Trace of an endomorphism as a wedge product

Question

Let $E$ be a finite dimensional vector space and $F=E^*$ its dual. We have that $Tr\in End(E)^*=(E\otimes F)^* = E^*\otimes F^* = \Lambda^1E^*\otimes \Lambda^1F^*\hookrightarrow \Lambda^2(E^*\oplus F^*)=\Lambda^2(E\oplus F)^*.$

I wonder what is the image of the trace under the last arrow, that is as a wedge product.

Answer 1

Let $(e_i)_{i=1}^n$ be a basis for $E$ and $e^i$ the corresponding dual basis. In what follows I will use the natural isomorphism $F^{*} = \left( E^{*} \right)^{*} \cong E$ and Einstein's summation convention.

1. As an element in $\left( E \otimes E^{*} \right)^{*}$, the trace acts on elementary tensors by $\operatorname{Tr} \left( e \otimes \varphi \right) = \varphi(e)$.
2. Under the isomorphism $\left( E \otimes E^{*} \right)^{*} \cong \left( E^{*} \otimes \left( E^{*} \right)^{*} \right) \cong E^{*} \otimes E$, the trace becomes the element $e^i \otimes e_i$. To see this, note that $$ \left( e^i \otimes e_i \right) \left( e \otimes \varphi \right) = e^i(e) \cdot e_i(\varphi) = e^i(e) \cdot \varphi(e_i) = \varphi \left( e^i(e) e_i \right) = \varphi(e). $$
3. Under the wedge product map $E^{*} \otimes E \hookrightarrow \Lambda^2 \left( E^{*} \oplus E \right)$, the trace corresponds to $(e^i,0) \wedge (0, e_i)$.
4. The natural isomorphism $E^{*} \oplus E \cong \left( E \oplus E^{*} \right)^{*}$ is given by $$(\varphi,e)(e',\varphi') = \varphi(e') + e(\varphi') = \varphi(e') + \varphi'(e).$$
5. Finally, inder the isomorphism $\Lambda^2 \left( E^{*} \oplus E \right) \cong \Lambda^2 \left( \left( E \oplus E^{*} \right)^{*} \right) \cong \Lambda^2 \left( E \oplus E^{*} \right)^{*}$, using the two previous items, we get that the trace acts as $$ \left( (e^i,0) \wedge (0,e_i) \right) \left( (e,\varphi) \wedge (e',\varphi') \right) = \left( (e^i,0)(e,\varphi) \right) \cdot \left( (0,e_i)(e',\varphi') \right) - \left( (e^i,0)(e',\varphi') \right) \cdot \left( (0,e_i)(e,\varphi) \right) = e^i(e) \cdot \varphi'(e_i) - e^i(e') \cdot \varphi(e_i) = \varphi' \left( e^i(e) e_i \right) - \varphi \left( e^i(e') e_i \right) = \varphi'(e) - \varphi(e').$$ In particular, if we identify elements of $E,E^{*}$ as elements in $E \oplus E^{*}$ we have $$ \operatorname{Tr}(e \wedge \varphi) = -\operatorname{Tr}(\varphi \wedge e) = \varphi(e), \\ \operatorname{Tr}(e \wedge e') = \operatorname{Tr}(\varphi \wedge \varphi') = 0.$$