Show that $\mathfrak{sp}_{2n}(\mathbb{C})$ is a sub lie-algebra of $\mathfrak{gl}(\mathbb{C}^{2n})$

Question

I want to show that $\mathfrak{sp}_{2n}(\mathbb{C})$ is a sub lie-algebra of $\mathfrak{gl}(\mathbb{C}^{2n})$. Where $$\mathfrak{sp}_{2n}(\mathbb{C})=\{X \in End(\mathbb{C}^{2n})| \Omega(Xv,w)+\Omega(v,Xw)=0, \text{for all} \ v,w \in \mathbb{C}^{2n}\},$$ where $\Omega(x,y)$ is given by $x^TMy$ with $M= \begin{pmatrix} 0 & I_n\\-I_n & 0\end{pmatrix} \in M_{2n}(\mathbb{C}). $ Furthermore, I need to know what form does $X \in \mathfrak{sp}_{2n}(\mathbb{C})$ explicitly have in regards to a block matrix.

To show that $\mathfrak{sp}_{2n}(\mathbb{C})$ is a sub lie-algebra, I tried to show that $[X,Y]=XY-YX \in \mathfrak{sp}_{2n}(\mathbb{C})$, So $$\Omega((XY-YX)+\Omega(v,(XY-YX)w)=\Omega(XYv-YXv,W)+\Omega(v,XYw-YXw)=$$

$$=\Omega(XYv,w)-\Omega(YXv,w)+\Omega(v,XYw)-\Omega(v,YXw).$$ However, I do not know how to proceed and to show that it is actually zero. Additionally, I am not sure how to derive a form for an arbitrarily $X\in \mathfrak{sp}_{2n}(\mathbb{C})$. Thanks for your help!!

Answer 1

This is more transparent if you make the following observations:

1. $X\in{\mathfrak g}{\mathfrak l}_{2n}$ belongs to ${\mathfrak s}{\mathfrak p}_{2n}$ iff $X^\dagger = -X$, where $(-)^\dagger$ means the transpose with respect to $\Omega$, i.e. $\Omega(X,-)=\Omega(-,X^\dagger)$. (If you replace $\Omega$ with the standard scalar product, this is the ordinary transpose)
2. The transpose has the property that $(XY)^\dagger=Y^\dagger X^\dagger$.

From that, you can calculate that for $X^\dagger = -X$ and $Y^\dagger = -Y$, also $[X,Y]^\dagger = -[X,Y]$.

Answer 2

Hint: If $X, Y \in \mathfrak{sp}_{2n}(\mathbb{C})$, then $\Omega(XYv,w) = -\Omega(Yv, Xw) = \Omega(v, YXw)$.