Expansions of Lie Algebras

Derivation. Derives Theorem 15.4. Well-definedness. The assignment \((a,X,b,Y)\mapsto(ab)\otimes\comm{X}{Y}\) is linear in each of its four arguments, because multiplication in \(A\) and the bracket of \(\mathfrak{g}\) are bilinear. A multilinear map factors uniquely through the tensor product in each pair of arguments, so it descends to a bilinear map \(\mathfrak{g}_{A}\times\mathfrak{g}_{A}\to\mathfrak{g}_{A}\), which is Equation (15.2). Substituting basis elements gives Equation (15.3), and the dimension count is that of a tensor product of vector spaces.

Antisymmetry. Using commutativity of \(A\) and antisymmetry of the bracket of \(\mathfrak{g}\),

Both hypotheses are used, and only these: associativity plays no part here.

Jacobi identity. Applying Equation (15.2) twice,

Because \(A\) is associative and commutative, the three products obtained by cycling \((a,X)\to(b,Y)\to(c,Z)\to(a,X)\) are equal,

so the common factor comes out of the cyclic sum of Equation (15.4):

the parenthesis vanishing by the Jacobi identity of \(\mathfrak{g}\). Since elementary tensors span \(\mathfrak{g}_{A}\) and both sides are trilinear, the identity holds throughout \(\mathfrak{g}_{A}\).

Derivation. Derives Proposition 15.6. On elementary tensors,

using that \(f\) is multiplicative; both sides are bilinear, so the identity extends. The kernel is \(\ker(f)\otimes\mathfrak{g}\) because tensoring the exact sequence \(0\to\ker f\to A\to\im f\to0\) with a vector space over a field is exact; surjectivity is inherited for the same reason.

Derivation. Derives Proposition 15.7. For \(u\in I\), \(b\in A\) and \(X,Y\in\mathfrak{g}\), \(\comm{u\otimes X}{b\otimes Y}=(ub)\otimes\comm{X}{Y}\) with \(ub\in I\), because an ideal absorbs multiplication. Bilinearity extends the containment, so \(\comm{I\otimes\mathfrak{g}}{\mathfrak{g}_{A}}\subseteq I\otimes\mathfrak{g}\). If \(A\) is not a field it has a proper nonzero ideal \(I\), and then \(I\otimes\mathfrak{g}\) is a proper nonzero ideal of \(\mathfrak{g}_{A}\).

Derivation. Derives Proposition 15.8. Write \(\mathfrak{s}=J\otimes\mathfrak{g}\), an ideal by Proposition 15.7. By Equation (15.2),

so by induction \(\mathfrak{s}^{(k)}\subseteq J^{2^{k}}\otimes\mathfrak{g}\). Choosing \(k\) with \(2^{k}>N\) gives \(J^{2^{k}}=0\) and hence \(\mathfrak{s}^{(k)}=0\): \(\mathfrak{s}\) is solvable. When \(\mathfrak{g}\) is perfect the second inclusion in Equation (15.6) is an equality, and the first is too, because \(J^{2m}\) is spanned by products \(uv\) with \(u,v\in J^{m}\); the induction then gives equality at every step.

Derivation. Derives Theorem 15.10. By Equation (15.2), the adjoint action of \(a\otimes X\) sends \(c\otimes Z\) to \((ac)\otimes\comm{X}{Z}\), that is

as an operator on \(A\otimes\mathfrak{g}\). Composing two of them,

where \(L_{a}L_{b}=L_{ab}\) is associativity of \(A\). The trace of a tensor product of operators is the product of the traces, which gives Equation (15.8).

For the second statement, Equation (15.8) exhibits \(\kappa_{A}\) as the tensor product of the bilinear forms \((\cdot,\cdot)_{A}\) on \(A\) and \(\kappa_{\mathfrak{g}}\) on \(\mathfrak{g}\). A tensor product of two bilinear forms is nondegenerate exactly when both factors are: in bases its Gram matrix is the Kronecker product of the two Gram matrices, of determinant \((\det G_{A})^{\dim\mathfrak{g}}(\det G_{\mathfrak{g}})^{\dim A}\).

Derivation. Derives Corollary 15.11. Cartan's criterion states that a finite-dimensional Lie algebra over a field of characteristic zero is semisimple if and only if its Killing form is nondegenerate; apply it to both sides of Theorem 15.10. For the last sentence: if \(x\in\operatorname{rad}(A)\) is nilpotent then \(L_{xy}\) is nilpotent for every \(y\), so \(\tau_{A}(xy)=0\) and \(x\) lies in the radical of the trace form; conversely \(\R^{r}\times\C^{s}\) has trace form \(\diag(1,\dots,1)\) on the real factors and, on each complex factor in the basis \(\set{1,\ii}\), the Gram matrix

both nondegenerate. The identification of the radical-free case with \(\R^{r}\times\C^{s}\) is Lemma 15.12.

Derivation. Derives Lemma 15.12. The argument is elementary commutative algebra and uses nothing else from this chapter. Write \(n=\dim A<\infty\) and assume \(A\neq0\).

Step 1: the nilpotents form a nilpotent ideal. Let \(N\subseteq A\) be the set of nilpotent elements. If \(u^{r}=v^{t}=0\) then \((u+v)^{r+t-1}=0\), since every term of the binomial expansion — which is available because \(A\) is commutative — carries either \(u^{r}\) or \(v^{t}\); and \((cu)^{r}=c^{r}u^{r}=0\) for every \(c\in A\). So \(N\) is an ideal. Pick a basis \(u_{1},\dots,u_{d}\) of \(N\), necessarily consisting of nilpotents, with \(u_{i}^{r_{i}}=0\), and set \(M:=1+\sum_{i}(r_{i}-1)\). Then \(N^{M}\) is spanned by products of \(M\) of the \(u_{i}\), and in any such product some \(u_{i}\) occurs at least \(r_{i}\) times; hence

Step 2: either \(A\) is local, or it splits. Fix \(a\in A\) and let \(p\in\R[X]\) be its minimal polynomial, the monic generator of the kernel of the evaluation \(\R[X]\to A\), \(X\mapsto a\); that kernel is nonzero because \(1,a,\dots,a^{n}\) are \(n+1\) vectors in an \(n\)-dimensional space. Factor \(p=p_{1}^{e_{1}}\cdots p_{s}^{e_{s}}\) into powers of distinct monic irreducibles of \(\R[X]\), and write \(q_{i}:=p/p_{i}^{e_{i}}\).

Suppose \(s\ge2\). The \(q_{i}\) have no common irreducible factor, so there are \(f_{i}\in\R[X]\) with \(\sum_{i}f_{i}q_{i}=1\); put \(\pi_{i}:=(f_{i}q_{i})(a)\). Then \(\sum_{i}\pi_{i}=1_{A}\); for \(i\neq j\) the product \(f_{i}q_{i}f_{j}q_{j}\) is divisible by \(p\), so \(\pi_{i}\pi_{j}=0\), and multiplying \(\sum_{j}\pi_{j}=1_{A}\) by \(\pi_{i}\) gives \(\pi_{i}^{2}=\pi_{i}\). No \(\pi_{i}\) vanishes: \(f_{i}q_{i}\equiv1\) modulo \(p_{i}^{e_{i}}\), because every other summand of \(\sum_{j}f_{j}q_{j}=1\) is divisible by \(p_{i}^{e_{i}}\); if \(\pi_{i}=0\) then \(p\), hence \(p_{i}^{e_{i}}\), would divide \(f_{i}q_{i}\), forcing \(p_{i}^{e_{i}}\) to divide \(1\). So the \(\pi_{i}\) are nonzero orthogonal idempotents summing to \(1_{A}\), the sum \(A=\bigoplus_{i}\pi_{i}A\) is direct (multiply a vanishing sum by \(\pi_{j}\)), and \(\pi_{i}A\cdot\pi_{j}A=0\) for \(i\neq j\): \(A\) is the product of the algebras \(\pi_{i}A\), each unital with unit \(\pi_{i}\) and each of dimension strictly less than \(n\).

