# What element of $\text{End}(V)$ does the trace correspond to?

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.

• In fact, I believe that you should find that $F:\operatorname{End}(V) \to \operatorname{End}(V)^*$ is the map $$F(\alpha)(\beta) = \operatorname{tr}(\alpha \beta)$$ Aug 19 '20 at 8:28

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
$$\text{id}_V = \sum_{i=1}^n e_i \otimes e_i^{\ast}.$$
• Cool, thanks so much!! I guess it had to be that, by Occam's razor or something :). So is there no "basis free" way of writing $1_V \in V \otimes V^*$? Or equivalently $\text{tr} \in V^* \otimes V^{**}$? Aug 18 '20 at 23:54
• You can almost formalize the intuition that it "has to be" the identity: whatever you get it has to be invariant under conjugation by automorphisms of $V$ (because every map you've written down is functorial wrt automorphisms) and the only elements of $\text{End}(V)$ with that property are the scalar multiples of the identity. Aug 19 '20 at 0:14
• As for writing $\text{id}_V \in V \otimes V^{\ast}$ in a basis-free way you can transport $\text{id}_V \in \text{End}(V)$ along the natural automorphism. There's a whole larger story to tell here about how duals work: ncatlab.org/nlab/show/dualizable+object Aug 19 '20 at 0:15