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The Math Thread

For each polytope or tiling, one can construct a dual one, and that one has a reversed Schläfli symbol. Since duality is its own inverse, one has either self-dual polytopes or pairs of duality-related polytopes. Here's what's what.

1D: line segment -- self-dual
2D: regular polygons -- self-dual

Infinite families:
Simplex -- self-dual
Hypercube and cross-polytope -- duality pair
Hypercube lattice -- self-dual

Extras:

2D: hexagonal and triangular plane tilings -- duality pair
3D: dodecahedron and icosahedron -- duality pair
4D: 24-cell -- self-dual
4D: 120-cell and 600-cell -- duality pair
4D: 16-cell and 24-cell space lattices -- duality pair


There are some polytopes that have non-integer symbol members.

Star polygons have (fraction), where the numerator is the number of vertices, and the denominator is the number of vertices from a vertex to a vertex that it is connected to. Thus, a pentagram is (5/2).

In 3D:  Kepler–Poinsot polyhedron
Small stellated dodecahedron: (5/2, 5)
Great stellated dodecahedron: (5/2, 3)
Great dodecahedron: (5, 5/2)
Great icosahedron: (3, 5/2)

In 4D:  Regular 4-polytope
(5/2, 3, 3), (5/2, 3, 5), (5/2, 5, 3)
(3, 5/2, 5), (5, 5/2, 3), (5/2, 5, 5)
(3, 3, 5/2), (3, 5, 5/2), (5, 3, 5/2)
(5/2, 5, 5/2)
 
Back to small finite groups, there is an interesting curiosity about the two order-8 nonabelian groups: Dih(4) and QD(2).

Dih(4) elements:
{{1,0},{0,1}}, {{0,-1},{1,0}}, {{-1,0},{0,-1}}, {{0,1},{-1,0}}, {{1,0},{0,-1}}, {{0,1},{1,0}}, {{-1,0},{0,1}}, {{0,-1},{-1,0}}

With an appropriate linear transform, this can be turned into:
{{1,0},{0,1}}, {{i,0},{0,-i}}, {{-1,0},{0,-1}}, {{-i,0},{0,i}}, {{0,1},{1,0}}, {{0,i},{-i,0}}, {{0,-1},{-1,0}}, {{0,-i},{i,0}}

Dnew = T.Dold.T-1 giving Dnew.T = T.Dold

QD(2) elements:
{{1,0},{0,1}}, {{i,0},{0,-i}}, {{-1,0},{0,-1}}, {{-i,0},{0,i}}, {{0,i},{i,0}}, {{0,-1},{1,0}}, {{0,-i},{-i,0}}, {{0,1},{-1,0}}

Or more symbolically, as elements a and b, where a.b = b.a3, a4 = e, and b2 = 1 for Dih(4) and a2 for QD(2). All the elements and their conjugacy classes:
{e}, {a2}, {a, a3}, {b, a2.b}, {a.b, a3.b}

This gives us 5 irreducible representations or irreps, and they share this character table (class multiplicities on top):

1, 1, 2, 2, 2
1, 1, 1, 1, 1
1, 1, -1, 1, -1
1, 1, 1, -1, -1
1, 1, -1, -1, 1
2, -2, 0, 0, 0

The first one is the identity character. The second one is for even number of a's being 1 and odd numbers of a's being -1. The third one is for even numbers of b's being 1 and odd numbers of b's being -1. The fourth one is the product of those two.

The fifth one is the trace of the group matrices -- same for both Dih(4) and QD(2).

Both groups have the same subgroup structure. Here, I'll give each subgroup with its cosets, followed by the quotient group and the subgroup's abstract structure:
{e, a, a2, a3}, {b, a.b, a2.b, a3.b} -- (Z2): Z4
{e, a.b, a2, a3.b}, {b, a, a2.b, a3} -- (Z2): Z2*Z2 or Z4
{e, b, a2, a2.b}, {a, a.b, a3, a3.b} -- (Z2): Z2*Z2
{e, a2}, {a, a3}, {b, a2.b}, {a.b, a3.b} -- (Z2*Z2): Z2
 
As I'd mentioned earlier, that's true in general for Dih(2*n) and QD(n) for n > 1.

I will now return to the issue of a generalized dihedral group, where an = e, bm = as, and a*b = b*ar.

