k-vectors being generalization of vectors I learnt that $k$-vectors are a generalization of vectors (also called $1$-vectors). For example, the following diagram shows the geometrical interpretations of a $0$-vector, a $1$-vector, a $2$-vector and a $3$-vector:

However, in manifold theory, apart from being interpreted as an arrow, a $1$-vector can also be interpreted as:
(1) an equivalence class of curves on a manifold;
(2) a partial differentiation operator on a function.
I wonder how these two interpretations can be generalized to a general $k$-vector. For example, can a $2$-vector be interpreted as an equivalence class of "curved parallelograms" on a manifold? Can it be interpreted as a partial differentiation operator on a pair of functions?
 A: Here is my conjecture for the answer to the second question. Now a 2-covector can be defined as $\alpha \wedge \beta = A(\alpha \otimes \beta)$ where $\otimes$ represents the tensor product and $A$ represents the "alternation" and for $\alpha \otimes \beta$, $A(\alpha \otimes \beta) = \alpha \otimes \beta - \beta \otimes \alpha$. Thus, for any two vectors $v$ and $w$, we have
\begin{eqnarray*}
\alpha \wedge \beta(v, w) & = & A(\alpha \otimes \beta)(v, w) \\
& = & (\alpha \otimes \beta - \beta \otimes \alpha)(v, w) \\
& = & \alpha(v)\beta(w) - \beta(v)\alpha(w) \\
& = & \left|\begin{array}{cc}
\alpha(v) & \alpha(w) \\
\beta(v) & \beta(w)
\end{array} \right|
\end{eqnarray*}
By analogy, if we view vectors as partial differentiation operators of functions, then we can define a 2-vector as $v \wedge w = A(v \otimes w)$, and so for any two functions $f$ and $g$, we have
\begin{eqnarray*}
v \wedge w(f, g) & = & A(v \otimes w)(f, g) \\
& = & (v \otimes w - w \otimes v)(f, g) \\
& = & v(f)w(g) - w(f)v(g) \\
& = & \left|\begin{array}{cc}
v(f) & v(g) \\
w(f) & w(g)
\end{array} \right|
\end{eqnarray*}
A: $
\newcommand\PD[2]{\frac{\partial#1}{\partial#2}}
\newcommand\Ext{{\bigwedge}}
$
This is a potential answer to your second question. Let $V$ be a real $n$-dimensional vector space. The identification of vectors $v$ with differential operators on functions over $V$ is via the map $v \mapsto v\cdot\nabla^*_x$, where
$$
  v\cdot w^* = w^*(v),\quad
  \nabla^*_x = \sum_{i=1}^n e^i\PD{}{x^n}
$$
for and $v \in V$, any $w^* \in V^*$, $x = x^1e_1 + \cdots + x^ne_n$, and $\{e^i\} \subset V^*$ is the basis dual to the standard basis $\{e_i\} \subset V$. There is a natural pairing between $\Ext V$ and $\Ext V^*$ given on simple multivectors by
$$
  (v_1\wedge\cdots\wedge v_k)\cdot(w^1\wedge\cdots\wedge w^l)
    = \delta_k^l\det\Bigl(w^i(v_j)\Bigr)_{i,j=1}^k.
$$
Hence, expressing $k$-vectors $X$ as
$$
  X = \sum_{1\leq i_1<\cdots<i_k\leq k} X^{i_1\cdots i_k}e_{i_1}\wedge\cdots\wedge e_{i_k},
$$
we define a $k$-vector derivative via
$$
  \nabla^*_X = \sum_{1\leq i_1<\cdots<i_k\leq k} e^{i_1}\wedge\cdots\wedge e^{i_k}\PD{}{X^{i_1\cdots i_k}}.
$$
We could stop here and map $Y \mapsto Y\cdot\nabla^*_X$, with such operators acting on functions over $\Ext^k V$. However, we can also come up with a mapping into differential operator acting on functions of $k$-vectors. When operating on linear functions $f$ over $\Ext^k V$, we can find that $\nabla^*_X$ is equivalent to the simplicial derivative
$$
  \partial^*_{(k)} = \frac1{k!}\nabla^*_{v_1}\wedge\cdots\wedge\nabla^*_{v_k},\quad v^i \in V^*
$$
in the sense that
$$
  [\nabla^*_X\cdot f(X)]_{X=v_{(k)}} = \partial^*_{(k)}\cdot f(v_{(k)}),\quad
  [\nabla^*_X\wedge f(X)]_{X=v_{(k)}} = \partial^*_{(k)}\wedge f(v_{(k)}),
$$
and various other expressions, where $v_{(k)} = v_1\wedge\cdots\wedge v_k$ is the simplicial variable. Hence we associate $k$-vectors $Y$ with differential operators
$$
  Y \mapsto Y\cdot\partial^*_{(k)}
    = \frac1{k!}\sum_{1\leq i_1<\cdots<i_k\leq k}Y^{i_1\cdots i_K}\det\left(\PD{}{x_q^{i_p}}\right)_{p,q=1}^k,
$$
where $v_i = x_i^1e^1 + x_i^2e^2 + \cdots + x_i^ne^n$. Such operators operate on functions of $k$ vector variables. Here are some examples:
When $k = 1$
$$
  Y\cdot\partial^*_{(1)} = Y\cdot\nabla^*_{v_1} = Y^1\PD{}{x_1^1} + Y^2\PD{}{x_1^2} + \cdots + Y^n\PD{}{x_1^n}.
$$
When $k = 2$ and $n = 3$
$$
  Y\cdot\partial^*_{(2)}
    = \frac1{k!}Y\cdot(\nabla^*_{v_1}\wedge\nabla^*_{v_2}) =
$$$$
    \frac12\left[Y^{12}\left(\PD{}{x_1^1}\PD{}{x_2^2} - \PD{}{x_1^2}\PD{}{x_2^1}\right) + Y^{13}\left(\PD{}{x_1^1}\PD{}{x_2^3} - \PD{}{x_1^3}\PD{}{x_2^1}\right) + Y^{23}\left(\PD{}{x_1^2}\PD{}{x_2^3} - \PD{}{x_1^3}\PD{}{x_2^2}\right)\right].
$$
When $k=3$ and $n=3$
$$
  Y\cdot\partial^*_{(3)} = Y\cdot(\nabla^*_{v_1}\wedge\nabla^*_{v_2}\wedge\nabla^*_{v_3}) =
$$$$
  \frac16Y^{123}\left(\PD{}{x_1^1}\PD{}{x_2^2}\PD{}{x_3^3} - \PD{}{x_1^1}\PD{}{x_2^3}\PD{}{x_3^2} + \PD{}{x_1^3}\PD{}{x_2^1}\PD{}{x_3^2} - \PD{}{x_1^3}\PD{}{x_2^2}\PD{}{x_3^1} + \PD{}{x_1^2}\PD{}{x_2^3}\PD{}{x_3^1} - \PD{}{x_1^2}\PD{}{x_2^1}\PD{}{x_3^3}\right).
$$
