Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a complex manifold and $Y \subset X$ a hypersurface. Let $x \in Y$ and $f$ a meromorphic function on $X$ near $x$. In Huybrecht's Complex Geometry the order of $f$ along $Y$ at $x$ is defined as $$ f = g^{ord_{Y,x}(f)} \cdot h $$ where $h \in \mathcal O_{X,x}^*$, the sheaf of germs of nonvanishing sections on $x$, and $g\in \mathcal O_{X,x}$ is irreducible and defines $Y$ near $x$.

I'm having trouble coming to grips with this definition. Consider the case $X = \mathbb C^2$ with coordinates $(z_1,z_2)$ and $Y = \{z_1 = 0\}$. Then what is $ord_{Y,(0,0)} z_2$? It seems impossible to write $z_2 = z_1^d h$ for some $h \in \mathcal O_{X,x}^*$. Further, it is stated that the order at any point on an irreducible hypersurface is independent of the point. But clearly $ord_{Y,x} z_2 = 0$ if $x \ne (0,0)$. So if the order is 0 at (0,0) as well, then this is saying that $z_2 \in \mathcal O_{X,(0,0)}^*$. What am I missing?

Thanks.

share|improve this question
    
In your example: isn't $z_2$ the uniformizer in the local ring of $Y$ at the origin? What do you mean by "defines" $Y$ near $x$: do you mean is the function giving the hypersurface (in your example $z_1$) or the function giving the coordinate on $Y$ (in your example $z_2$)? –  tkr Oct 27 '11 at 23:21
    
@tkr: i mean that around $x$, $Y$ is given as the zero locus of $g$. –  Eric O. Korman Oct 27 '11 at 23:25
2  
No wonder you're having trouble: Huybrechts's definition is incorrect. See the Edit to my answer. –  Georges Elencwajg Nov 2 '11 at 22:39

3 Answers 3

Everything becomes crystal-clear once you recall:

$$\large \mathcal O_{X,x} \text {is a UFD} $$
Then if the germ of $Y$ at $x$ is defined by the irreducible element $g\in \mathcal O_{X,x}$, you can write-as in every UFD- any non-zero $f\in \mathcal M_x=Frac(O_{X,x})$ uniquely as: $$f=g^n\Pi h^{n_h}=g^n (stuff) \quad (*)$$ where
- the $h$'s are pairwise non associated and run through the irreducibles of $\mathcal O_{X,x}$ not associated to $g$,
- $n,n_h\in \mathbb Z$ are almost all zero ( "unicity" is up to order of factors and up to invertible elements, as usual in a UFD) .

The order of $f$ along $Y$ at $x$ is then, very naturally, the exponent $n$ of $g$ in $(*)$.

In your example $z_1$ and $z_2$ are two non-asociated irreducibles so that if $Y$is defined locally by $z_1=0$, you write $f=z_2=(z^1)^0.( stuff)$ and you get that the order of vanishing of $z_2$ along $Y$ is $0$
[of course $(\text stuff)=z_2$, but that's irrelevant!]

Edit QiL's crisp and definitive comment made me want to check Huybrechts's definition.
His Definition 2.3.5.on page 78 states:

"Let $f$ be a meromorphic function in a neighbourhood of $x\in Y$. Then the order $ord_{Y,x}(f)$ of $f$ in $x$ with respect to$Y$ is given by the equality $f=g^{\text {ord}(f)}.h$ with $h\in \mathcal O^*_{X,x}$."

This is a completely wrong definition since it is impossible to find such a factorization of $f$ in general: it would imply in particular that any germ at $x$ of a holomorphic function on $X$ would have a zero set coinciding with $Y$ at $x$, an obviously preposterous conclusion.

share|improve this answer
    
My issue is how can we guarantee that $(stuff)$ is in $\mathcal O_{X,x}^*$, the germs of nonzero functions at $x$. –  Eric O. Korman Oct 29 '11 at 1:25
    
@Eric: in general, you can't. –  user18119 Nov 2 '11 at 19:56

I think there is actually a problem with this definition. The best one can do is to have $h$ meromorphic and belonging to $\mathcal O_{X,y}^*$ for some $y\in Y$ near $x$.

The idea is the order of $f$ at $(Y,x)$ is an integer $r$ such that the divisor of $f/g^r$ is not supported in $Y$ in a neighborhood of $x$.

share|improve this answer

The order function measures the extent to which a meromorphic function $f$ on a complex manifold $X$ vanishes along a codimension one subvariety $Y$. It appears that your notation for that is $ord_{X,Y}(f)$.

So in my mind, there are two sensible interpretations of your example. The less likely is that you want $ord_{X,Y}(z_2)$, which is $0$ because $z_2$ is holomorphic on $Y$ and does not vanish identically. More likely, you want $ord_{Y,(0,0)}(z_2)$, which is $1$ because $Y \cong \mathbb{C}$, with coordinate $z_2$, and of course, $z_2$ vanishes to order $1$. Said more verbosely, $z_2 = g^1 h$, where $g = z_2$ generates the maximal ideal of germs of functions at $(0,0)$ that vanish there (more colloquially, "$z_2$ defines $(0,0)$ in $Y = 0 \times \mathbb{C}$") and $h = 1$ is an invertible germ (i.e. $h(0,0) \neq 0$).

Please note that if $x \in X$, where $X$ is a complex manifold, then the defintion $ord_{X,x}(f)$ only makes sense if $\dim X = 1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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