If r = 1, then we get an abelian group
Dih(k) has n = k, m = 2, r = -1, s = 0
QD(k) has n = 2k, m = 2, r = -1, s = k

a*bm = bm*ar^m
giving us
rm - 1 = 0 mod n

Also, bm+1 = b*bm = bm*b = b*as = as*b
giving us
s*(r - 1) = 0 mod n

Given bm = as, let us consider related elements.
(ak*b)m = as + k*(r^(m-1)+r^(m-2)+...+1)

Note that if s satisfies (r-1)*s = 0 mod n, then so does s + k*(r^(m-1)+r^(m-2)+...+1). So one will get a family of related s's.
 
Let us see what conjugacy classes this family of groups has.

First, consider conjugacy for ak by b-1: b-1*ak*b = ar*k

So one has a class {ak, ar*k, ar^2*k, ..., ar^(t-1)*k}

where k*(rt - 1) = 0 mod n.

Then, consider conjugacy for bv*au by ak*bj: ak*bj+v*au*b-j*a-k

So one has a class {bv * ak*(r^v - 1) + u*r^(m-j) for values of j and k with distinct results}. Notice that the power of b is unchanged, even though the power of a changes in most cases.


Center of the group for r != 1 mod n: ak for k*(r-1) = 0 mod n -- k is like possible s values. That is also the group's commutator subgroup. Its order is n/gcd(n,r-1), and it is abelian. That subgroup's quotient group is the maximal abelian quotient group. That group is Z(m) * Z(gcd(n,r-1))
 
this is just a math riddle...what's the final number in the series 1,2, 3, 4,6, 11, 13, 22, ?


the first number is 1 in base10, then 2 in base9...so the last should be 9 in binary=1001

 
tantric said:
this is just a math riddle...what's the final number in the series 1,2, 3, 4,6, 11, 13, 22, ?


the first number is 1 in base10, then 2 in base9...so the last should be 9 in binary=1001

Click to expand...

I think you meant 1, 2, 3, 4, 5, 11, 13, 22, ?

Yeah, that's why I couldn't figure it out and had to look at the answer. Sure. ;)

Good riddle.
 
Bomb#20 said:
tantric said:
this is just a math riddle...what's the final number in the series 1,2, 3, 4,6, 11, 13, 22, ?


the first number is 1 in base10, then 2 in base9...so the last should be 9 in binary=1001

Click to expand...

I think you meant 1, 2, 3, 4, 5, 11, 13, 22, ?

Yeah, that's why I couldn't figure it out and had to look at the answer. Sure. ;)

Good riddle.
Click to expand...

Hmmph. The answer is obviously -347. :devil2:
 
Bomb#20 said:
tantric said:
this is just a math riddle...what's the final number in the series 1,2, 3, 4,6, 11, 13, 22, ?


the first number is 1 in base10, then 2 in base9...so the last should be 9 in binary=1001

Click to expand...

I think you meant 1, 2, 3, 4, 5, 11, 13, 22, ?

Yeah, that's why I couldn't figure it out and had to look at the answer. Sure. ;)

Good riddle.
Click to expand...

dammit, i had a hot mess in my text editor composing this, thanks, my bad. i usta have it so that the final number was 420, but i couldn't reconstruct it. please reuse and improve
 
Geometric/topological division by zero.

Can a 0d point surround an infinite volume?
 
Here's a nice new YouTube channel about mathematics: PBS Infinite Series - YouTube by mathematician Kelsey Houston-Edwards.

Like graphing relationships between people or groups or nations: Network Mathematics and Rival Factions | Infinite Series. It keeps relationships simple -- friend or enemy -- and considers which sorts of triangles of relationships are stable or unstable.

FFF = stable
FFE = unstable
FEE = stable
EEE = unstable

A weaker version has EEE = stable. What graphs of relationships are possible in the weak case? The strong case?


Here are some older ones: Numberphile - YouTube and Computerphile - YouTube (has some stuff on algorithms)
 
doi:10.1016/j.physd.2006.09.028 - dresden.pdf -- Social balance on networks: The dynamics of friendship and enmity -- T. Antal, P.L. Krapivsky, S. Redner

has these graphs for the major European nations in the years before World War I (Britain, Austria, Germany, Italy, Russia, France). Note: some of them are not fully connected.

