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Let $o(2l,F)$, with $l \ge2$ and $n=2l$ be the orthogonal lie algebra $\{L\in gl(n,F)|SL=-L^{t}S\}$ where $S=\begin{pmatrix} 0 &I_{l} \\ I_{l} & 0 \end{pmatrix}$. How can I show that $\dim(o(2l,F))=2l^{2}-l$?

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    @HenningMakholm, sorry for the confusion, I have edited the question.2012-01-31

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We write $L=\pmatrix{A&B\\ C&D}$. Then $L\in o(2l,F)$ iff $Sl=-L^tS$. We have $SL=\pmatrix{0&I\\ I&0}\pmatrix{A&B\\ C&D}=\pmatrix{C&D\\ A&B}$ and $-L^tS=\pmatrix{A^t&C^t\\ B^t&D^t}\pmatrix{0&I\\ I&0}=\pmatrix{-C^t&-A^t\\ -D^t&-B^t}$ so $C$ and $B$ have to be skew-symmetric and $D=-A^t$. So $L=\pmatrix{A&0\\ 0&-A^t}+\pmatrix{0&B\\ 0&0}+\pmatrix{0&0\\ C&0}$. Denoting $E_{rc}$ the matrix whose entries are $0$ except the term of the row $r$ and column $c$. Then $o(2l,F)$ is generated by $\left\{\pmatrix{E_{rc}&0\\\ 0&E_{cr}},1\leq c,r\leq l\right\}\cup\left\{\pmatrix{0&E_{rc}-E_{cr}\\\ 0&0},1\leq c which is linearly independent.

The dimension of $l\times l$ skew-symmetric matrices is $\frac{l(l-1)}2$ so $\dim o(2l,F)=2\frac{l(l-1)}2+l^2=l^2-l+l^2=2l^2-l.$