Living in infinite-dimensional space

[PatrickPowers]

W

Now consider a house in the shape of a dome, i.e., half of an N-sphere. Assuming the diameter is a unit long, it will also take just one unit of time to get from one end to the other. There are no corners, of course, but assuming you start at the point of the N-sphere in the same direction as the corner of the N-cube, and try to walk to the opposite side diagonally. In the case of the dome, it will always only take 1 unit of time, no matter how large N is. In the limiting case where N diverges to infinity, this remains invariant. However, something strange happens: the point at the N-sphere's diagonal, i.e., the intersection of the N-sphere with <1,1,1,1...>, is <1/N, 1/N, 1/N, ...>. The corresponding opposite side is <-1/N, -1/N, -1/N, ..>. In the limiting case, both points converge to <0,0,0,...>, and you get the paradoxical situation where walking from the origin to the origin takes a non-zero amount of time, because the distance between the origin and the origin is 1 rather than 0.

This apparent contradiction goes away if you admit infinitesimals into your number system, in which case the two opposite corners are <ε, ε, ε, ...> and <-ε, -ε, -ε, ...> where ε is an infinitesimal.

Let's say that the unit sphere has radius one. Then that intersection is actually [1/sqrt(N), 1/sqrt(N), 1/sqrt(N)...]. The distance between the positive and negative intersections is always 2 regardless of N, so that's the limit as well.

[quickfur]

W

Now consider a house in the shape of a dome, i.e., half of an N-sphere. Assuming the diameter is a unit long, it will also take just one unit of time to get from one end to the other. There are no corners, of course, but assuming you start at the point of the N-sphere in the same direction as the corner of the N-cube, and try to walk to the opposite side diagonally. In the case of the dome, it will always only take 1 unit of time, no matter how large N is. In the limiting case where N diverges to infinity, this remains invariant. However, something strange happens: the point at the N-sphere's diagonal, i.e., the intersection of the N-sphere with <1,1,1,1...>, is <1/N, 1/N, 1/N, ...>. The corresponding opposite side is <-1/N, -1/N, -1/N, ..>. In the limiting case, both points converge to <0,0,0,...>, and you get the paradoxical situation where walking from the origin to the origin takes a non-zero amount of time, because the distance between the origin and the origin is 1 rather than 0.

This apparent contradiction goes away if you admit infinitesimals into your number system, in which case the two opposite corners are <ε, ε, ε, ...> and <-ε, -ε, -ε, ...> where ε is an infinitesimal.

Let's say that the unit sphere has radius one. Then that intersection is actually [1/sqrt(N), 1/sqrt(N), 1/sqrt(N)...]. The distance between the positive and negative intersections is always 2 regardless of N, so that's the limit as well.

Correct, which means that when N=∞, the coordinates of the intersection must be infinitesimal, because otherwise the intersection would be <0,0,0,0...> in both cases and you get the contradiction that the distance between <0,0,0,...> and <0,0,0,...> is non-zero.

[quickfur]

More thoughts about houses in N-dimensional space, including the limiting case where N→∞:

For simplicity, we assume that houses are in the shape of an N-cube. You can add a sloping roof with various triangular shapes if you want (e.g., a triangular polyprism, or an (N-1)-simplex prism, or a pyramid, whatever, it's not really that important).

Let's say the vertical dimension is the first coordinate. This is somewhat unusual since people usually assign the Y or Z axes to the vertical, but standardizing on the 1st coordinate makes the discussion simpler for general N. Furthermore, let's say the front of the house is the side of the N-cube facing the <0,1,0,0,..> direction, i.e., the +Y direction, the 2nd coordinate. So the back of the house would face the <0,-1,0,0...> direction. Again, this choice is to simplify the discussion for general N.

Now we'd put the front door somewhere in the front of the house, generally in center of this wall. Well, in actual houses we usually offset it for aesthetic purposes, but I think it's safe to say that it's somewhere around the center of the wall -- it'd be strange for the front door of a house to be located right at the corner of the wall (even though this isn't entirely unheard of). But for the purposes of this discussion let's keep it simple and say it's at the center of the front wall. So the door would be installed on the point <0,1,½,½,½,...> (the first coordinate is 0 because that's where the ground is; we assume the door extends upwards from there).

In 3D, this gives us the coordinates <0,1,½>, i.e., standing on the ground, in the front wall of the house, and midway between either side. A pretty standard location for a front door. The distance from the door to either corner of the house that touches the front wall, is ½ units.

