The group $Sp(8)$ has 36 generators which obey the relation $J=\begin{pmatrix} 0 & I_{4}\\ -I_{4} & 0\\ \end{pmatrix}$ such that $G^TJG=J$ where $G$ are the generators of the group.
I would like to know the 8-dimensional representation of the generators. Does any one know how I can derive them?
Any help is appreciated.
There might be a language problem involved. By "generators" you mean Lie algebra elements, and not group elements that preserve the symplectic matrix, so your Gs are the exponentials of the 36 generators you are seeking.
The symplectic Lie algebras are for real symmetric matrices what the orthogonal Lie algebras are for antisymmetric ones. Consider the real 8×8 symmetric matrices S, 36 independent ones in all. Since $JJ=1\!\! 1$, The matrices $$ K\equiv JS$$ close under commutation, $$ J S ~ JS'- JS' ~ J S = J S'' \leftrightarrow S J S' - S' J S = \hbox {symmetric}. $$ Moreover, $$ \exp (-SJ ) ~ J ~ \exp (JS)= \exp (-SJ ) \exp (SJ) ~J= J, $$ by intercalating pairs of Js in-between powers in the schematic expansion of exponentials.
Thus, these Ks are a good basis of Lie algebra generators, in the fundamental (defining) representation.
Some people prefer the quaternionic parameterization, that is 8×8 antihermitean matrices of the form $$ K= 1\!\!1 \otimes A + i\vec{\sigma}\cdot \otimes \vec {S}, $$ where the left tensor factor is 2×2 matrices, while the right is 4×4 matrices. The As are all antisymmetric matrices in that space, so 6 independent ones in all, while each component of the $\vec S$ triplet is a symmetric matrix, with 10 independent components each. You then have 3×10+6=36 independent matrices.
First, check the closure of this set under commutation, and transpose your commutation relations to convince yourself that, indeed, you recover the same relations consistently. Nice fit!
The symplectic metric is $J=i\sigma_2\otimes 1\!\!1$, and, indeed, $$ e^{-1\!\!1 \otimes A + i\vec{\sigma}^T\cdot \otimes \vec {S} }~ ( \sigma_2\otimes 1\!\!1 ) ~e^{1\!\!1 \otimes A + i\vec{\sigma}\cdot \otimes \vec {S}}= \sigma_2\otimes 1\!\!1 .$$ Note how, in commuting the "conjugation matrix" $\sigma_2$ past the leftmost exponential to let it annihilate the rightmost one, you pick up a minus sign for the $\sigma_1$ and the $\sigma_3$, the symmetric ones, whereas the $\sigma_2$ term is unaffected, but it is antisymmetric, so the minus sign comes from that. As a consequence, commutation past $\sigma_2$ yields the exact inverse of the rightmost exponential, and -iJ is left unaffected.
There are other people who would insist on real forms, so I am just appending the customary real basis from Iachello's text, which, however, is substantially less intuitive: