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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 $V$.

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    $\begingroup$ This is called "Plücker relations". $\endgroup$ – Fedor Petrov Jun 12 '15 at 17:59
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    $\begingroup$ There is a MO about Plucker relations: math.stackexchange.com/questions/49190/pl%C3%BCcker-relations but it seems to be non related to this post. Namely, by Plucker relations people understand the equations satisfied by the Plucker embedding i.e. the equations that tell when a $k$-vector is decomposable. The linear combination of this OP does not seems to be such kind of Plucker relations, isn'it ?. $\endgroup$ – Holonomia Jun 12 '15 at 18:18
  • $\begingroup$ @Holonomia: I'm aware of Plucker relations, but I believe they're not related to "my" linear combination. For a simple reason: Plucker relations (whichever way you're suggesting to implement them) hold for any value of $n$ and $k$, whereas "my" relations begin to reveal themselves for $n\geq 4$ and only for special values of $k$, which also don't show any regularity... $\endgroup$ – Giovanni Moreno Jun 12 '15 at 18:41
  • $\begingroup$ Actually, I do not suggest implement nothing. I just wonder why Fedor Petrov called this "Plucker relations". $\endgroup$ – Holonomia Jun 12 '15 at 21:15
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This is about your specific question. For any vector space $V$ of dimension $n$, one has canonical decompositions $$ S^2(S^2V^*) \simeq S^4(V^*)\oplus K(V^*) $$ and $$ S^2(\Lambda^2V^*) \simeq \Lambda^4(V^*)\oplus K(V^*), $$ where $K(V^*)$ is an irreducible $\mathrm{GL}(V)$-module of dimension $n^2(n^2{-}1)/12$. It follows that, up to a constant multiple, there is only one nontrivial $\mathrm{GL}(V)$-equivariant linear mapping $$ S^2(S^2V^*) \longrightarrow S^2(\Lambda^2V^*), $$ and hence that the number of linearly independent minors of order $2$ of an $n$-by-$n$ symmetric matrix is $n^2(n^2{-}1)/12$.

You can answer similar questions by comparing $S^k(S^2V^*)$ with $S^2(\Lambda^kV^*)$ for higher values of $k$. This kind of question goes by the name 'plethysm' in representation theory. For example, when $k=3$, we find that $S^3(S^2V^*)$ and $S^2(\Lambda^3V^*)$ share only one irreducible $\mathrm{GL}(V)$-constituent, which has dimension $$ \frac{(n{-}2)(n{-}1)^2n^2(n{+}1)}{144}, $$ so this is the number of linearly independent $3$-by-$3$ minors of an $n$-by-$n$ symmetric matrix.

Update: Here is a conjectural answer to the big question: Recall that, for a vector space $V$ of dimension $n$, a fundamental system of irreducible $\mathrm{SL}(V)$ modules is given by $\Lambda^kV^*$ for $0<k<n$. If, as is traditional, one assigns highest weights so that $\Lambda^kV^*$ has highest weight $\alpha_k$, then any irreducible representation $W$ of $\mathrm{SL}(V)$ is determined by its highest weight $w = m_1\,\alpha_1 + \cdots + m_{n-1}\,\alpha_{n-1}$, where the $m_i$ are nonnegative integers.

Using this nomenclature, $S^2(V^*)$, which is irreducible, has highest weight $2\,\alpha_1$.

The following Conjecture is supported by explicit computations for $0<k< n\le 9$. (Presumably, this conjecture (if true) would be easy to prove for an expert combinatorialist or representation theorist, but I am neither.)

Conjecture: The $\mathrm{SL}(V)$-modules $S^k(S^2V^*)$ and $S^2(\Lambda^kV^*)$ have only one common irreducible $\mathrm{SL}(V)$-constituent, namely, the irreducible $\mathrm{SL}(V)$-module $W$ of highest weight $2\,\alpha_k$.

Since, by the Weyl Character formula, the dimension of the $W$ with highest weight $2\,\alpha_k$ is $$ \frac1{k{+}1}{n\choose k}{{n+1}\choose k}, $$ it would follow from the Conjecture that this is the number of linearly independent $k$-by-$k$ minors of a symmetric $n$-by-$n$ matrix.

