Proof. We clearly can assume that the degree of $H$ is greater than $1$ and that the higher homogeneous term of $H$ has the form

\[ r(x,y) (a x + b y)^{\tau}, \]

where $\tau \geq 1$ is an integer and $r(x,\,y)$ is a polynomial that does not have $a x + b y$ as a factor.

Clearly if $a b \neq 0$ and $\tau \geq 2$, then the higher homogeneous terms of $P$ and $Q$ have both the factor $a x + b y$, and we are done.

So it remains to consider the following three cases concerning the higher homogeneous term of $H$: (i) it has the form $r(x,\,y)(a x + b y)$ with $a b \neq 0$, i.e., $\tau = 1$; or (ii) it has the form $r(x,\,y) y^{\tau }$, i.e., $a = 0$; or (iii) it has the form $r(x,\,y) x^{\tau }$, i.e., $b = 0$. By changing $x$ and $y$ we do not need to consider case (iii) (observe that with a change like that, a $(s_1,\, s_2)$-weight-homogeneous polynomial is carried to a $(s_2,\, s_1)$-weight-homogeneous polynomial). Also, with a linear change of variable we can transform (i) into (ii) and consider just the later case (because the degree of $H$ is greater than $1$). Our conclusion will show that case (ii) with $\tau = 1$ (and so case (i)) cannot happen.

So we assume that case (ii) is in force. Letting $s = (s_1,\, s_2)$ be a given weight-exponent, we denote by $m$ and $n$ the weight degrees with respect to $s$ of $P$ and $Q$, respectively. In the sequel, we shall use notation on the Poincaré compactification of $X$. Observe that $q$ is the origin of the local chart $U_1$, that we will treat with the variables $(u,\,v)$, with the relation between $(x,\,y)$ and $(u,\,v)$ given by $(u,\,v) = (y/x,\,1/x)$. The equator of the Poincaré sphere, i.e., the infinite of $ {\mathbb {R}}^{2}$ is mapped in the straight line $v = 0$. Let $r_1$ and $r_2$ be the two separatrices of a hyperbolic sector $h$ of $(0,\,0)$ in $U_1$. Without loss of generality, we assume that the interior of $h$ is contained in the region $v > 0$. Suppose that $r_1$ is not contained in the infinite, i.e. in the straight line $v = 0$. We claim that $r_2$ is not contained in the infinite and that $r_1$ and $r_2$ have the same tangent line at $(0,\,0)$. Indeed, assume on the contrary that there exists a straight half-line $u = \lambda v$, with $v > 0$, between $r_1$ and $r_2$. Each obit of $X$ is contained in a level set of $H(x,\,y) = c$ of $H$. So, by letting

(3)\begin{equation} \tilde H(u,v) = v^{d+1} H\bigg(\frac{1}{v}, \frac{u}{v} \bigg) \end{equation}

and $\tilde G(u,\,v) = \tilde H(u,\,v)/v^{d+1}$, points $(u,\,v)$ of the compactified orbit will satisfy $\tilde G(u,\,v) = c$. Let $c$ be the value of $\tilde G$ in $r_1$. Since $h$ is an hyperbolic sector, each sequence $\{w_n\}$ in the interior of $h$ such that $\lim _{n \to \infty } w_n = (0,\,0)$ will satisfy $\lim _{n \to \infty }\tilde G(w_n) = c$. So

\[ \lim_{v \to 0} \tilde G( \lambda v, v) = c. \]

Then writing $\tilde H = \sum \nolimits _{i = 0}^{d+1} \tilde H_i$, with $\tilde H_i$ being the homogeneous part of degree $i$ of $\tilde H$, we get $\tilde H_0(\lambda,\,1) = \cdots = \tilde H_d(\lambda,\, 1) = 0$ and $\tilde H_{d+1}(\lambda,\, 1) = c$. Hence

\[ \tilde G(\lambda v, v) = \frac{\tilde H_{d+1}(\lambda v, v)}{v^{d+1}} = c, \]

meaning that the straight half-line $u = \lambda v$ is invariant by the flow, a contradiction. This proves the claim.

Clearly $\tilde G(u,\,v)$ have the same value $c$ in $r_1$ and $r_2$, by the continuity of $\tilde G$ in $v > 0$. Let $u = \lambda v$, $v > 0$, the common tangent of $r_1$ and $r_2$ at $(0,\,0)$. This line is contained in the tangent cone of the algebraic variety $\tilde H(u,\,v) - c v^{d+1} = 0$, with multiplicity at least two, i.e.,

\[ \tilde H(u,v) - c v^{d + 1} = \sum_{i = k}^{d+1}\tilde{ \tilde H}_i(u,v), \]

with $k \geq 2$ and $\tilde {\tilde H}_{k}(u,\,v) = (u - \lambda v)^{2} R(u,\,v)$, where $R(u,\,v)$ is a homogeneous polynomial of degree $k - 2$, and $\tilde { \tilde H}_i$ is the homogeneous part of degree $i$ of $\tilde H(u,\,v) - c v^{d+1}$. Therefore, from (3), it follows that

\[ H(x,y) = \frac{\tilde H(u,v)}{v^{d+1}} = c + \sum_{i = k}^{d+1} x^{d+1-i} \tilde{\tilde H}_i(y,1). \]

