# Math(s)! Currently: The Quantum 36 Officers Problem

Discussion in 'General Chatter' started by Exohedron, Mar 14, 2015.

1. ### TheSeer37 Bright Visionary Crushes The Doubtful

There's five platonic solids... is this because of the sides-to-corners relationships between cube/octahedron and icosahedron/dodecahedron? Each of those pairs only counts as one for this purpose? (And the tetrahedron has that relationship with itself.)

2. ### ExohedronDoesn't like words

Ack! You replied while I was busy editing. One moment. Restoring some stuff.

But yes, the group you get from the octahedron is identical to that of the cube, and same for the icosahedron and dodecahedron.

Last edited: Aug 14, 2016
3. ### ExohedronDoesn't like words

After complaining about how I never get to talk about the kind of math that I like, I'm going to attempt to talk about the kind of math that I like.

Consider a perfect sphere. We can rotate this sphere around in various ways about its center, and it still looks the same. We call the set of all rotations of the sphere SO(3). SO(3) is a group, in that you can do one rotation and then another, and you'll end up with a third rotation, and everything is reversible. Note that we're considering "don't do anything" to be a rotation by an angle of 0.

We can look for subsets of these symmetries that also form groups. In other words, we're looking for a set S of transformations where if A is in S and B is in S, then A followed by B is in S, and also the transformation that undoes A is also in S.

SO(3) has an infinite number of elements, because you can rotate by any angle you want. But infinite things are complicated, so we might want to look for finite subgroups of SO(3), or more explicitly subgroups of O(3) that have only a finite number of elements.
There are again an infinite number of these, but they're rather manageable. Moreover, we really only care about things up to isomorphism. In this case, that means that if you have two subgroups of SO(3), and changing coordinates (i.e. changing where you're looking at the sphere from) makes one subgroup look like the other, then we consider them to be equivalent.

So what are these subgroups?*

An: Pick a positive integer n and an axis of rotation. Now picture a regular n-sided polygon whose center is on that axis and which is perpendicular to that axis. An is the set of rotations around that axis that move a corner of the polygon to another corner of the polygon. Hence there are n distinct elements, because a rotation by 2π returns all the corners to their original positions.

Dn: Pick an integer n > 3 and an axis of rotation. Now picture a regular (n-2)-sided polygon whose center is on that axis and which is perpendicular to that axis. Dn is the set of rotations around that axis that move a corner of the polygon to another corner of the polygon, plus the rotations that flip the polygon over (still having corners get moved to corners). Dn has 2(n-2) elements: n-2 rotations around the axis, plus another n-2 flips.

E6: Consider a tetrahedron. E6 is the rotations that move the corners of the tetrahedron to corners. There are 12 such elements, because given any corner you can move it to any of 4 possible positions, and then there are 3 ways to rotate the tetrahedron while keeping that particular corner in place.

E7: Consider a cube. E7 is the rotations that move the corners of the cube to corners. There are 24 such elements, because given any corner you can move it to any of 8 possible positions, and then there are 3 ways to rotate the cube while keeping that particular corner in place.

E8: Consider a dodecahedron. E8 is the rotations that move the corners of the dodecahedron to corners. There are 60 such elements, because given any corner you can move it to any of 20 possible positions, and then there are 3 ways to rotate the dodecahedron while keeping that particular corner in place.

And that's it! The basic argument is that if there were any more, there would have to be more Platonic solids and there can't be.

Anyway, the reason I brought this up is that I wanted to just give a first glimpse of the ADE classification. This ADE classification shows up in a bunch of places, but this is probably the easiest one to understand.
And because there's no better place for a conspiracy theorist than in mathematics, yes, that is the same E8. Not the same manifestation (that's definitely not the E8 Lie group of physics fame) but the same place in the classification.

...Which is really super weird because the Lie groups E6. E7 and E8 exist because of the octonions and there fail to be more because of the octonions, but here we're talking about good old, 3-dimensional Platonic solids, not 8-dimensional nonassociative division algebras. Personally I think it must be because 8 = 23 but I have no idea what the actual reasoning is**.

*Okay, so normally this classification, called the Mckay Correspondence, is actually for the double covers of everything here: instead of SO(3), we look at the Spin(3) group, otherwise known as SU(2), and the corresponding "binary" versions of each group has twice as many elements.

**This is not all that ridiculous as far as serious mathematical statements go: for instance, there are still serious open questions surrounding the recently proven Poincare Conjecture (Perelman Theorem) whose intractability boils down to and can be easily explained by the fact that 1 + 3 = 4.

