This is where it starts to get really interesting – when we have two isometric spaces there is no way an inhabitant could actually know which one is his own reality.  A 2 dimensional being could be living in either the curved space model, or the field model and not know which was his true reality.

The difference between the 2 models is that in the first instance we accept an unexplained curvature of space that causes objects to travel in “straight” lines along great circles, and that in the second instance we accept an unexplained field which forces objects travelling in “straight” lines to follow curved paths.  Both of these ideas are fundamental to Einstein’s Theory of Relativity – where we must account for both the curvature of space-time and a gravitational force field. Interestingly, our own 3 dimensional reality is isomorphic to the projection onto a 4 dimensional sphere (hypersphere) – and so our 3 dimensional universe is indistinguishable from a a curved 3D space which is the surface of a hypersphere.  A hypersphere may be a bit difficult to imagine, but the video above is about as close as we can get.Such a scenario would allow for our space to be bounded rather than infinite, and for there to be an infinite number of 3D universes floating in the 4th dimension – each bounded by the surface of their own personal hypersphere.

Ammonites are a group of extinct cephalopod mollusks with ribbed spiral shells. They are exceptionally diverse and well known to fossil lovers. Researchers have developed the first biomechanical model explaining how these shells form and why they are so diverse. Their approach provides new paths for interpreting the evolution of ammonites and nautili, their smooth-shelled distant “cousins” that still populate the Indian and Pacific oceans.


i was tagged by the super gorgeous gamayuns for the 20 beautiful women challenge!  omg thank you lovely! 

pls ignore my super large and wrinkled shirt. somewhere my mother just got a chill at the fact that I’ve let people see me like this. 

tagging tyrannotina, lyrrium, mnemonicsweets alistairweekend, noblehouseoftargaryen, postcolonialprincess, sailorplu, just-us-carnivores, lanoraa, harleensahani, solesoffire, tropicwitch, operativelawsons, kiransingh, night-catches-us, morismako, melissabenoisttt, quibbler, isomorphic, suhoings

don’t feel obligated to share your pictures or post this yourself. (´▽`ʃƪ) this is just if you want to! you’re all awesome and gorgeous and amaziiiing~~ 

Some  Math Humor

Mathematicians have a sort of bizarre and subtle sense of humor. Example: this quote I recently found on MSE.

I’ve just recalled a funny bit of somewhat related trivia: in the first chapter of Serge Lang’s Algebra […] Lang gives two concrete examples of monoids. The first one consists of the natural numbers with addition. The second one consists of… homeomorphism classes of compact connected surfaces with connected sum

[If you don’t get the joke, it’s isomorphic* to this one. If that doesn’t help either, well, you can’t win them all.]

Although with Lang, it’s hard to know if this is was an intentional joke or he was just like “what are the two simplest monoids I can think of” and that’s all he could come up with.

* I definitely made the post yesterday just so that I could have something to link to for this post.

Self-Complementary Graphs II
A property of self-complementary graphs is that if n is the number of nodes in a self-complementary graph, either n or n-1 must be divisible by 4.
Proof: If graph G is isomorphic to its complement, ~G. The union of the set of edges of G and the set of edges of ~G is the complete graph with n vertices. The number of edges in the complete graph is n choose 2, which is n(n-1)/2.
Since G and ~G have the same number of edges, the number of edges in the complete graph in this case must be divisible by 2. This means that n(n-1) is divisible by 4.
Either n or n-1 is even, and the other is odd. 4=2·2, and an odd number cannot be divisible by either factor, so the even factor is divisible by 4. This completes the proof.

There has been a lot of talk about AIs taking over the world.

Much of that focuses on Yudkowsky’s description of a hypothetical Bayesian intelligence integrating sensory data to form an accurate model of the world much more quickly than a human would.

But what about problem domains that don’t require sensory data?

Examples: Chess! All the game state is visible, and the only hidden information is the algorithm being followed by the opponent. Poker! You know how many cards there are, but you don’t know which ones your opponents have, which requires a probabilistic approach. Rubik’s cube! Go! Prisoners dilemma!

All of these problems have a compact formal description that can be provided to the agent, so it has a perfect understanding of the laws of physics.

Alternatively they can be formulated as utility functions instead, and the physics can be a formal description of a virtual machine that will run the algorithm to solve the problem. That’s actually isomorphic, but might be easier to think about.

This eliminates a huge amount of complexity, while still being very difficult: can you write an algorithm that can solve arbitrary logic puzzles, that works better than the specialized algorithms that have been written for each puzzle?

(Hint: no you can’t, not yet, anyway. Maybe one day).

It may seem tempting at this point to say well, maybe we can’t write this algorithm, but we can write a simpler algorithm that can recursively self-improve to the point that it can write this algorithm!

I contest this assertion.

