## 4D visualization starting with 3d objects in 4D

Discussions about how to visualize 4D and higher, whether through crosseyedness, dreaming, or connecting one's nerves directly to a computer sci-fi style.

### 4D visualization starting with 3d objects in 4D

I think that trying to visualize 4D starting with tesseract or 3-sphere is rather hard. To really appreciate 4D we have to start with our familiar 3d objects, like a cube, a sphere, a tetrahedron. The surprising thing one notices right away is that the 3d objects appear rather flat in 4D. But not completely flat, like a square is flat. They still maintain their 3d space.

This may take some time, but once visualizing of a cube is accomplished, one can see a cube made of, say 27 cubes (3 per edge), which I described on one of the current threads. Then, a cube made of 8 cubes (2 per edge) becomes interesting too, because a tesseract is made of 8 cubes. So, the task becomes to arrange the 8 cubes in such a way that they make up a tesseract. To this effect I bumped into a paper today describing unfolding of the tesseract: http://unfolding.apperceptual.com/

The same goes for a tetrahedron. 5 tetrahedra make up a simplex.

I believe this is a superior method than using the analogies. This forum shows that analogies do not actually help people visualize 4D. Doing projections onto 3D can be misleading as well. This method works for me.
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### Re: 4D visualization starting with 3d objects in 4D

Found a better site where tesseract is folded: http://www.herenow4u.net/index.php?id=cd6129

When a tesseract is unfolded, 8 cubes are forming it.
The planes in the depiction, signified as a, b, c, d, e are laying on each other when folded.

Folding of a tesseract: One out of 8 cubes forming the tesseract stays now in 3D. The other 7 cubes of the tesseract are moved through 4D and folded

Another good one from wiki: http://www.search.com/reference/Tesseract

The "vertex first" is sort of how I see it.

But this is the best, from wiki again. Here you can see how those cubes are folded. Projected onto 3D, some cubes sort of intersect.

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### Re: 4D visualization starting with 3d objects in 4D

In fact, all cubes intersect. It's a squashed polytope, after all. If the cubes don't intersect with each other, then they are projected onto planes, as in the first three projections (face first, margin first, and edge first). The hexagon at the top is a squashed cube.

The vertex-first view is a tetrahedral antitegum.

There are similar figures in four dimensions, such as \3\ = m3o3o3m, which consists of five forshortened tesseracts, arranged completely around a point, and so forth. The general class \2\, \3\, \4\, and so forth, will always tile in N space.

In 3d the rhombic dodecahedron fills both roles of 1\Q double-cube and \2\ dual of SPC. In four dimensions, these are different: 1\1Q is the double-tesseract or 24choron, while \3\ has 20 faces only. They present in the identical fashion as the tesseract vertex-first.
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### Re: 4D visualization starting with 3d objects in 4D

Here is another good page with very good animations: http://mscoconut.blogspot.com/2009/10/tesseract.html Towards the bottom of the page, there is an animation that shows how the 8 cubes fold into tesseract.

The page has a good breakdown of the 8 cubes in color. Plus good animations that show each cube, like this one: http://www.flickr.com/photos/ai-momo/40 ... 6/sizes/o/

But this is the best by far! Exploring the 8 cubes of tesseract from inside out, a video from a gamer.
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### Re: 4D visualization starting with 3d objects in 4D

The particular video is actually a topological map of the surface of a 4d surface. It is no more 4D than supposing that a cube, unfolded like a chess-board, is three dimensions. You can see, for example, that the same room neighbours two rooms which is unmayly in three dimensions. This is because there are only three cubes around a square, not four.

Still, it is hardly 'four dimensions'. You have in these no more left the surface of the tesseract.

I suppose you should say what side of hyperspace you are viewing these from?
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### Re: 4D visualization starting with 3d objects in 4D

So, this gamer video, http://www.youtube.com/embed/lLg3oLbwRsQ. You can explore the tesseract from inside. Note the small model with 8 colored cubes in the upper right of the screen. It shows you where you are. And so, from inside the 3-space, the cubes look like regular cubes. It is their relationship to each other, i.e. which ones are adjacent to each other is interesting. From which cube you can get into which and in what sequence.

Also I was looking at this stereo projection

Note that there are 2 sets of 4 cubes each: one set is seen if you follow the blue central edges and the other set is seen if you follow red. And while on this projection the 2 sets intersect each other, the 4 cubes that make up each set do not intersect with each other in 3D. In fact, they are adjacent to each other. So, what we see are 2 hyperplanes, each consisting of 4 cubes. In reality they are parallel. Their intersection on the projection reflects the fact that one hyperplane is in front of another to our POV.

In any rate, the opposite pairs of cubes (right-most and left-most, or bottom and top) do not intersect each other and neither are adjacent to each other.

And so, the other very interesting thing to notice about the tesseract is that this 4d object apparently consists of 8 3d cubes. I.e. the cubes completely fill its 4d space. But they are 3d themselves. I find this fact fascinating.

