Can I have some clarification of the different meanings of $\otimes$ as in the unifying and separating implications in basic linear algebra and tensors?
Here is some of the overloading of this symbol...
1.1. Kronecker matrix product:
If $A$ is an $m \times n$ matrix and $B$ is a $p \times q$ matrix, then the Kronecker product A ⊗ B is the $mp \times nq$ block matrix:
$$A\color{red}{\otimes}B=\begin{bmatrix}a_{11}\mathbf B&\cdots&a_{1n}\mathbf B\\\vdots&\ddots&\vdots\\a_{m1}\mathbf B&\cdots&a_{mn}\mathbf B\end{bmatrix}$$
1.2. Outer product:
$\mathbf u \otimes \mathbf v = \mathbf{uv}^\top = \begin{bmatrix}u_1\\u_2\\u_3\\u_4\end{bmatrix}\begin{bmatrix}v_1&v_2&v_3&v_4\end{bmatrix}=\begin{bmatrix}u_1v_1&u_1v_2&u_1v_3\\u_2v_1&u_2v_2&u_2v_3\\u_3v_1&u_3v_2&u_3v_3\end{bmatrix}$
- Definition of the tensor space:
$$\begin{align}T^p_q\,V &= \underset{p}{\underbrace{V\color{darkorange}{\otimes}\cdots\color{darkorange}{\otimes} V}} \color{darkorange}{\otimes} \underset{q}{\underbrace{V^*\color{darkorange}{\otimes}\cdots\color{darkorange}{\otimes} V^*}}:=\{T\, |\, T\, \text{ is a (p,q) tensor}\}\\[3ex]&=\{T: \underset{p}{\underbrace{V^*\times \cdots \times V^*}}\times \underset{q}{\underbrace{V\times \cdots \times V}} \overset{\sim}\rightarrow K\}\end{align}$$
- Definition of the tensor product:
It takes $T\in T_q^p V$ and $S\in T^r_s V$ so that:
$$T\color{blue}{\otimes}S\in T_{q+s}^{p+r}V$$
defined as:
$$\begin{align}&(T\color{blue}{\otimes}S)(\underbrace{ \omega_1,\cdots,\omega_q,\cdots,\omega_{q+s}, v_1,\cdots,v_p,\cdots,v_{p+r}}_\text{'eats'})\\&:= T(\underbrace{\omega_1,\cdots,\omega_q, v_1,\cdots,v_p}_{\text{'eats up' p vec's + q covec's}\rightarrow \text{no.}})\underbrace{\cdot}_{\text{in the field}}S(\underbrace{\omega_{q+1},\cdots,\omega_{q+s}, v_{p+1},\cdots,v_{p+r}}_{\text{'eats up' p vec's and q covec's} \rightarrow\text{no.}})\end{align}$$
An example of, for instance, some operation like $\underbrace{e_{a_1}\color{blue}{\otimes}\cdots\color{blue}{\otimes}e_{a_p}\color{blue}{\otimes} \epsilon^{b_1}\color{blue}{\otimes}\cdots\color{blue}{\otimes}\epsilon^{b_q}}_{(p,q)\text{ tensor}}$ after settling for some basis could be helpful. For clarity this is a fragment of the more daunting expression:
$$ T=\underbrace{\sum_{a_1=1}^{\text{dim v sp.}}\cdots\sum_{b_1=1}^{\text{dim v sp.}}}_{\text{p + q sums (usually omitted)}}\underbrace{\color{green}{T^{\overbrace{a_1,\cdots,a_p}^{\text{numbers}}}_{\quad\quad\quad\quad\underbrace{b_1,\cdots,b_q}_{\text{numbers}}}}}_{\text{a number}}\underbrace{\cdot}_{\text{S-multiplication}}\underbrace{e_{a_1}\color{blue}{\otimes}\cdots\color{blue}{\otimes}e_{a_p}\color{blue}{\otimes} \epsilon^{b_1}\color{blue}{\otimes}\cdots\color{blue}{\otimes}\epsilon^{b_q}}_{(p,q)\text{ tensor}}$$
showing how to recuperate a tensor from its components.
I realize that there is a connection as stated here:
The Kronecker product of matrices corresponds to the abstract tensor product of linear maps. Specifically, if the vector spaces $V, W, X$, and $Y$ have bases
$\{v_1, \cdots, v_m\}, \{w_1,\cdots, w_n\}, \{x_1,\cdots, x_d\},$ and $\{y_1, \cdots, y_e\}$, respectively,
and if the matrices $A$ and $B$ represent the linear transformations $S : V \rightarrow X$ and $T : W \rightarrow Y$, respectively in the appropriate bases, then the matrix $A ⊗ B$ represents
the tensor product of the two maps, $S ⊗ T : V ⊗ W → X ⊗ Y$ with respect to
the basis $\{v_1 ⊗ w_1, v_1 ⊗ w_2, \cdots, v_2 ⊗ w_1, \cdots, v_m ⊗ w_n\}$ of $V ⊗ W$ and the similarly defined basis of $X ⊗ Y$ with the
property that $A ⊗ B(v_i ⊗ w_j) = (Av_i) ⊗ (Bw_j)$, where $i$ and $j$ are integers in the proper range.
But it is still elusive...