I am not sure if I can prevent this question from being too vague or with too large an overlap with other similar math.SE questions, but I will do my best…
A standard linear operation in tensor calculus is tensor contraction, which can be conveniently expressed in a given coordinate basis (or using Penrose-Rindler's abstract index notation) by means of Einstein's summation convention. Since this is a "pointwise" (i.e. algebraic) operation, we can restrict our discussion to (constant) mixed tensors of contravariant rank $r$ and covariant rank $s$ $$T\in\otimes^r_s V\doteq(\otimes^r V)\otimes(\otimes^s V^*)=\underbrace{V\otimes\cdots\otimes V}_{r\text{ times}}\otimes\underbrace{V^*\otimes\cdots\otimes V^*}_{s\text{ times}}$$ on a (finite dimensional) vector space $V$ over $\mathbb{R}$ or $\mathbb{C}$ with dual $V^*$. If $\{e_1\ldots,e_n\}$ is a basis of $V$ with dual basis $\{\theta^1,\ldots,\theta^n\}$ on $V^*$, i.e. $$\theta^j(e_i)=\delta^j_i=\begin{cases} 1 & (i=j) \\ 0 & (i\neq j) \end{cases}\ ,$$ and we expand $T$ in the corresponding basis of $\otimes^r_s V$ as $$T=\sum^n_{\substack{i_1,\ldots,i_r,\\ j_1,\ldots,j_s=1}} T^{i_1\cdots i_r}_{j_1\cdots j_s}e_{i_1}\otimes\cdots\otimes e_{i_r}\otimes\theta^{j_1}\otimes\cdots\otimes\theta^{j_s}\ ,$$ the tensor contraction of the $k$-th contravariant index with the $l$-th covariant index in this basis yields a tensor $S$ of contravariant rank $r-1$ and covariant rank $s-1$ whose components in the corresponding basis of $\otimes^{r-1}_{s-1}V$ are given by $$S^{i_1\cdots i_{r-1}}_{j_1\cdots j_{s-1}}=\sum^n_{m=1}T^{i_1\cdots i_{k-1}mi_k\cdots i_{r-1}}_{j_1\cdots j_{l-1}mj_l\cdots j_{s-1}}\text{ or simply }T^{i_1\cdots i_{k-1}mi_k\cdots i_{r-1}}_{j_1\cdots j_{l-1}mj_l\cdots j_{s-1}}$$ by Einstein's summation convention. For example, if $r=2$, $s=3$, $k=1$ and $l=2$, the tensor contraction of $T^{ab}_{cde}$ at the aforementioned indices would be $S^b_{ce}=T^{mb}_{cme}$ if one employs Penrose-Rindler's abstract index notation instead.
Since tensor contraction is a partial trace, it does not depend on a choice of basis. This can also be seen by the following, equivalent definition of tensor contraction (as done in this math.SE question): if $T$ is completely factorized, i.e. $T=X_1\otimes\cdots\otimes X_r\otimes\omega^1\otimes\cdots\otimes\omega^s$ with $X_i\in V$, $\omega^j\in V^*$, $i=1,\ldots,r$, $j=1,\ldots,s$, tensor contraction of the $k$-th contravariant index with the $l$-th covariant index yields $$S=\omega_l(X_k)X_1\otimes\cdots\otimes\widehat{X}_k\otimes\cdots\otimes X_r\otimes\omega_1\otimes\cdots\otimes\widehat{\omega}_l\otimes\cdots\otimes\omega_s\ ,$$ where the hats stand for omission. The above formula is then extended to general $T$ by linearity. In the particular case $r=s=k=l=1$, tensor contraction applied to $T$ is just the trace of the linear transformation $T:V\rightarrow V$, usually denoted by $\mathrm{Tr}\,T$.
Question: However, I have not found in the literature so far a good, general and coordinate-free notation (i.e. without
referring to components in a given basis) for tensor contraction
besides appealing to Penrose-Rindler's abstract index notation, which,
though computationally efficient (as physicists know well),
aesthetically speaking is kind of a crutch. Any ideas? I would
specially like to have references for such notation(s?), should they
exist.
(Remark: I am aware of this other closely related math.SE question, but the only answer given to it, based on the report by T.G. Kolda and B.W. Bader (Tensor Decompositions and Applications, Sandia Report 6702 (2007)), does not seem to cover my question)
Best Answer
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 $$ C^{(r,s)}_{p,q} := \otimes^{r-1}_{s-1} \circ \tilde{P}^{(r,s)}_{p,q} $$