Lie algebra associated with the orthogonal group $\operatorname{SO}(2n)$? How do I prove that the lie algebra associated with the even dimensional orthogonal group $\operatorname{SO}(2n)$ is given by matrices $B$ satisfying 
$B^\top  K + KB = 0$, where $K = U^\top U$, $U$ is another matrix, that is chosen.
 A: Any two symmetric positive definite inner products on a real vector space $V$ are equivalent.  Therefore the group $SO(V)$ may be defined (up to isomorphism) as the set of invertible linear maps $V \to V$ with determinant 1 preserving a positive definite inner product.
In this case you can consider the inner product on $\mathbb R^{2n}$ by $(v,w) = v^t K w$.  Since $K = U^t U$, $K$ is symmetric and positive definite (so that the inner product is also symmetric and positive definite).  Then $SO(2n)$ are all $A$ with determinant 1 such that $(Av,Aw) = (v,w)$ for all $v$ and $w$.  This happens if and only if $A^t K A = K$.  Now to find the Lie algebra, we differentiate this condition: let $A$ be a path in $SO(2n)$ with $A(0) = I$ and $A'(0) = B$.  Then differentiating $A(t)^t K A(t) = K$ at $t = 0 $ using the product rule shows that $B^t K + K B = 0$.  Conversely, if $B^t K + KB = 0$ then you need to show that $\exp(B) \in SO(2n)$.  But it's not too bad to work out that
\begin{align}\exp(B)^t K \exp(B) &=  \exp(B^t) K \exp (B) \\&= (1 + B^t + \frac{1}{2} {B^t}^2 + \dots) K (1 + B + \frac{1}{2} {B}^2 + \dots)\\& = 
K(1-B+ \frac{1}{2}{B}^2 - \dots) (1 + B + \frac{1}{2} {B}^2 + \dots)\\& = K\exp(-B) \exp (B)=K\end{align}
after using $B^t K + KB = 0$ repeatedly.
