Matrix representations of Lie groups of type $B_n$

Question

For the Lie algebra $\mathfrak{so}(2n+1, \mathbb{C})$, there is a matrix representation given by the following matrices: \begin{align} \left( \begin{matrix} 0 & x & y \\ -y^T & A & B \\ -x^T & C & -A^T \end{matrix} \right), \end{align} where $x, y$ are $1 \times n$ matrices, $A,B,C$ are $n \times n$ matrices. Are there similar matrix representations for Lie group $\mathrm{SO}(2n+1, \mathbb{C})$? Thank you very much.

Edit: I forgot to mention that $B,C$ are skew-symmetric, $B=-B^T, C=-C^T$.

Answer 1

The quadratic form whose matrix is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & I_n \\ 0 & I_n & 0 \end{pmatrix}$ gives an embedding of $\operatorname{SO}(2n + 1, \mathbb C)$ in $\operatorname{GL}(2n + 1, \mathbb C)$ whose derivative is your specified embedding $\mathfrak{so}(2n + 1, \mathbb C) \to \mathfrak{gl}(2n + 1, \mathbb C)$. Under this embedding,

• $\left\{\begin{pmatrix} 1 & 0 & 0 \\ 0 & A & 0 \\ 0 & 0 & A^{-\mathsf T} \end{pmatrix}\mathrel: \text{$A$ diagonal}\right\}$ is the image of a maximal (algebraic) torus in $\operatorname{SO}(2n + 1, \mathbb C)$,
• the image of a maximal unipotent subgroup of $\operatorname{SO}(2n + 1, \mathbb C)$ is generated by
  • $\exp\begin{pmatrix} 0 & 0 & 0 \\ 0 & E_{ij} & 0 \\ 0 & 0 & -E_{ji} \end{pmatrix}$, where $i$ is less than $j$, and
  • $\exp\begin{pmatrix} 0 & 0 & y \\ -y^{\mathsf T} & 0 & B \\ 0 & 0 & 0 \end{pmatrix}$, where $B$ is skew-symmetric, and
• the image of the Tits subgroup of $\operatorname{SO}(2n + 1, \mathbb C)$ is generated by
  • $\begin{pmatrix} 1 & 0 & 0 \\ 0 & A & 0 \\ 0 & 0 & A^{-\mathsf T} \end{pmatrix}$, where $A = I - E_{ii} - E_{jj} + E_{ij} - E_{ji}$ with $i \ne j$, and
  • $\begin{pmatrix} 1 & 0 & 0 \\ 0 & D' & D'' \\ 0 & D'' & D' \end{pmatrix}$, where $D'$ and $D''$ are diagonal $(0, 1)$-matrices such that $D' + D'' = I_n$.

(I originally made a comment to this effect, but I had the wrong quadratic form, and wrongly suggested that you had switched $x^{\mathsf T}$ and $y^{\mathsf T}$ in your embedding.)