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Derivatives with respect to a Supernumber?

  1. Jan 16, 2015 #1
    So I've been trying to think about some papers in Supersymmetry and I need to somehow define a derivative of a supernumber, with respect to another supernumber. I mean a supernumber to be a number with an ordinary "body" and a "soul" which is a product of an even number of Grassmann numbers. Call it [itex] z=z_{B}+z_{S} [/itex].

    I know what properties to expect from a derivative with respect to ordinary numbers and Grassmann numbers *separately* but I'm lost on how to combine them. In addition, I would like the body and soul to be complex.
    Thanks for any tips
     
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  3. Jan 17, 2015 #2

    mathman

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    "body" and "soul" for numbers?????????????????????????????????????
     
  4. Jan 17, 2015 #3
    haha indeed! Awesome names. A supernumber contains a "body" which is simply an ordinary complex number, and a "soul" which is a sum of even products of Grassmann numbers.
     
  5. Jan 17, 2015 #4
    Does this help?
    Chapter 3, section 3.2.2 page 23... here?
     
  6. Jan 17, 2015 #5
    That's almost exactly what I want. Except I know how to differentiate with respect to bosonic and fermionic variables separately; I'm just not sure how to define a derivative with respect to a number that is both bosonic and fermionic.
     
  7. Jan 19, 2015 #6
    My only idea would be to try to prove something analogous to the Cauchy-Riemann equations but I haven't had any luck. In other words, perhaps I could differentiate the body and soul of a function with respect to bodies and souls separately and then find some relation between them. Has anyone heard of anything along these lines?
     
  8. Jan 19, 2015 #7

    mathman

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    I am not at all familiar with these concepts. I suggest that you go back to the basic definition of derivative as the limit of difference quotients, in particular define a difference quotient.
     
  9. Jan 20, 2015 #8

    Demystifier

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    What turned out to be a problem?
     
  10. Jan 20, 2015 #9
    Probably the one where I'm just not clever enough :) In complex analysis, you consider a function [itex] f(z)=u(x,y)+i v(x,y) [/itex] and the Cauchy-Riemann Equations give relations on real valued functions u and v. I want to think of a supernumber's body and soul as analogous to x and y, respectively above. Likewise, I'd like to take a function of that supernumber and think of its ordinary and fermionic parts as analogous to u and v from above.

    Ultimately, I want something analogous to

    [tex]
    \frac{\partial}{\partial z}=\frac{1}{2}\bigg(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y}\bigg)
    [/tex]

    which would tell me exactly how to differentiate with respect to a supernumber in terms of the derivatives wrt the ordinary and fermionic parts, which I know how to do. However, the imaginary part of a holomorphic function is still a real-valued function whereas in my case, the "soul" of a superfunction may depend wildly on a bunch of Grassmann numbers.
     
  11. Jan 20, 2015 #10

    samalkhaiat

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    All these stuff are discussed nicely in the first chapter of the classic textbook “Supermanifolds” by Bryce DeWitt, Cam. Univ. Press (1992). And, by the way super-analytic functions have no body-soul decomposition, rather they admit even(commuting)-odd(anticommuting) decomposition.
     
  12. Jan 21, 2015 #11
    Right, I should be saying even vs. odd when it comes to super-analytic functions. However, from what I can see, DeWitt deals only with differentiation w.r.t even or odd variables separately but doesn't hint at how to combine them. Am I missing something? Have you seen this in there?
     
  13. Jan 21, 2015 #12

    Demystifier

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    Another book on the same subject is
    F.A. Berezin, Introduction to Superanalysis (1987)
     
  14. Jan 21, 2015 #13
    After all the help on this page, I think I've finally got this straightened out. Like someone pointed out, my main downfall seemed to be trying to think about super-analytic functions as having "bodies" and "souls" when in fact, only an even/odd decomposition is possible. That helps a lot. Also, maybe it's just me but I found Berezin a lot easier to digest than DeWitt. Thanks a lot everyone!
     
