Galois Groups and the Symmetries of Polynomials

https://www.quantamagazine.org/how-galois-groups-used-polynomial-symmetries-to-reshape-math-20210803/

by nsoonhui • 5 years ago

89 37 5 years ago

voldacar• 5 years ago

Color me impressed! A Quanta article with an equation.

I'm gonna self study group theory, at first using the book 'visual group theory' and figuring out the rest later. smart people of HN, give advice, resources please. thanks in advance

zant2• 5 years ago

Hey this is really cool. I want to start doing the same thing.

Visual group theory is a really nice book for intuition. Also, the YouTube series "Essence of Group Theory" can help in this same line.

What I also want to do while self learning is formalizing some theorems and definitions in Lean. Even just looking at how they're already defined in mathlib [1] can be of great help when internalizing the concepts.

https://github.com/leanprover-community/mathlib/blob/292e3fa...

jimsimmons2• 5 years ago

Side question: what do you think of Lean? I’m not sure what people who know this stuff think of it. I’ve read comments that say Lean is inferior to OCaml and it simply hyped by celebrity

iueijasdkn• 5 years ago

I recommend "How to Think About Abstract Algebra" by Lara Alcock and "A Book of Abstract Algebra" by Charles Pinter.

benrbray• 5 years ago

I second the book by Pinter -- the bulk of the book is a series of well-curated exercises meant to guide you through discovering the basic principles of abstract algebra for yourself. Excellent for self-study in the absence of a good professor.

Vattazi• 5 years ago

If you need a solid theoretic reference, Jackobson’s “Basic Algebra” books are recommended.

gmadsen2• 5 years ago

I've done the same, I'm a roboticist by training, but I've successfully self studied abstract algebra, topology, and complex analysis.

It certainly helped that I took proof based real analysis and linear algebra in school. Writing proofs is really hard to learn without a feedback loop of a professor. So if you don't have any experience with that, I'd start with an intro to proofs book first.

but the group theory material is generally easier than something like topology or functional analysis, which I am certain I was only able to do from finding like minded people in the mathematics sub reddit and did basically a mock semester class on discord.

specific to algebra, I really like the classic Pinter book, "A book of Abstract Algebra", which does a great job of mixing in exposition and history that was really beneficial for self study.

abetusk2• 5 years ago

My education on groups is by far and away deficient but one of the better books I came across is called "Fundamental Algorithms for Permutation Groups" by G. Butler [0]. It's kind of old (1991) but, in my opinion, excellent, especially for some one like me that's concerned with computation, run-time, algorithms, etc.

There's also a Akos Seress book, "Permutation Group Algorithms", which fills in some mathematical details and proofs but, notably, doesn't have one line of pseudo-code in the book.

Both focus on algorithms, and permutation groups in particular, which is just a small subset of group theory.

[0] https://www.springer.com/gp/book/9783540549550

sdenton4• 5 years ago

For the permutation group in particular, it's hard to beat Sagan's 'The Symmetric Group.' It's a bit more recent, and covers lots of really beautiful topics, and really shows off the nice connections between the combinatorics of S_n and the representation theory.

https://www.springer.com/gp/book/9780387950679

zomglings• 5 years ago

I really like Abstract Algebra by Dummit and Foote.

Get an old, used, international edition.

Do exercises!

graycat• 5 years ago

As a ugrad math major, I had some group theory. At the end, I wrote my honors paper on group representations in part to respond to a query from the chemistry department.

I got the first dose of group theory in a standard course in abstract algebra that also covered Galois theory, rings, integral domains, fields, vector spaces, even a little on quaternions.

In grad school I was forced into another course in abstract algebra, this time from the book I. Herstein, Topics in Algebra. Otherwise the prof was trying to understand algebraic geometry, as in Grothendieck. After his first few lectures on group theory, after class I asked if he was going to cover group representations, and he was intimidated and responded "That's deep". I explained that I'd just written my ugrad honors paper on that subject. So, I didn't bother to show up for his class, and at the end he gave me a little oral exam in Galois theory, which I had to review at the time.

One good thing from this prof: He was Italian; the university was in a small town awash in dimly lit, romantic pizza shops; and some of us students asked him which pizza was better, Italian or American. He hesitated and confessed "American". So, he was both a mathematician and a diplomat!

Apparently some people think that Herstein's book is still super good stuff: At

https://www.alibris.com

they want $36 for a used one and $200+ in some cases.

