Functional $Gram : S^m(S^2 V)^* \otimes S^2(\bigwedge^m V) \to \mathbb R$ and Gram determinant in invariant terms Let $V$ be an vector space over $\mathbb R$. Let $B \otimes ... \otimes B \in S^m(S^2 V)^* = ((S^2 V^*)^{\otimes m})^{S_m} \subset  (S^2 V^*)^{\otimes m}$ where $B \in (S^2 V)^*$ any symmetric bilinear form (not necessarily non-degenerate, not necessarily non-negative definite), and
$v_1\wedge...\wedge v_m \odot w_1\wedge...\wedge w_m \in S^2(\bigwedge^m V)$. Then we can define $$Gram(B \otimes ... \otimes B ,v_1\wedge...\wedge v_m \odot w_1\wedge...\wedge w_m) = \det (B(v_i,w_j))_{i,j=1..m}.$$ It is easy to check (I hope I'm not mistaken) that this construction gives us a linear functional
$$Gram : S^m(S^2 V)^* \otimes S^2(\wedge^m V) \to \mathbb R.$$ Is there some kind of "hidden algebra" around this functional (maybe $GL_n$ representations and so on) or more global picture? Can we describe it using symbols $\alpha \odot \beta$ instead $v_1\wedge...\wedge v_m \odot w_1\wedge...\wedge w_m$? Maybe some consequences for differential geometry (this gives us a section of the corresponding vector bundle)? It is strange for me that such functional even exist I got used to the fact that the naturally defined functionals appear as traces $W \otimes W^* \to \mathbb R$. Of course if $B$ is standart scalar product on $V=\mathbb R^n$ and $v_i=w_i$ we will obtain usual Gram determinant.
 A: More generally, if $B : V \otimes W \to k$ is bilinear (note the appearance of two vector spaces!), $k$ any field, then it induces a bilinear map on exterior powers
$$\wedge^n(B) : \wedge^n(V) \otimes \wedge^n(W) \to k$$
given by the determinant of the Gram matrix, which will be symmetric if $W = V$ and $B$ is symmetric. Note that we do not need to assume that either $V$ or $W$ is finite-dimensional here.
If $W$ is finite-dimensional $B$ corresponds to a map $B' : V \to W^{\ast}$ which has an exterior power in the usual sense $\Lambda^n(B') : \wedge^n(V) \to \wedge^n(W^{\ast})$, and then it should be true (but I haven't checked) that $\wedge^n(B)$ is the bilinear form corresponding to this map under a suitable natural identification
$$\wedge^n(W^{\ast}) \cong \wedge^n(W)^{\ast}.$$
In other words, in the finite-dimensional case the question should reduce by functoriality from an arbitrary bilinear form $B$ to the case of the dual pairing $\text{eval} : V^{\ast} \otimes V \to k$. So the question is to understand why there is a natural dual pairing
$$\wedge^n(V^{\ast}) \otimes \wedge^n(V) \to k.$$
Hopefully in this form things look a bit more intuitive, although personally I also find something confusing about this map; you can see me be confused about it here and here.
Edit: Okay, I'm happier now. In the second answer I linked to above I suggest the following strategy, which works: you can naturally and freely extend every linear functional $f \in V^{\ast}$ to a derivation of degree $-1$ on $\wedge^{\bullet}(V)$ by extending via the Leibniz rule. By the universal property of the exterior algebra this extends to an action of $\wedge^{\bullet}(V^{\ast})$ on $\wedge^{\bullet}(V)$ by "differential operators" and the pairing $\wedge^n(V^{\ast}) \otimes \wedge^n(V) \to k$ is a restriction of this map. The more general map is
$$\wedge^m(V^{\ast}) \otimes \wedge^n(V) \to \wedge^{n-m}(V)$$
for $m \le n$. For example when $m = 1$ it's given by
$$f(v_1 \wedge \dots \wedge v_n) = \sum_{i=1}^n (-1)^{i-1} f(v_i) \left( v_1 \wedge \dots \wedge \hat{v_i} \dots \wedge v_n \right)$$
where the hat means that $v_i$ is omitted. To get $m = 2$ we apply a second derivation to the above expression and we get a sum of $n(n-1)$ terms, etc. When $m = n$ we get
$$(f_1 \wedge \dots \wedge f_n)(v_1 \wedge \dots \wedge v_n) = (-1)^{ {n \choose 2} } \sum_{\pi \in S_n} \text{sgn}(\pi) f_i(v_{\sigma(i)})$$
which is, up to that pesky global sign, the full Leibniz formula for the Gram determinant, as desired. I am still a little unhappy that this description isn't obviously symmetric in $V$ and $V^{\ast}$: somehow we should be talking about a mutual action of each of $\wedge^{\bullet}(V)$ and $\wedge^{\bullet}(V^{\ast})$ on the other, or some kind of biderivation, or something...
