Step into multidimensional space.

The evolution of the human brain took place in three-dimensional space. Therefore, it is difficult for us to imagine spaces with dimensions greater than three. In fact, the human brain cannot imagine geometric objects with more than three dimensions. And at the same time, we can easily imagine geometric objects with dimensions not only three, but also with dimensions two and one.

The difference and analogy between one-dimensional and two-dimensional spaces, and the difference and analogy between two-dimensional and three-dimensional spaces allow us to slightly open the screen of mystery that fences us off from spaces of higher dimensions. To understand how this analogy is used, consider a very simple four-dimensional object - a hypercube, that is, a four-dimensional cube. Let, for definiteness, let's say we want to solve a specific problem, namely, to count the number of square faces of a four-dimensional cube. All consideration below will be very loose, without any evidence, purely by analogy.

To understand how a hypercube is built from an ordinary cube, one must first look at how an ordinary cube is built from an ordinary square. For the originality of the presentation of this material, we will here call an ordinary square SubCube (and we will not confuse it with a succubus).

To construct a cube from a subcube, it is necessary to extend the subcube in a direction perpendicular to the plane of the subcube in the direction of the third dimension. At the same time, a subcube will grow from each side of the original subcube, which is a two-dimensional side face of the cube, which will limit the three-dimensional volume of the cube from four sides, two perpendicular to each direction in the plane of the subcube. And along the new third axis, there are also two subcubes that limit the three-dimensional volume of the cube. This is the two-dimensional face where our subcube was originally located and the two-dimensional face of the cube where the subcube came at the end of the cube construction.

What you have just read is set out in excessive detail and with a lot of clarifications. And not casually. Now we will do such a trick, we will replace some words in the previous text formally in this way:
cube -> hypercube
subcube -> cube
plane -> volume
third -> fourth
2D -> 3D
four -> six
three-dimensional -> four-dimensional
two -> three
plane -> space

As a result, we get the following meaningful text, which no longer seems too detailed.

To build a hypercube from a cube, you need to stretch the cube in a direction perpendicular to the volume of the cube in the direction of the fourth dimension. At the same time, a cube will grow from each side of the original cube, which is the lateral three-dimensional face of the hypercube, which will limit the four-dimensional volume of the hypercube from six sides, three perpendicular to each direction in the space of the cube. And along the new fourth axis, there are also two cubes that limit the four-dimensional volume of the hypercube. This is the three-dimensional face where our cube was originally located and the three-dimensional face of the hypercube, where the cube came at the end of the hypercube construction.

Why do we have such confidence that we have received the correct description of the construction of the hypercube? Yes, because by exactly the same formal replacement of words we get a description of the construction of a cube from a description of the construction of a square. (Check it out for yourself.)

Now it is clear that if another three-dimensional cube should grow from each side of the cube, then a face must grow from each edge of the initial cube. In total, the cube has 12 edges, which means that there will be an additional 12 new faces (subcubes) for those 6 cubes that limit the four-dimensional volume along the three axes of three-dimensional space. And there are two more cubes that limit this four-dimensional volume from below and from above along the fourth axis. Each of these cubes has 6 faces.

In total we get that the hypercube has 12+6+6=24 square faces.

The following picture shows the logical structure of a hypercube. It is like a projection of a hypercube onto three-dimensional space. In this case, a three-dimensional frame of ribs is obtained. In the figure, of course, you see the projection of this frame also onto a plane.

On this frame, the inner cube is, as it were, the initial cube from which the construction began and which limits the four-dimensional volume of the hypercube along the fourth axis from the bottom. We stretch this initial cube up along the fourth dimension axis and it goes into the outer cube. So the outer and inner cubes from this figure limit the hypercube along the fourth dimension axis.

And between these two cubes, 6 more new cubes are visible, which are in contact with the first two by common faces. These six cubes limit our hypercube along three axes of three-dimensional space. As you can see, they are not only in contact with the first two cubes, which are internal and external on this three-dimensional frame, but they are still in contact with each other.

You can calculate directly in the figure and make sure that the hypercube really has 24 faces. But here comes the question. This 3D hypercube frame is filled with eight 3D cubes without any gaps. In order to make a real hypercube from this 3D projection of a hypercube, it is necessary to turn this frame inside out so that all 8 cubes limit the 4D volume.

