Contraction of a tensor over all indexes? In Chapter 2.3 of this book, let $F \in T^k_l(V)$ and $\{\omega^i\}_{i=1}^l \in V^*, \{X_i\}_{i=1}^k$, then it is clear that
$$
F \otimes \omega^1 \otimes \cdots \otimes w^l \otimes X_1 \otimes \cdots \otimes X_k \in T^{2k}_{2l}(V)
$$
so the contraction of this tensor over all indexes is
$$
\text{tr}(F \otimes \omega^1 \otimes \cdots \otimes w^l \otimes X_1 \otimes \cdots \otimes X_k) = F(\omega^1,\dotsc,\omega^l,X_1\dotsc,X_k)
$$
...which I don't understand at all. I know that $\text{tr}$ can be viewed as an operation that reduce the rank of the tensor by 2 via
$$  
  \text{tr}(F)(\omega^1,\dotsc,\omega^l,X_1,\dotsc,X_k) = \text{tr}(F(\omega^1,\dotsc,\omega^l, \star\ ,X_1,\dotsc,X_k, \star\ ))
  $$
but if we want to sum through all indexes we need to have $l = k $, so how is it possible to take the contraction over all indexes on a general tensor?
 A: $\newcommand\Tr{\operatorname{Tr}}$The most simple contraction operation is the trace :
$$\Tr : V\otimes V^* \to \mathbb K$$
Given a general tensor product $T^k_l(V) = V^{\otimes k} \otimes (V^*)^{\otimes l}$ (with $k,l\geq 1$), for each $i \in \{1,\ldots, k\}$ and $j\in \{1,\ldots,j\}$, we can define a contraction $C_{ij} : T^k_l(V) \to T^{k-1}_{l-1}(V)$ by contracting the $i$-th factor in $V$ and the $j$-th factor in $V^*$ using $\Tr$.
But we can also iterate this and contract more than one pair of factors. For example by taking the composite :
$$V\otimes V \otimes V^* \otimes V^* \simeq V \otimes (V\otimes V^*)\otimes V \overset{\Tr}\longrightarrow V\otimes V^*\overset\Tr\longrightarrow\mathbb K$$
Now, without using indices like physicists do, the notations quickly become nightmarish. The previous example would be $C_{11}\circ C{21} = C_{11}\circ C_{12}$. Therefore, when everything is clear, we prefer to write any contraction $\Tr$.
In your specific case, with $F \in T^{k}_{l}(V)$, we remark that : $$T = \omega_1\otimes \ldots\otimes \omega_k \otimes V_1\otimes \ldots \otimes V_l \in T_{k}^l(V) = (T^{k}_l(V))^*$$
Therefore, $F\otimes T \in T^k_l(V) \otimes (T^k_l(V))^*$ and we can take the trace over $T^k_l(V)$ instead of $V$, so it makes a lot of sense to write this full contraction as a trace. This can also be written in terms of the trace over $V$ (ie contractions over pairs of factors $(V,V^*)$). In this case, as $F \otimes \omega_1\otimes \ldots\otimes \omega_k \otimes V_1\otimes \ldots \otimes V_l$ belongs to :
$$\underset{k}{\underbrace{V\otimes \ldots \otimes V}}\otimes \underset{l}{\underbrace{V^*\otimes \ldots \otimes V^*}} \otimes \underset{k}{\underbrace{V^*\otimes \ldots \otimes V^*}} \otimes \underset{l}{\underbrace{V\otimes \ldots \otimes V}}$$
it is (implicitly) understood, that the $i$-th of the $k$ factors of $V$ is contracted with the $i$-th of the $k$ factors of $V^*$ and that the $j$-th of the $l$ factors of $V$ is contracted with the $j$-th of the $l$ factors of $V^*$.
A: Each $\omega$ and each $X$ has rank 1. Therefore $\omega^1\otimes\dots\otimes\omega^l\otimes X_1\otimes\dots\otimes X_k$ has rank $l+k$. Furthermore the tensor $F$ has also rank $l+k$. (So the tensor product of everything together has rank $2l+2k$). The "contraction over all indices" that the text talks about contracts each index of $F$ with one of the indices of $\omega^1\otimes\dots\otimes\omega^l\otimes X_1\otimes\dots\otimes X_k$. This works for any value of $k$ and $l$.
You are right, that if $k=l$, then you could also contract each of the $\omega$'s with one of the $X$'s. But thats a different operation with a different result, and probably not what is meant here.
The basic problem is this: Just writing "$\text{tr}(\dots)$" is often ambiguous. It sometimes means contracting exactly two indices (thus reducing the rank by 2), or it can mean contracting everything, thus creating a scalar. And in the second case there can be a myriad of ways how to match all the indices in pairs. Each gives a different result.
