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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 $\phi^{-1}(id_{V})$?

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I'd say no, because philosophically the way to see elements of $V^* \otimes V$ in a coordinate-free way is as elements of $Hom(V,V)$. But if anyone knows the answer it's Adam Coffman; see his book users.ipfw.edu/CoffmanA/pdf/book.pdf on coordinate-free linear algebra. (If nothing else, it'll make you respect the trace even more.) –  Gunnar Þór Magnússon Jul 19 '13 at 2:41

2 Answers 2

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.)

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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.

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