Lie Groups, Lie Algebras, and Fibre Bundles

Derives Proposition 14.3. Fix any submultiplicative norm on the operators of \(\mathbb{V}\), so that \(\norm{J^{n}} \le \norm{J}^{n}\). Then \(\sum_{n}\abs{t}^{n}\norm{J}^{n}/n! = \ee^{\abs{t}\norm{J}} < \infty\), so the series Equation (14.8) converges absolutely in the finite-dimensional — hence complete — space of operators, term by term differentiation is legitimate, and \(\dv{}{t}\ee^{tJ} = J\ee^{tJ} = \ee^{tJ}J\).

Differentiate Equation (14.7) with respect to \(s\) at \(s = 0\), holding \(t\) fixed. The left-hand side gives \(\dv{g}{t}(t)\) and the right-hand side gives \(g(t)\left.\dv{g}{s}\right|_{s=0} = g(t)J\), so

Put \(h(t) = g(t)\ee^{-tJ}\). Since \(J\) commutes with \(\ee^{-tJ}\),

so \(h\) is constant and equal to \(h(0) = \identity\). Multiplying on the right by \(\ee^{tJ}\) gives Equation (14.8). Finally \(g(t)g(-t) = g(0) = \identity\) by Equation (14.7).

Derives Lemma 14.11. The adjoint map is itself a representation, \(\ad_{\comm{X}{Y}} = \comm{\ad_{X}}{\ad_{Y}}\), which is the Jacobi identity rewritten. Hence

Adding the two lines, the terms \(\mp\tr(\ad_{Y}\ad_{X}\ad_{Z})\) cancel against each other, and the survivors \(\tr(\ad_{X}\ad_{Y}\ad_{Z})\) and \(-\tr(\ad_{Y}\ad_{Z}\ad_{X})\) cancel by cyclicity of the trace.

Derives Corollary 14.12. Antisymmetry in the first two indices is the antisymmetry of the bracket. For the last two, put \(X = T_{a}\), \(Y = T_{b}\), \(Z = T_{c}\) in Equation (14.16) and use the symmetry of \(\kappa\):

The transpositions \((ab)\) and \((bc)\) generate the whole symmetric group on three letters, so \(C_{abc}\) changes sign under every transposition.

Derives Theorem 14.16. In the associative algebra \(U(\mathfrak{g})\) the commutator with a fixed element obeys the Leibniz rule, \(\comm{XY}{Z} = X\comm{Y}{Z} + \comm{X}{Z}Y\), so

In the \(l\)-th term rename the two summation indices \(a_{l}\) and \(b\) into each other. This is a relabelling of dummy indices and moves no operator past another, so the ordered string becomes literally \(T_{a_{1}}T_{a_{2}}\cdots T_{a_{r}}\), the same string for every value of \(l\), while the coefficient becomes \(C^{a_{l}}{}_{bc}\,k^{a_{1}\cdots a_{l-1}\,b\,a_{l+1}\cdots a_{r}}\). The sum over \(l\) may therefore be taken inside:

and the parenthesis vanishes by Equation (14.18).

Derives Corollary 14.17. The statement is Theorem 14.16 for \(r=2\) and \(k^{ab} = \kappa^{ab}\), whose invariance is Corollary 14.12 with both free indices raised. It is worth seeing the two-index case done by hand, because it is the whole mechanism in three lines. By the Leibniz rule,

Fix \(c\) and set

using Equation (14.17). Exchanging the two free indices and then renaming the dummies \(b \leftrightarrow e\),

the last step by Corollary 14.12, since \(C_{ecb}\) differs from \(C_{bce}\) by the transposition of the first and third indices. The first term above is \(M^{ad}T_{a}T_{d}\); the second is \(\kappa^{ab}C^{d}{}_{ac}T_{d}T_{b} = M^{bd}T_{d}T_{b}\) by the symmetry of \(\kappa\), which is \(M^{ad}T_{d}T_{a}\) after renaming \(b \rightarrow a\). Hence

an antisymmetric array contracted with a symmetric one.

Derives Corollary 14.18. By Remark 14.6 the representation \(D\) extends to \(U(\mathfrak{g})\), so \(D(C)\) is a well-defined operator on \(\mathbb{V}\) and \(\comm{D(C)}{D(T_{a})} = D\left(\comm{C}{T_{a}}\right) = 0\) for every \(a\) by Equation (14.13). An operator commuting with \(D(T_{a})\) commutes with every power of \(D(T_{a})\), hence with the exponential \(\ee^{\theta^{a}D(T_{a})} = U\left(g(\theta^{a})\right)\) of Equation (14.6), that is, with \(U(g)\) for every \(g\) in a neighbourhood of the identity. A connected topological group is generated by any neighbourhood of its identity, so \(D(C)\) commutes with \(U(g)\) for every \(g \in G\). Schur's first lemma (Theorem 5.153), applied to the irreducible representation \(U\), gives \(D(C) = c\,\identity\).

Derives Proposition 14.22. Write \(r = \lfloor D/2 \rfloor\), work in the defining representation on \(\C^{D}\), and let \(\eta\) be diagonal as in Notation 14.1, so that \(X \in \mathfrak{so}(p,q)\) means

Diagonalizability in the defining representation is what is used throughout; that it agrees with the diagonalizability in the adjoint representation asked for by Definition 14.20 is standard structure theory, quoted here as Theorem 14.13 is. One direction is immediate — in a basis diagonalizing \(X\) the map \(Y \mapsto \comm{X}{Y}\) is diagonal on the matrix units, with eigenvalue the difference of two eigenvalues of \(X\) — and the other is that the semisimple and nilpotent parts of an element of a semisimple linear Lie algebra again lie in it, so that an element with diagonalizable adjoint action has central, hence zero, nilpotent part.

A commuting set of size \(r\). Take

one generator for each of the \(r\) disjoint index pairs \(\set{1,2}\), \(\set{3,4}\), and so on. No two of them share an index, so every \(\eta\) on the right of Equation (14.90) carries two distinct indices and vanishes, \(\eta\) being diagonal; hence \(\comm{H_{k}}{H_{l}} = 0\). In the defining representation \(J_{AB}\) is the matrix \(\left(J_{AB}\right)^{C}{}_{D} = \delta^{C}_{A}\eta_{BD} - \delta^{C}_{B}\eta_{AD}\), the identification already made for \(\mathfrak{so}(3)\) at Equation (14.32): it acts inside the plane of its own two coordinates and annihilates the others, and on that plane it is either the rotation generator Equation (14.27), with eigenvalues \(\pm\ii\), or a boost generator, with eigenvalues \(\pm1\); in both cases it is diagonalizable, and so is any linear combination of the \(H_{k}\), the blocks being independent. The span of the \(H_{k}\) is therefore an abelian subalgebra of diagonalizable elements, of dimension \(r\).

No larger one exists. Let \(\mathfrak{h}\) be any abelian subalgebra of \(\mathfrak{so}(p,q)\) all of whose elements are diagonalizable, and let \(\mathfrak{h}_{\C}\) be its complex span inside the complex \(D\times D\) matrices. Since the elements of \(\mathfrak{h}\) are real matrices, a real basis of \(\mathfrak{h}\) stays independent over \(\C\) — the real and imaginary parts of a vanishing complex combination vanish separately — so \(\dim_{\C}\mathfrak{h}_{\C} = \dim_{\R}\mathfrak{h}\). Commuting diagonalizable operators are simultaneously diagonalizable, and every complex combination of operators diagonal in one basis is diagonal in that basis, so there are a basis \(\set{v_{1},\ldots,v_{D}}\) of \(\C^{D}\) and complex-linear functionals \(\lambda_{i}\) on \(\mathfrak{h}_{\C}\) with \(Xv_{i} = \lambda_{i}(X)v_{i}\). Condition Equation (14.23) is complex-bilinear and therefore holds on \(\mathfrak{h}_{\C}\); applied to \(u = v_{i}\), \(v = v_{j}\) it gives

Collect the basis vectors by their functional: for each \(\alpha\) occurring among the \(\lambda_{i}\), let \(V_{\alpha}\) be the span of those \(v_{i}\) with \(\lambda_{i} = \alpha\). The display says \(\eta(V_{\alpha},V_{\beta}) = 0\) unless \(\beta = -\alpha\), so nondegeneracy of \(\eta\) forces \(-\alpha\) to occur whenever \(\alpha\) does, and forces the pairing of \(V_{\alpha}\) with \(V_{-\alpha}\) to be nondegenerate, whence \(\dim V_{\alpha} = \dim V_{-\alpha}\). The occurring functionals thus consist of \(k\) pairs \(\pm\alpha_{1},\ldots,\pm\alpha_{k}\) with \(\alpha_{i} \neq 0\), possibly together with \(0\); each pair accounts for at least two basis vectors, so \(2k \le D\) and \(k \le r\). Finally the occurring functionals span the dual of \(\mathfrak{h}_{\C}\), since an \(X\) annihilated by all of them acts as zero on every \(v_{i}\) and is therefore zero, and they span the same space as \(\alpha_{1},\ldots,\alpha_{k}\). Hence \(\dim_{\R}\mathfrak{h} = \dim_{\C}\mathfrak{h}_{\C} \le k \le r\).

The subalgebra Equation (14.24) is therefore maximal, and every maximal one has the same dimension \(r\).

Derives Proposition 14.27. Write \(R = \identity + \varepsilon X + O(\varepsilon^{2})\) in Equation (14.30). To first order \(R\transpose R = \identity\) gives \(X\transpose + X = 0\), and \(\det R = 1\) gives \(\tr X = 0\), which is implied by antisymmetry; conversely \(\exp\) of an antisymmetric matrix is orthogonal with unit determinant. The real antisymmetric \(3\times3\) matrices form a three-dimensional space, matching Equation (14.31), and the \(L_{k}\) of Equation (14.32) are three independent members of it. Note that \(L_{3}\) is the generator \(J\) of \(\SO(2,\R)\) found in Equation (14.27), embedded in the \(1\)–\(2\) block.

For the bracket, compute the \((m,n)\) entry using \(\left(L_{i}\right)_{mk} = -\epsilon_{imk}\):

where the contraction \(\epsilon_{kim}\epsilon_{knj} = \delta_{in}\delta_{mj} - \delta_{ij}\delta_{mn}\) was used twice after a cyclic relabelling. On the other hand

which is the same expression.

Derives Proposition 14.30. From Equation (14.32), \(\left(N\vect{x}\right)_{i} = -\epsilon_{kij}n_{k}x_{j} = \epsilon_{ikj}n_{k}x_{j} = \left(\hat{n}\times\vect{x}\right)_{i}\). Hence \(N^{2}\vect{x} = \hat{n}\times(\hat{n}\times\vect{x}) = \hat{n}(\hat{n}\cdot\vect{x}) - \vect{x}\) and \(N^{3}\vect{x} = \hat{n}\times\left(\hat{n}(\hat{n}\cdot\vect{x}) - \vect{x}\right) = -\hat{n}\times\vect{x} = -N\vect{x}\). Consequently \(N^{2k+1} = (-1)^{k}N\) and \(N^{2k+2} = (-1)^{k}N^{2}\) for \(k \in \N\cup\set{0}\), and splitting the exponential series into its odd and even parts, exactly as was done for \(\SO(2,\R)\) at Equation (14.28),

whose two coefficients are \(\sin\phi\) and \(1 - \cos\phi\). Substituting \(N^{2}\vect{x}\) gives Equation (14.37).

Surjectivity is Euler's theorem on rotations: \(R \in \SO(3,\R)\) is a real \(3\times3\) matrix, so its characteristic polynomial is a real cubic and has a real root; the roots have modulus \(1\) because \(R\) is orthogonal, and their product is \(\det R = 1\), so the real root is \(+1\) and there is a unit vector \(\hat{n}\) with \(R\hat{n} = \hat{n}\). In an orthonormal basis adapted to \(\hat{n}\), \(R\) acts as the identity on \(\hat{n}\) and as an element of \(\SO(2,\R)\) on the orthogonal plane, which is Equation (14.25) for some angle \(\phi\); replacing \(\hat{n}\) by \(-\hat{n}\) if necessary brings \(\phi\) into \([0,\pi]\).