Last edited: Aug 15, 2016
4. ### ExohedronDoesn't like words

Draw a bunch of circles and draw some lines connecting them so that for any two circles there's at most one line connecting them, and that from any other circle, you can get to any other circle just by following the connections.
Formally we'd ask for a connected graph.

The goal is to label the circles with positive integers such that for each circle, its label is half of the sum of the labels of the circles connected to it.

It turns out that this condition puts a very strong constraint on the number of circles and how they're connected, at least up to "graph isomorphism", or in other words, if you drag the circles around with the lines that are connected following them, you don't consider yourself to have a new graph. Can you figure out the full set of solutions up to graph isomorphism?

In fact, there's a full classification. And it looks a lot like the classification in the previous post. Although it isn't too bad to figure out the exact solutions for yourself so I'm putting them in a spoiler.
Now the real mystery is why the classifications look anything alike. I actually have a much better understanding of the reason that the ADE classification works here than I do of the previous one, but only in terms of rather high level machinery. There's some simple combinatorial reasoning that will also spit out the answers and would probably explain why it matches the classification above, but I actually don't think I can reconstruct it without some help.

The white dots are just for getting the spacing right; black dots are actual ellipsis, indicating omitted pieces of the diagram
An: Take a positive integer n and consider n circles connected in a loop.
1-1-...-1-1
|............|
1-1-...-1-1

Dn: Take a positive integer n and consider n-3 circles connected in a chain. To each of the circles on the ends of the chain, connect to more circles to form a Y shape.
....1................1
....|................|
1-2-2-2-...-2-2-1

E6:
.......1
.......|
.......2
.......|
1-2-3-2-1

E7:
...........2
...........|
1-2-3-4-3-2-1

E8:
.................3
.................|
1-2-3-4-5-6-4-2

Last edited: Aug 14, 2016
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5. ### TheSeer37 Bright Visionary Crushes The Doubtful

I think your spacing on the solutions got messed up.

Also, what about a cross of five nodes, with 4 in the center and 2 on each end?

Edit: nvm, that's D4

Edit again: wait, shouldn't it be D3? The sphere rotations had a D3. Is the central chain of 2s in the D group of length n-2, not n-3 as you said?

Last edited: Aug 14, 2016
6. ### ExohedronDoesn't like words

Yeah, I'll need to fix the spacing. Or more accurately I'll need to figure out how to fix the spacing.

And no, it's n-3. D4 has five nodes in total, i.e. a cross shape. The chain is one node long, and then you attach four nodes to it. Dn should end up with n+1 nodes in total, and should also technically have a tilde over the D.
Normally to get the diagrams I'm discussing, you really should be starting with a chain of n-2 nodes, add two more nodes to one end to get the regular diagram, and then one more node to the second-from-the-other-end node to get the extended diagram.

7. ### TheSeer37 Bright Visionary Crushes The Doubtful

In that case... in the rotating-sphere classification, the As started with A3 and the Ds started with D3, but in the graph classification it looks like the Ds start with D4? I notice that the numbering counts the number of lines between circles, so now I get why they're numbered that way, but the slight asymmetry is bugging me.

8. ### ExohedronDoesn't like words

The As start with A1 actually, which corresponds to a "1-gon", i.e. a point. The group is the trivial group.
For the D series, there's an indexing issue that doesn't seem to be consistent. According to the Mckay Correspondence, D2 should have the five-dot diagram, but in most other descriptions, D4 has the five dot diagram.
Let me fix the indices a bit.

Last edited: Aug 14, 2016
9. ### ExohedronDoesn't like words

Consider n-dimensional space. You want to pick n vectors such that the angles between any two vectors is either π/2 or 2π/3. You also want this selection of vectors to be "irreducible", in the sense that you can't divide the vectors into two sets where all the vectors in one set are perpendicular to all the vectors in the other set.

By now you can probably figure out where this is going. So to save you the trouble, I'll just say that the solutions are essentially the same as those of the previous version, with the translation being that the circles correspond to vectors; circles that are connected correspond to vectors where the angle between them is 2π/3, while circles that are not connected correspond to vectors where the angle between them is π/2.

Other instances of the ADE classification showing up include crystallographic lattices, simply-laced Lie groups, uniform polytopes, quivers of finite type, elementary catastrophes, and generalized quadrangles.*

Most of these are linked through the connected circles version, although the vector version is also quite useful in explaining a few. In fact, the simply-laced Lie groups (my favorite of these, by the way) are linked directly to both the circles and the vectors version!
Anyway, the point here is that there is definitely something going on with the ADE classification showing up in so many places. I can explain some of the connections, but definitely not all of them, and I'm not even sure good explanations are known.

