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> [!proposition] Proposition. ([[root space decomposition of sp4(C)]])
>
>
> Put $\mathfrak{g}:=\mathfrak{sp}_{4}(\mathbb{C})$.
>
> By [[diagonal matrices form Cartan subalgebras for special linear, orthogonal, symplectic Lie algebras]] we know that the [[Lie subalgebra|subalgebra]] $\mathfrak{t}$ of diagonal matrices is a [[Cartan subalgebra]] with [[basis]] $\{e_{1}= E_{11}-E_{33}, e_{2}=E_{22}-E_{44} \}.$ Let $\{ e^{1},e^{2} \} \subset \mathfrak{t}^{*}$ be the corresponding [[dual basis]].
>
> We want to find the set of $\alpha \in \mathfrak{t}^{*}$ [[root space decomposition of a Lie algebra|such that]]
> $\mathfrak{g}=\mathfrak{t} \oplus \bigoplus_{\alpha \in \Phi}\mathfrak{g}_{\alpha},$
> where $\mathfrak{g}_{\alpha}$ denotes the simultaneous eigenspace/weight space/root space $\mathfrak{g}_{\alpha}=\{ x \in \mathfrak{g}: [t, x] = \alpha(t) x \text{ for all } t \in \mathfrak{t} \}.$
> We know $\dim \mathfrak{g}=10$; since $\text{dim }\mathfrak{t}=2$ and for every root $\alpha$ its negation $-\alpha$ is also a root, we are looking for four roots $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}.$
>
> We need to find a [[simultaneously diagonalizable|simultaneous eigenbasis]] for the $\text{ad }t$, $t \in \mathfrak{t}$. Each eigenvector of this eigenbasis will give a weight.
>
> Let $t=c_{1}e_{1}+c_{2}e_{2} \in \mathfrak{t}$ be arbitrary. We have $\text{ad}_{t}(x)=[t,x]=tx-xt.$
> A general element of $\mathfrak{g}$ has the form $x=\begin{bmatrix}
> P & Q \\
> R & -P^{\top}
> \end{bmatrix}= \begin{bmatrix}
> p_{11} & p_{12} & q_{11} & q_{12} \\
> p_{21} & p_{22} & q_{12} & q_{22} \\
> r_{11} & r_{12} & -p_{11} & -p_{21} \\
> r_{12} & r_{22} & -p_{12} & -p_{22}
> \end{bmatrix}$
> where $Q$ and $R$ are symmetric. We may compute $[t,x]=tx - xt=\begin{bmatrix} 0 & c_1 p_{12} - c_2 p_{12} & 2c_1 q_{11} & c_1 q_{12} + c_2 q_{12} \\ -c_1 p_{21} + c_2 p_{21} & 0 & c_1 q_{12} + c_2 q_{12} & 2c_2 q_{22} \\ -2c_1 r_{11} & -c_1 r_{12} - c_2 r_{12} & 0 & c_1 p_{21} - c_2 p_{21} \\ -c_1 r_{12} - c_2 r_{12} & -2c_2 r_{22} & -c_1 p_{12} + c_2 p_{12} & 0 \end{bmatrix}$
> From this we can see the usual basis for $\mathfrak{sp}_{2 \ell}(\mathbb{C})$ (cf. example sheet 1):
> $\begin{align}
> E_{13} \\
> E_{14} + E_{23} \\
> E_{24} \\
> \ \\
> E_{31} \\
> E_{32} + E_{41} \\
> E_{42} \\
> \ \\
> E_{11}-E_{33} \\
> E_{22} - E_{44} \\
> E_{12} - E_{43} \\
> E_{21} - E_{34}
> \end{align}$
>
> is a simultaneous eigenbasis: $\begin{align*}
> [t, E_{13}] & = 2c_{1}E_{13} & & = 2e^{1}(t) \ E_{13} \\
> [t, E_{14}+E_{23}] & = (c_{1} + c_{2}) (E_{14}+E_{23}) & & =(e^{1}+e^{2})(t) \ (E_{14}+E_{23}) \\
> [t, E_{24}] & = 2c_{2}E_{24} & & = 2e^{2}(t) \ E_{24} \\
> [t, E_{31}] & = -2c_{1}E_{31} & & = -2e^{1}(t) \ E_{31} \\
> [t, E_{32} + E_{41}] & = -(c_{1} + c_{2}) (E_{32} + E_{41}) & & = -(e^{1}+e^{2})(t) \ (E_{32}+E_{41}) \\
> [t, E_{42}] & = -2c_{2}E_{42} & & = -2e^{2}(t) \ E_{42} \\
> [t, E_{11}- E_{33}] & = 0 & & = 0 \\
> [t, E_{22}-E_{44}] & = 0 & & = 0 \\
> [t, E_{12}- E_{43}] & = (c_{1}-c_{2}) (E_{12} - E_{43}) & & = (e^{1}-e^{2})(t) \ (E_{12}-E_{43}) \\
> [t, E_{21}-E_{34}] & = -(c_{1}-c_{2})(E_{21}-E_{34}) & & = -(e^{1}-e^{2})(t) \ (E_{21}-E_{34})
> \end{align*}$
> So we have nonzero roots $\begin{align}
> \pm 2 e^{1}, \pm 2 e^{2}, \pm (e^{1}+e^{2}), \pm (e^{1}-e^{2})
> \end{align}$whose corresponding root spaces are just the span of the provided basis elements for $\mathfrak{sp}_{4}(\mathbb{C})$. The zero root space $\mathfrak{t}$ is two-dimensional with basis $\{ E_{11} - E_{33}, E_{22}-E_{44} \}$ (as we already knew.) Thus, $\mathfrak{sp}_{4}(\mathbb{C}) = \mathfrak{t} \oplus \mathfrak{g}_{\pm e^{1}} \oplus \mathfrak{g}_{\pm e^{2}} \oplus \mathfrak{g}_{\pm(e^{1} + e^{2})} \oplus \mathfrak{g}_{\pm (e^{1}-e^{2})}.$
>
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####
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