$\newcommand{\id}{\text{id}}$ $\newcommand{\Hom}{\text{Hom}}$

This is a cross-post. Let $V$ be a $d$-dimensional real vector space, and let $2 \le k \le d-1$. Every inner product on $V$ induces an inner product on $\Lambda^k V$:

$$ \langle v_1 \wedge \dots \wedge v_k , w_1 \wedge \dots \wedge w_k \rangle:=\det (\langle v_i ,w_j \rangle). $$

What are necessary and sufficient conditions on a product on $\Lambda^k V$ to to be induced from a product on $V$?

For $k=d-1$ the answer is that every product on $\Lambda^{d-1} V$ is induced from a product on $V$.

**Edit 1:**

If there exist an inducing product at the base, this product is **unique** (details are provided under the "edit" here). Perhaps we can construct an "inverse map" which is defined on the space of products on $\Lambda^k V$, and see when the result is an honest inner product on $V$ (and not just a bilinear form).

**Edit 2:**

Here is an equivalent formulation of the question:

A choice of a product on $V$ is equivalent to a choise of a linear isomorphism $ g:V \to V^*$ that satisfies

$$ g(v)(w)=g(w)(v) \, \, \text{and}\, \,g(v)(v) \ge 0 \, \, \text{with equality only when } \, v=0. \tag{1}$$

The equivalence is via $g(v)(w):= \langle v,w \rangle$. Using this perspective, the metric on $\Lambda^{k} V$ induced by $g$ is $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong (\Lambda^{k} V)^*$.

So, the question becomes the following:

For which maps $h:\Lambda^{k} V \to (\Lambda^{k} V)^*$ which are symmetric and positive in the sense of $(1)$, there exist a symmetric and positive $g$ such that $h=\Lambda^kg$?

(The symmetry and positivity requirements on $g$ are in fact redundant-if there exist a "root" $g$ such that $h=\Lambda^kg$, then $g$ is symmetric and can be taken to be positive definite).

Peter Michor suggests using Plucker relations as a necessary condition. These relations give an equivalent conditions for an element $h\in \Lambda^k (V^* \otimes V^*)$ to be decomposable, i.e. $h=g_1\wedge g_2\wedge\dots\wedge g_k$, where $g_i \in V^* \otimes V^*$.

However, in the formulation above, $ g:V \to V^*$, and $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong (\Lambda^{k} V)^*$ is the induced map on exterior powers, that is an element of $\Lambda^{k} V^* \otimes \Lambda^{k}V^*$. Thus, in order to use the Plucker relations, we need a way to consider it as an element in $\Lambda^{k}(V^* \otimes V^*)$. Moreover, we need that this identification will map "power-elements" to "decomposable elements".

I am not sure such a map exists. Here is a precise question on this.

**Yet another equivalent formulation of the question...:**

Given $(\binom {d}k)^2$ numbers, indexed by ordered pairs $\big((i_1,\dots,i_k),(j_1,\dots,j_k)\big)$ where $1 \le i_1 <i_2 < \dots<i_k\le d$, under what conditions do they form the $k$ minors of some $d \times d$ matrix? i.e.

$$ b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}=k-\text{minor of a } d \times d \text{ matrix, corresponding to rows } (i_1,\dots,i_k), \text{and to colums} (j_1,\dots,j_k) $$

The equivalence is obtained by choosing a fixed basis $e_1,\dots,e_d$ for $V$, and setting $$b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}=\langle e_{i_1} \wedge \dots \wedge e_{i_k} , e_{j_1} \wedge \dots \wedge e_{j_k} \rangle.$$

Technically, we should also take care of the symmetry and positivity; however, it turns out that if the "upper matrix" $b$ is "symmetric*" and positive, then the underlying matrix $A$ (if exists) is symmetric and definite, and can always be chosen to be positive.

*The symmetry of the "matrix $b$" is $b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}=b_{(j_1,\dots,j_k),(i_1,\dots,i_k)}$. The positivity corresponds to $b_{(i_1,\dots,i_k),(i_1,\dots,i_k)} >0$.