I've seen in the literature the notation $C$ with some additional specifications for the contraction maps of all sorts, but the amount of decorations on the symbol $C$ varied depending on the context. See, e.g., A.Gray, Tubes, p.56, where these maps are used in the case of somewhat special tensors, and therefore the notation is simpler.
In general, there is a whole family of uniquely defined maps
$$
C^{(r,s)}_{p,q} \colon \otimes^{r}_{s} V \to \otimes^{r-1}_{s-1} V
$$
which are collectively called tensor contractions ($1 \le p \le r, 1 \le q \le s$).
These maps are uniquely characterized by making the following diagrams commutative:
$$
\require{AMScd}
\begin{CD}
\times^{r}_{s} V @> {P^{(r,s)}_{p,q}} >> \times^{r-1}_{s-1} V\\
@V{\otimes^{r}_{s}}VV @VV{\otimes^{r-1}_{s-1}}V \\
\otimes^{r}_{s} V @>{C^{(r,s)}_{p,q}}>> \otimes^{r-1}_{s-1} V
\end{CD}
$$
Explanations are in order.
Recall that the tensor products $\otimes^{r}_{s} V$ are equipped with the universal maps
$$
\otimes^{r}_{s} \colon \times^{r}_{s} V \to \otimes^{r}_{s} V
$$
where $\times^{r}_{s} V := ( \times^r V) \times (\times^s V^*)$.
Besides that, there is a canonical pairing $P$ between a vector space $V$ and its dual:
$$
P \colon V \times V^* \to \mathbb{R} \colon (v, \omega) \mapsto \omega(v)
$$
Notice that map $P$ is bilinear and can be extended to a family of multilinear maps
$$
P^{(r,s)}_{p,q} \colon \times^{r}_{s} V \to \times^{r-1}_{s-1} V
$$
by the formula:
$$
P^{(r,s)}_{p,q} (v_1, \dots, v_p, \dots, v_r, \omega_1, \dots, \omega_q, \dots, \omega_s) = \omega_q (v_p) (v_1, \dots, \widehat{v_p}, \dots, v_r, \omega_1, \dots, \widehat{\omega_q}, \dots, \omega_s)
$$
where a hat means omission.
Since maps $P^{(r,s)}_{p,q}$ are multilinear, the universal property of the maps $\otimes^{r}_{s}$ implies that there are uniquely defined maps
$$
\tilde{P}^{(r,s)}_{p,q} \colon \otimes^{r}_{s} V \to \times^{r-1}_{s-1} V
$$
and then the maps $C^{(r,s)}_{p,q}$ are given by
$$
C^{(r,s)}_{p,q} := \otimes^{r-1}_{s-1} \circ \tilde{P}^{(r,s)}_{p,q}
$$
Representing tensors using matrix notation is often confusing, but let's assume that
$Y = \begin{pmatrix}y^1_1&y^1_2\\v^2_1&y^2_2\end{pmatrix}$
and similarly for Z. If $W = Y \times Z$ then the components of $W$ are
$w^{ik}_{jl} = y^i_j z^k_l$
You have represented W as a 4x4 matrix, but it would be more accurate to represent it as a 2x2 matrix, each of whose entries is another 2x2 matrix.
Anyway, the four possible contractions of W are:
(1): $w^{ik}_{il} = y^i_i z^k_l$
(2): $w^{ik}_{jk} = y^i_j z^k_k$
(3): $w^{ik}_{ji} = y^i_j z^k_i$
(4): $w^{ij}_{jl} = y^i_j z^j_l$
In terms of matrix operations:
(1) is the component representation of $Tr(Y)Z$
(2) is the component representation of $Tr(Z)Y$
(3) is the component representation of $YZ$
(4) is the component representation of $ZY$
Best Answer
In mathematics there is no defined order in tensor product or contraction. Index order matters when 1) you want tensor to match some particular definition 2) you want to highlight the symmetry of the result, for example, $A_{ikjl}$ makes sense better if there is symmetry on $ik$ and $jl$, than $ij$ and $kl$. So I would leave this to user (see numpy.einsum ):
multiply(A, B, "ispk,sjkq->ijpq")Shortly, there is no convention. Tensors are loved because the order of terms doesn't matter and all the syntax information is in indices, so one can use the order of terms to convey some semantic information (like we do with simple multiplication). However, if you insist on fixed order, why use indices at all? You can define tensor product (append dimensions as they are in order) and contraction (cross out dimensions). So $C_{ijkl}=A_{ipjq}B_{pkql}$ is
Tcontract(Tproduct(A,B),2,5,4,7)