---- > [!definition] Definition. ([[orthogonal Lie algebra]]) > Let $\mathbb{F}$ be a [[field|field]] ($\text{char }\mathbb{F} \neq 2$) and $V$ an $\mathbb{F}$-[[vector space]] endowed with a [[symmetric multilinear map|symmetric]], [[nondegenerate bilinear form|nondegenerate]] [[bilinear map|bilinear form]] $\langle -, - \rangle: V \times V \to \mathbb{F}$. Then with $\mathfrak{gl}(V)$ denoting the [[general linear Lie algebra|general linear]] [[Lie algebra]] on $V$ define[^1] $\mathfrak{so}(V)=\{ f \in \mathfrak{gl}(V) : \langle f(v), w \rangle + \langle v, f(w) \rangle =0 \text{ for all } v,w \in V \}.$ This is a [[Lie subalgebra]] of $\mathfrak{gl}(V)$ called the **orthogonal Lie algebra**. The matrix version is denoted $\mathfrak{so}_{n}$. > If we choose an [[linear isomorphism|isomorphism]] $V \cong \mathbb{F}^{\oplus n}$, (i.e., [[free module|choose a basis]] of $V$), then $\langle -,- \rangle$ [[matrix of a bilinear form|corresponds to a]] [[matrix of a bilinear form|(nonsingular)]] [[symmetric matrix]] $A \in \text{Mat}_{n}(k)$, $\langle v,w \rangle=v^{\top}A w$ and we have $\mathfrak{so}(V, \langle -,- \rangle )\cong \{ X \in \mathfrak{gl}_{n}(\mathbb{F}) : X^{\top} A + AX = 0 \}.$ > Over $\mathbb{C}$, [[the isomorphism classes of orthogonal and symplectic Lie algebras do not depend on choice of bilinear form|it turns out]] that any two choices of [[bilinear map|bilinear form]] $\langle -,- \rangle$, $\langle -,- \rangle'$ yield [[Lie algebra homomorphism|isomorphic]] orthogonal Lie algebras. Often it is useful to consider the bilinear form determined by $A=\begin{cases} \begin{bmatrix} 0 & I_{\ell} \\ I_{\ell} & 0 \end{bmatrix} & n=2\ell \\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & I_{\ell} \\ 0 & I_{\ell} & 0 \end{bmatrix} & n=2\ell+1. \end{cases}$ By constructing a basis for the matrix space under this bilinear form, (see below), one sees that $\mathfrak{so}(V)$ has [[dimension]] $\frac{n(n-1)}{2}$. > Explicitly, $\mathfrak{so}_{2\ell}= \{ \begin{bmatrix} P & Q \\ R & -P^{\top} \end{bmatrix} : P,Q,S \in \mathfrak{gl}_{\ell}, Q=-Q^{\top}, R=-R^{\top} \}$ and $\mathfrak{so}_{2 \ell + 1}= \{ \begin{bmatrix} 0& \boldsymbol u & \boldsymbol v \\-\boldsymbol u^{\top} & P & Q \\ - \boldsymbol v^{\top} & R & -P^{\top} \end{bmatrix} : \alpha \in \mathbb{F}, \boldsymbol u, \boldsymbol v \in \mathbb{F}^{ 1 \times \ell}, P,Q,R \in \mathfrak{gl}_{\ell}, Q=-Q^{\top}, R=-R^{\top} \}.$ > [!justification] > To explicate upon the identification: we have $\begin{align} \mathfrak{so}(V) & =\{ f \in \mathfrak{gl}(V) : \langle f(v), w \rangle + \langle v, f(w) \rangle =0 \text{ for all } v,w \in V \} \\ &= \{ X \in \mathfrak{gl}_{n} : (Xv)^{\top} A w + v^{\top} A Xw = 0 \text{ for all } v, w \in \mathbb{F}^{n}\} \\ & = \{X \in \mathfrak{gl}_{n} : v^{\top} (X^{\top} A + AX) w = 0 \text{ for all } v , w \in \mathbb{F}^{n} \} \\ & =\{X \in \mathfrak{gl}_{n} : X^{\top} A + AX = 0 \}. \end{align}$ > We also