It is done like this. We invite a resident of the four-dimensional space to visit and ask him to help us. It grabs the inner cube of this framework and shifts it towards the fourth dimension, which is perpendicular to our 3D space. We in our three-dimensional space perceive it as if the entire inner frame had disappeared and only the frame of the outer cube remained.

Next, our 4D assistant offers to help in painless births, but our pregnant women are terrified at the prospect of the baby simply disappearing from the belly and ending up in a parallel 3D space. Therefore, the fourfold is politely refused.

And we're wondering if some of our cubes got unstuck when the hypercube frame was turned inside out. After all, if some three-dimensional cubes surrounding the hypercube touch their neighbors on the frame, will they also touch the same faces if the four-dimensional one turns the frame inside out.

Let us again turn to the analogy with spaces of lower dimension. Compare the image of the hypercube wireframe with the projection of the 3D cube onto the plane shown in the following picture.

Inhabitants of two-dimensional space built a framework of a cube projection onto a plane on a plane and invited us, three-dimensional residents, to turn this framework inside out. We take the four vertices of the inner square and move them perpendicular to the plane. At the same time, two-dimensional residents see the complete disappearance of the entire inner frame, and they only have the frame of the outer square. With such an operation, all the squares that were in contact with their edges continue to touch as before with the same edges.

Therefore, we hope that the logical scheme of the hypercube will also not be violated when the hypercube frame is turned inside out, and the number of square faces of the hypercube will not increase and will remain equal to 24. This, of course, is no proof at all, but purely a guess by analogy .

After everything read here, you can easily draw the logical framework of a five-dimensional cube and calculate how many vertices, edges, faces, cubes and hypercubes it has. It's not difficult at all.

Hypercube and Platonic Solids

Simulate a truncated icosahedron (“soccer ball”) in the “Vector” system
where each pentagon is bounded by hexagons

Truncated icosahedron can be obtained by cutting 12 vertices to form faces in the form of regular pentagons. In this case, the number of vertices of the new polyhedron increases by 5 times (12 × 5 = 60), 20 triangular faces turn into regular hexagons (in total faces becomes 20+12=32), A the number of edges increases to 30+12×5=90.

Steps for constructing a truncated icosahedron in the Vector system

Figures in 4-dimensional space.

--à

--à ?

For example, given a cube and a hypercube. There are 24 faces in a hypercube. This means that a 4-dimensional octahedron will have 24 vertices. Although no, the hypercube has 8 faces of cubes - in each center is a vertex. This means that a 4-dimensional octahedron will have 8 vertices of that one easier.

4-dimensional octahedron. It consists of eight equilateral and equal tetrahedra,
connected four at each vertex.

Rice. An attempt to simulate
hyperball-hypersphere in the "Vector" system

Front - back faces - balls without distortion. Another six balls - can be specified through ellipsoids or quadratic surfaces (through 4 contour lines as generators) or through faces (first defined through generators).

More tricks to "build" a hypersphere
- the same "soccer ball" in 4-dimensional space

Annex 2

For convex polyhedra, there is a property relating the number of its vertices, edges, and faces, proved in 1752 by Leonhard Euler, and called Euler's theorem.

Before formulating it, consider the polyhedra known to us and fill in the following table, in which B is the number of vertices, P - edges and G - faces of a given polyhedron:

It is directly seen from this table that for all the chosen polyhedra the equality B - P + T = 2 holds. It turns out that this equality is true not only for these polyhedra, but also for an arbitrary convex polyhedron.

Euler's theorem. For any convex polyhedron, the equality

V - R + G \u003d 2,

where B is the number of vertices, P is the number of edges, and G is the number of faces of the given polyhedron.

Proof. To prove this equality, imagine the surface of a given polyhedron made of an elastic material. Let's delete (cut out) one of its faces and stretch the remaining surface on a plane. We get a polygon (formed by the edges of the removed face of the polyhedron), divided into smaller polygons (formed by the remaining faces of the polyhedron).

Note that polygons can be deformed, enlarged, reduced, or even bent their sides, as long as the sides do not break. The number of vertices, edges and faces will not change.

Let us prove that the resulting partition of a polygon into smaller polygons satisfies the equality

(*) V - R + G "= 1,

where B is the total number of vertices, P is the total number of edges, and Г "is the number of polygons included in the partition. It is clear that Г" \u003d Г - 1, where Г is the number of faces of this polyhedron.

