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*When introduced to a new idea, always ask why you should care.\
Do not expect an answer right away, but demand one eventually.\
--- Ravi Vakil *
If you like this book and want to support me, please consider buying me a coffee!\
[{width="32ex"}](https://ko-fi.com/evanchen)\
<https://ko-fi.com/evanchen/>
*For Brian and Lisa, who finally got me to write it*.\
Evan Chen.\
Text licensed under [CC-by-SA-4.0](https://creativecommons.org/licenses/by-sa/4.0/). Source files licensed under [GNU GPL v3](https://choosealicense.com/licenses/gpl-3.0/).\
This is (still!) an **incomplete draft**. Please send corrections, comments, pictures of kittens, etc. to , or pull-request at <https://github.com/vEnhance/napkin>.\
Last updated 2026-02-25.
*When introduced to a new idea, always ask why you should care.\
Do not expect an answer right away, but demand one eventually.\
--- Ravi Vakil *
If you like this book and want to support me, please consider buying me a coffee!\
[{width="32ex"}](https://ko-fi.com/evanchen)\
<https://ko-fi.com/evanchen/>
*For Brian and Lisa, who finally got me to write it*.\
Evan Chen.\
Text licensed under [CC-by-SA-4.0](https://creativecommons.org/licenses/by-sa/4.0/). Source files licensed under [GNU GPL v3](https://choosealicense.com/licenses/gpl-3.0/).\
This is (still!) an **incomplete draft**. Please send corrections, comments, pictures of kittens, etc. to , or pull-request at <https://github.com/vEnhance/napkin>.\
Last updated 2026-02-25. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Title Embellishments | 01_title-embellishments.md | 0 | 436 | |
The origin of the name "Napkin" comes from the following quote of mine.
> I'll be eating a quick lunch with some friends of mine who are still in high school. They'll ask me what I've been up to the last few weeks, and I'll tell them that I've been learning category theory. They'll ask me what category theory is about. I tell them it's about abstracting things by looking at just the structure-preserving morphisms between them, rather than the objects themselves. I'll try to give them the standard example $\catname{Grp}$, but then I'll realize that they don't know what a homomorphism is. So then I'll start trying to explain what a homomorphism is, but then I'll remember that they haven't learned what a group is. So then I'll start trying to explain what a group is, but by the time I finish writing the group axioms on my napkin, they've already forgotten why I was talking about groups in the first place. And then it's 1PM, people need to go places, and I can't help but think:\
> *"Man, if I had forty hours instead of forty minutes, I bet I could actually have explained this all".*
This book was initially my attempt at those forty hours, but has grown considerably since then.
The origin of the name "Napkin" comes from the following quote of mine.
> I'll be eating a quick lunch with some friends of mine who are still in high school. They'll ask me what I've been up to the last few weeks, and I'll tell them that I've been learning category theory. They'll ask me what category theory is about. I tell them it's about abstracting things by looking at just the structure-preserving morphisms between them, rather than the objects themselves. I'll try to give them the standard example $\catname{Grp}$, but then I'll realize that they don't know what a homomorphism is. So then I'll start trying to explain what a homomorphism is, but then I'll remember that they haven't learned what a group is. So then I'll start trying to explain what a group is, but by the time I finish writing the group axioms on my napkin, they've already forgotten why I was talking about groups in the first place. And then it's 1PM, people need to go places, and I can't help but think:\
> *"Man, if I had forty hours instead of forty minutes, I bet I could actually have explained this all".*
This book was initially my attempt at those forty hours, but has grown considerably since then. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Preface | 02_preface.md | 0 | 682 | |
The *Infinitely Large Napkin* is a light but mostly self-contained introduction to a large amount of higher math.
I should say at once that this book is not intended as a replacement for dedicated books or courses; the amount of depth is not comparable. On the flip side, the benefit of this "light" approach is that it becomes accessible to a larger audience, since the goal is merely to give the reader a feeling for any particular topic rather than to emulate a full semester of lectures.
I initially wrote this book with talented high-school students in mind, particularly those with math-olympiad type backgrounds. Some remnants of that cultural bias can still be felt throughout the book, particularly in assorted challenge problems which are taken from mathematical competitions. However, in general I think this would be a good reference for anyone with some amount of mathematical maturity and curiosity. Examples include but certainly not limited to: math undergraduate majors, physics/CS majors, math PhD students who want to hear a little bit about fields other than their own, advanced high schoolers who like math but not math contests, and unusually intelligent kittens fluent in English.
The *Infinitely Large Napkin* is a light but mostly self-contained introduction to a large amount of higher math.
I should say at once that this book is not intended as a replacement for dedicated books or courses; the amount of depth is not comparable. On the flip side, the benefit of this "light" approach is that it becomes accessible to a larger audience, since the goal is merely to give the reader a feeling for any particular topic rather than to emulate a full semester of lectures.
I initially wrote this book with talented high-school students in mind, particularly those with math-olympiad type backgrounds. Some remnants of that cultural bias can still be felt throughout the book, particularly in assorted challenge problems which are taken from mathematical competitions. However, in general I think this would be a good reference for anyone with some amount of mathematical maturity and curiosity. Examples include but certainly not limited to: math undergraduate majors, physics/CS majors, math PhD students who want to hear a little bit about fields other than their own, advanced high schoolers who like math but not math contests, and unusually intelligent kittens fluent in English. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Preface | About this book | 02_preface.md | 1 | 689 |
The project is hosted on GitHub at <https://github.com/vEnhance/napkin>. Pull requests are quite welcome! I am also happy to receive suggestions and corrections by email.
The project is hosted on GitHub at <https://github.com/vEnhance/napkin>. Pull requests are quite welcome! I am also happy to receive suggestions and corrections by email.
As far as I can tell, higher math for high-school students comes in two flavors:
- Someone tells you about the hairy ball theorem in the form "you can't comb the hair on a spherical cat" then doesn't tell you anything about why it should be true, what it means to actually "comb the hair", or any of the underlying theory, leaving you with just some vague notion in your head.
- You take a class and prove every result in full detail, and at some point you stop caring about what the professor is saying.
Presumably you already know how unsatisfying the first approach is. So the second approach seems to be the default, but I really think there should be some sort of middle ground here.
Unlike university, it is *not* the purpose of this book to train you to solve exercises or write proofs,[^1] or prepare you for research in the field. Instead I just want to show you some interesting math. The things that are presented should be memorable and worth caring about. For that reason, proofs that would be included for completeness in any ordinary textbook are often omitted here, unless there is some idea in the proof which I think is worth seeing. In particular, I place a strong emphasis over explaining why a theorem *should* be true rather than writing down its proof. This is a recurrent theme of this book:
Natural explanations supersede proofs.
My hope is that after reading any particular chapter in Napkin, one might get the following out of it:
- Knowing what the precise definitions are of the main characters,
- Being acquainted with the few really major examples,
- Knowing the precise statements of famous theorems, and having a sense of why they *should* be true.
Understanding "why" something is true can have many forms. This is sometimes accomplished with a complete rigorous proof; in other cases, it is given by the idea of the proof; in still other cases, it is just a few key examples with extensive cheerleading.
Obviously this is nowhere near enough if you want to e.g. do research in a field; but if you are just a curious outsider, I hope that it's more satisfying than the elevator pitch or Wikipedia articles. Even if you do want to learn a topic with serious depth, I hope that it can be a good zoomed-out overview before you really dive in, because in many senses the choice of material is "what I wish someone had told me before I started".
As far as I can tell, higher math for high-school students comes in two flavors:
- Someone tells you about the hairy ball theorem in the form "you can't comb the hair on a spherical cat" then doesn't tell you anything about why it should be true, what it means to actually "comb the hair", or any of the underlying theory, leaving you with just some vague notion in your head.
- You take a class and prove every result in full detail, and at some point you stop caring about what the professor is saying.
Presumably you already know how unsatisfying the first approach is. So the second approach seems to be the default, but I really think there should be some sort of middle ground here.
Unlike university, it is *not* the purpose of this book to train you to solve exercises or write proofs,[^1] or prepare you for research in the field. Instead I just want to show you some interesting math. The things that are presented should be memorable and worth caring about. For that reason, proofs that would be included for completeness in any ordinary textbook are often omitted here, unless there is some idea in the proof which I think is worth seeing. In particular, I place a strong emphasis over explaining why a theorem *should* be true rather than writing down its proof. This is a recurrent theme of this book:
Natural explanations supersede proofs.
My hope is that after reading any particular chapter in Napkin, one might get the following out of it:
- Knowing what the precise definitions are of the main characters,
- Being acquainted with the few really major examples,
- Knowing the precise statements of famous theorems, and having a sense of why they *should* be true.
Understanding "why" something is true can have many forms. This is sometimes accomplished with a complete rigorous proof; in other cases, it is given by the idea of the proof; in still other cases, it is just a few key examples with extensive cheerleading.
Obviously this is nowhere near enough if you want to e.g. do research in a field; but if you are just a curious outsider, I hope that it's more satisfying than the elevator pitch or Wikipedia articles. Even if you do want to learn a topic with serious depth, I hope that it can be a good zoomed-out overview before you really dive in, because in many senses the choice of material is "what I wish someone had told me before I started". | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Preface | Source code | 02_preface.md | 2 | 1,460 |
The preface would become too long if I talked about some of my pedagogical decisions chapter by chapter, so contains those comments instead.
In particular, I often name specific references, and the end of that appendix has more references. So this is a good place to look if you want further reading.
The preface would become too long if I talked about some of my pedagogical decisions chapter by chapter, so contains those comments instead.
In particular, I often name specific references, and the end of that appendix has more references. So this is a good place to look if you want further reading.
I began writing this book in December 2014, after having finished my first semester of undergraduate at Harvard. It became my main focus for about 18 months after that, as I became immersed in higher math. I essentially took only math classes (gleefully ignoring all my other graduation requirements), and merged as much of it as I could (as well as lots of other math I learned on my own time) into the Napkin.
Towards August 2016, though, I finally lost steam. The first public drafts went online then, and I decided to step back. Having burnt out slightly, I then took a break from higher math, and spent the remaining two undergraduate years[^2] working extensively as a coach for the American math olympiad team, and trying to spend as much time with my friends as I could before they graduated and went their own ways.
During those two years, readers sent me many kind words of gratitude, many reports of errors, and many suggestions for additions. So in November 2018, some weeks into my first semester as a math PhD student, I decided I should finish what I had started. Some months later, here is what I have.
I began writing this book in December 2014, after having finished my first semester of undergraduate at Harvard. It became my main focus for about 18 months after that, as I became immersed in higher math. I essentially took only math classes (gleefully ignoring all my other graduation requirements), and merged as much of it as I could (as well as lots of other math I learned on my own time) into the Napkin.
Towards August 2016, though, I finally lost steam. The first public drafts went online then, and I decided to step back. Having burnt out slightly, I then took a break from higher math, and spent the remaining two undergraduate years[^2] working extensively as a coach for the American math olympiad team, and trying to spend as much time with my friends as I could before they graduated and went their own ways.
During those two years, readers sent me many kind words of gratitude, many reports of errors, and many suggestions for additions. So in November 2018, some weeks into my first semester as a math PhD student, I decided I should finish what I had started. Some months later, here is what I have. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Preface | More pedagogical comments and references | 02_preface.md | 3 | 813 |
I am indebted to countless people for this work. Here is a partial (surely incomplete) list.
- Thanks to all my teachers and professors for teaching me much of the material covered in these notes, as well as the authors of all the references I have cited here. A special call-out to , , , , , , , , , , which were especially influential.
- Thanks also to dozens of friends and strangers who read through preview copies of my draft, and pointed out errors and gave other suggestions. Special mention to Andrej Vuković and Alexander Chua for together catching over a thousand errors. Thanks also to Brian Gu and Tom Tseng for many corrections. (If you find mistakes or have suggestions yourself, I would love to hear them!) Thanks also to Royce Yao and `user202729` for their contributions of guest chapters to the document.
- Thanks to Jenny Chu and Lanie Deng for the cover artwork.
- I'd also like to express my gratitude for many, many kind words I received during the development of this project. These generous comments led me to keep working on this, and were largely responsible for my decision in November 2018 to begin updating the Napkin again.
Finally, a huge thanks to the math olympiad community, from which the Napkin (and me) has its roots. All the enthusiasm, encouragement, and thank-you notes I have received over the years led me to begin writing this in the first place. I otherwise would never have the arrogance to dream a project like this was at all possible. And of course I would be nowhere near where I am today were it not for the life-changing journey I took in chasing my dreams to the IMO. Forever TWN2!
[^1]: Which is not to say problem-solving isn't valuable; I myself am a math olympiad coach at heart. It's just not the point of this book.
[^2]: Alternatively: " ... and spent the next two years forgetting everything I had painstakingly learned". Which made me grateful for all the past notes in the Napkin!
I am indebted to countless people for this work. Here is a partial (surely incomplete) list.
- Thanks to all my teachers and professors for teaching me much of the material covered in these notes, as well as the authors of all the references I have cited here. A special call-out to , , , , , , , , , , which were especially influential.
- Thanks also to dozens of friends and strangers who read through preview copies of my draft, and pointed out errors and gave other suggestions. Special mention to Andrej Vuković and Alexander Chua for together catching over a thousand errors. Thanks also to Brian Gu and Tom Tseng for many corrections. (If you find mistakes or have suggestions yourself, I would love to hear them!) Thanks also to Royce Yao and `user202729` for their contributions of guest chapters to the document.
- Thanks to Jenny Chu and Lanie Deng for the cover artwork.
- I'd also like to express my gratitude for many, many kind words I received during the development of this project. These generous comments led me to keep working on this, and were largely responsible for my decision in November 2018 to begin updating the Napkin again.
Finally, a huge thanks to the math olympiad community, from which the Napkin (and me) has its roots. All the enthusiasm, encouragement, and thank-you notes I have received over the years led me to begin writing this in the first place. I otherwise would never have the arrogance to dream a project like this was at all possible. And of course I would be nowhere near where I am today were it not for the life-changing journey I took in chasing my dreams to the IMO. Forever TWN2!
[^1]: Which is not to say problem-solving isn't valuable; I myself am a math olympiad coach at heart. It's just not the point of this book.
[^2]: Alternatively: " ... and spent the next two years forgetting everything I had painstakingly learned". Which made me grateful for all the past notes in the Napkin! | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Preface | Acknowledgements | 02_preface.md | 4 | 1,112 |
As explained in the preface, the main prerequisite is some amount of mathematical maturity. This means I expect the reader to know how to read and write a proof, follow logical arguments, and so on.
I also assume the reader is familiar with basic terminology about sets and functions (e.g. "what is a bijection?"). If not, one should consult .
As explained in the preface, the main prerequisite is some amount of mathematical maturity. This means I expect the reader to know how to read and write a proof, follow logical arguments, and so on.
I also assume the reader is familiar with basic terminology about sets and functions (e.g. "what is a bijection?"). If not, one should consult .
There is no need to read this book in linear order: it covers all sorts of areas in mathematics, and there are many paths you can take. In , I give a short overview for each part explaining what you might expect to see in that part.
For now, here is a brief chart showing how the chapters depend on each other; again see for details. Dependencies are indicated by arrows; dotted lines are optional dependencies. **I suggest that you simply pick a chapter you find interesting, and then find the shortest path.** With that in mind, I hope the length of the entire PDF is not intimidating.
(The text in the following diagram should be clickable and links to the relevant part.)
