Clarification on explanation of mechanics of tensor product XylyXylyX has a great series of videos on manifolds and tensors. I would like to confirm a couple of points that are probably implied (or misunderstood (by me)). It makes reference to this point in the definition of tensor product:


... in the lower left-hand corner of the slide, he is illustrating the product tensor $e^1 \otimes e^2$ operating on a pair of vectors $A^\mu\,e_\mu, B^\nu\,e_\nu \in V,$ resulting in the real number $[e^1 \otimes e^2](A^\mu e_\mu, B^\nu e_\nu)=A^1 B^2.$
I would like to ask you to confirm two things: 


*

*Even though $A^1 B^2$ is a real number, if we were to carry out the complete tensor product $[*]$, we'd have to express it as 


$$A^0 B^0 e_0 \otimes e_0 + A^0 B^1 e_0 \otimes e_1 + \cdots + A^4 B^4 e_4 \otimes e_4.$$ 
In other words, it wouldn't be just a single real number, but rather more like a matrix. 
This seems consistent with the result in the Wikipedia example of the tensor product of $v = \begin{bmatrix}1& 2& 3 \end{bmatrix}$ and $w = \begin{bmatrix}1 & 0 & 0 \end{bmatrix}:$
$$v\otimes w=\hat x \otimes \hat x + 2 \hat y\otimes \hat x + 3 \hat z \otimes\hat x $$
although there are no apparent covectors in this Wikipedia example, possibly explaining the difference.

$[*]$ NOTE that this may be the main source of my misunderstanding: In the example on the slide posted, he picks out $e^1\otimes e^2$, but I don't know if one has to continue operating on the $15$ additional $e^i\otimes e^j$ pairs.



*When the covectors (linear functionals) are not just the basis of $V^*,$ and they have coefficients, the tensor product $\beta \otimes \gamma$ operating on these same two vectors would look like 


$$\beta_0 \gamma_0 A^0 B^0 e_0 \otimes e_0 + \beta_1 \gamma_0 A^1 B^0 e_1 \otimes e_0 + \cdots + \beta_4 \gamma_4 A^4 B^4 e_4 \otimes e_4$$
with $\beta_i, \gamma_i$ corresponding to the coefficients of the covectors $\beta, \gamma \in V^*.$
 A: I don't know what you mean by "carry out the complete tensor product" in 1. But the output should be a real number, regardless, if you apply a tensor of type $(2,0)$ to a pair of vectors. Similarly, in 2, the output should be a real number when you evaluate $(\beta\otimes\gamma)(\sum A^\mu e_\mu,\sum B^\nu e_\nu) = \sum \beta_\mu\gamma_\nu A^\mu B^\nu$.
(If you like, $(e^i\otimes e^j)(e_\mu,e_\nu) = \delta^i_\mu\delta^j_\nu$, as your slide says.)
EDIT: One thing to keep in mind is this. We use upper and lower indices precisely so that things that make sense will have compensating upper and lower indices. I.e., $\sum A^\mu e_\mu$ is a vector or $(0,1)$ tensor (one upper, one lower), $\sum A^\mu B^\nu e_\mu\otimes e_\nu$ is the $(0,2)$ tensor $v\otimes w$ (one $\mu$ up and one $\mu$ down, same for $\nu$), and $\sum \beta_\mu A^\mu$ is a real number (one $\mu$ up, one $\mu$ down), as is $\sum \beta_\mu\gamma_\nu A^\mu B^\nu$ (note compensating upper and lower indices for both $\mu$ and $\nu$). In 2. you have a surplus of lower indices, so it can't make sense. When we apply a $(2,0)$ tensor to a pair of vectors, we get a $(0,0)$ tensor, or scalar.
