Invariant Bilinear Forms on an Expanded Algebra

This appendix proves Theorem 19.54 (Expansions of Lie Algebras): on $\mathfrak{g}_{A}=A\otimes\mathfrak{g}$ every invariant symmetric bilinear form has the factorised shape Equation (19.61), so for $k=2$ the family constructed there is already the whole of it. This is the case the book uses — it is what makes Theorem 19.58 a statement about all invariant pairings and not merely about the ones written down. It is also the only case in which the statement is true: Proposition 19.55 exhibits an invariant symmetric $4$-linear form that is not factorised, and Remark A11.5 at the end says precisely which step of the argument below fails for $k\ge3$ and why no repair is possible.

The hypothesis on $\mathfrak{g}$ needs care, and the care is not academic: the Lorentz algebra of the observed $3{+}1$ dimensions is exactly the case where the naive hypothesis fails.

Definition A11.1 (Absolutely simple).

A real Lie algebra $\mathfrak{g}$ is absolutely simple if its complexification $\mathfrak{g}\otimes_{\R}\C$ is a simple complex Lie algebra. Every absolutely simple algebra is simple; the converse fails, the standard counterexample being a complex simple algebra regarded as a real one.

Lemma A11.2 (The invariant bilinear forms of an absolutely simple algebra form a line).

Let $\mathfrak{g}$ be absolutely simple with Killing form $\kappa_{\mathfrak{g}}$. Then every invariant bilinear form on $\mathfrak{g}$ is a real multiple of $\kappa_{\mathfrak{g}}$; in particular the space of such forms is one-dimensional and every one of them is symmetric. Rests on Definition A11.1.

Derivation. Derives Lemma A11.2. Let $B$ be an invariant bilinear form, that is

\begin{equation}\tag{A11.1} B\left(\comm{Z}{X},Y\right)+B\left(X,\comm{Z}{Y}\right)=0 \qquad\text{for all }X,Y,Z\in\mathfrak{g}\ep \end{equation}

Define $\beta:\mathfrak{g}\to\mathfrak{g}^{*}$ by $\beta(X)=B(X,\cdot)$. Equation (A11.1) says exactly that $\beta$ intertwines the adjoint representation on $\mathfrak{g}$ with the coadjoint representation on $\mathfrak{g}^{*}$. Since $\mathfrak{g}$ is simple it is semisimple, so $\kappa_{\mathfrak{g}}$ is nondegenerate by Cartan's criterion and the induced map $\kappa_{\mathfrak{g}}^{\flat}:\mathfrak{g}\to\mathfrak{g}^{*}$ is an isomorphism of representations. Hence

\begin{equation*} T:=\left(\kappa_{\mathfrak{g}}^{\flat}\right)^{-1}\circ\beta \in\operatorname{End}_{\mathfrak{g}}(\mathfrak{g})\ec \end{equation*}

the algebra of endomorphisms commuting with the adjoint action, and $B=\kappa_{\mathfrak{g}}(T\,\cdot,\cdot)$.

It remains to show $\operatorname{End}_{\mathfrak{g}}(\mathfrak{g})=\R$. Extending scalars,

\begin{equation*} \operatorname{End}_{\mathfrak{g}}(\mathfrak{g})\otimes_{\R}\C \cong\operatorname{End}_{\mathfrak{g}_{\C}} \left(\mathfrak{g}_{\C}\right)\ec \qquad\mathfrak{g}_{\C}:=\mathfrak{g}\otimes_{\R}\C\ec \end{equation*}

since forming the commutant of a set of operators commutes with extension of scalars. By hypothesis $\mathfrak{g}_{\C}$ is simple, so its adjoint representation is irreducible over $\C$, and Schur's lemma over an algebraically closed field gives $\operatorname{End}_{\mathfrak{g}_{\C}}(\mathfrak{g}_{\C})=\C$. Hence $\dim_{\R}\operatorname{End}_{\mathfrak{g}}(\mathfrak{g})=1$, so $T=\lambda\,\id$ and $B=\lambda\,\kappa_{\mathfrak{g}}$, which is symmetric.

∎

Theorem A11.3 (Completeness of the factorised bilinear forms).

Let $\mathfrak{g}$ be absolutely simple and let $A$ be a finite-dimensional commutative associative unital $\R$-algebra. Then for every invariant symmetric bilinear form $\Omega$ on $\mathfrak{g}_{A}$ there is a unique linear functional $\phi:A\to\R$ with

\begin{equation}\tag{A11.2} \boxed{\Omega\left(a\otimes X,\;b\otimes Y\right) =\phi(ab)\,\kappa_{\mathfrak{g}}(X,Y)} \qquad\text{for all }a,b\in A,\ X,Y\in\mathfrak{g}\ep \end{equation}

Equivalently $\Omega=\avg{\cdot,\cdot}_{\kappa_{\mathfrak{g}},\phi}$ in the notation of Equation (19.61), and $\phi\mapsto\avg{\cdot,\cdot}_{\kappa_{\mathfrak{g}},\phi}$ is a linear isomorphism from $A^{*}$ onto the space of invariant symmetric bilinear forms on $\mathfrak{g}_{A}$, which therefore has dimension $\dim A$. Rests on Definition A11.1, Lemma A11.2, Lemma 19.50, Theorem 19.51 and Equation (19.2).