Derivation. Derives Proposition 14.31. That the map is onto is Proposition 14.30, and it is injective on the open ball because \(R(\hat{n},\phi)\) determines its fixed axis and its rotation angle; on the boundary sphere \(\phi = \pi\) the only failure of injectivity is \(R(\hat{n},\pi) = R(-\hat{n},\pi)\), immediate from Equation (14.37) since \(\cos\pi = -1\) and \(\sin\pi = 0\) leave only terms even in \(\hat{n}\). A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so \(\SO(3,\R) \cong \R P^{3}\); connectedness and compactness follow. The statement about loops is Corollary 14.38 below, where it is read off the double cover.

Derives Proposition 14.33. Write

With \(\det U = 1\) the inverse is \(U^{-1} = \begin{pmatrix} d & -b\\ -c & a\end{pmatrix}\), while unitarity says \(U^{-1} = U^{\dagger} = \begin{pmatrix} \bar{a} & \bar{c}\\ \bar{b} & \bar{d}\end{pmatrix}\). Comparing entries, \(d = \bar{a}\) and \(c = -\bar{b}\), which is Equation (14.39) with \(\alpha = a\), \(\beta = b\); the determinant condition then reads \(\abs{\alpha}^{2} + \abs{\beta}^{2} = 1\). Conversely every such matrix is unitary with unit determinant. The correspondence and its inverse are given by polynomials in the real and imaginary parts, so both are continuous. Finally \(S^{3}\) is connected and simply connected by Lemma 14.34.

Derives Lemma 14.34. For \(p, q \in S^{n}\) with \(q \neq -p\) define the normalized chord

It is well defined, because \(\norm{(1-u)p+uq}^{2} = (1-u)^{2} + u^{2} + 2u(1-u)\,p\cdot q\) can vanish only if \(u = 1/2\) and \(p\cdot q = -1\), that is only if \(q = -p\); it is jointly continuous in \(p\), \(q\) and \(u\); it runs from \(p\) to \(q\); and its image lies in the plane spanned by \(p\) and \(q\). Path-connectedness follows at once: two points are joined by Equation (14.40) unless they are antipodal, and an antipodal pair is joined through any third point.

Every loop is homotopic to a piecewise-circular one. Let \(\gamma : [0,1] \longrightarrow S^{n}\) be continuous with \(\gamma(0) = \gamma(1)\). Each of the \(n+1\) components of \(\gamma\) is uniformly continuous by Theorem 7.25, whose proof applies verbatim to a map into \(\R^{n+1}\), so there is an \(N\) with \(\norm{\gamma(s) - \gamma(s')} < 1\) whenever \(\abs{s-s'} \le 1/N\). Put \(p_{k} = \gamma(k/N)\). Consecutive \(p_{k}\) are not antipodal, their distance being less than \(1 < 2\), so

is a well-defined continuous loop with \(\tilde{\gamma}(0) = p_{0} = \gamma(0)\) and \(\tilde{\gamma}(1) = p_{N} = \gamma(1)\). For \(s\) in the \(k\)-th subinterval both \(\norm{\gamma(s)-p_{k}} < 1\) and \(\norm{\gamma(s)-p_{k+1}} < 1\), and for unit vectors \(\norm{a-b}^{2} = 2 - 2\,a\cdot b\), so \(\gamma(s)\cdot p_{k} > 1/2\) and \(\gamma(s)\cdot p_{k+1} > 1/2\). Since \(\tilde{\gamma}(s)\) is a combination of \(p_{k}\) and \(p_{k+1}\) with nonnegative coefficients divided by a norm at most \(1\),

Hence \((1-t)\gamma(s) + t\tilde{\gamma}(s)\) never vanishes, its squared norm being at least \((1-t)^{2}+t^{2}\), and

is a homotopy from \(\gamma\) to \(\tilde{\gamma}\) fixing the base point.

A piecewise-circular loop misses a point. The image of \(\tilde{\gamma}\) lies in the union of the \(N\) sets \(C_{k} = S^{n}\cap\Pi_{k}\), where \(\Pi_{k}\) is a plane of dimension at most two containing \(p_{k}\) and \(p_{k+1}\); enlarging it if necessary, take \(\dim\Pi_{k} = 2\). Because \(n+1 \ge 3\), the planes \(\Pi' = \operatorname{span}\set{e_{1},\cos\alpha\,e_{2} + \sin\alpha\,e_{3}}\) are pairwise distinct for \(\alpha \in [0,\pi)\) and infinite in number, so some \(\Pi'\) differs from all \(N\) of the \(\Pi_{k}\). Two distinct two-dimensional planes meet in a subspace of dimension at most one, so \(S^{n}\cap\Pi'\cap\Pi_{k}\) has at most two points for each \(k\); the circle \(S^{n}\cap\Pi'\) is infinite, and therefore contains a point \(x\) lying on no \(C_{k}\).

The complement of a point is \(\R^{n}\). Stereographic projection from \(x\),

maps \(S^{n}\setminus\set{x}\) into the \(n\)-dimensional space \(x^{\perp} \cong \R^{n}\) and is continuous there. So is \(w \mapsto \left(2w + (\norm{w}^{2}-1)x\right)/(\norm{w}^{2}+1)\), which lands on \(S^{n}\) — its squared norm is \(\left(4\norm{w}^{2}+(\norm{w}^{2}-1)^{2}\right)/(\norm{w}^{2}+1)^{2} = 1\) — and a direct substitution shows the two maps are mutually inverse. Hence \(S^{n}\setminus\set{x}\) is homeomorphic to \(\R^{n}\), which is simply connected by Example 6.18. The loop \(\tilde{\gamma}\) therefore contracts to its base point inside \(S^{n}\setminus\set{x}\), hence inside \(S^{n}\); composing that homotopy with \(H\) contracts \(\gamma\).

Derives Proposition 14.35. Setting \(U = \identity + \varepsilon X + O(\varepsilon^{2})\) in Equation (14.38) gives \(X^{\dagger} + X = 0\) and \(\tr X = 0\). A traceless anti-Hermitian \(2\times2\) matrix is \(\begin{pmatrix} \ii u & v + \ii w\\ -v + \ii w & -\ii u\end{pmatrix}\) with \(u,v,w\) real, a three-dimensional real space, and the three matrices \(-\ii\sigma_{k}/2\) span it. Equation (14.43) is checked directly: \(\sigma_{i}^{2} = \identity\) for each \(i\) by inspection, and the three products \(\sigma_{1}\sigma_{2} = \ii\sigma_{3}\), \(\sigma_{2}\sigma_{3} = \ii\sigma_{1}\), \(\sigma_{3}\sigma_{1} = \ii\sigma_{2}\) are likewise immediate, the remaining cases following from \(\sigma_{j}\sigma_{i} = 2\delta_{ij}\identity - \sigma_{i}\sigma_{j}\). Antisymmetrizing Equation (14.43) gives \(\comm{\sigma_{i}}{\sigma_{j}} = 2\ii\epsilon_{ijk}\sigma_{k}\), that is Equation (14.44); taking the trace of Equation (14.43) and using \(\tr\sigma_{k} = 0\) gives \(\tr(\sigma_{i}\sigma_{j}) = 2\delta_{ij}\).

Derives Proposition 14.36. By Equation (14.43), \(\left(\hat{n}\cdot\vect{\sigma}\right)^{2} = n_{i}n_{j}\sigma_{i}\sigma_{j} = n_{i}n_{j}\delta_{ij}\identity = \identity\), the \(\epsilon\) term dropping because \(n_{i}n_{j}\) is symmetric. The even and odd powers of \(\hat{n}\cdot\vect{\sigma}\) therefore alternate exactly as \(J\) did in Equation (14.27), and the same resummation that produced Equation (14.28) produces Equation (14.45). Setting \(\psi = 2\pi\) gives \(\cos\pi = -1\), \(\sin\pi = 0\).

Derives Theorem 14.37. We check the four claims in turn.

\(\Phi(U)\) is a well-defined real linear map on \(\R^{3}\). The matrices \(\vect{x}\cdot\vect{\sigma}\) are exactly the traceless Hermitian \(2\times2\) matrices, since \(\set{\sigma_{1},\sigma_{2},\sigma_{3}}\) is a real basis of that space. If \(X\) is traceless Hermitian then so is \(UXU^{\dagger}\): Hermiticity because \(\left(UXU^{\dagger}\right)^{\dagger} = UX^{\dagger}U^{\dagger} = UXU^{\dagger}\), and tracelessness by cyclicity together with \(U^{\dagger}U = \identity\). So \(\Phi(U)\) maps \(\R^{3}\) to \(\R^{3}\) linearly.

\(\Phi(U)\) preserves the Euclidean norm. From Equation (14.43), \(\det\left(\vect{x}\cdot\vect{\sigma}\right) = -\vect{x}\cdot\vect{x}\), as one sees by writing out \(\begin{pmatrix} x_{3} & x_{1}-\ii x_{2}\\ x_{1}+\ii x_{2} & -x_{3}\end{pmatrix}\). Since \(\det U = 1\), \(\det\left(UXU^{\dagger}\right) = \det X\), so the norm of \(\vect{x}\) is unchanged and \(\Phi(U) \in \Ogrp(3,\R)\).

The image lies in \(\SO(3,\R)\), and \(\Phi\) is a homomorphism. \(\Phi(U_{1}U_{2}) = \Phi(U_{1})\Phi(U_{2})\) is immediate from Equation (14.46). The map \(U \mapsto \det\Phi(U) \in \set{+1,-1}\) is continuous on the connected space \(\SU(2)\) (Proposition 14.33) and equals \(+1\) at \(U = \identity\), hence equals \(+1\) everywhere.

The kernel. Suppose \(UXU^{\dagger} = X\) for every traceless Hermitian \(X\), i.e. \(U\) commutes with \(\sigma_{1},\sigma_{2},\sigma_{3}\). Writing \(U = \begin{pmatrix} a & b\\ c & d\end{pmatrix}\), commuting with \(\sigma_{3}\) forces \(b = c = 0\), and commuting with \(\sigma_{1}\) then forces \(a = d\). So \(U = a\identity\), and \(\det U = a^{2} = 1\) gives \(a = \pm1\). Both values do lie in the kernel, giving Equation (14.47).

Surjectivity, and the explicit form. It suffices to compute \(\Phi\left(U(\hat{n},\phi)\right)\) and find the rotation Equation (14.37), since those exhaust \(\SO(3,\R)\) by Proposition 14.30. Abbreviate \(\mathrm{c} = \cos(\phi/2)\), \(\mathrm{s} = \sin(\phi/2)\) and \(n = \hat{n}\cdot\vect{\sigma}\), \(X = \vect{x}\cdot\vect{\sigma}\). Using Equation (14.45),

Now Equation (14.43) gives \(\comm{X}{n} = x_{i}n_{j}\comm{\sigma_{i}}{\sigma_{j}} = 2\ii\,\epsilon_{ijk}x_{i}n_{j}\sigma_{k} = -2\ii\left(\hat{n}\times\vect{x}\right)\cdot\vect{\sigma}\), and, using \(nX + Xn = 2\left(\hat{n}\cdot\vect{x}\right)\identity\) from the symmetric part of the same identity together with \(n^{2} = \identity\),

Collecting the three pieces and using \(\mathrm{c}^{2}-\mathrm{s}^{2} = \cos\phi\), \(2\mathrm{c}\mathrm{s} = \sin\phi\), \(2\mathrm{s}^{2} = 1 - \cos\phi\),

which is exactly Equation (14.37).

Derives Lemma 14.41. From Equation (14.44), \(\comm{J_{3}}{J_{1}\pm\ii J_{2}} = \ii J_{2} \pm \ii(-\ii J_{1}) = \pm\left(J_{1} \pm \ii J_{2}\right)\), and \(\comm{J_{+}}{J_{-}} = -2\ii\comm{J_{1}}{J_{2}} = 2J_{3}\). Next

which rearranges to the first form of Equation (14.51); the second follows by exchanging the roles of \(J_{+}\) and \(J_{-}\), which flips the sign of the \(\comm{J_{1}}{J_{2}}\) term. For the Killing form use Equation (14.15) with \(C^{c}{}_{ab} = \ii\epsilon_{abc}\) from Equation (14.44): \(\kappa_{ab} = \left(\ii\epsilon_{adc}\right)\left(\ii\epsilon_{bcd}\right) = -\epsilon_{adc}\epsilon_{bcd} = \epsilon_{adc}\epsilon_{bdc} = 2\delta_{ab}\).

