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The number of Grassmann coordinates in ${\cal N}=1$, $3+1$ dimensional superspace is $4$. Let's call them: $\theta_1$ $\theta_2$ $\theta_3$ $\theta_4$.

The Grassmann variables can be represented by analogy with the matrix representation of complex numbers by $2^n\times2^n$ matrices where $n$ is the number of Grassmann variables.

For example for $n=2$ one can write:

\begin{equation*} \theta_1= \begin{bmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \end{bmatrix} \end{equation*}

\begin{equation*} \theta_2= \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 &-1 & 0 & 0 \end{bmatrix} \end{equation*}

My question is: if Grassmann numbers can be represented on $\mathbb{R}^{2^n}$ the same way that complex numbers are represented on $\mathbb{R}^{2}$? I'm not sure if $\mathbb{R}^{2^n}$ would be the minimum dimension needed for a global embedding, in case. But locally I assume it's the minimum dimension needed.

I know that Grassmann numbers do not make up a field and the analogy fails. But still, I want to know if such a geometrical embedding works out.

I mean, can it be interpreted that spacetime must be $20$ dimensional in principle but $16$ dimensions of it are constrained by Grassmann's algebra(given a specific algebraic structure)?

I know that it might sound stupid, but even something close to such an interpretation or a relation between different types of dimensions(fermionic, bosonic, both compact or large) can really help.

Bastam Tajik
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1 Answers1

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  1. That the Graßmann algebra has a representation as $2^n$ by $2^n$ matrices does not imply that it represents $2^n$ dimensions. The reason complex numbers correspond to two real dimensions is that the complex numbers are isomorphic to $\mathbb{R}^2$ as a vector space over the real numbers. The Graßmann algebra on $n$ generators actually does have dimension $2^n$ as a real vector space because the space of $k$ products of the generators has dimension $\binom{n}{k}$ and so the full algebra has dimension $\sum_k \binom{n}{k} = 2^n$, but this not a direct consequence of the representation being matrices over $\mathbb{R}^{2^n}$ - note that the space of all matrices of that size has dimension $(2^n)^2$.

  2. The representation as $2^n$ by $2^n$ matrices is the unique irreducible representation of a Graßmann algebra on $n$ generators because the Graßmann algebra is related to a Euclidean Clifford algebra of $2n$ $\gamma$-matrices by $$ \theta_i = \frac{1}{2}(\gamma_i -\mathrm{i}\gamma_{N+i})$$ and the unique irreducible representation of $2N$ $\gamma$-matrices is $2^n$-dimensional, see this answer by Qmechanic. Note that you can obtain the $\gamma$-matrices from the $\theta_i$ as $\theta_i\pm\theta_i^\dagger$ since the Euclidean $\gamma_i$ can be chosen Hermitian, so this representation of the Graßmann algebra is also unique.

  3. For supermanifolds, we usually do not use the notion of dimension as a single number, since the very different nature of bosonic and fermionic coordinates makes it rather useless to try and lump both of them into a single number. Instead one thinks of the bosonic and fermionic coordinates separately, saying that a supermanifold that is locally $\mathbb{R}^n$ with $m$ Graßmann coordinates attached has a dimension of $(n,m)$. The whole point of the fermionic coordinates is that they carry a very different kind of information than just being a list of real-valued numbers - the isomorphism as vector spaces to $\mathbb{R}^{n+2^m}$ ignores the whole algebra structure of the Graßmann numbers, and so forgets an essential structure. No one will accept calling this a manifold of dimension $n+2^m$ because that is just not a useful way to look at it - you've explicitly dropped the very thing that distinguishes a supermanifold from a manifold.

ACuriousMind
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