Simplicity of $\mathfrak{su}(3)$
This appendix proves what Lie Groups, Lie Algebras, and Fibre Bundles quotes in two places: that \(\mathfrak{su}(3)\) is a simple Lie algebra. The statement is used in Proposition 18.53, where the nondegeneracy of the Killing form is announced as the semisimplicity demanded by Theorem 18.13, and — more substantially — in Theorem 18.57, whose proof identifies the eight-dimensional summand of \(\vect{3}\otimes\bar{\vect{3}}\) with the adjoint representation and calls it irreducible because an invariant subspace would be an ideal of a simple Lie algebra. That last inference is what is discharged here.
The proof is the standard one and is short: complexify, split a putative ideal into eigenspaces of the two-dimensional Cartan subalgebra, and then walk around the root hexagon of Equation (18.80), using the fact that any two roots at \(120\) degrees add to a third. A single root vector inside the ideal therefore drags in all six, and with them the Cartan directions.
Two remarks on what is not used, because both matter for circularity. First, nothing below quotes the classification of simple Lie algebras; only Definition 18.51 and the multiplication table of the \(3\times3\) matrix units enter. Second, nothing below uses Proposition 18.53. That is deliberate: the chapter computes the Killing form of \(\mathfrak{su}(3)\) directly, precisely so as not to lean on the uniqueness of an invariant form on a simple algebra, and the present appendix repays the compliment by not leaning on the Killing form.
Statement
A Lie algebra \(\mathfrak{g}\) over a field is simple if it is not abelian and its only ideals are \(\set{0}\) and \(\mathfrak{g}\) itself; an ideal is a subspace \(\mathfrak{i}\) with \(\comm{\mathfrak{g}}{\mathfrak{i}} \subseteq \mathfrak{i}\). Rests on Definition 18.49.
The real Lie algebra \(\mathfrak{su}(3)\) of Definition 18.49 is simple, and so is its complexification \(\mathfrak{sl}(3,\C)\), the algebra of traceless complex \(3\times3\) matrices. Rests on Definition 18.49, Definition A35.1 and Proposition 18.50.
The adjoint representation of \(\mathfrak{su}(3)\) on itself, and equally the representation \(M \mapsto UMU^{\dagger}\) of \(\SU(3)\) on the traceless complex \(3\times3\) matrices, has no invariant subspace other than \(\set{0}\) and the whole space. Rests on Theorem A35.2 and Proposition 18.55.
Complexification
As a real vector space,
the sum being direct. If \(\mathfrak{i} \subseteq \mathfrak{su}(3)\) is an ideal of the real algebra, then \(\mathfrak{i}_{\C} = \mathfrak{i} + \ii\,\mathfrak{i}\) is a complex ideal of \(\mathfrak{sl}(3,\C)\) with \(\dim_{\C}\mathfrak{i}_{\C} = \dim_{\R}\mathfrak{i}\). Rests on Definition 18.49 and Proposition 18.50.
Derives Lemma A35.4. The splitting. Let \(X\) be traceless. Put \(X_{-} = \tfrac{1}{2}\left(X - X^{\dagger}\right)\) and \(X_{+} = \tfrac{1}{2}\left(X + X^{\dagger}\right)\), so that \(X = X_{-} + X_{+}\). Both are traceless, since \(\tr X^{\dagger} = \overline{\tr X} = 0\); \(X_{-}\) is anti-Hermitian, hence in \(\mathfrak{su}(3)\) by Definition 18.49; and \(X_{+}\) is Hermitian, so \(-\ii X_{+}\) is anti-Hermitian and \(X_{+} = \ii\left(-\ii X_{+}\right) \in \ii\,\mathfrak{su}(3)\). The sum is direct: if \(Y \in \mathfrak{su}(3) \cap \ii\,\mathfrak{su}(3)\), write \(Y = \ii Z\) with \(Z\) anti-Hermitian; then \(Y^{\dagger} = -\ii Z^{\dagger} = \ii Z = Y\), so \(Y\) is Hermitian as well as anti-Hermitian, and \(Y = 0\).