Derives Theorem 14.43. \(J_{3}\) is Hermitian on a finite-dimensional space, so it has an eigenvector; let \(\mu\) be an eigenvalue of largest real part, with eigenvector \(\ket{\mu}\). From Equation (14.50), \(J_{3}\left(J_{\pm}\ket{\mu}\right) = \left(\mu\pm1\right)J_{\pm}\ket{\mu}\), so \(J_{\pm}\) raises and lowers the \(J_{3}\) eigenvalue by one. Maximality of \(\mu\) forces \(J_{+}\ket{\mu} = 0\). Write \(j := \mu\) and \(\ket{j,j} := \ket{\mu}\) normalised. Then by Equation (14.51),

and since \(\vect{J}^{2}\) commutes with \(J_{3}\) and with \(J_{\pm}\) its eigenspaces are invariant under the whole algebra, so by irreducibility the value \(j(j+1)\) is taken on the entire representation. (This is Corollary 14.18 again, argued here directly so that nothing about the group is needed.)

Define \(\ket{j,m}\) recursively by lowering. Because \(J_{+}\) and \(J_{-}\) are adjoint to each other when the \(J_{a}\) are Hermitian,

using the second form of Equation (14.51); the same computation with \(J_{+}\) and the first form gives \(\norm{J_{+}\ket{j,m}}^{2} = j(j+1)-m(m+1)\), which is Equation (14.54) once phases are fixed by choosing the square roots positive. Since the space is finite-dimensional the chain of eigenvalues \(j, j-1, j-2, \ldots\) must terminate, which by the norm formula happens exactly at the value \(m_{\min}\) with \(j(j+1) - m_{\min}(m_{\min}-1) = 0\), that is \(m_{\min} \in \set{-j,\, j+1}\); the second is excluded because \(m_{\min} \le j\) and would need \(j+1 \le j\). Hence \(m_{\min} = -j\), and the chain runs from \(j\) down to \(-j\) in unit steps, so \(j - (-j) = 2j\) is a non-negative integer. This is the whole reason half-integer angular momentum is possible: nothing in the algebra forbids the chain from having an even number of steps. The span of the \(\ket{j,m}\) is invariant under \(J_{3}\) and \(J_{\pm}\), hence under the whole algebra, and is nonzero, so by irreducibility it is the whole space, of dimension \(2j+1\).

Derives Proposition 14.44. Let \(\mathbb{V}_{j}\) be the complex inner-product space with orthonormal basis \(\set{\ket{j,m}}\), \(m = -j,-j+1,\ldots,j\), of dimension \(2j+1\), and define operators on it by

with the convention that \(\ket{j,m}\) denotes the zero vector for \(\abs{m} > j\). The convention is consistent because \(j(j+1)-m(m+1) = (j-m)(j+m+1)\) and \(j(j+1)-m(m-1) = (j+m)(j-m+1)\), which are non-negative for \(-j \le m \le j\) and vanish exactly at \(c_{+}(j) = 0\) and \(c_{-}(-j) = 0\): the operators do not attempt to leave the basis. Put \(J_{1} = \tfrac{1}{2}(J_{+}+J_{-})\) and \(J_{2} = \tfrac{1}{2\ii}(J_{+}-J_{-})\).

The brackets hold. \(J_{3}J_{\pm}\ket{j,m} = (m\pm1)c_{\pm}(m)\ket{j,m\pm1}\) while \(J_{\pm}J_{3}\ket{j,m} = m\,c_{\pm}(m)\ket{j,m\pm1}\), so \(\comm{J_{3}}{J_{\pm}} = \pm J_{\pm}\). Since \(c_{+}(m-1) = c_{-}(m)\) by Equation (14.56), \(J_{+}J_{-}\ket{j,m} = c_{-}(m)^{2}\ket{j,m}\) and \(J_{-}J_{+}\ket{j,m} = c_{+}(m)^{2}\ket{j,m}\), whence

which is \(2J_{3}\ket{j,m}\). That is Equation (14.50), equivalent to Equation (14.44) by the computation in Lemma 14.41 run backwards.

The generators are Hermitian. The basis is orthonormal and the \(c_{\pm}\) are real, so the only nonzero matrix elements of the ladder operators are \(\bra{j,m+1}J_{+}\ket{j,m} = c_{+}(m)\) and \(\bra{j,m}J_{-}\ket{j,m+1} = c_{-}(m+1) = c_{+}(m)\). Hence \(J_{-} = J_{+}^{\dagger}\), and \(J_{1}\), \(J_{2}\), \(J_{3}\) are Hermitian.

It is irreducible. Let \(\mathbb{W} \neq 0\) be an invariant subspace. It is invariant under the Hermitian \(J_{3}\), whose eigenvalues on \(\mathbb{V}_{j}\) are the \(2j+1\) distinct numbers \(m\), so \(\mathbb{W}\) is spanned by those \(\ket{j,m}\) it contains and contains at least one of them. From any one of them the ladder operators reach every other, their coefficients \(c_{\pm}(m)\) vanishing only at the two ends of the chain. So \(\mathbb{W} = \mathbb{V}_{j}\).

Finally Equation (14.51) gives \(\vect{J}^{2}\ket{j,m} = \left(c_{+}(m)^{2}+m(m+1)\right)\ket{j,m} = j(j+1)\ket{j,m}\), so Equations (14.55) and (14.56) are Equations (14.53) and (14.54).

Derives Proposition 14.45. By Proposition 14.36, \(-\identity = U(\hat{e}_{3},2\pi) = \exp\left(-2\pi\ii\,T_{3}\right)\), so \(D^{(j)}(-\identity) = \exp\left(-2\pi\ii J_{3}\right)\) in the representation, and on \(\ket{j,m}\) this is \(\ee^{-2\pi\ii m}\) by Equation (14.53). The values of \(m\) are \(j, j-1,\ldots,-j\): all integers when \(j\) is an integer, all in \(\Z+\tfrac{1}{2}\) when \(j\) is half-odd-integer. In the first case \(\ee^{-2\pi\ii m} = 1\) for every \(m\), in the second \(\ee^{-2\pi\ii m} = -1\) for every \(m\); either way the operator is \((-1)^{2j}\identity\). A representation of \(\SU(2)\) factors through the quotient by \(\set{\pm\identity}\) precisely when it is trivial on that subgroup, which by Equation (14.57) happens exactly for integer \(j\).

Derives Theorem 14.47. That \(\vect{J}\) satisfies Equation (14.44) is immediate, the two factors commuting. Take \(j_{1} \ge j_{2}\) without loss of generality. In the product basis \(\ket{j_{1}m_{1}}\otimes\ket{j_{2}m_{2}}\) the operator \(J_{3}\) is diagonal with eigenvalue \(m = m_{1}+m_{2}\), so the multiplicity \(n(m)\) of the eigenvalue \(m\) is the number of pairs \((m_{1},m_{2})\) with \(m_{1}+m_{2} = m\), \(\abs{m_{i}} \le j_{i}\). Counting them,

Now the tensor product is a finite-dimensional representation, hence a direct sum of irreducibles \(V_{J}\) with multiplicities \(a_{J} \ge 0\); by Equation (14.53) each \(V_{J}\) contributes exactly \(1\) to \(n(m)\) for every \(\abs{m}\le J\) and nothing otherwise, so \(n(m) = \sum_{J \ge \abs{m}} a_{J}\). Subtracting consecutive values,

and Equation (14.61) makes this \(1\) for \(\abs{j_{1}-j_{2}} \le J \le j_{1}+j_{2}\) and \(0\) otherwise, which is Equation (14.59). For Equation (14.60) write \(J = j_{1}-j_{2}+k\) with \(k = 0,\ldots,2j_{2}\):

Derives Proposition 14.50. Linearizing Equation (14.62) as in Proposition 14.35 gives \(X^{\dagger} + X = 0\) and \(\tr X = 0\). The anti-Hermitian \(3\times3\) matrices form a real vector space of dimension \(9\) — three real diagonal entries times \(\ii\), and three complex entries above the diagonal — and the trace condition removes one, leaving \(8\). For the rank, a maximal abelian subalgebra consisting of diagonalizable elements may be taken to be the traceless diagonal anti-Hermitian matrices, of which there are two independent, and no larger commuting set of diagonalizable elements exists because a set of commuting Hermitian matrices is simultaneously diagonalizable and the diagonal traceless ones already exhaust that possibility.

Derives Proposition 14.52. The product \(T_{a}T_{b}\) is a \(3\times3\) matrix, and \(\set{\identity,T_{1},\ldots,T_{8}}\) is a basis of the complex \(3\times3\) matrices, since the \(T_{c}\) span the traceless ones. Expanding \(T_{a}T_{b}\) in it and separating the antisymmetric and symmetric parts in \((a,b)\) gives Equation (14.67), the coefficient of \(\identity\) in the symmetric part being \(\tfrac{1}{3}\tr\acomm{T_{a}}{T_{b}} = \tfrac{1}{3}\delta_{ab}\) by Equation (14.66). Multiplying either relation by \(T_{c}\) and taking the trace with Equation (14.66), and using \(\tr T_{c} = 0\), gives Equation (14.68). Total symmetry of \(d\) and total antisymmetry of \(f\) then follow from Equation (14.68) and the cyclicity of the trace: \(f\) is antisymmetric in \((a,b)\) by construction and

is invariant under cyclic permutation of \((a,b,c)\), which together with antisymmetry in the first pair gives total antisymmetry; the same argument with the anticommutator gives total symmetry of \(d\). The tables Equations (14.69) and (14.70) are then a finite computation from Equations (14.63), (14.64) and (14.65); two entries are done here as specimens. For \(f_{458}\): a direct multiplication gives \(\comm{\lambda_{4}}{\lambda_{5}} = \diag(2\ii,0,-2\ii)\), so

For \(d_{247}\): \(\acomm{\lambda_{2}}{\lambda_{4}} = -\lambda_{7}\) by direct multiplication, so \(d_{247} = \tfrac{1}{4}\tr\left(-\lambda_{7}\lambda_{7}\right) = -\tfrac{1}{4}\cdot2 = -\tfrac{1}{2}\).

Derives Proposition 14.53. With \(C^{c}{}_{ab} = \ii f_{abc}\) from Equation (14.67), Equation (14.15) gives \(\kappa_{ab} = \left(\ii f_{adc}\right)\left(\ii f_{bcd}\right) = -f_{adc}f_{bcd} = f_{adc}f_{bdc} = f_{acd}f_{bcd}\) after renaming.

The array is evaluated here directly. The shorter route — both \(\kappa_{ab}\) and \(\tr(T_{a}T_{b})\) are symmetric bilinear forms invariant under the adjoint action, and on a simple algebra such a form is unique up to a scale, so it is enough to evaluate one entry — would have to obtain that uniqueness from Schur's lemma applied to the adjoint representation, and Theorem 5.153 is stated for a representation on a complex space, whereas the adjoint representation of \(\mathfrak{su}(3)\) acts on the eight-dimensional real algebra. Over \(\R\) the conclusion of that lemma is false in general, so its hypotheses would first have to be recovered by complexifying and by quoting the irreducibility of the complexified adjoint — which is no cheaper than the computation that follows and rests on the same quoted simplicity.

Because \(\comm{T_{a}}{T_{c}} = \ii f_{acd}T_{d}\) and \(\tr(T_{d}T_{e}) = \tfrac{1}{2}\delta_{de}\) by Equation (14.66),

Expanding the two commutators gives four traces. Summed over \(c\) they collapse to two, by cyclicity of the trace together with \(\sum_{c}T_{c}T_{c} \propto \identity\), which Equation (14.74) below establishes:

Both sums follow from the completeness of the basis \(\set{\identity,T_{1},\ldots,T_{8}}\) of the complex \(3\times3\) matrices already used in Proposition 14.52. Expanding an arbitrary \(M\) in that basis, with the coefficients read off by Equation (14.66) and \(\tr T_{a} = 0\), gives \(M = \tfrac{1}{3}\left(\tr M\right)\identity + 2\,\tr\left(MT_{a}\right)T_{a}\), and taking for \(M\) the matrix whose only nonzero entry is a \(1\) in position \((j,k)\) turns that into

Setting \(l = k\) and summing gives \(\sum_{c}T_{c}T_{c} = \tfrac{1}{2}\left(3-\tfrac{1}{3}\right)\identity = \tfrac{4}{3}\identity\), so that \(\sum_{c}\tr(T_{a}T_{b}T_{c}T_{c}) = \tfrac{4}{3}\tr(T_{a}T_{b}) = \tfrac{2}{3}\delta_{ab}\); and contracting Equation (14.74) into \(\tr(T_{a}T_{c}T_{b}T_{c}) = \left(T_{a}\right)_{ij}\left(T_{c}\right)_{jk} \left(T_{b}\right)_{kl}\left(T_{c}\right)_{li}\) gives

Substituting both into Equation (14.73),

and Equation (14.72) gives \(\kappa_{ab} = 3\delta_{ab}\).

