Quick question: tensor product and dual of vector space Recall that for a finite dimensional vector space $V$ we have the natural isomorphism $\phi :V^{*} \otimes V \rightarrow Hom(V,V)$ given by $\alpha \otimes v \mapsto (x \mapsto \alpha (x)v)$.
Is there a coordinate-free way to express $\phi^{-1}(id_{V})$?
 A: Presumably you know that if $\{v_i\}$ is an arbitrary basis of $V$ and if $\{v^i\}$ denotes the dual basis, then $\phi^{-1}(\text{id}_V) = \sum_i v^i \otimes v_i$.
If by "coordinate-free" you mean "an expression that doesn't require choosing a basis of $V$", then no, there is no such way of writing $\phi^{-1}(\text{id}_V)$. An element of $V^* \otimes V$ is a linear combination of tensor products, after all. :)
Edited to add: My second paragraph was perhaps overly hasty. In the book by Adam Coffman linked by Gunnar Magnusson (see Definition 2.2, p. 23), $\phi^{-1}(\text{id}_V)$ is interpreted as the evaluation operator in $(V^* \otimes V)^*$, defined by $\lambda \otimes v \mapsto \lambda(v)$. (Coffman calls the covector "$\phi$"; I've used "$\lambda$" instead.)
A: The required object, $\phi^{-1}(id_V)$, doesn't exist when $V$ is infinite dimensional. So any way of expressing it has to apply only when $V$ is fin-dim. The definition of "finite dimensional" is "has a finite basis". So you are always going to need a basis somewhere.
