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It is classically known that every smooth plane quartic curve $C$ can be represented by an equation $q_1 q_3 = q_2^2,$ with $q_j\in\mathbb{C}[X,Y,Z]$, $1\leq j\leq 3$ quadratic forms, and the same is true for $C$ being the union of two smooth conics, cf. e.g. C.T.C.Wall's "Is every quartic a conic of conics?"

How about the union of a smooth conic and a smooth quartic, can it be represented by $q_1 q_3 = q_2^2,$ with $q_j\in\mathbb{C}[X,Y,Z]$, $1\leq j\leq 3$ cubics?


Edit. I do not know whether smoothness assumptions on the quartic and/or on the conic can be completely dropped without changing the outcome. E.g. in a similar setting of the union of a line and a cubic, a representation of the union by $q_1 q_3 = q_2^2,$ (with $\deg q_j=2$) need not exist if the cubic is cuspidal, cf. [loc.cit.].


There is also a natural counterpart to this for $\mathbb{R}$ instead of $\mathbb{C}$, where in the context of positive forms one should talk about the conics and the quartic represented by sums of 3 squares: let $a_i, c_i, b_i\in\mathbb{R}[X,Y,Z]$ be forms. Is it true that for any such $a_i$ of degree 1 and $c_i$ of degree 2, with $1\leq i\leq 3$, there always exist $b_i$ of degree 3 so that

$$(a_1^2+a_2^2+a_3^2)(c_1^2+c_2^2+c_3^2)=b_1^2+b_2^2+b_3^2$$

holds? Again, this is classically known to hold for $a_i$ and $c_i$ of degree 1 and $b_i$ of degree 2, the case extensively studied recently too, see e.g. C.Scheiderer's "Hilbert's theorem on positive ternary quartics : a refined analysis".

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  • $\begingroup$ In my opinion, that sounds unlikely. The locus of "quadrics of cubics" inside the full projective linear system of plane sextics forms a divisor. Why should that divisor contain the entire locus of reducible curves that are unions of a conic and a quartic? Presumably one could prove that it does not by computing the Zariski tangent space of this divisor at the point parameterizing a single reducible curve that has such a representation. If the Zariski tangent space of the divisor does not contain the tangent space of the reducible locus, then we are done. $\endgroup$ – Jason Starr Sep 8 '17 at 11:39
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Edit. Mistake in the analysis at end of answer. When I computed how $F$ extends to the blowing up, I ignored a particular subvariety contained in the exceptional locus that contains the intersection of the two exceptional divisors. That is a mistake. In fact, there is an affine curve in the image of $F$ that limits to the special point, yet whose lift to $U\otimes_{\mathbb{C}} S_3$ limits to the special subvariety, namely the image under $F$ of the following curve, $$(q_1(t),q_2(t),q_3(t)) = $$ $$t^0(X_1^3,X_1^2(X_2-X_1),X_1(X_2-X_1)^2) +$$ $$ t^1(2X_1(3X_2^2-3X_1X_2+X_1^2), (X_2-X_1)(3X_2^2-4X_1X_2+2X_1^2),2(X_1-X_2)^3) + $$ $$t^2(0,0,X_3^3).$$ So the argument below with Zariski tangent spaces is not complete. I need to work out the image under the extension of $F$ of this special subvariety. It appears, at first blush, that the reducible curves in the image all have a singular point (in the intersection of the two components) with analytic type $x^2 + y(y-x)(y-\lambda x).$ This is certainly not the analytic type of a typical intersection point of a plane conic and plane quartic (that analytic type is simply nodal). So hopefully I can complete this argument.

First edit. There was an issue with the answer below, but this has been addressed by the "analysis of stability" at the end of the answer. The induced morphism of quasi-projective varieties, $$\mathbb{P}F : \mathbb{P}(U\otimes_{\mathbb{C}}S_3) \setminus \text{Zero}(F) \to \mathbb{P}(S_6),$$ is not projective. Thus, I need to prove that the image of my special point, $F(q_1,q_2,q_3),$ is not also a "limit" of an affine curve in the image of $F$ whose inverse image under $F$ is a subvariety of the quasi-projective domain whose closure contains no point mapping to $F(q_1,q_2,q_3)$. That sounds complicated, but it almost exactly the analysis of the GIT unstable locus for the natural action of $\textbf{SO}_\beta$ on $U\otimes_{\mathbb{C}} S_3$. I will try to add this GIT analysis to the end of this problem soon.