Same goes for the tetrahedron. 5 tetrahedra, i.e. 5 3d things, make up one simplex, a 4d thing. How can it be? Only because they are arranged in 4D?

This is entirely different from planes in 3D. No matter how many planes we stack on top of each other in 3D, they still will amount to 0 thickness. But stacking 3d things in 4D, apparently, can make up a 4d object?

For example. If we cheat a bit and take 8 equal cubes with edge=8 units, then 8 to the 4th = 4096 and this is the 4-volume of a tesseract with edge = 8units. And 8 cubed = 512 == the 3-volume of one of the cubes. Since we have 8 of them, we multiply 512 by 8 and get 4096 as well. So, in this particular (and unique, lol) case where the edge = 8 units, the 8 3-volumes == 1 4-volume.
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### Re: 4D visualization starting with 3d objects in 4D

PS
Of course, this is of the same category as 6 squares with edges=6 units will equal to 6 cubed.

Area and volume are such different things... they are not compatible. But still, where is that bulk, that 4-volume in tesseract? The 8 cubes that bound it on 8 sides have their edges aligned in 4D. There is no space in between them. They are all adjacent to each other, cutting the 4D space into 8 3d subspaces. Which is different from faces of a cube in 3D. There is 3d space in between those squares.

Oh! is the bulk in between the 2 hyperplanes? In between the outer faces of the cubes? I can sort of see it in between the blue and red inner edges:
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### Re: 4D visualization starting with 3d objects in 4D

PPS
Still, from what I read, it appears to be true in regard to simplex. It does consist of 5 tetrahedra. There is nothing in between the edges. All faces are triangles. There is nothing in between.

So, it may be the case with the tesseract as well. I am not sure yet. It appears that simple stacking of 3d things in 4D will make up a 4d object. And by stacking, I mean like tiles in 2d, like 8 cubes making a larger cube in 3D. The 3d subspaces of 4D somehow amount to 4D bulk.

That's right. You can break the 4D space into 4 3d subspaces, which is not analogous to planes in 3D.
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### Re: 4D visualization starting with 3d objects in 4D

Hey it's me again.

I've just read this thread and, no offense, but the only thing I want to do is to put an image of a 4D facepalm.
You're missing a huge piece of the puzzle to understand 4D. And this piece is the 4th dimension itself.
I believe you have been partly misled by the commonly made mistake/missing step in the 3D->2D analogy that I have pointed out here.
Aside from human mistakes like that, the analogy is perfectly valid and you should stop being so wary of it. It's a great thing that we are able to use this downward analogy (if we were in a 2D or a 1D world, we couldn't since the analogy would involve 0D which is a special case) and it helps a lot in understanding 4D visualization.

So, this gamer video, http://www.youtube.com/embed/lLg3oLbwRsQ. You can explore the tesseract from inside.
Not at all, this is not a perspective or orthogonal projection of a tesseract. It's a self-wrapping 3D map of the surface of a tesseract. As the title of the video says too, you are exploring just the surface of the tesseract. The POV is in the surface. As if, by analogy with exploring the surface of a cube, the POV was in the faces, looking straight at the edges with a 1D display.

And so, the other very interesting thing to notice about the tesseract is that this 4d object apparently consists of 8 3d cubes. I.e. the cubes completely fill its 4d space. But they are 3d themselves. I find this fact fascinating.
It's so fascinating because you take a wrong assumption for reality. Though reality isn't much less fascinating. The eight cubes face to face constitute a surface with zero 'hyperdepth' enclosing a 4D space. You can't picture it? Well this is 4D! We can't picture it because it's unknown to us. The same way a person who sees only in shades of red and blue can understand the notion of other colors but can't get a faithful mental image of a colorful scene.

This is entirely different from planes in 3D. No matter how many planes we stack on top of each other in 3D, they still will amount to 0 thickness. But stacking 3d things in 4D, apparently, can make up a 4d object?
A cube is flat in 4D, it has zero length on the W axis. Stacking cubes doesn't fill up a 4D space. It's the same as stacking squares in 3D.

For example. If we cheat a bit and take 8 equal cubes with edge=8 units, then 8 to the 4th = 4096 and this is the 4-volume of a tesseract with edge = 8units. And 8 cubed = 512 == the 3-volume of one of the cubes. Since we have 8 of them, we multiply 512 by 8 and get 4096 as well. So, in this particular (and unique, lol) case where the edge = 8 units, the 8 3-volumes == 1 4-volume.
x(3-volume unit) = y(4-volume unit) is a false equation. The unit for 3-volumes and 4-volumes is different.
The 4-volume of a tesseract with cube faces of volume 8 unit³ is... 8 units. Simple enough.

Area and volume are such different things... they are not compatible.
The area is the 2D analog of the length of a 1D object and of the volume of a 3D object. They are the same thing in a different n-dimensional space, they have analog properties.