As a check on one entry, take \(a=b=3\) and read the nonzero \(f_{3cd}\) from Equation (14.69) — \(f_{312} = 1\), \(f_{345} = \tfrac{1}{2}\), \(f_{367} = -\tfrac{1}{2}\), each contributing twice because both index orders are summed:

Derives Proposition 14.54. For \(C_{2}\) this is Corollary 14.17 with \(\kappa^{ab} = \tfrac{1}{3}\delta^{ab}\) from Equation (14.71), up to the harmless factor of \(3\). For \(C_{3}\), apply Theorem 14.16 with \(r = 3\) and \(k^{abc} = d_{abc}\): the tensor is totally symmetric by Proposition 14.52, and its invariance in the sense of Equation (14.18) is Equation (14.148) for the \(G\)-invariant polynomial \(\gen{T_{a},T_{b},T_{c}} = \tfrac{1}{2}d_{abc}\) — symmetric and invariant because \(\tr\left(\acomm{T_{a}}{T_{b}}T_{c}\right)\) is a trace of a product of generators and therefore unchanged under \(T_{a} \mapsto gT_{a}g^{-1}\), which is Definition 14.109. Written out, the invariance reads

and it is exactly this identity that the proof of Theorem 14.16 consumes.

Derives Proposition 14.55. For Equation (14.77): \(C_{2}\) is central, so by Corollary 14.18 it acts as a number \(c\) on the irreducible \(\vect{3}\); taking the trace, \(3c = \sum_{a}\tr\left(T_{a}T_{a}\right) = 8 \cdot \tfrac{1}{2} = 4\) by Equation (14.66), so \(c = 4/3\). For Equation (14.78): the same argument with \(\tr\left(T_{a}^{\mathrm{ad}}T_{b}^{\mathrm{ad}}\right) = -f_{acd}f_{bdc} = f_{acd}f_{bcd} = 3\delta_{ab}\) from Equation (14.71) gives \(8c = \sum_{a}3 = 24\), so \(c = 3\).

That \(\bar{\vect{3}}\) is a representation is the observation that complex-conjugating \(\comm{T_{a}}{T_{b}} = \ii f_{abc}T_{c}\) and multiplying by \(-1\) reproduces the same brackets. That it is inequivalent to \(\vect{3}\) is read off the weights. Take the Cartan subalgebra spanned by \(T_{3}\) and \(T_{8}\), which are diagonal; the weights of a representation are the pairs of simultaneous eigenvalues of \(\left(T_{3},T_{8}\right)\), and being eigenvalues they are unchanged by any equivalence. For \(\vect{3}\) they are, from Equations (14.63) and (14.65),

summing to zero as tracelessness demands, and for \(\bar{\vect{3}}\) they are the three negatives of these. The two sets are different — an upward-pointing triangle and a downward-pointing one — so no equivalence can carry one representation to the other.

Derives Theorem 14.57. Identify \(\C^{3}\otimes\left(\C^{3}\right)^{\ast}\) with the space of complex \(3\times3\) matrices \(\operatorname{End}(\C^{3})\); the action induced by \(\vect{3}\) on the first factor and \(\bar{\vect{3}}\) on the second is \(M \mapsto UMU^{\dagger}\). The trace is invariant under it, so \(\operatorname{End}(\C^{3})\) splits as an \(\SU(3)\)-invariant direct sum

of dimensions \(1\) and \(8\). The first summand is the trivial representation \(\vect{1}\). The second is the complexification of \(\mathfrak{su}(3)\) — a traceless complex matrix splits uniquely into Hermitian and anti-Hermitian traceless parts — carrying the action \(M \mapsto UMU^{\dagger}\), which is the adjoint action; it is irreducible because an invariant subspace would be an ideal of a simple Lie algebra. Hence it is \(\vect{8}\).

The weights confirm the count independently: the weights of the tensor product are the sums of a weight of \(\vect{3}\) and a weight of \(\bar{\vect{3}}\), that is all differences \(w_{i} - w_{j}\) of the three weights Equation (14.79). The three coincident cases \(i = j\) give the weight zero three times, and the six cases \(i \neq j\) give exactly the six roots Equation (14.80). That is one zero for \(\vect{1}\) plus the six roots and two zeros of \(\vect{8}\), and nothing left over.

Derives Proposition 14.59. In the Cartesian coordinates \(x^{A}\) the metric components are constant, so the Christoffel symbols vanish and Killing's equation Equation (13.290) reads

(i) The field \(J_{AB}\) has components \(\xi^{C} = x_{A}\delta^{C}_{B} - x_{B}\delta^{C}_{A}\), hence \(\xi_{D} = x_{A}\eta_{BD} - x_{B}\eta_{AD}\), and with \(\pp_{C}x_{A} = \eta_{AC}\),

which is antisymmetric under \(C \leftrightarrow D\). So Equation (14.87) holds. That the translations satisfy it is immediate, their components being constant.

(ii) Let \(\xi\) satisfy Equation (14.87) and set \(T_{CDE} = \pp_{C}\pp_{D}\xi_{E}\). It is symmetric in its first two indices, because partial derivatives commute, and antisymmetric in its last two by Equation (14.87). Those two symmetries are incompatible unless \(T\) vanishes:

each step being one of the two symmetries applied once. Hence \(\pp_{C}\pp_{D}\xi_{E} = 0\), so each \(\xi_{E}\) is a polynomial of degree at most one,

with constant coefficients, and Equation (14.87) says exactly that \(\omega_{EC} = -\omega_{CE}\). Comparing with the components computed in (i), the constant part is \(a^{E}P_{E}\) and the linear part is \(-\tfrac{1}{2}\omega^{AB}J_{AB}\), a combination of the \(J_{AB}\). Conversely \(\xi\) determines \(a_{E}\) and \(\omega_{EC}\) through Equation (14.88), so distinct pairs \((a,\omega)\) give distinct fields and the \(D + D(D-1)/2 = D(D+1)/2\) listed fields are independent. That number is the maximum allowed by Proposition 13.140. The brackets of these fields are Equations (14.84), (14.90) and (14.97), computed below; they are those of the semidirect sum \(\mathfrak{t}(D)\,\bar{\oplus}\,\mathfrak{so}(p,q)\), which at \(q = 1\) is Equation (14.99).

(iii) Directly,

so \(J_{AB}\) is tangent to every level set of \(x^{C}x_{C}\) and its flow maps each such quadric to itself. That flow consists of isometries of \(\eta\) by (i), and an isometry of the ambient metric preserving a submanifold restricts to an isometry of the induced metric; by Definition 13.139 the restriction of \(J_{AB}\) is therefore a Killing field of the quadric.

Derives Proposition 14.60. \(\comm{P^{2}}{P_{C}} = 0\) is immediate from Equation (14.84). For the pseudo-rotations, use the Leibniz rule and Equation (14.97) in the form \(\comm{J_{AB}}{P^{C}} = \delta^{C}_{B}P_{A} - \delta^{C}_{A}P_{B}\):

each parenthesis vanishing separately by Equation (14.84).

Derives Lemma 14.62. The left-hand side is totally antisymmetric in \(ABCD\): exchanging any two of the four free indices swaps two of the terms with each other and reverses the sign of the remaining two through the antisymmetry of \(\epsilon\). A totally antisymmetric four-index array in four dimensions is a multiple of \(\epsilon_{ABCD}\), and contracting the left-hand side with \(\epsilon^{ABCD}\) shows that the multiple is proportional to \(\omega^{A}{}_{A} = \eta^{AB}\omega_{BA} = 0\), a symmetric array contracted with an antisymmetric one.

Derives Lemma 14.64. (i) By Equation (14.84) only the \(J\) factor contributes, and Equation (14.97) gives

since in each term \(\epsilon\) antisymmetrizes a pair of indices carried by the symmetric product \(P\,P\) of commuting translations.

(ii) \(W_{A}P^{A} = \tfrac{1}{2}\epsilon_{ABCD}J^{BC}P^{D}P^{A}\) vanishes for the same reason, the pair \((A,D)\) now being the symmetric one; the reversed order follows from (i).

(iii) Apply the Leibniz rule to Equation (14.103) using Equation (14.90) for the \(J\) factor and Equation (14.97) for the \(P\) factor. Both brackets are of the form “\(\eta\) times a generator with one index replaced”, that is, exactly the action of an infinitesimal pseudo-rotation on each index of the tensors \(J^{BC}\) and \(P^{D}\). The three contributions are therefore the four terms of Equation (14.102) with \(\omega\) the antisymmetrized index pair \((AB)\), less the one term in which the free index \(C\) of \(W\) is rotated; by Lemma 14.62 the first four cancel and the survivor is

In words: \(W\) is a contraction of the invariant tensor \(\epsilon\) with two tensors, so it is a tensor with the one uncontracted index it displays.

Derives Theorem 14.65. \(P^{2}\) is Proposition 14.60. For \(W^{2}\): \(\comm{W^{2}}{P_{E}} = 0\) follows from Lemma 14.64(i) by the Leibniz rule, and \(\comm{W^{2}}{J_{AB}} = 0\) is the computation already done for \(P^{2}\) in Proposition 14.60 with \(P\) replaced by \(W\), which is legitimate because Lemma 14.64(iii) says \(W\) carries its index the same way \(P\) does; the square of a vector is a scalar. Independence: \(W^{2}\) contains the generators \(J^{BC}\), which \(P^{2}\) does not, so neither is a function of the other.

Derives Lemma 14.68. Let \(\mathfrak{h} = \set{X \in \mathfrak{so}(D-1,1) \mid Xp = 0}\); it is a subalgebra, being the kernel of the linear map \(X \mapsto Xp\) and closed under the bracket because \(\comm{X}{Y}p = 0\) whenever \(Xp = Yp = 0\). For \(X \in \mathfrak{h}\) and any \(u\), the defining condition Equation (14.23) gives \(\eta(Xu,p) = -\eta(u,Xp) = 0\), so \(X\) maps the whole space into \(p^{\perp}\) and kills \(p\). Since \(\eta(p,p) \neq 0\) the space splits as \(\R p \oplus p^{\perp}\) with \(\eta\) nondegenerate on each summand, and \(X\) is determined by its restriction to \(p^{\perp}\), which by Equation (14.23) again is an element of the pseudo-orthogonal algebra of \(\eta\) restricted to \(p^{\perp}\). Conversely any such element, extended by \(Xp = 0\), satisfies Equation (14.23) on all of \(\R^{D-1,1}\): both sides vanish when either argument is \(p\). Removing one negative direction from the signature \((D-1,1)\) leaves \(\eta|_{p^{\perp}}\) positive definite in \(D-1\) dimensions, so \(\mathfrak{h} \cong \mathfrak{so}(D-1)\). Its rank is \(\lfloor (D-1)/2 \rfloor\) by Proposition 14.22, applicable because \(D-1 \ge 3\).

For Equation (14.108): if \(D = 2k\) then \(\lfloor (D-1)/2 \rfloor = k-1\) and \(\lceil D/2 \rceil = k\); if \(D = 2k+1\) then \(\lfloor (D-1)/2 \rfloor = k\) and \(\lceil D/2 \rceil = k+1\). In both cases the two sides differ by one.

Derives Proposition 14.69. Write \(f(X) = \eta_{MN}X^{M}X^{N}\), so that \(\dd f = 2\eta_{MN}X^{M}\dd X^{N}\) is nonzero at every point of either quadric; each is therefore a regular level set, a submanifold of dimension \(D\), with tangent space at \(X\) the orthogonal complement \(X^{\perp}\). Since \(f(X) \neq 0\) there, the ambient space splits as \(\R X \oplus X^{\perp}\) with \(\eta\) nondegenerate on each summand, so the signature of the induced metric is that of the ambient space less one direction: less a negative one for Equation (14.111) in signature \((D-1,2)\), and less a positive one for Equation (14.112) in signature \((D,1)\). Both give \((D-1,1)\).