Original answer. A computation of Zariski tangent spaces disproves this, as in my comment. Sorry for introducing more indices, but it is easier for me to organize the computation if I change the names of your variables. Denote by $S$ the polynomial ring $\mathbb{C}[X_1,X_2,X_3]$. Denote by $S_d$ the direct summand of homogeneous polynomials of degree $d.$ Denote the vector space $\mathbb{C}^{\oplus 3}$ by $U$. Denote by $\beta$ the quadratic form, $$\beta:U \to \mathbb{C}, \ \ (t_1,t_2,t_3) \mapsto t_1t_3-t_2^2.$$

There are two relevant homogeneous polynomial maps of $\mathbb{C}$-vector spaces. First, denote by $F$ the following homogeneous polynomial map of $\mathbb{C}$-vector spaces, $$F:U\otimes_{\mathbb{C}}S_3 \to S_6, \ \ (q_1,q_2,q_3)\mapsto q_1q_3-q_2^2.$$ For me, it is fastest to record the derivative of the map in the following way, $$dF_{(q_1,q_2,q_3)}:U\otimes_{\mathbb{C}}S_3 \to S_6, \ \ (\dot{q}_1,\dot{q}_2,\dot{q}_3) \mapsto q_3\dot{q}_1 -2q_2\dot{q}_2 + q_1\dot{q}_3.$$ This comes from expanding $F(q_1+\epsilon \dot{q}_1,q_2+\epsilon \dot{q}_2,q_3+\epsilon \dot{q}_3)$, and gathering the coefficient of $\epsilon$. Note that the kernel of $dF$ always contains the $3$-dimensional vector subspace that is a copy of $\mathfrak{so}_\beta \cong \mathfrak{so}_3$ inside $U\otimes_{\mathbb{C}}\text{span}(q_1,q_3,q_3) \subset U\otimes_{\mathbb{C}} S_3,$ because $F$ is composed with the quadratic form $\beta.$

Second, denote by $M$ the following homogeneous polynomial map of $\mathbb{C}$-vector spaces, $$M:S_2\oplus S_4 \to S_6, \ \ (b,h)\mapsto b\cdot h.$$ This is just a multiplication map. The derivation of this map equals, $$dM_{b,h}:S_2\oplus S_4 \to S_6, \ \ (\dot{b},\dot{h}) \mapsto h\dot{b} + b\dot{h}.$$ This comes from expanding $(b+\epsilon\dot{b})(h+\epsilon \dot{h})$, and gathering the coefficient of $\epsilon$.

Now consider the special point $(q_1,q_2,q_3) = (X_1^3,X_2^3,X_3^3)$. The image under $F$ is, $$F(X_1^3,X_2^3,X_3^3) = X_1^3X_3^3-X_2^6= b(X_1,X_2,X_3)h(X_1,X_2,X_3), $$ $$b(X_1,X_2,X_3) = X_1X_3-X_2^2, \ \ h(X_1,X_2,X_3) = X_1^2X_3^2 + X_1X_3X_2^2 + X_2^4.$$ The derivative map of $F$ at this special point is, $$ dF_{q_1,q_2,q_3}:S_3^{\oplus 3} \to S_6, $$ $$(\dot{q}_1,\dot{q}_2,\dot{q}_3) \mapsto X_3^3\dot{q}_1 -2X_2^3\dot{q}_2 + X_1^3\dot{q}_3.$$ By considering the exponent vectors of monomials, and in particular which monomials are divisible by at least one of $X_1^3$, $X_2^3$, and $X_3^3$, the kernel of the derivative map is precisely the copy of $\mathfrak{so}_\beta$. Thus, choosing an affine linear subspace of $S_3^{\oplus 3}$ containing $(q_1,q_2,q_3)$ and whose tangent space is complementary to this copy of $\mathfrak{so}_\beta$, the restriction of $F$ to this linear subspace is immersive with image (locally) equal to the image of $F$. Thus, to prove that the image of $F$ does not contain the locus of reducible curves, it suffices to prove that the image of $dF_{q_1,q_2,q_3}$ does not contain the Zariski tangent space to the locus of reducible curves.