Derivation. Derives Theorem A11.3. Step 1: for fixed $a,b$ the form is a multiple of the Killing form. Fix $a,b\in A$ and define the bilinear form

\begin{equation}\tag{A11.3} \omega_{a,b}(X,Y):=\Omega\left(a\otimes X,\;b\otimes Y\right) \end{equation}

on $\mathfrak{g}$. Invariance of $\Omega$, applied with the element $1_{A}\otimes Z$ and using $\comm{1_{A}\otimes Z}{a\otimes X}=a\otimes\comm{Z}{X}$ from Equation (19.2), gives

\begin{equation*} \Omega\left(a\otimes\comm{Z}{X},\;b\otimes Y\right) +\Omega\left(a\otimes X,\;b\otimes\comm{Z}{Y}\right)=0\ec \end{equation*}

that is, $\omega_{a,b}$ satisfies Equation (A11.1). By Lemma A11.2 there is a unique real number $\Psi(a,b)$ with

\begin{equation}\tag{A11.4} \omega_{a,b}=\Psi(a,b)\,\kappa_{\mathfrak{g}}\ep \end{equation}

Since $\Omega$ is bilinear in its two arguments and $\kappa_{\mathfrak{g}}\neq0$, the function $\Psi:A\times A\to\R$ is bilinear; since $\Omega$ and $\kappa_{\mathfrak{g}}$ are symmetric, $\Psi$ is symmetric.

Step 2: $\Psi$ is a $2$-trace. Now use invariance of $\Omega$ with a general element $c\otimes Z$. By Equation (19.2), $\comm{c\otimes Z}{a\otimes X}=(ca)\otimes\comm{Z}{X}$, so

\begin{equation*} \Omega\left((ca)\otimes\comm{Z}{X},\;b\otimes Y\right) +\Omega\left(a\otimes X,\;(cb)\otimes\comm{Z}{Y}\right)=0\ec \end{equation*}

which by Equation (A11.4) reads

\begin{equation}\tag{A11.5} \Psi(ca,b)\,\kappa_{\mathfrak{g}}\left(\comm{Z}{X},Y\right) +\Psi(a,cb)\,\kappa_{\mathfrak{g}}\left(X,\comm{Z}{Y}\right)=0\ep \end{equation}

The Killing form is itself invariant, so $\kappa_{\mathfrak{g}}(X,\comm{Z}{Y}) =-\kappa_{\mathfrak{g}}(\comm{Z}{X},Y)$, and Equation (A11.5) collapses to

\begin{equation}\tag{A11.6} \Bigl(\Psi(ca,b)-\Psi(a,cb)\Bigr)\, \kappa_{\mathfrak{g}}\left(\comm{Z}{X},Y\right)=0 \qquad\text{for all }X,Y,Z\in\mathfrak{g}\ep \end{equation}

The Killing factor cannot vanish identically: a simple algebra is perfect, so the elements $\comm{Z}{X}$ span $\mathfrak{g}$, and $\kappa_{\mathfrak{g}}$ is nondegenerate, so some choice of $X,Y,Z$ makes $\kappa_{\mathfrak{g}}(\comm{Z}{X},Y)\neq0$. Hence

\begin{equation}\tag{A11.7} \Psi(ca,b)=\Psi(a,cb)\qquad\text{for all }a,b,c\in A\ec \end{equation}

which is Equation (19.58) at $k=2$: $\Psi$ is a $2$-trace on $A$.

Step 3: a $2$-trace is a linear functional of the product. By Lemma 19.50 with $k=2$ there is a unique linear $\phi:A\to\R$, namely $\phi(x)=\Psi(x,1_{A})$, with $\Psi(a,b)=\phi(ab)$. Substituting into Equation (A11.4) gives Equation (A11.2).

Step 4: uniqueness and bijectivity. If $\phi$ and $\phi'$ both satisfy Equation (A11.2) then $\left(\phi(ab)-\phi'(ab)\right)\kappa_{\mathfrak{g}}(X,Y)=0$; choosing $X,Y$ with $\kappa_{\mathfrak{g}}(X,Y)\neq0$ and $b=1_{A}$ gives $\phi=\phi'$. Conversely every $\phi\in A^{*}$ produces an invariant symmetric bilinear form by the first part of Theorem 19.51, and the assignment is visibly linear in $\phi$, while Equation (A11.2) says it is onto. The two spaces are therefore isomorphic.