,45:Ch ,-[Abs Alg](#part:absalg) ,45:Ch ,-[Topology](#part:basictop) ,45:Ch -,[Lin Alg](#part:linalg) ,35:Ch [Grp Act](#part:groups) ,24:Ch [Grp Classif](#ch:sylow) ,35:Ch -[Rep Th](#part:repth) ,43:Ch -[Quantum](#part:quantum) ,38:Ch -[Calc](#part:calc) ,30:Ch -[Cmplx Ana](#part:cmplxana) ,20:Ch -[Measure/Pr](#part:measure) ,28:Ch -[Diff Geo](#part:diffgeo) ,10:Ch -[Alg NT 1](#part:algnt1) ,0:Ch -[Alg NT 2](#part:algnt2) ,10:Ch -[Alg Top 1](#part:algtop1) ,28:Ch -[Cat Th](#part:cats) ,0:Ch -[Alg Top 2](#part:algtop2) ,10:Ch -[Alg Geo 1](#part:ag1) ,0:Ch -[Alg Geo 2-3](#part:ag2) ,45:Ch -[Set Theory](#part:st1)
,45,30,35;0: ,45,33,45;0: ,45,30,35;0: ,45,45,43;0: ,45,20,28;0: ,45,20,28;0: ,45,20,28;80: ,45,20,28;-40: ,28,30,35;0: ,45,5,35;0: ,35,5,24;0: ,45,48,28;30: ,45,64,38;0: ,38,48,28;50: ,45,48,28;10: ,45,64,38;0: ,38,64,30;0: ,45,64,30;-10: ,10,64,30;0: ,45,55,20;0: ,38,55,20;50: ,10,6,0;0: ,45,6,10;0: ,45,6,10;-40: ,28,6,10;0: ,28,6,0;-18: ,10,23,0;0: ,45,23,10;0: ,35,23,10;20: ,10,20,28;20: ,28,23,0;-190: ,10,40,0;0: ,45,40,10;-40: ,45,40,10;50: ,35,40,0;10:
There is no need to read this book in linear order: it covers all sorts of areas in mathematics, and there are many paths you can take. In , I give a short overview for each part explaining what you might expect to see in that part.
For now, here is a brief chart showing how the chapters depend on each other; again see for details. Dependencies are indicated by arrows; dotted lines are optional dependencies. **I suggest that you simply pick a chapter you find interesting, and then find the shortest path.** With that in mind, I hope the length of the entire PDF is not intimidating.
(The text in the following diagram should be clickable and links to the relevant part.)
,45:Ch ,-[Abs Alg](#part:absalg) ,45:Ch ,-[Topology](#part:basictop) ,45:Ch -,[Lin Alg](#part:linalg) ,35:Ch [Grp Act](#part:groups) ,24:Ch [Grp Classif](#ch:sylow) ,35:Ch -[Rep Th](#part:repth) ,43:Ch -[Quantum](#part:quantum) ,38:Ch -[Calc](#part:calc) ,30:Ch -[Cmplx Ana](#part:cmplxana) ,20:Ch -[Measure/Pr](#part:measure) ,28:Ch -[Diff Geo](#part:diffgeo) ,10:Ch -[Alg NT 1](#part:algnt1) ,0:Ch -[Alg NT 2](#part:algnt2) ,10:Ch -[Alg Top 1](#part:algtop1) ,28:Ch -[Cat Th](#part:cats) ,0:Ch -[Alg Top 2](#part:algtop2) ,10:Ch -[Alg Geo 1](#part:ag1) ,0:Ch -[Alg Geo 2-3](#part:ag2) ,45:Ch -[Set Theory](#part:st1)
,45,30,35;0: ,45,33,45;0: ,45,30,35;0: ,45,45,43;0: ,45,20,28;0: ,45,20,28;0: ,45,20,28;80: ,45,20,28;-40: ,28,30,35;0: ,45,5,35;0: ,35,5,24;0: ,45,48,28;30: ,45,64,38;0: ,38,48,28;50: ,45,48,28;10: ,45,64,38;0: ,38,64,30;0: ,45,64,30;-10: ,10,64,30;0: ,45,55,20;0: ,38,55,20;50: ,10,6,0;0: ,45,6,10;0: ,45,6,10;-40: ,28,6,10;0: ,28,6,0;-18: ,10,23,0;0: ,45,23,10;0: ,35,23,10;20: ,10,20,28;20: ,28,23,0;-190: ,10,40,0;0: ,45,40,10;-40: ,45,40,10;50: ,35,40,0;10: | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Advice for the reader | Prerequisites | 03_advice.md | 0 | 1,205 |
In this book, there are three hierarchies:
- An inline **question** is intended to be offensively easy, mostly a chance to help you internalize definitions. If you find yourself unable to answer one or two of them, it probably means I explained it badly and you should complain to me. But if you can't answer many, you likely missed something important: read back.
- An inline **exercise** is more meaty than a question, but shouldn't have any "tricky" steps. Often I leave proofs of theorems and propositions as exercises if they are instructive and at least somewhat interesting.
- Each chapter features several trickier **problems** at the end. Some are reasonable, but others are legitimately difficult olympiad-style problems. Harder problems are marked with up to three chili peppers (), like this paragraph.
In addition to difficulty annotations, the problems are also marked by how important they are to the big picture.
- **Normal problems**, which are hopefully fun but non-central.
- **Daggered problems**, which are (usually interesting) results that one should know, but won't be used directly later.
- **Starred problems**, which are results which will be used later on in the book.[^1]
Several hints and solutions can be found in .
{width="14cm"}\
Image from [@img:exercise]
In this book, there are three hierarchies:
- An inline **question** is intended to be offensively easy, mostly a chance to help you internalize definitions. If you find yourself unable to answer one or two of them, it probably means I explained it badly and you should complain to me. But if you can't answer many, you likely missed something important: read back.
- An inline **exercise** is more meaty than a question, but shouldn't have any "tricky" steps. Often I leave proofs of theorems and propositions as exercises if they are instructive and at least somewhat interesting.
- Each chapter features several trickier **problems** at the end. Some are reasonable, but others are legitimately difficult olympiad-style problems. Harder problems are marked with up to three chili peppers (), like this paragraph.
In addition to difficulty annotations, the problems are also marked by how important they are to the big picture.
- **Normal problems**, which are hopefully fun but non-central.
- **Daggered problems**, which are (usually interesting) results that one should know, but won't be used directly later.
- **Starred problems**, which are results which will be used later on in the book.[^1]
Several hints and solutions can be found in .
{width="14cm"}\
Image from [@img:exercise] | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Advice for the reader | Questions, exercises, and problems | 03_advice.md | 1 | 764 |
At the risk of being blunt,
Read this book with pencil and paper.
Here's why:
{width="50%"}\
Image from [@img:read_with_pencil]
You are not God. You cannot keep everything in your head.[^2] If you've printed out a hard copy, then write in the margins. If you're trying to save paper, grab a notebook or something along with the ride. Somehow, some way, make sure you can write. Thanks.
At the risk of being blunt,
Read this book with pencil and paper.
Here's why:
{width="50%"}\
Image from [@img:read_with_pencil]
You are not God. You cannot keep everything in your head.[^2] If you've printed out a hard copy, then write in the margins. If you're trying to save paper, grab a notebook or something along with the ride. Somehow, some way, make sure you can write. Thanks.
I am pathologically obsessed with examples. In this book, I place all examples in large boxes to draw emphasis to them, which leads to some pages of the book simply consisting of sequences of boxes one after another. I hope the reader doesn't mind.
I also often highlight a "prototypical example" for some sections, and reserve the color red for such a note. The philosophy is that any time the reader sees a definition or a theorem about such an object, they should test it against the prototypical example. If the example is a good prototype, it should be immediately clear why this definition is intuitive, or why the theorem should be true, or why the theorem is interesting, et cetera.
Let me tell you a secret. Whenever I wrote a definition or a theorem in this book, I would have to recall the exact statement from my (quite poor) memory. So instead I often consider the prototypical example, and then only after that do I remember what the definition or the theorem is. Incidentally, this is also how I learned all the definitions in the first place. I hope you'll find it useful as well.
I am pathologically obsessed with examples. In this book, I place all examples in large boxes to draw emphasis to them, which leads to some pages of the book simply consisting of sequences of boxes one after another. I hope the reader doesn't mind.
I also often highlight a "prototypical example" for some sections, and reserve the color red for such a note. The philosophy is that any time the reader sees a definition or a theorem about such an object, they should test it against the prototypical example. If the example is a good prototype, it should be immediately clear why this definition is intuitive, or why the theorem should be true, or why the theorem is interesting, et cetera.
Let me tell you a secret. Whenever I wrote a definition or a theorem in this book, I would have to recall the exact statement from my (quite poor) memory. So instead I often consider the prototypical example, and then only after that do I remember what the definition or the theorem is. Incidentally, this is also how I learned all the definitions in the first place. I hope you'll find it useful as well. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Advice for the reader | Paper | 03_advice.md | 2 | 872 |
This part describes some of the less familiar notations and definitions and settles for once and for all some annoying issues ("is zero a natural number?"). Most of these are "remarks for experts": if something doesn't make sense, you probably don't have to worry about it for now.
A full glossary of notation used can be found in the appendix.
This part describes some of the less familiar notations and definitions and settles for once and for all some annoying issues ("is zero a natural number?"). Most of these are "remarks for experts": if something doesn't make sense, you probably don't have to worry about it for now.
A full glossary of notation used can be found in the appendix.
The set $\mathbb{N}$ is the set of *positive* integers, not including $0$. In the set theory chapters, we use $\omega = \{0, 1, \dots\}$ instead, for consistency with the rest of the book.
The set $\mathbb{N}$ is the set of *positive* integers, not including $0$. In the set theory chapters, we use $\omega = \{0, 1, \dots\}$ instead, for consistency with the rest of the book. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Advice for the reader | Conventions and notations | 03_advice.md | 3 | 306 |
This is brief, intended as a reminder for experts. Consult for full details.
An **equivalence relation** on a set $X$ is a relation $\sim$ which is symmetric, reflexive, and transitive. An equivalence relation partitions $X$ into several **equivalence classes**. We will denote this by $X / {\sim}$. An element of such an equivalence class is a **representative** of that equivalence class.
I always use $\cong$ for an "isomorphism"-style relation (formally: a relation which is an isomorphism in a reasonable category). The only time $\simeq$ is used in the Napkin is for homotopic paths.
I generally use $\subseteq$ and $\subsetneq$ since these are non-ambiguous, unlike $\subset$. I only use $\subset$ on rare occasions in which equality obviously does not hold yet pointing it out would be distracting. For example, I write $\mathbb{Q} \subset \mathbb{R}$ since "$\mathbb{Q} \subsetneq \mathbb{R}$" is distracting.
I prefer $S \setminus T$ to $S - T$.
The power set of $S$ (i.e., the set of subsets of $S$), is denoted either by $2^S$ or $\mathcal P(S)$.
This is brief, intended as a reminder for experts. Consult for full details.
An **equivalence relation** on a set $X$ is a relation $\sim$ which is symmetric, reflexive, and transitive. An equivalence relation partitions $X$ into several **equivalence classes**. We will denote this by $X / {\sim}$. An element of such an equivalence class is a **representative** of that equivalence class.
I always use $\cong$ for an "isomorphism"-style relation (formally: a relation which is an isomorphism in a reasonable category). The only time $\simeq$ is used in the Napkin is for homotopic paths.
I generally use $\subseteq$ and $\subsetneq$ since these are non-ambiguous, unlike $\subset$. I only use $\subset$ on rare occasions in which equality obviously does not hold yet pointing it out would be distracting. For example, I write $\mathbb{Q} \subset \mathbb{R}$ since "$\mathbb{Q} \subsetneq \mathbb{R}$" is distracting.
I prefer $S \setminus T$ to $S - T$.
The power set of $S$ (i.e., the set of subsets of $S$), is denoted either by $2^S$ or $\mathcal P(S)$. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Advice for the reader | Sets and equivalence relations | 03_advice.md | 4 | 608 |
This is brief, intended as a reminder for experts. Consult for full details.
Let $X \taking f Y$ be a function:
- By $f\pre(T)$ I mean the **pre-image** $$f\pre(T) := \left\{ x \in X \mid f(x) \in T \right\}.$$ This is in contrast to the $f^{-1}(T)$ used in the rest of the world; I only use $f^{-1}$ for an inverse *function*.
By abuse of notation, we may abbreviate $f\pre(\{y\})$ to $f\pre(y)$. We call $f\pre(y)$ a **fiber**.
- By $f\im(S)$ I mean the **image** $$f\im(S) := \left\{ f(x) \mid x \in S \right\}.$$ Almost everyone else in the world uses $f(S)$ (though $f[S]$ sees some use, and $f''(S)$ is often used in logic) but this is abuse of notation, and I prefer $f\im(S)$ for emphasis. This image notation is *not* standard.
- If $S \subseteq X$, then the **restriction** of $f$ to $S$ is denoted $f \restrict{S}$, i.e. it is the function $f \restrict{S} \colon S \to Y$.
- Sometimes functions $f \colon X \to Y$ are *injective* or *surjective*; I may emphasize this sometimes by writing $f \colon X \hookrightarrow Y$ or $f \colon X \twoheadrightarrow Y$, respectively.
This is brief, intended as a reminder for experts. Consult for full details.
Let $X \taking f Y$ be a function:
- By $f\pre(T)$ I mean the **pre-image** $$f\pre(T) := \left\{ x \in X \mid f(x) \in T \right\}.$$ This is in contrast to the $f^{-1}(T)$ used in the rest of the world; I only use $f^{-1}$ for an inverse *function*.
By abuse of notation, we may abbreviate $f\pre(\{y\})$ to $f\pre(y)$. We call $f\pre(y)$ a **fiber**.
- By $f\im(S)$ I mean the **image** $$f\im(S) := \left\{ f(x) \mid x \in S \right\}.$$ Almost everyone else in the world uses $f(S)$ (though $f[S]$ sees some use, and $f''(S)$ is often used in logic) but this is abuse of notation, and I prefer $f\im(S)$ for emphasis. This image notation is *not* standard.
- If $S \subseteq X$, then the **restriction** of $f$ to $S$ is denoted $f \restrict{S}$, i.e. it is the function $f \restrict{S} \colon S \to Y$.
- Sometimes functions $f \colon X \to Y$ are *injective* or *surjective*; I may emphasize this sometimes by writing $f \colon X \hookrightarrow Y$ or $f \colon X \twoheadrightarrow Y$, respectively. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Advice for the reader | Functions | 03_advice.md | 5 | 621 |
Additionally, a permutation on a finite set may be denoted in *cycle notation*, as described in say <https://en.wikipedia.org/wiki/Permutation#Cycle_notation>. For example the notation $(1 \; 2 \; 3 \; 4)(5 \; 6 \; 7)$ refers to the permutation with $1 \mapsto 2$, $2 \mapsto 3$, $3 \mapsto 4$, $4 \mapsto 1$, $5 \mapsto 6$, $6 \mapsto 7$, $7 \mapsto 5$. Usage of this notation will usually be obvious from context.
Additionally, a permutation on a finite set may be denoted in *cycle notation*, as described in say <https://en.wikipedia.org/wiki/Permutation#Cycle_notation>. For example the notation $(1 \; 2 \; 3 \; 4)(5 \; 6 \; 7)$ refers to the permutation with $1 \mapsto 2$, $2 \mapsto 3$, $3 \mapsto 4$, $4 \mapsto 1$, $5 \mapsto 6$, $6 \mapsto 7$, $7 \mapsto 5$. Usage of this notation will usually be obvious from context.
All rings have a multiplicative identity $1$ unless otherwise specified. We allow $0=1$ in general rings but not in integral domains.
**All rings are commutative unless otherwise specified.** There is an elaborate scheme for naming rings which are not commutative, used only in the chapter on cohomology rings:
Graded Not Graded
----------------------------------- ----------------------------- -------------
$1$ not required graded pseudo-ring pseudo-ring
Anticommutative, $1$ not required anticommutative pseudo-ring N/A
Has $1$ graded ring N/A
Anticommutative with $1$ anticommutative ring N/A
Commutative with $1$ commutative graded ring ring
On the other hand, an *algebra* always has $1$, but it need not be commutative.
