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I Complex Analysis holomorphic condition

  1. Jun 16, 2016 #1
    I understood the holomorphic condition this way.

    For a vector field

    F(x1, x2 . . ., xm) = <y1(x1, x2, x3 . . . , xm), y2(x1, x2, x3 . . . , xm), y3(x1, x2, x3 . . . , xm) . . . ,yn(x1, x2, x3 . . . , xm)>

    In a real analysis, its derivative is expressed as a Jacobian matrix because each field component would have a different incremental because the components are independent of each other.


    However, complex differentiation is just like the case of single real variable differentiation but defined on the complex number set, which imposes a stronger restriction. And this can be generalizd to quaternion, octanion and so on.


    Is this right?
     

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  3. Jun 16, 2016 #2

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    That sounds right. Because division of complex numbers is always defined, it is possible to mimic the definition of the derivative of a function on R. The definition of the derivative of an analytic function f(z) at z0 insists that the limit is a single complex number, a= limz->z0 (f(z)-f(z0))/(z-z0) = f'(z0), that is independent of how z approaches z0.
     
  4. Jun 18, 2016 #3

    Thank you. I'd really appreciate if you could help me out to figure out this last question.
    I am little confused of this relation

    J_F(x_0,y_0)*(H) = (du/dx-du/dy*i)(h1+h2i) = f'(z_0)h

    because J_F*(H) at here is actually T:R^2->R^2, Real to Real while (du/dx-du/dy*i)(h1+h2i) is in complex domain.

    I get that they are one to one but are they really EQUAL? because they are one to one?

    Thank you.
     

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  5. Jun 18, 2016 #4

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    After that equation, it says "where we have identified a complex number with the pair of real and imaginary parts". In that context, I guess it is Ok to say they are equal. It certainly is a 1-1 mapping.
     
  6. Jun 18, 2016 #5

    Thanks!
     
  7. Jun 23, 2016 #6

    Stephen Tashi

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    I'm curious where you've found a definition of "holomorphic" that applies to vector valued functions. The usual statement of the definition of "holomorphic" is for a function whose value is a single complex number.

    Are you defining a "holomorphic" vector valued function to be a function, each of whose component functions is a holomorphic function ?

    What kind of mathematical structure do you have in mind when you say
    ?

    What structure would you use for something that has higher dimension than an octonian ?
     
  8. Jun 23, 2016 #7

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    I think he is referring to geometric algebra, which extends the concept of analytic functions to higher dimensions and includes the (modified) quaternions, octonions(?), and higher dimensions. I don't know much about it, but it is something I have been trying to understand.
     
  9. Jun 23, 2016 #8




    I was trying to understand the definition of holomorphicity, and since I am an engineering student my approach and terminology may not make sense or may not be rigorous enough at all. I am learning complex analysis for my own interest. First, I apologize if I said something weird and wrong.



    According to the text

    1.

    J_F(x_0,y_0)*(H) = (du/dx-du/dy*i)(h1+h2i) = f'(z_0)h
    which connects Real Analysis (vector valued function) and Complex Analysis (complex function) in the sense that they are one to one.



    2.

    "And this can be generalized to quaternion, octanion and so on."
    This was my question. I was curious that if it is possible because quaternion is generalized complex number. if quaternion, octanion and higher order sets satisfy well-defined binary operations addition, multiplication and division, we can define derivative on the set.

    If such derivative has one single valued derivative at a point regardless of the direction of derivative, then we can call it holomorphic as it resembles complex function. We can call it a generalized derivative on a higher order set.

    These all together, were my questions.



    But obviously, I was wrong. I was just ignorant of the terminology. I looked for more information and a blogger told me that holomorphic is the same meaning as differentiable and analytic and analytic function must have a locally convergent power series.
    The blogger told me that there is no converging power series for quaternion and it will have left derivative and right derivative. Also, for the higher order sets, it would not share typical mathmatical properties.