Note that if $k = d+1$ then $H(x,\,y) = \tilde H_{d+1}(y)$ which is not possible because then $Q \equiv 0$. So, $k < d+1$, and $Q$ contains the term $(d + 1 - k) x^{d-k} \tilde {\tilde H}_k(y,\,1)$. Since $\tilde H_k=(y-\lambda )^{2} \tilde R(y,\,1)$ we get that $n$, which is the $s$-weight degree of $Q$, satisfies

(4)\begin{equation} n \ge (d-k) s_1 +2 s_2. \end{equation}

The $s$-weight-homogeneous part of weight degree $n$ of $Q$ writes

\[ Q_n = \sum_{\matrix{{i,j} \cr {i s_1 + j s_2 = n} \cr }} a_{i j} x^{i} y^{j}. \]

Since the maximum exponent of $x$ in $H$ is $d + 1 - k$, and so the maximum possible exponent of $x$ in $Q_n$ is $d - k$, it follows that if $a_{i j} \neq 0$ in the above sum, then $(d-k) s_1 + j s_2 \geq n \geq (d-k) s_1 + 2 s_2$, from (4), forcing that $j$ is at least $2$. This means that

\[ Q_n = y^{2} T(x,y), \]

where $T(x,\,y)$ is a suitable $s$-weight-homogeneous polynomial of weight degree $n - 2 s_2$. Since $Q_n = \partial H_{n + s_1}/\partial x$, where here $H_{n+s_1}$ means the $s$-weight-homogeneous term of weight degree $n + s_1$ of $H$ (recall that if $H$ has $s$-weight degree $\ell$ then $\partial H/\partial x$, if not zero, has $s$-weight degree $\ell - s_1$), it thus follows that

(5)\begin{equation} H_{n+ s_1} = y^{2} \int T(x,y) + G(y) \end{equation}

for some polynomial $G(y)$ which must be a factor of $y^{2}$, otherwise $H_{n + s_1}$ is not weight-homogeneous.

Now the higher $s$-weight-homogeneous term, $P_m$, of $P = -H_y$ comes from $H_{m + s_ 2}$. Clearly $m + s_2 \geq n + s_1$ by (5). In case $m + s_2 = n + s_1$, then $P_m$ has a factor $y$. On the other hand, if $m + s_2 > n + s_1$, it follows that $H_{m + s_2} = s y^{j}$, with $s_2 j = m + s_2$, otherwise the higher $s$-weight-homogeneous term of $Q$ is not $Q_n$ because $m + s_2 - s_1 > n$. In particular, $P_m$ has a factor $y$.

Observe that our proof shows in particular that the higher $s$-weight-homogeneous term of $H$ has the factor $(a x + b y)^{2}$ (homogeneous if $a b \neq 0$). In particular, case (i) of the beginning of the proof is not possible.

Proof. From Proposition 4, it follows that $q$ have no non-degenerate hyperbolic sectors. Thus, the number of hyperbolic sectors of $q$ is $h \leq 2$. By the index formula we conclude that the index of $q$ is greater than or equal to the number of elliptic sectors of $q$, and so greater than or equal to zero. If the index is zero, it thus clearly follows that there are no elliptic sectors and also that $q$ must have two degenerate hyperbolic sectors.

Proof. We let $m$ be the weight degree of $f$ and $g$ and write $f = f_0 + \cdots + f_m$ and $g = g_0 + \cdots + g_m$ the weight decomposition of $f$ and $g$. We first observe that $(f_m^{2} + g_m^{2})_x \not \equiv 0$ and $(f_m^{2} + g_m^{2})_y \not \equiv 0$, because if $(f_m^{2} +g_m^{2})_x \equiv 0$, for instance, then (we are dealing with polynomials) $f_m = a_{0\ell } y^{\ell }$ and $g_m = b_{0\ell } y^{\ell }$, with $\ell s_2 = m$ and $a_{0\ell },\, b_{0\ell } \in {\mathbb {R}}$, a contradiction.

Thus, the higher $s$-weigh-homogeneous parts of $f f_x + g g_x$ and $f f_y + g g_y$ are, respectively, $(f_m^{2} + g_m^{2})_x/2$ and $(f_m^{2} + g_m^{2})_y/2$. If there is a linear factor dividing the last polynomials, it will also divide

\[ s_1 x \frac{\partial \left(f_m^{2} + g_m^{2}\right)}{\partial x} + s_2 y \frac{\partial \left(f_m^{2} + g_m^{2} \right)}{\partial y} = m \left(f_m^{2} + g_m^{2}\right), \]

and so this factor will be common to $f_m$ and $g_m$, a contradiction.