05/11/2021, 08:53 PM
(05/11/2021, 12:37 PM)MphLee Wrote: omg... I don't feel ready for the integrals yet but... something is moving in the back of my head. I need to reread all of your posts at least 10 times more.
The bigger obstacle is that I'll need to study what contours, Jordan Curves and singularities really are... I need to study many key parts of your papers. But I'll get there.
........
Edit 2. That's amazing... it is beginning to make some sense. When the domain is simply connected the functions the loops around a point contracts themselves to identity (just like paths homotopic to the point) but if the domain is punctured (by the singularity) they can't contract just as the circle (a punctured disk) is not topological equivalent to the disk (that is really a point) because it has an hole! It's so marvelous... the residue classes (in your theory) are class of conjugate functions. There are many doubts but the bigger now is: how you define \( {\rm Rsd}(f,\zeta;z) \)?
Edit 3. I went back to your old post in the bullet notation thread. That method really comes from Euler method as you say... It is curious that there is no known link between the infinitesimal generator f and the solutions to y'=f(y).
In your anulus example you path \rho realizing that conjugate the two paths gamma is the interval [-\delta,\delta]. What happens if you measure the superfunctions of two elements of the residue class by some invariants attached to the paths (like the length)? Maybe finding the shortest path/curve?
The class \( \text{Rsd}(f,\zeta;z) \) is very readily just defined as: (this is taken word for word from the paper)
The residual class \( \text{Rsd}(f,\zeta;z) \) of a meromorphic function \( f: \mathcal{S} \times \mathbb{C} \to \widehat{\mathbb{C}} \) at a pole \( \zeta \) is defined to be:
\(
\text{Rsd}(f,\zeta;z) \ni \int_\gamma f(s,z)\,ds\bullet z\\
\)
Where:
\( \gamma:[a,b] \to \mathcal{S} \) is a Jordan curve oriented positively.
\( \zeta \) is the only pole enclosed within \( \gamma \).
When describing the conjugate property; it's difficult to explain it over the forum; as I spend about 20 pages building up a theory, lol. But it works differently than Cauchy because everything is non-abelian. The first thing is that the residual IS NOT independent of parameterization about a singularity. But it ALMOST is. The function:
\(
F(z) = \int_\gamma f(s,z)\,ds\bullet z\\
\)
Depends on the initial and final point of \( \gamma \). So the value \( \gamma(a) = \alpha = \gamma(b) \)--\( F \) depends on \( \alpha \), and the singularities within. We can of course reparameterize with a \( \gamma(\tau) \); this is no problem; but the integral will change depending on where our starting and endpoints are.
So say I have two contours \( \gamma,\varphi \) where \( \varphi(a) = \beta = \varphi(b) \) is the start/end point of \( \varphi \) (and they both encircle the same poles). Then take an arbitrary arc \( \sigma \) (which doesn't encircle any poles) such that \( \sigma(0) = \alpha \) and \( \sigma(1) = \beta \). Then this is our conjugate function. And luckily; if you take another arc \( \sigma' \) which satisfies the same initial point/end point (and doesn't encircle any poles):
\(
\int_{\sigma} = \int_{\sigma'}\\
\)
So the conjugate function IS UNIQUE; if it's written as a compositional integral about \( f(s,z) \).
Another way to think of the Residual class is; about a singularity; a family of functions,
\(
\text{Rsd}(f,\zeta) = \{F(\alpha,z)\,|\,F(\alpha,z) : \mathcal{S}/\{\zeta} \times \mathbb{C} \to \mathbb{C}\}\\
\)
And each compositional integral about an arc \( \sigma \); if it doesn't encircle any poles; only depends on the initial point \( \alpha \) and it's endpoint \( \beta \). We can surely think of this as an arrow \( \alpha \to \beta \). Then,
\(
(\alpha \to \beta) \bullet F(\beta,z) \bullet (\beta \to \alpha) \bullet z = F(\alpha,z)\\
\)
This is another way of thinking about it. Or if you prefer,
\(
\int_\beta^\alpha f(s,z)\bullet F(\beta,z) \bullet \int_{\alpha}^\beta f(s,z)\,ds\bullet z = F(\alpha,z)\\
\)
And we can choose whatever path we want from \( \beta \) to \( \alpha \) (as long as it doesn't encircle any poles).
Just as a brief FYI; there is a lot of things from Cauchy's analysis we inherit with the compositional integral; but we don't inherit everything. And a lot of the times, when we inherit something, we lose a couple properties because everything is non-abelian. Just be sure to remember that as you read about contour integration; there are somethings we begin to lose.
But the power player we get is The Compositional Integral Theorem; which is equivalent to Cauchy's integral theorem. Which is, if \( \phi(s,z) : \mathcal{S} \times \mathbb{C} \to \mathbb{C} \) is holomorphic, and \( \mathcal{S} \) is simply connected. Then for all jordan curves \( \gamma \),
\(
\int_\gamma \phi(s,z)\,ds\bullet z = z\\
\)
Everything in my book is essentially built around this theorem.