Let there be an antisymmetric tensor field $\Omega_{ab}(q)$ where $q^i$ are coordinates on a 2N dimensional manifold. For context, this is a general symplectic form on phase space. I want to find a coordinate transformation that takes this matrix to its canonical form $$\Omega^0=\begin{bmatrix}0&-\mathbb{I}_{N\times N}\\\mathbb{I}_{N\times N}&0\end{bmatrix}$$.
The general coordinate transformation will take the variable $q^i$ to a new variable $p^i$
$$q^i =p^i - \xi^i$$
Then the coordinate change affects the original matrix as
$$\Omega_{ab}(q)=\Omega_{cd}^0\frac{\partial p^c}{\partial q^a}\frac{\partial p^d}{q^b}$$
We can expand $p$ to get
\begin{align} \Omega_{ab}(q)&=\Omega_{cd}^0\frac{\partial p^c}{\partial q^a}\frac{\partial p^d}{q^b}\\ &=\Omega_{cd}^0\big(\delta^c_a+\frac{\partial\xi^c}{\partial q^a}\big)\big(\delta^d_b+\frac{\partial\xi^d}{\partial q^b}\big)\\ &=\Omega^0_{ab}+\Omega^0_{cb}\frac{\partial \xi ^c}{\partial q^a}+\Omega^0_{ac}\frac{\partial \xi ^c}{\partial q^b}+\Omega^0_{cd}\frac{\partial \xi ^c}{\partial q^a}\frac{\partial \xi ^d}{\partial q^b} \end{align} We can define a 1-form $\xi_a =\Omega^0_{ab}\xi^b$ to get
\begin{equation} \Omega_{ab}=\Omega^0_{ab}+\frac{\partial \xi_a}{\partial q^b}-\frac{\partial \xi_b}{\partial q^a}+\frac{\partial \xi ^c}{\partial q^a}\frac{\partial \xi_c}{\partial q^b} \end{equation}
My question regards this last term, the "quadratic piece" of the coordinate transformation. Is that term zero given the antisymmetry in the indices $ab$? If not, why not?