# 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!

• Anybody an idea? I am a bit stuck here.. – freddieknets Jul 19 '13 at 11:56
• The original sum depends on the order $abcde$ of the letters, which seems strange (although maybe not if there are gauge symmetries involved). Where does this come from? – Peter Shor Jul 21 '13 at 0:04
• It doesn't, because $\delta$ and $f$ are fully symmetric resp. antisymmetric. – freddieknets Jul 22 '13 at 13:41
• If the order is $abcde$, you get $\delta^{ab}f^{cde} + \delta^{ac}f^{bde} + \delta^{ad}f^{bce} + \delta^{ae}f^{bcd} + \delta^{bc}f^{ade} + \delta^{bd}f^{ace} + \delta^{be}f^{acd} + \delta^{cd}f^{abe} + \delta^{ce}f^{abd} + \delta^{de}f^{abc}$ If the original order is $cbade$, you get $\delta^{ab}f^{cde} + \delta^{ac}f^{bde} - \delta^{ad}f^{bce} -\delta^{ae}f^{bcd} + \delta^{bc}f^{ade} - \delta^{bd}f^{ace} -\delta^{be}f^{acd} -\delta^{cd}f^{abe} - \delta^{ce}f^{abd} - \delta^{de}f^{abc}$ – Peter Shor Jul 22 '13 at 14:44
• I don't see why you get the minus signs.. My original sum is $\delta^{ab}f^{cde}-\delta^{ab}f^{ced}-\delta^{ab}f^{dce}+\delta^{ab}f^{dec}+\delta^{ab}f^{ecd}-\delta^{ab}f^{edc}+\delta^{ac}f^{bde}-\delta^{ac}f^{bed}+\ldots$ which collapses to the 10 terms I mention in the start (ignoring constant factors in front) if you take the (anti) symmetry of $\delta$ and $f$ into account. – freddieknets Jul 22 '13 at 22:46

## 2 Answers

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 $f^{u}$ 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.

• Actually, I am not trying to decompose my tensor product in irreps, but rather I am looking for an straightforward notation, like one could write $\delta^{(ab)}$ as a shorthand for $\frac{1}{2}\left(\delta^{ab}+\delta^{ba}\right)$. But I'll try whether I can get something from your answer. Thanks anyway for your help! – freddieknets Jul 22 '13 at 13:54
• Btw, the two terms you mention won't cancel, as in my definition $\delta^{cd}f^{abe}$ and $\delta^{dc}f^{abe}$ do have the same sign. This is because the sign only depends on the number of permutations of the second list of indices. – freddieknets Jul 22 '13 at 13:55
• I just followed your definition $(ab|cde)$ where you control the order of $cde$ and not of the last three indices, tough this is what you meant as I can see from the comment. In any case, the multi-index language where $q^{aa}$ is a short-hand for $\frac12 (q^{a_1a_2}+q^{a_2a_1})$ is a natural solution – John Jul 22 '13 at 17:59
• How would you use it in tensor notation? For example when contracting indices (which is in the end what I will do), I can write $f^{[abc]}g_{bcd}$; how would I write this in your notation? It seems that $f^{a}g_{abc}$ cannot be resolved unambiguously – freddieknets Jul 22 '13 at 22:50
• $f^{abb} g_{bbd}$, surprisingly (if you mean that two indices remain free). One has to play a little bit with this notation, but it is really useful for a tensors of not very complicated symmetry type and many people use it. – John Jul 23 '13 at 18:53

" 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 $d$.

• That's not the point - I know that there isn't necessarily a full (block)-symmetry for the trace. Point is that when calculating the 5-generator trace, I get 123 terms, of which I can simplify 120 to 1 term using the existing symmetries and my short-hand notation. For example the 4-generators can also be written as $\frac{1}{4}N\left(d^{abx}d^{xcd}-f^{abx}f^{xcd}\right)+\frac{1}{2}\left( \delta^{ab} \delta^{cd}+3\delta^{(ab)}\delta^{(cd)}\right)$. It is the last term that in the case of the 5-generator trace correspondingly can be written as $10\delta^{(ab\,|}f^{cde]}$. – freddieknets Jul 25 '13 at 9:51
• Try math.exchange.com, with tags combinatorics, lie-groups, lie-algebras, symmetric-groups – Trimok Jul 25 '13 at 10:34