Starting from the definition of Lie groups,
In particular $Sp (p,q)$ : The group of matrices in $Sp(p+q,\mathbb{C})$ which leave invariant the Hermitian form \begin{align} & Z^T K_{p,q} \bar{Z} \qquad i.e., \qquad g^T K_{p,q} \bar{g} = K_{p,q}, \quad \det(g)=1 \\ & K_{p,q} = \begin{pmatrix} -I_p & 0 & 0 & 0\\ 0 & I_q & 0 & 0 \\ 0 & 0 & -I_p & 0 \\ 0 & 0 & 0 & I_q \end{pmatrix} \end{align} I want to find the expression of Lie algebra $sp(p,q)$.
So let $g = \exp[t Z]$ , $Z \in sp(p,q)$, $t\in \mathbb{R}$ then it satisfies $Z^T K_{p,q} + K_{p,q} \bar{Z}=0$ Thus we have
\begin{align} 0 =Z^T K_{p,q} + K_{p,q} \bar{Z}= \begin{pmatrix} - \bar{Z}_{11} - Z_{11}^T & - \bar{Z}_{12} + Z_{21}^T & - \bar{Z}_{13} - Z_{31}^T & - \bar{Z}_{14} + Z_{41}^T \\ - Z_{12}^T + \bar{Z}_{21} & \bar{Z}_{22} + Z_{22}^T & \bar{Z}_{23} - Z_{32}^T & \bar{Z}_{24} + Z_{42}^T \\ - Z_{13}^T - \bar{Z}_{31} & Z_{23}^T - \bar{Z}_{32} & - \bar{Z}_{33} - Z_{33}^T & - \bar{Z}_{34} + Z_{43}^T \\ - Z_{14}^T + \bar{Z}_{41} & Z_{24}^T + \bar{Z}_{42} & - Z_{34}^T + \bar{Z}_{43} & \bar{Z}_{44} + Z_{44}^T \end{pmatrix} \end{align}
The textbook says the desired form of element of $sp(p,q)$ is \begin{align} Z = \begin{pmatrix} Z_{11} & Z_{12} & Z_{13} & Z_{14} \\ \bar{Z}_{12}^T & Z_{22} & Z_{14}^T & Z_{24} \\ -\bar{Z}_{13} & \bar{Z}_{14} & \bar{Z}_{11} & -\bar{Z}_{12} \\ \bar{Z}_{14}^T & -\bar{Z}_{24} & -Z_{12}^T & \bar{Z}_{22} \end{pmatrix} \end{align} where $Z_{ij}$ are complex matrix. $Z_{11}$ and $Z_{13}$ of order $p$ and $Z_{12}$ and $Z_{14}$ are $p\times q$ matrices. $Z_{11}$ and $Z_{22}$ are skew-Hermitian and $Z_{13}$ and $Z_{24}$ are symmetric.
now i am confusing why $Z_{33}$, $Z_{44}$ terms are not needed. And again other terms like $Z_{23}$, $Z_{34}$ does not appear.
Am i missing some constraints? From $det(g)=1$ so $tr(Z)=0$, so this terms only related with diagonal terms, but what about others?
It seems that i am missing something...