Suppose instead that \(s=1\) for every \(a\in A\), so that each minimal polynomial is \(p_{1}^{e_{1}}\) with \(p_{1}\) irreducible. If \(p_{1}=X\) then \(a^{e_{1}}=0\) and \(a\in N\). Otherwise \(p_{1}(0)\neq0\) — an irreducible real polynomial with vanishing constant term is a multiple of \(X\) — so \(p\) has constant term \(c_{0}=p_{1}(0)^{e_{1}} \neq0\); writing \(p(X)=Xg(X)+c_{0}\) and evaluating at \(a\) gives \(a\cdot\left(-g(a)/c_{0}\right)=1_{A}\), so \(a\) is invertible. Thus every element is nilpotent or invertible. A nilpotent is not invertible, since \(a^{r}=0\) and \(ab=1_{A}\) would give \(1_{A}=(ab)^{r}=a^{r}b^{r}=0\); hence \(N\) is exactly the set of non-units. An ideal that contains every non-unit and consists of non-units is the unique maximal ideal, so \(A\) is local with \(\mathfrak{m}=N\), nilpotent by Equation (15.11).

Step 3: induction. If \(\dim A=1\) then \(A=\R\,1_{A}\) is a field, hence local with \(\mathfrak{m}=0\). If \(\dim A>1\), Step 2 either makes \(A\) local or splits it into factors of strictly smaller dimension, to which the inductive hypothesis applies. An idempotent of a factor is an idempotent of \(A\), and idempotents belonging to different factors are orthogonal, so the partitions of the units compose into a single partition \(1_{A}=\pi_{1}+\cdots+\pi_{k}\) with \(A_{i}=\pi_{i}A\) local.

Step 4: uniqueness. A local algebra has no idempotents but \(0\) and its unit: if \(e^{2}=e\) then \(e\) is a unit, whence \(e=1\) after multiplying \(e^{2}=e\) by \(e^{-1}\), or \(e\) is nilpotent, whence \(e=e^{m}=0\). Let \(1_{A}=\sum_{i}\pi_{i}=\sum_{j}\rho_{j}\) be the unit partitions of two decompositions into local factors. Each \(\pi_{i} \rho_{j}\) is an idempotent lying in \(\pi_{i}A\) and in \(\rho_{j}A\), so it is \(0\) or \(\pi_{i}\), and \(0\) or \(\rho_{j}\). Since \(\sum_{j}\pi_{i}\rho_{j}=\pi_{i}\neq0\), at least one \(j\) gives \(\pi_{i}\rho_{j}=\pi_{i}=\rho_{j}\); a second such \(j'\) is impossible, because then \(\pi_{i}=\pi_{i}\rho_{j}\) multiplied by \(\rho_{j'}\) would give \(\pi_{i}\rho_{j'}=\pi_{i}\rho_{j}\rho_{j'}=0\). The two partitions therefore coincide up to order, and with them the factors.

Step 5: the radical-free case. Suppose \(\mathfrak{m}_{i}=0\). Then the non-units of \(A_{i}\) are \(0\) alone, so \(A_{i}\) is a field \(K\), finite dimensional over \(\R\,1_{K}\cong\R\). For \(b\in K\) the minimal polynomial \(p_{b}\) is irreducible: a factorisation \(p_{b}=gh\) into nonconstant factors would give \(g(b)h(b)=0\) in a field, hence \(g(b)=0\) or \(h(b)=0\), contradicting minimality. By the fundamental theorem of algebra (Theorem 8.19) a real polynomial splits over \(\C\) into linear factors, and its non-real roots occur in conjugate pairs; pairing those factors gives a factorisation into real factors of degree \(1\) and \(2\), so an irreducible real polynomial has degree at most \(2\). If \(K\neq\R\,1_{K}\), choose \(b\in K\) outside \(\R\,1_{K}\): its minimal polynomial is an irreducible quadratic, and \(\R[b]\cong\R[X]/(p_{b})\cong\C\). Every \(c\in K\) then has a minimal polynomial over that copy of \(\C\) which is irreducible in \(\C[X]\), hence linear by Theorem 8.19, so \(c\in\R[b]\) and \(K\cong\C\). Otherwise \(K\cong\R\). Collecting the factors gives \(A\cong\R^{r}\times\C^{s}\).

Derivation. Derives Theorem 15.13. The splitting. With the orthogonal idempotents of Lemma 15.12, \(\pi_{i}\pi_{j}=\delta_{ij}\pi_{i}\), so Equation (15.2) gives

so the subspaces \(\pi_{i}A\otimes\mathfrak{g}=\mathfrak{g}_{A_{i}}\) are mutually commuting ideals whose sum is \(\mathfrak{g}_{A}\), which is Equation (15.12).

Branch (1). If \(A_{i}\cong\R\) then \(A_{i}\otimes\mathfrak{g}\cong \mathfrak{g}\) with the same bracket, since \(1\cdot1=1\). If \(A_{i}\cong\C\) then \(A_{i}\otimes_{\R}\mathfrak{g}\) is by definition the complexification of \(\mathfrak{g}\), a real Lie algebra of twice the dimension. Both are semisimple by Corollary 15.11, the trace forms of \(\R\) and \(\C\) over \(\R\) being nondegenerate as computed there.

Branch (2). By Proposition 15.8, \(\mathfrak{m}_{i}\otimes\mathfrak{g}\) is a nonzero solvable ideal, so \(\mathfrak{g}_{A_{i}}\) is not semisimple. Applying Proposition 15.6 to the quotient map \(A_{i}\to A_{i}/\mathfrak{m}_{i}\) gives a surjection \(\mathfrak{g}_{A_{i}}\to\mathfrak{g}_{A_{i}/\mathfrak{m}_{i}}\) with kernel \(\mathfrak{m}_{i}\otimes\mathfrak{g}\). Since \(A_{i}\) is local, \(A_{i}/\mathfrak{m}_{i}\) is a field, so the image is semisimple by branch (1) and its own radical is zero. Any solvable ideal of \(\mathfrak{g}_{A_{i}}\) therefore maps into that zero radical, hence lies in \(\mathfrak{m}_{i}\otimes\mathfrak{g}\); together with the solvability of \(\mathfrak{m}_{i}\otimes\mathfrak{g}\) itself this identifies the radical exactly.

Derivation. Derives Proposition 15.15. The bracket is Equation (15.2) with \(a=\varepsilon^{i}\), \(b=\varepsilon^{j}\) and the multiplication table of \(A_{N}\). Level \(0\) is \(1\otimes\mathfrak{g}\), closed because \(1\cdot1=1\), and \(X\mapsto X^{(0)}\) is an isomorphism onto it. The maximal ideal \(\mathfrak{m}=(\varepsilon)\) of \(A_{N}\) satisfies \(\mathfrak{m}^{N+1}=0\), so \(\mathfrak{m}\otimes\mathfrak{g}\) is a nilpotent, in particular solvable, ideal by Proposition 15.8.

