Grassmann Algebra

Let E be n-dimensional space and let \operatorname{Alg}E be the algebra over the set of real numbers generated by E. Elements P\in E are points and aP = Pa for a\in\mathbf{R} is a point with weight a. Note \mathbf{R} and E are subsets of \operatorname{Alg}E. If A,B\in\operatorname{Alg}E then A + B = B + A\in\operatorname{Alg}E whenever A,B\in\operatorname{Alg}E. The product AB\in\operatorname{Alg}E is not commutative.

Grassmann posited the rule PQ = 0 if and only if P = Q, for any P,Q\in E. This can be used to detect when geometric objects meet. An immediate consequence is PQ = -QP for any P,Q\in E.

Exercise. Prove PQ = -QP for P,Q\in E.

Hint: (P + Q)(P + Q) = 0.

The product PQ represents the oriented one dimensional space determined by the points P and Q when P\not=Q. We can interpret aP + bQ as the point having weight a + b that is the baricenter of aP and bQ. Let \langle PQ\rangle be the line consisting of the set of points R(t) = (1 - t)P + tQ, t\in R. Note R(0) = P and R(1) = Q.

Exercise Show \langle PQ\rangle = \langle QP\rangle.

Exercise Show \langle PQ\rangle = \langle R(t)R(u)\rangle if t\not= u.

Exercise Show Q - P does not belong to \langle PQ\rangle.

A finite set of points P_i\in E are independent if \sum a_i P_i = 0 implies a_i = 0 for all i.

Exercise. If P = \sum_{j = 1}^k a_j P_j, a_j\in\mathbf{R}, P_j\in E then P P_1\cdots P_k = 0.

Hint: P_j P_1 \cdots P_k = 0 for all j, 1\le j\le k.

Exercise. Given P_j\in E, 1\le j\le k and P P_1\cdots P_k = 0 then there exist a_j\in\mathbf{R} with P = \sum_{j = 1}^k a_j P_j.

The exercises show the extensor P_0 \cdots P_k \not= 0 if and only if P_0, …, P_k are independent. The order of the extensor is the number of points in the product. Let R(t_1,\ldots,t_k) = (1 - \sum_{j=1}^n t_j)P_0 + \sum_{j=1}^n t_j P_j and define \langle P_0 \cdots P_k\rangle to be the set of points R(t_1,\ldots,t_k), t_j\in\mathbf{R}.

Exercise If P,Q\in E belong to \langle P_0 \cdots P_k\rangle then so does (1 - t)P + tQ for all t\in\mathbf{R}.