TITLE: Why there is a relation among the second-order minors of a symmetric $4\times 4$ matrix?
QUESTION [35 upvotes]: A $4\times 4$ symmetric matrix
$$
\left(
\begin{array}{cccc}
 a_{11} & a_{12} & a_{13} & a_{14} \\
 a_{12} & a_{22} & a_{23} & a_{24} \\
 a_{13} & a_{23} & a_{33} & a_{34} \\
 a_{14} & a_{24} & a_{34} & a_{44} \\
\end{array}
\right)
$$
contains exactly 21 minors of order 2, but there is a linear combination of them, namely
$$
-(a_{13} a_{24}-a_{14} a_{23})+(a_{12} a_{34}-a_{14} a_{23})-(a_{12} a_{34}-a_{13} a_{24})\, ,
$$
which vanishes, as opposed as to the case of symmetric $3\times 3$ symmetric matrices, whose 6 minors of order two are linearly independent.

BIG QUESTION: Why is that? I mean, what is the theory behind such a phenomenon?
PHILOSOPHICAL QUESTION: What is the "true number" of minors (from the example above), 20 or 21?

I already gave myself an explanation, but I still cannot see the big picture. I would be grateful if anyone pointed out a reference, sparing me the efforts of reinventing the wheel.
If an $n\times n$ matrix $A$ is regarded as an element of $V\otimes_{\mathbb{K}} V^\ast$, with $V\equiv \mathbb{K}^n$, then there is an obvious way to extend $A$ to a $\mathbb{K}$-linear map
$$
A^{(k)}:\bigwedge^kV\longrightarrow\bigwedge^kV^\ast\, .
$$
If $A$ is symmetric, then so is $A^{(k)}$, i.e., there is a (polynomial of degree $k$) map
$$
S^2(V^\ast)\ni A\stackrel{p}{\longmapsto} A^{(k)}\in W^k:=S^2\left( \bigwedge^kV ^\ast \right)\, .
$$
Observe that $W_k$ has dimension $\frac{{n\choose k}\left({n\choose k}+1\right)}{2}$, which is 21 for $n=4$ and $k=2$. So, I was led to identify $W^k$ with the space of "formal minors" of order $k$ of a $n\times n$ matrix (indeed, the entries of $A^{(k)}$ are precisely such minors).
How to explain now the dependency of three of them?
There is a linear map
$$
S^2\left( \bigwedge^2V ^\ast \right)\ni\rho\odot\eta\stackrel{\epsilon}{\longmapsto}\rho\wedge\eta\in \bigwedge^4V ^\ast\equiv\mathbb{K}\, ,
$$
and it can be proved that $\epsilon\circ p=0$, i.e., that $p$ takes its values in the subspace
$$
W^2_0:=\ker\epsilon\, ,
$$
which has dimension 20 (vanishing of the above linear combination corresponds precisely to the equation $\epsilon(p(A))=0$).

SIDE QUESTION: How to define this $W_0^k$ for arbitrary $n$ and $k$? It should be the "linear envelope" of the image of $p$: but how to exhibit it explicitly?

Probably, the theory of representations may answer this: $W^k$ is not always an irreducible $\mathfrak{gl}(n,\mathbb{K})$-module, and $W_0^k$ is the only irreducible component which contains the image of $p$. Provided this guess is true, it doesn't help me finding the expression of $W_0^k$ in terms of tensors on $V$.

REPLY [14 votes]: This is a comment regarding Robert Bryant's conjecture: It is indeed true and not hard if you know the representation theory of $GL(V)$ well enough. In fact, the following stronger statement is true:

The only common constituent of $(S^2 V)^{\otimes k}$ and $(\bigwedge^k V)^{\otimes 2}$ is the representation with highest weight $2 \alpha_k$.

One must then check that this constituent lies in the symmetric parts of the tensor powers, which is probably most easily done by pointing out that the $k \times k$ minors of a symmetric matrix give a nonzero $GL(V)$ equivariant map $(\bigwedge^k V)^{\otimes 2} \longrightarrow (S^2 V)^{\otimes k}$.
Write weights as partitions $(\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n)$, so $S^2 V$ has height weight $(2,0,0,\ldots,0)$ and $\bigwedge^k V$ has highest weight $(\overbrace{1,1,\ldots,1}^k,0,0,\ldots,0)$. Tensoring with $S^k V$ and with $\bigwedge^k V$ is described by the Pieri rule and its transpose. From the Pieri rule, any irreducible constituent $V_{\lambda}$ of $(S^2 V)^{\otimes k}$ has $\sum \lambda_i=2k$ and $\lambda_{k+1} = \lambda_{k+2} = \cdots = \lambda_n=0$. From the transpose of the Pieri rule, any $V_{\lambda}$ in $(\bigwedge^k V)^{\otimes 2}$ has all $\lambda_i \leq 2$. The only way to satisfy these conditions is $\lambda = (\overbrace{2,2,\ldots,2}^k,0,0,\ldots,0)$.
More generally, the only common constituent of $(S^{\ell} V)^{\otimes k}$ and $(\bigwedge^k V)^{\otimes \ell}$ is the representation of highest weight $\ell \alpha_k$. 
It is definitely in $S^{\ell} (\bigwedge^k V)$. If $\ell$ is even, then it is in $S^k(S^{\ell} V)$, otherwise it is in $\bigwedge^k (S^\ell V)$.