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  • $\begingroup$ I can not help myself to say that $K(V^*)$ is the space of (algebraic) curvature tensors and $\frac{n^2(n^2-1)}{12}$ its dimension. $\endgroup$ – Holonomia Jun 13 '15 at 16:11
  • $\begingroup$ @Holonomia: Why do you think I called it "$K(V^*)$"? $\endgroup$ – Robert Bryant Jun 13 '15 at 16:50
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    $\begingroup$ @Robert: A nice conceptual approach. By the way, the standard spelling seems to be "plethysm" (derived from a Greek word and used in a paper by D.E. Littlewood around 1950 according to Wikipedia). It's my favorite odd mathematical word apart from "syzygy", I guess. $\endgroup$ – Jim Humphreys Jun 13 '15 at 21:36
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    $\begingroup$ Can you elaborate on why these are the right plethysms to look at? I'm afraid I don't quite see it. (In particular I'm confused about whether or not I need to introduce an inner product on $V$ to make sense of the question.) $\endgroup$ – Qiaochu Yuan Jun 14 '15 at 6:00
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    $\begingroup$ QiaochuYuan: A 'symmetric matrix' can be regarded as a linear map on an inner product space or as a quadratic form on a vector space (with no inner product assumed). Taking the latter interpretation (which I regard as more natural for most purposes), a quadratic form on $V$, i.e., $Q\in S^2(V^*)$, induces a quadratic form $Q^{(k)}$ on $\Lambda^k(V)$, i.e., $Q^{(k)}\in S^2(\Lambda^kV^*)$. (The pairing induced by $Q$ is symmetric because transpose doesn't affect determinants.) This mapping is polynomial of degree $k$, i.e., it is a natural linear map $\pi^{(k)}:S^k(S^2V^*)\to S^2(\Lambda^kV^*)$. $\endgroup$ – Robert Bryant Jun 14 '15 at 9:51
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This is only about the "big question".

The symmetric $n\times n$ matrices appear naturally as the big cell on the Lagrangian Grassmannian. Any projectively normal embedding (e.g. first include into the ordinary Grassmannian, then Pl\"ucker embed) of any $G/P$ is defined by degree $1$ and $2$ equations, a result one can find in the book [Brion-Kumar] for example.

Here, that result will give you a bunch of quadratic relations between the minors of size $n$.

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The answer to my question has been provided by Robert Bryant. This is just a long comment, containing the proof to his conjecture (I hope he'll be happy!).

The key is the diagram $\require{AMScd}$ \begin{CD} S^2V^* @= S^2V^*\\ @V \epsilon_k V V{\large\circlearrowleft} @VV p_k V\\ S^k(S^2 V^*) @>>\pi^{(k)}> S^2(\Lambda^kV^*)\, , \end{CD} where $p_k(Q):=Q^{(k)}$ is the natural extension of $Q$ (see Bryant's comment above), given by $$ Q^{(k)}(v_1\wedge\cdots\wedge v_k\, ,w_1\wedge\cdots\wedge w_k)=\sum_{\sigma\in S_k}(-1)^\sigma Q(v_1\, , w_{\sigma(1)})\cdots Q(v_k\, , w_{\sigma(k)})\, , $$ and $\epsilon_k$ is the "universal power map" (I wouldn't know how to call it otherwise): $$ \epsilon_k(Q)=Q^k\, . $$ As Bryant observed, being $p_k$ a homogeneous polynomial of degree $k$, it factors through $\epsilon_k$, and $\pi^{(k)}$ is a linear map.

So, $S^2(\Lambda^kV^*)$ is the "space of formal $k^\textrm{th}$ order minors", whereas $W:=\langle p_k(S^2V^*)\rangle$ is the "space of true $k^\textrm{th}$ order minors" (i.e., the object of my question).

Let us prove that $W=V_{2\alpha_k}$, i.e., the irreducible $\mathrm{SL}(V)$-module of highest weight $2\alpha_k$ which, accordingly to Bryant, has dimension $$ \frac{1}{k+1}{n\choose k}{n+1\choose k}\, . $$

Even though $\epsilon_k$ is not surjective, $\langle\epsilon_k(S^2V^*)\rangle=S^k(S^2V^*)$, so that $W=\pi^{(k)}(S^k(S^2V^*))$. This will help to understand the rest.

