[This is a question motivated by theoretical physics, so apologies if the language is rough...]


In three dimensions the spaces of invariant (or isotropic) tensors of rank $n$ have dimensions 1, 0, 1, 1, 3, 6, 15, 36, 91, 232, ... (see [MathWorld](https://mathworld.wolfram.com/IsotropicTensor.html)). Equivalently, this is the number of trivial representations in the $n$-fold tensor product of the fundamental representation of $SO(3)$.


Any invariant tensor can be expressed as a linear combination of products of Kronecker deltas and (for odd rank) one [Levi-Civita symbol](https://en.wikipedia.org/wiki/Levi-Civita_symbol) (see [this note](http://jeffreymepstein.com/IsotropicTensorsNote.pdf), which contains references to the original literture), but in general there will be linear independences.


The dimensions are given by the [Riordan numbers](https://www.sciencedirect.com/science/article/pii/S0012365X99000540?via%3Dihub), of which there are many combinatoric interpretations. My question is: **is there an explicit bijection between one of these interpretations and a linearly independent basis** of invariant tensors?


For example, in the case of $SU(2)$ there is such a bijection, expressed in terms of noncrossing pairings on a circle, with each pairing corresponding to $\epsilon\_{ab}$ (in quantum mechanics we say a two-spin singlet state). This is originally due to [Rumer, Teller, and Weyl](https://gdz.sub.uni-goettingen.de/id/PPN252457811_1932?tify=%7B%22view%22:%22info%22,%22pages%22:%5B503%5D%7D), and generalized by [Temperley and Lieb](https://royalsocietypublishing.org/doi/abs/10.1098/rspa.1971.0067). See [this paper](https://journals.jps.jp/doi/10.1143/JPSJ.59.482) for a more recent discussion



Basis of invariant tensors of rank n in three dimensions
Planar partitions with no singletons works. You need to pick for each $n>1$ some map with certain properties. One way to do this is to just fix a preferred trivalent tree of each size and interpret each vertex as a cross product. For example, one arbitrary choice gives the map $V^{\otimes n} \rightarrow \mathbb{C}$ given by $$v\_1 \otimes \ldots \otimes v\_n \mapsto (((v\_1 \times v\_2) \times v\_3) \ldots \times v\_{n-1})\ldots) \cdot v\_n.$$


It’s a little awkward that the above formula isn’t invariant under cyclic permutation of the variables, so you have to arbitrarily pick a "starting point" for each connected component. A better way is to write the non-crossing partition of 2n strands where each end connects to its neighbor and then attach a bunch of “2nd Jones-Wenzl projectors” (i.e. the projection from the square of the 2d to the 3d) along the boundary in an interleaving way (ie connecting each strand to its *other* neighbor).


As discussed in comments, the JW2 part can be interpreted as rewriting the vector in terms of the corresponding sum of Pauli matrices. If we let $\sigma\_v$ denote the sum of Pauli matrices corresponding to $v$, then the rotationally invariant map is just the trace of the product of Pauli matrices:
$$v\_1 \otimes \ldots v\_n \mapsto \mathrm{Tr}(\sigma\_{v\_1} \sigma\_{v\_2} \cdots \sigma\_{v\_n}).$$


This was the formula for a single connected component, so the general formula is a product over connected components of the trace of the product of Pauli matrices for each component.


Another approach, using a different set, is to take n pairs of boundary points and look at non-crossing partitions of 2n points such that no pair connects to itself. Then just interpret this in SU(2) by embedding each 3-dimensional irrep inside the square of the 2-dimensional irrep.


