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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?

Farther down in the same text, it is explained that 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?

And what is the interpretation of the $C^{\infty}$ functions, do they correspond to the superfields living in superspace?

So far I got no answer or comment to the same question here or here ...

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    $\begingroup$ If we interpret what is written about B literally, then B must be Spec(K[θ]), i.e., B=R^{0|1}, which is consistent with your later description if n=0, m=1. As to why this space is needed: in the nonsupersymmetric case we can ditch our smooth manifolds and work directly with their algebra of smooth functions, which preserves all information about them. Nevertheless, we find it useful to have a geometric language to talk about manifolds. The same applies to supermanifolds: although their algebras of functions contain all the information, a geometric language to talk about them is also useful. $\endgroup$ Apr 10, 2017 at 10:20

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