The 8 cubes that bound it on 8 sides have their edges aligned in 4D. There is no space in between them.
Really, get this idea out of your head. Don't think that you are somehow seeing in 4D when you look at a projection of a 4D object. There's no way to directly see the 4-volume when looking at a projection of a 4D object, since your POV is in the 3-plane of the projection. It's the same as looking edge-on at the drawing of a cube.
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### Re: 4D visualization starting with 3d objects in 4D

Ovo wrote:[...] I believe you have been partly misled by the commonly made mistake/missing step in the 3D->2D analogy that I have pointed out here.
[...] Don't think that you are somehow seeing in 4D when you look at a projection of a 4D object. There's no way to directly see the 4-volume when looking at a projection of a 4D object, since your POV is in the 3-plane of the projection. It's the same as looking edge-on at the drawing of a cube.

And the blame partly rests on me, for not making it clear that when we look at a (2D image of a) 3D projection of a 4D object, we aren't seeing the projection from a "native" 4D point of view at all, but from the side, as it were. A 2D image of a 3D projection of a 4D object is about as helpful in visualizing 4D as a 1D projection of a 2D projection of a 3D cube is in visualizing 3D.

When we draw a diagram of the cube, we may draw something like this (say on a piece of paper):

But, seen from the side, it looks like this:

This is about as helpful in understanding what a cube is, as a 2D image of a 3D projection of a 4D tesseract.
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### Re: 4D visualization starting with 3d objects in 4D

Ovo wrote:Hey it's me again.
So, this gamer video, http://www.youtube.com/embed/lLg3oLbwRsQ. You can explore the tesseract from inside.
Not at all, this is not a perspective or orthogonal projection of a tesseract. It's a self-wrapping 3D map of the surface of a tesseract. As the title of the video says too, you are exploring just the surface of the tesseract. The POV is in the surface. As if, by analogy with exploring the surface of a cube, the POV was in the faces, looking straight at the edges with a 1D display.

The video underscores that point that 3d subspace of 4D feels the same as our homey 3D. When you get into it, it does not look flat at all.

Ovo wrote:The eight cubes face to face constitute a surface with zero 'hyperdepth' enclosing a 4D space.

Yes, supposedly. While our router still worked, I spent whole night looking for cross sections of tesseract or simplex, or any 4d object. Hoping to see an arrow pointing at its middle with caption: here is where the 4d bulk lives. Everyone shows fancy projections and no one wants to do a boring cross-section through the middle of a tesseract to show its bulk.

If you see such a cross-section of tesseract or simplex, please let me know.

In 3D, the analogy is to cut a cube in the middle. If the cutting plane II to a pair of faces, we get a boring square. Just like a face of the same cube. Except that the edges of this square represent the cube's faces and its area is a section of 3d volume. That's where volume of a cube lives. Why no-one does the same for a 4-cube? Or a simplex? I want to see such a cross-section.

Ovo wrote:
This is entirely different from planes in 3D. No matter how many planes we stack on top of each other in 3D, they still will amount to 0 thickness. But stacking 3d things in 4D, apparently, can make up a 4d object?
A cube is flat in 4D, it has zero length on the W axis. Stacking cubes doesn't fill up a 4D space. It's the same as stacking squares in 3D.

See, here you're mistaken. That's where 4D is vastly different from 3D. Because you can break up a 4D space into 4 3d subspaces. So, take a tesseract and place its vertex at the origin of 4 axes. (It's like putting a cube's vertex at (0,0,0); thus its 3 adjacent faces coincide with XY, XZ and YZ planes.) With the tesseract the result is this: XYZ, XYW, XZW, YZW. Now, the difference with 3D is that you can put a cube into each of those 4 subspaces, and they will fill in the bulk. Because you will orient the additional cubes in the 4th direction.

It's like the 8 cubes of tesseract. The 8 cubes can be stacked in 3D or the same cubes can be stacked in 4D.

Draw a 4D graph. And break it into 4 3d subspaces for your convenience. (instead of drawing I recommend 4 corners of real boxes representing 3 planes; label them). Play with these real 3d subspaces and try to imagine all 4 of them fitting perfectly. Alas, not possible in 3D, but you can examine each pair of 3d subspaces separately, where their planes coinside, say XYZ and XYW are adjacent at XY plane. You realize that there is a cube sitting in each of subspaces: XYZ, XYW, XZW, YZW, with its vertex pointing at the origin of the XYZW axes. The faces of all those 4 cubes are adjacent in 4D and they are stacked continuously.

That's because volume IS a volume. But area is not volume. And even though, granted, 3-volume is not the same as 4-volume, and mathematically in both cases, going 2D->3D or 3D->4D it's just another power, but the result is not the same. You can stack n-volumes and they will make up (n+1) volumes for n>2. You simply stack them into the n+1 direction.