In 4D, the front door would be located at <0,1,½,½>, i.e., standing on the ground, in the front wall, which is a cube, right in the center of this cube. Now note that a 4D house has 2 different kinds of corners: a ridge corner where 2 walls meet, and an apical corner where 3 walls meet (and where a vertex of the cube is). The distance from the front door to a ridge corner on the front wall (there are 4 of these in 4D, compared to only 2 in 3D) is also ½. The distance to an apical corner, however, is √2/2, which is slightly larger than ½.

In 5D, the front door would be located at <0,1,½,½,½>, standing on the ground, in the middle of the front wall which is now a tesseract. In 5D, there are 3 kinds of corners: a ridge corner where 2 walls meet, an apical corner where 3 walls meet, and a verticial corner where 4 walls meet. The distance to the ridge corner is again ½; the distance to the apical corner is √2/2, and the distance to the verticial corner is √3/2.

When we go to 6D, we find that the front wall has 4 kinds of corners. The nearest kind, where 2 walls meet, remains at a distance of ½ units from the door. The farthest kind, where 5 walls meet, is now 1 unit away: twice the distance to the ridge corner.

If we now go to 7D, we find that the distance from the front door to the nearest ridge corner is still ½ units away, but the distance to the verticial corner is now √5/2 units away. When we get to 11D, the distance to the verticial corner is 3 times the distance to the ridge corner. In 18D, the distance is 4 times the distance to the ridge corner. In 27D, the distance is 5 times the distance to the ridge corner: you can now walk around the entire house faster than it takes to walk from the front door to one of the house's far corners.

In general, in N dimensions, the front wall has (N-2) kinds of corners; the nearest kind is where 2 walls meet, and is always ½ units from the front door. The farthest kind, where (N-1) walls meet, is √(N-2)/2 units away. As N grows without bound, this distance grows without bound. When N diverges to ∞, we find ourselves in the strange situation where we can walk around the house in only 4 units of time; however, we can never reach the corner of the house from the front door in finite time.

So in a ∞-dimensional city, if you arrive at a house in the direction of its front wall, you will be able to get into the house no problem. But if you arrive at the house from one of its corners, you'll never reach the front door.

[quickfur]

Furthermore, if you own a ∞-dimensional house, and your house is wide enough to accomodate 2 rooms along each wall (a pretty reasonable size IMO!), then you have more than enough space to run a Hilbert Hotel. (In fact, much more than that. But we'll get to that.) Here's how you do it: suppose again that your house is in the shape of a cube, between <0,0,0,...> and <1,1,1...>. Divide the interior of the house into rooms of dimensions <1,½,½,½,...> (remember, the first coordinate is the vertical; we want at least adequate head room in each room so that our guests don't have to crawl -- that would not be very nice). How many of these rooms can we fit into our house?

A lot. For example, we can fit a room between <0,0,0,...> and <1,½,½,½...>, and another room between <0,½,0,0,0...> and <1,1,½,½,½,...>. And a room between <0,0,½,0,0,...> and <1,½,1,½,½,½,...>. And another room between <0,0,0,½,0,0...> and <1,½,½,1,½,½,...>. For any natural number n, we can fit a room between <0, ... {n zeros}, ½, 0, 0, ....> and <1, ... {(n-1) ½'s}, 1, ½,½,½,...>. So we have at least 1 room per natural number, and so we rent them out and turn our house into a Hilbert Hotel.

But actually, we can run a hotel much bigger than a Hilbert Hotel, all within the same house! To see this, let's return to a finite N-dimensional space and look at our floor plan. The floor is basically an (N-1)-cube with 2^N-1 vertices. Since we can fit 2 rooms per wall, that means we can fit 1 room per vertex of the floor plan. (We're simply subdividing our floor in half along each axis.) Now let N diverge to ∞. Since the floor of our ∞-dimensional house is still an ∞-cube, that means we can fit as many rooms in our house as there are vertices in the ∞-cube, which is an uncountable number. So not only we can run a single Hilbert Hotel; we can even run an infinite number of them all within our humble little house!

And in fact, much more than that. A countably infinite number of Hilbert Hotels still would not fill up all our rooms. We can run an uncountable version of the Hilbert Hotel -- let's call it the Cantor Hotel, where we can accomodate an uncountable number of guests!! By comparison, the set of guests in a Hilbert Hotel has measure zero compared to the number of guests in our Cantor Hotel, meaning that our hotel is basically still empty after receiving all the guests from the nearby Hilbert Hotel when they closed down because of a nasty water leak that made all their rooms unoccupiable. We can take in the guests from all the Hilbert Hotels in our town, and we'd still be able to accomodate all of them -- assuming there are a countable number of Hilbert Hotels in town.