Ideals. The bracket of \(\mathfrak{sl}(3,\C)\) is \(\C\)-bilinear, so for \(X = A + \ii B\) with \(A,B \in \mathfrak{su}(3)\) and \(Y = P + \ii Q\) with \(P,Q \in \mathfrak{i}\),
and all four brackets lie in \(\mathfrak{i}\) because \(\mathfrak{i}\) is an ideal of \(\mathfrak{su}(3)\). So \(\comm{X}{Y} \in \mathfrak{i}_{\C}\), and \(\mathfrak{i}_{\C}\) is closed under multiplication by \(\ii\) hence a complex subspace. Finally \(\mathfrak{i} \cap \ii\,\mathfrak{i} = 0\), being contained in \(\mathfrak{su}(3) \cap \ii\,\mathfrak{su}(3) = 0\), so \(\mathfrak{i}_{\C} = \mathfrak{i} \oplus \ii\,\mathfrak{i}\) has real dimension \(2\dim_{\R}\mathfrak{i}\) and complex dimension \(\dim_{\R}\mathfrak{i}\).
∎The root decomposition
Write \(E_{jk}\) for the matrix unit with a single \(1\) in row \(j\) and column \(k\), so that \(E_{jk}E_{lm} = \delta_{kl}E_{jm}\). Let
the traceless diagonal matrices, which is two-dimensional and is the complex span of the \(T_{3}\) and \(T_{8}\) of Definition 18.51; it is the Cartan subalgebra whose existence Proposition 18.50 established, of rank \(\ell = 2\). For \(H = \diag(a_{1},a_{2},a_{3})\) a one-line computation with the matrix units gives
so that \(\mathfrak{sl}(3,\C)\) decomposes as
a direct sum of \(2 + 6 = 8\) pieces, each an eigenspace of every \(\ad_{H}\), and each of the six off-diagonal pieces one-dimensional. The linear functional \(\alpha_{jk} \in \mathfrak{h}^{*}\), \(\alpha_{jk}(H) = a_{j}-a_{k}\), is the root carried by \(E_{jk}\).
Evaluating \(\alpha_{jk}\) on the two Cartan generators of Definition 18.51 gives the coordinates used in Remark 18.56. With \(T_{3} = \tfrac{1}{2}\diag(1,-1,0)\) and \(T_{8} = \tfrac{1}{2\sqrt3}\diag(1,1,-2)\),
and \(E_{21}, E_{31}, E_{32}\) carry the negatives. These are exactly the six vectors of Equation (18.80): unit vectors at \(60\) degrees to one another, the regular hexagon. What the proof below uses is the one geometric property that hexagon has, namely that the sum of two roots at \(120\) degrees is again a root — \(|\gamma|=|\delta|=1\) and \(\gamma\cdot\delta = -\tfrac{1}{2}\) give \(\abs{\gamma+\delta}^{2} = 1\), and the sum bisects the angle. In matrix units that property reads \(\comm{E_{jk}}{E_{kl}} = E_{jl}\) for \(j,k,l\) distinct, which is the form in which it will be applied.
Let \(\mathfrak{i}\) be an ideal of \(\mathfrak{sl}(3,\C)\) and let \(x = h + \sum_{j\neq k}c_{jk}E_{jk} \in \mathfrak{i}\) be the decomposition of an element according to Equation (A35.4). Then \(h \in \mathfrak{i}\) and \(c_{jk}E_{jk} \in \mathfrak{i}\) for every pair; in particular each \(E_{jk}\) with \(c_{jk} \neq 0\) lies in \(\mathfrak{i}\). Rests on Equation (A35.4) and Definition A35.1.