Anyway, the point of this series of posts is to indicate that mathematics is a great place to be a conspiracy theorist, or as mathematicians are more likely to say, a category theorist.

*For some of these, the ADE classification is only a partial list, as evidenced by the missing letters. For instance, the full classification of uniform polytopes goes from A to I.

10. ### ExohedronDoesn't like words

Projective space

Imagine a pair of traintracks receding into the distance. They're actually parallel, but it looks like they get closer together, and if you can see all the way to the horizon then I envy you it looks like they might actually meet.
But parallel lines don't meet! That's how they're defined, right? Well, not necessarily. We could simply think of them as two lines that head off in the same direction. And certainly in our normal, flat, Euclidean plane, lines that start in the same direction never meet.

But this is mathematics, where the rules are made up and points can be added in if we want. In particular, we're going to add some extra points so that parallel lines can meet. These extra points can't be in our usual Euclidean plane, sure, but they can be beyond it. We mathematicians say that we have a bunch of points at infinity. This is a different infinity than any of the other infinities mentioned so far in this thread. This is a geometric infinity.

Let's go back to Euclid's axioms. We have that any two points determine a unique line that connects them. And normally we'd have that any two lines meet in one point, unless they're parallel. Only now we have that parallel lines meet at infinity.
But! Lines go off in two directions. So should we have them meet at two points at infinity? Well, we could, but that would make things awkward. We want every pair of lines to meet in a single point.

So we just have both ends of a line go to the same point at infinity. And now we have that all parallel lines go to the same point at infinity, in both directions. So now our "parallel" lines are just lines that go in the same direction, and we have a point at infinity for each set of parallel lines.
But now we have points at infinity, and any two points determine a line! So two points at infinity have to determine a line. That line can't be a normal line, because each normal line only goes to one point at infinity. So in addition to our new points at infinity, we need a new line at infinity, and all the points at infinity lie on this line.

Thus we end up with the projective plane, which is the real plane plus a bunch of points at infinity that form a line at infinity. One fun fact about the projective plane: any theorem of geometry that doesn't depend on angles or lengths, just on the question of whether various points lie on various lines, has a "dual" theorem where you replace all the points with lines and all the lines with points, since now the axioms for lines and points are dual to each other.

In terms of the Cayley-Klein classification of geometries listed a few posts ago, the projective plane would be Elliptic since it has no parallel lines and no parallel points.

I'm going to break this into several posts. The next one will be about sticking coordinates on our projective plane. In particular, it would be nice to have a way to talk about points without distinguishing between normal points and points at infinity, and a way to talk about lines without distinguishing between normal lines and the line at infinity.

Last edited: Dec 26, 2016
11. ### TheSeer37 Bright Visionary Crushes The Doubtful

How do you know there's only one line at infinity?

12. ### ExohedronDoesn't like words

Pick a normal point and a line at infinity. Every point on the line at infinity defines a line that goes through the normal point. If we want to make things continuous, then as we go along the line at infinity, the corresponding lines through the normal point have to sweep out some wedge of the normal plane. Since our line at infinity doesn't have endpoints, that wedge has to be the entire normal plane. Hence for each direction that you can go from the normal point, you end up at a point on our line at infinity. Since every normal line is parallel to some line through our normal point, every normal line hits our line at infinity.
Since our points at infinity correspond to sets of parallel normal lines, and every set of parallel lines hits our particular line at infinity, we can only have one line at infinity.

Note that there are a bunch of hidden assumptions here, which is why the actual axioms of projective geometry make no distinction between "normal" and "at infinity". We'd be much better off using either the fully synthetic version or the coordinate version of projective space.

13. ### TheSeer37 Bright Visionary Crushes The Doubtful

Got it. You were saying about coordinates?