By Proposition 14.59(iii) each \(J_{MN}\) is tangent to the quadric and restricts to a Killing field of the induced metric. The restrictions are still independent. Indeed a vanishing combination \(\omega^{MN}J_{MN}\) has ambient components proportional to \(\omega^{M}{}_{N}X^{N}\), so it vanishes on the quadric only if the matrix \(\omega^{M}{}_{N}\) annihilates every point of the quadric, hence the whole span of the quadric. That span is everything. In an indefinite signature the vectors of negative norm already span: a negative basis direction has negative norm, and a positive basis direction \(e\) is the half-sum of \(e + 2u\) and \(e - 2u\) for any negative unit \(u\), both of norm \(-3\). Every vector of negative norm is a real multiple of a point of Equation (14.111), since rescaling by \(\lambda\) multiplies the norm by \(\lambda^{2}\). Interchanging the two signs throughout gives the same conclusion for Equation (14.112). Hence \(\omega = 0\).

The count closes the argument. By Equation (14.89) with \(p + q = D+1\), both algebras have dimension \(D(D+1)/2\), which is the largest number of independent Killing fields a \(D\)-dimensional manifold can carry by Proposition 13.140. No further Killing field can be independent of them, so the isometry algebra is exactly what has been exhibited.

Derives Proposition 14.70. The fields are conformal Killing fields. In Cartesian coordinates the conformal Killing equation reads \(\pp_{B}\xi_{C} + \pp_{C}\xi_{B} = \tfrac{2}{D}\left(\pp^{E}\xi_{E}\right)\eta_{BC}\). For \(P_{A}\) and \(J_{AB}\) both sides vanish, by Proposition 14.59. For \(\Delta\), \(\xi_{C} = x_{C}\) gives \(\pp_{B}\xi_{C} + \pp_{C}\xi_{B} = 2\eta_{BC}\) and \(\pp^{E}\xi_{E} = D\), and the equation holds. For \(K_{A}\), \(\xi_{C} = 2x_{A}x_{C} - x^{2}\eta_{AC}\) gives

whose symmetric part is \(4x_{A}\eta_{BC}\) and whose trace is \(\pp^{E}\xi_{E} = 2x_{A} + 2Dx_{A} - 2x_{A} = 2Dx_{A}\); again the equation holds. That these are all of them is Equation (13.333), and the count is \(D + D(D-1)/2 + 1 + D = (D+1)(D+2)/2\).

The brackets. Every computation below uses only the derivation rule \(\comm{X}{fY} = X(f)\,Y + f\comm{X}{Y}\) for a function \(f\), the values

the last from Proposition 14.59(iii), and the identity \(x_{A}P_{B} - x_{B}P_{A} = J_{AB}\).

For \(\comm{\Delta}{P_{A}}\): \(\comm{x^{C}\pp_{C}}{\pp_{A}} = -\left(\pp_{A}x^{C}\right)\pp_{C} = -P_{A}\). For \(\comm{\Delta}{J_{AB}}\), the same computation on each of the two terms of \(J_{AB}\) gives \(x_{A}P_{B} - x_{A}P_{B}\) and so zero. For \(\comm{\Delta}{K_{A}}\),

For \(\comm{P_{A}}{K_{B}}\), using \(\comm{P_{A}}{\Delta} = P_{A}\) and \(P_{A}(x^{2}) = 2x_{A}\),

For \(\comm{J_{AB}}{K_{C}}\), using \(\comm{J_{AB}}{\Delta} = 0\), \(J_{AB}(x^{2}) = 0\) and Equation (14.97),

Finally, with \(K_{A}(x_{B}) = 2x_{A}x_{B} - x^{2}\eta_{AB}\), \(K_{A}(x^{2}) = 4x_{A}x^{2} - 2x_{A}x^{2} = 2x_{A}x^{2}\), \(\comm{K_{A}}{\Delta} = -K_{A}\) and \(\comm{K_{A}}{P_{B}} = -2\eta_{AB}\Delta - 2J_{AB}\),

The isomorphism. Let the indices \(M,N\) run over \(1,\ldots,D+2\), extend \(\eta\) by

all other new components vanishing, and set

the remaining \(J_{MN}\) being the \(J_{AB}\) already defined and \(J_{NM} = -J_{MN}\). This is a linear bijection onto the span, since \(P_{A}\) and \(K_{A}\) are recovered as \(J_{A,D+2} \mp J_{A,D+1}\). It remains to check that Equation (14.90) holds for every choice of the four indices among \(1,\ldots,D+2\). With four indices in \(1,\ldots,D\) it is Equation (14.90) itself. With exactly one index equal to \(D+1\) or \(D+2\) it is the bracket \(\comm{J_{AB}}{K_{C}}\) of Equation (14.116) together with Equation (14.97), both of which say that \(K\) and \(P\) carry their index as vectors. The case \(\comm{J_{AB}}{J_{D+1,D+2}} = \comm{J_{AB}}{\Delta} = 0\) agrees with Equation (14.90), every term of which carries a vanishing \(\eta\). There remain the pairs built from Equation (14.118). Using \(\comm{K_{A}}{P_{B}} = -2\eta_{AB}\Delta - 2J_{AB}\) and \(\comm{P_{A}}{K_{B}} = 2\eta_{AB}\Delta - 2J_{AB}\),

while Equation (14.90) demands, for these three, \(\eta_{D+1,D+1}J_{BA} = J_{AB}\), \(\eta_{D+2,D+2}J_{BA} = -J_{AB}\) and \(\eta_{AB}J_{D+2,D+1} = -\eta_{AB}\Delta\), which are the same three results. Likewise

against the demands \(\eta_{D+1,D+1}J_{D+2,A} = J_{A,D+2}\) and \(\eta_{D+2,D+2}J_{A,D+1} = J_{A,D+1}\) of Equation (14.90). All cases agree, so the fields Equation (14.115) satisfy the brackets of \(\mathfrak{so}\) of the extended metric Equation (14.117). That metric adds one positive and one negative direction, so its signature is \((p+1,q+1)\) after the coordinates are reordered to list the positive directions first.

Derives Proposition 14.73. Antisymmetry of \(c\) follows from antisymmetry of the bracket in Equation (14.120). Imposing the Jacobi identity on \(\tilde{\mathfrak{g}}\) for the three elements \(\tilde{\xi},\tilde{\eta},\tilde{\zeta}\), the terms not involving \(Z\) cancel because the Jacobi identity already holds in \(\mathfrak{g}\), and \(Z\) is central so it contributes nothing to a nested bracket; what survives is the coefficient of \(Z\), which is exactly Equation (14.121).

Derives Proposition 14.76. Write \(\partial b\) for the coboundary of Equation (14.122), \((\partial b)(\xi,\eta)=b\left(\comm{\xi}{\eta}\right)\). Three things have to be checked.

The class does not depend on the splitting. Two linear splittings of the projection differ by a linear map \(\mathfrak{g}\to\R\): if \(\xi\mapsto\tilde{\xi}\) and \(\xi\mapsto\tilde{\xi}'\) both project to \(\xi\), their difference lies in the kernel of the projection, which is the line spanned by \(Z\), so \(\tilde{\xi}'=\tilde{\xi}+b(\xi)Z\) with \(b\) linear. Since \(Z\) is central, \(\comm{\tilde{\xi}'}{\tilde{\eta}'}=\comm{\tilde{\xi}}{\tilde{\eta}}\), while the first term of Equation (14.120) shifts by \(b\left(\comm{\xi}{\eta}\right)Z\). Hence \(c'=c-\partial b\) and the two cocycles define the same class. An isomorphism of extensions in the sense of Definition 14.75 is the identity on \(\R\) and induces the identity on \(\mathfrak{g}\), so it acts on a splitting exactly as a change of splitting does; equivalent extensions therefore have equal classes.

Every class is realized. Given a cocycle \(c\), put \(\tilde{\mathfrak{g}}=\mathfrak{g}\oplus\R Z\) as a vector space, with \(Z\) central and the bracket Equation (14.120). It is antisymmetric because \(c\) is, and it satisfies the Jacobi identity: the \(\mathfrak{g}\)-component of the Jacobi expression is the Jacobi identity of \(\mathfrak{g}\), and the coefficient of \(Z\) is Equation (14.121). The obvious inclusion and projection make it a central extension, whose cocycle for the obvious splitting is \(c\) again.

Cohomologous cocycles give equivalent extensions, and only those. A map between the extensions built from \(c\) and from \(c'\) that commutes with the inclusion and the projection is the identity on \(\R Z\) and induces the identity on \(\mathfrak{g}\), so it sends \(\tilde{\xi}\mapsto\tilde{\xi}+\beta(\xi)Z\) for some linear \(\beta:\mathfrak{g}\to\R\) and is automatically bijective. Requiring it to respect the two brackets and reading off the coefficient of \(Z\) gives \(c'=c+\partial\beta\). So the two extensions are equivalent precisely when \(c'-c\) is a coboundary, that is, precisely when \(c\) and \(c'\) define the same class. The trivial extension is the one with \(c=0\), which is Definition 14.74, and it represents the zero class.

Derives Proposition 14.84. Throughout, write \(\comm{X}{Y}_{\varepsilon} = U(\varepsilon)^{-1}\comm{U(\varepsilon)X}{U(\varepsilon)Y}\) for the bracket at finite \(\varepsilon\). It is the pullback of the bracket of \(\mathfrak{g}\) along the linear isomorphism \(U(\varepsilon)\), so \(\left(\mathfrak{g},\comm{\cdot}{\cdot}_{\varepsilon}\right)\) is a Lie algebra isomorphic to \(\mathfrak{g}\) for every \(\varepsilon \neq 0\), and its structure constants \(C^{\gamma}{}_{\alpha\beta}(\varepsilon)\) converge, by hypothesis, to those of \(\mathfrak{g}_{0}\).

(i) Bilinearity and antisymmetry pass to the limit because they are linear conditions on the structure constants. The Jacobi identity is the system of quadratic equations \(C^{\delta}{}_{\alpha\beta}(\varepsilon)C^{\sigma}{}_{\delta\gamma} (\varepsilon) + C^{\delta}{}_{\beta\gamma}(\varepsilon)C^{\sigma}{}_{\delta\alpha} (\varepsilon) + C^{\delta}{}_{\gamma\alpha}(\varepsilon)C^{\sigma}{}_{\delta\beta} (\varepsilon) = 0\), valid for every \(\varepsilon \neq 0\); a polynomial is continuous, so the same equations hold in the limit. The underlying vector space never changed, whence the dimension.

(ii) A derivation is a solution of the linear system \(d\left(\comm{T_{\alpha}}{T_{\beta}}\right) = \comm{dT_{\alpha}}{T_{\beta}} + \comm{T_{\alpha}}{dT_{\beta}}\) in the unknown matrix \(d\), and the coefficients of that system are linear in the structure constants. Write \(L(\varepsilon)\) for its matrix. Since \(\operatorname{Der}\) of isomorphic algebras are conjugate, \(\dim\ker L(\varepsilon) = \dim\operatorname{Der}(\mathfrak{g})\) for all \(\varepsilon \neq 0\). Rank is lower semicontinuous: if \(L(0)\) carries a nonvanishing minor of size \(r\), that same minor is a continuous function of \(\varepsilon\) and so is nonvanishing for small \(\varepsilon\), whence \(\operatorname{rank}\,L(0) \le \operatorname{rank}\,L(\varepsilon)\) there, and the kernels satisfy the reverse inequality. That is the claim. If it is strict, then \(\mathfrak{g}_{0}\) is not isomorphic to \(\mathfrak{g}\), and a contraction the other way would give \(\dim\operatorname{Der}(\mathfrak{g}) \ge \dim\operatorname{Der}(\mathfrak{g}_{0})\) and hence equality, which is excluded.

(iii) From \(\ad^{\varepsilon}_{X} = U(\varepsilon)^{-1}\ad_{U(\varepsilon)X} U(\varepsilon)\) and cyclicity of the trace,

which in the fixed basis is the matrix \(\bigl(U\bigr)\transpose\kappa\,U\), of determinant \(\left(\det U\right)^{2}\det\kappa\). The Killing form is a quadratic polynomial in the structure constants by Equation (14.15), so \(\kappa_{0}\) is the limit of \(\kappa_{\varepsilon}\) and \(\det\kappa_{0}\) the limit of the determinants. If \(\det U \rightarrow 0\) then \(\det\kappa_{0} = 0\), the Killing form of \(\mathfrak{g}_{0}\) is degenerate, and Theorem 14.13 denies it semisimplicity.

