# Derivatives with respect to a Supernumber?

1. Jan 16, 2015

### "pi"mp

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 $z=z_{B}+z_{S}$.

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. Jan 17, 2015

### mathman

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

3. Jan 17, 2015

### "pi"mp

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.

4. Jan 17, 2015

### bahamagreen

Does this help?
Chapter 3, section 3.2.2 page 23... here?

5. Jan 17, 2015

### "pi"mp

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.

6. Jan 19, 2015

### "pi"mp

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?

7. Jan 19, 2015

### mathman

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.

8. Jan 20, 2015

### Demystifier

What turned out to be a problem?

9. Jan 20, 2015

### "pi"mp

Probably the one where I'm just not clever enough :) In complex analysis, you consider a function $f(z)=u(x,y)+i v(x,y)$ 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

$$\frac{\partial}{\partial z}=\frac{1}{2}\bigg(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y}\bigg)$$

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. Jan 20, 2015

### samalkhaiat

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.

11. Jan 21, 2015

### "pi"mp

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?

12. Jan 21, 2015

### Demystifier

Another book on the same subject is
F.A. Berezin, Introduction to Superanalysis (1987)

13. Jan 21, 2015

### "pi"mp

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!

14. Jan 21, 2015

### samalkhaiat

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

Sam

15. Jan 21, 2015

### "pi"mp

In your first line, should your z's be the $\theta^{\alpha_{i}}$? If not, I don't understand where they come from. Or are the z's general supernumbers?

16. Jan 21, 2015

### samalkhaiat

So, as I expected, you don't know even the elementary concepts involved in here. What does the equation $\mathbb{ R }^{ p | q } = \mathbb{ R }^{ p }_{ c } \times \mathbb{ R }^{ q }_{ a }$ mean? As I said before, the real superspace $\mathbb{ R }^{ p | q }$ is parametrized by the coordinates $z^{ A } = ( x^{ n } , \theta^{ \alpha } )$, where $x^{ n } \in \mathbb{ R }^{ p }_{ c } , \ \ n =1 , 2 , \cdots , p$ are the bosonic (commuting) coordinates [they are called c-numbers], and $\theta^{ \alpha } \in \mathbb{ R }^{ q }_{ a } , \ \ \alpha = 1 , 2 , \cdots , q$ are the fermionic (anticommuting) coordinates [these are called a-numbers].

17. Jan 22, 2015

### "pi"mp

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 $(x^{n}, \theta^{\alpha})$ 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 $\theta^{\alpha}$ where the coefficients are superfunctions of just the x's. The indices on the superfunctions must anti-commute since the $\theta^{\alpha}$ 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 $\theta^{\alpha}$, 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.

18. Jan 23, 2015

### samalkhaiat

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.

19. Jan 24, 2015

### "pi"mp

I'm curious, what happens if your coordinates parameterize *complex* superspace instead of real? How do the derivatives behave differently?

20. Jan 27, 2015

### samalkhaiat

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