All rings have a multiplicative identity $1$ unless otherwise specified. We allow $0=1$ in general rings but not in integral domains.
**All rings are commutative unless otherwise specified.** There is an elaborate scheme for naming rings which are not commutative, used only in the chapter on cohomology rings:
Graded Not Graded
----------------------------------- ----------------------------- -------------
$1$ not required graded pseudo-ring pseudo-ring
Anticommutative, $1$ not required anticommutative pseudo-ring N/A
Has $1$ graded ring N/A
Anticommutative with $1$ anticommutative ring N/A
Commutative with $1$ commutative graded ring ring
On the other hand, an *algebra* always has $1$, but it need not be commutative. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Advice for the reader | Cycle notation for permutations | 03_advice.md | 6 | 655 |
We accept the Axiom of Choice, and use it freely.
We accept the Axiom of Choice, and use it freely.
The appendix contains a list of resources I like, and explanations of pedagogical choices that I made for each chapter. I encourage you to check it out.
In particular, this is where you should go for further reading! There are some topics that should be covered in the Napkin, but are not, due to my own ignorance or laziness. The references provided in this appendix should hopefully help partially atone for my omissions.
[^1]: This is to avoid the classic "we are done by PSet 4, Problem 8" that happens in college sometimes, as if I remembered what that was.
[^2]: See also <https://blog.evanchen.cc/2015/03/14/writing/> and the source above.
The appendix contains a list of resources I like, and explanations of pedagogical choices that I made for each chapter. I encourage you to check it out.
In particular, this is where you should go for further reading! There are some topics that should be covered in the Napkin, but are not, due to my own ignorance or laziness. The references provided in this appendix should hopefully help partially atone for my omissions.
[^1]: This is to avoid the classic "we are done by PSet 4, Problem 8" that happens in college sometimes, as if I remembered what that was.
[^2]: See also <https://blog.evanchen.cc/2015/03/14/writing/> and the source above. | An Infinitely Large Napkin | napkin | general | advanced | Front Matter | Advice for the reader | Choice | 03_advice.md | 7 | 400 |
This chapter contains a pitch for each part, to help you decide what you want to read and to elaborate more on how they are interconnected.
For convenience, here is again the dependency plot that appeared in the frontmatter.
,45:Ch ,-[Abs Alg](#part:absalg) ,45:Ch ,-[Topology](#part:basictop) ,45:Ch -,[Lin Alg](#part:linalg) ,35:Ch [Grp Act](#part:groups) ,24:Ch [Grp Classif](#ch:sylow) ,35:Ch -[Rep Th](#part:repth) ,43:Ch -[Quantum](#part:quantum) ,38:Ch -[Calc](#part:calc) ,30:Ch -[Cmplx Ana](#part:cmplxana) ,20:Ch -[Measure/Pr](#part:measure) ,28:Ch -[Diff Geo](#part:diffgeo) ,10:Ch -[Alg NT 1](#part:algnt1) ,0:Ch -[Alg NT 2](#part:algnt2) ,10:Ch -[Alg Top 1](#part:algtop1) ,28:Ch -[Cat Th](#part:cats) ,0:Ch -[Alg Top 2](#part:algtop2) ,10:Ch -[Alg Geo 1](#part:ag1) ,0:Ch -[Alg Geo 2-3](#part:ag2) ,45:Ch -[Set Theory](#part:st1)
,45,30,35;0: ,45,33,45;0: ,45,30,35;0: ,45,45,43;0: ,45,20,28;0: ,45,20,28;0: ,45,20,28;80: ,45,20,28;-40: ,28,30,35;0: ,45,5,35;0: ,35,5,24;0: ,45,48,28;30: ,45,64,38;0: ,38,48,28;50: ,45,48,28;10: ,45,64,38;0: ,38,64,30;0: ,45,64,30;-10: ,10,64,30;0: ,45,55,20;0: ,38,55,20;50: ,10,6,0;0: ,45,6,10;0: ,45,6,10;-40: ,28,6,10;0: ,28,6,0;-18: ,10,23,0;0: ,45,23,10;0: ,35,23,10;20: ,10,20,28;20: ,28,23,0;-190: ,10,40,0;0: ,45,40,10;-40: ,45,40,10;50: ,35,40,0;10:
This chapter contains a pitch for each part, to help you decide what you want to read and to elaborate more on how they are interconnected.
For convenience, here is again the dependency plot that appeared in the frontmatter.
,45:Ch ,-[Abs Alg](#part:absalg) ,45:Ch ,-[Topology](#part:basictop) ,45:Ch -,[Lin Alg](#part:linalg) ,35:Ch [Grp Act](#part:groups) ,24:Ch [Grp Classif](#ch:sylow) ,35:Ch -[Rep Th](#part:repth) ,43:Ch -[Quantum](#part:quantum) ,38:Ch -[Calc](#part:calc) ,30:Ch -[Cmplx Ana](#part:cmplxana) ,20:Ch -[Measure/Pr](#part:measure) ,28:Ch -[Diff Geo](#part:diffgeo) ,10:Ch -[Alg NT 1](#part:algnt1) ,0:Ch -[Alg NT 2](#part:algnt2) ,10:Ch -[Alg Top 1](#part:algtop1) ,28:Ch -[Cat Th](#part:cats) ,0:Ch -[Alg Top 2](#part:algtop2) ,10:Ch -[Alg Geo 1](#part:ag1) ,0:Ch -[Alg Geo 2-3](#part:ag2) ,45:Ch -[Set Theory](#part:st1)
,45,30,35;0: ,45,33,45;0: ,45,30,35;0: ,45,45,43;0: ,45,20,28;0: ,45,20,28;0: ,45,20,28;80: ,45,20,28;-40: ,28,30,35;0: ,45,5,35;0: ,35,5,24;0: ,45,48,28;30: ,45,64,38;0: ,38,48,28;50: ,45,48,28;10: ,45,64,38;0: ,38,64,30;0: ,45,64,30;-10: ,10,64,30;0: ,45,55,20;0: ,38,55,20;50: ,10,6,0;0: ,45,6,10;0: ,45,6,10;-40: ,28,6,10;0: ,28,6,0;-18: ,10,23,0;0: ,45,23,10;0: ,35,23,10;20: ,10,20,28;20: ,28,23,0;-190: ,10,40,0;0: ,45,40,10;-40: ,45,40,10;50: ,35,40,0;10: | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | 04_salespitch.md | 0 | 749 | |
- [**.**]\
I made a design decision that the first part should have a little bit of both algebra and topology: so this first chapter begins by defining a **group**, while the second chapter begins by defining a **metric space**. The intention is so that newcomers get to see two different examples of "sets with additional structure" in somewhat different contexts, and to have a minimal amount of literacy as these sorts of definitions appear over and over.[^1]
- [**.**]\
The algebraically inclined can then delve into further types of algebraic structures: some more details of **groups**, and then **rings** and **fields** --- which will let you generalize $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$. So you'll learn to become familiar with all sorts of other nouns that appear in algebra, unlocking a whole host of objects that one couldn't talk about before.
We'll also come to **ideals**, which generalize the GCD in $\mathbb{Z}$ that you might know of. For example, you know in $\mathbb{Z}$ that any integer can be written in the form $3a+5b$ for $a,b \in \mathbb{Z}$, since $\gcd(3,5)=1$. We'll see that this statement is really a statement of ideals: "$(3,5)=1$ in $\mathbb{Z}$", and thus we'll understand in what situations it can be generalized, e.g. to polynomials.
- [**.**]\
The more analytically inclined can instead move into topology, learning more about spaces. We'll find out that "metric spaces" are actually too specific, and that it's better to work with **topological spaces**, which are based on the so-called **open sets**. You'll then get to see the buddings of some geometrical ideals, ending with the really great notion of **compactness**, a powerful notion that makes real analysis tick.
One example of an application of compactness to tempt you now: a continuous function $f \colon [0,1] \to \mathbb{R}$ always achieves a *maximum* value. (In contrast, $f \colon (0,1) \to \mathbb{R}$ by $x \mapsto 1/x$ does not.) We'll see the reason is that $[0,1]$ is compact.
- [**.**]\
I made a design decision that the first part should have a little bit of both algebra and topology: so this first chapter begins by defining a **group**, while the second chapter begins by defining a **metric space**. The intention is so that newcomers get to see two different examples of "sets with additional structure" in somewhat different contexts, and to have a minimal amount of literacy as these sorts of definitions appear over and over.[^1]
- [**.**]\
The algebraically inclined can then delve into further types of algebraic structures: some more details of **groups**, and then **rings** and **fields** --- which will let you generalize $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$. So you'll learn to become familiar with all sorts of other nouns that appear in algebra, unlocking a whole host of objects that one couldn't talk about before.
We'll also come to **ideals**, which generalize the GCD in $\mathbb{Z}$ that you might know of. For example, you know in $\mathbb{Z}$ that any integer can be written in the form $3a+5b$ for $a,b \in \mathbb{Z}$, since $\gcd(3,5)=1$. We'll see that this statement is really a statement of ideals: "$(3,5)=1$ in $\mathbb{Z}$", and thus we'll understand in what situations it can be generalized, e.g. to polynomials.
- [**.**]\
The more analytically inclined can instead move into topology, learning more about spaces. We'll find out that "metric spaces" are actually too specific, and that it's better to work with **topological spaces**, which are based on the so-called **open sets**. You'll then get to see the buddings of some geometrical ideals, ending with the really great notion of **compactness**, a powerful notion that makes real analysis tick.
One example of an application of compactness to tempt you now: a continuous function $f \colon [0,1] \to \mathbb{R}$ always achieves a *maximum* value. (In contrast, $f \colon (0,1) \to \mathbb{R}$ by $x \mapsto 1/x$ does not.) We'll see the reason is that $[0,1]$ is compact. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | The basics | 04_salespitch.md | 1 | 1,149 |
- [**.**]\
In high school, linear algebra is often really unsatisfying. You are given these arrays of numbers, and they're manipulated in some ways that don't really make sense. For example, the determinant is defined as this funny-looking sum with a bunch of products that seems to come out of thin air. Where does it come from? Why does $\det(AB) = \det A \det B$ with such a bizarre formula?
Well, it turns out that you *can* explain all of these things! The trick is to not think of linear algebra as the study of matrices, but instead as the study of *linear maps*. In earlier chapters we saw that we got great generalizations by speaking of "sets with enriched structure" and "maps between them". This time, our sets are **vector spaces** and our maps are **linear maps**. We'll find out that a matrix is actually just a way of writing down a linear map as an array of numbers, but using the "intrinsic" definitions we'll de-mystify all the strange formulas from high school and show you where they all come from.
In particular, we'll see *easy* proofs that column rank equals row rank, determinant is multiplicative, trace is the sum of the diagonal entries. We'll see how the dot product works, and learn all the words starting with "eigen-". We'll even have a bonus chapter for Fourier analysis showing that you can also explain all the big buzz-words by just being comfortable with vector spaces.
- [**.**]\
Some of you might be interested in more about groups, and this chapter will give you a way to play further. It starts with an exploration of **group actions**, then goes into a bit on **Sylow theorems**, which are the tools that let us try to *classify all groups*.
- [**.**]\
If $G$ is a group, we can try to understand it by implementing it as a *matrix*, i.e. considering embeddings $G \hookrightarrow \GL_n(\mathbb{C})$. These are called **representations** of $G$; it turns out that they can be decomposed into **irreducible** ones. Astonishingly we will find that we can *basically characterize all of them*: the results turn out to be short and completely unexpected.
For example, we will find out that there are finitely many irreducible representations of a given finite group $G$; if we label them $V_1$, $V_2$, ..., $V_r$, then we will find that $r$ is the number of conjugacy classes of $G$, and moreover that $$|G| = (\dim V_1)^2 + \dots + (\dim V_r)^2$$ which comes out of nowhere!
The last chapter of this part will show you some unexpected corollaries. Here is one of them: let $G$ be a finite group and create variables $x_g$ for each $g \in G$. A $|G| \times |G|$ matrix $M$ is defined by setting the $(g,h)$th entry to be the variable $x_{g \cdot h}$. Then this determinant will turn out to *factor*, and the factors will correspond to the $V_i$ we described above: there will be an irreducible factor of degree $\dim V_i$ appearing $\dim V_i$ times. This result, called the **Frobenius determinant**, is said to have given birth to representation theory.
- [**.**]\
If you ever wondered what **Shor's algorithm** is, this chapter will use the built-up linear algebra to tell you! | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | Abstract algebra | 04_salespitch.md | 2 | 892 |
- [**.**]\
In high school, linear algebra is often really unsatisfying. You are given these arrays of numbers, and they're manipulated in some ways that don't really make sense. For example, the determinant is defined as this funny-looking sum with a bunch of products that seems to come out of thin air. Where does it come from? Why does $\det(AB) = \det A \det B$ with such a bizarre formula?
Well, it turns out that you *can* explain all of these things! The trick is to not think of linear algebra as the study of matrices, but instead as the study of *linear maps*. In earlier chapters we saw that we got great generalizations by speaking of "sets with enriched structure" and "maps between them". This time, our sets are **vector spaces** and our maps are **linear maps**. We'll find out that a matrix is actually just a way of writing down a linear map as an array of numbers, but using the "intrinsic" definitions we'll de-mystify all the strange formulas from high school and show you where they all come from.
In particular, we'll see *easy* proofs that column rank equals row rank, determinant is multiplicative, trace is the sum of the diagonal entries. We'll see how the dot product works, and learn all the words starting with "eigen-". We'll even have a bonus chapter for Fourier analysis showing that you can also explain all the big buzz-words by just being comfortable with vector spaces.
- [**.**]\
Some of you might be interested in more about groups, and this chapter will give you a way to play further. It starts with an exploration of **group actions**, then goes into a bit on **Sylow theorems**, which are the tools that let us try to *classify all groups*.
- [**.**]\
If $G$ is a group, we can try to understand it by implementing it as a *matrix*, i.e. considering embeddings $G \hookrightarrow \GL_n(\mathbb{C})$. These are called **representations** of $G$; it turns out that they can be decomposed into **irreducible** ones. Astonishingly we will find that we can *basically characterize all of them*: the results turn out to be short and completely unexpected.
For example, we will find out that there are finitely many irreducible representations of a given finite group $G$; if we label them $V_1$, $V_2$, ..., $V_r$, then we will find that $r$ is the number of conjugacy classes of $G$, and moreover that $$|G| = (\dim V_1)^2 + \dots + (\dim V_r)^2$$ which comes out of nowhere!
The last chapter of this part will show you some unexpected corollaries. Here is one of them: let $G$ be a finite group and create variables $x_g$ for each $g \in G$. A $|G| \times |G|$ matrix $M$ is defined by setting the $(g,h)$th entry to be the variable $x_{g \cdot h}$. Then this determinant will turn out to *factor*, and the factors will correspond to the $V_i$ we described above: there will be an irreducible factor of degree $\dim V_i$ appearing $\dim V_i$ times. This result, called the **Frobenius determinant**, is said to have given birth to representation theory.
- [**.**]\
If you ever wondered what **Shor's algorithm** is, this chapter will use the built-up linear algebra to tell you! | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | Abstract algebra | 04_salespitch.md | 3 | 892 |
- [**.**]\
In this part, we'll use our built-up knowledge of metric and topological spaces to give short, rigorous definitions and theorems typical of high school calculus. That is, we'll really define and prove most everything you've seen about **limits**, **series**, **derivatives**, and **integrals**.
Although this might seem intimidating, it turns out that actually, by the time we start this chapter, *the hard work has already been done*: the notion of limits, open sets, and compactness will make short work of what was swept under the rug in AP calculus. Most of the proofs will thus actually be quite short. We sit back and watch all the pieces slowly come together as a reward for our careful study of topology beforehand.