To prove that $W\supseteq V_{2\alpha_k}$ it suffices to show that $W$ contains the highest weight vector $$ v_{2\alpha_k}=v_{\alpha_k}^2=(e_1\wedge\cdots\wedge e_k)^2\in S^2(\Lambda^kV^*)\, . $$ But this is true by the above formula for $Q^{(k)}$: $$ v_{2\alpha_k}=(e_1^2+\cdots+e_k^2)^{(k)}\, . $$ To prove that $W\subseteq V_{2\alpha_k}$, observe that, by $\mathrm{GL}(V)$-equivariancy of the whole framework, $V_{2\alpha_k}$ contains $Q^{(k)}$ for any $Q$ of rank $k$. Hence ($k$ is nonzero) it'll contain all $Q^{(k)}$ for any $Q$ of maximal rank, i.e., the $p_k$-image of a big cell of $S^2V^*$. By continuity, $V_{2\alpha_k}$ contains the whole image of $p_k$ and its linear envelope as well, which is our $W$.

Summing up, there is an exact sequence $$ \ker\pi^{(k)}\rightarrow W\oplus \ker\pi^{(k)}=S^k(S^2V^*)\stackrel{\pi^{(k)}}{\longrightarrow}S^2(\Lambda^kV^*)\rightarrow W\rightarrow 0\, , $$ showing that, as Bryant suspected, $W$ is a common irreducible constituent of both $S^k(S^2V^*)$ and $S^2(\Lambda^kV^*)$.

[In cooperation with D. Alekseevsky]

EDIT: this is a comment to @Allen Knutson answer

Yes, my question arose precisely in the context of Lagrangian Grassmannians. Take the Plucker embedding $$ \mathrm{Gr}(n,V\oplus V^*)\ni L=\langle l_1,\ldots, l_n\rangle\longmapsto \mathrm{vol}(L):=[l_1\wedge\cdots\wedge l_n]\in\mathbb{P}(\Lambda^n(V\oplus V^*)) $$ and observe that $$ \Lambda^n(V\oplus V^*)=\bigoplus_{k=0}^n\Lambda^k(V)\otimes\Lambda^{n-k}(V^*)=\bigoplus_{k=0}^n(\Lambda^k(V^*))^{\otimes 2}\, . $$ So, it is natural to ask: what is the smallest projective subspace of $$(*)\quad\quad\quad\mathbb{P}\left(\bigoplus_{k=0}^n(\Lambda^k(V^*))^{\otimes 2}\right)$$ containing the Plucker image of the Lagrangian Grassmannian $\mathrm{LGr}(n,V\oplus V^*)\subset \mathrm{Gr}(n,V\oplus V^*)$?

($V\oplus V^*$ is equipped with a canonical (up to a projective factor) symplectic form.)

The obvious answer, i.e., the one obtained by replacing the tensor product by the symmetric product in $(*)$, is not always correct, in view of the above linear relation among to minors. Instead of $S^2(\Lambda^KV^*)$, one has to use the irreducible submodule $W$.

Plucker relations for $\mathrm{Gr}(n,V\oplus V^*)$ constitute a set of homogeneous polynomial equations of degree $\leq n$ amongst the minors, whereas simmetricity conditions are linear in the entries ($1^\textrm{th}$ order minors): apparently, the ideal generated by them contains, in some cases, also linear conditions in the higher-than-one-order minors.

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    $\begingroup$ Yes, thanks for the argument. I was convinced that it was true, but was too lazy to go back and do the necessary computations with roots and weights in the general case. (It's easy to run a bunch of cases through a program such as Lie and convince yourself that it must be true.) $\endgroup$ – Robert Bryant Jun 17 '15 at 14:17
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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 $\bigwedge^k (S^\ell V)$.

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    $\begingroup$ Thanks for doing this. I probably should have dusted off my rusty memories of Pieri and Littlewood-Richardson and done it myself, but I was too lazy, and I was already convinced it was true. $\endgroup$ – Robert Bryant Jun 17 '15 at 14:21

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