The Lie algebra of eta-antisymmetric endomorphisms, its generators and commutators

Let $v,w\in V$ have coordinate columns $v,w$. By [L1] and [L2], $\eta (Kv,w)=v^T{M}^T{\eta }_{0}w$ and $\eta (v,Kw)=v^T{\eta }_{0}Mw$, so $\eta (Kv,w)+\eta (v,Kw)=v^T(M^T{\eta }_0+{\eta }_{0}M)w$. If (a) holds this bilinear form vanishes for all $v,w$, and testing $v,w$ among the standard basis vectors of ${\mathbb{R}}^4$ forces every entry of $M^T{\eta }_0+{\eta }_{0}M$ to vanish, which is (b); conversely (b) gives (a) for all $v,w$ by the same identity. Hence (a) and (b) are equivalent.

Write $M$ in the block form $M=\left(\begin{array}{cc}k& u^T\\ w& \mathrm{\Omega }\end{array}\right)$ with $k\in \mathbb{R}$, $u,w\in {\mathbb{R}}^3$ and $\mathrm{\Omega }\in M_3(\mathbb{R})$, and recall ${\eta }_0=\mathrm{diag}(-1,I_3)$. Then $M^T=\left(\begin{array}{cc}k& w^T\\ u& {\mathrm{\Omega }}^T\end{array}\right)$, so $M^T{\eta }_0=\left(\begin{array}{cc}-k& w^T\\ -u& {\mathrm{\Omega }}^T\end{array}\right)$ and ${\eta }_{0}M=\left(\begin{array}{cc}-k& -u^T\\ w& \mathrm{\Omega }\end{array}\right)$; the sum $M^T{\eta }_0+{\eta }_{0}M=\left(\begin{array}{cc}-2k& w^T-u^T\\ w-u& {\mathrm{\Omega }}^T+\mathrm{\Omega }\end{array}\right)$ vanishes exactly when $k=0$, $w=u$ and ${\mathrm{\Omega }}^T=-\mathrm{\Omega }$. Hence (b) and (c) are equivalent.

With the convention of [L2], $(M_{ab}{M}_{cd}{)}^{\mu }{}_{\nu }={\sum }_{\lambda }({\delta }_a^{\mu }{\eta }_{b\lambda }-{\delta }_b^{\mu }{\eta }_{a\lambda })({\delta }_c^{\lambda }{\eta }_{d\nu }-{\delta }_d^{\lambda }{\eta }_{c\nu })={\delta }_a^{\mu }{\eta }_{bc}{\eta }_{d\nu }-{\delta }_a^{\mu }{\eta }_{bd}{\eta }_{c\nu }-{\delta }_b^{\mu }{\eta }_{ac}{\eta }_{d\nu }+{\delta }_b^{\mu }{\eta }_{ad}{\eta }_{c\nu }$, because ${\sum }_{\lambda }{\eta }_{b\lambda }{\delta }_c^{\lambda }={\eta }_{bc}$ and ${\sum }_{\lambda }{\eta }_{b\lambda }{\delta }_d^{\lambda }={\eta }_{bd}$. Interchanging the pairs $(a,b)$ and $(c,d)$ and subtracting, the eight resulting terms regroup as $({\delta }_a^{\mu }{\eta }_{d\nu }-{\delta }_d^{\mu }{\eta }_{a\nu }){\eta }_{bc}-({\delta }_a^{\mu }{\eta }_{c\nu }-{\delta }_c^{\mu }{\eta }_{a\nu }){\eta }_{bd}-({\delta }_b^{\mu }{\eta }_{d\nu }-{\delta }_d^{\mu }{\eta }_{b\nu }){\eta }_{ac}+({\delta }_b^{\mu }{\eta }_{c\nu }-{\delta }_c^{\mu }{\eta }_{b\nu }){\eta }_{ad}$, which is exactly ${\eta }_{bc}{M}_{ad}-{\eta }_{bd}{M}_{ac}-{\eta }_{ac}{M}_{bd}+{\eta }_{ad}{M}_{bc}$; the computation is entrywise and therefore valid for all index values, including repeated indices and the diagonal generators $M_{aa}=0$.

