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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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- #1

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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

- #2

mathman

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"body" and "soul" for numbers?????????????????????????????????????

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- #4

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Does this help?

Chapter 3, section 3.2.2 page 23... here?

Chapter 3, section 3.2.2 page 23... here?

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mathman

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What turned out to be a problem?My only idea would be to try to prove something analogous to the Cauchy-Riemann equations but I haven't had any luck.

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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.

- #10

samalkhaiat

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All these stuff are discussed nicely in the first chapter of the classic textbook “

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- #12

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Another book on the same subject isclassic textbook “Supermanifolds” byBryce DeWitt,Cam. Univ. Press (1992).

F.A. Berezin, Introduction to Superanalysis (1987)

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- #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

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- #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].

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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.

- #18

samalkhaiat

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That is okay.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?

Supermatrices, superdeterminants(which also called Berezinian in modern works) and the operations on them are discussed in the above-mentioned two textbooks.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.

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- #20

samalkhaiat

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