I suspect that

S. Lang, Algebra is excellent although maybe not as a first or only source. I found a PDF at https://math24.files.wordpress.com/2013/02/algebra-serge-lan...

Just the basics of group theory are so well known, written about so often that finding good sources should be easy. Say, just go to the section of any research library, see where Herstein's or Lang's books are, and see what else is there.

Of course, here with Herstein and Lang I'm talking math, legitimately pure math, and not computing, computer science, or much if anything in recent applications.

The definition of a group and the early results are not just simple -- they are childishly, dirt simple, so simple that maybe an adult should be embarrassed to study them.

After some decades, here from just memory I will guess at the definition: A group is a non-empty set G together with an operation (here I omit a picky definition of an operation), say, + such that

(i) For any elements a, b in G, a + b is in G (likely redundant given a careful definition of an operation on G).

(ii) There is an element 0, the identity element, in G such that for any a in G we have

a + 0 = 0 + a = a

(iii) For any element a in G there exists an inverse of a, -a, such that

a + -a = -a + a = 0

(iv) The operation + is associative, that is, for any elements a, b, c in G, we have

(a + b) + c = a + (b + c)

So, let's check: Suppose a, b are in G and we have

a + b = 0

Then

-a + (a + b) = -a

Using associativity we have

(-a + a) + b = -a

Using the identity 0 we have

0 + b = -a

and finally

b = -a

and similarly for

b + a = 0

So, the inverse of a is unique.

How 'bout that! We have proved our first theorem in group theory!

For a little lesson, note the extreme care here where we use associativity and the identity. In a lot of math, we get just to slop through such details, but in group theory we have to be picky in the extreme.

To continue with group theory, we make another definition: Given group (G,+) (there is a style of math that likes such sparse, precise notation), if for any elements a, b in G, we have

a + b = b + a

then the group is commutative and Abelian (after the mathematician Abel).

Not all groups are Abelian, that is, they are not all commutative. For a simple example, given quickly (I'm omitting some picky details) in matrix theory in linear algebra matrix addition is commutative but matrix multiplication is not.

The topic group representations replaces each element of group G with (usually, maybe always) a matrix and the operation with matrix multiplication. An advantage is that we get to write down the matrices and multiply them; otherwise we are likely stuck with a big group multiplication (or in the case of calling the operation +, addition) table.

From all I can tell, the applications of group theory in quantum mechanics, particle physics, and chemistry make use of only the simplest parts of group theory.

I have a bias: In math, first I want to see an application. That is, a recipe for rabbit stew starts out "First catch a rabbit." Then maybe I can get interested in learning some math or doing some research in math that I can use for the application. I confess: This is a radical approach to math. This approach was easy enough to anticipate for someone early in their career long flooded with math topics and trying to make money to support a wife, needing to be selective, etc.

With this approach, I asked what were the applications of abstract algebra -- groups, rings, fields, etc. Not getting a good answer, I concluded that abstract algebra was abstract nonsense, tricky, picky stuff created for no good reason and returned to math analysis and its applications to physics, engineering, etc.

Well, surprising or not, that conclusion is wrong: Whatever Galois, Abel, etc. did/did not know about applications, abstract algebra got applied, has some applications. And abstract algebra is soon not childishly simple and, instead, has some deep/difficult questions and by now some deep/difficult answers.

To continue on a little, a group G might have only finitely many elements or infinitely many elements. For a group with infinitely many elements -- super easy, with R the set of real numbers, (R,+) is an example.

Given a group (G,+), maybe there is a proper subset H of G so that (H,+) (to be really picky, there is a slight abuse of notation here since an operation + cannot be the same on both G and H -- group theory is a picky subject) is also a group. Then (H,+) is a sub-group of (G,+).

Exercise: Fine a proper subgroup of (R,+) (uh, to be picky, proper here means not all of R).

So, if we let |G| denote the number of elements in a set, we have a theorem: In case G is finite, |H| factors |G| where factors means in the third grade sense divides with 0 remainder. Exercise: Prove it!

Some of the advanced results in group theory (Sylow's theorems, Jordan-Holder theory, and much more) are astounding -- tough to believe that they could be true, but they are.

chobytes• 5 years ago

That's definitely a good place to start. Having the right intuition for groups from the beginning makes everything much easier.

