end{pmatrix}, \] with the commutation relations \[ \begin{align*} [H, E] &= 2E, \\\\ [H, F] &= -2F, \\\\ [E, F] &= H. \end{align*} \]

These three relations are the entire content of the Lie algebra, and everything that follows is extracted from them. The element \(H\) is the one we will diagonalize, \(E\) will raise eigenvalues, and \(F\) will lower them.

A useful consequence is that representations of \(\mathfrak{sl}(2,\mathbb{C})\) are cheap to specify. Let \(V\) be a finite-dimensional complex vector space and let \(A, B, C\) be operators on \(V\) satisfying the same three relations \([A, B] = 2B\), \([A, C] = -2C\), and \([B, C] = A\). Consider the unique complex-linear map \(\pi : \mathfrak{sl}(2,\mathbb{C}) \to \mathfrak{gl}(V)\) determined by \[ \pi(H) = A, \quad \pi(E) = B, \quad \pi(F) = C. \] Both brackets are commutators, so both are skew-symmetric and complex-bilinear, the latter because composition of matrices, and of operators, is complex-linear in each factor. Both sides of the identity \(\pi([X, Y]) = [\pi(X), \pi(Y)]\) are therefore complex-bilinear in \((X, Y)\), and it suffices to verify it on pairs of basis elements. Skew-symmetry makes it read \(0 = 0\) on the diagonal pairs and reduces the rest to \((H, E)\), \((H, F)\), and \((E, F)\), where it says exactly the three relations we imposed on \(A, B, C\). So \(\pi\) is automatically a representation. Building a representation thus reduces to finding three operators with the right commutators.

The existence half is already in hand, since the \(\pi_m\) supply one irreducible representation in each dimension. What remains is the uniqueness, and it is the substantial part. We must show that an arbitrary irreducible representation, presented with no structure beyond the bracket relations, is forced into the shape of some \(\pi_m\). The strategy is to diagonalize \(\pi(H)\) and watch how \(E\) and \(F\) move its eigenvalues. The single lemma of the next section is the engine.

Raising and Lowering

When we built the polynomial representations, we saw the operators \(\pi_m(E)\) and \(\pi_m(F)\) shift the monomial basis up and down a ladder of \(\pi_m(H)\)-eigenvalues in steps of two. That was a feature of the explicit model. The following lemma shows it is not special to the model at all. In any representation, \(E\) and \(F\) move \(\pi(H)\)-eigenvalues by exactly \(+2\) and \(-2\). The fact is forced by the commutation relations alone, and it is the whole engine of the classification.

The mechanism is purely commutational. Conjugating \(\pi(E)\) past \(\pi(H)\) costs an extra \(+2\pi(E)\), and that surcharge is what bumps the eigenvalue up by two. Conjugating \(\pi(F)\) costs \(-2\pi(F)\) and bumps it down. So from a single \(\pi(H)\)-eigenvector we can manufacture a whole string of them, with eigenvalues marching in steps of two, until an application of \(E\) or \(F\) finally returns zero. Controlling where those strings start and stop is exactly what pins down the representation, and that is the next section.

Proof of the Classification

Let \(\pi\) be an irreducible complex representation of \(\mathfrak{sl}(2,\mathbb{C})\) acting on a finite-dimensional space \(V\). We prove it must take the explicit form of one of the \(\pi_m\). The argument has three movements: find a top of the ladder, descend it to discover that the eigenvalues are integers symmetric about zero, and check that the ladder spans an irreducible invariant subspace, which irreducibility of \(\pi\) then forces to be everything.

Finding the Top of the Ladder

Fix a basis of \(V\), so that operators on \(V\) become square complex matrices. Since \(V\) is nonzero, the characteristic polynomial of \(\pi(H)\) has positive degree, so it has a root by the fundamental theorem of algebra, and \(\pi(H)\) has at least one eigenvector \(u\), say with eigenvalue \(\alpha\). Applying the raising lemma repeatedly, \[ \pi(H)\,\pi(E)^k u = (\alpha + 2k)\,\pi(E)^k u, \] so each nonzero \(\pi(E)^k u\) is a \(\pi(H)\)-eigenvector with eigenvalue \(\alpha + 2k\). These eigenvalues \(\alpha, \alpha + 2, \alpha + 4, \dots\) are all distinct. But eigenvectors for distinct eigenvalues are linearly independent, a statement whose proof uses only field operations and so holds over \(\mathbb{C}\). A matrix of size \(\dim V\) therefore has at most \(\dim V\) eigenvalues, and the vectors \(\pi(E)^k u\) cannot all be nonzero. Let \(N\) be the largest index with \(\pi(E)^N u \neq 0\), so that \(\pi(E)^{N+1} u = 0\). Set \[ u_0 := \pi(E)^N u, \quad \lambda := \alpha + 2N. \]