Ovo wrote:
Area and volume are such different things... they are not compatible.
The area is the 2D analog of the length of a 1D object and of the volume of a 3D object. They are the same thing in a different n-dimensional space, they have analog properties.

Here you are too quick to draw wrong conclusions. I am off in search of a theorem that n-volumes can add to (n+1) volumes for n > 2.

Ovo wrote:
The 8 cubes that bound it on 8 sides have their edges aligned in 4D. There is no space in between them.

Show me where the bulk lives. If it is there, you can do a cross-section of a tesseract to demonstrate it.
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### Re: 4D visualization starting with 3d objects in 4D

PS
see, it turns out that 3D ≈ 0.25 of 4D for the same length l. But with 2D and 3D, 2D = 0 of 3D.

With 5D, 4D ≈ 0.2 of 5D
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### Re: 4D visualization starting with 3d objects in 4D

Here is the stereo 3D projection of 4 cubes stacked in 4D. I simply removed 4 edges from the tesseract projection above. What is left is 4 adjacent cubes that sit in the 4 3d subspaces of 4D, as discussed above:

See how all 4 cubes are stacked continuously? Two rows of this make up a tesseract. Unless there is 4-volume in between those 2 rows, which I don't think so. Because tesseract has 8 sides. Which means that I can take any set of continuous 4 cubes and call it a row. Just as there is no space in between the 4 cubes above, the same will apply for any other combination. Thus the is no 4d bulk sitting somewhere in between there. 4 3-volumes add up to a 4-volume. You just have to stack them right. That's because 3-volume is never 0. An area has 0 volume though. Thus there is a huge difference between 3D and 4D, and most analogies are moot.
Last edited by 4Dspace on Sat Jul 28, 2012 5:33 am, edited 3 times in total.
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### Re: 4D visualization starting with 3d objects in 4D

Here is another "row":

So, we stack 2 cubes one on top of another, just like in 3D (that's the front cubes on the projection, a smaller one on the bottom and a larger one on top). The 3rd cube is tricky: its 2 adjacent faces, which are perpendicular to each other, go along the straight wall made up of 2 continuous faces of the 2 stacked cubes (that's the largest cube behind the front 2; you can see one of its faces on the right). The 4th cube simply fits on the left into the concave niche that results after you put the first 3 cubes right (that's the squashed cube on the left, and all of its faces are obscured by the 2 front cubes. You can see its 3 edges though: it shares the bottom visible edge with the bottom front cube and its other 2 visible edges it shares with the top front cube).

Nice?

But the first, "red" row is easier. It's projection is less distorted than this one.
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### Re: 4D visualization starting with 3d objects in 4D

And here is some quotes from http://books.google.com/books?id=hZULAAAAYAAJ Geometry of Four Dimensions, in regard to a simplex (called pentahedroid there, which I changed) and its 5 tetrahedra, which is essentially the same as tesseract and its 8 cubes above, which confirms what I was saying above:

p 59

In a simplex each tetrahedron is adjacent to each of the other four.

It may not be very difficult to think of two adjacent tetrahedrons, even though they lie in different hyperplanes...

p 66

In a simplex, for example, there are 5 tetrahedra whose 20 faces fit together in pairs, each tetrahedron having a face in common with each of the other 4. We can take any one of these tetrahedra and place the other 4 upon it, all in one hyperplane, and then we can turn the 4 outside tertrahedra away from this hyperplane without separating them from the 5th or distorting them in any way, until we have brought together every pair of corresponding faces. The 5 tetrahedra together with their interiors now enclose a portion of hyperspace.

Another quote from the same book, underscoring the differences between spaces with even or odd number of dimensions, i.e. 3D and 4D:

p 14

There are also many properties in which space of an even number of dimensions differ from spaces of an odd number of dimensions, and these differences would hardly be recognized if we had only the ordinary geometries. Thus in space of an even number of dimensions roatation takes place around a point, a plane, or some other axis-space of an even number of dimensions, while in space of an odd number of dimensions the axis of a rotation is always of an odd number of dimensions (chap.IV)

So, there are more similarities between 3D and 5D than between 3D and 4D. That's why I say that most of the analogies used are moot.
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### Re: 4D visualization starting with 3d objects in 4D

I have this book, too.

The main similarites between odd dimensions vs even dimensions, do not affect most of the analogies.

What happens, for example, is that the 'hairy ball', must have a crown or vortex in an odd dimension, but not in an even dimension. Another difference is that solid space, for having an even dimension, is of the oppostit sign to its dual nulloid (of -1 D), so Euler's characteristic equation reduces to 0, rather than 2, eg

cube 8v - 12e + 6h = 2 vs tesseract 16v - 32e + 24h - 8c = 0.

One can not use euler's equation to solve the number of faces of a 4d polytope, since this equation for the tesseract is 2Fv - 4F e + 3F h - Fc = 0, for all values of F. This is similar to the polygon equation, where P v - P e = 0 for all P, and different to 3d, where one can derive v,e,h from the equity of 2.

