May 27, 2026
Euclidean geometry was put on an algebraic footing when René Descartes demonstrated how to replace ruler and compass constructions in Euclidean geometry by associating points in space with numerical coordinates, but Descartes only considered two and three dimensions. Grassmann extended this to Euclidean space of any number of dimensions with the audaciously simple rule: “The product of two points is zero if and only if they coincide.” Grassmann predated Einstein in treating all points in space of any dimension 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.
Alas, Grassmann was a lowly secondary-school teacher who wrote in an abstruse style that top mathematicians of the time failed to pay attention to. 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 in his attempt to popularize it.
Grassmann considered the set E of points in Euclidean 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 would be 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. We use a period for that, P.Q.
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 so the vector/point at infinity Q - P is “on” the line PQ.
If PQ = tST where t\in\mathbf{R} and {P,Q,S,T\in E} we say PQ and ST are congruent and write {{\frac{PQ}{ST}} = t}.
Exercise. Show \frac{(Q - P)Q}{PQ} = -1 and \frac{P(Q - P)}{PQ} = 1.
Exercise. For any R with PQR = 0 and PQ\not=0 we have R = \frac{PR}{PQ}P + \frac{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. See (Rouse Ball 1892) for an entertaining, but incorrect, proof that all triangles are isosceles. Using Grassmann algebra we can define R is between P and Q if and only if PQR = 0, \frac{PR}{PQ} > 0, and \frac{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 {(P + Q)/2} and {R(P + Q)/2} is the median from R to the midpoint of P and Q. 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 triangle meet at the centroid.
If there exist P,Q\in E with PQ\not=0 we consider the problem of when PQ = ST for S,T\in E. Since PQS = 0 and PQT = 0 we know both S and T are on the line determined by PQ so S = R(s) and T = R(t) for some s,t\in\mathbf{R}. Since ST = R(s)R(t) = (t - s)PQ we have t - s = 1.
Exercise. Show PQ = TS if and only if Q - P = T - S and either PQT = 0 or PQT = 0.
Hint: If PQT = 0 and Q - P = T - S then PQS = PQ(T + P - Q) = 0.
Exercise. Given P,Q\in E then \frac{TS}{PQ} is the length of the vector T - S.
Hint: If S = R(s) and T = R(t) the length of T - S is t - s.
Exercise. If E is 1-dimensional then PQ = ST if and only if Q - P = T - S.
Hint: For any S,T\in E we have PQS = 0 and PQT = 0.
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 we write t = \frac{P_0\cdots P_k}{Q_0\cdots Q_k} and say the extensors are congruent.
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 \frac{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.
We say an extensor \Omega\in\mathcal{G}(E) is a volume element if \Omega\not=0 and P\Omega = 0 for all P\in E. The order of E is the number of points in the extensor and the dimension of E is one less than its order.
The complement of an extensor A with respect to a volume element \Omega is the unique extensor \overline{A} for which A\overline{A} = \Omega.
TODO: Show \overline{A} is unique.
If E has dimension n and P_0\cdots P_n\not=0 then the for any subset I\subseteq\{0,\ldots,n\} the complement of P_I = \Pi_{i\in I} P_i is \overline{P_I} = \pm\Pi_{i\not\in I} P_i where the sign is \frac{P_I\overline{P_I}}{P_0\cdots P_n}.
Exercise. Show \overline{\overline{A}} = A.
This can be extended linearly to \mathcal{G}(E). The regressive product A.B is defined by \overline{A . B} = \overline{A}\,\overline{B}