By construction \(u_0\) is a nonzero \(\pi(H)\)-eigenvector at the top of its raising chain, \[ \pi(H)\,u_0 = \lambda\, u_0, \quad \pi(E)\,u_0 = 0. \] Raising it gives nothing. Such a vector deserves a name.

The vector \(u_0\) is a highest weight vector of weight \(\lambda\). What the bracket relations force on any such vector is the content of the next proposition. The three subsections that follow prove it, and they use only that \(V\) is finite-dimensional and that \(u_0\) is a highest weight vector, never the irreducibility of \(\pi\). Irreducibility enters only afterwards.

Descending the Ladder

Now lower repeatedly along the vectors \(u_k = \pi(F)^k u_0\) of the proposition. By the lowering half of the lemma, applied \(k\) times to \(u_0\), \[ \pi(H)\,u_k = (\lambda - 2k)\, u_k. \] The action of \(\pi(E)\) on this descending chain is governed by an identity we verify by induction: \[ \pi(E)\,u_k = k\,[\lambda - (k - 1)]\, u_{k-1} \quad (k \geq 1). \tag{1} \]

For \(k = 1\), use the bracket relation \([E, F] = H\) in the form \(\pi(E)\pi(F) = \pi(F)\pi(E) + \pi(H)\), applied to \(u_0\): \[ \begin{align*} \pi(E)\,u_1 = \pi(E)\pi(F)\,u_0 &= \pi(F)\,\pi(E)\,u_0 + \pi(H)\,u_0 \\\\ &= 0 + \lambda\, u_0 = 1 \cdot [\lambda - 0]\, u_0, \end{align*} \] using \(\pi(E) u_0 = 0\). This is \((1)\) at \(k = 1\).

Assume \((1)\) holds at \(k\). Then, again from \(\pi(E)\pi(F) = \pi(F)\pi(E) + \pi(H)\) applied to \(u_k\), \[ \begin{align*} \pi(E)\,u_{k+1} &= \pi(E)\pi(F)\,u_k = \pi(F)\,\pi(E)\,u_k + \pi(H)\,u_k \\\\ &= \pi(F)\bigl(k[\lambda - (k-1)]\,u_{k-1}\bigr) + (\lambda - 2k)\,u_k \\\\ &= k[\lambda - (k-1)]\,u_k + (\lambda - 2k)\,u_k \\\\ &= \bigl(k\lambda - k(k-1) + \lambda - 2k\bigr)\,u_k \\\\ &= (k + 1)\,[\lambda - k]\,u_k, \end{align*} \] where the last line factors \(k\lambda + \lambda - k^2 + k - 2k = (k+1)\lambda - k(k+1) = (k+1)(\lambda - k)\). This is \((1)\) at \(k + 1\), completing the induction.

The Eigenvalue Is a Non-Negative Integer

The descending eigenvalues \(\lambda, \lambda - 2, \lambda - 4, \dots\) are again distinct, so the \(u_k\) cannot all be nonzero. Once some \(u_k\) vanishes, so does every later \(u_{k'} = \pi(F)^{k'-k} u_k\). Let \(m\) be the largest index with \(u_m \neq 0\), so that \(u_0, \dots, u_m\) are all nonzero while \[ u_{m+1} = \pi(F)^{m+1} u_0 = 0, \] and \(u_k = 0\) for every \(k \gt m\).

Apply \(\pi(E)\) to this zero vector and use \((1)\) at \(k = m + 1\): \[ 0 = \pi(E)\,u_{m+1} = (m + 1)\,[\lambda - m]\, u_m. \] Since \(u_m \neq 0\) and \(m + 1 \neq 0\), we are forced to conclude \(\lambda - m = 0\), that is, \[ \lambda = m, \] a non-negative integer. The highest weight could not have been anything else. The bracket relations admit no representation with a non-integer or negative top eigenvalue. This is the crux of the classification, and it is Point 1 of the proposition. Which \(u_k\) vanish and which do not was settled above, so of Point 2 only the linear independence remains.

