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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 $\Lambda^k V$:

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

Question:

What are necessary and sufficient conditions on an inner 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:

Since the Plucker relations are equivalent to $h\in \Lambda^k (V^* \otimes V^*)$ being decomposable, i.e. $h=g_1\wedge g_2\wedge\dots\wedge g_k$, I think we can focus on the following question:

What are the conditions on a decomposable $h$, to have a "root", i.e to be of the form $\Lambda^k g$ for some $g \in V^* \otimes V^*$. (See the second reformulation below for the connection to this problem. The positivity and symmetry of such a root, if exists, "come for free" from the positivity and symmetry of $h$).

An equivalent formulation of the question is the following:

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 $b_{(i_1,\dots,i_k),(i_1,\dots,i_k)} >0$.

Yet another equivalent formulation...:

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 induced 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 $h=\Lambda^kg$? As mentioned before, if there exist a "root" $g$ such that $h=\Lambda^kg$, then $g$ is symmetric and can be taken to be positive definite.