Let us prove that the equality (*) does not change if we draw a diagonal in some polygon of the given partition (Fig. 5, a). Indeed, after drawing such a diagonal, the new partition will have B vertices, P + 1 edges, and the number of polygons will increase by one. Therefore, we have

V - (R + 1) + (G "+1) \u003d V - R + G" .

Using this property, we draw diagonals dividing the incoming polygons into triangles, and for the resulting partition we show that the equality (*) is satisfied (Fig. 5, b). To do this, we will consistently remove the outer edges, reducing the number of triangles. In this case, two cases are possible:

a) to remove the triangle ABC it is required to remove two ribs, in our case AB And BC;

b) to remove the triangleMKNit is required to remove one edge, in our caseMN.

In both cases, the equality (*) will not change. For example, in the first case, after removing the triangle, the graph will consist of B - 1 vertices, R - 2 edges and G "- 1 polygon:

(B - 1) - (P + 2) + (G "- 1) \u003d B - R + G".

Consider the second case for yourself.

Thus, removing one triangle does not change the equality (*). Continuing this process of removing triangles, we will eventually arrive at a partition consisting of a single triangle. For such a partition, B \u003d 3, P \u003d 3, Г "= 1 and, therefore, B - Р + Г" = 1. Hence, equality (*) also holds for the original partition, from which we finally obtain that for a given polygon partition equality (*) holds true. Thus, for the original convex polyhedron, the equality B - P + G = 2 is true.

An example of a polyhedron for which the Euler relation does not hold is shown in Figure 6. This polyhedron has 16 vertices, 32 edges and 16 faces. Thus, for this polyhedron, the equality B - P + G = 0 is satisfied.

Appendix 3

Movie Cube 2: Hypercube "(eng. Cube 2: Hypercube) - a fantasy film, a continuation of the movie" Cube ".

Eight strangers wake up in cube-shaped rooms. The rooms are inside a four-dimensional hypercube. The rooms are constantly moving by "quantum teleportation", and if you climb into the next room, then it is unlikely to return to the previous one. Parallel worlds intersect in the hypercube, time flows differently in some rooms, and some rooms are death traps.

The plot of the picture largely repeats the story of the first part, which is also reflected in the images of some characters. In the rooms of the hypercube, Nobel laureate Rosenzweig dies, who calculated the exact time of the destruction of the hypercube.

Criticism

If in the first part people imprisoned in a labyrinth tried to help each other, in this film it's every man for himself. There are a lot of extra special effects (they are also traps) that do not logically connect this part of the film with the previous one. That is, it turns out the film Cube 2 is a kind of labyrinth of the future 2020-2030, but not 2000. In the first part, all types of traps can theoretically be created by a person. In the second part, these traps are a program of some kind of computer, the so-called "Virtual Reality".

Let's start by explaining what a four-dimensional space is.

This is a one-dimensional space, that is, simply the OX axis. Any point on it is characterized by one coordinate.

Now let's draw the OY axis perpendicular to the OX axis. So we got a two-dimensional space, that is, the XOY plane. Any point on it is characterized by two coordinates - the abscissa and the ordinate.

Let's draw the OZ axis perpendicular to the axes OX and OY. You will get a three-dimensional space in which any point has an abscissa, an ordinate and an applicate.

It is logical that the fourth axis, OQ, should be perpendicular to the axes OX, OY and OZ at the same time. But we cannot accurately construct such an axis, and therefore it remains only to try to imagine it. Every point in four-dimensional space has four coordinates: x, y, z and q.

Now let's see how the four-dimensional cube appeared.

The picture shows a figure of one-dimensional space - a line.

If you make a parallel translation of this line along the OY axis, and then connect the corresponding ends of the two resulting lines, you get a square.

Similarly, if we make a parallel translation of the square along the OZ axis and connect the corresponding vertices, we get a cube.

And if we make a parallel translation of the cube along the OQ axis and connect the vertices of these two cubes, then we will get a four-dimensional cube. By the way, it's called tesseract.

To draw a cube on a plane, you need it project. Visually it looks like this:

Imagine that in the air above the surface hangs wireframe model cube, that is, as if "made of wire", and above it - a light bulb. If you turn on the light bulb, trace the shadow of the cube with a pencil, and then turn off the light bulb, then a projection of the cube will be shown on the surface.

