I am trying to clarify something with regard to tensor notation and I am hoping that somebody can set the record straight. As far as I understand:
If $V$ is a vector space over $\mathbb{K}$, with dual space $V^{\ast}$, a rank $(p,q)$-tensor $T$ is then a (multilinear) map $T$ that takes $p$ copies of $V$ and $q$ copies of $V^{\ast}$ and maps them into $\mathbb{K}$.
In other words, a vector is a rank $(0,1)$-tensor and a covector is a rank $(1,0)$-tensor. I hope I am right thus far? Now the question:
A rank $(p,q)$ tensor can be written in terms of its components (Einstien notation) as
$T_q^p :=T_{i_1, \dots, i_q}^{j_1, \dots, j_p} e_{j_1} \otimes \dots \otimes e_{j_p} \otimes \epsilon^{i_1} \otimes \dots \otimes \epsilon^{i_q} $
where $e_j$ is a basis on $V$ and $\epsilon^i$ is a basis on $V^{\ast}$.
So, from this notation, if I wanted to write a $(1,0)$ tensor in terms of its components, I would write it as $T^j e_j$. If I wanted to write a rank $(0,1)$-tensor in terms of its components, I would write it as $T_i \epsilon^i$. This seems completely nonsensical to me, since a rank $(0,1)$-tensor (a vector) is usually written as $T^i e_i$ (like when one wants to write the tangents vectors for the tangent space $TM$ on a manifold $M$), since $e_i$ is the basis on $V$? But the metric tensor $\textbf{g}$, which is a rank $(0,2)$-tensor, is usually written in terms of its components as $g_{ij} \epsilon^i \otimes \epsilon^j$, which would be in agreement with the general form.
Can anybody clarify this? Why this apparent inconsistency in notation?
This is not the usual definition. Indeed it should be:
With this definition, you'd conclude that a vector is a $(1,0)$-tensor and covector is a $(0,1)$-tensor. This is consistent with your second definition.