Lie algbra so(2n)?

Question

I am currently reading the book ``Symmetries, Lie Algebras and Representations: A graduate course for physicists'', by Jurgen Fuchs & Christoph Schweigert. On page 75 (unfortunately, the google book does not cover this page), I read:

Another simple subalgebra of $gl(2n)$ consists of those matrices of $M$ which fulfill the relation $M^tK+KM=0$, where $K$ is the $2n\times 2n$ -matrix $$K:=\begin{pmatrix} 0_n & 1_n \\ 1_n & 0_n \end{pmatrix}.$$

My question is how can such a definition be consistent with the normal definition of the lie algebra $so(2n)$ which consist of all $2n\times 2n$ antisymmetric matrices?

The authors also note after some sentences that

Note that, here the symbols $sp(n)$ and $so(n)$ refer to Lie algebras over complex numbers.

I am not sure is this remark relevant to my question.

Let's take $n=1$ for example. Let $M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then $M^tK+KM=0$ implies that $$\begin{pmatrix} a & c \\ b & d \end{pmatrix}\equiv M^t=-KMK^{-1}=-KMK=-\begin{pmatrix} d & c \\ b & a \end{pmatrix}.$$ That is, $a=-d$, $c=b=0$, meaning that $M=\begin{pmatrix} a & 0 \\ 0 & -a \end{pmatrix}$ which is inconsistent with the normal one $M=\begin{pmatrix} 0 & b \\ -b & 0 \end{pmatrix}$.

@DietrichBurde, Thank you very much for your remind. I will check the link. — Wein Eld, Sep 12 '16 at 12:56
You are right, this is confusing. But we obtain isomorphic Lie algebras, see the answer of Eric O. Korman. — Dietrich Burde, Sep 12 '16 at 13:00

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

Denote by $\mathfrak{g}(K)$ the Lie subalgebra of $\mathfrak{gl}(2n)$ defined as above. For $$K:=\begin{pmatrix} 0_n & 1_n \\ -1_n & 0_n \end{pmatrix}$$ we have $\mathfrak{g}(K)=\mathfrak{sp}(2n)$, the symplectic Lie algebra. For the orthogonal Lie algebras see this duplicate, and the book of J. Humphreys, where this is explained, too. For the field we would assume algebraically closed (of characteristic zero, although it works more general).

Lie algbra so(2n)?

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