That said, if you are willing to suspend belief, you can actually read most of the other parts without knowing the exact details of all the calculus here, so in some sense this part is "optional".
- [**.**]\
It turns out that **holomorphic functions** (complex-differentiable functions) are close to the nicest things ever: they turn out to be given by a Taylor series (i.e. are basically polynomials). This means we'll be able to prove unreasonably nice results about holomorphic functions $\mathbb{C} \to \mathbb{C}$, like
- they are determined by just a few inputs,
- their contour integrals are all zero,
- they can't be bounded unless they are constant,
- ....
We then introduce **meromorphic functions**, which are like quotients of holomorphic functions, and find that we can detect their zeros by simply drawing loops in the plane and integrating over them: the famous **residue theorem** appears. (In the practice problems, you will see this even gives us a way to evaluate real integrals that can't be evaluated otherwise.)
- [**.**]\
Measure theory is the upgraded version of integration. The Riemann integration is for a lot of purposes not really sufficient; for example, if $f$ is the function equals $1$ at rational numbers but $0$ at irrational numbers, we would hope that $\int_0^1 f(x) \; dx = 0$, but the Riemann integral is not capable of handling this function $f$.
The **Lebesgue integral** will handle these mistakes by assigning a *measure* to a generic space $\Omega$, making it into a **measure space**. This will let us develop a richer theory of integration where the above integral *does* work out to zero because the "rational numbers have measure zero". Even the development of the measure will be an achievement, because it means we've developed a rigorous, complete way of talking about what notions like area and volume mean --- on any space, not just $\mathbb{R}^n$! So for example the Lebesgue integral will let us integrate functions over any **measure space**.
- [**.**]\
Using the tools of measure theory, we'll be able to start giving rigorous definitions of **probability**, too. We'll see that a **random variable** is actually a function from a measure space of worlds to $\mathbb{R}$, giving us a rigorous way to talk about its probabilities. We can then start actually stating results like the **law of large numbers** and **central limit theorem** in ways that make them both easy to state and straightforward to prove.
- [**.**]\
Multivariable calculus is often confusing because of all the partial derivatives. But we'll find out that, armed with our good understanding of linear algebra, that we're really looking at a **total derivative**: at every point of a function $f \colon \mathbb{R}^n \to \mathbb{R}$ we can associate a *linear map* $Df$ which captures in one object the notion of partial derivatives. Set up this way, we'll get to see versions of **differential forms** and **Stokes' theorem**, and we finally will know what the notation $dx$ really means. In the end, we'll say a little bit about manifolds in general. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | Real and complex analysis | 04_salespitch.md | 4 | 1,091 |
- [**.**]\
In this part, we'll use our built-up knowledge of metric and topological spaces to give short, rigorous definitions and theorems typical of high school calculus. That is, we'll really define and prove most everything you've seen about **limits**, **series**, **derivatives**, and **integrals**.
Although this might seem intimidating, it turns out that actually, by the time we start this chapter, *the hard work has already been done*: the notion of limits, open sets, and compactness will make short work of what was swept under the rug in AP calculus. Most of the proofs will thus actually be quite short. We sit back and watch all the pieces slowly come together as a reward for our careful study of topology beforehand.
That said, if you are willing to suspend belief, you can actually read most of the other parts without knowing the exact details of all the calculus here, so in some sense this part is "optional".
- [**.**]\
It turns out that **holomorphic functions** (complex-differentiable functions) are close to the nicest things ever: they turn out to be given by a Taylor series (i.e. are basically polynomials). This means we'll be able to prove unreasonably nice results about holomorphic functions $\mathbb{C} \to \mathbb{C}$, like
- they are determined by just a few inputs,
- their contour integrals are all zero,
- they can't be bounded unless they are constant,
- ....
We then introduce **meromorphic functions**, which are like quotients of holomorphic functions, and find that we can detect their zeros by simply drawing loops in the plane and integrating over them: the famous **residue theorem** appears. (In the practice problems, you will see this even gives us a way to evaluate real integrals that can't be evaluated otherwise.)
- [**.**]\
Measure theory is the upgraded version of integration. The Riemann integration is for a lot of purposes not really sufficient; for example, if $f$ is the function equals $1$ at rational numbers but $0$ at irrational numbers, we would hope that $\int_0^1 f(x) \; dx = 0$, but the Riemann integral is not capable of handling this function $f$.
The **Lebesgue integral** will handle these mistakes by assigning a *measure* to a generic space $\Omega$, making it into a **measure space**. This will let us develop a richer theory of integration where the above integral *does* work out to zero because the "rational numbers have measure zero". Even the development of the measure will be an achievement, because it means we've developed a rigorous, complete way of talking about what notions like area and volume mean --- on any space, not just $\mathbb{R}^n$! So for example the Lebesgue integral will let us integrate functions over any **measure space**.
- [**.**]\
Using the tools of measure theory, we'll be able to start giving rigorous definitions of **probability**, too. We'll see that a **random variable** is actually a function from a measure space of worlds to $\mathbb{R}$, giving us a rigorous way to talk about its probabilities. We can then start actually stating results like the **law of large numbers** and **central limit theorem** in ways that make them both easy to state and straightforward to prove.
- [**.**]\
Multivariable calculus is often confusing because of all the partial derivatives. But we'll find out that, armed with our good understanding of linear algebra, that we're really looking at a **total derivative**: at every point of a function $f \colon \mathbb{R}^n \to \mathbb{R}$ we can associate a *linear map* $Df$ which captures in one object the notion of partial derivatives. Set up this way, we'll get to see versions of **differential forms** and **Stokes' theorem**, and we finally will know what the notation $dx$ really means. In the end, we'll say a little bit about manifolds in general. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | Real and complex analysis | 04_salespitch.md | 5 | 1,091 |
- [**.**]\
Why is $3+\sqrt5$ the conjugate of $3-\sqrt5$? How come the norm $\norm{a+b\sqrt5} = a^2-5b^2$ used in Pell's equations just happens to be multiplicative? Why is it we can do factoring into primes in $\mathbb{Z}[i]$ but not in $\mathbb{Z}[\sqrt{-5}]$? All these questions and more will be answered in this part, when we learn about **number fields**, a generalization of $\mathbb{Q}$ and $\mathbb{Z}$ to things like $\mathbb{Q}(\sqrt5)$ and $\mathbb{Z}[\sqrt{5}]$. We'll find out that we have unique factorization into prime ideals, that there is a real *multiplicative norm* in play here, and so on. We'll also see that Pell's equation falls out of this theory.
- [**.**]\
All the big buzz-words come out now: **Galois groups**, the **Frobenius**, and friends. We'll see quadratic reciprocity is just a shadow of the behavior of the Frobenius element, and meet the **Chebotarev density theorem**, which generalizes greatly the Dirichlet theorem on the infinitude of primes which are $a \pmod n$. Towards the end, we'll also state **Artin reciprocity**, one of the great results of **class field theory**, and how it generalizes quadratic reciprocity and cubic reciprocity.
- [**.**]\
Why is $3+\sqrt5$ the conjugate of $3-\sqrt5$? How come the norm $\norm{a+b\sqrt5} = a^2-5b^2$ used in Pell's equations just happens to be multiplicative? Why is it we can do factoring into primes in $\mathbb{Z}[i]$ but not in $\mathbb{Z}[\sqrt{-5}]$? All these questions and more will be answered in this part, when we learn about **number fields**, a generalization of $\mathbb{Q}$ and $\mathbb{Z}$ to things like $\mathbb{Q}(\sqrt5)$ and $\mathbb{Z}[\sqrt{5}]$. We'll find out that we have unique factorization into prime ideals, that there is a real *multiplicative norm* in play here, and so on. We'll also see that Pell's equation falls out of this theory.
- [**.**]\
All the big buzz-words come out now: **Galois groups**, the **Frobenius**, and friends. We'll see quadratic reciprocity is just a shadow of the behavior of the Frobenius element, and meet the **Chebotarev density theorem**, which generalizes greatly the Dirichlet theorem on the infinitude of primes which are $a \pmod n$. Towards the end, we'll also state **Artin reciprocity**, one of the great results of **class field theory**, and how it generalizes quadratic reciprocity and cubic reciprocity. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | Algebraic number theory | 04_salespitch.md | 6 | 677 |
- [**.**]\
What's the difference between an annulus and disk? Well, one of them has a "hole" in it, but if we are just given intrinsic topological spaces it's hard to make this notion precise. The **fundamental group** $\pi_1(X)$ and more general **homotopy group** will make this precise --- we'll find a way to define an abelian group $\pi_1(X)$ for every topological space $X$ which captures the idea there is a hole in the space, by throwing lassos into the space and seeing if we can reel them in.
Amazingly, the fundamental group $\pi_1(X)$ will, under mild conditions, tell you about ways to cover $X$ with a so-called **covering projection**. One picture is that one can wrap a real line $\mathbb{R}$ into a helix shape and then project it down into the circle $S^1$. This will turn out to correspond to the fact that $\pi_1(S^1) = \mathbb{Z}$ which has only one subgroup. More generally the subgroups of $\pi_1(X)$ will be in bijection with ways to cover the space $X$!
- [**.**]\
What do fields, groups, manifolds, metric spaces, measure spaces, modules, representations, rings, topological spaces, vector spaces, all have in common? Answer: they are all "objects with additional structure", with maps between them.
The notion of **category** will appropriately generalize all of them. We'll see that all sorts of constructions and ideas can be abstracted into the framework of a category, in which we *only* think about objects and arrows between them, without probing too hard into the details of what those objects are. This results in drawing many **commutative diagrams**.
For example, any way of taking an object in one category and getting another one (for example $\pi_1$ as above, from the category of spaces into the category of groups) will probably be a **functor**. We'll unify $G \times H$, $X \times Y$, $R \times S$, and anything with the $\times$ symbol into the notion of a product, and then even more generally into a **limit**. Towards the end, we talk about **abelian categories** and talk about the famous **snake lemma**, **five lemma**, and so on.
- [**.**]\
Using the language of category theory, we then resume our adventures in algebraic topology, in which we define the **homology groups** which give a different way of noticing holes in a space, in a way that is longer to define but easier to compute in practice. We'll then reverse the construction to get so-called **cohomology rings** instead, which give us an even finer invariant for telling spaces apart.
- [**.**]\
What's the difference between an annulus and disk? Well, one of them has a "hole" in it, but if we are just given intrinsic topological spaces it's hard to make this notion precise. The **fundamental group** $\pi_1(X)$ and more general **homotopy group** will make this precise --- we'll find a way to define an abelian group $\pi_1(X)$ for every topological space $X$ which captures the idea there is a hole in the space, by throwing lassos into the space and seeing if we can reel them in.
Amazingly, the fundamental group $\pi_1(X)$ will, under mild conditions, tell you about ways to cover $X$ with a so-called **covering projection**. One picture is that one can wrap a real line $\mathbb{R}$ into a helix shape and then project it down into the circle $S^1$. This will turn out to correspond to the fact that $\pi_1(S^1) = \mathbb{Z}$ which has only one subgroup. More generally the subgroups of $\pi_1(X)$ will be in bijection with ways to cover the space $X$!
- [**.**]\
What do fields, groups, manifolds, metric spaces, measure spaces, modules, representations, rings, topological spaces, vector spaces, all have in common? Answer: they are all "objects with additional structure", with maps between them.
The notion of **category** will appropriately generalize all of them. We'll see that all sorts of constructions and ideas can be abstracted into the framework of a category, in which we *only* think about objects and arrows between them, without probing too hard into the details of what those objects are. This results in drawing many **commutative diagrams**.
For example, any way of taking an object in one category and getting another one (for example $\pi_1$ as above, from the category of spaces into the category of groups) will probably be a **functor**. We'll unify $G \times H$, $X \times Y$, $R \times S$, and anything with the $\times$ symbol into the notion of a product, and then even more generally into a **limit**. Towards the end, we talk about **abelian categories** and talk about the famous **snake lemma**, **five lemma**, and so on.
- [**.**]\
Using the language of category theory, we then resume our adventures in algebraic topology, in which we define the **homology groups** which give a different way of noticing holes in a space, in a way that is longer to define but easier to compute in practice. We'll then reverse the construction to get so-called **cohomology rings** instead, which give us an even finer invariant for telling spaces apart. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | Algebraic topology | 04_salespitch.md | 7 | 1,432 |
- [**.**]\
We begin with a classical study of classical **complex varieties**: the study of intersections of polynomial equations over $\mathbb{C}$. This will naturally lead us into the geometry of rings, giving ways to draw pictures of ideals, and motivating **Hilbert's nullstellensatz**. The **Zariski topology** will show its face, and then we'll play with **projective varieties** and **quasi-projective varieties**, with a bonus detour into **Bézout's theorem**. All this prepares us for our journey into schemes.
- [**.**]\
We now get serious and delve into Grothendieck's definition of an **affine scheme**: a generalization of our classical varieties that allows us to start with any ring $A$ and construct a space $\Spec A$ on it. We'll equip it with its own Zariski topology and then a sheaf of functions on it, making it into a **locally ringed space**; we will find that the sheaf can be understood effectively in terms of **localization** on it. We'll find that the language of commutative algebra provides elegant generalizations of what's going on geometrically: prime ideals correspond to irreducible closed subsets, radical ideals correspond to closed subsets, maximal ideals correspond to closed points, and so on. We'll draw lots of pictures of spaces and examples to accompany this.
- [**.**]\
We begin with a classical study of classical **complex varieties**: the study of intersections of polynomial equations over $\mathbb{C}$. This will naturally lead us into the geometry of rings, giving ways to draw pictures of ideals, and motivating **Hilbert's nullstellensatz**. The **Zariski topology** will show its face, and then we'll play with **projective varieties** and **quasi-projective varieties**, with a bonus detour into **Bézout's theorem**. All this prepares us for our journey into schemes.
- [**.**]\
We now get serious and delve into Grothendieck's definition of an **affine scheme**: a generalization of our classical varieties that allows us to start with any ring $A$ and construct a space $\Spec A$ on it. We'll equip it with its own Zariski topology and then a sheaf of functions on it, making it into a **locally ringed space**; we will find that the sheaf can be understood effectively in terms of **localization** on it. We'll find that the language of commutative algebra provides elegant generalizations of what's going on geometrically: prime ideals correspond to irreducible closed subsets, radical ideals correspond to closed subsets, maximal ideals correspond to closed points, and so on. We'll draw lots of pictures of spaces and examples to accompany this. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | Algebraic geometry | 04_salespitch.md | 8 | 745 |
- [**.**]\
Why is **Russell's paradox** such a big deal and how is it resolved? What is this **Zorn's lemma** that everyone keeps talking about? In this part we'll learn the answers to these questions by giving a real description of the **Zermelo-Frankel** axioms, and the **axiom of choice**, delving into the details of how math is built axiomatically at the very bottom foundations. We'll meet the **ordinal numbers** and **cardinal numbers** and learn how to do **transfinite induction** with them.
- [**.**]\
The **continuum hypothesis** states that there are no cardinalities between the size of the natural numbers and the size of the real numbers. It was shown to be *independent* of the axioms --- one cannot prove or disprove it. How could a result like that possibly be proved? Using our understanding of the ZF axioms, we'll develop a bit of **model theory** and then use **forcing** in order to show how to construct entire models of the universe in which the continuum hypothesis is true or false.
[^1]: In particular, I think it's easier to learn what a homeomorphism is after seeing group isomorphism, and what a homomorphism is after seeing continuous map.
