The Grassmann analytic continuation principle and the expansion of a superfield in Grassmann coordinates

Question

The Grassmann analytic continuation principle says that a real function $f(x,\xi)$ that depends on $n$ real variables $x=(x_1,\ldots,x_n)$ and $n$ nilpotent Grassmann numbers $\xi=(\xi_1,\ldots,\xi_n)$ has the following Taylor expansion containing only a finit ($2n?$) number of terms

$$ f(x+\xi) = \sum_{\alpha}\frac{\xi^{\alpha}}{\alpha!}\frac{\partial^{\alpha}}{\partial x^{\alpha}}f(x) $$

where the $\xi_j \in \mathcal{G}_n$ which is the fermionic part of the supermanifold at hand are given by

$$ \xi_j =\sum_{|\epsilon|>0}a_{\epsilon}^j\theta^{\epsilon}$$

The $\theta_i$ are the generators of $\mathcal{G}_n$, $\alpha=(\alpha_1,\ldots,\alpha_n)\in\mathbb{N}^n$, and

$\xi^{\alpha}=\xi_1^{\alpha_1}\ldots\xi_n^{\alpha_n}$.

How is this related to expanding a superfield in Grassmann coordinates?

For example, a a scalar superfiled $\mathcal{F}(x,\theta_1,\theta_2)$ that depends on the ordinary
spacetime coordinates as well as on two Grassmann coordinates can be expanded as

$$\mathcal{F}(x,\theta_1,\theta_2) = A(x)+B(x)\theta_1+C(x)\theta_2+D(x)\theta_1\theta_2$$

Comparing this expansion to the Grassmann analytic continuation principle above I would think that in this specific case we have $n=2$, $\alpha=(\alpha_1,\alpha_2)$ and $\xi=\xi_1^{\alpha_1}\xi_2^{\alpha_2}$ as there are two Grassmann coordinates in addition to conventional spacetiem.

The first thing I dont understand is why in the Grassmann analytic continuation principle the function is not expanded directly in the $\theta_i$ but the $\xi$ are used instead which makes it hard for me to see what is going on.

Otherwise, I would have guessed that from setting $\alpha_1=0$ and $\alpha_2=0$ the coefficient $A(x)$ would correspond to $f(x)$, from setting $\alpha = (2,1)$ one obtains $B(x)=\frac{\partial f}{\partial\theta_1}$, from setting $\alpha = (1,2)$ $C(x)=\frac{\partial f}{\partial\theta_2}$ and from $\alpha = (1,1)$ one has $D(x)=\frac{\partial^2 f}{\partial\theta_1\partial\theta_2}$

Comments

Part of the issue is that I already strugle with the notations, as for example in the definition of the Grassmann numbers $ \xi_j =\sum_{|\epsilon|>0}a_{\epsilon}^j\theta^{\epsilon}$ I dont understand the role of the index $\epsilon$. Is it a label for the generators $\theta$ or does it denote to power with which $\theta$ appears in the sum?

Answer 1

I have only gone through an introductory supersymmetry course but here is my understanding of the matter.

An analytical superfield, even though formally written as $\Phi(x,\theta)$ is essentially only the expansion $\Phi = \phi(x) + \theta \psi(x) +...$.  This has nothing to do with Grassmannian analytical continuation per se.

The Grassmannian analytic continuation is simply a way to somehow build a superfield from a regular field $f(x)$ by formally going from the real space in the Grassmannian direction and expanding. But if you just naively define this expansion without the transformation of the basis, your grassmannian fibres with respect to which you expand are always aligned with the coordinate directions and the definition is actually not diffeomorphism invariant. So you need to have an invariant basis $\theta$ on the fibres and a transformation law into the coordinate one $\xi$.

I guess you could then derive e.g. the transformation law for the $a^j$ coefficients by requiring that 1) $\theta$ does not transform under $x$-coordinate changes and 2) $f(x)$ and its derivatives transform according to the usual rules. My other guess is that this is analogous to a choice of a gamma-matrix basis at every point of your underlying space and that this choice is thus essentially equivalent to a metric structure.

Comments

This answer seems to be exactly right as I learned from reading in the rather nicely written book of Rogers.

In particular, Theorem 4.2.4 therein states that the coefficient functions of the superfield expansion can be understood as Grassmann analytic continuations of ordinary $C^{\infty}$ functions and equation (4.16) in the proof explains how the expansion coefficients can be calculated by taking odd derivatives of the supersmooth function $G^{\infty}$.

Answer 2

A superfield on an $(m,n)$-dimensional superspace is a supersmooth function $f \in G^{\infty}$
and takes the form

\[
f(x^1,\ldots,x^m;\xi^1,\ldots,\xi^n) = \sum \limits_{\underline\mu\in M_n}
\widehat{ f_{\underline\mu}}(x)\xi^{\underline\mu}
\]

The $\hat{f_{\underline\mu}}(x)$ are Grassmann analytic continuations of $C^{\infty}$-functions in $\mathbb{R}^m$. If $f\in G^{\infty}$ they take values in $\mathbb{R}_S$ and if $f\in H^{\infty}$  they take value in $\mathbb{R}$. The $\underline\mu$ are multi indices and if $\underline\mu=\mu_1\ldots\mu_k$ then $\xi^{\underline\mu}=\xi^{\mu_1}\ldots \xi^{\mu_k}$

with

\[\hat{f}(x;\xi)=\sum\limits_{i_1=0,\ldots,i_m=0}^L \frac{1}{i_1!\ldots i_m!}\partial_1^{i_1}\ldots \partial_m^{i_m}f(\epsilon_{m,n}(x))\times s(x^1)^{i_1}\ldots  s(x^m)^{i_m}\]

where $s$ is the soul map and $\epsilon_{m,n}(x)$ the projection to the body.
Concerning superspaces, the body map $\epsilon_{m,n}(x)$ projects to the even coordinate part by setting the odd coordinates to zero

\[\epsilon_{m,n}(x): \mathbb{R}_{S[L]}^{m,n} \rightarrow \mathbb{R}^m\]

\[(x_1,\ldots,x_m;\xi_1,\ldots,\xi_n ) \mapsto (\epsilon(x^1),\ldots,\epsilon(x^m))\]

and the soul map projects to the Grassmann part of the superspace

\[s_{m,n}(x): \mathbb{R}_{S[L]}^{m,n} \rightarrow \mathbb{R}^m\]

\[(x_1,\ldots,x_m;\xi_1,\ldots,\xi_n ) \mapsto (s(x^1),\ldots,s(x^m);s(\xi_1),\ldots,s(\xi_n))\]

Comments

I think you'd explain what soul and body are in this context....