After reading the Wikipedia entry on Grassmann Numbers https://en.wikipedia.org/wiki/Grassmann_number , and seeing as how the vector product

$$ \mathbf{i} \wedge \mathbf{i} = 0,\ \text{("non-zero square root of zero"),}$$


$$\mathbf{a} \wedge \mathbf{b} = - \mathbf{b} \wedge \mathbf{a},$$

can one regard such as an elementary example of a Grassmann Number?

(I've been banging my head for a while trying to get a visual sense of Grassmann Numbers, but if this example is correct I can start modelling this simplest (?) of cases, mentally, in terms of the oriented area between two such entities - clearly zero in magnitude when taking the vector product of a vector with itself, and anticommuting with any other vector.)

  • $\begingroup$ Thanks for the edit, Aidan. $\endgroup$ – iSeeker Sep 17 '17 at 16:37

No: the cross product (the vector product that returns a vector) is not associative, whereas Grassmann multiplication is. The exterior algebra of the vector space is an essential component of the definition of the Grassmann algebra, because it is defined so that $\theta_i\theta_j$ as not the same type of object as $\theta_i$.

You have to consider the antisymmetric product $\wedge $, which takes two vectors $a,b \in V$ to a bivector $a \wedge b \in V \wedge V$, and bivectors to trivectors and so on (this is the ordinary exterior product). In the case of three dimensions, the exterior algebra $ \Lambda(\mathbb{R}^3) = \mathbb{R} \oplus (\mathbb{R}^3) \oplus \Lambda^2(\mathbb{R}^3) \oplus \Lambda^3 (\mathbb{R}^3) $ has a basis $$ 1 \\ e_1,e_2,e_3,\\ e_1 \wedge e_2, e_2 \wedge e_3, e_3 \wedge e_1, \\ e_1 \wedge e_2 \wedge e_3, $$ which has the required dimension $2^3 = 8$ as mentioned in the article you link. It is true that by its definition the exterior algebra does allow adding different types of object, so the geometric meaning can be obscured: this is not that different from polynomials, where one normally doesn't ask what the geometric meaning of, say, $1+X+X^2$ is.

(The cross product comes about by using the Hodge star to give an isomorphism $\Lambda^2(\mathbb{R}^3) \to (\mathbb{R}^3)$ which is natural in that it is basis-independent, provided that the orientation of the bases is the same (in that the transformation matrix has positive determinant). This use of the Hodge star is why cross products behave weirdly under reflections, and why the cross product is not associative.)

  • $\begingroup$ Great answer, Chappers - thank you so much. PS Any hints as to how one might visualise the simplest case of a Grassmann number? $\endgroup$ – iSeeker Sep 17 '17 at 16:36
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    $\begingroup$ That would be $u \wedge v$? One can visualise it as the parallelogram with sides $u$ and $v$, in that order. This has the appropriate properties of being bilinear and alternating in $u$ and $v$ (think about the side of your hand that faces up: it's opposite if your thumb and forefinger are the other way around). $\endgroup$ – Chappers Sep 17 '17 at 16:53
  • $\begingroup$ So, in this simplest of cases it's very like geometric algebra. $\endgroup$ – iSeeker Sep 17 '17 at 19:35
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    $\begingroup$ @iSeeker In the finite-dimensional case you can make a strong connection to projective geometry by treating the coefficients as homogeneous coordinates of various objects: elements of $\Lambda^1$ are points, $\Lambda^2$ planes and so on. The exterior product corresponds to the join of two objects. These coordinates will differ from the more conventional homogeneous coordinates by a sign that reflects the permutation required to bring the $\theta$s into canoncal order. Compare this to Plücker coordinates of lines, &c. $\endgroup$ – amd Sep 18 '17 at 19:06
  • $\begingroup$ @iSeeker I should also note that, at least when working with real coefficients, there are rank $k+1$ Grassmann numbers that don’t correspond to $k$-flats, that is, not every rank $k+1$ number is a product of rank-one numbers. $\endgroup$ – amd Sep 18 '17 at 20:37

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