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I am learning about Lie algebras and I do not understand the following subalgebra of $\mathfrak gl_{V}$. Let $V$ be a vector space and $\mathfrak gl_{V}$ be the Lie algebra of endomorphisms on $V$. Let $B$ be a bilinear form on a vector space V. Define

$o_{V,B}=\{ a\in \mathfrak{gl}_{V}| B(a(u),v)=-B(u,a(v))~~~ \forall u,v\in V \}$

Im not exactly sure how to ask this question, but I just do not understand this subalgebra. I would like to have some intuition, possibly with an example. Thanks

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up vote 1 down vote accepted

If $V$ is $\mathbb R^n$ and $B$ is the ordinary dot product, then your Lie algebra is the skew symmetric matrices. Take column vectors $u,v$ and their transposes $v', u'$ as row vectors, also take a skew symmetric matrix $A$ and its transpose $A' = -A.$ Then the value of your $B(a(u), v)$ means $$ Au \cdot v = v' A u. $$ Now the transpose of a 1 by 1 matrix is itself, so $$ Au \cdot v = v' A u = (v' A u)' = u' A' v = - u' A v = - u \cdot Av $$ where the final bit is your $-B(u,a(v)).$

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If $B(u,v)=u \;{\bf \cdot}\; v$ (the regular dot product) on $\mathbb{R}^n$, then $B(a(u),v)=-B(u,a(v)$ amounts to $u^TA^Tv=-u^TAv$ for all $u,v \in \mathbb{R}^n$. Thus $\mathfrak{o}=\{A \in \mathfrak{gl}_n(\mathbb{R}) \;|\; A^T+A=0\}$. This is the orthogonal Lie algebra which is associated with the orthogonal Lie group.

If the matrix of the bilinear form is $B = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \end{bmatrix}$. Then $\mathfrak{o}$ is the symplectic Lie algebra which is associated with the group of the same name.

The classification of finite dimensional simple Lie algebras (over $\mathbb{C}$), says that each simple algebra is isomorphic to one of $A_n$ ($n \geq 1$), $B_n$ ($n \geq 2$), $C_n$ ($n \geq 3$), $D_n$ ($n \geq 4$), $F_4$, $E_6$, $E_7$, $E_8$, or $G_2$.

The orthogonal algebras make up types $B$ and $D$ and the symplectic algebras make up type $C$. So this construction generalizes both and accounts for "most" of the finite dimensional simple Lie algebras.

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