Let's move on to something a little more complicated. Look again at the drawing with the light bulb: as you can see, all the rays converged at one point. It is called vanishing point and is used to build perspective projection(and sometimes parallel, when all the rays are parallel to each other. The result is that there is no sense of volume, but it is lighter, and if the vanishing point is far enough away from the projected object, then the difference between these two projections is hardly noticeable). To project a given point onto a given plane using a vanishing point, you need to draw a line through the vanishing point and the given point, and then find the intersection point of the resulting line and the plane. And in order to project a more complex figure, say, a cube, you need to project each of its vertices, and then connect the corresponding points. It should be noted that space-to-subspace projection algorithm can be generalized to 4D->3D, not just 3D->2D.

As I said, we can't imagine exactly what the OQ axis looks like, and neither can the tesseract. But we can get a limited idea of ​​it if we project it onto a volume and then draw it on a computer screen!

Now let's talk about the projection of the tesseract.

On the left is the projection of the cube onto the plane, and on the right is the tesseract onto the volume. They are quite similar: the projection of a cube looks like two squares, small and large, one inside the other, with corresponding vertices connected by lines. And the projection of the tesseract looks like two cubes, small and large, one inside the other, and whose corresponding vertices are connected. But we have all seen the cube, and we can say with confidence that both the small square and the large one, and the four trapezoids above, below, to the right and left of the small square, are in fact squares, moreover, they are equal. The same goes for the Tesseract. And a large cube, and a small cube, and six truncated pyramids on the sides of a small cube - these are all cubes, and they are equal.

My program can not only draw the projection of the tesseract onto the volume, but also rotate it. Let's see how this is done.

First, I'll tell you what is rotation parallel to the plane.

Imagine that the cube rotates around the OZ axis. Then each of its vertices describes a circle around the OZ axis.

A circle is a flat figure. And the planes of each of these circles are parallel to each other, and in this case they are parallel to the XOY plane. That is, we can talk not only about rotation around the OZ axis, but also about rotation parallel to the XOY plane. As you can see, for points that rotate parallel to the XOY axis, only the abscissa and ordinate change, while the applicate remains unchanged And, in fact, we we can talk about rotation around a straight line only when we are dealing with three-dimensional space. In 2D everything revolves around a point, in 4D everything revolves around a plane, in 5D space we are talking about rotation around a volume. And if we can imagine the rotation around a point, then the rotation around the plane and volume is something unthinkable. And if we talk about rotation parallel to the plane, then in any n-dimensional space a point can rotate parallel to the plane.

Many of you have probably heard of the rotation matrix. Multiplying a point by it, we get a point rotated parallel to the plane by an angle phi. For a two-dimensional space, it looks like this:

How to multiply: x of a point rotated by an angle phi = cosine of the angle phi*x of the original point minus the sine of the angle phi*y of the original point;
y of the point rotated by the angle phi=sine of the angle phi*x of the original point plus cosine of the angle phi*y of the original point.
Xa`=cosФ*Xa - sinФ*Ya
Ya`=sinФ*Xa + cosФ*Ya
, where Xa and Ya are the abscissa and ordinate of the point to be rotated, Xa` and Ya` are the abscissa and ordinate of the already rotated point

For a three-dimensional space, this matrix is ​​generalized as follows:

Rotation parallel to the XOY plane. As you can see, the Z coordinate does not change, but only X and Y change.
Xa`=cosФ*Xa - sinФ*Ya + Za*0
Ya`=sinФ*Xa + cosФ*Ya + Za*0
Za`=Xa*0 + Ya*0 + Za*1 (essentially Za`=Za)

Rotation parallel to the XOZ plane. Nothing new,
Xa`=cosФ*Xa + Ya*0 - sinФ*Za
Ya`=Xa*0 + Ya*1 + Za*0 (essentially Ya`=Ya)
Za`=sinФ*Xa + Ya*0 + cosФ*Za

And the third matrix.
Xa`=Xa*1 + Ya*0 + Za*0 (essentially Xa`=Xa)
Ya`=Xa*0 + cosФ*Ya - sinФ*Za
Za`=Xa*0 + sinФ*Ya + cosФ*Za

And for the fourth dimension, they look like this:


I think you already understood what to multiply by, so I won’t paint it again. But I note that it does the same as the matrix for rotating parallel to the plane in three-dimensional space! Both that and this one change only the ordinate and the applicate, and the rest of the coordinates are not touched, therefore it can be used in the three-dimensional case, simply ignoring the fourth coordinate.