Derivation. Derives Lemma 15.19. For \(X\in V_{i}\), \(Y\in V_{j}\) one has \(\comm{X}{Y}\in V_{i+j}\) by Equation (15.17), so \(\theta\comm{X}{Y}=(-1)^{i+j}\comm{X}{Y} =\comm{(-1)^{i}X}{(-1)^{j}Y}=\comm{\theta X}{\theta Y}\): \(\theta\) is an automorphism, and \(\theta^{2}=\id\). The Killing form is invariant under every automorphism, so for \(X\in V_{0}\), \(Y\in V_{1}\),

which forces Equation (15.18) in characteristic zero. If \(X\in V_{0}\) satisfies \(\kappa_{\mathfrak{g}}(X,V_{0})=0\) then by Equation (15.18) also \(\kappa_{\mathfrak{g}}(X,V_{1})=0\), so \(X\) lies in the radical of \(\kappa_{\mathfrak{g}}\) and vanishes; likewise for \(V_{1}\).

Derivation. Derives Proposition 15.21. Take \(X\in V_{[i]}\) and \(Y\in V_{[j]}\), so \(\varepsilon^{i}\otimes X\) and \(\varepsilon^{j}\otimes Y\) lie in \(\mathfrak{R}_{N}^{\sigma}\). By Equation (15.15) their bracket is \(\varepsilon^{i+j}\otimes\comm{X}{Y}\) when \(i+j\le N\) and zero otherwise; the zero case is trivially in the subspace. In the surviving case Equation (15.17) gives \(\comm{X}{Y}\in V_{[i]+[j]}\), and \([i]+[j]=[i+j]\) because parity is additive, so the bracket lies in \(\varepsilon^{i+j}\otimes V_{[i+j]}\subseteq\mathfrak{R}_{N}^{\sigma}\). The dimension is the number of even levels in \(\set{0,\dots,N}\) times \(\dim V_{0}\) plus the number of odd ones times \(\dim V_{1}\).

Derivation. Derives Proposition 15.23. Each bracket is Equation (14.90) with the indices substituted. For instance

using \(J_{DD}=0\) and \(\eta_{iD}=\eta_{jD}=0\); and

The remaining four are the same computation. The identification of \((\gamma,\mu,\nu)\) is read off by comparing with Equation (14.128), whose second and third relations are \(\comm{G_{i}}{H}=\mu P_{i}\) and \(\comm{P_{i}}{H}=\nu G_{i}\).

Derivation. Derives Proposition 15.25. Read Equations (15.22) and (15.23). For \(\sigma_{\mathrm{st}}\), the set \(V_{0}=\gen{J_{ij},G_{i}}\) is the Lorentz subalgebra, so \(\comm{V_{0}}{V_{0}}\subseteq V_{0}\); \(\comm{G_{i}}{H}=P_{i}\) and \(\comm{J}{P}\sim P\) give \(\comm{V_{0}}{V_{1}}\subseteq V_{1}\); and \(\comm{P_{i}}{P_{j}}=-\varsigma J_{ij}\), \(\comm{P_{i}}{H}=\varsigma G_{i}\) give \(\comm{V_{1}}{V_{1}}\subseteq V_{0}\).

For \(\sigma_{\mathrm{ss}}\): \(\comm{J}{J}\sim J\) and \(\comm{J_{ij}}{H}=0\) close \(V_{0}\); \(\comm{J}{G}\sim G\), \(\comm{J}{P}\sim P\), \(\comm{H}{G_{i}}=-P_{i}\) and \(\comm{H}{P_{i}}=-\varsigma G_{i}\) keep \(\comm{V_{0}}{V_{1}}\) in \(V_{1}\); and \(\comm{G_{i}}{G_{j}}=J_{ij}\), \(\comm{G_{i}}{P_{j}}=\delta_{ij}H\), \(\comm{P_{i}}{P_{j}}=-\varsigma J_{ij}\) all land in \(V_{0}\).

For \(\sigma_{\mathrm{sT}}\): \(\comm{P_{i}}{P_{j}}=-\varsigma J_{ij}\) closes \(V_{0}\); \(\comm{P_{i}}{H}=\varsigma G_{i}\) and \(\comm{P_{j}}{G_{i}}=-\delta_{ij}H\) send \(\comm{V_{0}}{V_{1}}\) into \(V_{1}\); and \(\comm{G_{i}}{G_{j}}=J_{ij}\), \(\comm{G_{i}}{H}=P_{i}\) land in \(V_{0}\).

For \(\sigma_{\mathrm{gen}}\) the third condition fails at once: \(\comm{G_{i}}{P_{j}}=\delta_{ij}H\in V_{1}\), not \(V_{0}\).

Derivation. Derives Proposition 15.26. Compute the three cases of Equation (14.125) using Equation (15.17). For \(X,X'\in V_{0}\) the map \(U(\lambda)\) is the identity on the arguments and, since \(\comm{X}{X'}\in V_{0}\), on the result too, so \(U^{-1}\comm{UX}{UX'}=\comm{X}{X'}\), independent of \(\lambda\). For \(X\in V_{0}\), \(Y\in V_{1}\) we have \(\comm{X}{Y}\in V_{1}\), so

again independent of \(\lambda\). For \(Y,Y'\in V_{1}\) we have \(\comm{Y}{Y'}\in V_{0}\), on which \(U^{-1}\) acts trivially, so

The limit exists and \(\mathfrak{g}_{0}\) has the brackets of \(\mathfrak{g}\) except that \(\comm{V_{1}}{V_{1}}=0\).

Define \(\varphi:\mathfrak{g}_{0}\to\mathfrak{R}_{1}^{\sigma}\) by \(\varphi(X)=1\otimes X\) for \(X\in V_{0}\) and \(\varphi(Y)=\varepsilon\otimes Y\) for \(Y\in V_{1}\), extended linearly; it is a linear bijection, both spaces having dimension \(\dim V_{0}+\dim V_{1}\). It is a homomorphism by Equation (15.15): the \(V_{0}\)–\(V_{0}\) bracket lands at level \(0\) with value in \(V_{0}\), the \(V_{0}\)–\(V_{1}\) bracket at level \(1\) with value in \(V_{1}\), and the \(V_{1}\)–\(V_{1}\) bracket at level \(2>1\), hence zero — matching the three cases just computed.

Derivation. Derives Corollary 15.27. For \(\sigma_{\mathrm{st}}\), Proposition 15.26 sets \(\comm{V_{1}}{V_{1}}=0\), that is \(\comm{P_{i}}{P_{j}}= \comm{P_{i}}{H}=0\) by Equation (15.23), leaving the Lorentz algebra acting on an abelian ideal of translations: this is Equation (14.99). For \(\sigma_{\mathrm{ss}}\) applied to the resulting Poincaré algebra, \(\comm{V_{1}}{V_{1}}=0\) now reads \(\comm{G_{i}}{G_{j}}=\comm{G_{i}}{P_{j}}=\comm{P_{i}}{P_{j}}=0\) while \(\comm{G_{i}}{H}=P_{i}\) survives, which is the Galilei algebra \((\gamma,\mu,\nu)=(0,1,0)\) of Table 14.1.