need to justify that this is indeed a [[Lie subalgebra]] of $\mathfrak{gl}_{2}(V)$. Clearly $0 \in \mathfrak{sl}_{2}(V)$ and $\mathfrak{sl}_{2}(V)$ is stable under addition and scalar multiplication due to bilinearity of the form $\langle -,- \rangle$. To see that it is stable under the Lie bracket we may choose an isomorphism $V \cong \mathbb{F}^{\oplus n}$ and work with matrices $X \in \mathfrak{gl}_{n}$ satisfying $X^{\top}A + AX=0$, where $A$ is the matrix of $\langle -,- \rangle$ in our chosen coordinates. Given $X, Y$ satisfying this property, compute >$\begin{align} [X,Y]^{\top} A + A[X,Y] & = Y^{\top}X^{\top}A - X^{\top}Y^{\top}A + AXY - AYX \\ &= Y^{\top}(-AX) - X^{\top}(-AY) + AXY - AYX \\ & = AYX -AXY +AXY - AYX \\ &=0 \end{align}$ as required. ^justification > [!justification] On basis and dimension. > > **$n=2\ell$.** For $n=2 \ell$ even, one has $\begin{align} > \mathfrak{so}(V, \langle -,- \rangle_{\text{preferred}} ) & \cong \{ X \in \mathfrak{gl}_{n}: X^{\top} A + A^{\top }X = 0 \} \\ > &= \{ \begin{bmatrix}P &Q \\ R & S\end{bmatrix} \in \mathfrak{gl}_{n} : \begin{bmatrix}P^{\top} & R^{\top} \\ Q^{\top} & S^{\top}\end{bmatrix} \begin{bmatrix}0 & I_{\ell} \\I_{\ell} & 0 > \end{bmatrix} +\begin{bmatrix}0 & I_{\ell} \\I_{\ell} & 0 > \end{bmatrix} \begin{bmatrix}P & Q \\ R & S\end{bmatrix} = 0 \}. > \end{align}$ > We see that $\begin{align} > \begin{bmatrix}P^{\top} & R ^{\top}\\ Q^{\top} & S^{\top}\end{bmatrix} \begin{bmatrix}0 & I_{\ell} \\I_{\ell} & 0 > \end{bmatrix} + \begin{bmatrix}0 & I_{\ell} \\I_{\ell} & 0 > \end{bmatrix}\begin{bmatrix}P & Q \\ R & S\end{bmatrix} & = \begin{bmatrix} > R^{\top} & P^{\top} \\ S^{\top} & Q ^{\top} > \end{bmatrix} + \begin{bmatrix} > R & S \\ P & Q > \end{bmatrix} > \end{align}$ > and so the condition $X^{\top}A+A^{\top}X=0$ in this case amounts to asserting $R^{\top}=-R, Q^{\top}=-Q, P^{\top}=-S, S^{\top}=-P,$ > in other words, we have that $R,Q$ are [[skew-symmetric matrix|skew-symmetric]] while $P^{\top}=-S$. Finding a [[basis]] for this space amounts to find a 'basis for each block'; the skew-symmetric blocks have the usual skew-symmetric basis $E_{ij}-E_{ji}$ which has $\ell(\ell-1) / 2$ entries. Then there are $\ell^{2}$ options for basis elements of $P$, which enforces $S$, this means there are $\ell^{2} + \ell(\ell - 1) / 2 + \ell(\ell - 1) / 2=2 \ell^{2} - \ell=n(n-1) / 2$ basis elements in total. > > > **$n=2 \ell+1$.** Suppose $n=2 \ell +1$ is odd. A similar computation to that above shows $\mathfrak{so}_{2 \ell + 1}= \{ \begin{bmatrix} > 0 & \boldsymbol u & \boldsymbol v \\-\boldsymbol u^{\top} & P & Q \\ - \boldsymbol v^{\top} & R & -P^{\top} > \end{bmatrix} : \boldsymbol u, \boldsymbol v \in \mathbb{F}^{ 1 \times \ell}, P,Q,R \in \mathfrak{gl}_{\ell}, Q=-Q^{\top}, R=-R^{\top} \}.