title: Grassmann Algebra
author: Keith A. Lewis
institute: KALX, LLC
classoption: fleqn
fleqn: true
abstract: The algebra of space
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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 $\RR$ 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 $t\in\RR$.
This is an example of incidence involve three points. It means $R(t)$
meets the line determined by $P$ and $Q$ for all $t\in\RR$ if $P\not=Q$.
Exercise Show $Q - P\not= R(t)$ for any $t\in\RR$.
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 $t\in\RR$ 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 $P_0\cdots P_nR(t_1,\ldots,t_n) = 0$, $t_i\in\RR$.
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 $P = \sum{j = 0}^k t_j P_j$, $t_j\in\RR$, $P_j\in E$ 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 $t_j\in\RR$ 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 $\langle A\vee B\rangle = 0$.