The Ladder Is Everything

With \(\lambda = m\), the vectors \(u_0, u_1, \dots, u_m\) are nonzero eigenvectors of \(\pi(H)\) with the distinct eigenvalues \(m, m - 2, \dots, -m\), and eigenvectors for distinct eigenvalues are linearly independent. So the \(u_k\) are independent, which completes Point 2, and their span \(W = \langle u_0, \dots, u_m \rangle\) has dimension \(m + 1\). Collecting the three actions gives exactly the formulas \((2)\). The first line is the eigenvalue computation above, the second is the definition of \(u_{k+1}\) together with \(u_{m+1} = 0\), and the third is \((1)\) with \(\lambda = m\) together with \(\pi(E)\,u_0 = 0\). By \((2)\), \(W\) is carried into itself by \(\pi(H)\), \(\pi(E)\), and \(\pi(F)\), hence by \(\pi(Z)\) for every \(Z \in \mathfrak{sl}(2,\mathbb{C})\), because \(\pi\) is linear and \(H, E, F\) span the algebra. So \(W\) is an invariant subspace, which is Point 3.

For Point 4, note first that \(\pi_W(X) := \pi(X)|_W\) is again a representation of \(\mathfrak{sl}(2,\mathbb{C})\), since the bracket identity holds on all of \(V\) and in particular on \(W\). Let \(U \subseteq W\) be a nonzero invariant subspace and pick \(w = \sum_{k=0}^{m} c_k u_k \in U\) with some \(c_j \neq 0\). The operator \[ P_j := \prod_{k \neq j} \bigl(\pi(H) - (m - 2k)\,I\bigr) \] is a polynomial in \(\pi(H)\), so it preserves \(U\). It kills every \(u_k\) with \(k \neq j\), and it multiplies \(u_j\) by \[ \prod_{k \neq j} \bigl((m - 2j) - (m - 2k)\bigr) = \prod_{k \neq j} 2(k - j) \neq 0. \] Hence \(P_j w\) is a nonzero multiple of \(u_j\), and \(u_j \in U\). Lowering then gives \(u_{j+1} = \pi(F)\,u_j, \dots, u_m \in U\), and raising gives \(u_{j-1}, \dots, u_0 \in U\), because \(\pi(E)\,u_k = k\,[m - (k-1)]\,u_{k-1}\) with coefficient \(k(m - k + 1) \neq 0\) for \(1 \leq k \leq m\). Thus \(U = W\), and \(\pi_W\) is irreducible. This proves the proposition.

Return now to the irreducible \(\pi\). The highest weight vector \(u_0\) constructed above generates the invariant subspace \(W\) of Point 3, which is nonzero. Because \(\pi\) is irreducible, \(W = V\). In particular \(\dim V = m + 1\), and \(\pi\) is given on the basis \(u_0, \dots, u_m\) by the formulas \((2)\).

The formulas \((2)\) determine \(\pi\) completely once \(m\) is fixed, and \(m\) is fixed by \(\dim V = m + 1\). So if \(\pi'\) is a second irreducible complex representation of dimension \(m + 1\), with the chain \(u_0', \dots, u_m'\) that the same argument produces, consider the linear map sending \(u_k\) to \(u_k'\) for each \(k\). This map carries the action of each of \(\pi(H)\), \(\pi(E)\), \(\pi(F)\) to that of \(\pi'(H)\), \(\pi'(E)\), \(\pi'(F)\), hence by linearity the action of \(\pi(Z)\) to that of \(\pi'(Z)\) for every \(Z\). Being invertible, it is an isomorphism of representations, so \(\pi\) and \(\pi'\) are isomorphic. The polynomial representation \(\pi_m\) is irreducible of dimension \(m + 1\), so the argument applies to it as well and exhibits it in the form \((2)\). Hence every irreducible complex representation of dimension \(m + 1\) is isomorphic to \(\pi_m\). This proves the classification.

General Representations

The classification described the irreducible representations. For applications we need a few facts about finite-dimensional representations that need not be irreducible. Those applications include locating copies of \(\mathfrak{sl}(2,\mathbb{C})\) inside larger algebras and reading the weight structure of a representation that has not been decomposed. The facts follow from the classification by decomposing into irreducibles, but several can be seen directly.