Three Emperors' League: 1872-81
Friends: AG, AR, GR
Enemies: BA, BR, BF, FA, FG, FR

Triple Alliance: 1882
Friends: AG, AI, AR, GI, GR
Enemies: BA, BR, BF, FA, FR

German-Russian Lapse: 1890
Friends: AG, AI, GI
Enemies: BA, BR, BF, FA, FG, RG

French-Russian Alliance: 1891-94
Friends: AG, AI, GI, FR
Enemies: BA, BR, BF, FA, FG, RA, RG

Entente Cordiale: 1904
Firends: BF, FR, AG, AI, GI
Enemies: BA, BR, FA, FG, RG

British-Russian Alliance: 1907
Friends: BF, BR, FR, AG, AI, GI
Enemies: BA, BG, BI, FA, FG, FI, RA, RG, RI

In WWI, Italy went neutral, then changed alliances:
Friends: BF, BR, FR, AG,
Enemies: BA, BG, FA, FG, RA, RG

Friends: BF, BR, BI, FR, FI, RI, AG
Enemies: BA, BG, FA, FG, RA, RG, IA, IG

I'll have to make diagrams of all of these, like in the paper.
 
lpetrich said:
Here's a nice new YouTube channel about mathematics: PBS Infinite Series - YouTube by mathematician Kelsey Houston-Edwards.

Like graphing relationships between people or groups or nations: Network Mathematics and Rival Factions | Infinite Series. It keeps relationships simple -- friend or enemy -- and considers which sorts of triangles of relationships are stable or unstable.

FFF = stable
FFE = unstable
FEE = stable
EEE = unstable

A weaker version has EEE = stable. What graphs of relationships are possible in the weak case? The strong case?


Here are some older ones: Numberphile - YouTube and Computerphile - YouTube (has some stuff on algorithms)
Click to expand...

Check out Mathologer, 3Blue1Brown, singingbanana, and standupmaths too.
 
This is not a high level problem but it is tricky if you are rusty

11374471776231847160.png
First, this is a square

hint of what methods are probably useless in the spoiler box. I wasted a lot of time and made lots of diagrams following would seems to be a circular goose chase where an equation A substituted with data M and N leads to equation B and then C which when solved gives equation A again.


This does not require any geometry constructions at all, only simple algebra.

second hint



Labels
side length s
short side of area 2 triangle a
short side of area 4 triangle b



 
repoman said:
This is not a high level problem but it is tricky if you are rusty

View attachment 11794
First, this is a square

hint of what methods are probably useless in the spoiler box. I wasted a lot of time and made lots of diagrams following would seems to be a circular goose chase where an equation A substituted with data M and N leads to equation B and then C which when solved gives equation A again.


This does not require any geometry constructions at all, only simple algebra.

second hint



Labels
side length s
short side of area 2 triangle a
short side of area 4 triangle b



Click to expand...

I like brilliant.org, they have some fun puzzles. My (geometric) approach:


If we take the legs of the yellow triangle, their product is 6 and their difference, times the side length of the square, is 4. Since the red triangle has half the area of the green triangle, the side length of the square is twice the longer yellow leg less the shorter yellow leg. Putting those together, we get that their difference is 1, and so the square side length is 4, and the blue area is 7.

Alternatively:


The three triangles add to 9, and the next perfect square is 16, which makes everything work!

 
So, every complex number has a direction associated with it.

What direction is 0 associated with?

How about with analytic continuation, what direction is 0 associated with?
 
Is there a 3d (filled) Riemann sphere? I'm thinking of one in which j (z) is constrained to positives.


RiemannKugel3dsmall.jpg
 

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Kharakov said:
So, every complex number has a direction associated with it.

What direction is 0 associated with?

How about with analytic continuation, what direction is 0 associated with?
Click to expand...

Every complex number has a direction except for 0, which will have indeterminate direction. Arg(z) isn't analytic, so analytic continuation might be a bit of a challenge. ;)

Kharakov said:
Is there a 3d (filled) Riemann sphere? I'm thinking of one in which j (z) is constrained to positives.


View attachment 12665
Click to expand...

Stereographic projection will still work for the n-sphere in (n+1)-dimensional space. You could also view the Riemann sphere as the complex projective line and then generalize to the complex projective plane (more generally, complex projective space).
 
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