Is this map $F:V^*\times V\to \mathbb R$ a basis-free definition?

Question

Let $V$ be a vector space of dimension $2$.

Let $\{w_1,w_2\}$ be a basis of $V$ and $\{\beta_1,\beta_2\}\subset V^*$ its dual basis.

Let $T:V^*\times V^*\times V\times V \to \mathbb R$ be a multiliear map.

Let $F:V^*\times V\to \mathbb R$ be the multiliear map defined by

$$F(u,v)=\sum_{k=1}^2T(\beta_k,u,w_k,v).$$

Question: Is the definition of $F$ indepentend of the fixed basis $\{w_1,w_2\}$ of $V$? In other words, does any basis give the same $F$?

If so, is it trivial? This is a remark in some notes but I couldn't prove it by myself. I would appreciate an explanation of the proof.

Answer 1

Yes it is. You just have to check that the value agree for all basis. That is if the basis transform as $$ w_k = \sum_{\alpha = 1}^{2} A_{\alpha k} w'_{\alpha} $$ and by condition $\beta_k(w_l) = \delta_{wl}$, the dual basis transform as $$ \beta_{k} = \sum_{\mu =1}^2 A^{-1}_{k \mu }\beta'_{\mu}, $$ then for any $u \in V^*, v \in V$ \begin{align} F(u,v) &= \sum_{k=1}^{2} T(\beta_k,u,w_k,v) \\ &= \sum_{k=1}^{2} T\Big(\sum_{\mu =1}^2 A^{-1}_{k \mu }\beta'_{\mu},u,\sum_{\alpha = 1}^{2} A_{\alpha k} w'_{\alpha},v\Big) \\ &= \sum_{\mu =1}^2 \sum_{\alpha = 1}^{2} \, \Big(\sum_{k=1}^{2} A^{-1}_{k \mu } A_{\alpha k}\Big) T(\beta'_{\mu},u,w'_{\alpha},v) \\ &= \sum_{\mu =1}^2 \sum_{\alpha = 1}^{2}\delta_{\mu \alpha} T(\beta'_{\mu},u,w'_{\alpha},v) \\&= \sum_{\alpha=1}^{2} T(\beta'_{\alpha},u,w'_{\alpha},v). \end{align} As @Ivo Terek says, this work for any (finite) dimension of $V$.

Answer 2

This solution can be used to prove the same result for an arbitrary finite-dimensional vector space $V$ over an arbitrary field $\mathbb{K}$, without any modification.

We consider $\phi:X\to Y$, where $X:= V\otimes V\otimes V^*\otimes V^*$ and $Y:V\otimes V^*$, to be the linear map given by the contraction via the natural paring of the first factor and the third factor. That is, $$\phi(x\otimes y\otimes f\otimes g)=f(x)\,(y\otimes g)$$ for every $x,y\in V$ and $f,g\in V^*$. The definition of $\phi$ is, of course, independent of the bases of $V$ and $V^*$ (i.e., $\phi$ is a canonical map).

Since $V$ is a finite-dimensional vector space (whence so is $V^*$), the natural embeddings $$V\otimes V\otimes V^*\otimes V^* \hookrightarrow V^{**}\otimes V^{**}\otimes V^*\otimes V^*$$ and $$V^{**}\otimes V^{**}\otimes V^*\otimes V^*\hookrightarrow \left(V^*\otimes V^*\otimes V\otimes V\right)^*$$ become natural isomorphisms (the first one is induced by the canonical inclusion $V\hookrightarrow V^{**}$ and the second one is a consequence of the existence of a natural inclusion $U^*\otimes W^*\hookrightarrow \left(U\otimes W\right)^*$ for vector spaces $U$ and $W$). That is, their composition $$V\otimes V\otimes V^*\otimes V^* \hookrightarrow \left(V^*\otimes V^*\otimes V\otimes V\right)^*\,.$$ is a natural isomorphism of finite-dimensional vector spaces. Thus, we can canonically (i.e., independent of bases) identify $$X=V\otimes V\otimes V^*\otimes V^* \boldsymbol{=\!=}\left(V^*\otimes V^*\otimes V\otimes V\right)^*\,.$$ Similarly, there is a canonical identification $$Y=V\otimes V^* \boldsymbol{=\!=} \left(V^*\otimes V\right)^*\,.$$ (The canonical identifications are denoted by long equal signs $\boldsymbol{=\!=}$.)

Note that the map $\phi$ takes each multilinear function $T\in \left(V^*\otimes V^*\otimes V\otimes V\right)^*\boldsymbol{=\!=}X$ to its associate $F\in \left(V^*\otimes V\right)^*\boldsymbol{=\!=}Y$. Since $\phi$ is a canonical map, $F$ is independent of the choice of the basis of $V$.