By the form of $dF_{q_1,q_2,q_3}$, the image is precisely the vector space spanned by monomials that are divisible either by $X_1^3$, by $X_2^3$, or by $X_3^3$. On the other hand, the Zariski tangent space to the locus of reducible plane curves contains the image of $$dM_{b,h}:S_2\oplus S_4 \mapsto S_6, \ \ (\dot{b},\dot{h}) \mapsto (X_1^2X_3^2 + X_1X_3X_2^2 + X_2^4)\dot{b} + (X_1X_3-X_2^2)\dot{h}.$$ For $(\dot{b},\dot{h}) = (0,X_1X_3X_2^2)$, among many other possibilities, the image tangent vector is not a polynomial whose monomials are each divisible by $X_1^3,$ by $X_2^3,$ or by $X_3^3.$ Indeed, the image is, $$dM_{b,h}(0,X_1X_3X_2^2) = X_1^2X_2^2X_3^2 - X_1X_3X_2^4.$$ The second term is divisible by $X_2^3$, however, the first term is divisible by none of $X_1^3,$ $X_2^3,$ nor $X_3^3.$

Analysis of Stability. The zero locus of $F$ inside $U\otimes_{\mathbb{C}} S_3$ consists of two irreducible components. The simpler component consists of triples $(a_1,a_1,a_3)\otimes q$ for $(a_1,a_2,a_3)\in U$ an isotropic vector of $\beta$. These triples are all GIT unstable for the natural action of $\textbf{SO}_\beta$ on $U\otimes_{\mathbb{C}} S_3$. That means that when we consider $\mathbb{P}F$ as a rational map on the GIT quotient, $$X = \mathbb{P}(U\otimes_{\mathbb{C}} S_3)^{\text{ss}}/\textbf{SO}_\beta,$$ then these triples are not even in $X$. So we can ignore these triples.

The second component is more tricky. It is the image of the morphism, $$e:\mathbb{P}(S_1)\times \mathbb{P}(S_1\oplus S_1) \to \mathbb{P}(U\otimes_{\mathbb{C}}S_3), $$ $$ ([L_2],[(L_1,L_3)]) \mapsto [(L_2L_1^2, L_2L_1L_3,L_2L_3^2)].$$ The image of this morphism is singular along the locus of triples where $L_2$ is in the span of $(L_1,L_3)$. So we need to blow up this locus. Then the strict transform of the image is smooth, but $F$ is still zero on the strict transform. So we again need to blow up the strict transform. Once we do this, $F$ is zero only on points of the exceptional divisors that map into the unstable locus. Removing this locus, $\mathbb{P}F$ extends to a regular morphism. Equivalently, performing the two blowings up in the GIT quotient $X$, the morphism $\mathbb{P}F$ extends to a regular morphism. The domain of this extended morphism is projective.

Finally, the restriction of the extension of $\mathbb{P}F$ to each exceptional divisor maps into the locus in $\mathbb{P}(S_6)$ of polynomials that are divisible by a linear factor. Since $X_1^3X_3^3-X_2^6$ is not divisible by a linear factor (it is a product of three irreducible quadratic factors), the image of the special point is not contained in the image of the exceptional divisors. That settles the issue raised at the top of the post.

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  • $\begingroup$ By the way, how did you guess the special point? $\endgroup$ – Dima Pasechnik Sep 8 '17 at 15:05
  • $\begingroup$ I have asked for a smooth quartic in the question, while the quartic you specify in your special point is highly singular (the union of two conics intersecting in two double points). It, naively at least, looks like a highly singular point on $S_2\oplus S_4$, and I won't be surprised if its tangent space exhibited an unexpected weirdness. $\endgroup$ – Dima Pasechnik Sep 8 '17 at 21:04
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    $\begingroup$ Dear Dima, The tangent bundle is a coherent sheaf. So if we compute that the tangent space to the image of $F$ is the appropriate dimension at a special point, yet the tangent space does not contain the tangent space of the locus of reducible plane curves, then that means that the same is true for generizations of that special point. This is part of the method of "specialization" in algebraic geometry. $\endgroup$ – Jason Starr Sep 8 '17 at 21:12
  • $\begingroup$ Oops, sorry, you are right---I'm too used analytic definition of tangent spaces... $\endgroup$ – Dima Pasechnik Sep 8 '17 at 21:58
  • $\begingroup$ The problem I see here is that one cannot take the full $S_4$ in $S_2\oplus S_4$, but only an open set: the smooth part (as requested in the question) or a bigger part, that can be represented by $s_1 s_3 -s_2^2$, with $s_i$ degree 2 forms. Otherwise your approach might merely confirm the fact that one cannot take the full $S_4$ to hope for a positive answer to the original question, something we know all along. $\endgroup$ – Dima Pasechnik Sep 10 '17 at 9:11

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