Feb 22, 2026
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 Euclidean geometry by associating points in space with numerical coordinates, but Descartes only considered two and three dimensions. Grassmann predated Einstein 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 was a secondary-school teacher who wrote in an abstruse style that top mathematicians of the time ignored. It wasn’t until (Peano 1888) wrote an exposition of the theory that his ideas began to gain traction. Peano, like Descartes, only developed the theory in two and three dimensions but used a notation more accessible to mathematicians of the time.
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 addition and multiplication allows writing x + y + z and 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.
The product in \mathcal{G}(E) is Grassmann’s exterior product. Italian differential geometers decided to call it the wedge product, written as P\wedge Q. This was an unfortunate historical accident. The exterior product is analogous to a union of sets so P\vee Q is a more appropriate notation. Grassmann’s regressive product, defined below, is analogous to the intersection of sets so P\wedge Q would be the natural notation for that.
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 barycentre of aP and bQ.
Let R(t) = P + t(Q - P) = (1 - t)P + t Q, 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 congruent 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 coordinate-free calculation using the Grassmann algebra.
David Hilbert noticed Euclid’s axioms were insufficient to determine when a point lies between two points. Using Grassmann algebra we can define R is between P and Q if and only if PQR = 0, PR/PQ > 0, and RQ/PQ < 1.
A classical result from Euclidean geometry is the three medians of a triangle meet at the same point. A median is the line from a vertex of a triangle to the midpoint of the opposite side.
Three points P,Q,R\in E determine a triangle. The midpoint of PQ is {(1/2)(P + Q)} and {(1/2)(P + Q)R} is the median from the midpoint of P and Q to R. The centroid (1/3)(P + Q + R) 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 triangle meet at the centroid.
Grassmann’s algebra works in any number of dimensions. Any non-zero product of points is an extensor (also called a blade, or flag). 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 extensor 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 P = \sum_0^n (P_0\cdots P_{i-1} P P_{i + 1}\cdots P_n/P_0\cdots P_n)P_i.
Exercise. Show the coefficients sum to 1.
This is essentially Cramer’s rule. TODO: ref
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 P belongs to the convex hull of the points if and only if all coefficients in the sum are non-negative.
Rene Descartes invented a method to reduce the straightedge and compass constructions of Euclid to algebraic computation by using a coordinate system. Although Descartes only considered two and three dimensional space, any point in n-dimensional space \mathbf{R}^n can be described by a tuple (x_1,\ldots,x_n) where x_i\in\mathbf{R}, 1\le i\le n. In La Géométrie (1637), Descartes showed any length constructed from a unit length must come from a finite sequence of additions, subtractions, multiplications, divisions, and square roots. A famous problem from antiquity is “squaring the circle”: given a circle construct a square having the same area. This is equivalent to constructing \sqrt{\pi}. It took 245 years until (Lindemann 1882) showed \pi is not the root of any non-zero polynomial (transcendental) when Descartes result can be used to prove squaring the circle was impossible using Euclidean geometry.
Exercise. Show if x\in\mathbf{R} is transcendental then so is \sqrt{x}.
Hint: ???
Grassmann greatly generalizes cartesian geometry. If we choose any independent points (P_i) and origin O in E$ then the Decartes’ tuple x = (x_i) corresponds to the point P(x) = (1 - \sum_i x_i)O + \sum_i x_i P_i.
Calculation can be greatly simplified by noting O\Pi_i (x_iO + P_i) = O\Pi_i P_i_.
The complement of an extensor A with respect to a volume extensor \Omega is the unique extensor \overline{A} for which A\overline{A} = \Omega.
If E has dimension n and P_0\cdots P_n\not=0 then the complement of \Pi_{j\in\{0,\ldots,n}
Grassmann used the notation \overline{A} for the unique solution to A\overline{A} = \Omega whenever A is an extensor.
Exercise. Show \overline{\overline{A}} = A.
This can be extended linearly to \mathcal{G}(E). The regressive product is defined by A\vee B = \overline{\overline{A}\overline{B}}