ad-hoc basis in Frobenius Lie algebras Let $\mathfrak{g}$ be a Frobenius Lie algebra, that is, there exist $f \in \mathfrak{g}^{*}$ such that the bilinear form defined by $b(x,y)=f([x,y])$ is non-degenerate. My question is : Why can we assure that there exist some basis $\lbrace e_1, \ldots e_{2n} \rbrace$ such that the corresponding dual basis $\lbrace e^1,\ldots , e^{2n} \rbrace$ satisfies $f=e^1$ and $b=- \left( e^1 \wedge e^2 + e^3 \wedge e^4 + \cdots + e^{2n-1} \wedge e^{2n} \right)$ ?
I think the second part is clear, since $b$ is non-degenerate then $\hat{b}=-b$ is also non-degenerate and therefore there exist some basis $\lbrace e_1, \ldots e_{2n} \rbrace$ of $\mathfrak{g}$ such that $\hat{b}(e_{2i-1},e_{2i})=1$ (and thus $\hat{b}(e_{2i},e_{2i-1})=-1$) for $i=1, \ldots , n$ and $\hat{b}(e_i,e_j)=0$ otherwise.
Now, since
$$
\hat{b}= \sum_{i<j}^{}{a_{ij}e^{i} \wedge e^{j}}
$$
and $a_{ij}=\hat{b}(e_i,e_j)$ it follows from the last paragraph that
$$
\hat{b}= e^1 \wedge e^2 + e^3 \wedge e^4 + \cdots + e^{2n-1} \wedge e^{2n} 
$$
but I don't see why would we have $f=e^1$.
Any help?
In advance thank you.
 A: I thought I would expand my comment into a full answer. Firstly, lets note that we are being asked to find a basis $e_1, e_2, \dots, e_{2n}$ such that $e_2, \dots, e_{2n}$ span $\ker f$, $f(e_1) = 1$ and such that $b(e_{2i},e_{2i+1}) = -1$ and $b(e_k,e_l) = 0$ otherwise. We can fix some $e_1$ with $f(e_1) = 1$ without loss of generality at this point.
Note that we will be done if we can decompose $\mathfrak{g}$ into $\langle e_1,e_2 \rangle \oplus V$ where V is a nondegenerate subspace in $\ker f$ and is the orthocomplement of $\langle e_1,e_2 \rangle$ as then we can just fix an arbitrary Darboux basis of $V$ as you have done in your question.
In fact, these requirements force $V = e_1^\perp \cap \ker f$. This will have the right dimension since $e_1^\perp$ and $\ker f$ are both hyperplanes and they cannot coincide since $e_1 \in e_1^\perp$ but  $e_1 \notin\ker f$.
Now we need some of the theory of Frobenius Lie algebras. There is an element $F$ called the principal element $F \in \mathfrak{g}$ such that $f(x) = f([F,x])$ for all $x\in \mathfrak{g}$. This comes straight from the fact that $b$ is nondegenerate. We note that $b(F,e_1) = f(e_1) = 1$ and $b(F,V) = f(V) \equiv 0$. So take $e_2 := F$ and we're done:  $$\mathfrak{g} = \langle e_1,F \rangle \oplus V$$
