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Note: I am not sure if this relates to representation theory, of which I am mostly ignorant -- the only result on Google was phrased in terms of representation theory, which I did not understand. These two questions on Math.SE (1)(2) might also be relevant? In other words, please feel free to change the tags as appropriate.

Background: Let $V$ be a vector space, I don't think it matters but just in case let's say finite-dimensional and over $\mathbb{F}\in \{\mathbb{R}, \mathbb{C}\}$. Then we naturally have a bilinear map $V^* \times V \to \mathbb{F}$ defined by $$e:(f,v) \mapsto f(v) \,,$$ which for obvious reasons I choose to call the evaluation map. Note in particular that any function formed by fixing the first argument is an element of $V^*$, and any function formed by fixing the second argument is an element of $V^{**}$. I am not sure if this is relevant.

Anyway, taking the tensor product $V^* \otimes V$, it follows from the universal property of tensor products that a linear map $V^* \otimes V \to \mathbb{F}$ is induced, call it $\tilde{e}$.

Because $V^* \otimes V$ is supposed to be canonically isomorphic to $\operatorname{End}(V)=\operatorname{Hom}(V,V)$, $\tilde{e}$ can be considered a linear function from the space of $n \times n$ matrices over $\mathbb{F}$ to $\mathbb{F}$ itself.

Question: What is $\tilde{e}$? Is it important or as important as $e$? What is significant about the fact that its kernel is $(n^2-1)$-dimensional? If anything? Can you point me to resources with which I could learn how to compute coordinate representations of this and similar maps?

It can't be determinant or trace, since those aren't linear functions. However, I am not sure how to compute it, since I am not sure how to compute a coordinate representation for $e$, and I don't think that $e$ can be written as the Cartesian product of two different maps, so as a result $\tilde{e}$ can't be written as the tensor product of two maps either, making it impossible to apply this similar question.

EDIT: What I said above was dumb -- the trace is linear, it's only the determinant which isn't linear. Is $\tilde{e}$ just the trace?

EDIT 2: Based on the fact that $\tilde{e}$ is the trace, its kernel is just $\mathfrak{sl}(n, \mathbb{F})$, the space of matrices with zero trace. The above (using the Rank-Nullity theorem on trace) is one way to derive its dimension; a perhaps more common way is to note that it is canonically isomorphic to the tangent space of $\operatorname{SL}(n, \mathbb{F})$ (space of matrices with determinant $1$), thus has the same dimension as $\operatorname{SL}(n, \mathbb{F})$, namely $n^2-1$ (the dimension of $\operatorname{SL}(n,\mathbb{F})$ following from the implicit function theorem and the fact that it is equal to $\det^{-1}(1)$).

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    $\begingroup$ Yes, it is the trace. $\endgroup$ – Nate Jan 11 '17 at 21:57
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Fix dual basis $v_1,\dots,v_n$ and $\alpha^1,\dots,\alpha^n$ in $V$ and $V^*$ respectively.

If $T:V\to V$ then trace of $T$ by definition is $$tr(T)=\sum_i T_i^i,$$ where $T^i_i$ are such that $T(v_j)=\sum_k T^k_j v_k.$

T can be equivalently written as $$T=\sum_k\sum_j T_j^k\alpha^j \cdot v_k$$

But if you consider canonical isomorphism (as you mentioned) $\phi:V^*\otimes V\to End(V)$ then $T$ can be read as $$T=\sum_k\sum_j T_j^k\alpha^j \otimes v_k.$$

Since by definition $\bar{e}(\alpha\otimes v)=\alpha(v)$ for any $\alpha,v$ then $$\bar{e}(T)=\bar{e}(\sum_k\sum_j T_j^k\alpha^j \otimes v_k)=\sum_k\sum_j T_j^k\alpha^j(v_k)=\sum_k T^k_k.$$ And it is exactly the trace of $T$.

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  • $\begingroup$ This is the type of detailed, rigorous calculation I was looking for -- I really appreciate it! I'll have to try this with some other maps now -- thank you for your help! $\endgroup$ – Chill2Macht Jan 12 '17 at 11:00
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As I said in a comment it is indeed the trace map. I'd like to elaborate a bit more though.

This map $V^* \otimes V \to \mathbb{F}$ is in fact a map of $GL(V)$ representations, where the right hand side is the trivial representation. Under the identification of $V^* \otimes V$ with $n \times n$ matrices with $GL(n)$ acting by conjugation, the fact that this is a map of representations is just the fact that conjugating a matrix does not change its trace. This is the unique such map up to a scalar.

There is also an adjoint map the other direction $\mathbb{F}^* \cong \mathbb{F} \to (V^* \otimes V)^* \cong V^* \otimes V$, called the coevaluation map. You may wonder what this map is in terms of matrices. It turns out this map is just the inclusion of the space of scalar matrices inside the space of $n \times n$ matrices. It's a map of $GL(n)$ representations as scalar matrices are invariant under conjugation. Again this is the unique such map up to a scalar.

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