(iv) Take \(U(\varepsilon) = \varepsilon\,\identity\). Then \(\comm{X}{Y}_{\varepsilon} = \varepsilon^{-1}\comm{\varepsilon X}{\varepsilon Y} = \varepsilon\comm{X}{Y} \rightarrow 0\), so every algebra contracts to the abelian one of the same dimension; and \(\det U = \varepsilon^{n} \rightarrow 0\), consistently with (iii). The second cohomology of an abelian algebra is computed in Example 14.79 and is the whole space of antisymmetric bilinear forms, nonzero as soon as \(n \ge 2\).

Derives Lemma 14.86. For \(X, Y \in \mathfrak{h}\), \(U(\varepsilon)\) acts as the identity and \(U(\varepsilon)^{-1}\) multiplies the \(\mathfrak{m}\)-part by \(\varepsilon^{-1}\), so \(\comm{X}{Y}_{\varepsilon} = \left(1-\pi\right)\comm{X}{Y} + \varepsilon^{-1}\pi\comm{X}{Y}\), which converges if and only if \(\pi\comm{X}{Y} = 0\), that is if and only if \(\comm{\mathfrak{h}}{\mathfrak{h}} \subseteq \mathfrak{h}\); the limit is then \(\comm{X}{Y}\). For \(X \in \mathfrak{h}\) and \(Z \in \mathfrak{m}\),

and for \(Z, Z' \in \mathfrak{m}\), \(\comm{Z}{Z'}_{\varepsilon} = \varepsilon^{2}\left(1-\pi\right)\comm{Z}{Z'} + \varepsilon\pi\comm{Z}{Z'} \rightarrow 0\). Both limits exist unconditionally, so the condition on \(\mathfrak{h}\) is the whole condition. Equation (14.126) then puts \(\comm{\mathfrak{g}_{0}}{\mathfrak{m}}_{0} \subseteq \mathfrak{m}\) and \(\comm{\mathfrak{m}}{\mathfrak{m}}_{0} = 0\), and the quotient bracket is that of \(\mathfrak{h}\).

Derives Lemma 14.87. Because \(U(\varepsilon)\) is an isomorphism of Lie algebras from \(\left(\mathfrak{g},\comm{\cdot}{\cdot}_{\varepsilon}\right)\) to \(\mathfrak{g}\), it extends to an isomorphism \(\hat{U}(\varepsilon)\) of the corresponding enveloping algebras, which carries the centre to the centre. Hence \(\comm{C_{\varepsilon}}{T_{\alpha}}_{\varepsilon} = 0\) for every \(\alpha\) and every \(\varepsilon \neq 0\), the scalar \(\lambda(\varepsilon)\) being irrelevant to that.

Reduce both sides to the ordered basis of Theorem 14.7. Every reordering step uses the relations Equation (14.12) for the bracket \(\comm{\cdot}{\cdot}_{\varepsilon}\), so the coefficients of \(\comm{C_{\varepsilon}}{T_{\alpha}}_{\varepsilon}\) in that basis are polynomials in the coefficients of \(C_{\varepsilon}\) and in the structure constants \(C^{\gamma}{}_{\alpha\beta}(\varepsilon)\), with the same polynomials for every \(\varepsilon\) — the degree of \(C_{\varepsilon}\) being fixed. Both arguments converge, the polynomials are continuous, and the value is identically zero; therefore \(\comm{C_{0}}{T_{\alpha}}_{0} = 0\) for every \(\alpha\), which is Equation (14.13) for \(\mathfrak{g}_{0}\).

Derives Proposition 14.98. If \(P \cong M\times G\), then \(x \mapsto (x,e)\) is a global section. Conversely let \(s\) be one and define

It is smooth, satisfies \(\pi\circ\Phi = \operatorname{pr}_{1}\), and is bijective: surjective because \(G\) acts transitively on each fibre, so every \(p\) with \(\pi(p) = x\) is \(s(x)g\) for some \(g\); injective because the action is free. Its inverse is smooth, as one sees in any local trivialization \(\varphi_{i}\): writing \(\varphi_{i}(s(x)) = (x,\sigma_{i}(x))\) with \(\sigma_{i}\) smooth, \(\varphi_{i}\circ\Phi(x,g) = (x,\sigma_{i}(x)g)\), whose inverse \((x,h) \mapsto \left(x,\sigma_{i}(x)^{-1}h\right)\) is smooth because inversion and multiplication in \(G\) are. Hence \(\Phi\) is a diffeomorphism commuting with the right action, and \(P\) is trivial.

Derives Lemma 14.102. For the first, the graded Leibniz rule gives \(\dd\left(\alpha^{a}\wedge\alpha^{b}\right) = \dd\alpha^{a}\wedge\alpha^{b} - \alpha^{a}\wedge\dd\alpha^{b}\), and in the second term a \(1\)-form and a \(2\)-form commute, so \(-\alpha^{a}\wedge\dd\alpha^{b}\comm{\vect{T}_{a}}{\vect{T}_{b}} = -\dd\alpha^{b}\wedge\alpha^{a}\comm{\vect{T}_{a}}{\vect{T}_{b}} = +\dd\alpha^{a}\wedge\alpha^{b}\comm{\vect{T}_{a}}{\vect{T}_{b}}\) after renaming \(a \leftrightarrow b\) and using the antisymmetry of the bracket; the two terms are equal. For the second,

and the \(3\)-form \(\alpha^{c}\wedge\alpha^{a}\wedge\alpha^{b}\) is totally antisymmetric in \((c,a,b)\), so only the totally antisymmetric part of the coefficient survives. That part is the cyclic sum \(C^{d}{}_{ab}C^{e}{}_{cd} + C^{d}{}_{bc}C^{e}{}_{ad} + C^{d}{}_{ca}C^{e}{}_{bd}\), which vanishes: it is the Jacobi identity for \(\mathfrak{g}\) written in the structure constants of Equation (14.5).

Derives Proposition 14.104. For \(v \in T_{p}P\) write \(v = \left(\vect{A}(v)\right)^{\sharp} + hv\) with \(hv\) defined by that equation. The first term is vertical by construction, and \(\vect{A}(hv) = \vect{A}(v) - \vect{A}\left(\left(\vect{A}(v)\right)^{\sharp}\right) = \vect{A}(v) - \vect{A}(v) = 0\) by the first condition of Equation (14.138), so \(hv \in H_{p}\). The sum is direct: a vector in \(V_{p}\cap H_{p}\) is \(X^{\sharp}\) for some \(X\), and \(0 = \vect{A}(X^{\sharp}) = X\) forces it to vanish.

Derives Proposition 14.106. By Equation (14.136), \(\tfrac{1}{2}\comm{\vect{A}}{\vect{A}}(v,w) = \comm{\vect{A}(v)}{\vect{A}(w)}\). By bilinearity it is enough to check Equation (14.140) when each argument is either horizontal or vertical, using Proposition 14.104, and the invariant formula \(\dd\vect{A}(V,W) = V\vect{A}(W) - W\vect{A}(V) - \vect{A}\left(\comm{V}{W}\right)\) for extensions to vector fields. That formula is the coordinate identity \(\dd\alpha(V,W) = \left(\pp_{\mu}\alpha_{\nu} - \pp_{\nu}\alpha_{\mu}\right)V^{\mu}W^{\nu}\) of Section 13.6.4, rearranged: the derivatives of the components of \(V\) and \(W\) cancel against the bracket term.

Both horizontal. Then \(\vect{A}(v) = \vect{A}(w) = 0\), the bracket term vanishes and \(hv = v\), \(hw = w\), so both sides are \(\dd\vect{A}(v,w)\); extending to horizontal fields and using the invariant formula gives Equation (14.141).

Both vertical, \(v = X^{\sharp}\), \(w = Y^{\sharp}\). The bracket term is \(\comm{X}{Y}\). For the fundamental fields \(\comm{X^{\sharp}}{Y^{\sharp}} = \comm{X}{Y}^{\sharp}\), because \(X \mapsto X^{\sharp}\) is the differential of a right action, and \(\vect{A}(X^{\sharp}) = X\) is constant along \(P\), so the invariant formula gives \(\dd\vect{A}(X^{\sharp},Y^{\sharp}) = -\comm{X}{Y}\). The two cancel, and the right-hand side of Equation (14.140) vanishes because \(hv = hw = 0\).

Mixed, \(v\) horizontal and \(w = Y^{\sharp}\). The bracket term vanishes, since \(\vect{A}(v) = 0\). Extend \(v\) to the horizontal lift \(V\) of a vector field on \(M\), which is invariant under every \(R_{g}\) by the second condition of Equation (14.138); then \(\comm{Y^{\sharp}}{V} = 0\), being the derivative along the flow of \(Y^{\sharp}\) — that is, along the action of \(\ee^{tY}\) — of an invariant field. The invariant formula leaves \(V\vect{A}(Y^{\sharp}) = V(Y) = 0\). Both sides vanish, the right-hand one because \(hw = 0\).

Derives Theorem 14.107. Differentiate Equation (14.139). Since the exterior derivative squares to zero, and by the first identity of Equation (14.137),

On the other hand, by the second identity of Equation (14.137),

Finally \(\comm{\vect{A}}{\dd\vect{A}} = A^{a}\wedge\dd A^{b}\comm{\vect{T}_{a}}{\vect{T}_{b}} = \dd A^{b}\wedge A^{a}\comm{\vect{T}_{a}}{\vect{T}_{b}} = -\comm{\dd\vect{A}}{\vect{A}}\), moving the \(1\)-form past the \(2\)-form at no cost in sign and then renaming \(a \leftrightarrow b\). The two contributions cancel.

Derives Theorem 14.110. We prove the four statements in turn.

1. Expanding \(\vect{B}_{1},\ldots,\vect{B}_{r}\) in the basis \(\set{\vect{T}_{A}}\) of the Lie algebra \(\mathfrak{g}\) of the group \(G\) and using the linearity of the polynomial, we have \(\dd\gen{\vect{B}_{1},\ldots,\vect{B}_{r}} = \dd\left(B_{1}^{A_{1}}\cdots B_{r}^{A_{r}}\right) \gen{\vect{T}_{A_{1}},\ldots,\vect{T}_{A_{r}}}\). By the graded Leibniz rule for the exterior derivative,

   \begin{align*} \dd\left(B_{1}^{A_{1}}\cdots B_{r}^{A_{r}}\right) &= \dd B_{1}^{A_{1}}\,B_{2}^{A_{2}}\cdots B_{r}^{A_{r}} + (-1)^{p_{1}}B_{1}^{A_{1}}\,\dd B_{2}^{A_{2}}\,B_{3}^{A_{3}}\cdots B_{r}^{A_{r}}\\ &\quad+ \cdots + (-1)^{p_{1}+\cdots+p_{r-1}}B_{1}^{A_{1}}\cdots B_{r-1}^{A_{r-1}}\, \dd B_{r}^{A_{r}}\\ &= \sum_{l=1}^{r}(-1)^{p_{1}+\cdots+p_{l-1}} B_{1}^{A_{1}}\cdots B_{l-1}^{A_{l-1}}\,\dd B_{l}^{A_{l}}\, B_{l+1}^{A_{l+1}}\cdots B_{r}^{A_{r}}\ec \end{align*}

   and hence

   \begin{align*} &\dd\gen{\vect{B}_{1},\ldots,\vect{B}_{r}}\\ &= \sum_{l=1}^{r}(-1)^{p_{1}+\cdots+p_{l-1}} B_{1}^{A_{1}}\cdots B_{l-1}^{A_{l-1}}\,\dd B_{l}^{A_{l}}\, B_{l+1}^{A_{l+1}}\cdots B_{r}^{A_{r}} \gen{\vect{T}_{A_{1}},\ldots,\vect{T}_{A_{r}}}\\ &= \sum_{l=1}^{r}(-1)^{p_{1}+\cdots+p_{l-1}} \gen{\vect{B}_{1},\ldots,\vect{B}_{l-1},\dd\vect{B}_{l}, \vect{B}_{l+1},\ldots,\vect{B}_{r}}\ep \end{align*}

   The expression Equation (14.147) for the covariant exterior derivative is proved by an entirely analogous procedure, yielding the claim.

