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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?

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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.

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