I am trying to understand tensors and one particular question have caused me a great deal of confusion. The particular example with the metric tensor below is an attempt to highlight where my confusion is.
In my understanding, one can view the metric tensor $g$ as a bilinear map $g:V \times V \rightarrow \mathbb{R}$. And one can construct it using the tensor product of dual vectors from $V^{*}$ as below: $$ g=g_{ij} \mathbf{\epsilon^{i}} \otimes \mathbf{\epsilon^{j}} $$
If we now put a vector $\mathbf{v}=v^{k}\mathbf{e}_{k}$ in the first argument:
$$ g_{ij} \mathbf{\epsilon^{i}}(v^{k}\mathbf{e_{k}}) \otimes \mathbf{\epsilon^{j}} $$
$$\Rightarrow g_{ij} v^{k} \mathbf{\epsilon^{i}}(\mathbf{e_{k}}) \otimes \mathbf{\epsilon^{j}} $$
$$\Rightarrow g_{ij} v^{k} \delta _{k}^{i} \otimes \mathbf{\epsilon^{j}} $$
$$\Rightarrow g_{ij} v^{i} \mathbf{\epsilon^{j}} $$
Now the question: How can we in the last step remove the tensor product sign? How can we have the $\delta _{k}^{i}$ (a scalar) "tensor multiplied" by a dual vector? Surely you cant put a scalar on one side of the $\otimes$-sign and a vector on the other?... Thanks on beforehand.
To prove that $$ g_{ij} e^i \otimes e^j (x,-) = g_{ij} e^i(x) e^j(-) $$ start with the definition of the tensor product, $$ e^i \otimes e^j (x,y) := e^i(x) e^j(y) $$ So, \begin{equation} g(x,y)=g_{ij}e^i \otimes e^j (x,y) = g_{ij} e^i(x) e^j(y) \tag{I} \label{I} \end{equation} Since $e^i(x)=x^i$ and $g_{ij}x^i=x_j$, \begin{equation} g(x,y)=x_j e^j(y) \tag{II} \label{II} \end{equation} The r.h.s. of the last equation is a linear map $x_j e^j(-)$ i.e. a covector. Therefore $\tilde x \in V^*$ is, \begin{equation} \tilde x (y) = g(x,y) \end{equation} which is a nice property of inner product spaces.
Since, $$ g(x,-) = g_{ij} e^i \otimes e^j (x,-) $$ we can combine (\ref{I}) and (\ref{II}) to get, $$ x_j e^j(-) = g_{ij} e^i \otimes e^j (x,-) $$ We have already shown that $x_j = g_{ij} e^i(x)$. It therefore follows that, $$ g_{ij} e^i(x) e^j(-) = g_{ij}e^i \otimes e^j (x,-) $$ and this is only possible if $e^i \otimes e^j (x,-)=e^i(x)e^j(-)$.