Is it possible to map each point on a hypercube onto a simple plane...

Is it possible to map each point on a hypercube onto a simple plane? So that each unique point on the 4d cube has a unique point on the plane, and has the same neighboring points as it would on the hypercube? Except the edges of the plane obviously. Is it possible? If so, what would that look like?

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m.youtube.com/watch?v=BVo2igbFSPE
unfolding.apperceptual.com/
en.wikipedia.org/wiki/Simplicial_complex
en.wikipedia.org/wiki/Triangulation_(geometry)
twitter.com/SFWRedditImages

>on a hypercube

How do you define a point being "on" a hypercube?

no

On the surface of it. Like if you unfold a 3d cube, you get this. What would you get with a hypercube if you unfold it into 2d, and is that even possible without losing any surfaces?

If you unfold a 3d cube, you get 2d cubes (squares). If you unfold a 4d cube, you get 3d cubes. So in this sense, it's impossible what you're asking for.

How are those 3d cubes connected when you unfold a hypercube? Could you maybe unfold those into a 2d surface?

Very similarly what we have in the 3d case. Check this: m.youtube.com/watch?v=BVo2igbFSPE

The surface of a hypercube is homeomorphic to the 3-sphere via, say, [math]f[/math] from the sphere to the surface. Suppose you could have a continuous injection [math]g[/math] from the hypercube's surface to [math]\mathbb{R}^2[/math], and let [math]i \colon \mathbb{R}^2 \to \mathbb{R}^3[/math] be the inclusion [math]x \mapsto (x, 0)[/math]. Define [math]h=i \circ g \circ f[/math]. Then [math]h \colon S^3 \to \mathbb{R}^3[/math] is a continuous injection contradicting Borsuk-Ulam.

So you get 8 cubes, each of which unfold into 6 squares? I can't visualize what the layout of squares would look like though.

So, no? Not possible?

Since [math]\mathbb{R}^1\cong\mathbb{R}^2[/math], yes there exists a set isomorphism from the hypercube onto the plane, but you preserve no geometry.

You can easily give projection mappings onto the plane as well but you lose perspective. In fact the image in the OP is a projection of the hypercube onto a 2d raster image.

A hypercube is a solid, you cannot 'unfold' it. Center and surface are one and the same. Now imagine a hypersphere in stationary motion.

Sorry, but not possible.

You are asking this question, so for you, no

unfolding.apperceptual.com/

I think I found what I need, I will try to work through it.

Are you asking if you can draw a hybercube with a vector function on an R3 (or R2) plane?

No, not directly at least.

I ask for this, but better. This is what I got with some putting cubes together, if you would fold this together you would get these 8 cubes, and then you could fold it in 4 dimensions and get a hypercube.

Is there a better shape of the 2d surface, better meaning less empty space, more compact, and the cubes which would later fold together in 4d touching surfaces as many as possible?

That breaks the condition
>has the same neighboring points as it would on the hypercube
on the borders of those coloured areas, as the surface gets ripped on some points.

Is there any piece of math that deals with this? How would I calculate all the possible combination of squares that if folded together in 4 dimensions give me a hypercube?

I cut out the shape from paper and put it together into 8 cubes, it works. But there must be more shapes that do. Are there infinitely many?

>except the edges of the plane obviously

You have triangulations for polyhedra, and a hypercube would be a polyhedron. These are topology.

Oops, I missed that part completely. Then you can discard my proof of impossibility, too. It holds only for the whole surface, not for a surface with edges and vertices removed.

The only thing I know about topology is what the awesome klein bottle dude talks about in his videos. How would I calculate all the possible surfaces that fold up into hypercubes?

that's the neatest thing I've seen all week

You construct one of these en.wikipedia.org/wiki/Simplicial_complex and prove the underlying space is homeomorphic (that is, topologically equivalent) to the space, in this case the hypercube. To calculate how many things you need, is to consider the amount of simplices required to do it.Here's more info en.wikipedia.org/wiki/Triangulation_(geometry)

Homeomorphic, are you sure? Isn't a hypercube homeomorphic to any closed 4d space without any holes in it, so a 4d sphere, a 4d pyramid (does that exist?) and so on?

The interior is an open 4-simplex, yes. The boundary is a union of smaller dimensional simplices.

I still really don't know how I would do it mathematically, how would I get all the simplices that form a hypercube? If I do it by proving homeomorphy, wouldn't I also get stuff that folds together to other 4d shapes?

Thanks for all the help btw

2d and 4d spaces have the same cardinality.

Okay, I'll give you examples of spaces homeomorphic to (open) simplices.A singleton of one point is homeomorphic to the 0-simplex, an open interval is homeomorphic to the 1-simplex, an open disk and the interior of a square are homeomorphic to the 2-simplex, an open ball or the interior of an open ball is homeomorphic to a 3-simplex, etc. Try to construct a square using only vertices (0-simplices), edges (1-simplices) and the interior (a 2-simplex). Then do the same for a cube. These are triangulations for those, and a triangulation for the hypercube is constructed similarly.