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Consider a Riemann surface $X$ given by function $F(z,w) = w^2 - z^2 - a^2$, $a \neq 0$: $$ X = \left\{ (z,w) \in \mathbb{C}^2 \colon F(z,w) = 0 \right\}. $$ To find its points at infinity we first compactify $X$. We can write $$ F \left( \frac{\xi}{\zeta}, \frac{\eta}{\zeta} \right) = \frac{\eta^2 - \xi^2 - a^2 \zeta^2}{\zeta ^2} = \frac{ Q(\xi,\eta,\zeta)}{\zeta^2}. $$ Then compactification $\overline{X}$ of surface $X$ is given by homogenous polynomial $Q$: $$ \overline{X} = \left\{ (\xi : \eta : \zeta ) \in \mathbb{C}P^2 : Q(\xi,\eta,\zeta)=0 \right\} $$ Points at infinity are such points of $\overline{X}$ that $\zeta = 0$. So if $\zeta = 0$ we have $\eta^2 = \xi^2$ from what we can derive two points at infinity: $(1:1:0)$ and $(1:-1:0)$.

I try to use the same method to find points at infinity of Riemann surface $$ Y = \left\{ (z,w) \in \mathbb{C}^2 \colon w^2 - z^4 + 1 = 0 \right\}. $$ Its compactification is a surface $$ \overline{Y} = \left\{ (\xi:\eta:\zeta) \in \mathbb{C}P^2 \colon \eta^2\zeta^2 - \xi^4 + \zeta^4 = 0 \right\}. $$ In order to find points at infinity we find solutions of system $\eta^2 \zeta^2 - \xi^4 + \zeta^4 = 0$, $\zeta = 0$. It follows that $\xi = 0$ and the unique point at infinity is $(0:1:0)$ but it is a contradiction with theory: surface $Y$ has two points at infinity. Help me please, where is my mistake?

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You have to assume $a\neq 0$. –  Georges Elencwajg Sep 22 '12 at 9:50
    
@GeorgesElencwajg of course, thank you –  Nimza Sep 22 '12 at 9:54
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First the bad news: the theory says that you have $4$ points at infinity, since your curve has degree $4$, so that the situation is twice worse than you think !
But the good news is that there are indeed four points at infinity, if you take the multiplicities into account!

More precisely, in the open neighbourhood $\eta \neq 0$ of $P_\infty=[0:1:0]$ take as coordinates $x=\xi/\eta, y=\zeta/\eta$ so that $P_\infty$ has coordinates $x=y=0$.
The line at infinity is $y=0$ and the curve has equation $y^2-x^4+y^4=0$ in our coordinates.
Their intersection is given by the ideal $\langle y, y^2-x^4+y^4\rangle=\langle y, x^4\rangle \subset \mathbb C [x,y]$ and the multiplicity of the intersection is $\text{dim} _\mathbb C \frac {\mathbb C [x,y]}{\langle y, x^4\rangle}=4$ : all is well that ends well!

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That's right, thank you. But I would like to know where is my mistake. I used method given in Dubrobin's book on Riemann surfaces and nonlinear PDEs (and in one more book). –  Nimza Sep 22 '12 at 15:06
    
Dear Nimza, it seems difficult to tell where you made a mistake if you write absolutely nothing about what you did... –  Georges Elencwajg Sep 22 '12 at 17:19
    
I found a homogenous polynomial $Q(\xi,\eta,\zeta)$ such that $\overline{Y}$ is given by equation $Q(\xi,\eta,\zeta) = 0$. Dubrovin writes that points at infinity are given then by $\zeta = 0$, $Q(\xi,\eta,0) = 0$. –  Nimza Sep 22 '12 at 17:30
    
Dear Nimza, yes that is correct: there is only one point at infinity set-theoretically, but intersection theory is more refined and attributes a multiplicity to that point. The calculation in the answer shows that this multiplicity is 4. Also, my advice is to forget about non-linear PDE's in this context... –  Georges Elencwajg Sep 22 '12 at 17:42
    
It is excellent, I understanded! Great thanks, Georges. I would like to know, if set-theoretically it has one point at infinity, intersection-theoretically there are 4 points at infinity then what does mean author, saying that this surface and any surface of type $w^2 = P_{2m}(z)$ (where $P_{2m}$ is a polynomial of degree $2m$, $m>1$) has two points at infinity? Is it possible to draw a curve around one point at infinity but not around the others in this case? –  Nimza Sep 22 '12 at 17:55
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