As for other materials, what kind of math background do you have?

shreyshnaccount2• 5 years ago

just graduated high school in india

alanbernstein• 5 years ago

The Abel–Ruffini theorem, that general quintics and higher have no solutions via radicals, feels to me like such a fundamental result, that there must be some intuitive explanation for it. I have tried to understand the topic at a level that shows me that intuition, but I have never quite reached it. Any suggestions?

go_elmo• 5 years ago

The proof reduces it to galois-groups and shows some easy properties, which are intuitive in the galois-group domain (only divisor of S_5 is A_5 which is not of prime-order). Maybe studying the relationship / reduction to / from polynomial to galois will help

jstx1• 5 years ago

It's apparently the topic of 3blue1brown's next video - https://twitter.com/3blue1brown/status/1420461376896520193

gmadsen2• 5 years ago

I think the intuition only comes from understanding the arguments intuitively. so basically an entire algebra class, you need to understand field extensions and PIDs.

kortex• 5 years ago

Gonna butcher this, but it's good practice, so I'm gonna take a stab at it. The roots of polynomials are intimately related to the Alternating symmetry groups. These describe the ways you can permute a set, specifially A_n describes the even permutations of a set of n elements. The even permutations of a sequence are those obtained by making an even number of transpositions ("moves").

I can make 2 swaps to go 1,2,3,4,5 -> 1,4,3,2,5 -> 1,4,5,2,3. That's an even permutation.

The order (size) of A_n is the number of even permutations of n unique elements. The orders are as follows:

A2: 1

A3: 3

A4: 12

A5: 60

A6: 360

A_n: n! / 2

The "solvability" of a polynomial depends on the the degrees of freedom and number of constraints. Quintic is where the number of degrees of freedom blows up and you can no longer factor out roots in such a way that you can uniquely factor the polynomial. Quadratic is basically trivial. Cubic cannot be directly factored, because you have 3 variables and 2 equations, but you can "factor out" a quadratic that you can solve, and then solve the cubic. Quartic is the same, you end up with 4 variables and 3 unique equations, you can factor out a cubic, which you can solve, etc. Quartic, you have 4 equations to solve and...something like 30 variables (or maybe it's 60. It's some way more than we can factor!).

https://en.wikipedia.org/wiki/Alternating_group

https://math.stackexchange.com/questions/550401/intuitive-re...

paulpauper2• 5 years ago

Sorta. There are obviously infinite quintics that can be solved despite this. For example any polynomial in which the roots are arranged on a circle on the real,complex plane

anon_tor_123452• 5 years ago

arnold's proof is the most intuitive and what made it click after years of being responsible for understanding it (several years of algebra at the undergrad and grad level):

https://www.youtube.com/watch?v=zeRXVL6qPk4

mjreacher• 5 years ago

Is there any difference between that video and this one?: https://www.youtube.com/watch?v=RhpVSV6iCko seems to be the same author but that video is 10 minutes longer.

paulpauper2• 5 years ago

The original proofs are quite long. Not exacyly something you can compress to a comment or an article. I dunno

kevinventullo• 5 years ago

So first, consider nth roots. These correspond to particularly simple field extensions. If I take two choices of nth root, their ratio is always an nth root of unity. This more or less forces the symmetry group to be a subgroup of the nth roots of unity, i.e. Z/n.

Now if you have a general expression built out of extracting nth roots for various n, that might generate some big complicated field extension. But it corresponds to a tower of field extensions which are all just extracting nth roots. Thus the intermediate symmetry groups are still all subsets of Z/n for various n.

The Galois correspondence tells you that the symmetry group of the entire extension, call it G, has a corresponding decomposition: there is an increasing sequence of subgroups 1 < G_1 < … < G_k = G such that each intermediate quotient group G_{i+1}/G_i is a subgroup of Z/n for some n. Groups with such a decomposition are called solvable.

Insolvability of the quintic then boils down to this: the symmetry group of any degree k polynomial must be a subgroup of S_k. For k <= 4, S_k is solvable; for k >= 5, it is not.

This last sentence is the closest thing I can offer to intuition. It is a very finite and tangible fact that one can write down and computationally verify for themselves. Like, you can write down the corresponding solvable decomposition of S_4, and you can see that S_5 is not solvable by verifying that A_5 is simple which itself boils down to a small number of cases.

gradschool• 5 years ago

"For instance, the Galois group immediately tells you whether a polynomial can be solved at all ..."