∎

Remark A11.4 (The hypothesis is sharp, and $3{+}1$ is where it bites).

Lemma A11.2 needs $\mathfrak{g}$ absolutely simple, not merely simple, and the difference is not a technicality in this book. The Lorentz algebra of four-dimensional spacetime, $\mathfrak{so}(3,1)$, is simple as a real Lie algebra but not absolutely simple: its complexification is $\mathfrak{sl}_{2}(\C)\oplus\mathfrak{sl}_{2}(\C)$, which is not simple — this is the self-dual/anti-self-dual splitting. Accordingly it carries a two-dimensional space of invariant symmetric bilinear forms, spanned by

\begin{equation}\tag{A11.8} \avg{J_{\mu\nu},J_{\rho\sigma}}_{1} =\eta_{\mu\rho}\eta_{\nu\sigma}-\eta_{\mu\sigma}\eta_{\nu\rho}\ec \qquad \avg{J_{\mu\nu},J_{\rho\sigma}}_{2}=\epsilon_{\mu\nu\rho\sigma}\ec \end{equation}

the first proportional to the Killing form and the second available only because $D=4$ supplies a rank-four invariant alternating symbol. Both are invariant, and the two Chern–Weil $4$-forms they generate are the two topological densities of four-dimensional gravity: the Killing form gives $R^{\mu\nu}\wedge R_{\mu\nu}$, the Pontryagin density, and $\epsilon_{\mu\nu\rho\sigma}$ gives $\epsilon_{\mu\nu\rho\sigma}R^{\mu\nu}\wedge R^{\rho\sigma}$, the Euler density of the Gauss–Bonnet term. That a four-dimensional gauge theory of the Lorentz group admits two such terms where a general dimension admits one is exactly this failure of absolute simplicity. For such a $\mathfrak{g}$ the conclusion of Theorem A11.3 must be replaced by a sum over a basis of the invariant forms of $\mathfrak{g}$, one functional $\phi$ for each. The algebras to which Theorem 19.29 attaches the three signs of the cosmological constant — $\mathfrak{so}(3,2)$ and $\mathfrak{so}(4,1)$ in $3{+}1$ — are absolutely simple, their complexifications being $\mathfrak{sp}(4,\C)$ and $\mathfrak{so}(5,\C)$, so Theorem A11.3 applies to them verbatim.

Remark A11.5 (Why the argument stops at $k=2$, and why that is not a gap).

It is worth being explicit about where the proof above uses $k=2$, because the manuscript this chapter follows [Gonzalez:2026] states the converse for every $k$, and that statement is false.

Step 1 generalises without difficulty: evaluating invariance on $1_{A}\otimes Z$ shows, for any $k$, that $(X_{1},\dots,X_{k})\mapsto \Omega(a_{1}\otimes X_{1},\dots,a_{k}\otimes X_{k})$ is an invariant $k$-linear form on $\mathfrak{g}$. Step 2 does not. It works here because the invariance relation for a bilinear form has exactly two terms, and the invariance of $\kappa_{\mathfrak{g}}$ itself collapses them into the single factor Equation (A11.6); for $k\ge3$ the same manipulation leaves a sum of $k-1$ terms with no reason to vanish separately. That is not a defect of the argument: by Proposition 19.55 there is an invariant symmetric $4$-linear form on $\mathfrak{so}(2,1)\gen{1}=\mathfrak{iso}(2,1)$ — the square of the standard pairing $\avg{J_{a},P_{b}}=\eta_{ab}$ — which is not factorised at all, so no argument could have succeeded.

The structural reason is Theorem 19.53: the factorised forms are exactly the balanced ones, and a product of two invariant bilinear forms is not balanced, because $\phi(a_{1}a_{2})\phi(a_{3}a_{4})$ is not a function of $a_{1}a_{2}a_{3}a_{4}$. At $k=2$ no product is available, since $\mathrm{Inv}^{1}(\mathfrak{g})=(\mathfrak{g}^{*})^{\mathfrak{g}}$ vanishes for perfect $\mathfrak{g}$ — which is exactly what Theorem A11.3 exploits without saying so, and is the whole of the difference between the two cases.

Invariant Bilinear Forms on an Expanded Algebra discharges the proof obligation of Theorem 19.54. Its yield for the rest of the book is that Theorem 19.58 classifies all invariant pairings on a resonant subalgebra and not merely the constructed ones, which is what makes the statement that the Galilei algebra admits none, while the Bargmann algebra does, a theorem rather than a failure to find one. Its other yield is negative and equally worth having: together with Proposition 19.55 it fixes the exact reach of the factorisation, which is $k=2$ and no further.