I was missed $Z \in sp(p,q)$ is also element in $sp(p+q,\mathbb{C})$
So Write \begin{align} &Z = \begin{pmatrix} Z_{11} & Z_{12} & Z_{13} & Z_{14} \\ Z_{21} & Z_{22} & Z_{23} & Z_{24} \\ Z_{31} & Z_{32} & Z_{33} & Z_{34} \\ Z_{41} & Z_{42} & Z_{43} & Z_{44} \end{pmatrix} = \begin{pmatrix} X_1 & Z_2 \\ X_3 & X_4 \end{pmatrix} \\ & X_1 = \begin{pmatrix} Z_{11} & Z_{12} \\ Z_{21} & Z_{22} \end{pmatrix}, \quad X_2 = \begin{pmatrix} Z_{13} & Z_{14} \\ Z_{23} & Z_{24} \end{pmatrix}, \quad X_3 = \begin{pmatrix} Z_{31} & Z_{32} \\ Z_{41} & Z_{42} \end{pmatrix}, \quad X_4 =\begin{pmatrix} Z_{33} & Z_{34} \\ Z_{43} & Z_{44} \end{pmatrix} \end{align}
\begin{align} 0 =Z^T K_{p,q} + K_{p,q} \bar{Z}= \begin{pmatrix} - \bar{Z}_{11} - Z_{11}^T & - \bar{Z}_{12} + Z_{21}^T & - \bar{Z}_{13} - Z_{31}^T & - \bar{Z}_{14} + Z_{41}^T \\ - Z_{12}^T + \bar{Z}_{21} & \bar{Z}_{22} + Z_{22}^T & \bar{Z}_{23} - Z_{32}^T & \bar{Z}_{24} + Z_{42}^T \\ - Z_{13}^T - \bar{Z}_{31} & Z_{23}^T - \bar{Z}_{32} & - \bar{Z}_{33} - Z_{33}^T & - \bar{Z}_{34} + Z_{43}^T \\ - Z_{14}^T + \bar{Z}_{41} & Z_{24}^T + \bar{Z}_{42} & - Z_{34}^T + \bar{Z}_{43} & \bar{Z}_{44} + Z_{44}^T \end{pmatrix} \label{estar} \end{align}
Want the form \begin{align} Z = \begin{pmatrix} Z_{11} & Z_{12} & Z_{13} & Z_{14} \\ \bar{Z}_{12}^T & Z_{22} & Z_{14}^T & Z_{24} \\ -\bar{Z}_{13} & \bar{Z}_{14} & \bar{Z}_{11} & -\bar{Z}_{12} \\ \bar{Z}_{14}^T & -\bar{Z}_{24} & -Z_{12}^T & \bar{Z}_{22} \end{pmatrix} \end{align} where $Z_{ij}$ are complex matrix. $Z_{11}$ and $Z_{13}$ of order $p$ and $Z_{12}$ and $Z_{14}$ are $p\times q$ matrices. $Z_{11}$ and $Z_{22}$ are skew-Hermitian and $Z_{13}$ and $Z_{24}$ are symmetric.
This can be achieved considering $Z \in sp(p+q,\mathbb{C})$, so \begin{align} Z = \begin{pmatrix} X_1 & X_2 \\ X_3 & X_4 \end{pmatrix} = \begin{pmatrix} X_1 & X_2 \\ X_3 & - X_1^T \end{pmatrix} \end{align} where $X_2$ and $X_3$ are symmetric. Thus \begin{align} & X_4 = - X_1^T \\ & \qquad \Rightarrow \qquad Z_{33} = -Z_{11}^T = \bar{Z}_{11} \\ & \qquad \Rightarrow \qquad Z_{34} = - Z_{21}^T = - \bar{Z}_{12} \\ & \qquad \Rightarrow \qquad Z_{44} = Z_{22}^T = \bar{Z}_{22} \\ & \qquad \Rightarrow \qquad Z_{43} = -Z_{12}^T \\ & X_2 = X_2^T \\ & \qquad \Rightarrow \qquad Z_{13} = Z_{13}^T, \qquad Z_{24} = Z_{24}^T \\ & \qquad \Rightarrow \quad Z_{13} = - \bar{Z}_{31}^T \quad \Rightarrow \quad Z_{31} = - \bar{Z}_{13}^T = -\bar{Z}_{13}\\ & \qquad \Rightarrow \quad Z_{24} = - \bar{Z}_{42}^T \quad \Rightarrow \quad Z_{42} = - \bar{Z}_{24}^T = -\bar{Z}_{24} \\ & \qquad \Rightarrow \quad Z_{14} = Z_{23}^T \quad \Rightarrow \quad Z_{23} = Z_{14}^T \\ & \qquad \textrm{from} \quad Z_{14} = \bar{Z}_{41}^T \quad \Rightarrow \quad Z_{41} = \bar{Z}_{14}^T \end{align} Note in the process we used above constraint, then it provide desired result
Note the traceless condition comes naturally from
\begin{align} tr(A+\bar{A})= tr(A) + tr(\bar{A}) = tr(A) + tr(-A^T) = 0 \end{align} where $A$ is skew-hermitian matrix