I wish she wrote one more sentence before changing the subject. If it's immediate, Can anyone explain the decision procedure?

gizmo686• 5 years ago

A polynomial is solvable if and only if its corresponding Galois group is "solvable". Solvable groups are motivating primarily by the problem of determining if polynomials are solvable. In general, it takes some work to know if a group is solvable, and I don't think there is a way to determine that immidiatly.

The most charitable reading of the statement is that there are relativly few groups of small order. This means that for polynomials of relativly small degree you have probably seen their Galois group before so can go off of memory.

tacomonstrous• 5 years ago

Not sure what this statement is supposed to mean. Every polynomial can be solved tautologically by adding a zero for it into the picture. For instance, every complex polynomial admits a complex zero (this is the Fundamental theorem of algebra), so you can find the zeros of any rational polynomial somewhere. You can of course look at the smallest 'field' in which these zeros live: basically, look at everything you get by writing down all possible arithmetic expressions involving those zeros. For the polynomial x^2-2, you get numbers of the form a+\sqrt(2)b where a and b are rational numbers.

The Galois group tells you how complicated this new field is, and how involved its symmetries are. In the example above there is only one nontrivial symmetry which switches a+sqrt(2)b with a-sqrt(2)b. In general there are many more, and when you consider the symmetries of the zeros of all possible rational polynomials, you get the absolute Galois group of the rational numbers. It can be said without exaggeration that number theory is the study of this Galois group, which is still a rather mysterious object in its full glory.

xyzzyz• 5 years ago

The intended (and, in fact, widely understood within mathematical community) meaning of that statement is whether the roots of polynomials can be expressed using radicals, ie. as algebraic expressions involving four standard arithmetic operations and taking roots.

I’d rather quibble with the part where they say that this is “immediately” seen from Galois group, as showing solvability of a group is not trivial matter at all.

IngoBlechschmid• 5 years ago

The big picture is the following:

There is a certain kind of groups called "solvable groups". There is a theorem stating that a polynomial equation can be solved by radicals if and only if its Galois group is a solvable group. And whether a given finite group (such as a Galois group) is solvable is an entirely finite combinatorial question which can be mechanically checked.

Even better: A witness for the solvability of the Galois group can be transformed -- again nontrivially but mechanically -- into explicit formulas for the solutions of the polynomial equation.

Wikipedia has some examples on solvable groups and their connection to Galois theory, though I hope that someone has written some blog post or similar explaining this beautiful connection with less prerequisites. https://en.wikipedia.org/wiki/Solvable_group

jordigh• 5 years ago

If you're into this stuff, I really like this book on Galois theory:

https://www.worldcat.org/title/classical-introduction-to-gal...

The reason I like this book is that it is very computational, in the spirit of Galois's original discoveries. Modern treatment of Galois theory has been greatly coloured by Emil Artin's reformulation of the subject: the Galois group is nowadays primarily taught to undergrads as being a group of automorphisms of a field extension instead of being primarily the computational object of the permutations of the roots of a polynomial.

Artin's approach has merit, of course. The extra level of abstraction generalises to much more than just simple algebraic extensions of the rationals. However, I think the historical approach in this case is more instructive. Seeing lots of calculations before the abstractions gives you a better sense of what's being abstracted away.

dmch-1• 5 years ago

There is Artin's approach, and then there is a more abstract approach by Grothendieck relating to the fundamental group of algebraic topology.

In the movie Beautiful mind there is a scene, where a student tells John Nash that he can proof that 'Galois extensions are the same as covering spaces'. This follows from the Grothendieck's approach. However the analogy between Galois extessions and the fundamental group was known even before Grothendieck.

Then there are even more general approaches in the category theory setting.

Today these generalisations are taught indeed without much regard to the computational spirit of 19th century mathematics. They have their merit as you say, but I agree understanding the computational aspects are instructive in fully appreciating the generalisations and analogies.

cmpb• 5 years ago

Reaching Galois groups (and field extensions, rings, and other related tools) in class was the point in my math undergrad that I realized that math had become too subtle for my brain to easily comprehend. I quickly realized that I'd either have to start studying outside of class to learn the motivations and connections or accept that I would not be pursuing math as a career.

And that's when I got a job in software.

nobody0• 5 years ago

You can still porting those algebraic structures to day to day software construction.

https://corecursive.com/050-sam-ritchie-portal-abstractions-...

Galois Groups and the Symmetries of Polynomials

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