How Can Abelian Integrals Be Simplified for Beginners?

[bolbteppa]

Hi guys!

Looking at the wiki page for abelian integrals I get no intuition on these scary monsters, & since I'm still not 100% ready as regards all the material in the chapters preceeding sections on abelian integrals in the reference books mentioned on that page I'd think I'd have problems trying to learn about such a topic at the moment, however after reading two pages of this paper you see Arnold talking about Abelian integrals as associating areas of algebraic oval's to lines intersecting such ovals! Do you guys know of any other reference that would go more in depth on these matters with this kind of unbelievably clear geometric perspective?

 

[jackmell]

What's not intutitive about it? We have the algebraic function, w(z), defined implicitly as:

f(z,w)=a_0(z)+a_1(z)w+\cdots+a_n(z)w^n=0

which is n-valued and then consider integrals of the type:

\int_{z_0}^z G(z,w(z)) dz

which since w(z) is n-valued, surely the integral is so depending upon which determination of w(z) the integration is taken over.

In my opinion, the best way to cultivate an intutitive understanding of this topic is to actually plot w(z), observe the sheeting of its cycles, create an abelian test integral, create a particular path over the sheeting, and then numerically integrate it over various paths over w(z) and observe how the integral changes.

 

[bolbteppa]

Could you give me an explicit example or two that you can both calculus & geometrically interpret?

Something like G(z,w(z)) already looks like it'll get so crazy it'll be completely unmanageable, unintegrable & geometrically makes no sense yet I seen that little picture Arnold drew & had a faint glimmer of hope there was a way to understand this

 

[jackmell]

How about the algebraic function
f(z,w)=z+(3w^2-w+2)=0
and the abelian integral
\int_0^1 \frac{z}{w}dz

In this case we can solve explicitly for w:
w(z)=1/6(1+\sqrt{-23-12z})
and insert it directly into the integral
\int_0^1 \frac{z}{1/6(1+\sqrt{-23-12z})}dz
keeping in mind the root is actually multivalued so the integral actually represents two integrals which we can solve explicitly for antiderivatives A_i(z) and if we are careful to evaluate A(z) along the path in an analytically-continuous fashion, we may write:
\int_0^1 \frac{z}{w}dz=A_i(z)\biggr|_0^1,\quad i=1,2

For more complicated functions, w(z), we cannot compute a simple antiderivative like above but the principle is the same: we map the integration contour over a particular determination of w(z) and insert those values of w(z) into the integrand and evaluate the integral. It's not difficult to integrate them numerically also as long as the algorithm computes correctly the desired determination of w(z) to integrate over.

And there are other ways of solving these integrals. For example, a modified version of the Residue Theorem can be used in some cases.

 

[bolbteppa]

Yeah that's exactly the kind of thing I've been looking for, thanks!

 

[jackmell]

The whole theory I believe is presented in terms of Riemann surfaces and the (single-valued) meromorphic functions over these surfaces. Like G(z,w)=\frac{z}{w}. That's a single-valued, analytic, nicely-behaved, meromorphic function over the Riemann surface of w(z) even though it's a multi-valued function because of w(z).

Me personally, the subject is a formadiable undertaking but interesting nevertheless.

 

[bolbteppa]

Thanks a lot for the help, if I get stuck I might pop back again if you don't mind.