Venn diagram method of seeing intersections in n-space...

Higher-dimensional geometry (previously "Polyshapes").

Venn diagram method of seeing intersections in n-space...

Postby Paul » Sat Sep 09, 2006 11:20 pm

Hello all,

Haven't posted here in awhile...

I posted this message, Venn diagram method of seeing intersections in n-space... bivectors in 4-space... on the Math Message Board.

Unfortunately, few there show much interest in higher-dimensional geometry. No one has responded to my post.

I've answered some of my questions myself, but I'm still confused about what precisely Lounestro might have meant concerning that decomposition of a non-simple 4D bivector...?

There's just something odd about his statement... and I'm just unsure precisely what he meant. For one thing, does one need to perform the decomposition on the right-hand side in order to have two orthogonal bivectors? Aren't the two bivectors on the left-hand side also orthogonal?

And, also... as I mention in one of my follow on responses... I'm also pretty sure Lounestro is using the convention that a bivector doesn't refer to a specific parallogram... but rather, that all we keep is its directed magnitude, so we only know that the bivector represents some directed area in the 2D-plane spanned by its component 1-vectors.

Anyway... the long and the short of it is that I don't Lounestro's assertion is true for general 2D-planes in 4-space, which utilizing the convention I wish to use make reference to a specific parallogram by keeping the information on the component 1-vectors and defining that area as the specific area defined by the bivector.

Further, to define arbitrary 1-vectors one needs two position vectors... or else all 1-vectors are limited to going through the origin. That is, there's no distinction between a 1-vector and all it's translates.

One way or another, I don't believe Lounestro's assertion could hold with bivectors describing planes not going through the origin...

Does anyone have ideas about what precisely Lounestro meant? And/or any comments, or observations, on my Venn diagram method of seeing these intersections in n-space...?
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Postby pat » Sun Sep 10, 2006 3:39 am

For now, I just want to get myself on the notification list for this topic... I have a bunch to say, but I'm going to have to sort through some things first.

For now, let me say that I love the Venn diagram description. It reminds me a bit of Gale diagrams, but yours is much more intuitive. I can't find a satisfactory link on Gale diagrams either, so unless you have Gunther M. Zeigler's Lectures on Polytopes, it'll have to wait until I get some more time.
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Postby Paul » Sun Sep 10, 2006 4:37 am

Hello Pat,

Thanks for your response.

Yes... I do have that book. I recall it's also like almost the only book that I have that discusses Face Lattices of polytopes too.

I'll take a look in "Lectures on Polytopes" about the Gale Diagrams.
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Postby wendy » Sun Sep 10, 2006 7:49 am

I looked at this, and i could not see the relevance of venn diagrams.

In four dimensions, it is possible for two hedrices (2-flats) to cross at a point. These can be done to make clifford-parallels, or at some askew angle.

It is useful to consider the space of great arrows. A great arrow is a great circle with a direction. One considers the set of great arrows around a point, which consists of in turn, all circles of fixed diameter centered on that point, and thence every hedrix through that point.

The topology of the points representing great arrows forms a "bi-glomohedric prism". This is a terix (4-fabric) in 6d. It can be read as a grid, where a point corresponds to an "x" sphere, and a "y" sphere.

The shape is then x1^2 + x2^2 + x3^2 = y1 ^2 + y2^2 + y3 ^2 = r^2

Two points on the same x axis represent left-parallel curves. Two points on the y axis represent right-parallel curves.

For any two great arrows, A,a and B,b there is a pair of great arrows that one is left-parallel to, and one is right parallel to, ie A,b and B,a.

If you have a hedrix of fixed orientation (as happens in complex numbers and quarterions), then one of these vectors is fixed, and the other divises into A1 and 1a. These are of course not orthogonal, but you can construct any rotation by a left-parallel clifford rotation, followed by a right-parallel one. This is what the e1e2 stuff means.

