Tensor Contraction producing a trace. I was reading a meta-paper and came across this notation. 
Given V as a vector space, then there exists a dual space V*. V*⊗V is isomorphic to the set of linear transformation that map V into V, which is the set of n by n matrices with entries in the scalar field, where n is the dimension of V. Then there is a mapping from V*⊗V into a scalar field which takes an element f⊗V to f(v). Thanks.This is the mapping that sends a matrix to the trace of that matrix.
I think I've found my answer in this article: The Duality Functor in Linear Algebra
So, this is a tensor contraction. How, in this case, does this tensor contraction produce a trace? I am not sure about that. Good books on tensors might be helpful since I'm having trouble with this. Thanks.
 A: If $V$ is $n$-dimensional, then let $\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n$ be a basis for $V$, and let $\pi_1,\pi_2,\ldots,\pi_n$ be a basis for $V^*$, defined by
$$\pi_i(\mathbf e_j) = \begin{cases} 1 & \text{if }i=j \\ 0 & \text{otherwise} \end{cases}$$
$V^*\otimes V$ has a basis consisting of all elements of the form $\pi_i\otimes \mathbf e_j$, so every element of $V^*\otimes V$ is a linear combination of such things. The way it can be interpreted as a linear mapping is by the rule
$$ (\pi_i\otimes \mathbf e_j)(v) = (\pi_i(v))\mathbf e_j $$
and extend to the full $V^*\otimes V$ by linearity.
The matrix corresponding to $F\in V^*\otimes V$ this mapping is the one whose element at position $j,i$ is the coefficient of $\pi_i\otimes \mathbf e_j$ when $F$ is written in this basis. In other words, $\pi_i\otimes \mathbf e_j$ corresponds to a matrix with $1$ at index $j,i$ and zeroes everywhere else, and these matrices combine linearly.
Now contraction in $V^*\otimes V$ is defined by the rule $v^*\otimes v \mapsto v^*(v)$, which means that its effect on basis elements is
$$ \pi_i\otimes\mathbf e_j \mapsto \pi_i(\mathbf e_j) = \begin{cases} 1 & \text{if }i=j \\ 0 & \text{otherwise} \end{cases} $$
So when we write some $F\in V^*\otimes V$ as a linear combination of $(\pi_i\otimes \mathbf e_j)$s, all of the ones with $i\ne j$ drop out when we contract, and the coefficients of $\pi_i\otimes \mathbf e_i$ are all just added. But these coefficients are just the elements on position $i,i$ in the matrix representation, and their sum is the trace of the matrix!
