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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.

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

$$\text{id}_V = \sum_{i=1}^n e_i \otimes e_i^{\ast}.$$

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  • $\begingroup$ 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 $\text{tr} \in V^* \otimes V^{**}$? $\endgroup$ Aug 18 '20 at 23:54
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    $\begingroup$ 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. $\endgroup$ Aug 19 '20 at 0:14
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    $\begingroup$ 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 $\endgroup$ Aug 19 '20 at 0:15

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