It is not the case PQ is the line segement from the point P to the point Q. It is more like the line containing P and Q having magnitude Q – P. If PQ = ST, then PQS = 0 and so S = P + s(Q – P) for some s. Likewise, T = Q + t(Q – P) for some t. Hence ST = PQ + (t – s)PQ so s = t and Q – P = T – S.
This gives the first example of the boundary operator, ∂. It shows ∂PQ = Q – P is well defined.
Math is all about following your nose. Grassman had the notion that P2 = 0, where P is a point, was how to detect intersections and somehow that idea generalized to higher dimensions. Don’t forget that in Grassmann’s time the of notion working in general n-dimensional space was a novelty. He ran into trouble when he tried to define a product that went beyond the dimension of the space. We’ll get back to that later.
Given points P and Q, what is Q – P? Note that Q – P is not equal to R(t) = (1 – t)P + tQ for any t. Since PQ(Q – P) = 0, it should be on the line determined by P and Q. In this sense, it has a direction. If Q – P = T – S, then T = S + (Q – P). In this sense, it has a magnitude. Differences of points are vectors.
I tried various LaTeX WordPress math plugins and didn’t like any of them. In fact, I thought they were all pretty awful. Wedging ancient TeX technology into the world of HTML seems to be an impedance mismatch. Don’t get me started on MathML. That is a nonstarter.
Every time I tried a new WordPress theme things turned ugly so I rolled my own that uses straight HTML. I’m no Donald Knuth, but it does what I need. Here are some tests:
Use standard HTML entities for anything other than sub and superscripts
α^2^ + β^2^ = γ^2^
α2 + β2 = γ2
Use curly braces for grouping
a_{b_1_ + 2}_
ab1 + 2
Feel free to hardwire standard HTML if you think that will make things look better.
∫_a_^b^ f(x) dx
∫ab f(x) dx
Hermann Grassmann had the notion that it was possible to use algebra to prove geometric facts. Instead of using scalar Cartesian coordinates, he used actual points in space. Let E be Euclidean space and call the elements of E points. If P and Q are two points then PQ should represent something like the line from P to Q. If P, Q, and R are three points then PQR should resemble the triangle having vertices P, Q, and R. And so on.
If t is a scalar tP is the point P with weight t. Let R(t) = (1 – t)P + tQ, where P and Q are points and t is a scalar, then R(0) = P and R(1) = Q. In general aP + bQ is the point (a/(a + b) P + b/(a + b) Q = R(b/(a + b)) having weight a + b.
What is PQ if P = Q? If PQ resembles the line segment from P to Q, then the line from P to P is the point P. Assuming P = P2 we have P + Q = (P + Q)(P + Q) = PP + PQ + QP + QQ = P + PQ + QP + Q so 0 = PQ + QP and hence 0 = PP + PP = 2P for any point P. That might have been the first theory Hermann discarded before he came up with the idea PQ = 0 if P = Q. We have 0 = (P + Q)(P + Q) = PQ + QP so PQ = -QP. Since 0 = -0 we don’t have to discard this theory yet.
Note that PQR = 0 if P = Q or Q = R or R = P. More generally PQR(t) = 0 for all t with R(t) as above. In general, PQR = 0 if and only if the three points are colinear.
For example, P(Q + R) is the median of the triangle PQR from P to the midpoint of QR. The barycenter of the triangle is P + Q + R. Note P(Q + R)(P + Q + R) = PQP + PQQ + PQR + PRP + PRQ + PRR = PQR + PRQ = 0. This proves the medians of any triangle meet at a point, if you think about the symmetry for a second.
Grassman’s rule that P2 = 0 enables the detection of points not being in general position, i.e., they don’t determine an n – 1 dimensional object. If n points belong to a subspace of dimension less than n – 1 their product vanishes, not just for n = 2