On the other hand, the application of equalities to define planes (ie a N-1 space has one equal sign, and thus divides space, while an N-2 space has two equal signs, and doesn't divide allspace), applies throughout.
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### Re: 4D visualization starting with 3d objects in 4D

Ovo wrote:Hey it's me again.
And so, the other very interesting thing to notice about the tesseract is that this 4d object apparently consists of 8 3d cubes. I.e. the cubes completely fill its 4d space. But they are 3d themselves. I find this fact fascinating.
It's so fascinating because you take a wrong assumption for reality. Though reality isn't much less fascinating. The eight cubes face to face constitute a surface with zero 'hyperdepth' enclosing a 4D space. You can't picture it? Well this is 4D! We can't picture it because it's unknown to us. The same way a person who sees only in shades of red and blue can understand the notion of other colors but can't get a faithful mental image of a colorful scene.

This is entirely different from planes in 3D. No matter how many planes we stack on top of each other in 3D, they still will amount to 0 thickness. But stacking 3d things in 4D, apparently, can make up a 4d object?
A cube is flat in 4D, it has zero length on the W axis. Stacking cubes doesn't fill up a 4D space. It's the same as stacking squares in 3D.

Area and volume are such different things... they are not compatible.
The area is the 2D analog of the length of a 1D object and of the volume of a 3D object. They are the same thing in a different n-dimensional space, they have analog properties.

The 8 cubes that bound it on 8 sides have their edges aligned in 4D. There is no space in between them.
Really, get this idea out of your head. Don't think that you are somehow seeing in 4D when you look at a projection of a 4D object. There's no way to directly see the 4-volume when looking at a projection of a 4D object, since your POV is in the 3-plane of the projection. It's the same as looking edge-on at the drawing of a cube.

Here, Ovo, I found the reference I was looking for in the same great book, that's Geometry of 4D, 1914, available from google books. I have not found the theorem itself yet. But the quote is very interesting and totally in line with what I was saying. The quote about 8 3d volumes making up a hypervolume, just as I saw.

p 210 - 211

A spherical tetrahedron has 6 edges, each lying in the edge of a spherical dihedral angle whose interior contains the interior of the tetrahedton. The interior of one of these spherical dihedral angles contains also the interiors of 3 of the 15 tetrahedrons associated with the given tetrahedron as explained above, and its volume is = to the sum of the volumes of the 4 tetrahedrons whose interors are within it.

Writing A' for the opposite point to A, the other extremity of the diameter to A, and so for other points, we let T denote the volume of the tetrahedron ABCD, T1 the volume of A'BC,T12 the volume of A'B'CD, and so on. ABC'D' is congruent to A'B'CD, and we have T34 = T12 etc.

The interior of the dihedral angle C-AB-D contains the interiours of the 4 tetrahedrons whose volumes are T, T1, T2 and T12. If Ф12 is the measure of the dihedral angle AB in terms of a right dihedral angle, and if we take fo unit of volume 1/16 of the volume of the hypersphere, we shall have the relation

T + T1 + T2 + T12 = 4 Ф12

There are 6 if these equations, and in addition one equation expressing the fact that the sum of the 8 different volumes is equal to the volume of a half-hypersphere, namely,

T + Sum(T1) + Sum(T12) = 8

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### Re: 4D visualization starting with 3d objects in 4D

The quote on 211 to 212 is actually a reference to solid angle, which is measured as surface in all dimensions. In 4D, the surface is a 3-space.

It's like saying 3d things are 2D, because the sphere and photos are 2d surfaces. It entirely misses the point. I seriously think that you're trolling the list. In any case, what you're presenting as 4D is not what is commonly recieved as 4D.
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### Re: 4D visualization starting with 3d objects in 4D

Hello everyone! years later here I'm again

this topic keeps bugging me from time to time. If true, it has far-reaching consequences in many areas of science, especially physics.

For those who missed it, it appears, one can stack 3d objects in 4D in such a way (stacking them into the W direction) so that in the end they entirely fill the 4D space.

With 8 3d cubes, you can stack them in 3D to make a cube with twice as long edge, or you can stack them in 4D, to make up a tesseract of the same edge. That's what 4D allows: to stack 3d cubes into W direction. Those who bothered to play with actual cubes, trying to arrange them in 3D as if they were in 4D, labeling boxes and such, as if they were the 4 4D planes, could see for themselves that there is no gaps between faces, edges or vertices of the adjacent cubes. And why, it's obvious that 3D is a subspace of 4D, i.e. it's a part of it, and simply adding cubes or simplexes in all 4 directions will fill the 4D volume.

