Components of an $(k,l)$-tensor I'm watching this lecture by Frederic Schuller on YouTube trying to understand the structure of the vector space $T^k_l(V)=\underbrace{V^* \otimes \dots \otimes V^*}_{l} \otimes \underbrace{V \otimes \dots \otimes V}_{k}$ and I have some difficulties with the components.
If $(e_1, \dots ,e_n)$ is a basis for $V$ and $(\epsilon^1, \dots, \epsilon^n)$ is the dual basis for $V^*$, then the basis for $T^k_l(V)$ is of the form $$e_{j_1} \otimes \dots \otimes e_{j_l} \otimes \epsilon^{i_1} \otimes \dots \otimes \epsilon^{i_k}.$$
Now any tensor $F \in T^k_l(V)$ can be written in the form $$F=F^{j_1 \dots j_l}_{i_1 \dots i_k}e_{j_1} \otimes \dots \otimes e_{j_l} \otimes \epsilon^{i_1} \otimes \dots \otimes \epsilon^{i_k}$$ where $F^{j_1 \dots j_l}_{i_1 \dots i_k}=F(\epsilon^{j_1}, \dots, \epsilon^{j_l}, \dots e_{i_1}, \dots, e_{i_k}).$

I'm trying to verify this by computation, but I don't yet get it right. Given vectors $a_1, \dots, a_k$ and covectors $b^1, \dots, b^l$ I should be able to start computing from $$F(b^1, \dots, b^l, a_1, \dots,a^k)$$ but I'm stuck in how I can express $b^i$ as a linear combination of the dual basis vectors. I think I have that $b^i = b^1\epsilon^1 + \dots +b^n\epsilon^n$, but now I'm having issues with the indexes as $b^i$ represents a covector and at the same time a scalar component of the covector.
Is there some way around this as I cannot figure this out with the indexes here?
 A: You're mixing up $k$ and $l$ several times. Also, where is it said that $b^i$ is both a covector and also a scalar component of the covector? It is just a covector! Its basis expansion is $b^i=b^i(e_j)\epsilon^j$, i.e the components of the covector $b^i$ relative to the basis $\{\epsilon^1,\dots,\epsilon^n\}$ of $V^*$ are the numbers $b^i(e_1),\dots, b^i(e_n)$, which if you feel so inclined, you can denote $b^i_{\,j}$.
Anyway, the basis for $T^k_l(V)=\underbrace{V^* \otimes \dots \otimes V^*}_{l} \otimes \underbrace{V \otimes \dots \otimes V}_{k}$ is given by the set
\begin{align}
B_{(k,l)}=\{\epsilon^{i_1}\otimes\cdots\otimes \epsilon^{i_l}\otimes e_{j_1}\otimes\cdots\otimes e_{j_k}\,|\,\, i_1,\dots, i_l,j_1,\dots, j_k\in\{1,\dots, n\}\}.
\end{align}
Note the ordering and the indices (the $l$ covectors appear in the tensor product first, followed by the $k$ vectors). Given any $F\in T^k_l(V)$, this can be thought of as a multilinear map $F:\underbrace{V^*\times \cdots \times V^*}_{k}\times\underbrace{V\times \dots \times V}_{l}\to\Bbb{R}$. Note the $k$ and $l$ and where the $*$ appears. The claim now is that
\begin{align}
F&=F(\epsilon^{j_1},\dots, \epsilon^{j_k},e_{i_1},\dots, e_{i_l})\cdot\epsilon^{i_1}\otimes \cdots\otimes\epsilon^{i_1}\otimes e_{j_1}\otimes\cdots\otimes e_{j_k}.
\end{align}

Anyway, I would suggest first writing out the proof in a few special cases such as $(k,l)\in\{(1,0), (0,1), (1,1),(1,2),(2,1)\}$, because the general case is just a zillion more indices.
