# Which Slots Are Which? Tensors

I'm trying to verify a calculation in Mathematica and I'm confused about how Mathematica arranges tensor "slots" for contractions.

I have the matrix $$J_1 = \begin{pmatrix} 0 & -ih & 0 \\ ih & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

and I'm defining a tensor

$$(J_2)^{ab}_{ij} = (J_1)^a_i \otimes I^b_j + (J_1)^b_j \otimes I^a_i$$

where $$I$$ is the identity matrix.

I've defined a tensor $$J^{(2)}$$ with Mathematica as such

J1 = {{0, -I*h, 0}, {I*h, 0, 0}, {0, 0, 0}};
Id = IdentityMatrix[3];
J2 = TensorProduct[J1, Id] +TensorProduct[Id, J1]


where I is $$i$$.

Now I'm trying to contract $$(J_2)^{ab}_{ij}$$ with

$$\epsilon^{ij} = \frac{1}{2} \begin{pmatrix} 1 & i & 0 \\ i & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

to get $$(J_2)^{ab}_{ij}\epsilon^{ij}.$$ I know this $$\epsilon$$ matrix is an "eigenmatrix" of our tensor, and that it should contract to a multiple of itself

So I tried writing

> \[CurlyEpsilon]A = (1/2)*{{1, I, 0}, {I, -1, 0}, {0, 0, 0}};
> TensorContract[TensorProduct[J2, \[CurlyEpsilon]A], {{2, 5}, {3, 6}}]


thinking that maybe I have constructed the tensor with indices

$$J^{ab}_{ij}\epsilon^{kl}$$ and that the indexes were numbered $$(a,i,b,j,k,l) = (1,2,3,4,5,6)$$ so that contracting slots (2,5) and (3,6) would do the trick, though I was really unsure. This gave the wrong answer of 0s, and after some testing I found that contracting "slots" (1,5) and (3,6) gave the desired result.

How do I know what I'm contracting exactly in this case? What do the "slots" in Mathematica's tensor contraction function even mean? How can I keep track of indexes while doing calculations like this, especially since I'm defining things using tensor products and such.

UPDATE:

I realize now I made an error. $$(a,i,b,j,k,l) = (1,2,3,4,5,6)$$ was right - I was just to contract 2 and 5 and 4 and 6.

(tp = TensorProduct[Array[Subscript[a, #1, #2] &, {2, 2}],

Therefore the element a1,2 b2,1 has the indices 1,2,2,1: tp[[1, 2, 2, 1]]. The first 2 indices belong to a and the second 2 indices belong to b.