Getting A Lot Of: “No But Like You Don’t Get It Math Is Stupid And Useless And Reading Books Is Actually

One of the worst parts about being a mathematician is that sometimes your working on a project that is relatively simple and your talking about it to a friend and in an effort to convey what your doing you simplify exactly what it is and somehow others end up more confused.

This is very interesting! It makes sense that, multiplication being repeated addition, it should be able to arise from just the group structure of the integers, but it never occurred to me before reading this. Thanks for the write-up.

Okay so to get the additive group of integers we just take the free (abelian) group on one generator. Perfectly natural. But given this group, how do we get the multiplication operation that makes it into the ring of integers, without just defining it to be what we already know the answer should be? Actually, we can leverage the fact that the underlying group is free on one generator.

So if you have two abelian groups A,B, then the set of group homorphisms A -> B can be equipped with the structure of an abelian group. If the values of homorphisms f and g at a group element a are f(a) and g(a), then the value of f + g at a is f(a) + g(a). Note that for this sum function to be a homomorphism in general, you do need B to be abelian. This abelian group structure is natural in the sense that Hom(A ⊗ B,C) is isomorphic in a natural way to Hom(A,Hom(B,C)) for all abelian groups A,B,C, where ⊗ denotes the tensor product of abelian groups. In jargon, this says that these constructions make the category of abelian groups into a monoidal closed category.

In particular, the set End(A) = Hom(A,A) of endomorphisms of A is itself an abelian group. What's more, we get an entirely new operation on End(A) for free: function composition! For f,g: A -> A, define f ∘ g to map a onto f(g(a)). Because the elements of End(A) are group homorphisms, we can derive a few identities that relate its addition to composition. If f,g,h are endomorphisms, then for all a in A we have [f ∘ (g + h)](a) = f(g(a) + h(a)) = f(g(a)) + f(h(a)) = [(f ∘ g) + (f ∘ h)](a), so f ∘ (g + h) = (f ∘ g) + (f ∘ h). In other words, composition distributes over addition on the left. We can similarly show that it distributes on the right. Because composition is associative and the identity function A -> A is always a homomorphism, we find that we have equipped End(A) with the structure of a unital ring.

Here's the punchline: because ℤ is the free group on one generator, a group homomorphism out of ℤ is completely determined by where it maps the generator 1, and every choice of image of 1 gives you a homomorphism. This means that we can identify the elements of ℤ with those of End(ℤ) bijectively; a non-negative number n corresponds to the endomorphism [n]: ℤ -> ℤ that maps k onto k added to itself n times, and a negative number n gives the endomorphism [n] that maps k onto -k added together -n times. Going from endomorphisms to integers is even simpler: evaluate the endomorphism at 1. Note that because (f + g)(1) = f(1) + g(1), this bijection is actually an isomorphism of abelian groups

This means that we can transfer the multiplication (i.e. composition) on End(ℤ) to ℤ. What's this ring structure on ℤ? Well if you have the endomorphism that maps 1 onto 2, and you then compose it with the one that maps 1 onto 3, then the resulting endomorphism maps 1 onto 2 added together 3 times, which among other names is known as 6. The multiplication is exactly the standard multiplication on ℤ!

A lot of things had to line up for this to work. For instance, the pointwise sum of endomorphisms needs to be itself an endomorphism. This is why we can't play the same game again; the free commutative ring on one generator is the integer polynomial ring ℤ[X], and indeed the set of ring endomorphisms ℤ[X] -> ℤ[X] correspond naturally to elements of ℤ[X], but because the pointwise product of ring endomorphisms does not generally respect addition, the pointwise operations do not equip End(ℤ[X]) with a ring structure (and in fact, no ring structure on Hom(R,S) can make the category of commutative rings monoidal closed for the tensor product of rings (this is because the monoidal unit is initial)). We can relax the rules slightly, though.