A mathematician has developed a new way to uncover simple patterns that might underlie apparently complex systems, such as clouds, cracks in materials or the movement of the stockmarket. The method, named fractal Fourier analysis, is based on new branch of mathematics called fractal geometry. The method could help scientists better understand the complicated signals that the body gives out, such as nerve impulses or brain waves.

Anyway I’m going up to nine vertices for these hypergraphs and ordinary graph isomorphism is not known to be in P, and obviously hypergraph isomorphism is going to be harder so I figure the only way is to brute-force it. Only 9! is like 360k so I end up running through over a hundred billion hypergraphs if I force it the most brutal way possible. The shortcuts are:

  • match on number of edges, vertex degrees, and edge ranks before trying to find isomorphism
  • only try to map vertices to other vertices if they have the same degree (so never run through more than 8! maps, and usually much less than that)

the lovely becomewords tagged me! which was perfect, because i needed something else to help me procrastinate packing for the uk. hashtag stressed.

i’m tagging alyciadeb, dicktective, foxsmoulders, isomorphic, jamesbonds, mr-satchmo, ravensrays, sh0pgirl, silllystyles, and spareourworld!

Alphabetical Ordering

Hey the math people who follow me. I know you guys work with students more often than I. Is there a better approach to this proof than my basic approach?

Theorem: The natural numbers under alphabetical order is wellordered.
Proof: Let 26 be the finite ordinal with 26 members. Then, 26^ω is order-isomorphic the set of all finite length sequences with values in 26. Since the ordinals are closed under exponentiation, 26^ω is an ordinal, so all of its subsets are wellordered. Any collection of words in English can be thought of as a subset of 26^ω, so the natural numbers under alphabetical order are a subset, and thus wellordered.

I bet there’s a clearer way, isn’t there?


Stronger Than You claims, “Garnet is not a single mineral, but a group contains closely related, isomorphous minerals that form a series with each other. The Garnet members form intermediary minerals between each member, and may even intergrow within a single crystal. “

I just think that’s pretty cool as seeing how well Ruby and Saphire blend into Garnet.

Tryna find some hypergraph isomorphisms and after deciding that my code was running too slowly (I monitor the progress by making it print checkpoints when it reaches certain milestones), I added one line to check that the graphs match on edge rank (initially I was just checking the number of hyperedges and the vertex degrees) and it’s going like 20 times faster now.

Self-Complementary Graphs I
If G is a graph, its complement is a graph with the same vertices, but with only every edge not in the graph. For instance, the complement of a complete graph, which has every pair of vertices connected, is the empty graph, where no vertices are connected. Some graphs are self-complementary, meaning they are isomorphic to (a rearrangement of) their complements. Two such graphs are the path of length 4, and the cycle of length 5.

Day 103

last semester I initiated that a math course would be held so that we students of electro-acoustic composition get a better understanding of certain mathematical and physical aspects of sound-synthesis and musical programming. so today at university in our first lesson our prof spoke with us about the basic mathematic operations one learns in elementary school up to the more complex one like logarithms and square-roots, considering their aspect of abstraction from real world phenomenons and viewed from a more or less philosophical point of view. That was quite interesting, as it always is when light is shed unto things we tend to do a lot but not think to much about as to why and how. I also learned a new word, which is “isomorphic”. 

Later I ate some vegan apple crumble with mango-passionfruit sorbet at Sorelli’s in Werden.

btw: the video by bowriders string quartet I helped filming can now be watched and listened to:

16.I.1, 2, 3, 4, 5

Let H be a subset of K where K is a normal subgroup of G. Let P send G/H to G/K where P(Ha) = Ka [a is an element of G]

1. P is defined on cosets.

If Ha = Hb, P(Ha) = P(Hb), Ka = Kb (i.e. ab-1 is in K)

2. Let Ha and Hb be elements of G/H, then for some coset arithmetic…

P((Ha)(Hb)) = P(H(ab)) = K(ab)
P(Ha)P(Hb) = (Ka)(Kb) = K(ab)

3. H is a subset of K and any Ka is in the image of Ha under P.

4. Ker(P) = K/H?
Let e = the identity element on G/H,
The kernel of P is all elements in G/H such that P(Ha) = e.
But if P(Ha) = Ka, the kernel of P is all cosets of K in G such that Ka = K.

But if Ka = K, then a is an element of K, so the kernel of P is K/H.

[Recall that H is a normal subgroup of K, so K/H is the set of all cosets of H in K.]

This could be more clear:
ker(P) = {Ha in G/K such that P(Ha) = e}
= {Ha in G/H such that Ka = K}
= {Ha in G/H such that a is in K}
= {Hb, Hc, Hd, … such that b, c, d etc. is in K}
= K/H

5. We know P is a (well-defined, surjective…) homomorphism from G/H to G/K with kernel K/H. By FHT, then G/K is isomorphic to (G/H)/(K/H).

[Note: the notation also suggests this. If G/K is iso. to (G/H)/(K/H), cancel H and you get back G/K]