Of course, this same argument can be applied to any ∞-dimensional house that's big enough to be subdivided into smaller rooms, so essentially every house in town can run a Hilbert Hotel.

But by the same argument, if our town consists of a map with blocks of N×N×N×... houses, for some small number N>1, that means there are an uncountable number of houses in town, and therefore an uncountable population of townsfolk. So a Hilbert Hotel wouldn't be able to accomodate the entire population of the town, but we don't really expect the number of visitors to match the number of local townsfolk anyway, so we may reasonably assume that we'd only get a countable number of visitors, and they can all stay at the local Hilbert Hotel. But in the event that a disaster destroys the next town over, there'd be an influx of an uncountable number of fleeing refugees. In which case we just have to designate one Safe House -- which has an uncountable number of rooms -- to accomodate them all, without upsetting the rest of the local townsfolk. No problem.

[steelpillow]

Is this why all infinite-dimensional buildings have curved walls - and floors? Any flat bits can have a cube erected on them and take forever to cross!

[PatrickPowers]

A surface usually has N-1 dimensions. In infinite dimensions there is no fixed N so a surface might seem not possible. Not!

Let's start by constructing an infinite dimensional rectangeloid with finite positive volume. We know that the series 1/n^2 converges. So all we have to do is have the series e^(1/n^2) for the widths of the rectangleoid. The sum of the logarithms of this series converges. So the product of the series also converges to a positive number. To get any positive volume you like use the series e^(x/n^2).

Now that we have this we know the values of all the vertices. Given that, we can determine whether any point in our space is inside, outside, or on the surface of the rectangloid. The surface is well-defined.

We know all the vertexes of the unit cube so it is also well-defined. The unit sphere, we just calculate the distance from any point to the center, so that's well-defined too. How this is done, not my problem.

[PatrickPowers]

By comparison, the set of guests in a Hilbert Hotel has measure zero compared to the number of guests in our Cantor Hotel, meaning that our hotel is basically still empty after receiving all the guests from the nearby Hilbert Hotel when they closed down because of a nasty water leak that made all their rooms unoccupiable.

I know on good authority that this occurred when infinite dimensional Keith Moon exploded a cherry bomb in his room's toilet.

[quickfur]

By comparison, the set of guests in a Hilbert Hotel has measure zero compared to the number of guests in our Cantor Hotel, meaning that our hotel is basically still empty after receiving all the guests from the nearby Hilbert Hotel when they closed down because of a nasty water leak that made all their rooms unoccupiable.

I know on good authority that this occurred when infinite dimensional Keith Moon exploded a cherry bomb in his room's toilet.

Due to the infinite rate of dissipation of an expanding cloud of gas in ∞-dimensional space, an ∞-dimensional bomb requires an infinite amount of explosive power in order to destroy anything outside a radius of 0. Therefore, cherry bombs are inherently safe in ∞-dimensional space.

[PatrickPowers]

By comparison, the set of guests in a Hilbert Hotel has measure zero compared to the number of guests in our Cantor Hotel, meaning that our hotel is basically still empty after receiving all the guests from the nearby Hilbert Hotel when they closed down because of a nasty water leak that made all their rooms unoccupiable.

I know on good authority that this occurred when infinite dimensional Keith Moon exploded a cherry bomb in his room's toilet.

Due to the infinite rate of dissipation of an expanding cloud of gas in ∞-dimensional space, an ∞-dimensional bomb requires an infinite amount of explosive power in order to destroy anything outside a radius of 0. Therefore, cherry bombs are inherently safe in ∞-dimensional space.

What makes you think they don't have an infinite amount of explosive power?

[quickfur]

[...]
Due to the infinite rate of dissipation of an expanding cloud of gas in ∞-dimensional space, an ∞-dimensional bomb requires an infinite amount of explosive power in order to destroy anything outside a radius of 0. Therefore, cherry bombs are inherently safe in ∞-dimensional space.

What makes you think they don't have an infinite amount of explosive power?

Because I'm assuming they're approximately in the shape of a ∞-sphere of some finite radius, which has zero volume (or at most infinitesimal). Therefore they only contain at most a negligible amount of firepower.