But with the projection formula, not everything is so simple. No matter how much I read the forums, none of the projection methods suited me. Parallel did not suit me, since the projection will not look three-dimensional. In some projection formulas, to find a point, you need to solve a system of equations (and I don’t know how to teach a computer to solve them), I simply didn’t understand others ... In general, I decided to come up with my own way. Consider for this the projection 2D->1D.

pov means "Point of view" (point of view), ptp means "Point to project" (the point to be projected), and ptp` is the desired point on the OX axis.

Angles povptpB and ptpptp`A are equal as corresponding (dashed line is parallel to axis OX, line povptp is secant).
The x of ptp` is equal to the x of ptp minus the length of segment ptp`A. This segment can be found from the triangle ptpptp`A: ptp`A = ptpA/tangent of angle ptpptp`A. We can find this tangent from triangle povptpB: tangent of angle ptpptp`A = (Ypov-Yptp)(Xpov-Xptp).
Answer: Xptp`=Xptp-Yptp/tangent of angle ptpptp`A.

I did not describe this algorithm in detail here, since there are a lot of special cases where the formula changes somewhat. Who cares - look in the source code of the program, everything is written in the comments.

In order to project a point in three-dimensional space onto a plane, we simply consider two planes - XOZ and YOZ, and solve this problem for each of them. In the case of a four-dimensional space, it is necessary to consider already three planes: XOQ, YOQ and ZOQ.

And finally, about the program. It works like this: initialize sixteen vertices of the tesseract -> depending on the commands entered by the user, rotate it -> project onto the volume -> depending on the commands entered by the user, rotate its projection -> project onto a plane -> draw.

Projections and rotations I wrote myself. They work according to the formulas that I just described. The OpenGL library draws lines and also mixes colors. And the coordinates of the vertices of the tesseract are calculated in this way:

Line vertex coordinates centered at the origin and length 2 - (1) and (-1);
- "-" - a square - "-" - and an edge of length 2:
(1; 1), (-1; 1), (1; -1) and (-1; -1);
- " - " - cube - " - " -:
(1; 1; 1), (-1; 1; 1), (1; -1; 1), (-1; -1; 1), (1; 1; -1), (-1; 1; -1), (1; -1; -1), (-1; -1; -1);
As you can see, the square is one line above the OY axis and one line below the OY axis; a cube is one square in front of the XOY plane, and one behind it; a tesseract is one cube on the other side of the XOYZ volume, and one on this side. But it is much easier to perceive this alternation of units and minus units if they are written in a column

1; 1; 1
-1; 1; 1
1; -1; 1
-1; -1; 1
1; 1; -1
-1; 1; -1
1; -1; -1
-1; -1; -1

In the first column, one and minus one alternate. In the second column, first there are two pluses, then two minuses. In the third - four plus one, and then four minus one. These were the tops of the cube. The tesseract has twice as many of them, and therefore it was necessary to write a cycle for declaring them, otherwise it is very easy to get confused.

My program also knows how to draw anaglyph. Happy owners of 3D glasses can watch a stereoscopic picture. There is nothing tricky in drawing a picture, it just draws two projections on a plane, for the right and left eyes. But the program becomes much more visual and interesting, and most importantly - gives a better idea of ​​the four-dimensional world.

Less significant functions - highlighting one of the faces in red, so that you can better see the turns, as well as minor conveniences - adjusting the coordinates of the "eye" points, increasing and decreasing the speed of rotation.

Archive with the program, source code and instructions for use.