What you have here, then is trying to imagine the horoterix (E4) by using quarterions, in much the same way as you might imagine the euclidean plane (horohedrix) by using complex numbers. The latter is decisively a better fit, though, because the E4 assumes that there is a kind of directed rotation, when in fact, there is not.

The actual complete space of rotations in 4d is actually a petix (5-fabric) in 7 dimensions. The name for this shape is "bi-glomohedric pyramid". One must imagine that the representation of SO4 does not represent the complete rotation space in four dimensions for this reason. SO4 is the set of orthogonal matrices.
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Postby Paul » Tue Sep 19, 2006 7:45 am

Hello Wendy,

Thanks for your response.

I'm sorry I didn't answer sooner. I've been busy, but also, many of your terms are unfamiliar to me.

First of all, I thought all directions of 4-space could be mapped to all the points of 3-space... just as S1 can be mapped to the real line, and S2 can be mapped to the plane. It may have to be projective space... perhaps to include vertical direction.

Clifford parallels...? I'm not sure what you mean by this. Does it have anything to do with Hamiltion choosing a left-handed system for quaternions? Of course, I believe the right-handed system produces the unfortunate ijk = 1.

Great arrows...? That kinda sounds like spherical vectors?

"hedrix"... like polyhedron? Does it mean an n-hyperplane? So, perhaps a polyhedron is the enclosed hull of intersectioning planes?

Do you still have your glossary of terms online? Do you get these terms from old math books?

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Postby wendy » Wed Sep 20, 2006 8:25 am

There are not enough words to go around, so i invented a lot. The basic notion of space and subspace is that it is a "fabric" (ie -ix), from which you cut "patches" (-on). So a "hedron" is a 2d patch, of which "poly" is (many with closure). A "poly hedron" is then "many 2d patches with closure".

One then has names for 2d fabric in any dimension, eg "hedrix". Plane suggests division (cf plain = what you stand on). The idea is that the terminology is consistant in higher dimensions, and therefore each stem must be applied to one meaning only, and not to several.

Part of the richness of the PG is that there are explicit words for, say the surface of an 5d sphere, regardless of whether this is S4, or any 5d sphere (in 5 or higher dimensions). The surface is a "glomoterix" (globe-shaped 4-fabric), and the body is a "glomoteron" (globe-shaped 4-patch). Only patches bound.

Note also we can describe a tiling of patches, by using 'aperi' (a = without), peri = bounds of referenced points in a space. So a tiling of hexagons (2-patches or 'hedra') is an aperihedron.

Much of the confusion with the 'tiger' (tri-circular torus), is because the thinking that the 2d margin (ie N-4 fabric), somehow divides space, and therefore can not be enclosed.

You can indeed produce a map of directions in 3d, of all 4d vectors, by using the same sorts of mapping that render S2 onto E2. You can, for example, use azithmal projection (projected through the centre of a sphere, giving a half-sphere, but preserving great circles to lines), or stereographic (preserves angles, the centre of projection is the opposite pole), or orthogonal.

Projections like mercartor's do not make much sense, because the nature of rotation is different in four dimensions, to three.

Clifford-parallels does in fact correspond to hamilton's quarterions. If you pick a quarterion, say q, the rotation from 1 to q, will rotate every point by an identical angle, from x to qx. You can reverse the all-rotation, and make it into a great-arrow rotation, by using x => qxq'.

Great arrows are great circles with directions on them.

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Postby Paul » Wed Sep 20, 2006 9:53 pm

Hello Wendy,

I'm wondering... do you see yourself as more a scientist, or an artist?

Take care,
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Postby wendy » Thu Sep 21, 2006 7:57 am

The maths is rather hard, and i use clever tricks to avoid it.

Visualisation is easy, and the numbers talk to me, so i guess a bit of an artist. More often deep insight rather than serious plodding, but deep insight is sort of like a 4×2. If you get beaten over the head enough, you will see the answer!

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