  15. Jan 21, 2015 #14

    samalkhaiat

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    Now, I am not sure how much you know about the subject! The whole exercise is to parametrize the real superspace [itex]\mathbb{ R }^{ p | q } = \mathbb{ R }^{ p }_{ c } \times \mathbb{ R }^{ q }_{ a }[/itex] by [itex]p[/itex] c-number coordinates [itex]x^{ n }[/itex] and [itex]q[/itex] a-number coordinates [itex]\theta^{ \alpha }[/itex]. A supernumber-valued function [itex]f : \mathbb{ R }^{ p | q } \to \Lambda_{ \infty }[/itex] is called super-analytic if the following Taylor series converges: [tex]f ( z ) \equiv f ( x , \theta ) = \sum_{ k = 0 }^{ \infty } f_{ A_{1} A_{ 2 } \cdots A_{ k } } \ z^{ A_{ 1 } } z^{ A_{ 2 } } \cdots z^{ A_{ k } } , \ \ f_{ A_{ 1 } \cdots A_{ k } } \in \Lambda_{ \infty } .[/tex] With a bit of work, you can rewrite the series as [tex]f ( x , \theta ) = f_{ 0 } ( x ) + \sum_{ k = 1 }^{ q } \frac{ 1 }{ k ! } f_{ [ \alpha_{ 1 } \cdots \alpha_{ k } ] } ( x ) \ \theta^{ \alpha_{ 1 } } \theta^{ \alpha_{ 2 } } \cdots \theta^{ \alpha_{ k } } ,[/tex] with all the [itex]f[/itex]’s being superfunctions of [itex]\mathbb{ R }^{ p }_{ c }[/itex]. Now, if you want to work with the first expansion, be my guest and use the following properties [tex]\partial_{ B } z^{ A } = \delta_{ B }{}^{ A } , \ \ \ \partial_{ A } \partial_{ B } = ( - 1 )^{ \epsilon ( A ) \ \epsilon ( B ) } \partial_{ B } \partial_{ A } ,[/tex] where [itex]\epsilon ( A ) \equiv \epsilon ( z^{ A } )[/itex] is Grassmann parity of the coordinates: [itex]\epsilon ( \alpha ) = 1, \ \ \epsilon ( n ) = 0[/itex]. And for superfunctions [itex]f[/itex] and [itex]g[/itex], you have [tex]\partial_{ A } ( f \ g ) = ( \partial_{ A } f ) \ g + ( - 1 )^{ \epsilon ( A ) \ \epsilon ( f ) } f \ ( \partial_{ A } g ) ,[/tex] where [itex]\epsilon ( f_{ \mbox{ even } } ) = 0[/itex], [itex]\epsilon ( f_{ \mbox{ odd } } ) = 1[/itex] and [itex]\epsilon ( \partial_{ A } f ) = \epsilon ( A ) + \epsilon ( f ) \ \ \mbox{ mod } \ 2[/itex]. You can also show that the action of complex conjugation on derivatives is given by [tex]( \partial_{ A } f )^{ * } = ( - 1 )^{ \epsilon ( A ) \left( 1 + \epsilon ( f ) \right) } \ \partial_{ A } f^{ * } .[/tex]

    Sam
     
  16. Jan 21, 2015 #15
    In your first line, should your z's be the [itex] \theta^{\alpha_{i}} [/itex]? If not, I don't understand where they come from. Or are the z's general supernumbers?
     
  17. Jan 21, 2015 #16

    samalkhaiat

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    So, as I expected, you don't know even the elementary concepts involved in here. What does the equation [itex]\mathbb{ R }^{ p | q } = \mathbb{ R }^{ p }_{ c } \times \mathbb{ R }^{ q }_{ a }[/itex] mean? As I said before, the real superspace [itex]\mathbb{ R }^{ p | q }[/itex] is parametrized by the coordinates [itex]z^{ A } = ( x^{ n } , \theta^{ \alpha } )[/itex], where [itex]x^{ n } \in \mathbb{ R }^{ p }_{ c } , \ \ n =1 , 2 , \cdots , p[/itex] are the bosonic (commuting) coordinates [they are called c-numbers], and [itex]\theta^{ \alpha } \in \mathbb{ R }^{ q }_{ a } , \ \ \alpha = 1 , 2 , \cdots , q[/itex] are the fermionic (anticommuting) coordinates [these are called a-numbers].
     
  18. Jan 22, 2015 #17
    Well, you're right; I certainly don't understand this subject like I need to. I was really just referring to earlier, when I was trying to talk about differentiating a super-analytic function by a body and a soul. So let me see if I'm getting this: the coordinates on real superspace are [itex] (x^{n}, \theta^{\alpha}) [/itex] with the first being bosonic and the last being bosonic. A general super-analytic function F is a linear combination of all possible products of the [itex] \theta^{\alpha} [/itex] where the coefficients are superfunctions of just the x's. The indices on the superfunctions must anti-commute since the [itex] \theta^{\alpha} [/itex] do. Then, to take a derivative with respect to the x's, you simply take ordinary derivatives of these coefficient superfunctions within the expression. To take a derivative with respect to the [itex] \theta^{\alpha} [/itex], you use the usual right/left derivatives discussed in Berezin and DeWitt. Am I starting to get on the right track?

    Really, all I need to do is compute Jacobians using superdeterminants, which of course requires a matrix of derivatives. I think Berezin hopefully helped me understand how to take these but I would like to understand the basics. So I appreciate your comments.
     
  19. Jan 23, 2015 #18

    samalkhaiat

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    That is okay.

    Supermatrices, superdeterminants(which also called Berezinian in modern works) and the operations on them are discussed in the above-mentioned two textbooks.
     
  20. Jan 24, 2015 #19
    I'm curious, what happens if your coordinates parameterize *complex* superspace instead of real? How do the derivatives behave differently?
     
  21. Jan 27, 2015 #20

    samalkhaiat

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    The rules of the derivatives do not change if the superfunctions are complex-valued functions.
     
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