Tesseract - a four-dimensional hypercube - a cube in four-dimensional space.
According to the Oxford Dictionary, the word tesseract was coined and used in 1888 by Charles Howard Hinton (1853-1907) in his book A New Age of Thought. Later, some people called the same figure a tetracube (Greek τετρα - four) - a four-dimensional cube.
An ordinary tesseract in Euclidean four-dimensional space is defined as the convex hull of points (±1, ±1, ±1, ±1). In other words, it can be represented as the following set:
[-1, 1]^4 = ((x_1,x_2,x_3,x_4) : -1 = A tesseract is bounded by eight hyperplanes x_i= +- 1, i=1,2,3,4 , whose intersection with the tesseract itself defines it 3D faces (which are regular cubes) Each pair of non-parallel 3D faces intersect to form 2D faces (squares), etc. Finally, the tesseract has 8 3D faces, 24 2D, 32 edges, and 16 vertices.
Popular Description
Let's try to imagine how the hypercube will look without leaving the three-dimensional space.
In one-dimensional "space" - on a line - we select a segment AB of length L. On a two-dimensional plane at a distance L from AB, we draw a segment DC parallel to it and connect their ends. You will get a square CDBA. Repeating this operation with a plane, we get a three-dimensional cube CDBAGHFE. And by shifting the cube in the fourth dimension (perpendicular to the first three) by a distance L, we get the CDBAGHFEKLJIOPNM hypercube.
The one-dimensional segment AB serves as a side of the two-dimensional square CDBA, the square is the side of the cube CDBAGHFE, which, in turn, will be the side of the four-dimensional hypercube. A straight line segment has two boundary points, a square has four vertices, and a cube has eight. Thus, in a four-dimensional hypercube, there will be 16 vertices: 8 vertices of the original cube and 8 vertices shifted in the fourth dimension. It has 32 edges - 12 each give the initial and final positions of the original cube, and 8 more edges "draw" eight of its vertices that have moved into the fourth dimension. The same reasoning can be done for the faces of the hypercube. In two-dimensional space, it is one (the square itself), the cube has 6 of them (two faces from the moved square and four more will describe its sides). A four-dimensional hypercube has 24 square faces - 12 squares of the original cube in two positions and 12 squares from twelve of its edges.
As the sides of a square are 4 one-dimensional segments, and the sides (faces) of a cube are 6 two-dimensional squares, so for the "four-dimensional cube" (tesseract) the sides are 8 three-dimensional cubes. The spaces of opposite pairs of tesseract cubes (that is, the three-dimensional spaces to which these cubes belong) are parallel. In the figure, these are cubes: CDBAGHFE and KLJIOPNM, CDBAKLJI and GHFEOPNM, EFBAMNJI and GHDCOPLK, CKIAGOME and DLJBHPNF.
In a similar way, we can continue the reasoning for hypercubes of a larger number of dimensions, but it is much more interesting to see how a four-dimensional hypercube will look like for us, inhabitants of three-dimensional space. Let us use for this the already familiar method of analogies.
Let's take the wire cube ABCDHEFG and look at it with one eye from the side of the face. We will see and can draw two squares on the plane (its near and far faces), connected by four lines - side edges. Similarly, a four-dimensional hypercube in three-dimensional space will look like two cubic "boxes" inserted into each other and connected by eight edges. In this case, the "boxes" themselves - three-dimensional faces - will be projected onto "our" space, and the lines connecting them will stretch in the direction of the fourth axis. You can also try to imagine a cube not in projection, but in a spatial image.
Just as a three-dimensional cube is formed by a square shifted by the length of a face, a cube shifted into the fourth dimension will form a hypercube. It is limited by eight cubes, which in the future will look like some rather complex figure. The four-dimensional hypercube itself consists of an infinite number of cubes, just as a three-dimensional cube can be “cut” into an infinite number of flat squares.
By cutting six faces of a three-dimensional cube, you can decompose it into a flat figure - a net. It will have a square on each side of the original face, plus one more - the face opposite to it. A three-dimensional development of a four-dimensional hypercube will consist of the original cube, six cubes that "grow" from it, plus one more - the final "hyperface".
The properties of a tesseract are an extension of the properties of geometric figures of a smaller dimension into a four-dimensional space.

Teachings about multidimensional spaces began to appear in the middle of the 19th century in the works of G. Grassmann, A. Cayley, B. Riemann, W. Clifford, L. Schläfli and other mathematicians. At the beginning of the 20th century, with the advent of A. Einstein's theory of relativity and the ideas of G. Minkowski, physics began to use a four-dimensional space-time coordinate system.

Then science fiction writers borrowed the idea of ​​four-dimensional space from scientists. In their works, they told the world about the amazing wonders of the fourth dimension. The heroes of their works, using the properties of four-dimensional space, could eat the contents of the egg without damaging the shell, drink a drink without opening the cork of the bottle. The kidnappers retrieved the treasure from the safe through the fourth dimension. The links of the chain can be easily disconnected, and the knot on the rope can be untied without touching its ends. Surgeons performed operations on the internal organs without cutting the tissues of the patient's body. The mystics placed the souls of the dead in the fourth dimension. For an ordinary person, the idea of ​​four-dimensional space remained incomprehensible and mysterious, and many generally consider four-dimensional space to be the fruit of the imagination of scientists and science fiction writers, which has nothing to do with reality.