- [**.**]\
Why is **Russell's paradox** such a big deal and how is it resolved? What is this **Zorn's lemma** that everyone keeps talking about? In this part we'll learn the answers to these questions by giving a real description of the **Zermelo-Frankel** axioms, and the **axiom of choice**, delving into the details of how math is built axiomatically at the very bottom foundations. We'll meet the **ordinal numbers** and **cardinal numbers** and learn how to do **transfinite induction** with them.
- [**.**]\
The **continuum hypothesis** states that there are no cardinalities between the size of the natural numbers and the size of the real numbers. It was shown to be *independent* of the axioms --- one cannot prove or disprove it. How could a result like that possibly be proved? Using our understanding of the ZF axioms, we'll develop a bit of **model theory** and then use **forcing** in order to show how to construct entire models of the universe in which the continuum hypothesis is true or false.
[^1]: In particular, I think it's easier to learn what a homeomorphism is after seeing group isomorphism, and what a homomorphism is after seeing continuous map. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Sales pitches | Set theory | 04_salespitch.md | 9 | 672 |
A group is one of the most basic structures in higher mathematics. In this chapter I will tell you only the bare minimum: what a group is, and when two groups are the same.
A group is one of the most basic structures in higher mathematics. In this chapter I will tell you only the bare minimum: what a group is, and when two groups are the same. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | 05_grp-intro.md | 0 | 98 | |
A group consists of two pieces of data: a set $G$, and an associative binary operation $\star$ with some properties. Before I write down the definition of a group, let me give two examples.
The pair $(\mathbb{Z}, +)$ is a group: $\mathbb{Z} = \left\{ \dots,-2,-1,0,1,2,\dots \right\}$ is the set and the associative operation is *addition*. Note that
- The element $0 \in \mathbb{Z}$ is an *identity*: $a+0=0+a = a$ for any $a$.
- Every element $a \in \mathbb{Z}$ has an additive *inverse*: $a + (-a) = (-a) + a = 0$.
We call this group $\mathbb{Z}$.
Let $\mathbb{Q}^\times$ be the set of *nonzero rational numbers*. The pair $(\mathbb{Q}^\times, \cdot)$ is a group: the set is $\mathbb{Q}^\times$ and the associative operation is *multiplication*.
Again we see the same two nice properties.
- The element $1 \in \mathbb{Q}^\times$ is an *identity*: for any rational number, $a \cdot 1 = 1 \cdot a = a$.
- For any rational number $x \in \mathbb{Q}^\times$, we have an inverse $x^{-1}$, such that $$x \cdot x^{-1} = x^{-1} \cdot x = 1.$$
From this you might already have a guess what the definition of a group is.
A **group** is a pair $G = (G, \star)$ consisting of a set of elements $G$, and a binary operation $\star$ on $G$, such that:
- $G$ has an **identity element**, usually denoted $1_G$ or just $1$, with the property that $$1_G \star g = g \star 1_G = g \text{ for all $g \in G$}.$$
- The operation is **associative**, meaning $(a \star b) \star c = a \star (b \star c)$ for any $a,b,c \in G$. Consequently we generally don't write the parentheses.
- Each element $g \in G$ has an **inverse**, that is, an element $h \in G$ such that $$g \star h = h \star g = 1_G.$$
Some authors like to add a "closure" axiom, i.e. to say explicitly that $g \star h \in G$. This is implied already by the fact that $\star$ is a binary operation on $G$, but is worth keeping in mind for the examples below.
It is not required that $\star$ is commutative ($a \star b = b \star a$). So we say that a group is **abelian** if the operation is commutative and **non-abelian** otherwise.
- The pair $(\mathbb{Q}, \cdot)$ is NOT a group. (Here $\mathbb{Q}$ is rational numbers.) While there is an identity element, the element $0 \in \mathbb{Q}$ does not have an inverse.
- The pair $(\mathbb{Z}, \cdot)$ is also NOT a group. (Why?)
- Let $\mathrm{Mat}_{2 \times 2}(\mathbb{R})$ be the set of $2 \times 2$ real matrices. Then $(\mathrm{Mat}_{2 \times 2}(\mathbb{R}), \cdot)$ (where $\cdot$ is matrix multiplication) is NOT a group. Indeed, even though we have an identity matrix $$\begin{bmatrix}
1 & 0 \\ 0 & 1
\end{bmatrix}$$ we still run into the same issue as before: the zero matrix does not have a multiplicative inverse.
(Even if we delete the zero matrix from the set, the resulting structure is still not a group: those of you that know some linear algebra might recall that any matrix with determinant zero cannot have an inverse.)
Let's resume writing down examples. Here are some more **abelian examples** of groups:
Let $S^1$ denote the set of complex numbers $z$ with absolute value one; that is $$S^1 := \left\{ z \in \mathbb{C} \mid \left\lvert z \right\rvert = 1 \right\}.$$ Then $(S^1, \times)$ is a group because
- The complex number $1 \in S^1$ serves as the identity, and
- Each complex number $z \in S^1$ has an inverse $\frac 1z$ which is also in $S^1$, since $\left\lvert z^{-1} \right\rvert = \left\lvert z \right\rvert^{-1} = 1$.
There is one thing I ought to also check: that $z_1 \times z_2$ is actually still in $S^1$. But this follows from the fact that $\left\lvert z_1z_2 \right\rvert = \left\lvert z_1 \right\rvert \left\lvert z_2 \right\rvert = 1$.
Here is an example from number theory: Let $n > 1$ be an integer, and consider the residues (remainders) modulo $n$. These form a group under addition. We call this the **cyclic group of order $n$**, and denote it as $\mathbb{Z}/n\mathbb{Z}$, with elements $\overline 0, \overline 1, \dots$. The identity is $\overline 0$.
Let $p$ be a prime. Consider the *nonzero residues modulo $p$*, which we denote by $(\mathbb{Z}/p\mathbb{Z})^\times$. Then $\left( (\mathbb{Z}/p\mathbb{Z})^\times, \times \right)$ is a group.
Why do we need the fact that $p$ is prime?
(Digression: the notation $\mathbb{Z}/n\mathbb{Z}$ and $(\mathbb{Z}/p\mathbb{Z})^\times$ may seem strange but will make sense when we talk about rings and ideals. Set aside your worry for now.)
Here are some **non-abelian examples**:
Let $n$ be a positive integer. Then $\GL_n(\mathbb{R})$ is defined as the set of $n \times n$ real matrices which have nonzero determinant. It turns out that with this condition, every matrix does indeed have an inverse, so $(\GL_n(\mathbb{R}), \times)$ is a group, called the **general linear group**.
(The fact that $\GL_n(\mathbb{R})$ is closed under $\times$ follows from the linear algebra fact that $\det (AB) = \det A \det B$, proved in later chapters.)
Following the example above, let $\SL_n(\mathbb{R})$ denote the set of $n \times n$ matrices whose determinant is actually $1$. Again, for linear algebra reasons it turns out that $(\SL_n(\mathbb{R}), \times)$ is also a group, called the **special linear group**. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Definition and examples of groups | 05_grp-intro.md | 1 | 1,488 |
Let $S_n$ be the set of permutations of $\left\{ 1,\dots,n \right\}$. By viewing these permutations as functions from $\left\{ 1,\dots,n \right\}$ to itself, we can consider *compositions* of permutations. Then the pair $(S_n, \circ)$ (here $\circ$ is function composition) is also a group, because
- There is an identity permutation, and
- Each permutation has an inverse.
The group $S_n$ is called the **symmetric group** on $n$ elements.
The **dihedral group of order $2n$**, denoted $D_{2n}$, is the group of symmetries of a regular $n$-gon $A_1A_2 \dots A_n$, which includes rotations and reflections. It consists of the $2n$ elements $$\left\{ 1, r, r^2, \dots, r^{n-1}, s, sr, sr^2, \dots, sr^{n-1} \right\}.$$ The element $r$ corresponds to rotating the $n$-gon by $\frac{2\pi}{n}$, while $s$ corresponds to reflecting it across the line $OA_1$ (here $O$ is the center of the polygon). So $rs$ means "reflect then rotate" (like with function composition, we read from right to left).
In particular, $r^n = s^2 = 1$. You can also see that $r^k s = sr^{-k}$.
Here is a picture of some elements of $D_{10}$.
size(12cm); picture aoeu(string a, string b, string c, string d, string e, string x) draw(dir(0)--dir(72)--dir(144)--dir(216)--dir(288)--cycle); MP(a, dir(0), dir(0)); MP(b, dir(72), dir(72)); MP(c, dir(144), dir(144)); MP(d, dir(216), dir(216)); MP(e, dir(288), dir(288)); MP(x, origin, origin); return CC(); picture one = aoeu(\"1\", \"2\", \"3\", \"4\", \"5\", \"1\"); picture r = aoeu(\"5\", \"1\", \"2\", \"3\", \"4\", \"r\"); picture s = aoeu(\"1\", \"5\", \"4\", \"3\", \"2\", \"s\"); picture sr = aoeu(\"5\", \"4\", \"3\", \"2\", \"1\", \"sr\"); picture rs = aoeu(\"2\", \"1\", \"5\", \"4\", \"3\", \"rs\"); add(shift( (0,0) ) \* one); add(shift( (3,0) ) \* r); add(shift( (6,0) ) \* s); add(shift( (9,0) ) \* sr); add(shift( (12,0) ) \* rs);
Trivia: the dihedral group $D_{12}$ is my favorite example of a non-abelian group, and is the first group I try for any exam question of the form "find an example...".
More examples:
Let $(G, \star)$ and $(H, \ast)$ be groups. We can define a **product group** $(G \times H, {\cdot})$, as follows. The elements of the group will be ordered pairs $(g,h) \in G \times H$. Then $$(g_1, h_1) \cdot (g_2, h_2) = (g_1 \star g_2, h_1 \ast h_2) \in G \times H$$ is the group operation.
What are the identity and inverses of the product group?
The **trivial group**, often denoted $0$ or $1$, is the group with only an identity element. I will use the notation $\{1\}$.
Which of these are groups?
(a) Rational numbers with odd denominators (in simplest form), where the operation is addition. (This includes integers, written as $n/1$, and $0 = 0/1$).
(b) The set of rational numbers with denominator at most $2$, where the operation is addition.
(c) The set of rational numbers with denominator at most $2$, where the operation is multiplication.
(d) The set of nonnegative integers, where the operation is addition. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Definition and examples of groups | 05_grp-intro.md | 2 | 853 |
A group consists of two pieces of data: a set $G$, and an associative binary operation $\star$ with some properties. Before I write down the definition of a group, let me give two examples.
The pair $(\mathbb{Z}, +)$ is a group: $\mathbb{Z} = \left\{ \dots,-2,-1,0,1,2,\dots \right\}$ is the set and the associative operation is *addition*. Note that
- The element $0 \in \mathbb{Z}$ is an *identity*: $a+0=0+a = a$ for any $a$.
- Every element $a \in \mathbb{Z}$ has an additive *inverse*: $a + (-a) = (-a) + a = 0$.
We call this group $\mathbb{Z}$.
Let $\mathbb{Q}^\times$ be the set of *nonzero rational numbers*. The pair $(\mathbb{Q}^\times, \cdot)$ is a group: the set is $\mathbb{Q}^\times$ and the associative operation is *multiplication*.
Again we see the same two nice properties.
- The element $1 \in \mathbb{Q}^\times$ is an *identity*: for any rational number, $a \cdot 1 = 1 \cdot a = a$.
- For any rational number $x \in \mathbb{Q}^\times$, we have an inverse $x^{-1}$, such that $$x \cdot x^{-1} = x^{-1} \cdot x = 1.$$
From this you might already have a guess what the definition of a group is.
A **group** is a pair $G = (G, \star)$ consisting of a set of elements $G$, and a binary operation $\star$ on $G$, such that:
- $G$ has an **identity element**, usually denoted $1_G$ or just $1$, with the property that $$1_G \star g = g \star 1_G = g \text{ for all $g \in G$}.$$
- The operation is **associative**, meaning $(a \star b) \star c = a \star (b \star c)$ for any $a,b,c \in G$. Consequently we generally don't write the parentheses.
- Each element $g \in G$ has an **inverse**, that is, an element $h \in G$ such that $$g \star h = h \star g = 1_G.$$
Some authors like to add a "closure" axiom, i.e. to say explicitly that $g \star h \in G$. This is implied already by the fact that $\star$ is a binary operation on $G$, but is worth keeping in mind for the examples below.
It is not required that $\star$ is commutative ($a \star b = b \star a$). So we say that a group is **abelian** if the operation is commutative and **non-abelian** otherwise.
- The pair $(\mathbb{Q}, \cdot)$ is NOT a group. (Here $\mathbb{Q}$ is rational numbers.) While there is an identity element, the element $0 \in \mathbb{Q}$ does not have an inverse.
- The pair $(\mathbb{Z}, \cdot)$ is also NOT a group. (Why?)
- Let $\mathrm{Mat}_{2 \times 2}(\mathbb{R})$ be the set of $2 \times 2$ real matrices. Then $(\mathrm{Mat}_{2 \times 2}(\mathbb{R}), \cdot)$ (where $\cdot$ is matrix multiplication) is NOT a group. Indeed, even though we have an identity matrix $$\begin{bmatrix}
1 & 0 \\ 0 & 1
\end{bmatrix}$$ we still run into the same issue as before: the zero matrix does not have a multiplicative inverse.
(Even if we delete the zero matrix from the set, the resulting structure is still not a group: those of you that know some linear algebra might recall that any matrix with determinant zero cannot have an inverse.)
Let's resume writing down examples. Here are some more **abelian examples** of groups:
Let $S^1$ denote the set of complex numbers $z$ with absolute value one; that is $$S^1 := \left\{ z \in \mathbb{C} \mid \left\lvert z \right\rvert = 1 \right\}.$$ Then $(S^1, \times)$ is a group because
- The complex number $1 \in S^1$ serves as the identity, and
- Each complex number $z \in S^1$ has an inverse $\frac 1z$ which is also in $S^1$, since $\left\lvert z^{-1} \right\rvert = \left\lvert z \right\rvert^{-1} = 1$.
There is one thing I ought to also check: that $z_1 \times z_2$ is actually still in $S^1$. But this follows from the fact that $\left\lvert z_1z_2 \right\rvert = \left\lvert z_1 \right\rvert \left\lvert z_2 \right\rvert = 1$.
Here is an example from number theory: Let $n > 1$ be an integer, and consider the residues (remainders) modulo $n$. These form a group under addition. We call this the **cyclic group of order $n$**, and denote it as $\mathbb{Z}/n\mathbb{Z}$, with elements $\overline 0, \overline 1, \dots$. The identity is $\overline 0$.
Let $p$ be a prime. Consider the *nonzero residues modulo $p$*, which we denote by $(\mathbb{Z}/p\mathbb{Z})^\times$. Then $\left( (\mathbb{Z}/p\mathbb{Z})^\times, \times \right)$ is a group.
Why do we need the fact that $p$ is prime?
(Digression: the notation $\mathbb{Z}/n\mathbb{Z}$ and $(\mathbb{Z}/p\mathbb{Z})^\times$ may seem strange but will make sense when we talk about rings and ideals. Set aside your worry for now.)
Here are some **non-abelian examples**:
Let $n$ be a positive integer. Then $\GL_n(\mathbb{R})$ is defined as the set of $n \times n$ real matrices which have nonzero determinant. It turns out that with this condition, every matrix does indeed have an inverse, so $(\GL_n(\mathbb{R}), \times)$ is a group, called the **general linear group**.
(The fact that $\GL_n(\mathbb{R})$ is closed under $\times$ follows from the linear algebra fact that $\det (AB) = \det A \det B$, proved in later chapters.)