And yes, the same will work for all higher volumes. This implies that any n-volume with n>=3 can be expressed in terms of a 3d volume. This has huge implications. First of all, for visualisations: imagine, a 5-cube (that's a 5d cube) apparently it consists of 10 tesseracts, which implies that, roughly speaking, it consists of 80 cubes. Not only that! these 80 cubes are arranged in such a way that each of them shows one of its 6 faces outward and the other 5 it shares with the other cubes. And! all 80 cubes (a row of 40 and a row of 40) meet at a point with one of their their 8 vertices! That's the power of 5D.

But 5D is the last with such properties, because with a 6-cube, which consists of 12 5-cubes, we now have 960 3d cubes, with only 240 of them showing a face outward. This implies that 6-cube has a shell, which consists of 240, and a core of 720 cubes.

If you continue this progression, you will see pretty soon that 9D and 10D are unlikely to exist as "real" objects (they can live in the land of math though) and 11D looks impossible. This is mainly because the bounding area, or the "skin" of an object, grows much slower than the bulk. Soon it makes a monstrosity with enormous volume enclosed in a relatively tiny area "skin". This thing is sorta "waiting" for a first opportunity to implode as it sorta turns itself inside out, spewing its bulk into surrounding space in a manner of a supernova.

ah! I got carried away again

Funny that in years that passed no one found a cross-section of a tesseract showing where its 4d bulk lives. maybe that's because there is no such thing?

re Wendy, don't know if she's still around, but the point of that quote was that the sum of 8 different 3d volumes make up half a 4d volume. The context in which it came up was not important.
Last edited by 4Dspace on Thu Jan 18, 2018 4:40 am, edited 2 times in total.
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### Re: 4D visualization starting with 3d objects in 4D

4Dspace wrote:For those who missed it, it appears, one can stack 3d objects in 4D in such a way (stacking them into the W direction) so that in the end they entirely fill the 4D space.

4Dspace wrote: And why is it not obvious that 3D is a subspace of 4D, i.e. it's a part of it, and simply adding 3d cubes or 3d trapezoids in all 4 directions will fill the 4D volume.

Welcome back to the forum. Almost there, but not exactly. You can definitely stack 3D objects in 4-space, i.e., along the w-axis. Nothing is wrong there.

But, you can't fill a 4D space with 3D objects, unless you use and infinite number of them. That's a fundamental concept of a higher dimensional space. There are an infinite amount of 3D cubes stacked within a tesseract (from one cube-face to another).

The cubes take up 3 of the 4 dimensions of the tesseract, while the extra direction you stack them along is the 4th and final dimension of the tesseract.

The concept extends further to stacking an infinite number of 4D cubes along a 5th axis, to fill in a 5D hypercube's volume. And, so on.

If this doesn't make sense, then ask yourself: how many squares do I need to build up a cube? Assume these squares are purely two dimensional, having absolutely no 3D thickness to them.

How many do you need, to fill up a 3D volume, into a cube? 10? 1,000? 10google? Nope. It's even more than that. You need an infinite amount of those zero-thickness 2D squares, to fill in a true 3D volume, along the 3rd axis. That's how it works!
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### Re: 4D visualization starting with 3d objects in 4D

Hi ICN5D, the tone of confidence in your post made me think that I have missed a novel trend in math and 0 times infinity is no longer 0, as I thought. So I checked it out. Take a look:

https://math.oregonstate.edu/home/progr ... _zero.html

math.oregonstate.edu says it still is!

I see nothing changed here in the past years. Wendy claims to see in 8D, you in 5D, yet you can't see what I see in 4

I tell you what:
Why don't you show me, point on the diagram, or explain in words, where exactly you see a gap in between those 8 cubes ?

I can point it in a cube diagram: 3D lives in between the planes that bound it. There is empty space there, between the planes, and it is plainly seen.

I already described how you too could actually take a look and determine for yourself that there is no gaps whatsoever between 8 cubes that form a tesseract. NO GAPS: they align vertex to vertex, edge to edge and face to face. This can mean only one thing: that 3d volumes can be tiled in such a way that they fill completely a 4D space. And it does not take infinity of them. 8 will do.

If you disagree, please point out where that extra space in between those 8 cubes is.
Last edited by 4Dspace on Thu Jan 18, 2018 4:47 am, edited 1 time in total.
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### Re: 4D visualization starting with 3d objects in 4D

4Dspace wrote: .... no gaps whatsoever between 8 cubes that form a tesseract. NO GAPS: they align vertex to vertex, edge to edge and face to face. This can mean only one thing: that 3d volumes can be tiled in such a way that they fill completely a 4D space. And it does not take an infinity of them. 8 will do.

You almost have it. But, it sounds like you're thinking of two different things at once. That's where the hiccup is.

A finite number of 3D things (in this case, 8 cubes) can be arranged in 4D, to make up the surface of a 4D object (in this case, a 4D cube). But, you will need an infinite number of those 3D cubes to fill in the interior of the 4D volume.

