I am currently working on a paper that requires using the theory of invariants of binary quartic forms. Playing around, I have found an interesting identity that gives the Hessian from the minors of the catalecticant matrix of the binary quartic curve. I have not seen such relations in the reviews I have read so far, but I hope it might have a simple explanation for experts in that area. If possible, I would like to have a reference to quote in my paper and to read to educate myself.

QuestionIs there a simple way to see that the coefficients of the determinant of the Hessian of a binary quartic form depend linearly on the minors of the catalecticant matrix of the same binary quartic form (see theorem below)? Does it generalize to other binary form? I would be very thankful for a reference.

**Background and definitions**

I work over a field of characteristic zero. A binary quartic form is a homogeneous polynomial $$ f(x,y)= a x^4 + b x^3 y + c x^2 y^2 + d x y^3 + e y^4, $$ with coefficient in the field. The vector space of its covariants is generated by

$$ f(x,y), I, J, H(x,y), T(x,y), $$

where

$$ I=12 a e-3 bd +c^2, \quad J= 72 a c e+ 9 b cd -27 a d^2 -27 e b^2 -2 c^3. $$ are invariants and $H(x,y)$ and $T(x,y)$ are polynomial in (x,y,a,b,c,d,e). The polynomial $T(x,y)$ is of degree 6 in $x,y$ and will not be relevant here. The polynomial $H(x,y)$ is the determinant of the Hessian (with a convenient overall coefficient):

$H(x,y)=\frac{1}{6} \Big(\frac{\partial^2 f}{\partial^2 x} \frac{\partial^2 f}{\partial^2 y}-(\frac{\partial^2 f}{\partial x \partial y})^2 \Big).$

The invariant $J$ was discovered by Cayley and is called the *catalecticant* of the binary quartic form $f$. The invariant $J$ can be expressed as a determinant of the following symmetric matrix

$$ J=det\ Cat(f)=-\frac{1}{432}\det \begin{pmatrix} \check{a} & \check{b} &\check{c}\\ \check{b} & \check{c} &\check{d}\\ \check{c} & \check{d} &\check{e} \end{pmatrix}, $$ where $ \quad \check{a} =a, \quad \check{b}=\frac{1}{4} b, \quad \check{c}=\frac{1}{6} c, \quad \check{d}=\frac{1}{4} d, \quad \check{e}= e.$

The following definition is introduced to simplify the theorem below.

Given a $3\times 3$ matrix M, I associate a binary quartic $f_M$ in the following way: $ g_M(x,y)=\sum_{i,j} M_{i,j} y^{6-i-j} x^{i+j-2}. $ In particular, if $M$ is the symmetric matrix $$ M=\begin{pmatrix} q_1 & q_2 & q_3 \\ q_2 & q_4 & q_5 \\ q_3 & q_5 & q_6 \end{pmatrix},$$ we have $$g_M(x,y)= q_1 y^4 +2 q_2 y^3 x + (2q_3+q_4) y^2 x^2 + 2q_5 yx^3 + q_6 x^4. $$

Theorem: Let $f$ be a binary quartic form, let $M$ be the matrix of minors of $Cat(f)$, let H$(x,y)$ be the determinant of the Hessian matrix of $f(x,y)$, then $H(x,y)=48 g_M (x,y)$.

The theorem can be proven by direct inspection since

$$M=\left( \begin{array}{ccc} \frac{1}{48} \left(8 c e-3 d^2\right) & \frac{1}{48} (12 b e-2 c d) & \frac{1}{144} \left(9 b d-4 c^2\right) \\ \frac{1}{48} (12 b e-2 c d) & \frac{1}{36} \left(36 a e-c^2\right) & \frac{1}{24} (6 a d-b c) \\ \frac{1}{144} \left(9 b d-4 c^2\right) & \frac{1}{24} (6 a d-b c) & \frac{1}{48} \left(8 a c-3 b^2\right) \\ \end{array} \right) $$ and $$ H(x,y)=x^4 (8 a c-3 b^2)-2 x^2 y^2 (-24 a e-3 b d+2 c^2)-4 x^3 y (b c-6 a d)-4 x y^3 (c d-6 b e)-y^4 (3 d^2-8 c e). $$