Fierz identity for symplectic group For the fundamental representation of $SU(N)$, there is a Fierz identity:
$$
\sum_iT^i_{ab}T^i_{cd}=\frac{1}{2}\left(\delta_{ad}\delta_{bc}-\frac{1}{N}\delta_{ab}\delta_{cd}\right)
$$
where $T^i$ is the $i$th generator of $SU(N)$ normalized as
$$
{\rm Tr}(T^iT^j)=\frac{1}{2}\delta^{ij}.
$$
My question is: For the symplectic group $Sp(N)$, what is the Fierz identity? Namely, suppose $T^i$ is the $i$th generator of $Sp(N)$, can 
$$
\sum_iT^i_{ab}T^i_{cd}
$$
be written as something like the identity above?
 A: The main idea of the Fierz identities is to find an orthogonal basis for the space of matrices, or a subspace like the space of hermitian traceless matrices in the case of $SU(N)$ since the generators are traceless and hermitian. For $Sp(2n)$ the generators satisfy
$$
T_{I}^{K}\Omega_{KJ}+T_{J}^{K}\Omega_{IK}=0\qquad \text{with}\qquad\Omega_{IJ}=-\Omega_{IJ}
$$
If we use the symplectic metric $\Omega_{IJ}$, and its inverse $\Omega_{IJ}\Omega^{JK}=\delta_{I}^{K}$, to raise and lower indices we can solve this constraint by imposing 
$$T_{IJ}=T_{JI}$$
so the generators of $Sp(2n)$ are all the symmetrical matrices (in contrasts with the $SO(n)$ where the generators are the anti-symmetriccal ones). 
Fixing an orthogonal basis for these symmetrical matrices, say $(T_a)_{IJ}$, with $a=1$ to $(2n^{2}+n)$, by the inner product 
$$
tr(T_{a}T_{b})\equiv(T_{a})^{I}_{J}(T_{b})_{I}^{J}=-(T_{a})^{IJ}(T_{b})_{IJ}=\delta_{ab}
$$
Using the fact that $(T_{a})_{IJ}$ span the space of symmetrical matrices, for a general matrix $M_{IJ}$ we can write
$$
M_{(IJ)}=\frac{1}{2}(M_{IJ}+M_{JI})=M^{a}(T_{a})_{IJ}
$$
where $M^{a}$ can be obtained by contracting the indices $IJ$ with $(T_b)^{IJ}$ on both sides, giving 
$$
M^{a}=-\delta^{ab}(T_b)^{IJ}M_{IJ}\implies M_{(IJ)}=-\delta^{ab}(T_b)^{KL}M_{KL}(T_{a})_{IJ}
$$
Since this is true for general matrix $M_{KL}$ we have:
$$
\delta^{ab}(T_{a})^{KL}(T_{b})_{IJ}=-\delta_{(I}^{K}\delta_{J)}^{L}
$$
Raising and lowering indices we obtain more two identities:
$$
\delta^{ab}(T_{a})_{KL}(T_{b})_{IJ}=\Omega_{K(I}\Omega_{J)L}
$$
and
$$
\delta^{ab}(T_{a})_{K}^{L}(T_{b})_{I}^{J}=\frac{1}{2}(\delta_{K}^{J}\delta_{I}^{L}-\Omega_{KI}\Omega^{JL})
$$
up to some signs that I might be missing.
A: First, let me say that a good place to look at would be [1]. I haven't checked there so I'm not sure, but it contains technology for computing such things even for exceptional groups. It uses a so-called "birdtrack" notation that treats tensor contractions as graphs.
Here however is a way to brute force it. I didn't try it myself so I don't know if it's doable.
The generators of $\mathrm{Sp}(2n)$ may be taken as matrices
$$
T^a = \left\lbrace\begin{aligned}
&A^a & 1 \leq &a \leq n^2 \\
&B^a & n^2+1 \leq &a \leq \tfrac12n(3n + 1) \\
&C^a & \tfrac32n(n + 1) \leq &a \leq n(2n + 1)
\end{aligned}\right.
$$
where $A^a,B^a$ and $C^a$ are defined as $2\times 2$ block matrices with $n\times n$ blocks
$$
A^a=\left(\begin{matrix}m^a & 0 \\0& -m^{a\,\mathsf{T}}\end{matrix}\right)\,,\qquad
B^a=\left(\begin{matrix}0 & s^{a-n^2} \\0& 0\end{matrix}\right)\,,\qquad
C^a=\left(\begin{matrix}0 & 0 \\s^{a-\frac12n(3n+1)}& 0\end{matrix}\right)\,.
$$
with $m^a$ being a matrix with all zeros and a $1$ in the $a$th entry and $s^a$ being the $a$th matrix in the basis of symmetric matrices. If we let $a$ be a multi index $a \to (\mu,\nu)$ we can write those generators explicitly in terms of Kronecker deltas
$$
\begin{aligned}
(m^{\mu\nu})_{ij} &= \delta^\mu_{i} \delta^\nu_j\,,\\
(s^{\mu\nu})_{ij} &= \tfrac{1}{2\sqrt{2}}\left(\delta^\mu_{i} \delta^\nu_j+\delta^\mu_{j} \delta^\nu_i\right)\,.
\end{aligned}
$$
I'm not entirely sure about the normalization.${}^1$ Now you can write your sum as (sorry if I named the indices differently)
$$
\sum_{a=1}^{n(2n+1)} (T^a)_{IJ}\,(T^a)_{KL} = \sum_{\mu,\nu = 1}^n (T^{\mu\nu})_{IJ}\,(T^{\mu\nu})_{KL}\,,
$$
and now you'll have many cases according to whether $I,J,K,L$ is smaller or larger than $n$. If it's smaller then $I\to i$ and if it's bigger $I \to i - n$, and the appropriate blocks need to be considered. In the end you'll end up with contractions involving only $\delta^\mu_i$ which can be done easily for general $N = 2n$.

[1] Cvitanović, Predrag. Group Theory: Birdtracks, Lie's, and Exceptional Groups, Chapter 12.

$\qquad{}^1$ Anyway you can trace everything in the end and see if it's correct.