Derivation. Derives Proposition 15.28. \(t>0\). Put \(\pi_{\pm}:=\tfrac{1}{2}(1\pm x/\sqrt{t})\). Using \(x^{2}=t\),

and \(\pi_{+}+\pi_{-}=1\): two orthogonal idempotents, so \(A^{(t)}\) splits into two copies of \(\R\).

\(t<0\). Put \(\jmath:=x/\sqrt{-t}\); then \(\jmath^{2}=x^{2}/(-t)=-1\), so \(\R[\jmath]\cong\C\), and \(A^{(t)}\cong\C\) by dimension count.

\(t=0\). The relation is \(x^{2}=0\): the dual numbers verbatim.

For the trace form, \(L_{1}=\id\) has trace \(2\), \(L_{x}\) interchanges \(1\mapsto x\mapsto tx^{0}\) and has trace \(0\), and \(L_{x^{2}}=L_{t}\) has trace \(2t\); the Gram matrix of \(\tau(ab)\) in \(\set{1,x}\) is therefore \(\diag(2,2t)\), of determinant \(4t\). Corollary 15.11 then gives semisimplicity of \(\mathfrak{g}_{A^{(t)}}\) for \(t\neq0\) and its failure at \(t=0\).

Derivation. Derives Theorem 15.29. Closure and brackets. For \(J,\tilde{J}\in V_{0}\) and \(P,\tilde{P}\in V_{1}\) write \(J':=1\otimes J\) and \(P':=x\otimes P\). Then Equation (15.2) gives

the last using \(x^{2}=t\) together with \(\comm{V_{1}}{V_{1}}\subseteq V_{0}\). All three stay in \(\mathfrak{R}^{(t)}\), so it is a subalgebra; and since \(\comm{P}{\tilde{P}}\) is given by Equation (15.23) with \(\varsigma=-1\), the brackets of the primed generators are those of Equations (15.22) and (15.23) with \(\varsigma\) replaced by \(-t\). Explicitly

with all other brackets independent of \(t\).

The three cases. For \(t>0\) rescale \(P_{i}'\mapsto P_{i}'/\sqrt{t}\), \(H'\mapsto H'/\sqrt{t}\): an invertible change of basis which turns Equation (15.31) into \(\comm{P}{P}=J\), \(\comm{P}{H}=-G\), that is Equation (15.23) with \(\varsigma=-1\), the anti-de Sitter algebra \(\mathfrak{so}(D-1,2)\). For \(t<0\) rescale by \(\sqrt{-t}\) instead, obtaining \(\comm{P}{P}=-J\), \(\comm{P}{H}=+G\), which is Equation (15.23) with \(\varsigma=+1\): by Proposition 15.23 this is the algebra with \(\eta_{DD}=+1\), namely \(\mathfrak{so}(D,1)\). For \(t=0\) both brackets in Equation (15.31) vanish and the translations become an abelian ideal, giving \(\mathfrak{iso}(D-1,1)\) — which is also Corollary 15.27, since \(A^{(0)}\) is the algebra of dual numbers and \(\mathfrak{R}^{(0)}=\mathfrak{R}_{1}^{\sigma_{\mathrm{st}}}\).

Derivation. Derives Proposition 15.31. Trace Equation (15.33) with \(g^{\mu\nu}\), using \(g^{\mu\nu}g_{\mu\nu}=D\):

valid for \(D\neq2\). Substituting back,

From Equation (15.33) (trace with \(g^{\mu\nu}\) and substitute the resulting scalar curvature back). Contracting Equation (15.34) with \(g^{\mu\rho}\),

Comparing with Equation (15.37) gives Equation (15.35), and Equation (15.36) follows on setting \(\abs{K}=1/\ell^{2}\). The isometry algebras are those of the hyperboloids \(\eta_{AB}X^{A}X^{B}=\pm\ell^{2}\) in \(\R^{D,1}\) and \(\R^{D-1,2}\) respectively, which is how \(\mathfrak{so}(D,1)\) and \(\mathfrak{so}(D-1,2)\) were introduced in Lie Groups, Lie Algebras, and Fibre Bundles.

Derivation. Derives Corollary 15.32. Theorem 15.29 makes \(\mathfrak{R}^{(t)}\) the anti-de Sitter algebra for \(t>0\) and the de Sitter algebra for \(t<0\); Proposition 15.31 attaches \(\Lambda<0\) to the first and \(\Lambda>0\) to the second, with \(\abs{\Lambda}=(D-1)(D-2)/(2\ell^{2})\) in both cases. Since the rescaling used in the proof of Theorem 15.29 sends \(\abs{t}\) to \(1\), the parameter \(t\) carries the units of \(\abs{K}=1/\ell^{2}\) and the sign that distinguishes the branches, which is Equation (15.38) once Equation (15.35) is used.

Derivation. Derives Theorem 15.34. (1) By Proposition 15.6, \(\pi_{N}\otimes\id\) is a Lie homomorphism \(\mathfrak{g}\gen{N+1}\to\mathfrak{g}\gen{N}\); it maps \(\varepsilon^{k}\otimes V_{[k]}\) identically onto itself for \(k\le N\) and kills the top level, so it carries \(\mathfrak{R}_{N+1}^{\sigma}\) onto \(\mathfrak{R}_{N}^{\sigma}\) with kernel Equation (15.40). That kernel is abelian because

since \(2N+2>N+1\) for every \(N\ge0\).

(2) Let \(X\in V_{[N+1]}\) and let \(\varepsilon^{j}\otimes Y\) be an arbitrary element of \(\mathfrak{R}_{N+1}^{\sigma}\), so \(Y\in V_{[j]}\). Then

which vanishes automatically for every \(j\ge1\), since then \(N+1+j>N+1\). Only \(j=0\) can survive; there \(Y\in V_{0}\) and Equation (15.44) equals \(\varepsilon^{N+1}\otimes\comm{X}{Y}\). By Equation (15.17), \(\comm{X}{Y}\in V_{[N+1]}\), and \(\varepsilon^{N+1}\otimes Z\) vanishes for \(Z\in V_{[N+1]}\) only if \(Z=0\). So the bracket vanishes for all \(Y\) exactly when \(\comm{X}{V_{0}}=0\).

(3) The sequence Equation (15.42) is exact by (1) and central by (2), so it is a central extension. Define the splitting \(s:\mathfrak{R}_{N}^{\sigma}\to\mathfrak{R}_{N+1}^{\sigma}\) by \(s(\varepsilon^{i}\otimes X)=\varepsilon^{i}\otimes X\) for \(i\le N\), a linear section of \(p_{N}\). For \(i,j\le N\),

in \(\mathfrak{R}_{N+1}^{\sigma}\), whereas the same bracket computed in \(\mathfrak{R}_{N}^{\sigma}\) and then lifted by \(s\) gives \(\varepsilon^{i+j}\otimes\comm{X}{Y}\) when \(i+j\le N\) and \(0\) when \(i+j>N\). The two agree except when \(i+j=N+1\), where the difference is \(\varepsilon^{N+1}\otimes\comm{X}{Y}\in\mathfrak{k}_{N}\); for \(i+j>N+1\) both vanish. Comparing with Equation (14.120), the defect is the cocycle Equation (15.43), which satisfies Equation (14.121) automatically by Proposition 14.73.

Derivation. Derives Corollary 15.35. Apply Theorem 15.34(2) with \(N\) in place of \(N+1\): an element \(\varepsilon^{N}\otimes X\) with \(X\in V_{[N]}\) is central precisely when \(\comm{X}{V_{0}}=0\). For \(N\) even, \(V_{[N]}=V_{0}\) and the condition defines \(\mathfrak{z}(V_{0})\); for \(N\) odd, \(V_{[N]}=V_{1}\) and it defines \(V_{1}^{V_{0}}\). For the last sentence, in each grading of Equation (15.24) the space \(V_{1}\) is spanned by generators carrying a free spatial index, on which the rotations \(J_{ij}\subseteq V_{0}\) act as on the vector representation of \(\mathfrak{so}(D-1)\); that representation has no trivial summand for \(D\ge3\), so \(V_{1}^{V_{0}}=0\).

Derivation. Derives Corollary 15.36. The grading. The Poincaré brackets are Equations (15.22) and (15.23) with \(\varsigma=0\), that is \(\comm{P_{i}}{P_{j}}=\comm{P_{i}}{H}=0\), the rest unchanged (Corollary 15.27). Then \(\comm{G_{i}}{G_{j}}=J_{ij}\), \(\comm{G_{i}}{P_{j}}=\delta_{ij}H\) and \(\comm{P_{i}}{P_{j}}=0\) lie in \(V_{0}\); \(\comm{J}{G}\sim G\), \(\comm{J}{P}\sim P\), \(\comm{H}{G_{i}}=-P_{i}\) and \(\comm{H}{P_{i}}=0\) lie in \(V_{1}\); and \(\comm{J}{J}\sim J\), \(\comm{J_{ij}}{H}=0\) lie in \(V_{0}\). Equation (15.17) holds, and \(V_{0}=\mathfrak{so}(D-1)\oplus\gen{H}\) as a direct sum of ideals.

(1). \(\mathfrak{R}_{1}^{\sigma_{\mathrm{ss}}} =(1\otimes V_{0})\oplus(\varepsilon\otimes V_{1})\), and by Equation (15.15) every bracket of two level-\(1\) generators lands at level \(2>1\) and vanishes, so \(\comm{G_{i}}{G_{j}}=\comm{G_{i}}{P_{j}}=\comm{P_{i}}{P_{j}}=0\) while \(\comm{H}{G_{i}}=-P_{i}\) survives: the Galilei algebra \((\gamma,\mu,\nu)=(0,1,0)\). It has no central charge because Corollary 15.35 with \(N=1\) odd gives \(V_{1}^{V_{0}}=0\) for \(D\ge3\).

(2). By Equation (15.43) with \(i=j=1\) and \(N+1=2\),

which is Equation (15.46). That \(M\) is central follows from Equation (15.41): \(\comm{H}{V_{0}}=0\) because \(\comm{H}{J_{ij}}=0\) and \(\comm{H}{H}=0\). By Corollary 15.35 the full central content of the top level is \(\varepsilon^{2}\otimes\mathfrak{z}(V_{0})\), and \(\mathfrak{z}(V_{0})=\gen{H}\oplus\mathfrak{z}(\mathfrak{so}(D-1)) =\gen{H}\) for \(D-1\ge3\), since \(\mathfrak{so}(n)\) has trivial centre for \(n\ge3\). The subspace \(\mathfrak{i}:=\varepsilon^{2}\otimes \gen{J_{ij}}\) is an ideal: its brackets with levels \(\ge1\) vanish for degree reasons, and at level \(0\) one has \(\comm{J_{ij}}{J_{kl}}\in \gen{J}\) and \(\comm{J_{ij}}{H}=0\). Quotienting by \(\mathfrak{i}\) kills \(\comm{\varepsilon\otimes G_{i}}{\varepsilon\otimes G_{j}} =\varepsilon^{2}\otimes J_{ij}\) and leaves the generators \(J_{ij},H,G_{i},P_{i},M\) with \(M\) central and Equation (15.46) the only new bracket: the Bargmann algebra.

(3). For \(D=3\) the rotation sector is one-dimensional, spanned by a single \(J\), so \(\comm{V_{0}}{V_{0}}=0\) and \(\mathfrak{z}(V_{0})=V_{0}\). Then \(S:=\varepsilon^{2}\otimes J\) is central as well and \(\comm{\varepsilon\otimes G_{1}}{\varepsilon\otimes G_{2}} =\varepsilon^{2}\otimes J_{12}\) is a multiple of \(S\): the two-fold central extension called the extended Bargmann algebra. That \(D=3\) is the only such case is the computation of \(\mathfrak{z}(V_{0})\) in part (2).

The cocycle. Reading Equation (15.46) against Equation (14.120) with central generator \(Z=M\) gives \(c(G_{i},P_{j})=\delta_{ij}\) and \(c=0\) on every other pair, which is Equation (25.18) with the mass scaled to unity.

Derivation. Derives Lemma 15.40. Evaluate \(\theta_{\mathrm{st}}\theta_{\mathrm{ss}}\) on each generator, using \(\theta=+1\) on \(V_{0}\) and \(-1\) on \(V_{1}\) for the corresponding splitting of Equation (15.24):

which is \(+1\) on \(\gen{J_{ij},P_{i}}\) and \(-1\) on \(\gen{G_{i},H}\): that is \(\theta_{\mathrm{sT}}\), proving Equation (15.47). Each \(\theta\) squares to the identity, so the group they generate is \(\Z_{2}\times\Z_{2}\).

Label a character of \(\Gamma\) by its values on \((\theta_{\mathrm{st}},\theta_{\mathrm{ss}})\), writing \(0\) for \(+1\) and \(1\) for \(-1\). Reading the signs above gives Equation (15.48). That this is a grading is Equations (15.22) and (15.23) read modulo two: \(\comm{G}{P}\propto H\) is \(01+11=10\); \(\comm{G}{H}\propto P\) is \(01+10=11\); \(\comm{H}{P}\propto G\) is \(10+11=01\); \(\comm{G}{G}\) and \(\comm{P}{P}\) are proportional to \(J\), which is \(01+01=00\) and \(11+11=00\); and \(\comm{J}{\,\cdot\,}\) preserves each component because \(J\) has degree \(00\).

A surjection \(\Gamma\to\Z_{2}\) has a two-element kernel, so it makes exactly one of the three nontrivial components even together with \(\mathfrak{g}_{00}\); the three kernels \(\set{00,01}\), \(\set{00,10}\), \(\set{00,11}\) give \(V_{0}=\gen{J,G}\), \(\gen{J,H}\), \(\gen{J,P}\), which are \(\sigma_{\mathrm{st}}\), \(\sigma_{\mathrm{ss}}\), \(\sigma_{\mathrm{sT}}\). The splitting \(\sigma_{\mathrm{gen}}\) has \(V_{0}=\gen{J_{ij}}\) alone; no subgroup of \(\Gamma\) has order one and index two, so it is not the kernel of any homomorphism, and Proposition 15.25 showed directly that it is not a symmetric coset.

Derivation. Derives Proposition 15.41. By Equation (15.2),

the middle inclusion being the two grading hypotheses. Proposition 15.21 is the case \(\Gamma=\Z_{2}\) with \(A=A_{N}\) graded by the parity of the exponent of \(\varepsilon\).