$ > We get analogues to all the basis elements from before, plus ones for $\boldsymbol u$ ($\ell$-many), and $\boldsymbol v$ ($\ell$-many). Hence the dimension is $\frac{\ell(\ell-1)}{2} + \frac{\ell(\ell-1)}{2}+\ell^{2}+ \ell + \ell=2 \ell^{2}+\ell$. > > > For example, for $\mathfrak{so}_{5}$ a full basis would be > > ![[Pasted image 20250515192613.png|400]] > > > > $\begin{aligned} > \boldsymbol u :\;& > E_{12}-E_{21},\; > E_{13}-E_{31};\\[4pt] > \boldsymbol v :\;& > E_{14}-E_{41},\; > E_{15}-E_{51};\\[4pt] > P :\;& > \textcolor{Skyblue}{E_{22}-E_{44}},\; > \textcolor{Skyblue}{E_{33}-E_{55}},\; > E_{23}-E_{54},\; > E_{32}-E_{45};\\[4pt] > Q :\;& > E_{25}-E_{34};\\[4pt] > R :\;& > E_{43}-E_{52}. > \end{aligned} > $ > [[root space decomposition of a Lie algebra|Root space decomposition]] of $\mathfrak{so}_{5}(\mathbb{C})$: the [[Cartan subalgebra|CSA]] is $\mathfrak{t}= \text{span}(\textcolor{Skyblue}{E_{2 2}-E_{44}}, \textcolor{Skyblue}{E_{3 3}-E_{5 5}}).$ Let $e^{1},e^{2}$ be the [[dual basis]]. The form $A$ in the definition of $\mathfrak{so}_{n}$ was chosen so that the natural [[basis]] derived in the justification block about is in fact a [[simultaneously diagonalizable|simultaneous]] [[eigenbasis]] for the [[adjoint representation|adjoint]] [[Lie algebra representation|action]] of $\mathfrak{t}$. The corresponding simultaneous [[eigenvalue|eigenvalues]] $\alpha \in \mathfrak{t}^{*}$ correspond to the [[root system of a Lie algebra|roots]] of $\mathfrak{so}_{5}$. Recalling the [[vector space of m-by-n matrices|general fact]] $E_{ij}E_{k \ell}=\delta_{jk}E_{i \ell}$, it is easy to compute ($t \in \mathfrak{t}$): $\begin{align} \boldsymbol{u}&: \begin{cases} \textcolor{red}{[t, (E_{12}-E_{21})] = e^1(t)(E_{12}-E_{21})} \\ \textcolor{blue}{[t, (E_{13}-E_{31})] = e^2(t)(E_{13}-E_{31})} \end{cases} \\ \boldsymbol{v}&: \begin{cases} \textcolor{red}{[t, (E_{14}-E_{41})] = -e^1(t)(E_{14}-E_{41})} \\ \textcolor{blue}{[t, (E_{15}-E_{51})] = -e^2(t)(E_{15}-E_{51})} \end{cases} \\ P&: \begin{cases} [t, (E_{22}-E_{44})] = 0 \\ [t, (E_{33}-E_{55})] = 0 \\ \textcolor{green}{[t, (E_{23}-E_{54})] = (e^1-e^2)(t)(E_{23}-E_{54})} \\ \textcolor{green}{[t, (E_{32}-E_{45})] = (e^2-e^1)(t)(E_{32}-E_{45})} \end{cases} \\ Q&: \begin{cases} \textcolor{orange}{[t, (E_{25}-E_{34})] = (e^1+e^2)(t)(E_{25}-E_{34})} \end{cases} \\ R&: \begin{cases} \textcolor{orange}{[t, (E_{43}-E_{52})] = -(e^1+e^2)(t)(E_{43}-E_{52})} \end{cases} \end{align}$ From these calculations, we can identify the following root pairs of $\mathfrak{so}_5(\mathbb{C})$: 1. $\textcolor{red}{\pm e^1}$ (from $\boldsymbol{u}$ and $\boldsymbol{v}$ components) 2. $\textcolor{blue}{\pm e^2}$ (from $\boldsymbol{u}$ and $\boldsymbol{v}$ components) 3. $\textcolor{green}{\pm(e^1-e^2)}$ (from $P$ components) 4. $\textcolor{orange}{\pm(e^1+e^2)}$ (from $Q$ and $R$ components) This is the [[root system]] $B_{2}$. - [ ] $\mathfrak{so}_{6}$ ---- #### [^1]: Think about this condition as 'the derivative of preserving the bilinear form'. To me, it looks [[derivation|Leibniz rule]]-esque. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```