Let $\Lambda^k(V)$ be the space of alternating $k$-linear tensors on $V$. Consider the map $f: \left(\mathbb{R}^n\right)^{n-k} \to \Lambda^k(\mathbb{R}^n)$ given by $\left(v\_1,v\_2, ..., v\_{n-k}\right) \mapsto \text{Det}(v\_1,v\_2,...,v\_{n-k},\cdot)$, i.e. the $k$-form you get is a partially applied determinant.


Is there a name for the image of $f$? I am having trouble computing even basic things like the dimension of this space. I would imagine this space must have been studied classically.


I am thinking about this because the surjectivity of the map for $n=3$ seems to underlie a lot of multivariable calculus in $3$ dimensions (it lets you translate 2 forms into vector fields for instance, which is what the textbooks all seem to do).



Variety determined by interior product of the determinant?
Using the determinant is something of a distraction. In completely coordinate-free terms, and after applying the observation in abx's answer, the map you're trying to understand is the wedge product map
$$V^k \ni (v\_1, ..., v\_k) \mapsto v\_1 \wedge ... \wedge v\_k \in \Lambda^k(V)$$


where $V$ is a finite-dimensional vector space (of dimension, say, $n$). So the variety you're trying to understand is the variety of pure tensors sitting inside $\Lambda^k(V)$. 


This variety has definitely been studied classically. The pure tensors are characterized by a set of quadratic relations called the [Plücker relations](http://en.wikipedia.org/wiki/Pl%C3%BCcker_coordinates#Quadratic_relation). They cut out a subvariety of the projective space $\mathbb{P}(\Lambda^k(V))$ into which the Grassmannian $\text{Gr}\_k(V)$ embeds via the [Plücker embedding](http://en.wikipedia.org/wiki/Pl%C3%BCcker_embedding)


$$\text{Gr}\_k(V) \ni \text{span}(v\_1, ..., v\_k) \mapsto v\_1 \wedge ... \wedge v\_k \in \mathbb{P}(\Lambda^k(V)).$$


It follows that the pure tensors are the affine cone of $\text{Gr}\_k(V)$ in this embedding, and so the space of pure tensors has dimension one greater than $\text{Gr}\_k(V)$, hence $k(n-k) + 1$.


*Example.* The first nontrivial Plücker relation occurs when $\dim V = 4$ and $k = 2$. We can describe a pure tensor $v \wedge w$ using a $2 \times 4$ matrix, say 


$$\left[ \begin{array}{ccc} c\_1 & c\_2 & c\_3 & c\_4 \\ d\_1 & d\_2 & d\_3 & d\_4 \end{array} \right],$$


where the first row gives the components of $v$ in some basis and the second row gives the components of $w$. The components of $v \wedge w$ are then given by the $2 \times 2$ minors of this matrix, and the only nontrivial Plücker relation, given by expanding $(v \wedge w) \wedge (v \wedge w) = 0$, is


$$p\_{12} p\_{34} - p\_{13} p\_{24} + p\_{14} p\_{23} = 0$$


where $p\_{ij} = \det \left[ \begin{array}{cc} c\_i & c\_j \\ d\_i & d\_j \end{array} \right]$. Hence the pure tensors have dimension $6 - 1 = 2(4 - 2) + 1 = 5$ in this case. 




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Identifying forms and vector fields is a somewhat different matter. The canonical identifications are as follows. As abx says, the wedge product gives a nondegenerate pairing $\Lambda^k(V) \otimes \Lambda^{n-k}(V) \to \Lambda^n(V)$. The partial determinant gives a nondegenerate pairing $\Lambda^k(V) \otimes \Lambda^k(V^{\ast}) \to 1$ ($1$ the ground field) which allows us to commute taking duals with taking exterior powers. This gives identifications


$$\Lambda^k(V) \cong \Lambda^n(V) \otimes \Lambda^{n-k}(V)^{\ast} \cong \Lambda^n(V) \otimes \Lambda^{n-k}(V^{\ast}).$$


Hence to identify $\Lambda^k(V)$ with $\Lambda^{n-k}(V^{\ast})$ requires precisely the data of a nonzero element of $\Lambda^n(V)$. Globally this says that to identify vector fields with $n-1$-forms on an $n$-dimensional manifold requires precisely the data of a volume form. In particular you do not need a metric. 