Perception problem

It is traditionally believed that a person cannot perceive and represent four-dimensional figures, since he is a three-dimensional being. The subject perceives three-dimensional figures with the help of the retina, which is two-dimensional. To perceive four-dimensional figures, a three-dimensional retina is needed, but a person does not have such an opportunity.

To get a visual representation of four-dimensional figures, we will use analogies from lower-dimensional spaces for extrapolation to higher-dimensional figures, use the modeling method, apply system analysis methods to search for patterns between elements of four-dimensional figures. The proposed models should adequately describe the properties of four-dimensional figures, do not contradict each other and give a sufficient idea of ​​a four-dimensional figure and, first of all, of its geometric shape. Since there is no systematic and visual description of four-dimensional figures in the literature, but only their names indicating some properties, we propose to start the study of four-dimensional figures with the simplest - a four-dimensional cube, which is called a hypercube.

Hypercube definition

hypercubea regular polytope is called, the cell of which is a cube.

Polytop is a four-dimensional figure, the boundary of which consists of polyhedra. An analogue of a cell of a polytope is a face of a polyhedron. The hypercube is analogous to a three-dimensional cube.

We will have an idea about the hypercube if we know its properties. The subject perceives some object, representing it in the form of some model. Let's use this method and present the idea of ​​a hypercube in the form of various models.

Analytical Model

We will consider a one-dimensional space (straight line) as an ordered set of pointsM(x), Where xis the coordinate of an arbitrary point on the straight line. Then the unit segment is given by specifying two points:A(0) and B(1).

A plane (two-dimensional space) can be viewed as an ordered set of points M(x; y). The unit square will be completely defined by its four vertices: A(0; 0), B(1; 0), C(1; 1), D(0; 1). The coordinates of the vertices of the square are obtained by adding zero to the coordinates of the segment, and then one.

Three-dimensional space - an ordered set of points M(x; y; z). Eight points are required to define a 3D cube:

A(0; 0; 0), B(1; 0; 0), C(1; 1; 0), D(0; 1; 0),

E(0; 0; 1), F(1; 0; 1), G(1; 1; 1), H(0; 1; 1).

The cube coordinates are obtained from the square coordinates by adding zero and then one.

Four-dimensional space is an ordered set of points M(x; y; z; t). To specify a hypercube, you need to determine the coordinates of its sixteen vertices:

A(0; 0; 0; 0), B(1; 0; 0; 0), C(1; 1; 0; 0), D(0; 1; 0; 0),

E(0; 0; 1; 0), F(1; 0; 1; 0), G(1; 1; 1; 0), H(0; 1; 1; 0),

K(0; 0; 0; 1), L(1; 0; 0; 1), M(1; 1; 0; 1), N(0; 1; 0; 1),

O(0; 0; 1; 1), P(1; 0; 1; 1), R(1; 1; 1; 1), S(0; 1; 1; 1).

The hypercube coordinates are obtained from the coordinates of the 3D cube by adding a fourth coordinate equal to zero and then one.

Using the formulas of analytic geometry for the four-dimensional Euclidean space, one can obtain the properties of a hypercube.
As an example, consider the calculation of the length of the main diagonal of a hypercube. Let it be required to find the distance between points A(0, 0, 0, 0) and R(1, 1, 1, 1). To do this, we use the distance formula in four-dimensional Euclidean space.

In two-dimensional space (on a plane), the distance between points A(x 1 , y 1) and B(x 2 , y 2) is calculated by the formula

This formula follows from the Pythagorean theorem.

The corresponding formula for the distance between points A(x 1 , y 1 , z 1) and B(x 2 , y 2 , z 2) in three-dimensional space has the form

And in one-dimensional space (on a straight line) between points A( x 1) and B( x 2) you can write the corresponding distance formula:

Similarly, the distance between points A(x 1 , y 1 , z 1 , t 1) and B(x 2 , y 2 , z 2 , t 2) in four-dimensional space will be calculated by the formula:

For the proposed example, we find

Thus, the hypercube exists analytically, and its properties can be described no worse than the properties of a three-dimensional cube.

Dynamic Model

The analytical model of a hypercube is very abstract, so let's consider another model - a dynamic one.

