Well, Iâve been through it all again (hence the delay) and you are right: what we think of as n-1 dimensional Euclidean hyperplanes are the grade-1 primitives of the Geometric Algebra G(n,0,1), and since thatâs a 1-up projective algebra, they are actually 0-dimensional geometric objects. Further, since G(n,0,1) is the minimum algebra that encodes them, it is justified to claim that such planes ARE 0-dimensional, and that points ARE n-1 dimensional. But in this, we need to decouple the notion of how many numbers it takes to specify a point within a linear subspace (our Linear Algebra definition of a subspaceâs dimensionâŚ), from the notion of the grade of the Euclideanly-transforming Geometric Algebra blade that represents that subspace within the GA. Normally in GAâs these two numbers go up together, but in the PGA case they run oppositely: thatâs what I meant in the statement you quoted.

Still, itâs a bit of a Copernican operation on our minds. How is it the case that maths gives such a different answer on dimension and primitivity to the one our minds conceive? It must be because there are these two different definitions of dimension at play. In normal linear algebra, a point has no substructure but a plane does because I cannot project onto linear subspaces from within a point. But in a Euclidean PGA, a plane has no substructure because it is not the wedge of anything. Would it be helpful to decouple the notions of dimensionality and primitivity?

I was intrigued by Eric Lengyalâs point: since meet and join are in both the standard algebra and its dual, why not also both dot and antidot, and antireversion, and thus the antiproduct too; and indeed it may be convenient, notationally, educationally, and otherwise, to have them all. This enables Euclidean geometric calculations with points as vectors and so on, and seems to confer equal status on the standard algebra P(R(3,0,1)). But properly speaking G(3,0,1) is generated by its own geometric product alone, not by its anti-product, and that product, when used to exponentiate bivectors, moves vectors around as if they are Euclidean planes, and trivectors around as if they are Euclidean points, exactly as you say. Since in a GA there is only the grade of the object and the space it inhabits to indicate its dimension, and since the Euclidean object encoded by a blade in an n-dimensional Euclidean PGA can be said to have dimension n-1, it follows that, according to GAâs, the planes in the space we live in are 0-dimensional. Furthermore they are primitives, because the constructive operation in a GA is the wedge, and these planes are the base vectors. It follows, in a further mind-bend, that the primitives of G(3,1,1), what you might call the projective Minkowski/space-time algebra, are 0-dimensional volumes. Interestingly there we have a set of 10 bivectors (dual to 10 trivectorsâŚ) available to exponentiate and move things around STA-style. I project that Lorentz will transform in his grave.

In the above version of your argument, which I wrote to help myself understand (sorry if it makes things worse for anyone else!), I focused on the geometric product and stayed away from the issue of when and whether meet or join is the constructive operation, since that confusingly shifts around in the way you indicate. I found your 2011 paper to be very helpful in understanding your perspective.

But still, we are left grasping for further intuition about why all this is the case. In your

GAME talk at 59:20, you say âthe dual metric is less degenerateâ, when talking about the issue of the differences between the standard P(R(3,0,1)) and dual P(R*(3,0,1)) algebras. What does this mean? I see what it means to projectivise the exterior algebra of a vector space or its dual, but I still donât really see in my mind why the isometries, or angle and distance preserving transformations, arise in the latter rather than the former. Maybe the answer is âthey just doâ, but I know you have a more precise notion of the form of the asymmetry at play. In what exact sense is the dual metric more degenerate? And what is it about this extra degeneracy of the dual metric that makes the exponentiated bivectors perform correct Euclidean transformations on the dual objects of G(n,0,1) but not on the standard ones? It must have something to do with the different rotational symmetry groups that points, lines and planes have, and how this meshes with the two algebras. It also must relate to your statement at 59:35 that the motion group (of exponentiated bivectors) is generated by reflections in planes, not points.

NOTE ADDED: I did find some helpful material on my last paragraph of questions, in sections 3.6.1 to 3.7.1 of this paper. I didnât quite see the answer to my questions though. By the way, just to make sure I am correct: I know that if I want to implement P(R*(3,0,1)) in G(3,0,1) I transform trivectors, using exponentiated bivectors, while imagining them to be points (etc). But if I want to implement P(R(3,0,1)) as a GA, I use exactly the same algebra G(3,0,1) but I just imagine that the vectors are points, and so on, instead? ie: from the GA point of view there is only one algebra, and the two schemes arise from the interpretations of the blades.

(My apologies for writing at such length. It just seems these are very important things to understand. By the way, Iâm still thinking about physics and the role of the weights and I will get back on this topic at some point.)