14. ### ExohedronDoesn't like words

Projective Space in Coordinates

Let's go back to our normal plane. We can describe points via coordinates: (0, 0), (1, 4), (3.23, 451.659), etc. So each point takes two numbers to describe.
And we can describe lines by three numbers. Namely, given numbers a, b and c where at least one of a or b is nonzero we say that the corresponding line is all of the points (X, Y) such that
aX + bY + c = 0
If two lines have the same a and b but different values of c, they are parallel. Having different values for a and b doesn't quite net you nonparallel lines, because you can see that if you multiply a, b and c by the same nonzero factor, the resulting equation describes the same set of points. In other words,
paX + pbY + pc = 0
describes the same set of points as the previous equation.
But if we have (a, b, c) and (r, s, t) where there is no factor p such that pa = r and pb = s and pc = t, then the two triples do describe different lines, and if the ratio of a to b is different from the ratio of s to t, then the lines aren't parallel. So instead we write the line as [a: b : c] to denote that what we're concerned about is the ratios.
So points take two numbers, and lines take three numbers only you can scale the numbers by a common nonzero factor to get the same line, and the statement "the point (X, Y) is on the line [a: b: c] if and only if ax + by + c = 0".
But we want points and lines to act similarly. So we tack on a third coordinate to our points, and write
[X: Y: 1]
so that our equation becomes aX + bY + c1 = 0
And now we note that the triple [pX: pY: p] satisfies the equation apX + bpY + cp = 0 if [X: Y: 1] satisfies aX + bY + c1 = 0. So now our points also have three coordinates and the scaling property. We can now start writing them as [x: y: z], with the translation to normal coordinates X = x/z, Y = y/z, as long as z isn't 0.

What about a point [x: y: 0]? Where does that lie?
Well, suppose we have a line [a: b: c] where ax + by + c0 = 0. Then it doesn't actually matter what c is. So [x: y: 0] lies on a bunch of parallel lines, and hence must be a point at infinity.
What about the line [0: 0: 1]? We would have to look for points such that 0x + 0y + 1z = 0, i.e. z = 0. So the line with coordinates [0: 0: 1] must be the line at infinity.

So to recap: a point has coordinates [x: y: z] where at least one coordinate is nonzero, and a line has coordinates [a: b: c] where at least one coordinate is nonzero, and we're actually only concerned about the ratios of the coordinates, and we say that the point [x: y: z] is on the line [a: b: c] if and only if ax + by + cz = 0.
Note that in our original description we distinguished between normal stuff and stuff at infinity by saying that the last coordinate acts differently than the first two, but if we don't consider the last coordinate to be special then we get just three ratios that work for describing all of the points and all of the lines and the condition for a point to be on a line is nice and symmetrical in all of the coordinates involved. Also now the computation for finding the line that connects two points looks exactly like the computation for finding the point where two lines meet.
Also of note: we're not considering [0: 0: 0], because then there are no meaningful ratios anywhere. The corresponding "line" would contain all points, and the corresponding "point" would lie on all lies, and neither of those are possible for any line or point, be they normal or at infinity. So we don't consider [0: 0: 0] for anything; we always demand that at least one coordinate be nonzero.

What about geometric shapes other than points and lines? Polygons still sort of make sense because those are built out of points and line segments, but what about, say, circles?
We usually describe a circle centered at the origin by X2 + Y2 = r2. Using our substitution of X = x/z and Y = y/z, we would get (x/z)2 + (y/z)2 = r2, but that only works if z is nonzero. So we'll just multiply everything by z2 to get
x2 + y2 = r2z2
for our projective circle.
A more general circle would look like (X - a)2 + (Y - b)2 = r2, and again we can substitute and then multiply by z2. Same for ellipses.
What about a parabola? Starting with Y = X2, we substitute and get (y/z) = (x/z)2. Again multiplying by z2 gives
yz = x2
Notably, the point [0: 1: 0] is now on the parabola! Our parabola actually becomes an ellipse with one point on the line at infinity.
If we look at hyperbolas, a similar thing happens. Given X2 - Y2 = r2, we substitute and multiply by z2 to get
x2 - y2 = r2z2
Note that the points [1: 1: 0] and [1: -1: 0] are both on the hyperbola! So a hyperbola is just like a circle that touches the line at infinity twice.
So our three types of (nondegenerate) conic sections, which look quite different in normal Euclidean space, are really the same kind of object when we move to projective space; the only difference is how many times they touch the line at infinity. Ellipses never touch, parabolas touch once, and hyperbolas touch twice.

Last edited: Aug 18, 2016
15. ### Wiwaxiaproblematic taxon

geologist says hi. can you go over this one for me?

16. ### ExohedronDoesn't like words

From what I know of crystallography, all the interesting behavior comes from impurities, but mathematicians usually don't deal with that; we only talk about where the "atoms" are and not what they are, so we assume uniform behavior everywhere.