Derivation. Derives Proposition 15.42. The three products Equation (15.50) are forced to lie in the stated one-dimensional pieces by the grading, since \(01+11=10\), \(01+10=11\) and \(10+11=01\); the mnemonic is that \(\lambda_{X}\) is the constant of the product that lands on \(e_{X}\). A square has degree \(00\), so \(e_{X}^{2}\) is a multiple of \(1\).

For Equation (15.51), impose associativity on \(e_{G}\,e_{G}\,e_{P}\):

whence \(e_{G}^{2}=\lambda_{H}\lambda_{P}\); the other two squares follow by the same computation on \(e_{H}e_{H}e_{G}\) and \(e_{P}e_{P}e_{G}\). With Equation (15.51) in force every remaining triple product is associative: the only case not already used is \(\left(e_{G}e_{H}\right)e_{P}=\lambda_{P}e_{P}^{2} =\lambda_{G}\lambda_{H}\lambda_{P}\) against \(e_{G}\left(e_{H}e_{P}\right)=\lambda_{G}e_{G}^{2} =\lambda_{G}\lambda_{H}\lambda_{P}\), which agree.

For the trace form, \(L_{e_{G}}\) sends \(1\mapsto e_{G}\), \(e_{G}\mapsto e_{G}^{2}\cdot1\), \(e_{H}\mapsto\lambda_{P}e_{P}\) and \(e_{P}\mapsto\lambda_{H}e_{H}\): no basis vector is sent to a multiple of itself, so \(\tau_{A}(e_{G})=0\), and likewise for \(e_{H},e_{P}\). Hence \(\tau_{A}(e_{X}e_{Y})=0\) for \(X\neq Y\), while \(\tau_{A}(1)=\tr(\id)=4\) and \(\tau_{A}(e_{X}^{2})=4e_{X}^{2}\). The determinant is \(4^{4}\left(\lambda_{G}\lambda_{H}\lambda_{P}\right)^{2}\) by Equation (15.51), and Corollary 15.11 converts nondegeneracy into semisimplicity.

Derivation. Derives Theorem 15.43. Write the resonant generators as \(J':=1\otimes J_{ij}\), \(G':=e_{G}\otimes G_{i}\), \(H':=e_{H}\otimes H\), \(P':=e_{P}\otimes P_{i}\); by Proposition 15.41 they span a subalgebra, of dimension \(\dim\mathfrak{g}\) because each graded piece of \(A(\lambda)\) is one-dimensional. Now apply Equation (15.2) to each bracket of Equation (14.128), using Equation (15.50):

which is Equation (15.52). For the two remaining brackets, use Equation (15.51):

which are exactly Equation (14.129) written for the image constants. The rotations act on \(G_{i}'\) and \(P_{i}'\) as before, since \(1\cdot e_{X}=e_{X}\), so hypothesis (H1) of Theorem 14.90 survives, and the parity and time-reversal automorphisms of hypothesis (H2) commute with the grading because they act on each of \(J,G,H,P\) by a sign.

The consistency of the last display is the point worth noting: the squares Equation (15.51) were forced on \(A(\lambda)\) by associativity, and the relations Equation (14.129) were forced on \(\mathfrak{g}\) by the Jacobi identity. That the first reproduces the second is the factorisation of Equation (15.3) doing its work.

Derivation. Derives Corollary 15.44. The first sentence is Equation (15.52). For the identification of the splittings, Proposition 15.26 sets \(\comm{V_{1}}{V_{1}}=0\) and leaves the other brackets alone. Reading Equation (15.24) against Equations (14.128) and (14.129): for \(\sigma_{\mathrm{st}}\), \(V_{1}=\gen{H,P_{i}}\) and the brackets inside it are \(\comm{P_{i}}{H}=\nu G_{i}\) and \(\comm{P_{i}}{P_{j}}=\gamma\nu \epsilon_{ijk}J_{k}\), so exactly \(\nu\to0\), while \(\comm{G}{P}\) and \(\comm{G}{H}\) are \(\comm{V_{0}}{V_{1}}\) brackets and are untouched. For \(\sigma_{\mathrm{ss}}\), \(V_{1}=\gen{G_{i},P_{i}}\) contains \(\comm{G_{i}}{P_{j}}=\gamma\delta_{ij}H\), so \(\gamma\to0\). For \(\sigma_{\mathrm{sT}}\), \(V_{1}=\gen{G_{i},H}\) contains \(\comm{G_{i}}{H}=\mu P_{i}\), so \(\mu\to0\). For \(\sigma_{\mathrm{gen}}\), \(V_{1}=\gen{G_{i},H,P_{i}}\) contains all three of Equation (14.128), so all three vanish together.

Derivation. Derives Corollary 15.46. Under \(e_{X}\mapsto s_{X}e_{X}\) the products Equation (15.50) give \(\lambda_{H}\mapsto\lambda_{H}s_{G}s_{P}/s_{H}\) and cyclically, so no \(\lambda\) has an invariant sign; but \(e_{X}^{2}\mapsto s_{X}^{2}e_{X}^{2}\), so the three signs Equation (15.56) are invariant, and by Equation (15.51) they are the pairwise sign products of the \(\lambda\) — which is the same invariant that Corollary A.59 extracts from the generators. Claim (1) is Equation (15.51) multiplied out.

(2) By Theorem 15.43 the boosts of \(\mathfrak{R}(\lambda)\) satisfy \(\comm{G_{i}'}{G_{j}'}=-\left(\gamma\lambda_{H}\right) \left(\mu\lambda_{P}\right)\epsilon_{ijk}J_{k}' =-\gamma\mu\,e_{G}^{2}\,\epsilon_{ijk}J_{k}'\). Together with \(\comm{J}{J}\sim J\) and \(\comm{J}{G'}\sim G'\), the span \(\gen{J_{ij},G_{i}'}\) closes on \(\mathfrak{so}(4)\) when the coefficient of \(\epsilon_{ijk}J_{k}'\) is positive and on \(\mathfrak{so}(3,1)\) when it is negative; in the first case the boosts generate a circle and in the second a line, and when the coefficient vanishes they span an abelian \(\R^{3}\), again noncompact. So (H3) holds precisely when \(-\gamma\mu\,e_{G}^{2}\le0\). Normalising the parent by Proposition 15.23, which has \(\gamma\mu=1>0\), this is Equation (15.57).

(3) Go through the eight vertices, using Equation (15.51) to see which of the three signs survives. Where all three \(\lambda\) are nonzero, (H3) fixes \(\sgn(e_{G}^{2})=+\) and then (1) forces \(\sgn(e_{H}^{2})=\sgn(e_{P}^{2})\): one free sign, two algebras (de Sitter and anti-de Sitter). Where \(\lambda_{G}=0\) we have \(e_{H}^{2}=e_{P}^{2}=0\) and \(e_{G}^{2}=\lambda_{H}\lambda_{P}\), whose sign is fixed by (H3): one algebra (Poincaré). Where \(\lambda_{H}=0\), only \(e_{H}^{2}=\lambda_{G}\lambda_{P}\) survives and (H3) says nothing about it: two (Newton–Hooke and anti-Newton–Hooke). Where \(\lambda_{P}=0\), only \(e_{P}^{2}=\lambda_{G}\lambda_{H}\) survives, again unconstrained: two (para-Poincaré and inhomogeneous \(\SO(4)\)). At each of the three vertices with two vanishing \(\lambda\) all three squares vanish, so no sign is left: one each (Galilei, Carroll, para-Galilei). At the origin, one (static). The total is \(2+1+2+2+1+1+1+1=11\), and the three vertices carrying two entries are the three at which a sign survives unconstrained — which is Theorem 14.90.

