I just learn the concept of elementary k form: on an open set $A$. That is if $x\in A$ , then we have $$\begin{align*} \phi_I(x)=\phi_{i_1}(x) \wedge \cdots \wedge \phi_{i_k}(x) \end{align*}$$ where $I=(i_1,\ldots ,i_k)$ is an increasing tuple, $\phi_{i_1},\ldots,\phi_{i_k}$ are elementrary 1 form on $A$, and $\phi_{I}$ is a elementary $k$ form on $A$, and hence $\phi_I(x)$ is an alternating $k$ tensor on the tangent space of $x$ denoted as ${\scr T}_x(\mathbb{R}^n)$ . That is $$\begin{align*} \phi_I(x): {\scr T_x}(\mathbb{R}^n) \to \mathbb{R} \end{align*}$$ Then if I use $(x,v)$ denote vectors in the tangent space of $x$ , I'm wandering what's the result of $$\begin{align*} \phi_I(x)((x,v_1),\ldots ,(x,v_k)) \end{align*}$$ I think the result is not easy to write out. In other words, for elementary $k$ tensors $$\begin{align*} \psi_I(v_1,\cdots ,v_k)&= [\phi_{i_1}\cdots \phi_{i_k}](v_1,\cdots ,v_k)\\ &= \phi_{i_1}(v_1)\cdots \phi_{i_k}(v_k) \\ &= v_1^{i_1}\cdots v_k^{i_k} \end{align*}$$ where $v_1^{i_1}$ is the $i_1$ coordinate of vector $v_1$. But if we have $$\begin{align*} \phi_{I}(v_1,\cdots ,v_k) \end{align*}$$ where $\phi_{I}=\phi_{i_1} \wedge \cdots \wedge \phi_{i_k}$, I don't think one can get the answer easily, since wedge product involves the sign of permutations. Is this true?
Edit:
Here the definition of wedge product is the following:
If we have a vector space $V$ with dimension $n$, and $f$ is a $k$ alternating tensor and $g$ is a $l$ alternating tensor. Then we have there wedge product： $$\begin{align*} (f \wedge g)(v_1,\ldots ,v_{k+l})=\sum\limits_{\sigma\in S_{k,l}} \text{sgn } (\sigma) f(v_{\sigma(1)},\ldots ,v_{\sigma(k)}) g(v_{\sigma(k+1)},\ldots ,v_{\sigma(k+l)}) \end{align*}$$ where $$\begin{align*} S_{k,l}=\{\sigma\in S_{k+l}: \sigma(1)<\cdots <\sigma(k);\sigma(k+1)<\cdots <\sigma(k+l)\} \end{align*}$$ $\sigma$ is a permutation.
This is from "Munkres-Analysis on Manifolds"
$ \newcommand\sgn{\mathrm{sgn}} \newcommand\alt{\mathrm{alt}} $It helps to unravel the set $S_{k,l}$. Suppose we drop the $\sigma(1) < \dotsb < \sigma(k)$ requirement. This gives a set $S_{[k],l}$ which we can express as $$ S_{[k],l} = \{\sigma'\circ\sigma \;:\; \sigma \in S_{k,l},\: \sigma'\in S_{(k),l}\},\quad S_{(k),l} = \{\sigma \in S_{k+l} \;:\; \sigma(i) = i,\: k+1\leq i\leq k+l\} $$ where $S_{k+l}$ is the set of all permutations on $\{1,\dotsc,k+l\}$. For every $\sigma \in S_{k,l}$ there are $k!$ elements of the form $\sigma'\circ\sigma$ in $S_{[k],l}$, and $|S_{[k],l}| = k!|S_{k,l}|$. Now we note that $$ f(v_{\sigma'\circ\sigma(1)},\dotsc,v_{\sigma'\circ\sigma(k)}) = \sgn(\sigma')f(v_{\sigma(1)},\dotsc, v_{\sigma(k)}) $$ and then $$\begin{aligned} (f\wedge g)(v_1,\dotsc,v_{k+l}) &= \sum_{\sigma\in S_{k,l}} \sgn(\sigma) f(v_{\sigma(1)},\dotsc,v_{\sigma(k)})g(v_{\sigma(k+1)},\dotsc,v_{\sigma(k+l)}) \\ &= \frac1{k!}\sum_{\sigma'\in S_{(k),l}}\sum_{\sigma\in S_{k,l}} \sgn(\sigma')\sgn(\sigma) f(v_{\sigma'(\sigma(1))},\dotsc,v_{\sigma'(\sigma(1))}) g(v_{\sigma(k+1)},\dotsc,v_{\sigma(k+l)}) \\ &= \frac1{k!}\sum_{\delta\in S_{[k],l}} \sgn(\delta) f(v_{\delta(1)},\dotsc,v_{\delta(k)}) g(v_{\delta(k+1)},\dotsc,v_{\delta(k+l)}). \end{aligned}$$ A similar argument for the $k+1,\dotsc,k+l$ portion yields $$ (f\wedge g)(v_1,\dotsc,v_{k+l}) = \frac1{k!l!}\sum_{\sigma\in S_{k+l}} \sgn(\sigma) f(v_{\sigma(1)},\dotsc,v_{\sigma(k)}) g(v_{\sigma(k+1)},\dotsc,v_{\sigma(k+l)}). $$ Define the tensor product as $$ (f\otimes g)(v_1,\dotsc,v_{k+l}) = f(v_1,\dotsc,v_k)g(v_{k+1},\dotsc,v_{k+l}) $$ and the (unnormalized) alternation map for 1-forms $\phi_1,\dotsc,\phi_k$ by $$ \alt(\phi_1\otimes\dotsb\otimes\phi_k) = \sum_{\sigma\in S_k} \sgn(\sigma)\phi_{\sigma(1)}\otimes\dotsb\otimes\phi_{\sigma(k)} $$ which we extend by linearity to all tensors. Then $$ f\wedge g = \frac1{k!l!}\alt(f\otimes g). $$ By the fact that $$ \sgn(\sigma)\alt(\phi_{\sigma(1)}\otimes\dotsb\otimes\phi_{\sigma(k)}) = \alt(\phi_1\otimes\dotsb\otimes\phi_k) $$ for any $\sigma \in S_k$, we can argue by induction on $k$ that $$ \phi_1\wedge\dotsb\wedge\phi_k = \alt(\phi_1\otimes\dotsb\otimes\phi_k). $$ The RHS applied to $(v_1,\dotsc,v_k)$ is precisely the determinant $$\begin{vmatrix} \phi_1(v_1) & \phi_1(v_2) & \cdots & \phi_1(v_k) \\ \phi_2(v_1) & \phi_2(v_2) & \cdots & \phi_2(v_k)\\ \vdots & \vdots & \ddots & \vdots \\ \phi_k(v_1) & \phi_k(v_2) & \cdots & \phi_k(v_k). \end{vmatrix}$$