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  Clarifying the role of the mysterious space underlying the definition of a superspace

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In the intro to chapter 12.3 of this book about the applications of coherent states, it says that

classical spaces for bosons are real or complex vector spaces or manifolds, whereas classical spaces for fermions are Grassmann algebras $\mathbb{K}[\theta_1, ... ,\theta_n], \mathbb{K}=\mathbb{R}$ or $\mathbb{C}$.
More precisely the classical space for fermions is given by the space of smooth
mappings $f$ from $B$, a certain set of dimension $0$, to the Grassmann
algebra $\mathbb{K}[\theta]$

$$
f: B \rightarrow \mathbb{K}[\theta]
$$

What exactly is the role of this ominous space $B$ in this definition? Why is it even needed, why can one not directly work with the Grassmann algebra for the fermionic part of superspace?

As an example, for a system with $n$ bosons and $m$ fermions (or $n$ bosonic and $m$
fermionic coordinates? The bosons and fermions should be obtained as the coefficients
when expanding a superfield in the Grassmann coordinates!) the configuration
space is called a superspace $\mathbb{R}^{n|m}$ and is defined as

$$C^{\infty}(\mathbb{R}^{n|m})
=C^{\infty}(\mathbb{R}^n)[\theta_1, ... ,\theta_n]
=C^{\infty}(\mathbb{R}^n)\otimes \mathcal G_m
$$
where $\mathcal G_m$ is the fermionic part parameterized by $n$ Grassmann variables $\theta_i$.

Does $\mathbb{R}^{n|m}$ here take the role of the space $B$ that was rather abstractly alluded to above?

asked Mar 31, 2017 in Mathematics by Dilaton (6,240 points) [ revision history ]
edited Apr 1, 2017 by Dilaton

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