Let $P_1,\dots,P_4$ be four projective plane conics without a common point. A dimension count tells us that there is a 9-dimensional space of quadratic relations among the defining equations $p_j=0$ for $P_j.$ Indeed, we have a $24=4\times 6$-dimensional vector space of coefficients that is mapped to the 15-dimensional space of plane quartics via the map $(q_1,\dots,q_4)\mapsto \sum_{k=1}^4 q_k p_k$.

Thus, apart from the 6 obvious Koszul relations $(p_2,-p_1,0,\dots,0)$, there should be 3 more. Do they have a good description? (At least, in the generic case of no triple of $P_j$ having a point in common, too?)

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    $\begingroup$ What do you mean by a "good description"? For a $3$-dimensional vector space $V$, the kernel $K(V)$ of the map $\text{Sym}^2(\text{Sym}^2(V))\to \text{Sym}^4(V)$ is a representation of $\textbf{GL}(V)$ that can be explicitly described via Schur functors. For a $4$-dimensional subspace $U$ of $\text{Sym}^2(V)$ with $2$-dimensional quotient $W$, there is an induced surjection $T_W:K(V)\to \text{Sym}^2(W)$. The $3$-dimensional space you describe is the kernel of $T_W$. $\endgroup$ – Jason Starr Jul 25 '17 at 12:51
  • $\begingroup$ By "good" I meant "explicit", say, like Koszul ones, sorry. $\endgroup$ – Dima Pasechnik Jul 25 '17 at 13:18

Maybe this is something like what you have in mind:

Let $V$ be a (complex) vector space of dimension $3$. It is easy to show that a generic subspace $P\subset S^2(V^*)$ of dimension $4$ can be written as $$ P = \mathrm{span}\{\ {x_1}^2,\ {x_2}^2,\ {x_3}^2,\ x_1x_2{+}x_2x_3{+}x_3x_1\ \} $$ for some basis $x_1,x_2,x_3$ of $V^*$. (There are only a finite number of other 'nongeneric' cases to be handled.)

It seems that you are asking for a 'natural' basis for the kernel of the multiplication map $$ \mu:P\otimes S^2(V^*)\longrightarrow S^4(V^*) $$ Now, $P\subset S^2(V^*)$ is invariant under permutations of the basis $x_1,x_2,x_3$, a group isomorphic to $S_3$. The action of $S_3$ on $S^2(V^*)$ then has a natural complement to $P$, which is the space $$ W = \mathrm{span}\{\ \ x_1x_2{-}x_2x_3,\ x_2x_3{-}x_3x_1\ \}. $$ As representations of $S_3$, $W$ is irreducible and $P \simeq \mathbb{C}\oplus\mathbb{C}\oplus W$.

Now, we have the natural decomposition $$ P\otimes S^2(V^*) = P\otimes (P \oplus W) = \Lambda^2(P)\oplus S^2(P)\oplus P\otimes W. $$ Clearly, $\mu\bigl(\Lambda^2(P)\bigr) = 0$ (these are the relations you are calling the 'Koszul relations'), so we need to examine the other two pieces. It is easy to check that $\mu$ is surjective, so, indeed, there is a $3$-dimensional kernel of the map $$ \mu: S^2(P)\oplus P\otimes W\to S^4(V^*). $$ It's obvious that $\mu$ is injective on $S^2(P)$ and it's easy to check that it's injective on $P\otimes W$. The two $\mu$-images intersect in a space of dimension $3$ that is invariant under $S_3$, and it is easy to see from this that the kernel $K$ splits as an $S_3$-module as a sum $K\simeq \mathbb{C}\oplus W$. Hence, there is one relation that is invariant under the action of $S_3$ and the complement is an irreducible $S_3$-module of dimension $2$. You can write them out without difficulty, but they all involve a significant number of terms.

  • $\begingroup$ This is a totally unexpected and mysterious for me. In what sense $P$ can be chosen as the span of $x_i^2$ etc. as above? Is there a group acting on $S^2(V^*)$, and you pick an orbit representative? $\endgroup$ – Dima Pasechnik Jul 25 '17 at 15:11
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    $\begingroup$ @DimaPasechnik:Take the 2-dimensional space (i.e., pencil) of conics $P^\perp\subset S^2(V)$ annihilated by $P\subset S^2(V^*)$. Generically, this pencil will be the set of conics that pass through $4$ points in general position in $\mathbb{P}V^*\simeq \mathbb{P}^2$; take these points to be $[x_1],[x_2],[x_3], [x_1+x_2+x_3]$. Then $P^\perp$ is spanned by $x^1x^2-x^2x^3$ and $x^2x^3-x^3x^1$, so $P\subset S^2(V^*)$ has the basis claimed. When the pencil is degenerate, there are special cases to consider, but not very many, and your hypotheses rule out the pencil having nonzero perfect squares. $\endgroup$ – Robert Bryant Jul 25 '17 at 15:47

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