Derives Lemma A35.6. Take the particular Cartan element \(H_{0} = \diag(3,-1,-2) \in \mathfrak{h}\). Its six root values are
together with \(-4,-5,-1\): six distinct and nonzero numbers, which is the only property of \(H_{0}\) that is used. Enumerate the six ordered pairs as \(\mu = 1,\ldots,6\), write \(E_{\mu}\) for the corresponding matrix unit and \(\lambda_{\mu} \in \set{\pm1,\pm4,\pm5}\) for its root value, so that \(x = h + \sum_{\mu}c_{\mu}E_{\mu}\). Because \(\mathfrak{i}\) is an ideal and \(\comm{H_{0}}{h} = 0\), all six elements
lie in \(\mathfrak{i}\), by Equation (A35.3). The \(6\times6\) coefficient matrix \(\left(\lambda_{\mu}^{n}\right)\) has determinant \(\left(\prod_{\mu}\lambda_{\mu}\right)\) times the Vandermonde determinant \(\prod_{\mu<\nu}(\lambda_{\nu}-\lambda_{\mu})\), which is nonzero because the \(\lambda_{\mu}\) are nonzero and pairwise distinct. Inverting it expresses each \(c_{\mu}E_{\mu}\) as a linear combination of the \(\ad_{H_{0}}^{n}x\), so \(c_{\mu}E_{\mu} \in \mathfrak{i}\); and then \(h = x - \sum_{\mu}c_{\mu}E_{\mu} \in \mathfrak{i}\). Since \(\C E_{\mu}\) is one-dimensional, \(c_{\mu} \neq 0\) forces \(E_{\mu} \in \mathfrak{i}\).
∎Walking round the hexagon
Let \(\mathfrak{i}\) be an ideal of \(\mathfrak{sl}(3,\C)\) containing \(E_{jk}\) for one pair \(j \neq k\). Then \(\mathfrak{i} = \mathfrak{sl}(3,\C)\). Rests on Equations (A35.3) and (A35.4).
Derives Lemma A35.7. Let \(l\) be the remaining index, so that \(\set{j,k,l} = \set{1,2,3}\). Every bracket below is a bracket of an element of \(\mathfrak{sl}(3,\C)\) with an element of \(\mathfrak{i}\), and therefore lies in \(\mathfrak{i}\).
The opposite root vector. From \(E_{kj}E_{jk} = E_{kk}\) and \(E_{jk}E_{kj} = E_{jj}\),
and \(H_{jk}\) is the diagonal matrix with entries \(+1\) at \(j\), \(-1\) at \(k\) and \(0\) at \(l\), so Equation (A35.3) gives \(\comm{H_{jk}}{E_{kj}} = (-1-1)E_{kj} = -2E_{kj}\). Hence \(E_{kj} \in \mathfrak{i}\). In the language of Remark A35.5: the root \(\gamma\) and the coroot direction it generates carry the ideal to \(-\gamma\), the opposite vertex of the hexagon.
The two neighbouring roots. Because \(k \neq l\) and \(l \neq j\),
the second term of each commutator vanishing — \(E_{jk}E_{lj} = 0\) since \(k \neq l\), and \(E_{kl}E_{jk} = 0\) since \(l \neq j\). These are the two roots at \(60\) degrees from the starting one, reached by adding the roots of \(E_{lj}\) and of \(E_{kl}\), each at \(120\) degrees to it.
Closing. Apply the first step to \(E_{lk}\) and to \(E_{jl}\): it returns \(E_{kl}, H_{lk} \in \mathfrak{i}\) and \(E_{lj}, H_{jl} \in \mathfrak{i}\). All six matrix units are now in \(\mathfrak{i}\), which is the whole hexagon. The diagonal part follows from Equation (A35.7): \(H_{jk} = E_{jj}-E_{kk}\) and \(H_{lk} = E_{ll}-E_{kk}\) are linearly independent elements of the two-dimensional \(\mathfrak{h}\) of Equation (A35.2), so they span it. By Equation (A35.4), \(\mathfrak{i} = \mathfrak{sl}(3,\C)\).
∎Proof of the theorem
Proof of Theorem A35.2. Derives Theorem A35.2. The complex algebra. \(\mathfrak{sl}(3,\C)\) is not abelian, since \(\comm{E_{12}}{E_{21}} = E_{11}-E_{22} \neq 0\). Let \(\mathfrak{i} \neq \set{0}\) be an ideal and pick \(x \in \mathfrak{i}\) nonzero. By Lemma A35.6 either some \(E_{jk}\) lies in \(\mathfrak{i}\) — in which case \(\mathfrak{i} = \mathfrak{sl}(3,\C)\) by Lemma A35.7 — or every \(c_{jk}\) vanishes for every element of \(\mathfrak{i}\), so that \(\mathfrak{i} \subseteq \mathfrak{h}\) and \(x = h \neq 0\) is a nonzero traceless diagonal matrix. In that second case Equation (A35.3) gives \(\comm{h}{E_{jk}} = \alpha_{jk}(h)\,E_{jk} \in \mathfrak{i}\); the right-hand side lies in \(\C E_{jk}\), which meets \(\mathfrak{h}\) only in \(0\), so \(\alpha_{jk}(h) = 0\) for all six pairs. Writing \(h = \diag(a_{1},a_{2},a_{3})\) this reads \(a_{1} = a_{2} = a_{3}\), which with \(\sum a_{i} = 0\) gives \(h = 0\) — contradicting the choice of \(x\). (Equivalently: the roots Equation (A35.5) span \(\mathfrak{h}^{*}\), since \((1,0)\) and \(\left(-\tfrac{1}{2},\tfrac{\sqrt3}{2}\right)\) are independent, so a Cartan element annihilated by all of them is zero.) The second case is therefore empty, and \(\mathfrak{sl}(3,\C)\) is simple.
