# Tensor (anti)symmetrization for non-adjacent indices: how can I notate $T^{(ab)c}$ but symmetrizing over $a$ and $c$ instead?

Refer to the following passage from Robert Wald's General Relativity:

More generally, for a tensor $T_{a_1\cdots a_l}$ of type $(0,l)$ we define \begin{align} T_{(a_1\cdots a_l)} & = \frac{1}{l!}\sum_\pi T_{a_{\pi(1)} \cdots a_{\pi(l)}} \tag{2.4.3}\\ T_{[a_1\cdots a_l]} & = \frac{1}{l!}\sum_\pi \delta_\pi T_{a_{\pi(1)} \cdots a_{\pi(l)}} \tag{2.4.4} \end{align} where the sum is taken over all permutations, $\pi$, of $1,\ldots,l$ and $\delta_\pi$ is $+1$ for even permutations and $-1$ for odd permutations. Similar definitions apply for any group of bracketed covariant or contravariant indices; e.g., we have $$T^{(ab)c}_{\phantom{(ab)c}[de]} = \frac14\left[ T^{abc}_{\phantom{abcc}de} + T^{bac}_{\phantom{abcc}de} - T^{abc}_{\phantom{abcc}ed} - T^{bac}_{\phantom{abcc}ed} \right]. \tag{2.4.5}$$

I am confused with the last notation. Say if I got a tensor ${T^{abc}}_{de}$ and I would like to denote a new tensor which is defined by permuting the indices $a$ and $c$. But you just can't add an open bracket before $a$ and a closed bracket after $c$, because in the notation, it means permuting $a,b,c$ instead of just $a,c$ only.

So how should I do? Thank you.

Remark: Someone suggest the notation ${T^{(a|b|c)}}_{de}$, but I found a situation in which this notation doesn't work. For example, if I want to do something like this: Given ${T^{abcdef}}_{gh}$, and I want to permute $a,e,f$ and $b,d$. If this notation is employed, then the new tensor should be denoted by ${T^{(a|(b|c|d)|ef)}}_{de}$. But this will confuse with the permutation of $a,d$ and $b,e,f$, so how can I deal with this?

• @Emil But what does it have to do with the notation? – Jerry Aug 16 '18 at 16:37
• @Emil Are you looking at this wiki page (en.wikipedia.org/wiki/…)? – Jerry Aug 16 '18 at 16:41
• @Jerry: sorry, I thought you wanted to come up with your own notation. Misread question. I looked at the one about braided monodial category. – Emil Aug 16 '18 at 16:41
• I've never seen notation that does this. From the perspective of someone who doesn't have much cause to use tensor-symmetrization notation on a day-to-day basis, the obvious conclusion is that non-adjacent symmetrization is something that comes up sufficiently rarely (and is easy enough to avoid when it does, by defining your index positions correctly to begin with) that there hasn't been enough of a need to notate it separately. – Emilio Pisanty Aug 16 '18 at 16:55
• The best approach I can think of is to use a Young tableau. – AccidentalFourierTransform Aug 16 '18 at 17:29

(This is not an answer for voting. It's a notation suggestion that could not be placed in the space of a comment)

1. Symmetrization : We select the first group of symmetrization upper indices, label it 1 and enter the superscript 1 to the left of every one of the indices of the group 1. If there exists a second group of symmetrization upper indices we label it 2 and enter the superscript 2 to the left of every one of the indices of the group 2. We continue for all groups of symmetrization upper indices if any. We repeat the same notation for the groups of symmetrization lower indices if any.

2. Antisymmetrization : We select the first group of antisymmetrization upper indices, label it 1 and enter the subscript 1 to the left of every one of the indices of the group 1. If there exists a second group of antisymmetrization upper indices we label it 2 and enter the subscript 2 to the left of every one of the indices of the group 2. We continue for all groups of antisymmetrization upper indices if any. We repeat the same notation for the groups of antisymmetrization lower indices if any.

So, any index (upper or lower) will have to its left exclusively: either (i) a superscript indicating its group for symmetrization or (ii) a subscript indicating its group for antisymmetrization or (iii) nothing, that is not belonging to any group for symmetrization or antisymmetrization.

Under this suggestion for the tensor of equation (2.4.5) we have $$T^{(ab)c}_{\hphantom{(ab)c}[de]}=T^{^1 a^1b\, c}_{\hphantom{^1 a^1 b\, c}_1 d_1e} \tag{A}\label{A}$$ If you want to symmetrize with respect to the upper indices $a,c$

$$T^{^1 ab^1 c}_{\hphantom{^1 ab^1 c}_1 d_1e} \tag{B}\label{B}$$

In the remark : If you want to symmetrize $\;{T^{abcdef}}_{gh}\;$ with respect to the two groups $\;a,e,f\;$ and $\;b,d\;$ $${T^{^1a^2bc\,^2d\,^1\! e\,^1\! f}}_{gh} \tag{C}\label{C}$$ different from symmetrization with respect to the two groups $\;a,d\;$ and $\;b,e,f\;$
$${T^{^1 a\,^2b\,c\,^1d\,^2e\,^2 f}}_{gh} \tag{C}\label{D}$$

But I don't dare to imagine how terribly would look an equation with a few tensors like these.

Good question which, I am afraid, has no good answer unless one is willing to use a graphical notation like Feynman diagrams. Here is a snapshot from the book Group Theory by Cvitanovic:

Note that when using tensors in an irreducible representation for a Young diagram other than a single row or column, the $$[ab]$$, $$(cd)$$ notation becomes completely useless. By definition, there are two set partitions for the index locations: one for the symmetrizations (white rectangles above) and one for the antisymmetrizations (black recangles). Moreover, these two partitions are "maximally transverse" meaning that a block of one partition can only intersect a block of the other at one index location. This is about as non-adjacent as possible.