Let $s\mapsto L(s)$ be a ${\mathcal{C}}^1$ curve in $O(V,\eta )$ with $L(0)={\mathrm{id}}_V$, and fix $v,w\in V$. The function $s\mapsto \eta (L(s)v,L(s)w)$ is constant equal to $\eta (v,w)$ because each $L(s)$ satisfies the defining identity of [L4], and $s\mapsto L(s)v$, $s\mapsto L(s)w$ are differentiable with derivatives $L^{\mathrm{\prime }}(0)v$, $L^{\mathrm{\prime }}(0)w$ by [L5]. Differentiating the constant function at $s=0$ with the product rule for the bilinear form $\eta$ gives $0=\eta (L^{\mathrm{\prime }}(0)v,w)+\eta (v,L^{\mathrm{\prime }}(0)w)$, so $K:=L^{\mathrm{\prime }}(0)$ satisfies condition (a) and lies in $\mathfrak{g}$; and $L(s)=L(0)+sK+o(s)={\mathrm{id}}_V+sK+o(s)$ is the definition of the derivative in the finite-dimensional space $L(V,V)$ recalled in [L5]: each scalar matrix entry has remainder divided by $s$ tending to $0$ by The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set, and the coordinates of $L(s)v$ are finite linear combinations of these entries.

By (c), the maps in $\mathfrak{g}$ are precisely those whose matrices have the form $\left(\begin{array}{cc}0& u^T\\ u& \mathrm{\Omega }\end{array}\right)$ with $u\in {\mathbb{R}}^3$ and $\mathrm{\Omega }$ skew-symmetric; this set is closed under addition and scalar multiplication, so $\mathfrak{g}$ is a linear subspace of $L(V,V)$ by [L3]. A skew-symmetric $3\times 3$ matrix is determined by its three entries above the diagonal and conversely every such triple occurs, so the skew-symmetric matrices form a $3$-dimensional subspace; adding the three free coordinates of $u$ gives $\dim \mathfrak{g}=3+3=6$. For $K,H\in \mathfrak{g}$, $\eta (KHv,w)=-\eta (Hv,Kw)=\eta (v,HKw)$ and $\eta (HKv,w)=\eta (v,KHw)$, so $\eta ([K,H]v,w)=-\eta (v,[K,H]w)$; thus the commutator is closed on $\mathfrak{g}$. Composition is bilinear, hence so is the commutator, and $[K,H]=-[H,K]$. Finally $[K,[H,Z]]=KHZ-KZH-HZK+ZHK$; adding the two cyclic permutations cancels all twelve terms by associativity. This proves the Lie algebra axioms stated above.

For $0\le a<b\le 3$ define $M_{ab}$ by $(M_{ab}{)}^{\mu }{}_{\nu }={\delta }_a^{\mu }{\eta }_{b\nu }-{\delta }_b^{\mu }{\eta }_{a\nu }$. Reading off the matrix entries: $M_{0i}$ has $(M_{0i}{)}^{\mu }{}_{\nu }={\delta }_0^{\mu }{\delta }_{i\nu }+{\delta }_i^{\mu }{\delta }_{0\nu }$, the block form with $u=e_i$ and $\mathrm{\Omega }=0$, while for $1\le i<j\le 3$ the matrix $M_{ij}$ has entries $(M_{ij}{)}^{\mu }{}_{\nu }={\delta }_i^{\mu }{\delta }_{j\nu }-{\delta }_j^{\mu }{\delta }_{i\nu }$, the block form with $u=0$ and $\mathrm{\Omega }=e_i{e}_j^T-e_j{e}_i^T$ (the $i$-th row of $\eta$ restricted to spatial indices being $e_i^T$). Hence each $M_{ab}$ belongs to $\mathfrak{g}$ by (c), and the correspondence $M_{ab}\leftrightarrow (u,\mathrm{\Omega })$ maps the three $M_{0i}$ onto the three coordinate directions of $u$ and the three $M_{ij}$ onto the three standard skew matrices $e_i{e}_j^T-e_j{e}_i^T$, which are independent; therefore the six $M_{ab}$ are linearly independent and, since $\dim \mathfrak{g}=6$ by step 2.1, they form a basis of $\mathfrak{g}$.

Steps 1.1 and 1.2 establish the equivalence of (a), (b), (c); steps 2.1 and 3.1 establish that $\mathfrak{g}$ is a $6$-dimensional linear subspace with basis $M_{ab}$; step 1.3 establishes the commutator identity; and step 1.4 establishes the derivative statement. This proves every clause of the statement.