Who says we need the multiplication (or addition, for that matter) on End(ℤ[X])? We still have the bijection ℤ[X] ↔ End(ℤ[X]), so we can just give ℤ[X] the composition operation by transfering along the correspondence anyway. If p and q are polynomials in ℤ[X], then p ∘ q is the polynomial you get by substituting q for every instance of X in p. By construction, this satisfies (p + q) ∘ r = (p ∘ r) + (q ∘ r) and (p × q) ∘ r = (p ∘ r) × (q ∘ r), but we no longer have left-distributivity. Furthermore, composition is associative and the monomial X serves as its unit element. The resulting structure is an example of a composition ring!

The composition rings, like the commutative unital rings, and the abelian groups, form an equational class of algebraic structures, so they too have free objects. For sanity's sake, let's restrict ourselves to composition rings whose multiplication is commutative and unital, and whose composition is unital as well. Let C be the free composition ring with these restrictions on one generator. The elements of this ring will look like polynomials with integers coefficients, but with expressions in terms of X and a new indeterminate g (thought of as an 'unexpandable' polynomial), with various possible arrangements of multiplication, summation, and composition. It's a weird complicated object!

But again, the set of composition ring endomorphisms C -> C (that is, ring endomorphisms which respect composition) will have a bijective correspondence with elements of C, and we can transfer the composition operation to C. This gets us a fourth operation on C, which is associative with unit element g, and which distributes on the right over addition, multiplication, and composition.

This continues: every time you have a new equational class of algebraic structures with two extra operations (one binary operation for the new composition and one constant, i.e. a nullary operation, for the new unit element), and a new distributivity identity for every previous operation, as well as a unit identity and an associativity identity. We thus have an increasing countably infinite tower of algebraic structures.

Actually, taking the union of all of these equational classes still gives you an equational class, with countably infinitely many operations. This too has a free object on one generator, which has an endomorphism algebra, which is an object of a larger equational class of algebras, and so on. In this way, starting from any equational class, we construct a transfinite tower of algebraic structures indexed by the ordinal numbers with a truly senseless amount of associative unital operations, each of which distributes on the right over every previous operation.

mathematical revelation so great i almost became religious

Euclidean Geometry in Mathematical Olympiads

Geometric proofs… might anyone have a good resource for learning them more formally? I need them for a project I’ve recently begun, and hope to find a guide stronger than those reserved for high school geometry courses haha.

Specifically, i’m looking to understand lines, line intersections, things of that type, in a deeper way.

Thank you! :)

I've been pulled into watching the Gotham Chess recaps of the International Chess Championship and here are my thoughts:

This would make a great pro wrestling story

(hearing him actually recap a chess strategy) oh my god i never want to be good at chess

BASELESS ACCUSATION CHEATING GUY?????

ANAL BEADS CHEATING GUY???????????

Ding nooooooo you can do it I believe in you (I'm kinda cheering for both of them but I'm tickled by the concept of "incumbent champ who is somehow the neurotic underdog")

I REALLY never want to be good at chess

probably the sickos on Ao3 have written rpf about this (they have)

God Eidolon is really just a victim of the most ridiculous moral consequences ever. Like “Aha, you thought it was a good idea to neglect your own mental health in order to achieve a greater good and save other people’s lives? Well think again, because just maybe giant hell demons will emerge and destroy Japan.” Poor guy never had a chance. The guy got his brain connected directly to the “summon monsters that kill people” button with absolutely no warning that the button existed or that he could press it. Oh also if you reach a certain threshold of stress the button presses itself. Consequences of drinking alien brain goo i guess

oh youre a math phd? Name 5 equations

1. The functional equation for the completed L-function associated with a modular form.

2. The Analytic Class Number Formula

3. 196883 + 1 = 196884

4. Chinese Remainder Theorem (gives an “equality” of rings)

5. Prime Number Theorem (gives an asymptotic equality of two functions)

(Can you tell what kind of mathematician I am from this? I feel like it’s pretty obvious lmao)

math people scare me. math people will be like "math works in mysterious ways TO YOU. i get it though." and they do and it's fucking terrifying.

born to go to beautiful libraries and study, forced to hustle in my room <3

You took this photo? I've loved it for a long time!

when people are like “he’s not even attractive you could find a guy that looks like him at any gas station” i’m like….. well you see there’s beauty everywhere actually