# 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.

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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 $$\exp(B)^t K \exp(B) = \exp(B^t) K \exp (B) = (1 + B^t + \frac{1}{2} {B^t}^2 + \ldots) K (1 + B + \frac{1}{2} {B}^2 + \ldots) = K$$ after using $B^t K + KB = 0$ repeatedly.