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 $\varepsilon_{i} = \theta_i$)

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.)

• Do I get this right: The notation of a Grassmann number like $\theta_i$ does not imply that $\theta$ is a vector with $i$ components. Instead $\theta_i$ is a scalar number like e.g. the imaginary number $i$, correct? So the product of two Grassmann numbers $\theta_i\theta_j$ is not an outer product of two vectors, but just a product of two scalars. – asmaier Jan 24 '15 at 15:26
• No, Grassmann numbers (and their nontrivial products) are basis elements of a vector space (the exterior algebra). They are not like ordinary scalars in any sense. Each $\theta_i$ is characterized by $n$ real components, where $n$ is the number of distinct $\theta_i$. How could they be scalars? Scalar multiplication commutes. – Muphrid Jan 24 '15 at 15:35
• Ok. But then isn't the product $\theta_i\theta_j$ a tensor (more precisely a pseudotensor, since it anticommutes)? – asmaier Jan 24 '15 at 15:49
• Yeah, bivectors can be identified with 2-index tensors. Though in pure exterior algebra, I'm not sure you can actually write out how to take a bivector and two vectors and return a scalar (that isn't zero). – Muphrid Jan 24 '15 at 15:53
• Wouldn't it then be much less confusing to call these $\theta_i$s "Grassmann vectors" instead of "Grassmann numbers"? – asmaier Jan 26 '15 at 11:35