A point (a zero-dimensional figure), moving in one direction, generates a segment (a one-dimensional figure). The segment, moving in a direction perpendicular to itself, creates a square (two-dimensional figure). The square, moving in a direction perpendicular to the plane of the square, creates a cube (three-dimensional figure).

The cube, moving perpendicular to the three-dimensional space in which it was originally located, generates a hypercube (four-dimensional figure).

The hypercube boundary is three-dimensional, finite and closed. It consists of a three-dimensional cube in the initial position, a three-dimensional cube in the final position, and six cubes formed by moving the squares of the original cube in the direction of the fourth dimension. The entire boundary of the hypercube consists of 8 three-dimensional cubes (cells).

When moving in the initial position, the cube had 8 vertices and in the final position also 8 vertices. Therefore, the hypercube has a total of 16 vertices.

Four mutually perpendicular edges emanate from each vertex. In total, the hypercube has 32 edges. In the initial position, it had 12 edges, in the final position also 12 edges, and 8 edges formed the tops of the cube when moving in the fourth dimension.

Thus, the border of the hypercube consists of 8 cubes, which consist of 24 squares. Namely, 6 squares in the initial position, 6 in the final position, and 12 squares formed by moving 12 edges in the direction of the fourth dimension.

geometric model

The dynamic model of a hypercube may seem insufficiently clear. Therefore, consider the geometric model of the hypercube. How do we get the geometric model of a 3D cube? We unfold it, and from the unfold we “glue” the cube model. The development of a three-dimensional cube consists of a square, to the sides of which is attached a square plus one more square. We turn adjacent squares around the sides of the square, and connect the adjacent sides of the squares to each other. And we close the remaining four sides with the last square (Fig. 1).

Similarly, consider the unfolding of the hypercube. Its development will be a three-dimensional figure, consisting of the original three-dimensional cube, six cubes adjacent to each face of the original cube, and one more cube. There are eight three-dimensional cubes in total (Fig. 2). In order to obtain a four-dimensional cube (hypercube) from this development, each of the adjacent cubes must be rotated by 90 degrees. These adjoining cubes will be located in a different 3D space. Connect adjacent faces (squares) of cubes to each other. Embed the eighth cube with its faces into the remaining unfilled space. We get a four-dimensional figure - a hypercube, the boundary of which consists of eight three-dimensional cubes.

Hypercube image

It was shown above how to “glue” a hypercube model from a three-dimensional sweep. We get images using projection. The central projection of a three-dimensional cube (its image on a plane) looks like this (Fig. 3). Inside the square is another square. The corresponding vertices of the square are connected by segments. Adjacent squares are depicted as trapezoids, although they are squares in 3D space. The inner and outer squares are different sizes, but in real 3D space they are equal squares.

Similarly, the central projection of a four-dimensional cube onto three-dimensional space will look like this: inside one cube is another cube. The corresponding vertices of the cubes are connected by segments. The inner and outer cubes have different sizes in 3D space, but they are equal cubes in 4D space (Figure 4).

Six truncated pyramids are images of equal six cells (cubes) of a four-dimensional cube.

This three-dimensional projection can be drawn on a plane and you can verify the truth of the properties of the hypercube obtained using the dynamic model.

The hypercube has 16 vertices, 32 edges, 24 faces (squares), 8 cells (cubes). Four mutually perpendicular edges emanate from each vertex. The boundary of the hypercube is a three-dimensional closed convex figure, the volume of which (the side volume of the hypercube) is equal to eight unit three-dimensional cubes. Inside itself, this figure contains a unit hypercube, the hypervolume of which is equal to the hypervolume of the unit hypercube.

Conclusion

In this work, the goal was to give an initial acquaintance with four-dimensional space. This was done on the example of the simplest figure - the hypercube.

The world of four-dimensional space is amazing! In it, along with similar figures in three-dimensional space, there are also figures that have no analogues in three-dimensional space.

Many phenomena of the material world, the macrocosm and the megaworld, despite the grandiose successes in physics, chemistry and astronomy, have remained inexplicable.

There is no single theory that explains all the forces of nature. There is no satisfactory model of the Universe that explains its structure and excludes paradoxes.

By knowing the properties of four-dimensional space and borrowing some ideas from four-dimensional geometry, it will be possible not only to build more rigorous theories and models of the material world, but also to create tools and systems that function according to the laws of the four-dimensional world, then human capabilities will be even more impressive.