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Centralizer of $\mathfrak{so}_{p,q}(\mathbb{R})$

This notebook describe a SageMath script designed to compute the centralizer of the real Lie algebra $\mathfrak{so}{p,q}$ considered as a sub-algebra of $\mathfrak{gl}{p+q}(\mathbb{R})$.

$C_{\mathfrak{gl}_n}(\mathfrak{so}_{p,q})=\{X\in \mathfrak{gl}_{n}(\mathbb{R}),\ \forall S\in\mathfrak{so}_{p,q}\ \ [X,S] = 0\}$

We drop the notation $\mathbb{R}$ and set $n=p+q$ for simplicity.

# Define dimensions
p = 2; q = 2;
m = p + q; d = (m*(m-1))/2;


Let $E_{i,j}\in\mathfrak{gl}n$ be the matrix whose $(i,j)$ entry is $1$ and all others are $0$. A common basis of $\mathfrak{so}{p,q}$ is given by,

$\left\{\begin{array}{cll} A_{i,j} &= E_{i,j} - E_{j,i}\ &:\ 1\leq i< j\leq p, \\ B_{i,j} &= E_{i,j} + E_{j,i}\ &:\ 1\leq i\leq p,\ p+1\leq j\leq m, \\ D_{i,j} &= E_{i,j} - E_{j,i}\ &:\ p+1\leq i< j\leq m. \end{array}\right\}$
S = {} # basis for so(p,q)
k = 1

for i in range(0, p-1): # A_{i,j} = E_{i,j} - E_{j,i} \forall 1 \leq i < j \leq p
for j in range(i+1, p):
A = matrix(m)
A[i,j] = 1
S[k] = A - A.transpose()
k = k + 1

for i in range(p, m-1): # D_{i,j} = E_{i,j} - E_{j,i} \forall p+1 \leq i < j \leq m
for j in range(i+1, m):
A = matrix(m)
A[i,j] = 1
S[k] = A - A.transpose()
k = k + 1

for i in range(0, p): # B_{i,j} = E_{i,j} + E_{j,i} \forall 1\leq i\leq p, p+1\leq j\leq m
for j in range(p, m):
A = matrix(m)
A[i,j] = 1
S[k] = A + A.transpose()
k = k + 1


Let $X$ be an element of the centralizer, then certainly $X$ commutes with every element of the basis.

# Define commutators equations
x = var(['x_%d%d' %(i,j) for (i,j) in cartesian_product([range(1,m+1), range(1,m+1)])]);
X = matrix(SR, m, m, x)

E = {} # set of equations
J = {} # set of jacobian

for k in range(1, d+1):
E[k] = X*S[k] - S[k]*X
J[k] = jacobian(E[k].list(), x)

eqns = [] # list of equations
for k in range(1, d+1):
eqns = eqns + [(J[k]*vector(SR, x))[n] == 0 for n in range(m*m)]



We now just solve the equations and substitute the solution in $X$ to get a common representation of the centralizer.

# solve equations
sol = solve(eqns, x)
print X.subs(sol)

[r2  0  0  0]
[ 0 r2  0  0]
[ 0  0 r2  0]
[ 0  0  0 r2]


This code is consistent with the following result,

$C_{\mathfrak{gl}_2}(\mathfrak{so}_2) = \left\{\begin{pmatrix}x & y \\ -y&x\end{pmatrix}\ :\ x,y\in\mathbb{R}\right\} = \mathfrak{so}_2\oplus \text{span}\{\text{Id}_2\} \\ C_{\mathfrak{gl}_2}(\mathfrak{so}_{1,1}) = \left\{\begin{pmatrix}x & y \\ y&x\end{pmatrix}\ :\ x,y\in\mathbb{R}\right\} = \mathfrak{so}_{1,1}\oplus \text{span}\{\text{Id}_2\} \\ C_{\mathfrak{gl}_n}(\mathfrak{so}_{p,q}) = \text{span}\{\text{Id}_{n}\}\cong \mathbb{R},\quad n \geq 3.$ Download script

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