In particular, a mathematician's lattice is a set of points in n-dimensional space where you can shift the entire configuration of points around and it doesn't look like you've done anything. In other words, it looks like linear combinations of some set of basis vectors where the coefficients in the linear combination are integers.
In particular, for any two points in the configuration, you can slide the entire configuration over so that the first point ends up where the second point originally was, and the resulting configuration looks exactly like the original configuration. Notably, a mathematician's lattice is infinite, and all the points look the same.
Also we demand that there is some minimum distance between distinct points, because things get rather complicated otherwise.
The lattices that we're concerned with here are ones where you can have reflections and the configuration looks the same. In particular, pick a point P. For any point Q that is of minimal distance to P, there should be another point R that is directly opposite Q with respect to P. In other words, the vector PQ is negative the vector PR. Then we demand that there be a reflection that keeps P in place, moves Q to R and vice-versa, and leaves the configuration looking the same.
So this rules out, say, the body-centered cubic lattice, which is a perfectly good crystal lattice but doesn't have the reflection property. This comes from mathematicians sort of going at this backwards; they were trying to figure out stuff about groups of reflections, and then started asking what sets of reflections could be extended to lattices, and so ended up with crystallographic reflection groups and the lattices that were obtained from crystallographic reflection groups, as opposed to geologists who are probably more interested in the crystals themselves.

Anyway, we can ask what kind of lattices we can have with the reflection property.

An is an n-dimensional lattice, although the easiest description lives in n+1 dimensional space: you take the points (x0, x1,...,xn) where the coordinates are all integers and their sum is 0. So A2 is a triangular lattice, but is easiest to see as living in a plane sitting at an angle in 3-space. A3 is the face-centered cubic. A1 is just a line of evenly spaced dots.

Dn is an n-dimensional lattice where you take the points (x1, ..., xn) where all the coordinates are integers and their sum is an even number. D4 is a body-centered hypercube lattice at a slightly strange angle.

E6, E7 and E8 then give three more lattices, in dimensions 6, 7, and 8 respectively, but they're a little difficult to describe. Basically, you take the vectors you get from the third version of the ADE classification I gave you and take all possible linear combinations with integer coefficients.

All the other lattices end up being "products" of these lattices. For instance, the product of A1 and A2 would be parallel layers of triangular lattice; A2 gives the triangular lattice, A1 gives the layers. Similarly, what we would call D2 is equivalent to the product of A1 with itself, i.e. a square lattice, and the product of three copies of A1 is a simple cubic lattice.

As a technical note: my personal training tells me that there is a Bn. a Cn, an F4 and a G2, except that those correspond to root systems, i.e. what kinds of reflections we want to have. The Bn lattice is just the product of n copies of the A1 lattice, the Cn lattice is just the Dn lattice, the F4 lattice is just the D4 lattice, and the G2 lattice is just the A2 lattice.

Note: the E8 lattice was recently proven to be the most efficient way to pack identically-sized spheres in 8-dimensional space, which is pretty cool.

Last edited: Aug 17, 2016
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18. ### ExohedronDoesn't like words

So consider the body-centered cubic. You have a point in the center of the cube, and you have eight nearest neighbors at the corners of the cube. Reflecting one of the corners across the center to the opposite corner would screw up the other six corners. You'd still have a body centered cubic lattice, but you wouldn't have the same body-centered cubic lattice. The planes of reflection are perpendicular to the coordinate axes or perpendicular to the face diagonals of the cube, but not perpendicular to the body diagonals.

19. ### ExohedronDoesn't like words

Projective spaces

Our coordinate description of the projective plane RP2 allows us to talk about more general projective spaces RPn. For instance, we can talk about the projective line, RP1. Points are labeled by pairs [x: y], where x and y are not simultaneously 0, and we really only care about the ratio. A bit of doodling will tell you that our projective line is just a regular line plus a single point at infinity. So it's a circle.
Moreover, you can get the projective plane by taking a regular plane and attaching it to a projective line at infinity. In particular, we have points in the projective plane being written as [x: y: z], with the points "at infinity" having coordinates [x: y: 0], which obey the same conditions as the points of the projective line.
You can think of this as taking a disk and attaching its edge to a circle, but note that each point on the circle gets glued to two points on the boundary of the disk, points which are normally on opposite sides of the disk.
Similarly, we can talk about projective 3-space, RP3, which have [x: y: z: w], with [x: y: z: 1] being normal points, and the stuff at infinity is described by [x: y: z: 0], so the stuff at infinity forms a projective plane. But let's put the w in front, so that [1: x: y: z] are the normal points (or indeed any set of coordinates with w being nonzero) and [0: x: y: z] are the points at infinity.
And now you can think of this as taking a solid ball and gluing its surface to a projective plane, where each point on the projective plane glues to two antipodal points on the surface of the ball.
And in general, we have points in the n-dimensional projective space RPn being described by [x0: x1: ... : xn], where you only care about the ratio. And then the points [0: x1: x2: ... : xn] form a (n-1)-dimensional projective space RPn-1. Again we can think of it as taking an n-dimensional ball and gluing its surface to a RPn-1, where each point of the RPn-1 gets glued to two points on the surface of the n-dimensional ball.

