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For any homogeneous polynomial $f \in \mathbb R [x,y]$, define the homogeneous polynomial

$$H(f) := \partial_yf^2\partial_x\partial_xf-2\partial_xf\partial_yf\;\partial_x\partial_yf+\partial_xf^2\partial_y\partial_yf$$

which is the Hessian of $f$ applied to the tangent of its level set. I came across it while studying the asymptotics of certain complete open surfaces.

From numerical tests, it appears that any homogeneous polynomial $f(x,y)$ factors as follows

$$H(f) = G(x,y) \, f$$

for some homogeneous polynomial $G$.

Question: Is there a simple proof of, or a counterexample to, this factorization?

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Let $f$ be a homogeneous polynomial of degree $n$. Define: $$Q:= f_{xx} f_y^2 - 2 f_{xy} f_x f_y + f_{yy} f_x^2.$$

Consider $(n-1)^2 Q$ in the following way $$(n-1)^2 Q = f_{xx} \Big((n-1)^2 f_y^2\Big) - 2 f_{xy} \Big((n-1)f_x\Big) \Big((n-1)f_y\Big) + f_{yy} \Big((n-1)^2 f_x^2\Big)$$

By Euler's Lemma: $$(n-1) f_y = x f_{xy} + yf_{yy}, \quad (n-1) f_x = x f_{xx} + yf_{xy}$$ So, replacing these expressions and simplifying the products we obtain

\begin{align*} (n-1)^2 Q &= -x^2 f_{xx}f_{xy}^2 -y^2 f_{yy}f_{xy}^2 -2 xy f_{xy}^3 + f_{xx} f_{yy}\Big(y^2 f_{yy} +2xy f_{xy} + x^2 f_{xx} \Big) \\ & = - f_{xy}^2 \Big( x^2 f_{xx} +2xy f_{xy} + y^2 f_{yy} \Big) + n(n-1) f \Big(f_{xx} f_{yy}\Big)\\ & = - n(n-1) f f_{xy}^2 + n(n-1) f (f_{xx} f_{yy})\\ & = n(n-1) f \Big(f_{xx} f_{yy} - f_{xy}^2\Big) \\ & = n(n-1) f \Big( \mbox{Hess} f \Big) \end{align*}

Finally, $$ Q = \frac{n}{n-1} f \Big( \mbox{Hess} f \Big) .$$

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  • $\begingroup$ Very nice answer! $\endgroup$ – Zach Teitler Jun 20 '18 at 4:25
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Assume $f(x,y)=x^{n} P(\frac{y}{x})$ for some $n\geq 0$, and $\mathrm{deg}\, P(x)\leq n$. Then $$ f_{x}^{2}f_{xx} + f_{y}^{2}f_{yy}-2f_{x}f_{y}f_{xy} = x^{3n-4}nP(t)\left(-P'(t)^{2}(n-1)+n P''(t)P(t) \right) $$ where $t=y/x$. Then $G(x,y)=x^{2n-4} n\left(-P'(t)^{2}(n-1)+n P''(t)P(t) \right)$ works provided that $n \geq 2$ because one can easily check that if $\mathrm{deg}\,P\leq n$ then $\mathrm{deg}(-P'(t)^{2}(n-1)+n P''(t)P(t))\leq 2n-4$. Besides notice that the assertion trivially holds for $n=0$ and $n=1$ homogeneous polynomials $f(x,y)$, we can just take $G=0$

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