[quickfur]

More strange characteristics of ∞-space:

1) CD diagrams break down. While it is theoretically possible to extend CD diagrams to contain a non-finite number of nodes, it's not clear how to interpret such a diagram, and it runs into the problem that the resulting infinite diagrams cannot distinguish between the ∞-cross and the ∞-simplex. There's also the interesting question of what a mesotruncated ∞-cube might be (there are no obvious answers; there are two distinct limits that one might construct that truncate to each other, but it's not clear if these limits actually exist within ∞-space, and in any case they are unreachable via a countable number of operations starting from either the ∞-cube or the ∞-cross).

2) If we assume that a convex ∞-polytope is the convex hull of a set of coordinates (it's not clear this actually holds, because it's not clear how one would go about defining a convex hull in ∞ dimensions, but let's assume it works), then there are (at least) 3 distinct classes of uniform ∞-polytopes with ∞-cube symmetry:

(a) The "small" uniform polytopes: those whose outradius is finite. This includes the ∞-cross and some of its truncates. These have the property that there exists a ∞-sphere of finite radius that contains all of the vertices of the uniform polytope.

(b) The "large" uniform polytopes: those whose outradius is non-finite, but whose vertex coordinates are bounded. This includes the ∞-cube, some of its truncates, and some of the truncates of the ∞-cross. These have a finite bounding box: you can scale the ∞-cube by a finite factor such that it contains all the vertices of the uniform polytope. However, they do not have an inscribing sphere, because no finite scaling of the ∞-sphere can contain all of their vertices.

(c) the "very large" uniform polytopes: those whose vertex coordinates are unbounded, including some of the truncates of the ∞-cube, for example, the postulated omnitruncated ∞-cube whose vertex coordinates are all permutations of coordinate and changes of sign of <1, 1+√2, 1+2√2, ... 1+n√2, ...>. These also have infinite outradius, but since they have unbounded coordinates, they do not have a finite bounding box (no finite scaling of the ∞-cube can contain all the vertices of the uniform polytope). Obviously, no finite scaling of the ∞-sphere is able to bound them either.

The polytopes in category (a) exist in Hilbert space, and are relatively well-behaved. The polytopes in category (b) are outside Hilbert space, and have some pathological properties such as a non-finite outradius and the consequences of that (such as a house whose front door is unreachably far from its corners).

It's not clear whether the postulated polytopes in category (c) are even polytopes in any meaningful sense, because it's not clear how you'd define their bounding hyperplanes. Their bounding boxes are so large they span the entire ∞-dimensional space(!). Their dual polytopes either do not exist (if we restrict ourselves to ∞-space with finite coordinates), or they must be "extremely large" (the duals have vertices with infinite coordinate values). In the former case, it's impossible to define bounding hyperplanes for the polytope so it can't be a proper polytope in the ordinary sense; in the latter case the bounding hyperplanes lie outside of ∞-space (with finite coordinates), so it's not clear whether this is even well-defined.

This conundrum can only be resolved if the underlying field of coordinate values is extended from the real numbers to something like the hyperreals or the surreal numbers. But if we were to go that route, then that would imply also the existence of infinitesimal polytopes: those who coordinates have infinitesimal, but non-zero, values. Some of them may have non-infinitesimal volume and bounding boxes in ∞-space. And this immediately leads to a lot of very strange, even pathological consequences (and it's not even clear whether the resulting geometry can even be reasonably called a geometry anymore, or whether the whole thing will even be consistent!).

[quickfur]

Even more weirdness in ∞-space:

In finite-dimensional space, rotations and projections are distinct. In ∞-space, however, this is not necessary true. Consider, for example, a rotation R₁ that rotates the X axis 90° into the Y axis. Let's say we have a point P = <1, 2, 3, ...>. After applying this rotation, we have R₁·P = <-2, 1, 3, 4, ...>. Consider another rotation R₂ that rotates the Y axis 90° into the Z axis. If we apply this to the image of P under R₁, we get: R₂·R₂·P = <-2, -3, 1, 4, 5, ...>. We can invent a series of rotations R_n that rotates the n'th axis 90° into the (n+1)'th axis. If we chain all of these rotations together, we get a transformation that maps <1, 2, 3, ...> to <-2, -3, -4, ..>. What happened to the 1? It's gone! We've effectively dropped it from P. I.e., the limit of an infinite sequence of rotations is equal to a projection. (Don't worry about the minus signs; those can be fixed with a simple reflection.) This is only possible in ∞-dimensional space; in any N-dimensional space of finite N, no matter how large N and no matter how many rotations you chain together, you'll never end up with a projection. In ∞-dimensional space, however, rotations asymptotically approach a projection!

