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Worldsheet and complex analysis stuff

  1. May 9, 2007 #1

    nrqed

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    This really is a question on complex analysis but is about Polchinski's introduction to worlsdheet physics, so I am sure people here will answer this easily. I know it is a very basic question.

    Polchinski considers a field which is analytic and then says that because of this, one may write it as a Laurent expansion, for example

    [tex] L(z) = \sum_{m= -\infty}^{\infty} \frac{L_m}{z^{m+2}} [/tex]

    But I thought that a function which is analytic in some domain can be written as an expansion with positive exponents (because it is well defined at all points of the domain and is infinitely differentiable).

    But Polchinski seems to assume that the function is not analytic at z=0 from the get go, but only in some annulus. Is he making an extra assumption there or is there some theorem that says that an analytic function in some domain D can always be written as a Laurent expansion in some domain D' where there is some relation between the two domains?

    Thanks for the help.
     
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  3. May 9, 2007 #2
    In the book Complex Analysis by Brown and Churchill, 6th ed, pg 150 they present a Theorem that states:

    "Suppose that a function f is analytic throughout an annular domain
    R1 < |z-z0| <R2, and let C denote any positively oriented simple closed contour around z0 and lying that domain. Then, at each point z in the domain f(z) has the series representation:

    [tex]
    \begin{document}

    \dispSFinmath{
    \MathBegin
    \sum _{n=0}^{\infty }{a_n}{{(z-{z_0})}^n}+\sum _{n=1}^{\infty }\frac{{b_n}}{{{(z-{z_0})}^n}} \\
    \noalign{\vspace{1.03125ex}} \\
    {R_1}<|z-{z_0}|<{R_2}
    \MathEnd
    }


    \end{document}
    [/tex]

    Hence any analytic function (even ones defined at z0) may be expanded as a Laurent Series if you restrict its domain to an annulus. The book actually makes it clear on pg 151 that if f(z) is actually analytic through the disk |z-z0|<R2, then all of the coefficients bn are zero and you recover the Taylor expansion. Hence you may view the Laurent series as a generalization of Taylor Series. Polchinski's notation does allow for terms with positive exponents since his sum runs from negative infinity to positive infinity.
     
    Last edited: May 9, 2007
  4. May 9, 2007 #3

    matt grime

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    You're confusing meromorphic with holomorphic - singularities are perfectly acceptable: we're just using Cu{infinity}, the extended complex plane.
     
  5. May 9, 2007 #4

    nrqed

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    My confusion is more about analytic vs holomorphic. Some sources use the two terms as being equivalent, others say that analytic is more general than holomorphic.

    I have a textbook that says "A function is said to be analytic in adomain D if f(z) is defined and differentiable at all points of D".
    Now, a function with a singularity is not defined at that point , is it? So is this definition wrong? and what is the relation with holomorphic?

    Thanks
     
  6. May 9, 2007 #5

    matt grime

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    For me, poles are prefectly acceptable (as long as they are not essential singularities). But since you're the only person with the text book in question if front of you, you're the only person here who can say what conventions are used in it.
     
  7. May 9, 2007 #6
    The key idea is how do you define the domain D? If you exclude the singular points from your domain then your function is analytic on the modified domain. This is the reason why the domain of a Laurent expansion is restricted to an annular ring, you exclude the singular point.

    Mathworld has a pretty good explanation:

    http://mathworld.wolfram.com/AnalyticFunction.html

    Analytic is synonymous with holomorphic according to most sources. I suppose that different books may define analytic and holomorphic differently. In mathematics you can classify and define things however you want as long as it is useful, I suppose those authors found some use in defining analytic and holomorphic differently.
     
  8. May 9, 2007 #7

    nrqed

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    Thanks, this is useful. I want to make sure I understand.

    First, how do you define analyticity? (In your own words)

    Now, this said that if the function is analytic in the annulus, it may be written as the Laurent expansion. Then, you say that if it is analytic in the entire disk , then all the coefficients bn are zero. Am I getting this right?


    EDIT: Just to make it clear, a function is therefore analytic in a domain D if it has no pole in that domain, right? I guess that my confusion arises when people just say that a function is analytic, period. This means that that it is analytic except at some points. right?
     
    Last edited: May 9, 2007
  9. May 9, 2007 #8

    nrqed

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    Ok. I guess that my confusion came from the distinction between saying that a function is "analytic" (with no further specification) and saying that it is "analytic in a domain D" which, as far as I can tell, means that it does not have poles in the domain. This is th eonly way for me to make sense of several statements I have read.

    And sorry fo rthe quote. I did not realize that "is defined" or "is differentiable" was book-dependent.

    Thanks for the help
     
  10. May 9, 2007 #9

    matt grime

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    There is (apparently) some debate as to what one ought to mean by holomorphic or analytic. Seems entirely pointless, really, but troublesome nonetheless. As I say, personally, I'm happy with anything that has a Laurent expansion with only finitely many negative powers. Others may have their own views, and mine are certainly not trustworthy.
     
  11. May 9, 2007 #10
    A complex function is analytic in an open set if it has a derivative at each point in the set.

    Yes exactly right. If the function is analytic at every point in the disk |z-z0|<R2 all the coefficients bn are zero. Basically you can make the annulus into a disk.