Derivation. Derives Lemma 15.50. If \(\Phi\) has the form Equation (15.59) then Equation (15.58) holds because \((ab)c_{2}\cdots c_{k}=a(bc_{2})c_{3}\cdots c_{k}\) by associativity, and \(\Phi\) is symmetric by commutativity.

Conversely let \(\Phi\) be a \(k\)-trace and put \(\phi(x):=\Phi(x,1_{A},\dots,1_{A})\). Using Equation (15.58) with \(b=a_{2}\) and \(c_{2}=1_{A}\),

which absorbs one argument into the first slot and replaces it by the unit. Symmetry of \(\Phi\) then allows the next argument to be moved into the second slot, and repeating \(k-1\) times gives \(\Phi(a_{1},\dots,a_{k})=\Phi(a_{1}\cdots a_{k},1_{A},\dots,1_{A}) =\phi(a_{1}\cdots a_{k})\). Uniqueness follows by evaluating Equation (15.59) at \(a_{2}=\dots=a_{k}=1_{A}\).

Derivation. Derives Theorem 15.51. Symmetry and \(k\)-linearity of Equation (15.61) are clear, the first using commutativity of \(A\). For invariance take \(Y=a\otimes X\) and compute with Equation (15.2):

By commutativity and associativity of \(A\) the argument of \(\phi\) equals \(a\,a_{1}\cdots a_{k}\) for every \(i\), so that scalar factors out of the sum and leaves Equation (15.60), which vanishes.

Derivation. Derives Theorem 15.53. Only if. For \(\Omega=\avg{\cdot}_{\omega,\phi}\) both sides of Equation (15.62) equal \(\phi\left(a\,a_{1}\cdots a_{k}\right)\omega(X_{1},\dots,X_{k})\), by commutativity and associativity of \(A\); sums of such forms inherit the property.

If. For \(x\in A\) define a \(k\)-linear form on \(\mathfrak{g}\) by

It is symmetric: balancedness moves \(x\) from the first slot to the second, and the symmetry of \(\Omega\) then exchanges the two slots entirely, so \(\omega^{(x)}\) is unchanged under the transposition \(X_{1}\leftrightarrow X_{2}\); the same argument applies to any transposition, and transpositions generate the symmetric group. It is invariant: since \(\comm{1_{A}\otimes Z}{a\otimes X}=a\otimes\comm{Z}{X}\), the invariance of \(\Omega\) under \(1_{A}\otimes Z\), evaluated on the arguments of Equation (15.64), is literally \(\sum_{i}\omega^{(x)}(X_{1},\dots,\comm{Z}{X_{i}},\dots,X_{k})=0\). So \(x\mapsto\omega^{(x)}\) is a linear map \(A\to\mathrm{Inv}^{k}(\mathfrak{g})\), and expanding in the basis gives functionals \(\phi_{j}\in A^{*}\) with \(\omega^{(x)}=\sum_{j}\phi_{j}(x)\,\omega_{j}\).

It remains to see that \(\Omega\) is recovered from \(\omega^{(\cdot)}\) by multiplying the \(A\)-arguments. Fix \(X_{1},\dots,X_{k}\) and set \(\Phi(a_{1},\dots,a_{k}):=\Omega(a_{1}\otimes X_{1},\dots, a_{k}\otimes X_{k})\), a \(k\)-linear map \(A^{k}\to\R\). Balancedness is exactly Equation (15.58), so \(\Phi\) is a \(k\)-trace, and Lemma 15.50 gives \(\Phi(a_{1},\dots,a_{k}) =\Phi(a_{1}\cdots a_{k},1_{A},\dots,1_{A}) =\omega^{(a_{1}\cdots a_{k})}(X_{1},\dots,X_{k})\), which is Equation (15.63).

Derivation. Derives Proposition 15.55. \(Q\) is the full polarisation of the invariant polynomial \(u\mapsto B(u,u)^{2}\); a product of invariant polynomials is invariant, because the invariants form a ring, so \(Q\) is an invariant symmetric \(4\)-linear form. Now evaluate it on \(u_{1}=1\otimes X_{1}\), \(u_{2}=1\otimes X_{2}\), \(u_{3}=\varepsilon\otimes X_{3}\), \(u_{4}=\varepsilon\otimes X_{4}\). From \(\phi_{1}(1)=0\), \(\phi_{1}(\varepsilon)=1\) and \(\phi_{1}(\varepsilon^{2})=\phi_{1}(0)=0\),

and similarly for the last pairing, so that

which is nonzero — take \(X_{3}=X_{1}\) and \(X_{4}=X_{2}\) with \(\kappa_{\mathfrak{g}}(X_{1},X_{1})\neq0\neq \kappa_{\mathfrak{g}}(X_{2},X_{2})\), possible because \(\kappa_{\mathfrak{g}}\) is nondegenerate. But the product of the four algebra elements is \(1\cdot1\cdot\varepsilon\cdot\varepsilon=\varepsilon^{2}=0\), so any form of the shape Equation (15.61) would give \(\phi(0)\,\omega=0\) there. Hence \(Q\) is not of that shape; equivalently, by Theorem 15.53, \(Q\) is not balanced.

Derivation. Derives Theorem 15.58. By Equation (15.61), for \(X\in V_{[i]}\) and \(Y\in V_{[j]}\),

From Equation (15.61) (take \(\omega=\kappa_{\mathfrak{g}}\) and \(a_{1},a_{2}=\varepsilon^{i},\varepsilon^{j}\)). Two things can make this vanish. By Lemma 15.19, \(\kappa_{\mathfrak{g}}(V_{0},V_{1})=0\), so Equation (15.69) is zero unless \([i]=[j]\), that is unless \(i+j\) is even; and \(\phi(\varepsilon^{i+j})=0\) whenever \(i+j>N\).

Suppose \(N\) is odd and take \(i=N\), so \([N]=1\). Any \(j\) with \([j]=[N]\) is odd, whence \(i+j\ge N+1>N\) and Equation (15.69) vanishes for every \(j\). The whole top level \(\varepsilon^{N}\otimes V_{1}\) lies in the radical of the form, which is therefore degenerate.

Suppose \(N\) is even and \(\phi(\varepsilon^{N})\neq0\). Order the levels \(0,1,\dots,N\) and consider the block matrix of the form. Every entry with \(i+j>N\) vanishes, and the anti-diagonal entries \(i+j=N\) are \(\phi(\varepsilon^{N})\,\kappa_{\mathfrak{g}}|_{V_{[i]}}\), which are nondegenerate: \([i]=[N-i]\) because \(N\) is even, so the two levels pair within the same graded piece, and \(\kappa_{\mathfrak{g}}\) restricted to \(V_{0}\) and to \(V_{1}\) is nondegenerate by Lemma 15.19. A block matrix whose entries vanish strictly beyond the anti-diagonal and whose anti-diagonal blocks are invertible has determinant equal, up to sign, to the product of those blocks' determinants, hence is invertible.