The real algebra. \(\mathfrak{su}(3)\) is not abelian: the generators \(T_{1},T_{2},T_{3}\) of Definition 18.51 satisfy \(\comm{T_{1}}{T_{2}} = \ii T_{3} \neq 0\) by Equation (18.69), and \(\ii T_{a}\) is the anti-Hermitian element of \(\mathfrak{su}(3)\) attached to the Hermitian generator \(T_{a}\). Let \(\mathfrak{i} \neq \set{0}\) be an ideal of \(\mathfrak{su}(3)\). By Lemma A35.4, \(\mathfrak{i}_{\C}\) is a nonzero complex ideal of \(\mathfrak{sl}(3,\C)\), hence all of it by the previous paragraph, so
the last equality being Proposition 18.50. A subspace of full dimension is the whole space, so \(\mathfrak{i} = \mathfrak{su}(3)\).
∎Proof of Corollary A35.3. Derives Corollary A35.3. An invariant subspace \(V\) of the adjoint representation of a Lie algebra \(\mathfrak{g}\) on itself satisfies \(\comm{\mathfrak{g}}{V} \subseteq V\) by definition, which is exactly the statement that \(V\) is an ideal (Definition A35.1); so Theorem A35.2 leaves only \(\set{0}\) and \(\mathfrak{su}(3)\).
For the group statement, note first that the space of traceless complex \(3\times3\) matrices is \(\mathfrak{sl}(3,\C)\) and that \(M \mapsto UMU^{\dagger} = UMU^{-1}\) is the adjoint action of \(\SU(3)\) on it, as Proposition 18.55 and the proof of Theorem 18.57 record. A subspace \(V\) invariant under the group is invariant under the algebra: for \(X \in \mathfrak{su}(3)\) and \(M \in V\), the curve \(t \mapsto \ee^{tX}M\ee^{-tX}\) lies in \(V\), which is a closed subspace of a finite-dimensional space, and its derivative at \(t=0\) is \(\comm{X}{M}\), so \(\comm{X}{M} \in V\). Complex-linearity extends this from \(\mathfrak{su}(3)\) to \(\mathfrak{sl}(3,\C)\) by Equation (A35.1), and the first paragraph applies.
∎Theorem A35.2 discharges the two quotations named at the head of this section. In Proposition 18.53 it justifies the phrase “which is simple and hence semisimple”: the Killing form \(\kappa_{ab} = 3\delta_{ab}\) computed there is nondegenerate, as Theorem 18.13 requires of a semisimple algebra, and the implication now runs in the direction the chapter states rather than resting on an unproved assertion. In Theorem 18.57 it supplies the irreducibility of the eight-dimensional summand of \(\vect{3}\otimes\bar{\vect{3}}\), so that the decomposition \(3\times3 = 1+8\) of Equation (18.81) is proved and not merely counted; Corollary A35.3 is the form in which it is used there.
One limitation is worth stating plainly. What is proved is the simplicity of this one algebra, by an argument that inspects its root system directly. The general theorem — that a complex simple Lie algebra is determined by its root system, and that the possible root systems are the classified Dynkin diagrams — is not proved here and is not needed: the whole of Lie Groups, Lie Algebras, and Fibre Bundles uses \(\mathfrak{su}(2)\) and \(\mathfrak{su}(3)\) and nothing else, and both are handled by hand.