    Simply because a function doesn't have a pole doesn't mean it is analytic, there exist functions which do not have poles yet are not analytic at any point in their domain, e.g. f(z)=|z|^2. For a function to be analytic you have prove/show that in an open set (usually an open set which is useful or of interest) the function has a derivative at every point. So when someone says f(z) is analytic they must specify the set in which the function is analytic and for every point in that set the function can't have poles the function may have poles outside the set, but within the set it can't. If they don't specify the set then f(z) analytic has no meaning.
     
  12. May 9, 2007 #11
    Holomorphic functions are analytic functions of z while antiholomorphic functions are analytic functions of the complex conjugate z*. Meromorphic functions are analytic everywhere.
     
  13. May 10, 2007 #12

    nrqed

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    Thanks again for all the help. Yes, this clarifies things quite a bit.
    I was under the incorrect impression that if one would write a Laurent expansion in an annulus centered at z=0 say, this implied that the function had to have a pole at that point. You have clarified that this was not true. This was part of my confusion since I was wondering how Polchinski knew that the energy-momentum tensor, say, had to have a pole at z=0. You have made me realize that this was not required in order to write the Laurent expansion in the first place.

    Ok, that is a very good point! Before reading your reply, I realized that in a footnote my book says that they say that a function is analytic if it is analytic in *some domain*. That made it clear to me what they meant when they kept referring to a function being analytic later on.

    Just a last thing in order to make sure I understand Polchinski's approach. First, he shows that a certain quantity (the energy-momentum tensor T for
    example, although it does not matter for my question) obeys the Cauchy-Riemann equations, [itex] \partial_{{\bar z}} T = 0 [/itex]. This tells him that T is holomorphic. Then he says that because of that, he may write it as a Laurent series (in an annulus centered at the origin).

    If I understand correctly, this is always possible as long as the annulus does not contain any pole, correct? I mean, in some sense this is purely formal because one cannot a priori know what are the bounds of the annulus (the radii) such that this expansion is possible. One needs extra information in order to define the radii of the annulus such that the expansion is allowed. Am I correct in that interpretation? I think that my confusion in reading his presentation was twofold: first I thought incorrectly that the Laurent expansion necessarily assumed that there was a pole at the origin which si not necessarily implied, as you said. Secondly, I thought that it was implicit that his Laurent expansion was valid for any z (except at the origin) and I was wondering how he knew that there would be no other poles. But I guess that this is not the case..he does not mention the domain in which his Laurent expansion is valid but that does not mean it is meant to be valid for all z.

    Thanks a lot for your input.
     
  14. May 10, 2007 #13
    The reason why Polchinski showed that T satisfies the C-R conditions is so that he can then state that T is analytic/holomorphic/differentiable at some point z0 in an epsilon neighborhood where T satisfies those conditions. It is guaranteed by a Theorem, for reference; Brown/Churchill, Complex analysis pg 51, 6ed. Actually Polchinski leaves out many other conditions necessary for differentiability, for instance, let [tex]\dispSFinmath{
    f(z)=u(x,y)+\Mvariable{iv}(x,y)
    }
    [/tex] be the function in question, it must be defined throughout some epsilon neighborhood of a point [tex]\dispSFinmath{
    {z_0}={x_0}+{{\Mvariable{iy}}_0}
    }[/tex]
    The partial derivatives of u and v with respect to x and y exist everywhere in that neighborhood and are continuous at [tex]\dispSFinmath{
    ({x_0},{y_0})
    }[/tex] these conditions will be useful.

    Now, I suppose (I may be wrong) that he wanted to make his expansion apply to as many different T functions as possible and to as many different open sets as possible. Though we don't know where the poles are we make the assumption that as long as T is a well behaved function with respect to poles we can easily find annuli or disks that exclude any pole and allow us expand T. Any crazy function T that has poles that you can't isolate you'll deal with separately. The actual radii of the disk or annuli is a technicality that is important only when you know the specific form of T. This is why he didn't specify the radius or domain of differentiability, you determine it on a case by case basis. So let us break it down into two cases:

    Case 1:
    Let us say that for some reason we wish to expand T about a point z0 that is NOT singular, T satisfies the conditions for differentiability at that point, furthermore we can find an open disk centered at z0 where T is analytic for every point in the disk, then we can simply use a Taylor Expansion valid for points within the disk. We must make some assumptions about T around point z0 namely that if any poles do exist they must be isolated (no other poles infinitely close to z0) as such if our disk happens to contain a singular point all we have to do is make the disk small enough so that it does not contain any poles.

    Case 2:
    Let us say that for some reason we wish to expand T about a point z0 that IS singular, obviously T is not differentiable at z0 BUT as long as the poles are isolated (no other poles close to z0) we can find a region around the point where T has no poles. So we exclude the pole(s) and make a little annulus small enough so that is contains no poles, then you show T satisfies all the conditions for differentiability for every point in the annulus, then you can expand T in a Laurent Series.

    How can we combine both cases into one mathematical entity, well why not consider a Taylor Series as a Laurent Series where we let the inner radius of the annulus approach 0, in fact this actually works a Laurent series turns into a Taylor series if z0 is NOT singular. Great we only need to work with one type of expansion and we can develop all of our mathematics around this expansion, which takes care of both singular and non-singular points.
     
  15. May 11, 2007 #14

    nrqed

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    Yes, this makes perfect sense. This clarifies completely all my confusion. Thank you so much for spelling this out!!

    I will be posting questions on conformal and Weyl transformations soon (I will hae to decide what subforum is the most appropriate). I hope you will have time to help me out with that as well.

    Best regards


    Patrick
     
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