Following the example above, let $\SL_n(\mathbb{R})$ denote the set of $n \times n$ matrices whose determinant is actually $1$. Again, for linear algebra reasons it turns out that $(\SL_n(\mathbb{R}), \times)$ is also a group, called the **special linear group**. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Definition and examples of groups | 05_grp-intro.md | 3 | 1,488 |
Let $S_n$ be the set of permutations of $\left\{ 1,\dots,n \right\}$. By viewing these permutations as functions from $\left\{ 1,\dots,n \right\}$ to itself, we can consider *compositions* of permutations. Then the pair $(S_n, \circ)$ (here $\circ$ is function composition) is also a group, because
- There is an identity permutation, and
- Each permutation has an inverse.
The group $S_n$ is called the **symmetric group** on $n$ elements.
The **dihedral group of order $2n$**, denoted $D_{2n}$, is the group of symmetries of a regular $n$-gon $A_1A_2 \dots A_n$, which includes rotations and reflections. It consists of the $2n$ elements $$\left\{ 1, r, r^2, \dots, r^{n-1}, s, sr, sr^2, \dots, sr^{n-1} \right\}.$$ The element $r$ corresponds to rotating the $n$-gon by $\frac{2\pi}{n}$, while $s$ corresponds to reflecting it across the line $OA_1$ (here $O$ is the center of the polygon). So $rs$ means "reflect then rotate" (like with function composition, we read from right to left).
In particular, $r^n = s^2 = 1$. You can also see that $r^k s = sr^{-k}$.
Here is a picture of some elements of $D_{10}$.
size(12cm); picture aoeu(string a, string b, string c, string d, string e, string x) draw(dir(0)--dir(72)--dir(144)--dir(216)--dir(288)--cycle); MP(a, dir(0), dir(0)); MP(b, dir(72), dir(72)); MP(c, dir(144), dir(144)); MP(d, dir(216), dir(216)); MP(e, dir(288), dir(288)); MP(x, origin, origin); return CC(); picture one = aoeu(\"1\", \"2\", \"3\", \"4\", \"5\", \"1\"); picture r = aoeu(\"5\", \"1\", \"2\", \"3\", \"4\", \"r\"); picture s = aoeu(\"1\", \"5\", \"4\", \"3\", \"2\", \"s\"); picture sr = aoeu(\"5\", \"4\", \"3\", \"2\", \"1\", \"sr\"); picture rs = aoeu(\"2\", \"1\", \"5\", \"4\", \"3\", \"rs\"); add(shift( (0,0) ) \* one); add(shift( (3,0) ) \* r); add(shift( (6,0) ) \* s); add(shift( (9,0) ) \* sr); add(shift( (12,0) ) \* rs);
Trivia: the dihedral group $D_{12}$ is my favorite example of a non-abelian group, and is the first group I try for any exam question of the form "find an example...".
More examples:
Let $(G, \star)$ and $(H, \ast)$ be groups. We can define a **product group** $(G \times H, {\cdot})$, as follows. The elements of the group will be ordered pairs $(g,h) \in G \times H$. Then $$(g_1, h_1) \cdot (g_2, h_2) = (g_1 \star g_2, h_1 \ast h_2) \in G \times H$$ is the group operation.
What are the identity and inverses of the product group?
The **trivial group**, often denoted $0$ or $1$, is the group with only an identity element. I will use the notation $\{1\}$.
Which of these are groups?
(a) Rational numbers with odd denominators (in simplest form), where the operation is addition. (This includes integers, written as $n/1$, and $0 = 0/1$).
(b) The set of rational numbers with denominator at most $2$, where the operation is addition.
(c) The set of rational numbers with denominator at most $2$, where the operation is multiplication.
(d) The set of nonnegative integers, where the operation is addition. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Definition and examples of groups | 05_grp-intro.md | 4 | 853 |
From now on, we'll often refer to a group $(G, \star)$ by just $G$. Moreover, we'll abbreviate $a \star b$ to just $ab$. Also, because the operation $\star$ is associative, we will omit unnecessary parentheses: $(ab)c = a(bc) = abc$.
From now on, for any $g \in G$ and $n \in \mathbb{N}$ we abbreviate $$g^n
=
\underbrace{g \star \dots \star g}_{\text{$n$ times}}.$$ Moreover, we let $g^{-1}$ denote the inverse of $g$, and $g^{-n} = (g^{-1})^n$.
In mathematics, a common theme is to require that objects satisfy certain minimalistic properties, with certain examples in mind, but then ignore the examples on paper and try to deduce as much as you can just from the properties alone. (Math olympiad veterans are likely familiar with "functional equations" in which knowing a single property about a function is enough to determine the entire function.) Let's try to do this here, and see what we can conclude just from knowing .
It is a law in Guam and 37 other states that I now state the following proposition.
Let $G$ be a group.
(a) The identity of a group is unique.
(b) The inverse of any element is unique.
(c) For any $g \in G$, ${(g^{-1})}^{-1} = g$.
*Proof.* This is mostly just some formal manipulations, and you needn't feel bad skipping it on a first read.
(a) If $1$ and $1'$ are identities, then $1 = 1 \star 1' = 1'$.
(b) If $h$ and $h'$ are inverses to $g$, then $1_G = g \star h
\implies h' = (h' \star g) \star h = 1_G \star h = h$.
(c) Trivial; omitted.
◻
Now we state a slightly more useful proposition.
Let $G$ be a group, and $a,b \in G$. Then $(ab)^{-1} = b^{-1} a^{-1}$.
*Proof.* Direct computation. We have $$(ab)(b^{-1} a^{-1})
= a (bb^{-1}) a^{-1} = aa^{-1} = 1_G.$$ Similarly, $(b^{-1} a^{-1})(ab) = 1_G$ as well. Hence $(ab)^{-1} = b^{-1} a^{-1}$. ◻
Finally, we state a very important lemma about groups, which highlights why having an inverse is so valuable.
Let $G$ be a group, and pick a $g \in G$. Then the map $G \to G$ given by $x \mapsto gx$ is a bijection.
Check this by showing injectivity and surjectivity directly. (If you don't know what these words mean, consult .)
Let $G = (\mathbb{Z}/7\mathbb{Z})^\times$ (as in ) and pick $g=3$. The above lemma states that the map $x \mapsto 3 \cdot x$ is a bijection, and we can see this explicitly: $$\begin{align*}
1 &\overset{\times 3}{\longmapsto} 3 \pmod 7 \\
2 &\overset{\times 3}{\longmapsto} 6 \pmod 7 \\
3 &\overset{\times 3}{\longmapsto} 2 \pmod 7 \\
4 &\overset{\times 3}{\longmapsto} 5 \pmod 7 \\
5 &\overset{\times 3}{\longmapsto} 1 \pmod 7 \\
6 &\overset{\times 3}{\longmapsto} 4 \pmod 7.
\end{align*}$$
The fact that the map is injective is often called the **cancellation law**. (Why do you think so?)
You don't need to worry about this for a few chapters, but I'll bring it up now anyways. In most of our examples up until now the operation $\star$ was thought of like multiplication of some sort, which is why $1 = 1_G$ was a natural notation for the identity element.
But there are groups like $\mathbb{Z} = (\mathbb{Z},+)$ where the operation $\star$ is thought of as addition, in which case the notation $0 = 0_G$ might make more sense instead. (In general, whenever an operation is denoted $+$, the operation is almost certainly commutative.) We will eventually start doing so too when we discuss rings and linear algebra. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Properties of groups | 05_grp-intro.md | 5 | 956 |
First, let me talk about what it means for groups to be isomorphic. Consider the two groups
- $\mathbb{Z} = (\left\{ \dots,-2,-1,0,1,2,\dots \right\}, +)$.
- $10\mathbb{Z} = (\left\{ \dots, -20, -10, 0, 10, 20, \dots \right\}, +)$.
These groups are "different", but only superficially so -- you might even say they only differ in the names of the elements. Think about what this might mean formally for a moment.
Specifically the map $$\phi \colon \mathbb{Z} \to 10 \mathbb{Z} \text{ by } x \mapsto 10 x$$ is a bijection of the underlying sets which respects the group operation. In symbols, $$\phi(x + y) = \phi(x) + \phi(y).$$ In other words, $\phi$ is a way of re-assigning names of the elements without changing the structure of the group. That's all just formalism for capturing the obvious fact that $(\mathbb{Z},+)$ and $(10 \mathbb{Z}, +)$ are the same thing.
Now, let's do the general definition.
Let $G = (G, \star)$ and $H = (H, \ast)$ be groups. A bijection $\phi \colon G \to H$ is called an **isomorphism** if $$\phi(g_1 \star g_2) = \phi(g_1) \ast \phi(g_2) \quad
\text{for all $g_1, g_2 \in G$}.$$ If there exists an isomorphism from $G$ to $H$, then we say $G$ and $H$ are **isomorphic** and write $G \cong H$.
Note that in this definition, the left-hand side $\phi(g_1 \star g_2)$ uses the operation of $G$ while the right-hand side $\phi(g_1) \ast \phi(g_2)$ uses the operation of $H$.
Let $G$ and $H$ be groups. We have the following isomorphisms.
(a) $\mathbb{Z} \cong 10 \mathbb{Z}$, as above.
(b) There is an isomorphism $$G \times H \cong H \times G$$ by the map $(g,h) \mapsto (h,g)$.
(c) The identity map $\mathrm{id} \colon G \to G$ is an isomorphism, hence $G \cong G$.
(d) There is another isomorphism of $\mathbb{Z}$ to itself: send every $x$ to $-x$.
As a nontrivial example, we claim that $\mathbb{Z}/6\mathbb{Z} \cong (\mathbb{Z}/7\mathbb{Z})^\times$. The bijection is $$\phi(\text{$a$ mod $6$}) = \text{$3^a$ mod $7$}.$$
- This map is a bijection by explicit calculation: $$(3^0, 3^1, 3^2, 3^3, 3^4, 3^5)
\equiv (1,3,2,6,4,5) \pmod 7.$$ (Technically, I should more properly write $3^{0 \bmod 6} = 1$ and so on to be pedantic.)
- Finally, we need to verify that this map respects the group operation. In other words, we want to see that $\phi(a+b) = \phi(a) \phi(b)$ since the operation of $\mathbb{Z}/6\mathbb{Z}$ is addition while the operation of $(\mathbb{Z}/7\mathbb{Z})^\times$ is multiplication. That's just saying that $3^{a+b \bmod 6} \equiv 3^{a \bmod 6} 3^{b \bmod 6} \pmod 7$, which is true.
More generally, for any prime $p$, there exists an element $g \in (\mathbb{Z}/p\mathbb{Z})^\times$ called a **primitive root** modulo $p$ such that $1, g, g^2, \dots, g^{p-2}$ are all different modulo $p$. One can show by copying the above proof that $$\mathbb{Z}/c\mathbb{Z}{p-1} \cong (\mathbb{Z}/p\mathbb{Z})^\times \text{ for all primes $p$}.$$ The example above was the special case $p=7$ and $g=3$.
Assuming the existence of primitive roots, establish the isomorphism $\mathbb{Z}/c\mathbb{Z}{p-1} \cong (\mathbb{Z}/p\mathbb{Z})^\times$ as above.
It's not hard to see that $\cong$ is an equivalence relation (why?). Moreover, because we really only care about the structure of groups, we'll usually consider two groups to be the same when they are isomorphic. So phrases such as "find all groups" really mean "find all groups up to isomorphism". | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Isomorphisms | 05_grp-intro.md | 6 | 972 |
First, let me talk about what it means for groups to be isomorphic. Consider the two groups
- $\mathbb{Z} = (\left\{ \dots,-2,-1,0,1,2,\dots \right\}, +)$.
- $10\mathbb{Z} = (\left\{ \dots, -20, -10, 0, 10, 20, \dots \right\}, +)$.
These groups are "different", but only superficially so -- you might even say they only differ in the names of the elements. Think about what this might mean formally for a moment.
Specifically the map $$\phi \colon \mathbb{Z} \to 10 \mathbb{Z} \text{ by } x \mapsto 10 x$$ is a bijection of the underlying sets which respects the group operation. In symbols, $$\phi(x + y) = \phi(x) + \phi(y).$$ In other words, $\phi$ is a way of re-assigning names of the elements without changing the structure of the group. That's all just formalism for capturing the obvious fact that $(\mathbb{Z},+)$ and $(10 \mathbb{Z}, +)$ are the same thing.
Now, let's do the general definition.
Let $G = (G, \star)$ and $H = (H, \ast)$ be groups. A bijection $\phi \colon G \to H$ is called an **isomorphism** if $$\phi(g_1 \star g_2) = \phi(g_1) \ast \phi(g_2) \quad
\text{for all $g_1, g_2 \in G$}.$$ If there exists an isomorphism from $G$ to $H$, then we say $G$ and $H$ are **isomorphic** and write $G \cong H$.
Note that in this definition, the left-hand side $\phi(g_1 \star g_2)$ uses the operation of $G$ while the right-hand side $\phi(g_1) \ast \phi(g_2)$ uses the operation of $H$.
Let $G$ and $H$ be groups. We have the following isomorphisms.
(a) $\mathbb{Z} \cong 10 \mathbb{Z}$, as above.
(b) There is an isomorphism $$G \times H \cong H \times G$$ by the map $(g,h) \mapsto (h,g)$.
(c) The identity map $\mathrm{id} \colon G \to G$ is an isomorphism, hence $G \cong G$.
(d) There is another isomorphism of $\mathbb{Z}$ to itself: send every $x$ to $-x$.
As a nontrivial example, we claim that $\mathbb{Z}/6\mathbb{Z} \cong (\mathbb{Z}/7\mathbb{Z})^\times$. The bijection is $$\phi(\text{$a$ mod $6$}) = \text{$3^a$ mod $7$}.$$
- This map is a bijection by explicit calculation: $$(3^0, 3^1, 3^2, 3^3, 3^4, 3^5)
\equiv (1,3,2,6,4,5) \pmod 7.$$ (Technically, I should more properly write $3^{0 \bmod 6} = 1$ and so on to be pedantic.)
- Finally, we need to verify that this map respects the group operation. In other words, we want to see that $\phi(a+b) = \phi(a) \phi(b)$ since the operation of $\mathbb{Z}/6\mathbb{Z}$ is addition while the operation of $(\mathbb{Z}/7\mathbb{Z})^\times$ is multiplication. That's just saying that $3^{a+b \bmod 6} \equiv 3^{a \bmod 6} 3^{b \bmod 6} \pmod 7$, which is true.
More generally, for any prime $p$, there exists an element $g \in (\mathbb{Z}/p\mathbb{Z})^\times$ called a **primitive root** modulo $p$ such that $1, g, g^2, \dots, g^{p-2}$ are all different modulo $p$. One can show by copying the above proof that $$\mathbb{Z}/c\mathbb{Z}{p-1} \cong (\mathbb{Z}/p\mathbb{Z})^\times \text{ for all primes $p$}.$$ The example above was the special case $p=7$ and $g=3$.
Assuming the existence of primitive roots, establish the isomorphism $\mathbb{Z}/c\mathbb{Z}{p-1} \cong (\mathbb{Z}/p\mathbb{Z})^\times$ as above.
It's not hard to see that $\cong$ is an equivalence relation (why?). Moreover, because we really only care about the structure of groups, we'll usually consider two groups to be the same when they are isomorphic. So phrases such as "find all groups" really mean "find all groups up to isomorphism". | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Isomorphisms | 05_grp-intro.md | 7 | 972 |
As is typical in math, we use the word "order" for way too many things. In groups, there are two notions of order.
The **order of a group** $G$ is the number of elements of $G$. We denote this by $\left\lvert G \right\rvert$. Note that the order may not be finite, as in $\mathbb{Z}$. We say $G$ is a **finite group** just to mean that $\left\lvert G \right\rvert$ is finite.