2. From the definition of a \(G\)-invariant polynomial we have

   \begin{equation}\tag{14.151} \gen{\vect{T}_{A_{1}},\ldots,\vect{T}_{A_{r}}} = \gen{[g]\vect{T}_{A_{1}},\ldots,[g]\vect{T}_{A_{r}}} = \gen{g\vect{T}_{A_{1}}g^{-1},\ldots,g\vect{T}_{A_{r}}g^{-1}}\ec \end{equation}

   where \(g = \ee^{t^{A}\vect{T}_{A}}\) is an element of the Lie group \(G\). Furthermore, since the generators \(\vect{T}_{A}\) do not depend on the parameters of the infinitesimal transformations, neither does the polynomial \(\gen{\vect{T}_{A_{1}},\ldots,\vect{T}_{A_{r}}}\); therefore \(\left.\frac{\dd}{\dd t^{A}} \gen{\vect{T}_{A_{1}},\ldots,\vect{T}_{A_{r}}}\right|_{t^{A}=0} = 0\), and from Equation (14.151) we arrive at

   \begin{equation}\tag{14.152} \left.\frac{\dd}{\dd t^{A}} \gen{g\vect{T}_{A_{1}}g^{-1},\ldots,g\vect{T}_{A_{r}}g^{-1}} \right|_{t^{A}=0} = 0\ep \end{equation}

   From part 1 — analogously to the procedure used to prove Equation (14.146) — one obtains

   \begin{align} &\frac{\dd}{\dd t^{A}} \gen{g\vect{T}_{A_{1}}g^{-1},\ldots,g\vect{T}_{A_{r}}g^{-1}}\notag\\ &= \sum_{l=1}^{r} \gen{g\vect{T}_{A_{1}}g^{-1},\ldots,g\vect{T}_{A_{l-1}}g^{-1}, \frac{\dd}{\dd t^{A}}\left(g\vect{T}_{A_{l}}g^{-1}\right), g\vect{T}_{A_{l+1}}g^{-1},\ldots,g\vect{T}_{A_{r}}g^{-1}}\ec \tag{14.153} \end{align}

   but \(\left.g\vect{T}_{A}g^{-1}\right|_{t^{A}=0} = \left.\ee^{t^{B}\vect{T}_{B}}\vect{T}_{A}\, \ee^{-t^{C}\vect{T}_{C}}\right|_{t^{A}=0} = \vect{T}_{A}\), and moreover

   \begin{align*} \left.\frac{\dd}{\dd t^{A}} \left(g\vect{T}_{B}g^{-1}\right)\right|_{t^{A}=0} &= \left.\left(\frac{\dd}{\dd t^{A}} \left(\ee^{t^{C}\vect{T}_{C}}\right)\vect{T}_{B}\, \ee^{-t^{E}\vect{T}_{E}} + \ee^{t^{C}\vect{T}_{C}}\vect{T}_{B}\, \frac{\dd}{\dd t^{A}}\left(\ee^{-t^{E}\vect{T}_{E}}\right) \right)\right|_{t^{A}=0}\\ &= \left.\left(\ee^{t^{C}\vect{T}_{C}}\vect{T}_{A}\vect{T}_{B}\, \ee^{-t^{E}\vect{T}_{E}} - \ee^{t^{C}\vect{T}_{C}}\vect{T}_{B}\, \ee^{-t^{E}\vect{T}_{E}}\vect{T}_{A}\right)\right|_{t^{A}=0}\\ &= \comm{\vect{T}_{A}}{\vect{T}_{B}}\\ &= C_{AB}{}^{C}\,\vect{T}_{C}\ec \end{align*}

   so that we finally conclude from Equation (14.152) and Equation (14.153) that the claim holds.

3. In the basis \(\set{\vect{T}_{A}}\) we have

   \begin{align*} &\gen{\vect{B}_{1},\ldots,\vect{B}_{l-1},\comm{\vect{B}}{\vect{B}_{l}}, \vect{B}_{l+1},\ldots,\vect{B}_{r}}\\ &= B_{1}^{A_{1}}\cdots B_{l-1}^{A_{l-1}}\,B^{A}B_{l}^{A_{l}}\, B_{l+1}^{A_{l+1}}\cdots B_{r}^{A_{r}} \gen{\vect{T}_{A_{1}},\ldots,\vect{T}_{A_{l-1}}, \comm{\vect{T}_{A}}{\vect{T}_{A_{l}}}, \vect{T}_{A_{l+1}},\ldots,\vect{T}_{A_{r}}}\\ &= (-1)^{p(p_{1}+\cdots+p_{l-1})} B^{A}B_{1}^{A_{1}}\cdots B_{l-1}^{A_{l-1}}B_{l}^{A_{l}} B_{l+1}^{A_{l+1}}\cdots B_{r}^{A_{r}}\\ &\quad\times \gen{\vect{T}_{A_{1}},\ldots,\vect{T}_{A_{l-1}}, C_{AA_{l}}{}^{B}\vect{T}_{B}, \vect{T}_{A_{l+1}},\ldots,\vect{T}_{A_{r}}}\ec \end{align*}

   where the sign arises from commuting the \(p\)-form components \(B^{A}\) past the forms \(B_{1}^{A_{1}},\ldots,B_{l-1}^{A_{l-1}}\) of total degree \(p_{1}+\cdots+p_{l-1}\). Hence

   \begin{align*} &(-1)^{p(p_{1}+\cdots+p_{l-1})} \gen{\vect{B}_{1},\ldots,\vect{B}_{l-1},\comm{\vect{B}}{\vect{B}_{l}}, \vect{B}_{l+1},\ldots,\vect{B}_{r}}\\ &= B^{A}B_{1}^{A_{1}}\cdots B_{r}^{A_{r}}\,C_{AA_{l}}{}^{B} \gen{\vect{T}_{A_{1}},\ldots,\vect{T}_{A_{l-1}},\vect{T}_{B}, \vect{T}_{A_{l+1}},\ldots,\vect{T}_{A_{r}}}\ec \end{align*}

   so that, summing over all values of \(l = 1, \ldots, r\), part 2 yields the desired result.

4. Since \(\mathrm{D} = \dd + \comm{\vect{A}}{\ \cdot\ }\), from part 1 we see that

   \begin{equation*} \mathrm{D}\gen{\vect{B}_{1},\ldots,\vect{B}_{r}} = \dd\gen{\vect{B}_{1},\ldots,\vect{B}_{r}} + \sum_{l=1}^{r}(-1)^{p_{1}+\cdots+p_{l-1}} \gen{\vect{B}_{1},\ldots,\vect{B}_{l-1},\comm{\vect{A}}{\vect{B}_{l}}, \vect{B}_{l+1},\ldots,\vect{B}_{r}}\ep \end{equation*}

   From part 3 with \(\vect{B} = \vect{A}\) and \(p = 1\) — since the connection is a \(1\)-form — the second term vanishes.

Derives Theorem 14.111. Theorem 14.110 is the principal tool.

1. Consider part 1 of Theorem 14.110, in its covariant form Equation (14.147), for \(\vect{B}_{1},\ldots,\vect{B}_{r} = \vect{F}\), so that \(p_{1},\ldots,p_{r} = 2\). We obtain

   \begin{align} \mathrm{D}\gen{\vect{F},\ldots,\vect{F}} &= \sum_{l=1}^{r}(-1)^{2+\cdots+2} \gen{\vect{F},\ldots,\vect{F},\mathrm{D}\vect{F}, \vect{F},\ldots,\vect{F}}\notag\\ &= \sum_{l=1}^{r} \gen{\vect{F},\ldots,\vect{F},\mathrm{D}\vect{F}, \vect{F},\ldots,\vect{F}}\notag\\ &= r\gen{\vect{F},\ldots,\vect{F},\mathrm{D}\vect{F}, \vect{F},\ldots,\vect{F}}\ep\tag{14.157} \end{align}

   From the Bianchi identity we have \(\mathrm{D}\vect{F} = \dd\vect{F} + \comm{\vect{A}}{\vect{F}} = 0\), and \(\gen{\vect{F},\ldots,\vect{F},0,\vect{F},\ldots,\vect{F}} = 0\) as a direct consequence of multilinearity:

   \begin{align*} \gen{\vect{F},\ldots,\vect{F},0,\vect{F},\ldots,\vect{F}} &= \gen{\vect{F},\ldots,\vect{F},\vect{B}-\vect{B}, \vect{F},\ldots,\vect{F}}\\ &= \gen{\vect{F},\ldots,\vect{F},\vect{B},\vect{F},\ldots,\vect{F}} + \gen{\vect{F},\ldots,\vect{F},-\vect{B},\vect{F},\ldots,\vect{F}}\\ &= \gen{\vect{F},\ldots,\vect{F},\vect{B},\vect{F},\ldots,\vect{F}} - \gen{\vect{F},\ldots,\vect{F},\vect{B},\vect{F},\ldots,\vect{F}}\\ &= 0\ep \end{align*}

   Together with part 4 of Theorem 14.110, which gives \(\dd\gen{\vect{F},\ldots,\vect{F}} = \mathrm{D}\gen{\vect{F},\ldots,\vect{F}} = 0\), this proves the statement.

2. Write \(\vect{\theta} = \vect{A} - \bar{\vect{A}}\), so that \(\vect{A}_{0} = \bar{\vect{A}}\) and \(\vect{A}_{1} = \bar{\vect{A}} + \vect{\theta} = \vect{A}\), and denote by \(\bar{\mathrm{D}} = \dd + \comm{\bar{\vect{A}}}{\ \cdot\ }\) the covariant exterior derivative of the connection \(\bar{\vect{A}}\). The field strength corresponding to \(\vect{A}_{t}\) is given by \(\vect{F}_{t} = \dd\vect{A}_{t} + \frac{1}{2}\comm{\vect{A}_{t}}{\vect{A}_{t}}\), so that \(\vect{F}_{0} = \bar{\vect{F}}\) and \(\vect{F}_{1} = \vect{F}\). We have

   \begin{equation*} \gen{\vect{F},\ldots,\vect{F}} - \gen{\bar{\vect{F}},\ldots,\bar{\vect{F}}} = \gen{\vect{F}_{1},\ldots,\vect{F}_{1}} - \gen{\vect{F}_{0},\ldots,\vect{F}_{0}} = \int_{0}^{1}\frac{\dd}{\dd t} \gen{\vect{F}_{t},\ldots,\vect{F}_{t}}\,\dd t\ep \end{equation*}

   Analogously to the procedure used to obtain Equation (14.146), one finds

   \begin{equation*} \gen{\vect{F},\ldots,\vect{F}} - \gen{\bar{\vect{F}},\ldots,\bar{\vect{F}}} = \int_{0}^{1}\sum_{l=1}^{r} \gen{\vect{F}_{t},\ldots,\vect{F}_{t}, \frac{\dd\vect{F}_{t}}{\dd t}, \vect{F}_{t},\ldots,\vect{F}_{t}}\,\dd t\ep \end{equation*}

   Since the summand does not depend on the index \(l\), and since a \(G\)-invariant polynomial is by definition totally symmetric, we obtain \(r\) equal terms, arriving at

   \begin{equation*} \gen{\vect{F},\ldots,\vect{F}} - \gen{\bar{\vect{F}},\ldots,\bar{\vect{F}}} = r\int_{0}^{1} \gen{\frac{\dd\vect{F}_{t}}{\dd t}, \vect{F}_{t},\ldots,\vect{F}_{t}}\,\dd t\ep \end{equation*}

