February 16, 2025
In the middle of the 19th century Hermann Grassmann came up with a simple axiom to reduce geometric truth to algebra: the product of two points is 0 if and only if the points coincide. René Descartes demonstrated how to prove results in Euclidian geometry by associating points in space with numerical coordinates, but Descartes only considered two and three dimensions. Grassmann, living in the tenor of his time, extended this to any number of dimensions. He predated Einstien, in one aspect, by treating all points in space on equal footing. Grassmann’s audaciously simple notion also predated the modern definition of a vector space. Any point in space can be chosen as an origin and a vector is just the difference of a point with the origin.
Grassmnn considered the set E of points in a space of any dimension. The Grassmann algebra of E, \mathcal{G}(E), is the algebra of generated by the points in E over the real numbers \mathbf{R} with the axiom PQ = 0 if and only if P = Q. The boundary of PQ is \partial(PQ) = Q - P.
Exercise. Show \partial(PQ) = 0 if and only if P = Q.
Exercise. Prove PQ = -QP for P,Q\in E.
Hint: Use (P + Q)(P + Q) = 0.
The product PQ represents the oriented one dimensional line segement 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 R(t) = P + t(Q - P) = (1 - t)P + tQ, t\in R. Note R(t) has weight 1, R(0) = P, and R(1) = Q.
Exercise Show PQR(t) = 0 for t\in\mathbf{R}.
This is an example of incidence involve three points. It means R(t) meets the line determined by P and Q for all t\in\mathbf{R} if P\not=Q.
Exercise Show Q - P\not= R(t) for any t\in\mathbf{R}.
Since R(t)/t = (1/t - 1)P + Q and 1/t - 1 tends to -1 as t goes to infinity, we can think of Q - P as a point at infinity with 0 weight. It represents the vector from P to Q. We can interpret R(t) = P + t(Q - P) as the point from P along the vector Q - P with length t.
If tPQ = ST for some t\in\mathbf{R} we write ST/PQ for t..
Exercise. For any R with PQR = 0 and PQ\not=0 we have R = (PR/PQ)P + (RQ/PQ)Q.
Hint: Every such R = R(t) for some t. Consider PR(t) and R(t)Q.
David Hilbert noticed Euclid’s axioms were insufficent to determine when a point lies between two other points. It is clear that R(t) is between P and Q if and only if 0 < t < 1. Grassmann’s algebra provides a coordinate-free way of detecing that.
Exercise. Show if {PQR = 0} and PQ\not=0 then R is between P and Q if and only if {PR/PQ > 0} and {RQ/PQ > 0}.
Three points P,Q,R\in E with PQR\not=0 determine a triangle. The midpoint of the side from P to Q is {(1/2)P + (1/2)Q} so {(1/2)(P + Q)R} is the median from R to the midpoint of P and Q. The centroid (1/3)(P + Q + R) lies on the median.
Exercise. Show (1/2)(P + Q)R(1/3)(P + Q + R) = 0.
Replacing P,Q,R with Q,R,P and R,P,Q shows all medians of a triange meet at the centroid.
This generalizes to any number of dimensions. Let P_j be points in E, 0\le j\le n, with P_0\cdots P_n\not=0. Define R(t_1,\ldots,t_n) = (1 - \sum_1^n t_i)P_0 + \sum_1^n t_i P_i.
Exercise. Show P_0\cdots P_nR(t_1,\ldots,t_n) = 0, t_i\in\mathbf{R}.
Exercise. Show if P_0\cdots P_nR = 0 and P_0\cdots P_n\not=0 then R = \sum_0^n (P_0\cdots R\cdots P_n/P_0\cdots P_n)P_i where P_i is replaced by R in the coefficents of the sum.
The smallest convex set containing P_0,\ldots,P_n is \{\sum_i t_i P_i\mid t_i\ge0,\sum_i t_i = 1\}. The above exercise show R belongs to the convex hull if and only if all coefficients in the sum are non-negative.
A finite set of points P_i\in E are independent if \sum t_i P_i = 0 implies t_i = 0 for all i.
Exercise. If P = \sum_{j = 0}^k t_j P_j, t_j\in\mathbf{R}, P_j\in E then P P_0\cdots P_k = 0.
Hint: P_j P_0 \cdots P_k = 0 for all j, 1\le j\le k.
Exercise. Given P_j\in E, 0\le j\le k and P P_0\cdots P_k = 0 then there exist t_j\in\mathbf{R} with P = \sum_{j = 1}^k t_j P_j.
The exercises show the extensor P_0 \cdots P_k \not= 0 if and only if {P_0,\ldots, P_k} are independent. The order of the extensor is the number of points in the product.
If A and B are extensors with \langle A\rangle \cap \langle B\rangle = 0 then \langle A\vee B\rangle = \langle A\rangle \oplus \langle B\rangle. If \langle A\rangle \cap \langle B\rangle \not= 0 then \langle A\vee B\rangle = 0.
If A and B are extensors with \langle A\rangle \cap \langle B\rangle \not= 0 then \langle A\wedge B\rangle = \langle A\rangle \oplus \langle B\rangle. If \langle A\rangle \cap \langle B\rangle \not= 0 then \langle A\vee B\rangle = 0.