Mixed symmetrization and antisymmetrization / Combinatorics I have the following sum of 10 terms:
$$
\delta^{ab}f^{cde} + \delta^{ac}f^{bde} + \delta^{ad}f^{bce} + \delta^{ae}f^{bcd} +
\delta^{bc}f^{ade} + \delta^{bd}f^{ace} + \delta^{be}f^{acd} + \delta^{cd}f^{abe} +
\delta^{ce}f^{abd} + \delta^{de}f^{abc}
$$
In other words I consider all permutations of 5 indices, but only use those for which the first two indices and the last three are ordered (at the same time).
On top op that, $\delta$ is symmetric and $f$ is fully antisymmetric.
What I am looking for is some short-hand notation which would evaluate exactly to this sum. Consider the following sum as an easy example:
$$
\delta^{ab}\delta^{cd} + \delta^{ac}\delta^{bd} + \delta^{ad}\delta^{bc} =
3\,\delta^{(ab}\delta^{cd)}
$$
Normally $\delta^{(ab}\delta^{cd)}$ would evaluate to 24 terms, but because of the symmetry property of $\delta$, these simplify to three. I am looking for a similar notation for the first sum.
Because $\delta$ is symmetric and $f$ antisymmetric, one has $\delta^{(ab}f^{cde)}=0$ and $\delta^{[ab}f^{cde]}=0$, so these don't fit. And $\delta^{(ab)}f^{[cde]}$ is incorrect as it doesn't mix the two sets of indices. I came up with some kind of "mixed symmetrization":
$$\delta^{(ab\,|}f^{cde]} $$
where I defined:
$$
\begin{align}
T^{(a_1 \cdots a_m \, |\, a_{m+1} \cdots a_n]} &= \text{sum of all } n! \text{ permutations, where each permutation gets a sign depending} \\ &\text{ on the number of permutations needed to put } \mathcal{P}\left(a_{m+1} \cdots a_n \right) \text{ in canonical} \\ &\text{ order.}
\end{align}
$$
This indeed evaluates (up to a factor 10) to the first sum, but it feels a bit awkward to introduce a notation that is not generally usable (and for which properties have to re-derived). As these kind of "ordered" sums are for sure not uncommon, I expect them to be treated in some corner of combination theories..
Does anybody know whether such 'mixed symmetrisation" already exists in literature?
Or even better, does anybody know of a simple way to rewrite the first sum, maybe in some combinatorics notation?
Many thanks in advance!
 A: What you are doing amounts to computing a tensor product and decomposing it into irreducible components. There is a standard way of doing this with Littlewood-Richardson rule, Young symmetrizers etc. and there are nice pictures, Young diagrams (http://en.wikipedia.org/wiki/Young_tableau), that help to visualize different types of symmetries. In many cases it is sufficient to use Young diagrams, so there is no need to write indices directly. For example, $\delta$ is a rank two symmetric, it is depicted by a diagram made of two boxes in a row, we can say it is $(2,0,0,...)$ with the meaning that all other rows are of zero length. Your $f$ is a rank three antisymmetric, it is depicted by a diagrams with three boxes in a column, $(1,1,1,0,...)$. Then $\delta \otimes f$ contains two irreducible components, $(3,1,1,0,...)$ and $(2,1,1,1)$. Each box corresponds to one index and different arrangements of the same number of boxes are known to correspond to different types of tensors one can have with the same number of indices. You can look in Hamermesh or Fulton&Harris.
There are several notations that can be of use for you and are actually used by people. First of all it is convenient to denote all indices that belong to the group of indices in which a tensor is symmetric or antisymmetric by the same letter. For example $f^{uuu}$ or $f^{u[3]}$ instead of your $f^{abc}$ and $\delta^{aa}$ or $\delta^{a(2)}$ for your $\delta^{ab}$ and I used round(square) brackets to indicate the number of indices and whether they are symmetric or antisymmetric. But this works for rather simple types of symmetries, like the one you need.
In the case of $T^{a(n)|u[m]}$ that you gave ($n$ symmetric indices and $m$ antisymmetric) there are still only two irreducible components one is given by $T^{a(s-1)u|u[m]}$ and I assume antisymmetrization over all $u$ indices. Since it is already antisymmetric in the last $u[m]$ this requires $m+1$ terms. The second irreducible components is given by $t^{a(n)|au[m-1]}$ and it requires $n+1$ symmetric permutations. This is just a shorthand notation to same time.
The symmetrization operator you defined is strange and I cannot see that you followed your own recipe in the first formula, for example the 8th term $\delta^{cd}f^{abe}$ must be accompanied by $-\delta^{dc}f^{abe}$, this is another permutation that belongs to $5!$ and there is a sign needed according to your procedure. But the two just cancel each other. Actually, this is always true if you try to move antisymmetric indices to a tensor that is symmetric and vice-verse. Hope this is helpful.
A: " I got this sum from calculating the trace of $5 SU(N)$ generators in the adjoint representation $\text{tr}\,T^aT^bT^cT^dT^e$"
There is no reason why there could be a simple block-symmetry or block-antisymmetry. For instance, looking at Trace and adjoint representation of $SU(N)$, you have the trace of $4$ $SU(N)$ generators in the adjoint representation.
$$\mathrm{tr}(t^a_Gt^b_Gt^c_Gt^d_G)=\delta^{ab}\delta^{cd}+\delta^{ad}\delta^{bc}+\frac{N}{4}(d^{abe}d^{cde}-d^{ace}d^{bde}+d^{ade}d^{bce})$$where the tensor $d^{abc}$ is defined with the fundamental representation : $\{t^a_N,t^b_N\}=\frac{1}{N}\delta^{ab}+d^{abc}t^c_N$
($d^{abc}$ is symmetric in its $2$ first indices and $d^{aac}=0$)
This $4$-generator trace expression is symmetric in $a$ and $c$, and is symmetric in $b$ and $d$, and of course we have  the cyclic symmetries. But that's all. For instance, the expression is not symmetric in $a$ and $b$, and is not symmetric in $c$ and $d$.
