November 25, 2025
In the middle of the 19th century (Grassmann 1844) 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 predated Einstien in 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.
Grassmann considered the set E of points in a space of any (finite) dimension. The Grassmann algebra of E, \mathcal{G}(E), is the algebra generated by the points in E over the real numbers \mathbf{R} with the axiom PQ = 0 if and only if P = Q. More precisely, the Grassmann algebra is the quotient algebra \mathcal{G}(E) = \operatorname{Alg}(E)/\operatorname{Alg}(\{PP\mid P\in E\}).
The (associative) algebra over the real numbers generated by any set S, \operatorname{Alg}(S), is defined by
Note that we do not require xy = yx. The associativity of multliplication allows us to write xyz unambiguously.
Exercise. Show if x\in\operatorname{Alg}(\{PP\mid P\in E\}) and y\in\operatorname{Alg}(S) then xy and yx are in Alg(\{PP\mid P\in E\}).
This shows \operatorname{Alg}(\{PP\mid P\in E\}) is a two-sided ideal of \operatorname{Alg}(S) so the quotient space \mathcal{G}(E) is well defined.
Before introducing the general theory let’s consider some simple examples.
Exercise. Prove PQ = -QP for P,Q\in E.
Hint: Use (P + Q)(P + Q) = 0.
The product PQ represents the oriented line determined by 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}.
Hint: PQR(t) = PQ((1 - t)P + tQ).
This is an example of incidence involve three points. It indicates 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.
Note PQ(Q - P) = 0.
Exercise. Show \frac{(Q - P)Q}{PQ} = -1 and \frac{P(Q - P)}{PQ} = 1.
If PQ = tST where t\in\mathbf{R} and P,Q,S,T\in\mathcal{G}(E) we say PQ and ST are conguent and write {\frac{PQ}{ST}} = t.
Exercise. For any R with PQR = 0 and PQ\not=0 we have R = (PR/PQ)P + (RQ/PQ)Q.
Hint: If R = aP + bQ then PR = bPQ and QR = aQP.
This is an example of coordnate-free calculation using the Grassmann algebra.
David Hilbert noticed Euclid’s axioms were insufficent to determine when a point lies between two points. Using Grassman algebra we can define R is between P and Q if and only if PQR = 0, PR/PQ > 0, and RQ/PQ < 1.
Three points P,Q,R\in E with PQR\not=0 determine an oriented plane. The midpoint of the side from P to Q is {(P + Q)/2} so {(P + Q)R/2} is the median from the midpoint of P and Q to R. The centroid (P + Q + R)/3 lies on the median.
Exercise. Show (P + Q)R(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.
Grassmann’s algebra works in any number of dimensions. Any non-zero product of points is an extensor. The number of points is the order of the extensor.
If P_0\cdots P_k = tQ_0\cdots Q_k for some non-zero t\in\mathbf{R} then write P_0\cdots P_k/Q_0\cdots Q_k = t and say the products are congruent.
Exercise. Show if P_0\cdots P_j and Q_0\cdots Q_k are congruent then j = k.
If two extensors are congruent then they have the same order.
The points P_0,\ldots,P_k are independent if P_0\cdots P_k\not=0.
Lemma. If P_0,\ldots,P_k are independent and \sum_j a_j P_j = 0 for some a_j\in\mathbf{R} then a_i = 0 for 0\le i\le k.
Proof. Multiply both sides by the product of \Pi_{j\not=i} P_j. This results in \pm a_i P_0\cdots P_k = 0 so a_i = 0 for all 0\le i\le k.
If P_0\cdots P_k=0 we say P_0,\ldots,P_k are dependent.
Lemma. If P_0,\ldots,P_k are dependent then there exists i and a_j\in\mathbf{R}, j\not=i, with P_i = \sum_{j\not=i} a_jP_j.
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 of the points if and only if all coefficients in the sum are non-negative.
The product of
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.
The boundary of PQ is \partial(PQ) = Q - P.
Exercise. Show \partial(PQ) = 0 if and only if P = Q.