It could be the way I read your posts, but it sounds like you use these two interchangeably (surface vs. interior), which can be confusing.
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### Re: 4D visualization starting with 3d objects in 4D

you speak of "interior of the 4D volume". I'm asking you yet again to please point where in between those 8 cubes it is hiding.

The other thing you do not see, because you're confused by analogies, is that 3D volume IS NOT 0 volume, like 2D is. See the difference? Granted, 3D is by one order of magnitude lower than 4D, but is still a volume. I.e. it fills space. Planes don't fill space.
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### Re: 4D visualization starting with 3d objects in 4D

4Dspace wrote:Planes don't fill space.

Exactly. A 'plane' is an area, not a 3D volume. And, as you just pointed out, an area cannot fill in a 3D volume. You can arrange 6 of these planes to build the surface of a 3D cube. But this will still be a hollow cube, with no filled-in 3D 'stuff' inside.

Likewise, a 3D volume is not a 4D volume. A 3-volume cannot fill in a 4-volume in the same way as the 2d square --> cube analogy works. You can arrange 8 of these 3-volumes in 4D, to build up the surface of a hollow tesseract. But it will still be hollow, with no filled-in 4D stuff to speak of. A hollow tesseract is just a bunch of 3D volumes arranged in 4-space like a box. They enclose the 4D volume, but do not fill it in, like water.

There is a difference here. Surfaces are not interiors.
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### Re: 4D visualization starting with 3d objects in 4D

Honestly man, I think it's great that you still think about this stuff. Even if you did leave this discussion alone six years ago.
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### Re: 4D visualization starting with 3d objects in 4D

ICN5D wrote:
4Dspace wrote:Planes don't fill space.

Likewise, a 3D volume is not a 4D volume. A 3-volume cannot fill in a 4-volume in the same way as the 2d square --> cube analogy works. You can arrange 8 of these 3-volumes in 4D, to build up the surface of a hollow tesseract. But it will still be hollow, with no filled-in 4D stuff to speak of. A hollow tesseract is just a bunch of 3D volumes arranged in 4-space like a box. They enclose the 4D volume, but do not fill it in, like water.

There is a difference here. Surfaces are not interiors.

Again, it would be much simpler and productive, if you could point out WHERE in between those 8 cubes lies the hollow 4D volume. If you cannot do this, it means only that those 8 cubes already OCCUPY the 4D volume.

Why can't you see that all 8 cubes are aligned face to face, edge to edge and vertex to vertex? All 8. There is no hollow. You say there is. Where?

All you do is parrot the analogy. This is a concrete thing.
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### Re: 4D visualization starting with 3d objects in 4D

Well, that's another problem, one that gets a lot of people. The image you speak of is a 3D image. You cannot portray a true 4D image of what's really going on in a 3D way. A hollow tesseract will look the same as a solid (and transparent) one, if you use the cube-within-cube projection, like this one:

The logic will always fail if you take this image too literally. You have to understand that this is a flat 3D shadow of the true 4D body. If it were a solid tesseract, the 4D volume is in between the large cube and small cube in the center. The problem is, we can't see it from the perspective we need (from true 4D) to really get this part. We have to overcome this little comprehension barrier, before getting ahead of ourselves.

Actually, here's a thought for you: if you cast a 3D shadow of a solid, opaque tesseract (read: not transparent) , the shadow would be a solid 3D cube. You wouldn't see anything inside of it, no cube within cube.

Does that help any? Again, 3D images of 4D things have limitations that make it harder to get past some concepts. Also remembering that using 2D squares to enclose a 3D space (as in a cube) does not fill in the 3D space like a solid cube would.
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### Re: 4D visualization starting with 3d objects in 4D

Can you point to where the 3D volume is, in between these 6 squares? Trick question : it's a 2D image of a 3D body. There is no 3D volume here in the image. The concept is that the 3D space is in between the squares, same as the 4D volume is in between the 8 cubes, in an imperceptible-at-first way.

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### Re: 4D visualization starting with 3d objects in 4D

Even on this rotating projection you see exactly 8 cubes. The same 8 cubes. You're saying that two central ones represent the 4D of the hollow tesseract -? I say it's not enough, you need all 8 for that.

And no, I'm not confused by seeing 8 cubes. They remind me what you forget with your analogies, namely that objects don't change into something else only because you place them in a higher dimensional space. 3d does not become flat in 4D. It becomes "flatter", but not entirely flat, like planes are flat. And just like a plane is always a plane, no matter in what D you put it, 3D is always 3D, no matter in what space you put it. Yes, it looks flattish to us in 4D, especially if we look at its 3 planes from W position -- it's because it is orthogonal to all 3 of them at once (the 3 closest you see edge on) but they are not flat, 'cause they are still orthogonal to each other. They still occupy a 3d volume in 4D. That's where you're misled by the analogy. They occupy a 3d subspace of a 4D. They fill a part of it.

looking at this moving projection is even easier to see how each of the cubes shares each of its edges and faces and vertices with its neighbours. ..even as they wrap around themselves, it's still the same 8 cubes. Wait! there is something ... as if space is scooped from outside -- as if 4D is outside really.. ..but nah, it's still the same 8 cubes no matter which way you rotate the tesseract.