For a prime $p$, $\left\lvert (\mathbb{Z}/p\mathbb{Z})^\times \right\rvert = p-1$. In other words, the order of $(\mathbb{Z}/p\mathbb{Z})^\times$ is $p-1$. As another example, the order of the symmetric group $S_n$ is $n!$ and the order of the dihedral group $D_{2n}$ is $2n$.
The **order of an element** $g \in G$ is the smallest positive integer $n$ such that $g^n = 1_G$, or $\infty$ if no such $n$ exists. We denote this by $\ord g$.
The order of $-1$ in $\mathbb{Q}^\times$ is $2$, while the order of $1$ in $\mathbb{Z}$ is infinite.
Find the order of each of the six elements of $\mathbb{Z}/6\mathbb{Z}$, the cyclic group on six elements. (See if you've forgotten what $\mathbb{Z}/6\mathbb{Z}$ means.)
If you know olympiad number theory, this coincides with the definition of an order of a residue mod $p$. That's why we use the term "order" there as well. In particular, a primitive root is precisely an element $g \in (\mathbb{Z}/p\mathbb{Z})^\times$ such that $\ord g = p-1$.
You might also know that if $x^n \equiv 1 \pmod p$, then the order of $x \pmod p$ must divide $n$. The same is true in a general group for exactly the same reason.
If $g^n = 1_G$ then $\ord g$ divides $n$.
Also, you can show that any element of a finite group has a finite order. The proof is just an olympiad-style pigeonhole argument. Consider the infinite sequence $1_G, g, g^2, \dots$, and find two elements that are the same.
Let $G$ be a finite group. For any $g \in G$, $\ord g$ is finite.
What's the last property of $(\mathbb{Z}/p\mathbb{Z})^\times$ that you know from olympiad math? We have Fermat's little theorem: for any $a \in (\mathbb{Z}/p\mathbb{Z})^\times$, we have $a^{p-1} \equiv 1 \pmod p$. This is no coincidence: exactly the same thing is true in a more general setting.
Let $G$ be any finite group. Then $x^{\left\lvert G \right\rvert} = 1_G$ for any $x \in G$.
Keep this result in mind! We'll prove it later in the generality of .
As is typical in math, we use the word "order" for way too many things. In groups, there are two notions of order.
The **order of a group** $G$ is the number of elements of $G$. We denote this by $\left\lvert G \right\rvert$. Note that the order may not be finite, as in $\mathbb{Z}$. We say $G$ is a **finite group** just to mean that $\left\lvert G \right\rvert$ is finite.
For a prime $p$, $\left\lvert (\mathbb{Z}/p\mathbb{Z})^\times \right\rvert = p-1$. In other words, the order of $(\mathbb{Z}/p\mathbb{Z})^\times$ is $p-1$. As another example, the order of the symmetric group $S_n$ is $n!$ and the order of the dihedral group $D_{2n}$ is $2n$.
The **order of an element** $g \in G$ is the smallest positive integer $n$ such that $g^n = 1_G$, or $\infty$ if no such $n$ exists. We denote this by $\ord g$.
The order of $-1$ in $\mathbb{Q}^\times$ is $2$, while the order of $1$ in $\mathbb{Z}$ is infinite.
Find the order of each of the six elements of $\mathbb{Z}/6\mathbb{Z}$, the cyclic group on six elements. (See if you've forgotten what $\mathbb{Z}/6\mathbb{Z}$ means.)
If you know olympiad number theory, this coincides with the definition of an order of a residue mod $p$. That's why we use the term "order" there as well. In particular, a primitive root is precisely an element $g \in (\mathbb{Z}/p\mathbb{Z})^\times$ such that $\ord g = p-1$.
You might also know that if $x^n \equiv 1 \pmod p$, then the order of $x \pmod p$ must divide $n$. The same is true in a general group for exactly the same reason.
If $g^n = 1_G$ then $\ord g$ divides $n$.
Also, you can show that any element of a finite group has a finite order. The proof is just an olympiad-style pigeonhole argument. Consider the infinite sequence $1_G, g, g^2, \dots$, and find two elements that are the same.
Let $G$ be a finite group. For any $g \in G$, $\ord g$ is finite.
What's the last property of $(\mathbb{Z}/p\mathbb{Z})^\times$ that you know from olympiad math? We have Fermat's little theorem: for any $a \in (\mathbb{Z}/p\mathbb{Z})^\times$, we have $a^{p-1} \equiv 1 \pmod p$. This is no coincidence: exactly the same thing is true in a more general setting.
Let $G$ be any finite group. Then $x^{\left\lvert G \right\rvert} = 1_G$ for any $x \in G$.
Keep this result in mind! We'll prove it later in the generality of . | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Orders of groups, and Lagrange's theorem | 05_grp-intro.md | 8 | 1,330 |
Earlier we saw that $\GL_n(\mathbb{R})$, the $n \times n$ matrices with nonzero determinant, formed a group under matrix multiplication. But we also saw that a subset of $\GL_n(\mathbb{R})$, namely $\SL_n(\mathbb{R})$, also formed a group with the same operation. For that reason we say that $\SL_n(\mathbb{R})$ is a subgroup of $\GL_n(\mathbb{R})$. And this definition generalizes in exactly the way you expect.
Let $G = (G, \star)$ be a group. A **subgroup** of $G$ is exactly what you would expect it to be: a group $H = (H, \star)$ where $H$ is a subset of $G$. It's a **proper subgroup** if $H \neq G$.
To specify a group $G$, I needed to tell you both what the set $G$ was and the operation $\star$ was. But to specify a subgroup $H$ of a given group $G$, I only need to tell you who its elements are: the operation of $H$ is just inherited from the operation of $G$.
(a) $2\mathbb{Z}$ is a subgroup of $\mathbb{Z}$, which is isomorphic to $\mathbb{Z}$ itself!
(b) Consider again $S_n$, the symmetric group on $n$ elements. Let $T$ be the set of permutations $\tau \colon \{1, \dots, n\} \to \{1, \dots, n\}$ for which $\tau(n) = n$. Then $T$ is a subgroup of $S_n$; in fact, it is isomorphic to $S_{n-1}$.
(c) Consider the group $G \times H$ () and the elements $\left\{ (g, 1_H) \mid g \in G \right\}$. This is a subgroup of $G \times H$ (why?). In fact, it is isomorphic to $G$ by the isomorphism $(g,1_H) \mapsto g$.
For any group $G$, the trivial group $\{1_G\}$ and the entire group $G$ are subgroups of $G$.
Next is an especially important example that we'll talk about more in later chapters.
Let $x$ be an element of a group $G$. Consider the set $$\left<x\right> = \left\{ \dots, x^{-2}, x^{-1}, 1, x, x^2, \dots \right\}.$$ This is also a subgroup of $G$, called the subgroup generated by $x$.
If $\ord x = 2015$, what is the above subgroup equal to? What if $\ord x = \infty$?
Finally, we present some non-examples of subgroups.
Consider the group $\mathbb{Z} = (\mathbb{Z}, +)$.
(a) The set $\left\{ 0,1,2,\dots \right\}$ is not a subgroup of $\mathbb{Z}$ because it does not contain inverses.
(b) The set $\{ n^3 \mid n \in \mathbb{Z} \}
= \{ \dots, -8, -1, 0, 1, 8, \dots \}$ is not a subgroup because it is not closed under addition; the sum of two cubes is not in general a cube.
(c) The empty set $\varnothing$ is not a subgroup of $\mathbb{Z}$ because it lacks an identity element.
Earlier we saw that $\GL_n(\mathbb{R})$, the $n \times n$ matrices with nonzero determinant, formed a group under matrix multiplication. But we also saw that a subset of $\GL_n(\mathbb{R})$, namely $\SL_n(\mathbb{R})$, also formed a group with the same operation. For that reason we say that $\SL_n(\mathbb{R})$ is a subgroup of $\GL_n(\mathbb{R})$. And this definition generalizes in exactly the way you expect.
Let $G = (G, \star)$ be a group. A **subgroup** of $G$ is exactly what you would expect it to be: a group $H = (H, \star)$ where $H$ is a subset of $G$. It's a **proper subgroup** if $H \neq G$.
To specify a group $G$, I needed to tell you both what the set $G$ was and the operation $\star$ was. But to specify a subgroup $H$ of a given group $G$, I only need to tell you who its elements are: the operation of $H$ is just inherited from the operation of $G$.
(a) $2\mathbb{Z}$ is a subgroup of $\mathbb{Z}$, which is isomorphic to $\mathbb{Z}$ itself!
(b) Consider again $S_n$, the symmetric group on $n$ elements. Let $T$ be the set of permutations $\tau \colon \{1, \dots, n\} \to \{1, \dots, n\}$ for which $\tau(n) = n$. Then $T$ is a subgroup of $S_n$; in fact, it is isomorphic to $S_{n-1}$.
(c) Consider the group $G \times H$ () and the elements $\left\{ (g, 1_H) \mid g \in G \right\}$. This is a subgroup of $G \times H$ (why?). In fact, it is isomorphic to $G$ by the isomorphism $(g,1_H) \mapsto g$.
For any group $G$, the trivial group $\{1_G\}$ and the entire group $G$ are subgroups of $G$.
Next is an especially important example that we'll talk about more in later chapters.
Let $x$ be an element of a group $G$. Consider the set $$\left<x\right> = \left\{ \dots, x^{-2}, x^{-1}, 1, x, x^2, \dots \right\}.$$ This is also a subgroup of $G$, called the subgroup generated by $x$.
If $\ord x = 2015$, what is the above subgroup equal to? What if $\ord x = \infty$?
Finally, we present some non-examples of subgroups.
Consider the group $\mathbb{Z} = (\mathbb{Z}, +)$.
(a) The set $\left\{ 0,1,2,\dots \right\}$ is not a subgroup of $\mathbb{Z}$ because it does not contain inverses.
(b) The set $\{ n^3 \mid n \in \mathbb{Z} \}
= \{ \dots, -8, -1, 0, 1, 8, \dots \}$ is not a subgroup because it is not closed under addition; the sum of two cubes is not in general a cube.
(c) The empty set $\varnothing$ is not a subgroup of $\mathbb{Z}$ because it lacks an identity element. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Subgroups | 05_grp-intro.md | 9 | 1,383 |
Just for fun, here is a list of all groups of order less than or equal to ten (up to isomorphism, of course).
1. The only group of order $1$ is the trivial group.
2. The only group of order $2$ is $\mathbb{Z}/2\mathbb{Z}$.
3. The only group of order $3$ is $\mathbb{Z}/3\mathbb{Z}$.
4. The only groups of order $4$ are
- $\mathbb{Z}/4\mathbb{Z}$, the cyclic group on four elements,
- $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, called the Klein Four Group.
5. The only group of order $5$ is $\mathbb{Z}/5\mathbb{Z}$.
6. The groups of order six are
- $\mathbb{Z}/6\mathbb{Z}$, the cyclic group on six elements.
- $S_3$, the permutation group of three elements. This is the first non-abelian group.
Some of you might wonder where $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ is. All I have to say is: Chinese remainder theorem!
You might wonder where $D_6$ is in this list. It's actually isomorphic to $S_3$.
7. The only group of order $7$ is $\mathbb{Z}/7\mathbb{Z}$.
8. The groups of order eight are more numerous.
- $\mathbb{Z}/8\mathbb{Z}$, the cyclic group on eight elements.
- $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.
- $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.
- $D_8$, the dihedral group with eight elements, which is not abelian.
- A non-abelian group $Q_8$, called the *quaternion group*. It consists of eight elements $\pm 1$, $\pm i$, $\pm j$, $\pm k$ with $i^2=j^2=k^2=ijk=-1$.
9. The groups of order nine are
- $\mathbb{Z}/9\mathbb{Z}$, the cyclic group on nine elements.
- $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.
10. The groups of order $10$ are
- $\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (again Chinese remainder theorem).
- $D_{10}$, the dihedral group with $10$ elements. This group is non-abelian.
Just for fun, here is a list of all groups of order less than or equal to ten (up to isomorphism, of course).
1. The only group of order $1$ is the trivial group.
2. The only group of order $2$ is $\mathbb{Z}/2\mathbb{Z}$.
3. The only group of order $3$ is $\mathbb{Z}/3\mathbb{Z}$.
4. The only groups of order $4$ are
- $\mathbb{Z}/4\mathbb{Z}$, the cyclic group on four elements,
- $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, called the Klein Four Group.
5. The only group of order $5$ is $\mathbb{Z}/5\mathbb{Z}$.
6. The groups of order six are
- $\mathbb{Z}/6\mathbb{Z}$, the cyclic group on six elements.
- $S_3$, the permutation group of three elements. This is the first non-abelian group.
Some of you might wonder where $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ is. All I have to say is: Chinese remainder theorem!
You might wonder where $D_6$ is in this list. It's actually isomorphic to $S_3$.
7. The only group of order $7$ is $\mathbb{Z}/7\mathbb{Z}$.
8. The groups of order eight are more numerous.
- $\mathbb{Z}/8\mathbb{Z}$, the cyclic group on eight elements.
- $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.
- $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.
- $D_8$, the dihedral group with eight elements, which is not abelian.
- A non-abelian group $Q_8$, called the *quaternion group*. It consists of eight elements $\pm 1$, $\pm i$, $\pm j$, $\pm k$ with $i^2=j^2=k^2=ijk=-1$.
9. The groups of order nine are
- $\mathbb{Z}/9\mathbb{Z}$, the cyclic group on nine elements.
- $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.
10. The groups of order $10$ are
- $\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (again Chinese remainder theorem).
- $D_{10}$, the dihedral group with $10$ elements. This group is non-abelian. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Groups of small orders | 05_grp-intro.md | 10 | 1,070 |
A common question is: why these axioms? For example, why associative but not commutative? This answer will likely not make sense until later, but here are some comments that may help.
One general heuristic is: Whenever you define a new type of general object, there's always a balancing act going on. On the one hand, you want to include enough constraints that your objects are "nice". On the other hand, if you include too many constraints, then your definition applies to too few objects.
So, for example, we include "associative" because that makes our lives easier and most operations we run into are associative. In particular, associativity is required for the inverse of an element to necessarily be unique. However we don't include "commutative", because examples below show that there are lots of non-abelian groups we care about. (But we introduce another name "abelian" because we still want to keep track of it.)
Another comment: a good motivation for the inverse axioms is that you get a large amount of *symmetry*. The set of positive integers with addition is not a group, for example, because you can't subtract $6$ from $3$: some elements are "larger" than others. By requiring an inverse element to exist, you get rid of this issue. (You also need identity for this; it's hard to define inverses without it.)
Even more abstruse comment: shows that groups are actually shadows of symmetric groups. This makes rigorous the notion that "groups are very symmetric".
What is the joke in the following figure? (Source: [@img:snsd].)
{height="8cm"}
Orders.
The point is that $\heartsuit$ is a group, $G \subsetneq \heartsuit$ a subgroup and $G \cong \heartsuit$. This can only occur if $\left\lvert \heartsuit \right\rvert = \infty$; otherwise, a proper subgroup would have strictly smaller size than the original.
Prove Lagrange's theorem for orders in the special case that $G$ is a finite abelian group.
Copy the proof of Fermat's little theorem, using .
Let $\{g_1, g_2, \dots, g_n\}$ denote the elements of $G$. For any $g \in G$, this is the same as the set $\{gg_1, \dots, gg_n\}$. Taking the entire product and exploiting commutativity gives $g^n \cdot g_1g_2 \dots g_n = g_1g_2 \dots g_n$, hence $g^n=1$.
Show that $D_6 \cong S_3$ but $D_{24} \not\cong S_4$.
For the former, decide where the isomorphism should send $r$ and $s$, and the rest will follow through. For the latter, look at orders.
One can check manually that $D_6 \cong S_3$, using the map $r \mapsto (1 \; 2 \; 3)$ and $s \mapsto (1 \; 2)$. (The right-hand sides are in "cycle notation", as mentioned in .) On the other hand $D_{24}$ contains an element of order $12$ while $S_4$ does not.