(And BTW, this is one characteristic of ∞-space that you cannot get by studying an N-dimensional space for large N. The limit works only when N=∞; for all finite N there's nothing that would indicate that this is the case.)

Now consider the projection S we've constructed as the limit of ...R₃·R₂·R₁. Applying S twice to P drops 2 coordinates off the front: S·S·P = S²·P= <3, 4, 5, ...>. Applying it n times drops n coordinates off P. Now the million dollar question: what's the limit of Sⁿ·P as n→∞? Note that the coordinates of P are unbounded, and that after n applications of S, the image of P looks like ±<n, n+1, n+2, ...>. So it's not clear what the limiting case would be, or whether there would be a limit at all. But we can get an idea of this by considering the infinite (pseudo)transformation matrix that represents S. It's basically a matrix with -1's down its diagonal, except shifted by one column so that there's a column of 0's on the left. If you compose S with itself, you get an infinite (pseudo)matrix with two columns of 0's on the left, and then the off-diagonal diagonal of 1's. Take that to the limit, and what do you get? The zero matrix. IOW, the result of projecting P along an infinite number of axes collapses it to the origin, i.e., a point in 0D space.

There are problems with this interpretation, though. For one thing, where did those 0 coordinates of P's image come from? Every finite series of applications of S produce a vector with non-zero coordinates, because P does not contain any 0's. Furthermore, each further application of S increases the minimum absolute value of the coordinates of the image of P. By that logic, at the limit we should get a point whose coordinates are infinitely large, rather than 0!

Furthermore, there's the issue of what happens to P just before it got projected to the 0 vector. Intuitively we'd expect that sometime before we dropped all the coordinates of P (via projection), there should be a finite number of non-zero coordinates left, and this number should gradually decrease to 0 at the limit. However, every finite series of compositions of S yields a transformation that maps P to a vector with decidedly non-zero coordinates. The number of non-zero coordinates in P never decreases, in spite of the projection. It's only at the limit that it leaps from an ∞-vector of very large coordinates suddenly into the zero vector. There is no in-between state; something discontinuous happens when we take the limit and it just leaps from a vector of non-zero coordinates to a vector of all zero coordinates.

There's also the issue of a different kind of projection operator: the kind that actually projects a point by setting one of its coordinates to zero. I.e., T₁·P = <0, 2, 3, 4, ...>. We can again, construct a sequence of these projection operators: T₁·T₂·P = <0, 0, 3, 4, ...> and so on. Taking the limit to ∞, we also get the zero matrix that maps P to the zero vector. Except, in this case, there's a gradual reduction of non-zero coordinates in P, until all coordinates become zero. There's still a quantum leap at the end -- any finite iteration of T still yields an image of P with an infinite number of non-zero coordinates, and the limiting transformation gets rid of all of 'em in one fell swoop. But at least there's an increasing prefix of 0's that makes the limit that projects to the zero vector seem more sensible. So T_i is a different kind of projection than S_i, that behaves more like the usual projection operator in finite-dimensional spaces. So we may call T_i the "proper" projection operators, whereas S_i are the "improper" projection operators. Both have the zero matrix at the limit. The "improper" projection operators behave like some weird intermediate class of operators between proper projections and rotation/reflections, that only exist in ∞-space.

(And BTW, you can combine the iterated T's with the iterated S's in such a way that the number of 0 coordinates grow at a different rate than the non-zero coordinates are "consumed". AFAIK there is no analogue of such a strange transformation in any finite-dimensional space. Another way to think of "improper projections" is that they behave like "deletion" operators, that delete, or forget, coordinates from an ∞-vector. A kind of weird boundary-case transformation the sits at the limit of infinite sequences of rotations that resembles both rotations and projections but behaves somewhat differently from either, in a way only possible in an ∞-space.)

[PatrickPowers]

You're right, I don't see how you can do without infinitesimal numbers. I guess ix is the number that when summed together infinite times is equal to x. I imagine there is a better way to do it, but I know zilch about infinitesimals.

As for the infinite Keith Moon cherry bomb, it is in direct contact with water so the infinite explosive dissipation doesn't apply. Can an infinite amount of explosive power can be confined in a finite cherry bomb? No, but the bomb could have infinite volume because any nonzero length of the pipe also has infinite volume. Just make a cylindrical bomb an inch long that fits inside the pipe. Then could water hammer cause countably infinite damage? I guess we require experimental results.