Background:
Let $V$ be a vector space over a field $k$. Let me describe several different canonical maps which we shall compose in the question.
- There is a canonical bilinear map $V \times V^* \to \text{End}(V)$ sending $v , \varphi \mapsto [w \mapsto \varphi(w) v]$, so the universal property of tensor product gives a linear map $\Phi: V \otimes V^* \to \text{End}(V)$. If $V$ is finite-dimensional (f.d.), this is an isomorphism. Its dual map $\Phi^* : \text{End}(V)^* \to (V \otimes V^*)^*$ is then also an isomorphism.
- If $W$ is another $k$-vector space and there is a canonical bilinear map $V^* \times W^* \to (V \otimes W)^*$ sending $\varphi , \psi \mapsto [v \otimes w \mapsto \varphi(v)\psi(w)]$. Again if $V$ and $W$ are f.d., the induced map is also an isomorphism. In the special case when $W = V^*$ ($V$ f.d.), let's name this isomorphism $\Psi: V^* \otimes V^{**} \to (V \otimes V^*)^*$.
- There is a canonical map $V \to V^{**}$ sending $v \mapsto \text{eval}_v$. Again when $V$ is f.d. this map is an isomorphism, hence we obtain an isomorphism $\Theta: V^* \otimes V \to V^* \otimes V^{**}$.
- Finally, to be completely pedantic, there is a canonical isomorphism $\Gamma: V \otimes V^* \to V^* \otimes V$ given by swapping the order of the simple tensors.
- Composing maps (f.d. case), we have a canonical isomorphism $F : \text{End}(V) \to \text{End}(V)^*$:
$$ \text{End}(V) \overset{\Phi^{-1}}{\longrightarrow} V \otimes V^* \overset{\Gamma} {\longrightarrow} V^* \otimes V \overset{\Theta}{\longrightarrow} V^* \otimes V^{**} \overset{\Psi}{\longrightarrow} (V \otimes V^*)^* \overset{(\Phi^*)^{-1}}{\longrightarrow} \text{End}(V)^*$$
- In the f.d. case, there is a special element of $\text{End}(V)^*$, namely the trace. As an element of $(V \otimes V^*)^*$ it is given by tensor contraction: $\Phi^*(\text{tr})(v \otimes \varphi) = \varphi(v)$.
Actual Question:
This seems like it should be totally obvious, but I'm kinda stumped! What the heck element of $\text{End}(V)$ does the trace correspond to under the isomorphism $F$? i.e. what is $F^{-1}(\text{tr})$? And actually, while we're at it (or perhaps along the way), what is $\Psi^{-1}(\Phi^*(\text{tr}))$? It feels strange to have a distinguished element of $V^* \otimes V^{**}$. Well I suppose the image of $1_V \in \text{End}(V)$ is also distinguished… Hm.
Best Answer
As you say, the only distinguished element of $\text{End}(V)$ is $\text{id}_V$ and that's what you end up getting. I haven't checked but you should be able to verify this by writing everything out in terms of a basis $e_i$ of $V$ and the corresponding dual basis $e_i^{\ast}$ of $V^{\ast}$. You get
$$\text{id}_V = \sum_{i=1}^n e_i \otimes e_i^{\ast}.$$