I am a bit confused about some of the things that Griffiths says on page 62 of his book, Introduction to Algebraic Curves. I am not sure how I can reproduce the text here. I can see that GoogleBooks does have those pages on display.

  • Firstly I felt that the points $(x,y_\nu(x))$ are points on the algebraic curve $f$ which can have a non-trivial topology but from the way the line $\Lambda$ or the set $\Omega$ is talked about it seems that one is working on $\mathbb{C}\times \mathbb{C}$. What is the right picture? Are these local analytic elements on the generically non-trivial Riemann surface or are they on just $\mathbb{C}\times \mathbb{C}$?

Now these function elements $y_\nu(x)$ were obtained by using the implicit function theorem locally around each of the roots of an algebraic curve $f(x,y)=0$ This could be done only on the complement of the non-singular points. Now one removes a semi-infinite line from the plane which passes through the finite number of singular points. Then what is left is still simply connected.

  • Now to extend these local analytic elements to the complement of the line that has been removed, does one necessarily need to use the Riemannn monodromy theorem or isn't it possible to simply argue that from the property of analytic continuation? (..that if two analytic functions agree on a "large" enough open set then they are the same..)

  • The explicit need for a path and continuation along it comes only when one introduces the equivalence relation whereby one calls two of the $y_\nu$s to be equivalent if there exists a path in the full space (without deleting any of the points unlike at first) along which analytic continuation will convert one into the other...this again doesn't seem to need the monodromy theorem...Am I missing something?

  • What exactly is this "heredity property" that is being invoked to say that if $f$, a second degree homogeneous polynomial in two complex variables evaluated on the point $(x,y_\nu(x)$ went to zero then it should evaluate to $0$ on any extension of the function $y_\nu(x)$.

  • I guess that given any set of $n$ locally holomorphic functions $y_i(x)$ one can remove a semi-infinite line in the plane and repeat the above argument with some specific choice of path that intersects the line.

Hence any line segment and a path intersecting it induces a permutation of the $y_i$s?

  • Lastly I want to know if that entire argument can be made on $\mathbb{C}$ or is this entire thing something typical over $\mathbb{C}\times \mathbb{C}$ ? May be by replacing the holomorphic functions by power-series expandable real functions...they too will satisfy the analytic continuation property like the holomorphics..I am not clear about what happens to the "heredity" property.

  • Or more generally is there an extension of this argument/proof/technique for $\mathbb{C}^n$ or $\mathbb{R}^n$ (..looking at the zero-sets of homogeneous polynomials on appropriate number of variables..)...possibly by restricting the class of functions appropriately.

I would like to know if or what is the essential crux or main idea behind this technique of proof.


The general case

  • Let $X$ be a compact Riemann surface and $f\in \mathcal{M}_X[t]$, say $$f(x,t)=t^d+t^{d-1}f_{d-1}(x)+\cdots+f_0(x)$$ If $x_0$ avoids a finite bad set $\Delta\subseteq X$, $f(x_0,-)$ will have $n$ zeroes; $y_1,...,y_n$. The implicit function theorem says that there are then functions $$Y_i:U_i\longrightarrow \mathbb{C}$$ on open sets $U_i$ containing $x_0$ with $Y_i(x_0)=y_i$. Then for all $x$ in $U=\bigcap U_i$, $$Y_1(x), \ . \ . \ . \ , Y_n(x)$$ are the zeroes of $f(x,-)$.
  • This shows that for small enough open set $U\subseteq X\setminus \Delta$ , $$\mathcal{S}(U)=\{\text{holomorphic solutions }s:U\to \mathbb{C}\text{ to }0=f(-,s(-)):U\to \mathbb{C}\}$$ has size $n$. This can only decrease for larger $U$.
  • If $U,V$ two open sets on which there are $n$ solutions, and $U\cap V$ is connected, then $\mathcal{S}(U\cup V)$ has size $n$ too. If $u_1,...,u_n$ and $v_1,...,v_n$ are solutions on $U$ and $V$, then by connectedness, for each $u_i$ there is a $v_{\tau(i)}$ equal to it on $U\cap V$. The identity principle (uniqueness of analytic continuation) guarantees that this is a bijection.

What it means here

In our case, $X=\mathbb{CP}^1$ and the $f_i$ are polynomials in $x$. If $\Omega\subseteq \mathbb{CP}^1\setminus \Delta$ is simply connected and open, then one can write $$\Omega=B_1\cup B_1\cup \cdots$$ as the union of balls, with $B_k\cap (B_{k-1}\cup \cdots \cup B_1)$ connected, and each $\mathcal{S}(B_k)$ having size $n$.


We have proven that for each polynomial $f\in \mathbb{C}[x,y]$ and open simply connected set $\Omega \subseteq \mathbb{C}$ avoiding the critical values of $x$, there are $n$ functions ($n$ is the $x$-degree of $f$) $$s_1,...,s_n:\Omega\longrightarrow \mathbb{C}$$ with $f(-,s_i(-))=0$ as functions $\Omega \to \mathbb{C}$. The proof only uses the implicit function theorem and the uniqueness of analytic continuation.

$$\text{}$$ $$\text{}$$

Hereditary property

I think the ''hereditary property" means that if $$s:U\longrightarrow \mathbb{C}$$ is a solution to $f(-,s(-))=0$ on open set $U$, and $\widetilde{s}:\widetilde{U}\to \mathbb{C}$ is a holomorphic extension of $s$ to a bigger open set $\widetilde{U}\supseteq U$ then $$f(-,\widetilde{s}(-))=0$$ on $\widetilde{U}$. This is easy: the holomorphic function $$f(-,\widetilde{s}(-)):\widetilde{U}\longrightarrow \mathbb{C}$$ which is zero on $U$, hence on $\widetilde{U}$, by the identity principle (uniqueness of analytic continuation). In particular, I would not expect this sort of thing to work in $\mathbb{R}^n$, since there are counterexamples to the identity principle.

$$\text{}$$ $$\text{}$$

Why thinking about $\mathcal{S}$ is useful

By picturing it, the sheaf of solutions $\mathcal{S}(U)$ should give a degree $n$ cover $$Y'\longrightarrow X\setminus\Delta$$ since on small opens it is the $n$-point set. To do this formally, topologise $$Y'=\bigcup_{x\in X\setminus \Delta} \mathcal{S}_x$$ by declaring basic opens to be $(U,s_i)$ for solutions $s_i$ on small open sets $U$.

Lifting the Riemann surface structure to make $Y'\to X\setminus \Delta$ an unbranched cover of Riemann surfaces. We can glue over $\Delta$ to get a degree $n$ branched cover $$Y\longrightarrow X$$ Call $Y$ the ''the Riemann surface associated to $f$''; this gives one direction of the equivalence $$\{\text{degree }n\text{ extensions of }\mathcal{M}_X\}\longleftrightarrow \{\text{degree }n \text{ branched covers of }X\} $$ which sends $$\text{splitting field of }f(t)\in \mathcal{M}_X[t] \longrightarrow Y$$


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.