Evaluating the isomorphism $\wedge^{2}V\cong\mathfrak{so}(2n)$ on basis elements. Im back with another question about section 20 of Fulton and Harris (a little earlier than I had hoped).
Let $V$ be an even dimensional complex vector space with a quadratic form $Q$ and an orthonormal basis $\{e_{1},\cdots,e_{2n}\}$. We have the following isomorphism of Lie algebras:
$$
\phi:\wedge^{2}V\to\mathfrak{so}(2n)\subseteq\text{End}(V)\text{ , }\ a\wedge b\mapsto\phi_{a\wedge b}
$$
where
$$
\phi_{a\wedge b}(v)=2(\langle b,v \rangle a-\langle a, v\rangle b)
$$
Fulton and harris claim that this isomorphism maps the standard basis elements $\frac{1}{2}e_{i}\wedge e_{n+j}$ of $\wedge^{2}V$ to $E_{i,j}-E_{n+j,n+i}$, where the $E_{i,j}$ are $2n\times 2n$ matrices with the $i$-th $j$-th entry $1$ and all other entries $0$.
This should be an easy enough calculation, but it is not working out. I proceed as follows:
$$
\phi_{\frac{1}{2}e_{i}\wedge e_{n+j}}(e_{k})=\langle e_{n+j},e_{k}\rangle e_{i}-\langle e_{i}, e_{k}\rangle e_{n+j}
=
\delta_{n+j,k}e_{i}-\delta_{i,k}e_{n+j}
$$
While on the other hand:
$$
(E_{i,j}-E_{n+j,n+i})e_{k}
=
\delta_{j,k}e_{i}
-
\delta_{n+i,k}e_{n+j}
$$
These two expressions are clearly different, and so either I have made an elementary mistake somewhere, or I have a fundamental misunderstanding of something. I would really appreciate it if somebody could point out my mistake, or help to reconcile these two expressions.
 A: I have managed to resolve my issue.
In the above calculation I had been using the standard bilinear form $\langle x,y \rangle=x^{T}y$ which is associated with the quadratic from given by the identity matrix. On the other hand Fulton and Harris were using the bilinear form given by $\langle x,y \rangle=x^{T}Sy$, where $S$ is the block anti-diagonal matrix $S=\begin{pmatrix}0&\mathbb{1}_{n}\\\mathbb{1}_{n}&0\end{pmatrix}$.
Both these bilinear forms are symmetric and non-degenerate and so (at least over $\mathbb{C}$), they define isomorphic copies of the special orthogonal group and the associated Lie algebra.
Despite both forms leading to isomorphic Lie algebras, the elements look different. With the first convention the elements are antisymmetric matrices, whereas for the second convention they are matrices $A$ such that $A^{T}S+SA=0$. In particular, the given $H_{i}$ form a Cartan subalgebra in the second convention, but do not even belong to $\mathfrak{so}(2n)$ in the first convention.
Using the second convention we have that for $1\leq i,j\leq n$, $\langle e_{i},e_{j}\rangle=\langle e_{n+i},e_{n+j}\rangle=0$ and $\langle e_{i},e_{n+j}\rangle=\langle e_{n+i},e_{j}\rangle=\delta_{ij}$, which leads to the desired result.