Finally, if \(N\) is even but \(\phi(\varepsilon^{N})=0\) the whole anti-diagonal vanishes and the top level is again in the radical: pairing \(\varepsilon^{N}\otimes X\) with \(\varepsilon^{j}\otimes Y\) requires \(j=0\) and gives \(\phi(\varepsilon^{N})\kappa(X,Y)=0\).

Derivation. Derives Proposition 15.60. \(\R[S]\) is commutative because \(S\) is abelian and associative because \(S\) is, and it is unital after adjoining a unit if \(S\) has none. Its structure constants in the basis \(\set{\lambda_{\alpha}}\) are \(\mu_{\alpha\beta}{}^{\gamma}=K_{\alpha\beta}{}^{\gamma}\), so Equation (15.3) is Equation (15.70).

Derivation. Derives Proposition 15.62. Suppose \(\R[S]\cong\C\). Then \(\abs{S}=2\) and the image of \(S\) in \(\C\) is an \(\R\)-basis closed under multiplication. In a finite semigroup every element has an idempotent power, and the only idempotents of \(\C\) are \(0\) and \(1\), so each element of the image is \(0\) or a root of unity. If \(0\) lies in the image the two elements span a real line, not \(\C\). So the image is \(\set{z,w}\) with \(z,w\) roots of unity, \(\R\)-independent, and \(z^{2}\in\set{z,w}\).

If \(z^{2}=z\) then \(z=1\), and \(w^{2}\in\set{1,w}\) forces \(w=\pm1\), dependent on \(z\). If \(z^{2}=w\) then \(w^{2}=z^{4}\in\set{z,w}\): from \(z^{4}=z\) we get \(z^{3}=1\) with \(z\neq1\), so \(w=z^{2}\) and \(zw=z^{3}=1\notin\set{z,w}\), contradicting closure; and from \(z^{4}=w=z^{2}\) we get \(z^{2}=1\), hence \(w=1\) and \(z=-1\), again dependent. No case survives.

Derivation. Derives Lemma 15.67. \(B_{\psi}(au,v)=\psi((au)v)=\psi(u(av))=B_{\psi}(u,av)\), by commutativity and associativity of \(A\). If \(\psi(uA)=0\) then for every \(b\in A\) also \(\psi((bu)A)=\psi(u(bA))\subseteq\psi(uA)=0\), so \(I_{\psi}\) absorbs multiplication.

Derivation. Derives Theorem 15.68. (2)\(\Rightarrow\)(1). Multiplicativity of each \(\chi_{i}\) gives

(1)\(\Rightarrow\)(2). If \(\psi=0\) take \(r=0\). Otherwise let \(I_{\psi}\) be the radical of \(B_{\psi}\), an ideal by Lemma 15.67. For a positive semidefinite form the radical coincides with the null cone — \(B_{\psi}(u,u)=0\) forces \(u\in I_{\psi}\) by the Cauchy–Schwarz inequality, which holds for every positive semidefinite symmetric form — so the induced form \(\bar{B}\) on \(\bar{A}:=A/I_{\psi}\) is positive definite, and \(\bar{A}\) is again commutative, associative and unital (\(1_{A}\notin I_{\psi}\) because \(\psi\neq0\)). Each multiplication operator \(\bar{M}_{a}\) on \(\bar{A}\) is \(\bar{B}\)-symmetric by Lemma 15.67, and the family \(\set{\bar{M}_{a}:a\in A}\) commutes, so the spectral theorem for commuting symmetric operators on a real inner-product space provides a \(\bar{B}\)-orthogonal basis \(\bar{u}_{1},\dots,\bar{u}_{r}\) of \(\bar{A}\) and functions \(\chi_{i}:A\to\R\) with \(\bar{M}_{a}\bar{u}_{i}=\chi_{i}(a)\,\bar{u}_{i}\). Each \(\chi_{i}\) is linear in \(a\), multiplicative because \(\bar{M}_{ab}=\bar{M}_{a}\bar{M}_{b}\), and unital because \(\bar{M}_{1_{A}}=\mathrm{id}\). Writing \(\bar{1}=\sum_{i}c_{i}\bar{u}_{i}\) and using \(\psi(a)=B_{\psi}(a,1_{A})\),

with every \(w_{i}\ge0\); discard the terms with \(w_{i}=0\) and merge repeated characters, which keeps the surviving weights positive and the characters distinct.

Distinct characters are linearly independent: a shortest nontrivial relation \(\sum_{i}\lambda_{i}\chi_{i}=0\) with all \(\lambda_{i}\neq0\) and \(r\ge2\) admits some \(a_{0}\) with \(\chi_{1}(a_{0})\neq\chi_{2}(a_{0})\); evaluating the relation at \(a_{0}a\) and subtracting \(\chi_{1}(a_{0})\) times the relation at \(a\) kills the first term but not the second, producing a shorter nontrivial relation. Hence \(B_{\psi}(a,b)=\sum_{i}w_{i}\chi_{i}(a)\chi_{i}(b)\) has rank exactly \(r\).

Finally, a character kills every nilpotent element, since \(\chi(n)^{k}=\chi(n^{k})=0\), hence kills \(\operatorname{rad}(A)\). And let \(A_{j}\) be a factor of the Artin decomposition, with unit the idempotent \(\pi_{j}\) and residue field \(\C\). From \(\chi(\pi_{j})^{2}=\chi(\pi_{j})\) either \(\chi(\pi_{j})=0\), and then \(\chi\) kills \(A_{j}=\pi_{j}A_{j}\) outright, or \(\chi(\pi_{j})=1\), and then \(\chi\) restricts to a unital \(\R\)-algebra homomorphism \(A_{j}\to\R\), which kills the maximal ideal of \(A_{j}\) (it is nilpotent) and so induces an \(\R\)-algebra homomorphism \(\C\to\R\) — impossible, since \(\chi(i)^{2}=\chi(-1)=-1\) has no real solution. So every \(\psi\) of the form Equation (15.73) vanishes on \(\operatorname{rad}(A)\) and on every factor whose residue field is \(\C\).

Derivation. Derives Corollary 15.69. (1) is Theorem 15.68 applied to \(A_{N}\): an \(\R\)-character must send \(\varepsilon\) to a real root of \(x^{N+1}=0\), that is to \(0\), so the only character is evaluation at \(\varepsilon=0\) and Equation (15.73) reads \(\psi=w_{1}\operatorname{ev}_{0}\) with \(w_{1}=\psi(1)\), allowing also \(w_{1}=0\) for \(\psi=0\). (2): in the basis \(1,\varepsilon,\dots,\varepsilon^{N}\) the matrix of \(B_{\psi}\) has entries \(\psi(\varepsilon^{i+j})\), zero for \(i+j>N\) and equal to \(\psi(\varepsilon^{N})\neq0\) along the anti-diagonal \(i+j=N\), so \(\det B_{\psi}=\pm\psi(\varepsilon^{N})^{N+1}\neq0\) and the form is nondegenerate. The span of \(\set{\varepsilon^{i}:2i>N}\) has all pairwise products of total degree exceeding \(N\), hence is totally isotropic for \(B_{\psi}\), of dimension \(N/2\); a nondegenerate symmetric form on a space of dimension \(N+1\) with a totally isotropic subspace of dimension \(N/2\) has at least \(N/2\) positive and \(N/2\) negative directions, which fixes the signature up to the placement of the one remaining unit.

Derivation. Derives Corollary 15.70. Diagonalise each factor and count the signs of the products of diagonal entries.