I see while i was typing this you posted a cute picture of a cube... Thanks for the effort.

I have a trick question for you in turn: do you agree that all 8 cubes of tesseract are aligned face to face, vertex to vertex and edge to edge? Yes or no?
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### Re: 4D visualization starting with 3d objects in 4D

Welcome back, 4Dspace.

1) A square consists of 4 sides, correct? Thus you can construct 4 unit lengths within 1D, then cut them out, bend them into the next direction about 90 degrees each at any "vertex", and thus the endpoint of the 4th "side" would connect to the startpoint of the first, thus closing the - hollow - circuit.
You can embed the 1D space into 2D space, e.g. as the extension of one of the sides of this circuit of sides of that square. Let's use the "bottom" one. Then you could consider an affine subspaces too, that is a parallel line to that extended bottom line. As long as it would be below that bottom one, the intersection with the square is empty. As soon those 2 lines become coincident, the intersection with the square is exactly that very bottom side. When that parallel intersecting line moves further up, say the distance would be lesser than 1, then the intersection with the square happens to be still a unit sized Segment. But wait. The intersection with the 2D interior is the (topologically) open unit segment, while the intersection with the vertical sides then are just the 2 endpoints. These endpoints in fact then are internal points of the 2 vertical sides. Then, moving further up, you get again a distance, where the opposite side, the top side, becomes fully incident to the moving line, and beyond that level the intersection would be empty again. Thus the 4 sides truely are just the boundary of the square. The 2D body of the square is still empty, the square happens to be hollow. You do not fill the square by 4 line segments. But by means of the moving intersecting line you get a paramtrizable continuum of "levels" where you could intersect. Those are the infinite amount of line segments, which would be required to fill the interior - in contrast to the 4 line segments to enclose that empty interior. - So far you agree?

2) Now consider a flat plan or net of a cube. Say 4 squares in a column, and one additional square like its arms attached on either side. So far this is a truely 2D concept. In order to become what you visually consider a cube, you would have to cut it out, fold it at the square-connecting sides about 90 degrees. Thus you get an envelope or surface of the true 3D cube. Again only the boundary is supported, the 3D interior still is empty. Again you could consider an affine plane parallel to the "bottom" face. Again the intersection with the cube either is empty (below or above the cube), is exactly the bottom face or the top face, or it intersects the 3D interior of the cube. That intersection will have still the shape of a square. But in fact you have again to distinguish between internal points of that square, which would represent the hollow 3D body, and the 4 sides, which represent sections of the vertical side squares. Thus again, you just need 6 squares to encompass the surface of the cube, but again you'll have a parametrizable (level height) continuum of infinitely thin squares to fill the 3D body of the cube like the pages in a very thick book. - So far you agree?

3) Now let's turn towards the dimensions of interest. Again we will have to build the floor plan or net of a tesseract first. This can be done within 3D by stacking 4 cubes in a column, and now attach 4 more cubes one each on either side of that pillar, like the arms of a guidepost. In order to become 4D, you once more would have to cut that out, and fold it at the connecting squares about 90 degrees into the 4th new direction. Doing this, the net would Close. That is all cubes would connect truely side by side. All sides would be used, none remains unconnected. But still this represents not the full 4D tesseract, it just describes its 3D surface only! Again we can do our experiment with an affine hyperplane (which itself now is an affine 3D space!!!) parallel to the bottem 3D space, where the tesseract stands on. Within the intermediate cases you'll see that this Hyperspace would intersect the vertical sides of the tesseract, i.e. those 6 cubes, each in a medial square only. In fact, the intersection of that hyperplane with the tesseract is a cube. The body of this cubical intersection again represents the intersection region of the 4D body of the tesseract. Only the 6 side squares of this intersection happen to be the intersections with the vertical sidescubes of the 3D envelope. And again we count a total of 8 cubes to be used for the boundary of the tesseract. But we once more would Need a parametrizable continuum of cubical intersections to stack along the 4th new direction, in order to fill the 4D interior of the tesseract.

4) As you could have seen in 1): all sides themselves and also their connecting points/corners/vertices are fully at the outside of the square. They just encompass its surface. The same holds true in 2): The full 2D body of the squares, as well as the connecting sides/edges and the points/corners/vertices are fully at the outside of the cube. They just encompass its surface. And this once more is true in 3): The full 3D body of the cubes, as well as the connecting squares, the edges and the vertices all happen to be at the outside of the tesseract. They just encompass its surface. - In fact the same argument would hold in any dimension! There will never be a limiting dimensional number, where you would get internal shells. By construction, you always will describe only the surfice. The (n-1)D net becomes the envelope of the nD (thus hollow!) figure.

--- rk
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