Variety determined by interior product of the determinant?

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).

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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. 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

$$\text{Gr}_k(V) \ni \text{span}(v_1, ..., v_k) \mapsto v_1 \wedge ... \wedge v_k \in \mathbb{P}(\Lambda^k(V)).$$

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.

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.

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Awesome, thanks a lot. –  Steven Gubkin Nov 15 '13 at 12:05

I think everything you ask boils down to the following: if $V$ is a vector space over a field $K$, say of dimension $n$, the wedge product $\ \wedge^pV\times \wedge^{n-p}V\rightarrow \wedge^nV\cong K\$ is a perfect pairing, i.e. it identifies one space to the dual of the other.

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How does it do this exactly? $Im(f)$ is not all of $\Lambda^k(\mathbb{R}^n)$, since the domain of $f$ is $n(n-k)$ dimensional, and $\Lambda^k(\mathbb{R}^n)$ is $\binom{n}{k}$ dimensional... –  Steven Gubkin Nov 14 '13 at 18:58
@abx: it does not quite identify one space with the dual of the other, e.g. as $\text{GL}(V)$-representations we do not get duality because the action of $\text{GL}(V)$ on $\Lambda^n(V)$ is nontrivial. –  Qiaochu Yuan Nov 14 '13 at 19:20
@Qiaochu Yuan: Right. But the OP is working with $\Bbb{R}^n$, so there is a canonical identification $\wedge^n\Bbb{R}^n\cong \Bbb{R}$. –  abx Nov 14 '13 at 19:34
@QiaochuYuan I understand that $f$ is a multilinear map, and not a linear one. It is still a function from a $n(n-k)$ dimensional vector space to a $binom{n}{k}$ dimensional vector space, so I do not see how it is going to be a surjective map. –  Steven Gubkin Nov 14 '13 at 20:27
@Steven: oh, I see, you're not looking at the induced linear map but at the image of the pure tensors. My apologies. I think the image is described by Plucker relations of some kind. –  Qiaochu Yuan Nov 14 '13 at 22:02