3
$\begingroup$

It is known that any meromorphic function $f$ on projective plane curve $C=Z(P)\subset\mathbb{P}^{2}$ is rational, i.e. $$ f([x:y:z])=\frac{Q(x,y,z)}{R(x,y,z)} $$ for some homogeneous polynomial of same degree $m$ and $P\nmid Q$. If $p$ is a point on $C$, then order of $f$ at $p$ is given by $$ \mathrm{ord}_{p}(f)=I_{p}(P, Q)-I_{p}(P, R)=I_{p}(C, Z(Q))-I_{p}(C, Z(R)). $$ where $I_{p}(C, D)$ is local multiplicity of curves $C$ and $D$ at $p$. Intuitively, I think this statement is true but I can't prove this. (I really hope this is true.) How to prove this? Or... are there any counterexamples? Thanks in advance.

$\endgroup$
  • $\begingroup$ Some questions: Do you assume your curve to be smooth? How do you define $\operatorname{ord}_p$ and $I_p(C,D)$? $\endgroup$ – Jesko Hüttenhain Dec 16 '16 at 13:11
  • $\begingroup$ @JeskoHüttenhain Yes, $C$ is a smooth curve. I use definition in Kirwan's Complex Algebraic curve book : For local multiplicity $I_{p}(C, D)$, I use definition in here (en.wikipedia.org/wiki/Intersection_number) and $\mathrm{ord}_{p}(f)=m$ iff there is a chart $(\phi, U)$ near $p$ s.t. $f\circ\phi^{-1}:\phi(U)\to C$ has order $m$ zero (or order $-m$ pole) near $\phi(p)$. $\endgroup$ – Seewoo Lee Dec 16 '16 at 14:35
  • $\begingroup$ @JeskoHüttenhain There is another definition of local multiplicity : if $p=[a:b:c]$, the local multiplicity is the largest integer $k$ such that $(bx-cy)^{k}$ divides the resultant $\mathcal{R}_{P, Q}(x, y)$. $\endgroup$ – Seewoo Lee Dec 16 '16 at 18:57
1
$\begingroup$

I found a proof with my friend, so I'll introduce our proof.

We will follow the proof of the theorem 3.22 in Kirwan's "Complex Algebraic curves". Assume $p=[0:0:1]$ and $\frac{\partial P}{\partial x}(0, 0, 1)\neq 0$. By implicit function thoerem for complex polynomials, there is a holomorphic function $\lambda_{1}:U\to V$ where $U$ and $V$ are open neighborhoods of $0$ in $\mathbb{C}$ s.t. $\lambda_{1}(0)=0$ and for $(x, y)\in U\times V$, $P(x, y, 1)=0$ iff $x=\lambda_{1}(y)$. Moreover, $$P(x, y, 1)=(x-\lambda_{1}(y))l(x, y)$$ where $l(x, y)$ is a polynomial in $x$ whose coefficients are holomorphic functions on $y$.
If we assume $P(1, 0, 0)=1$, then $$l(x, y)=\prod_{i=2}^{n}(x-\lambda_{i}(y))$$ where $\lambda_{1}(y), \dots, \lambda_{n}(y)$ are the roots of $P(x, y, 1)$ regarded as a polynomial in $x$ with $y$ fixed. Similarly if $U$ and $V$ are chosen small enough there is a holomorphic function $\mu_{1}:U\to V$ s.t. $\mu_{1}(0)=0$ and we can write \begin{align*} Q(x, y, 1)=(x-\mu_{1}(y))m(x, y) \\ R(x, y, 1)=(x-\eta_{1}(y))k(x, y) \end{align*} where \begin{align*} m(x, y)=\prod_{i=2}^{m}(x-\mu_{i}(y)) \\ k(x, y)=\prod_{i=2}^{m}(x-\eta_{i}(y)) \end{align*} is a polynomial in $x$ whose coefficients are holomorphic functions of $y$. Then the tangent lines to $C$, $D=Z(Q)$ and $E=Z(R)$ at $p=[0:0:1]$ are defined by the equations $$ x=\lambda_{1}'(0)y, \quad x=\mu_{1}'(0)y, \quad x=\eta_{1}'(0)y $$ If $y\in U$ then \begin{align*} \mathcal{R}_{P, Q}(y, 1)=(\mu_{1}(y)-\lambda_{1}(y))S(y) \\ \mathcal{R}_{P, R}(y, 1)=(\eta_{1}(y)-\lambda_{1}(y))T(y) \end{align*} where \begin{align*} S(y)=\prod_{(i, j)\neq (1, 1)}(\mu_{i}(y)-\lambda_{i}(y)) \\ T(y)=\prod_{(i, j)\neq (1, 1)}(\eta_{i}(y)-\lambda_{i}(y)) \end{align*} (Here $\mathcal{R}_{P, Q}$ is resultant of two polynomials $P$ and $Q$.) Since $\lambda_{1}(0)=0=\mu_{1}(0)$ we obtain on differentiating above equation that $$ \frac{\partial \mathcal{R}_{P, Q}}{\partial y}(0, 1)=(\mu_{1}'(0)-\lambda_{1}'(0))S(0) $$ So $I_{p}(C, D)>1$ iff $\lambda_{1}'(0)-\mu_{1}'(0)$. Similarly, we can show that $I_{p}(C, D)=e$ iff $$\lambda_{1}'(0)-\mu_{1}'(0)=\cdots=\lambda_{1}^{(e-1)}(0)-\mu_{1}^{(e-1)}(0)=0, \quad \lambda_{1}^{(e)}(0)-\mu_{1}^{(e)}(0)\neq 0$$ inductively by differentiating the equation repeatedly. Same thing holds for $I_{p}(C, E)$ and $\lambda_{1}^{(j)}(0)-\eta_{1}^{(j)}(0)$. Since local chart of $P$ near $p$ is given by $\phi:[\lambda_{1}(w):w:1]\mapsto w$, the function $f$ can be written as $$ f\circ\phi^{-1}(w)=\frac{Q(\lambda_{1}(w), w, 1)}{R(\lambda_{1}(w), w, 1)}=\frac{(\lambda_{1}(w)-\mu_{1}(w))m(\lambda_{1}(w), w)}{(\lambda_{1}(w)-\eta_{1}(w))k(\lambda_{1}(w), w)} $$ By previous observation, $ord_{0}(\lambda_{1}(w)-\mu_{1}(w))=I_{p}(C, D)$ and $ord_{0}(\lambda_{1}(w)-\eta_{1}(w))=I_{p}(C, E)$. By differentiating the equation $Q(x, y, 1)=(x-\mu_{1}(y))m(x, y)$ w.r.t. $x$, we have $$ \frac{\partial Q}{\partial x}(\lambda_{1}(w), w, 1)=m(\lambda_{1}(w), w)+(\lambda_{1}(w)-\mu_{1}(w))\frac{\partial m}{\partial x}(\lambda_{1}(w), w) $$ and $m(0, 0)=\frac{\partial Q}{\partial x}(0, 0, 1)\neq 0$, so $m(\lambda_{1}(w), w)$ is nonvanishing near $w=0$. Similarly, $k(\lambda_{1}(w), w)\neq 0$ near $w=0$, so $$ ord_{p}f=ord_{0}(f\circ\phi^{-1}) =I_{p}(C, D)-I_{p}(C, E). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.