Relationship between Levi-Civita symbol and Grassmann numbers? The multiplication rule for Grassmann numbers $\theta_i$ is
$$
\theta_i\theta_j = - \theta_j \theta_i
$$
so that $\theta_i\theta_i = 0$. Multiplying three Grassmann numbers yields
$$
\theta_i\theta_j\theta_k = - \theta_j\theta_i\theta_k = \theta_j\theta_k\theta_i,
$$
so everytime we interchange two Grassmann numbers, the sign of the product changes. If two Grassmann numbers are the same within a product of three Grassmann numbers the product is zero $\theta_i\theta_i\theta_k = 0$ (see also Hermann Grassmanns Algebra). 
Doesn't this resemble the rules for index exchange in the Levi-Civita Symbol? E.g. in 2 dimensions we have
$$
\varepsilon_{ij} = - \varepsilon_{ji}
$$
with $\varepsilon_{ii} = 0$. The usual Levi-Civita Symbol in three dimensions follows the same rules as a product of three Grassmann numbers
$$
\varepsilon_{ijk} = - \varepsilon_{jik} = \varepsilon_{jki}.
$$
If any of the indexes is the same we have $\varepsilon_{iik} = 0$ as for the product of three Grassmann numbers.
Is this all an accident? Or can one really say that the Levi-Civita symbol is equal to a product of Grassmann numbers like
$$
\varepsilon_{ij} = \theta_i\theta_j
$$
(which might also imply that the one dimensional Levi-Civita symbol should be defined to be a Grassmann number $\varepsilon_{i} = \theta_i$)
 A: Grassmann numbers can be embedded in a clifford algebra.  Let $g(a,b)$ be some scalar-valued, symmetric, bilinear function for $a, b$ that are linear combinations of $\theta_i$--call $a,b$ "vectors".  Then the multiplication laws could be tweaked to read
$$\theta_i \theta_j + \theta_j \theta_i  = g(\theta_i, \theta_j)$$
True grassmann numbers are the case $g(\theta_i, \theta_j) = 0$.
Suppose $g(\theta_i, \theta_j) = 0$ for $i \neq j$.  Then the quantity $\epsilon = \theta_1 \theta_2 \ldots \theta_n$ has components like those of the Levi-Civita.
But how do you define components?  You can't extract scalars here without either (a) a set of "dual" vectors $\theta^1, \theta^2, \ldots$ such that $\theta_1 \theta^1 = 1$, or (b) the function $g$ that serves the purpose of a metric on the vector space of $\theta_i$.  In the latter case, you get extract components like so:
$$\epsilon \theta_1 \theta_2 \theta_3 = \theta_1 \theta_2 \theta_3 \theta_1 \theta_2 \theta_3 = -g(\theta_1, \theta_1) g(\theta_2, \theta_2) g(\theta_3, \theta_3) = \epsilon_{321}$$
(Nb: if you're accustomed to treating Grassmann numbers through their matrix representation, then by "scalar" you should think "the set of matrices such that multiplication commutes", which would all just be multiples of the identity.)