    Also, this is the text i am using.
    http://www.fing.edu.uy/~cerminar/Complex_Analysis.pdf
     
    Last edited: Jun 23, 2016
  10. Jun 23, 2016 #9

    Stephen Tashi

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    Let me rephrase the question in your original post.

    According to the text "Complex Analysis" by Stein and Shakarchi ( http://www.fing.edu.uy/~cerminar/Complex_Analysis.pdf ) pages 11-13, the existence of the derivative of a complex valued function ##F(z) = u(z) + iv(z) ## of a single complex variable ##z = x + i y ## implies a relation between the partial derivatives of the real valued functions ## u(x,y) ## and ## v(x,y)## given by the Cauchy-Riemann equations.

    Are there analogous relations to Cauchy-Riemann equations when we consider quaternion valued functions of quaternions ? - or octonion valued functions of octonions ? etc.

    I'm interested to read what the blogger told you.

    The approach of the text (pages 11-13) is to define the differential of the function ##F(z)## evaluated at ## z_0 = x_0 + i y_0 ## as a 2x2 matrix that comes from evaluating the Jacobian matrix of ##F## when we considered ##F## as a vector valued function ##( u(z), v(z) ) = ( u(x,y), v(x,y) )##. So I agree it is natural to ask whether we can define the differential of an quaternion valued function of a quaternion as a Jacobian, and so forth.

    A derivative involves the limit of "difference quotient" ##\frac{F(z+h) - F(z)}{h} ## so it involves division by ##h##. There is a theorem (https://en.wikipedia.org/wiki/Hurwitz's_theorem_(composition_algebras)) that says we run into problems defining division for "higher dimensional" numbers. Notice that the multiplication of complex numbers is not defined as the component-by-component multiplication of the real and imaginary parts. (i.e. taking ## (a,b) ## to denote the complex number ## a + i b ##, we don't do multiplication as ## ( (a,b) ) ((c,d)) = (ac,bd)##. Likewise, we don't define division of complex numbers as a component-by-component operation.
     
  11. Jun 23, 2016 #10

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    Geometric Algebra is a Clifford Algebra of a vector space. Every element has a multiplicative inverse. This allows the definition of the difference quotient, derivatives, and analytic functions for higher dimensions. Geometric algebra includes complex numbers, quaternians, and extends to higher dimensions.
     
  12. Jun 24, 2016 #11

    Stephen Tashi

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    Is Geometric Algebra an algebra on a vector space? The scalars of an abstract vector space are assumed to be a field and thus have a commutative multiplication. So we can't have a vector space "over the quaternions". I suppose Geometric Algebra treats quaternions as a 4 dimensional vector space over the field of real numbers. It would be interesting to see if that approach overcomes the objections (e.g. the thread http://mathforum.org/kb/message.jspa?messageID=4084898) that the derivative of a quaternion valued function of a quaternion is not uniquely defined.
     
  13. Jun 24, 2016 #12

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    I wish I knew enough about it to answer your questions. It's something I ran into just a couple of months ago. It is not a vector space over the quaternians. I am not sure how to describe it in the usual terms. It is a vector space where multiplication is defined so that the it is closely related to rotations in 2 and 3 dimensions. That guarantees that every non-zero element has a multiplicative inverse (just as every rotation has a rotation in the opposite direction). In spite of my best intentions, I will probably never have the energy and discipline it takes to figure it out. Hopefully someone else can give better answers.
     
  14. Jun 25, 2016 #13

    Stephen Tashi

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    I probably won't either. However, this is the age of video. Have you tried listening to any of the Youtubes about the subject ? I started to watch a video by Hestenes on Geometric Calculus and concluded that wasn't the way to start.
     
  15. Jun 26, 2016 #14

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    I looked at Hestenes video. I agree that it is not introductory and was not much use to me. I just discovered a series of 6 videos by Alan Macdonald that I think is very good (). It doesn't get into calculus, but he has some other material that does. I plan to look it up.

    PS. I am afraid that I may have hijacked this thread. Sorry. I will abstain from any further comments.
     
    Last edited: Jun 26, 2016
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