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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.

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Anirbit, Your accept rate is a bit low (considering you've asked 40 questions other than this). Please consider accepting answers to other questions if you found them helpful. Thanks. –  Srivatsan Sep 26 '11 at 20:43
@Srivatsan Since Anirbit has accepted some answers, it seems pretty clear that he doesn't need pointing out that this feature exists. When and how to use it is entirely up to him. If he feels like better answers to his questions might appear some day, there is no pressure on him to accept answers that he finds deficient in some ways. You can feel free not to answer his questions if you fear that you will not be duly rewarded. –  Alex B. Sep 27 '11 at 1:03
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