   To compute the derivative of the interpolating field strength \(\vect{F}_{t}\) with respect to \(t\), observe that

   \begin{align*} \vect{F}_{t} &= \dd\vect{A}_{t} + \frac{1}{2}\comm{\vect{A}_{t}}{\vect{A}_{t}}\\ &= \dd\bar{\vect{A}} + t\,\dd\vect{\theta} + \frac{1}{2}\comm{\bar{\vect{A}}+t\vect{\theta}} {\bar{\vect{A}}+t\vect{\theta}}\\ &= \dd\bar{\vect{A}} + t\,\dd\vect{\theta} + \frac{1}{2}\comm{\bar{\vect{A}}}{\bar{\vect{A}}} + \frac{t}{2}\comm{\bar{\vect{A}}}{\vect{\theta}} + \frac{t}{2}\comm{\vect{\theta}}{\bar{\vect{A}}} + \frac{t^{2}}{2}\comm{\vect{\theta}}{\vect{\theta}}\\ &= \dd\bar{\vect{A}} + \frac{1}{2}\comm{\bar{\vect{A}}}{\bar{\vect{A}}} + t\,\dd\vect{\theta} + t\comm{\bar{\vect{A}}}{\vect{\theta}} + \frac{t^{2}}{2}\comm{\vect{\theta}}{\vect{\theta}}\ec \end{align*}

   that is,

   \begin{equation}\tag{14.158} \vect{F}_{t} = \bar{\vect{F}} + t\,\bar{\mathrm{D}}\vect{\theta} + \frac{t^{2}}{2}\comm{\vect{\theta}}{\vect{\theta}}\ec \end{equation}

   so that \(\frac{\dd\vect{F}_{t}}{\dd t} = \bar{\mathrm{D}}\vect{\theta} + t\comm{\vect{\theta}}{\vect{\theta}}\), and thus

   \begin{equation*} \gen{\vect{F},\ldots,\vect{F}} - \gen{\bar{\vect{F}},\ldots,\bar{\vect{F}}} = r\int_{0}^{1} \gen{\vect{F}_{t},\ldots,\vect{F}_{t}, \bar{\mathrm{D}}\vect{\theta} + t\comm{\vect{\theta}}{\vect{\theta}}, \vect{F}_{t},\ldots,\vect{F}_{t}}\,\dd t\ep \end{equation*}

   Since the polynomial is totally symmetric we may write

   \begin{equation}\tag{14.159} \gen{\vect{F},\ldots,\vect{F}} - \gen{\bar{\vect{F}},\ldots,\bar{\vect{F}}} = r\int_{0}^{1} \gen{\bar{\mathrm{D}}\vect{\theta} + t\comm{\vect{\theta}}{\vect{\theta}}, \vect{F}_{t},\ldots,\vect{F}_{t}}\,\dd t\ep \end{equation}

   On the other hand, applying part 1 of Theorem 14.110 — for the connection \(\bar{\vect{A}}\) — with \(\vect{B}_{1} = \vect{\theta}\) and \(\vect{B}_{2},\ldots,\vect{B}_{r} = \vect{F}_{t}\), so that \(p_{1} = 1\) and \(p_{2},\ldots,p_{r} = 2\), together with part 4, we observe that

   \begin{align} \dd\gen{\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}} &= \bar{\mathrm{D}} \gen{\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}}\notag\\ &= \gen{\bar{\mathrm{D}}\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}} + \sum_{l=2}^{r}(-1)^{1+2+\cdots+2} \gen{\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}, \bar{\mathrm{D}}\vect{F}_{t}, \vect{F}_{t},\ldots,\vect{F}_{t}}\notag\\ &= \gen{\bar{\mathrm{D}}\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}} - (r-1)\gen{\vect{\theta},\bar{\mathrm{D}}\vect{F}_{t}, \vect{F}_{t},\ldots,\vect{F}_{t}}\ep\tag{14.160} \end{align}

   Recall that the difference of two connections is a tensor, and that a connection plus a tensor defines a new connection. In this way \(\vect{A}_{t}\) determines a connection on the bundle, to which corresponds a covariant derivative \(\mathrm{D}_{\vect{A}_{t}} = \dd + \comm{\vect{A}_{t}}{\ \cdot\ }\). For the field strength \(\vect{F}_{t}\) we have the Bianchi identity \(\mathrm{D}_{\vect{A}_{t}}\vect{F}_{t} = 0\), which implies

   \begin{align*} \dd\vect{F}_{t} + \comm{\vect{A}_{t}}{\vect{F}_{t}} &= 0\ec\\ \dd\vect{F}_{t} + \comm{\bar{\vect{A}}+t\vect{\theta}}{\vect{F}_{t}} &= 0\ec\\ \dd\vect{F}_{t} + \comm{\bar{\vect{A}}}{\vect{F}_{t}} + t\comm{\vect{\theta}}{\vect{F}_{t}} &= 0\ec\\ \implies \bar{\mathrm{D}}\vect{F}_{t} &= -t\comm{\vect{\theta}}{\vect{F}_{t}}\ec \end{align*}

   so that

   \begin{equation}\tag{14.161} \dd\gen{\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}} = \gen{\bar{\mathrm{D}}\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}} + t(r-1)\gen{\vect{\theta},\comm{\vect{\theta}}{\vect{F}_{t}}, \vect{F}_{t},\ldots,\vect{F}_{t}}\ep \end{equation}

   Finally, applying part 3 of Theorem 14.110 with \(\vect{B} = \vect{\theta}\), \(\vect{B}_{1} = \vect{\theta}\) and \(\vect{B}_{2},\ldots,\vect{B}_{r} = \vect{F}_{t}\), so that \(p = 1\), \(p_{1} = 1\) and \(p_{2},\ldots,p_{r} = 2\), we obtain

   \begin{align*} \gen{\comm{\vect{\theta}}{\vect{\theta}}, \vect{F}_{t},\ldots,\vect{F}_{t}} + \sum_{l=2}^{r}(-1)^{1\cdot(1+2+\cdots+2)} \gen{\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}, \comm{\vect{\theta}}{\vect{F}_{t}}, \vect{F}_{t},\ldots,\vect{F}_{t}} &= 0\ec\\ \implies \gen{\comm{\vect{\theta}}{\vect{\theta}}, \vect{F}_{t},\ldots,\vect{F}_{t}} - (r-1)\gen{\vect{\theta},\comm{\vect{\theta}}{\vect{F}_{t}}, \vect{F}_{t},\ldots,\vect{F}_{t}} &= 0\ec \end{align*}

   from which we see that Equation (14.161) becomes \(\dd\gen{\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}} = \gen{\bar{\mathrm{D}}\vect{\theta} + t\comm{\vect{\theta}}{\vect{\theta}}, \vect{F}_{t},\ldots,\vect{F}_{t}}\). Substituting this result into Equation (14.159), we obtain

   \begin{align*} \gen{\vect{F},\ldots,\vect{F}} - \gen{\bar{\vect{F}},\ldots,\bar{\vect{F}}} &= r\int_{0}^{1} \dd\gen{\vect{\theta},\vect{F}_{t},\ldots,\vect{F}_{t}}\,\dd t\\ &= \dd\left(r\int_{0}^{1} \gen{\vect{A}-\bar{\vect{A}},\vect{F}_{t},\ldots,\vect{F}_{t}}\, \dd t\right)\\ &= \dd Q^{(2r-1)}(\vect{A},\bar{\vect{A}})\ep \end{align*}

Derives Proposition 14.113. Abbreviate \(\vect{G}_{t} = \vect{F} + \tfrac{1}{2}(t-1)\comm{\vect{A}}{\vect{A}}\), which is the entry appearing in every slot but the first of Equation (14.164). For \(r = 2\) there is one such slot, and linearity gives

and the two integrals are \(1/2\) and \(1/3-1/2 = -1/6\), which is Equation (14.165). For \(r = 3\) there are two such slots, and by total symmetry

so that the three coefficients of Equation (14.166) are \(3\int_{0}^{1}t^{2}\dd t = 1\), \(3\int_{0}^{1}t^{2}(t-1)\dd t = 3\left(\tfrac{1}{4}-\tfrac{1}{3}\right) = -\tfrac{1}{4}\) and \(\tfrac{3}{4}\int_{0}^{1}t^{2}(t-1)^{2}\dd t = \tfrac{3}{4}\left(\tfrac{1}{5}-\tfrac{1}{2}+\tfrac{1}{3}\right) = \tfrac{1}{40}\).

For Equation (14.167), matrix-valued \(1\)-forms obey \(\comm{\vect{A}}{\vect{A}} = 2\,\vect{A}\wedge\vect{A}\) by Equation (14.135), so \(\gen{\vect{A},\vect{F}} = \tr\left(\vect{A}\wedge\dd\vect{A}\right) + \tr\left(\vect{A}\wedge\vect{A}\wedge\vect{A}\right)\) and \(\tfrac{1}{6}\gen{\vect{A},\comm{\vect{A}}{\vect{A}}} = \tfrac{1}{3}\tr\left(\vect{A}\wedge\vect{A}\wedge\vect{A}\right)\).

Derives Proposition 14.114. Pullback commutes with \(\dd\) and with the bracket Equation (14.135), and \(g\) is a fixed group element, so

using Equation (14.138) in the first step. Hence \(R_{g}^{\ast}\gen{\vect{F},\ldots,\vect{F}} = \gen{g^{-1}\vect{F}g,\ldots,g^{-1}\vect{F}g} = \gen{\vect{F},\ldots,\vect{F}}\) by Definition 14.109: the form is invariant. It is also horizontal, being a wedge of the component forms of \(\vect{F}\), each of which vanishes as soon as one argument is vertical by Proposition 14.106.

An invariant horizontal form descends. Define \(\rho_{x}(v_{1},\ldots,v_{2r}) = \gen{\vect{F},\ldots,\vect{F}}_{p} \left(\tilde{v}_{1},\ldots,\tilde{v}_{2r}\right)\) for any \(p \in \pi^{-1}(x)\) and any \(\tilde{v}_{i}\) with \(\dd\pi\,\tilde{v}_{i} = v_{i}\). Two lifts at the same \(p\) differ by vertical vectors, which horizontality kills, so the value does not depend on the choice; two points of the same fibre differ by some \(g\), and invariance makes the two values agree. Smoothness follows by computing \(\rho\) on a local section, \(\rho|_{U} = s^{\ast} \gen{\vect{F},\ldots,\vect{F}}\). Uniqueness holds because \(\dd\pi\) is onto. Finally \(\dd\rho\) pulls back to \(\dd\gen{\vect{F},\ldots,\vect{F}} = 0\) by Theorem 14.111, and \(\pi^{\ast}\) is injective, so \(\rho\) is closed.

Derives Lemma 14.116. Invariance is \(\det\left(\identity+\lambda\,g\vect{X}g^{-1}\right) = \det\left(g\left(\identity+\lambda\vect{X}\right)g^{-1}\right) = \det\left(\identity+\lambda\vect{X}\right)\), true coefficient by coefficient in \(\lambda\); each \(\sigma_{r}\) is a homogeneous polynomial of degree \(r\) in the entries of \(\vect{X}\), and the associated symmetric multilinear form is its polarization. For antisymmetric \(\vect{X}\),

so the polynomial is even in \(\lambda\) and the odd \(\sigma_{r}\) vanish. For the Pfaffian, an orthogonal \(g\) has \(g\vect{X}g^{-1} = g\vect{X}g\transpose\), each factor in Equation (14.170) carries two of the \(g\)'s, and

which is the definition of the determinant. The stated transformation follows, and \(\det g = \pm1\) for an orthogonal matrix.

Derives Proposition 14.117. In the defining representation the generators are the matrices \(\left(J_{AB}\right)^{M}{}_{N} = \delta^{M}_{A}\eta_{BN} - \delta^{M}_{B}\eta_{AN}\) already used in Proposition 14.22, so

which is symmetric under the exchange of the index pairs and invariant, being a trace of a product of generators. Contracting,

the two \(2\)-forms commuting. For the second equality in Equation (14.171): the elements of \(\mathfrak{so}(p,q)\) are traceless, so \(\sigma_{1}(\vect{X}) = \tr\vect{X}\) vanishes, and expanding Equation (14.169) to second order gives \(\sigma_{2}(\vect{X}) = \tfrac{1}{2}\left[(\tr\vect{X})^{2} - \tr(\vect{X}\vect{X})\right] = -\tfrac{1}{2}\tr(\vect{X}\vect{X})\); at \(\vect{X} = \vect{R}/2\pi\), the value prescribed for the Pontryagin forms in Definition 14.115, this is \(-\gen{\vect{R},\vect{R}}/8\pi^{2}\). The same substitution in the matrix components gives \(\left(\vect{\omega}\right)^{M}{}_{N} = \omega^{M}{}_{N}\), so Equation (14.167) reads \(Q^{(3)} = \omega^{M}{}_{N}\wedge\dd\omega^{N}{}_{M} + \tfrac{2}{3}\omega^{M}{}_{N}\wedge\omega^{N}{}_{P} \wedge\omega^{P}{}_{M}\), and lowering the indices in the first term turns \(\omega^{M}{}_{N}\wedge\dd\omega^{N}{}_{M}\) into \(-\omega^{AB}\wedge\dd\omega_{AB}\). That \(\dd Q^{(3)}\) is \(\gen{\vect{R},\vect{R}}\) is Definition 14.112.