We also see another way to describe things in n-dimensional projective space, looking at the geometric interpretation of our coordinate description. A projective point corresponds to a set of points in Rn+1 that form most of a line, missing only the origin. So really the projective point corresponds to the entire line in Rn+1.
We can then expect projective lines to correspond to planes through the origin in Rn+1. In the case of our projective plane, this now gives us our duality theorem, because lines through the origin in R3 are dual to planes through the origin in R3 space in a precise way: for each plane through the origin, there is a unique line through the origin perpendicular to the plane, and for each line through the origin there is a unique plane through the origin perpendicular to the line. A projective point being on a projective line is equivalent to a line through the origin in R3 being on a plane through the origin, and if a line L is on a plane P, then the plane perpendicular to L contains the line that is perpendicular to P. So hence all theorems in RP2 that only talk about whether various points are on various lines all have duals.

In general, a lot of geometers like projective spaces because we get nicer behavior than in regular Euclidean spaces. The same process that makes ellipses, parabolas and hyperbolas all the same thing in projective space makes a lot of other families of geometric objects nicer in projective space, because those points at infinity fill in holes that aren't visible in Euclidean space.

Finally, we can replace the real numbers with any other field, i.e. any other set where you can add, subtract, multiply and divide by nonzero things. We can look at complex projective spaces, where the coordinates are allowed to be complex numbers. For instance, CPn is the complex plane plus a point at infinity: [x: 1] is a normal complex number, and [x: 0] corresponds to the single point at infinity. Note that, from the point of view of the complex numbers themselves, the "complex plane" is 1-dimensional, because any point on it can be described by a single complex number. A single line gets a single point at infinity. So if you're only the complex plane and you head off infinitely in any direction, you reach the same point at infinity. What do you get when you attach a single point to the complex plane? You get a sphere. CP1 is a sphere, just as RP1 is a circle.
We can even do this for the quaternions: HP1 is a 4-dimensional sphere.* Although you have to be careful, because "ratios" has two meanings in the quaternions: when you say the ratio of y over x, do you mean x-1y or yx-1? As long as you pick one you'll be all right, but you have to pick one.

Things go kind of badly for OPn. In particular, for FPn to work for some object F, we need to be able to say that [x0: x1: ... : xn] and [px0: px1: ... : pxn] have the same ratios, but while for quaternions we have that
(px0)-1px1 = x0-1p-1px1 = x0-1x1
we don't have that for the octonions because the octonions aren't associative; there's an implicit moving of parentheses there that doesn't work in the octonions!

There are some objects that by some standards might be called OP1 and OP2, but they're defined rather indirectly, and the lack of further examples is why the En series in the ADE classification peters out at E8.
OP1 is an 8-dimensional sphere.

Of interest to a branch of mathematics called "arithmetic geometry" are projective spaces where the coordinates are from the various fields that number theorists like, for instance the rational numbers Q, or finite fields Fq, or p-adic fields Qp.
I'll let @Emu maybe explain what those fields are, because I don't work with them unless I really have to. As far as I'm concerned, FqPn falls under the heading of "really weird" for me.

*Note: a 4-sphere is the stuff at distance 1 from the origin in R5, just as the 2-dimensional sphere is the stuff at distance 1 from the origin in R3. The idealized surface of the Earth is a 2-sphere, not a 3-sphere, and a circle is a 1-sphere

Last edited: Aug 19, 2016
20. ### evilasSure, I'll put a custom title here

It's weird how this is the second time I'm just about to ask a question then I figure it out for myself. Like, a few days ago I wanted to ask whether the line at infinity basically has a point for each possible angle a line can point at in R2, and figured that one out on my own. Now I was just about to ask why was it that RP2 has a line but CP1 has a point when C can be mapped onto R2... and figured that one out on my own as well. All through just dividing stuff by other stuff.

Projective spaces are awesome. I can see why geometers love them. I definitely loved the Riemann Sphere (which I now recognize as CP1) when I took complex analysis.

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