Let $L(V_1,V_2,...,V_n;\mathbb{R})$ the set of multilinear functions where $V_i$ are finite vector fields. Find a basis for $L(V_1,V_2,...,V_n;\mathbb{R})$.

I've proved that $L(V_1,V_2,...,V_n;\mathbb{R})$ is a vector field and for any $G, F \in L(V_1,V_2,...,V_n;\mathbb{R}) \implies G\otimes F\in L(V_1,V_2,...,V_n;\mathbb{R})$

But i don't know how to find a basis for that vector space, how i asociate the basis of each $V_i$ with the basis of $L(V_1,V_2,...,V_n;\mathbb{R})$?

  • $\begingroup$ Is an element of $L(V_1 \times \cdots \times V_n; \mathbb{R})$ a function of $V_1 \times \cdots \times V_n$ to itself or to $\mathbb{R}$? $\endgroup$ – Connor Harris Sep 30 '19 at 17:49
  • $\begingroup$ @ConnorHarris to $\mathbb{R}$, question fixed. $\endgroup$ – sango Sep 30 '19 at 17:50
  • $\begingroup$ en.wikipedia.org/wiki/Dual_basis $\endgroup$ – Connor Harris Sep 30 '19 at 17:54

I just consider two $\mathbb{R}$-linear spaces $V$ and $W$ in order to simplify the notation, but this works in general.

Let $\{e_1,...,e_n \}$ be a basis of $V$ and let $\{f_1,...,f_m\}$ be a basis of $W$.

For any $i \in \{1,...,n \}$ and any $j \in \{1,...,m \}$, let $F_{i,j}$ be the element of $L(V,W;\mathbb{R})$ such that $$F_{i,j}(v,w)=v_iw_j$$ for every $(v,w) \in V\times W$, being $v_i$ the $i$-th component of $v$ (with respect to the given basis) and being $w_j$ the $j$-th component of $w$ (with respect to the given basis).

Then $\{F_{i,j}: i \in \{1,...,n \}, j \in \{1,...,m \}\}$ is a basis of $L(V,W;\mathbb{R})$.

For the linear independence, observe that, if $\lambda_{1,1},\lambda_{1,2},...\lambda_{n,m}$ are real numbers such that $\sum_{i,j}\lambda_{i,j} F_{i,j}$ is the null element of $L(V,W,\mathbb{R})$, then $0=\sum_{i,j}\lambda_{i,j}F_{i,j}(e_{i'},f_{j'})=\lambda_{i',j'}$ for every $i' \in \{1,...,n\}$ and every $j' \in \{1,...,m\}$.

For the generation, observe that, for every $F\in L(V,W,\mathbb{R})$, it is the case that: \begin{aligned} F(v,w)&=F(v_1e_1+...+v_ne_n,w_1f_1+...+w_mf_m) \\ &=F(e_1,f_1)v_1w_1+...+F(e_n,f_m)v_nw_m\\ &=F(e_1,f_1)F_{1,1}(v,w)+...+F(e_n,f_m)F_{n,m}(v,w) \end{aligned} for every $(v,w)\in V\times W$. Hence $F=F(e_1,f_1)F_{1,1}+...+F(e_n,f_m)F_{n,m}$.

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