Let $p$ be a prime. Show that if $G$ is a group of order $p$ then $G \cong \mathbb{Z}/p\mathbb{Z}$.
Generated groups.
Let $G$ be a group of order $p$, and $1 \neq g \in G$. Look at the group $H$ generated by $g$ and use Lagrange's theorem.
Find a subgroup $H$ of $S_8$ which is isomorphic to $D_8$, and write the isomorphism explicitly.
Let $G$ be a finite group.[^1] Show that there exists a positive integer $n$ such that
(a) (Cayley's theorem) $G$ is isomorphic to some subgroup of the symmetric group $S_n$.
(b) (Representation Theory) $G$ is isomorphic to some subgroup of the general linear group $\GL_n(\mathbb{R})$. (This is the group of invertible $n \times n$ matrices.)
Use $n = \left\lvert G \right\rvert$.
The idea is that each element $g \in G$ can be thought of as a permutation $G \to G$ by $x \mapsto gx$.
Find the smallest integer $n$ such that the symmetric group $S_n$ has a subgroup isomorphic to the dihedral group $D_{2018}$ of order $2018$.
For the lower bound, consider orders (note that $1009$ is prime). For the upper bound, consider a $1009$-gon.
The answer is $n = 1009$. This solution uses the fact that $1009$ is prime.
To show that no smaller $m$ is possible, note that $D_{2018}$ has elements of order $1009$, a prime. Since $S_n$ has no elements of this order for $n < 1009$, we need $n \ge 1009$.
To give a construction from $n = 1009$, note that $D_{2018}$ can be thought of the symmetries of a $1009$-gon. If one labels the vertices of the $1009$-gon by $S \coloneqq \{1,2,\dots,1009\}$, then elements of $D_{2018}$ induces permutations on $S$, and the set of permutations achieved is the desired subgroup.
There are $n$ markers, each with one side white and the other side black. In the beginning, these $n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it.
Prove that if $n \equiv 1 \pmod 3$ it's impossible to reach a state with only two markers remaining. (In fact the converse is true as well.)
Draw inspiration from $D_6$.
We have $www = bb$, $bww = wb$, $wwb = bw$, $bwb = ww$. Interpret these as elements of $D_6$.
Let $p$ be a prime and $F_1 = F_2 = 1$, $F_{n+2} = F_{n+1} + F_n$ be the Fibonacci sequence. Show that $F_{2p(p^2-1)}$ is divisible by $p$.
Look at the group of $2 \times 2$ matrices mod $p$ with determinant $\pm 1$. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Unimportant long digression | 05_grp-intro.md | 11 | 1,508 |
Look at the group $G$ of $2 \times 2$ matrices mod $p$ with determinant $\pm 1$ (whose entries are the integers mod $p$). Let $g = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ and then use $g^{\left\lvert G \right\rvert} = 1_G$.
[^1]: In other words, permutation groups can be arbitrarily weird. I remember being highly unsettled by this theorem when I first heard of it, but in hindsight it is not so surprising. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Unimportant long digression | 05_grp-intro.md | 12 | 118 |
A common question is: why these axioms? For example, why associative but not commutative? This answer will likely not make sense until later, but here are some comments that may help.
One general heuristic is: Whenever you define a new type of general object, there's always a balancing act going on. On the one hand, you want to include enough constraints that your objects are "nice". On the other hand, if you include too many constraints, then your definition applies to too few objects.
So, for example, we include "associative" because that makes our lives easier and most operations we run into are associative. In particular, associativity is required for the inverse of an element to necessarily be unique. However we don't include "commutative", because examples below show that there are lots of non-abelian groups we care about. (But we introduce another name "abelian" because we still want to keep track of it.)
Another comment: a good motivation for the inverse axioms is that you get a large amount of *symmetry*. The set of positive integers with addition is not a group, for example, because you can't subtract $6$ from $3$: some elements are "larger" than others. By requiring an inverse element to exist, you get rid of this issue. (You also need identity for this; it's hard to define inverses without it.)
Even more abstruse comment: shows that groups are actually shadows of symmetric groups. This makes rigorous the notion that "groups are very symmetric".
What is the joke in the following figure? (Source: [@img:snsd].)
{height="8cm"}
Orders.
The point is that $\heartsuit$ is a group, $G \subsetneq \heartsuit$ a subgroup and $G \cong \heartsuit$. This can only occur if $\left\lvert \heartsuit \right\rvert = \infty$; otherwise, a proper subgroup would have strictly smaller size than the original.
Prove Lagrange's theorem for orders in the special case that $G$ is a finite abelian group.
Copy the proof of Fermat's little theorem, using .
Let $\{g_1, g_2, \dots, g_n\}$ denote the elements of $G$. For any $g \in G$, this is the same as the set $\{gg_1, \dots, gg_n\}$. Taking the entire product and exploiting commutativity gives $g^n \cdot g_1g_2 \dots g_n = g_1g_2 \dots g_n$, hence $g^n=1$.
Show that $D_6 \cong S_3$ but $D_{24} \not\cong S_4$.
For the former, decide where the isomorphism should send $r$ and $s$, and the rest will follow through. For the latter, look at orders.
One can check manually that $D_6 \cong S_3$, using the map $r \mapsto (1 \; 2 \; 3)$ and $s \mapsto (1 \; 2)$. (The right-hand sides are in "cycle notation", as mentioned in .) On the other hand $D_{24}$ contains an element of order $12$ while $S_4$ does not.
Let $p$ be a prime. Show that if $G$ is a group of order $p$ then $G \cong \mathbb{Z}/p\mathbb{Z}$.
Generated groups.
Let $G$ be a group of order $p$, and $1 \neq g \in G$. Look at the group $H$ generated by $g$ and use Lagrange's theorem.
Find a subgroup $H$ of $S_8$ which is isomorphic to $D_8$, and write the isomorphism explicitly.
Let $G$ be a finite group.[^1] Show that there exists a positive integer $n$ such that
(a) (Cayley's theorem) $G$ is isomorphic to some subgroup of the symmetric group $S_n$.
(b) (Representation Theory) $G$ is isomorphic to some subgroup of the general linear group $\GL_n(\mathbb{R})$. (This is the group of invertible $n \times n$ matrices.)
Use $n = \left\lvert G \right\rvert$.
The idea is that each element $g \in G$ can be thought of as a permutation $G \to G$ by $x \mapsto gx$.
Find the smallest integer $n$ such that the symmetric group $S_n$ has a subgroup isomorphic to the dihedral group $D_{2018}$ of order $2018$.
For the lower bound, consider orders (note that $1009$ is prime). For the upper bound, consider a $1009$-gon.
The answer is $n = 1009$. This solution uses the fact that $1009$ is prime.
To show that no smaller $m$ is possible, note that $D_{2018}$ has elements of order $1009$, a prime. Since $S_n$ has no elements of this order for $n < 1009$, we need $n \ge 1009$.
To give a construction from $n = 1009$, note that $D_{2018}$ can be thought of the symmetries of a $1009$-gon. If one labels the vertices of the $1009$-gon by $S \coloneqq \{1,2,\dots,1009\}$, then elements of $D_{2018}$ induces permutations on $S$, and the set of permutations achieved is the desired subgroup.
There are $n$ markers, each with one side white and the other side black. In the beginning, these $n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it.
Prove that if $n \equiv 1 \pmod 3$ it's impossible to reach a state with only two markers remaining. (In fact the converse is true as well.)
Draw inspiration from $D_6$.
We have $www = bb$, $bww = wb$, $wwb = bw$, $bwb = ww$. Interpret these as elements of $D_6$.
Let $p$ be a prime and $F_1 = F_2 = 1$, $F_{n+2} = F_{n+1} + F_n$ be the Fibonacci sequence. Show that $F_{2p(p^2-1)}$ is divisible by $p$.
Look at the group of $2 \times 2$ matrices mod $p$ with determinant $\pm 1$. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Unimportant long digression | 05_grp-intro.md | 13 | 1,508 |
Look at the group $G$ of $2 \times 2$ matrices mod $p$ with determinant $\pm 1$ (whose entries are the integers mod $p$). Let $g = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ and then use $g^{\left\lvert G \right\rvert} = 1_G$.
[^1]: In other words, permutation groups can be arbitrarily weird. I remember being highly unsettled by this theorem when I first heard of it, but in hindsight it is not so surprising. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Groups | Unimportant long digression | 05_grp-intro.md | 14 | 118 |
At the time of writing, I'm convinced that metric topology is the morally correct way to motivate point-set topology as well as to generalize normal calculus.[^1] So here is my best attempt.
The concept of a metric space is very "concrete", and lends itself easily to visualization. Hence throughout this chapter you should draw lots of pictures as you learn about new objects, like convergent sequences, open sets, closed sets, and so on.
At the time of writing, I'm convinced that metric topology is the morally correct way to motivate point-set topology as well as to generalize normal calculus.[^1] So here is my best attempt.
The concept of a metric space is very "concrete", and lends itself easily to visualization. Hence throughout this chapter you should draw lots of pictures as you learn about new objects, like convergent sequences, open sets, closed sets, and so on.
A **metric space** is a pair $(M, d)$ consisting of a set of points $M$ and a **metric** $d \colon M \times M \to \mathbb R_{\ge 0}$. The distance function must obey:
- For any $x,y \in M$, we have $d(x,y) = d(y,x)$; i.e. $d$ is symmetric.
- The function $d$ must be **positive definite** which means that $d(x,y) \ge 0$ with equality if and only if $x=y$.
- The function $d$ should satisfy the **triangle inequality**: for all $x,y,z \in M$, $$d(x,z) + d(z,y) \ge d(x,y).$$
Just like with groups, we will abbreviate $(M,d)$ as just $M$.
(a) The real line $\mathbb{R}$ is a metric space under the metric $d(x,y) = \left\lvert x-y \right\rvert$.
(b) The interval $[0,1]$ is also a metric space with the same distance function.
(c) In fact, any subset $S$ of $\mathbb{R}$ can be made into a metric space in this way.
(a) We can make $\mathbb{R}^2$ into a metric space by imposing the Euclidean distance function $$d\left( (x_1, y_1), (x_2, y_2) \right) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}.$$
(b) Just like with the first example, any subset of $\mathbb{R}^2$ also becomes a metric space after we inherit it. The unit disk, unit circle, and the unit square $[0,1]^2$ are special cases.
It is also possible to place the **taxicab distance** on $\mathbb{R}^2$: $$d\left( (x_1, y_1), (x_2, y_2) \right) =
\left\lvert x_1-x_2 \right\rvert + \left\lvert y_1-y_2 \right\rvert.$$ For now, we will use the more natural Euclidean metric.
We can generalize the above examples easily. Let $n$ be a positive integer.
(a) We let $\mathbb{R}^n$ be the metric space whose points are points in $n$-dimensional Euclidean space, and whose metric is the Euclidean metric $$d\left(
\left( a_1, \dots, a_n \right), \left( b_1, \dots, b_n \right)
\right)
= \sqrt{(a_1-b_1)^2 + \dots + (a_n-b_n)^2}.$$ This is the $n$-dimensional **Euclidean space**.
(b) The open **unit ball** $B^{n}$ is the subset of $\mathbb{R}^n$ consisting of those points $\left( x_1, \dots, x_n \right)$ such that $x_1^2 + \dots + x_n^2 < 1$.
(c) The **unit sphere** $S^{n-1}$ is the subset of $\mathbb{R}^n$ consisting of those points $\left( x_1, \dots, x_n \right)$ such that $x_1^2 + \dots + x_n^2 = 1$, with the inherited metric. (The superscript $n-1$ indicates that $S^{n-1}$ is an $n-1$ dimensional space, even though it lives in $n$-dimensional space.) For example, $S^1 \subseteq \mathbb{R}^2$ is the unit circle, whose distance between two points is the length of the chord joining them. You can also think of it as the "boundary" of the unit ball $B^n$.
We can let $M$ be the space of continuous functions $f \colon [0,1] \to \mathbb{R}$ and define the metric by $d(f,g) = \int_0^1 \left\lvert f-g \right\rvert \; dx$. (It admittedly takes some work to check $d(f,g) = 0$ implies $f=g$, but we won't worry about that yet.)
Here is a slightly more pathological example.
Let $S$ be any set of points (either finite or infinite). We can make $S$ into a **discrete space** by declaring $$d(x,y)
=
\begin{cases}
1 & \text{if $x \neq y$} \\
0 & \text{if $x = y$}.
\end{cases}$$ If $\left\lvert S \right\rvert = 4$ you might think of this space as the vertices of a regular tetrahedron, living in $\mathbb{R}^3$. But for larger $S$ it's not so easy to visualize...
Any connected simple graph $G$ can be made into a metric space by defining the distance between two vertices to be the graph-theoretic distance between them. (The discrete metric is the special case when $G$ is the complete graph on $S$.)
Check the conditions of a metric space for the metrics on the discrete space and for the connected graph.
From now on, we will refer to $\mathbb{R}^n$ with the Euclidean metric by just $\mathbb{R}^n$. Moreover, if we wish to take the metric space for a subset $S \subseteq \mathbb{R}^n$ with the inherited metric, we will just write $S$. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | 06_metric-top.md | 0 | 1,341 | |
A **metric space** is a pair $(M, d)$ consisting of a set of points $M$ and a **metric** $d \colon M \times M \to \mathbb R_{\ge 0}$. The distance function must obey:
- For any $x,y \in M$, we have $d(x,y) = d(y,x)$; i.e. $d$ is symmetric.
- The function $d$ must be **positive definite** which means that $d(x,y) \ge 0$ with equality if and only if $x=y$.
- The function $d$ should satisfy the **triangle inequality**: for all $x,y,z \in M$, $$d(x,z) + d(z,y) \ge d(x,y).$$
Just like with groups, we will abbreviate $(M,d)$ as just $M$.
(a) The real line $\mathbb{R}$ is a metric space under the metric $d(x,y) = \left\lvert x-y \right\rvert$.
(b) The interval $[0,1]$ is also a metric space with the same distance function.
(c) In fact, any subset $S$ of $\mathbb{R}$ can be made into a metric space in this way.
(a) We can make $\mathbb{R}^2$ into a metric space by imposing the Euclidean distance function $$d\left( (x_1, y_1), (x_2, y_2) \right) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}.$$
(b) Just like with the first example, any subset of $\mathbb{R}^2$ also becomes a metric space after we inherit it. The unit disk, unit circle, and the unit square $[0,1]^2$ are special cases.
It is also possible to place the **taxicab distance** on $\mathbb{R}^2$: $$d\left( (x_1, y_1), (x_2, y_2) \right) =
\left\lvert x_1-x_2 \right\rvert + \left\lvert y_1-y_2 \right\rvert.$$ For now, we will use the more natural Euclidean metric.
We can generalize the above examples easily. Let $n$ be a positive integer.
(a) We let $\mathbb{R}^n$ be the metric space whose points are points in $n$-dimensional Euclidean space, and whose metric is the Euclidean metric $$d\left(
\left( a_1, \dots, a_n \right), \left( b_1, \dots, b_n \right)
\right)
= \sqrt{(a_1-b_1)^2 + \dots + (a_n-b_n)^2}.$$ This is the $n$-dimensional **Euclidean space**.
(b) The open **unit ball** $B^{n}$ is the subset of $\mathbb{R}^n$ consisting of those points $\left( x_1, \dots, x_n \right)$ such that $x_1^2 + \dots + x_n^2 < 1$.
(c) The **unit sphere** $S^{n-1}$ is the subset of $\mathbb{R}^n$ consisting of those points $\left( x_1, \dots, x_n \right)$ such that $x_1^2 + \dots + x_n^2 = 1$, with the inherited metric. (The superscript $n-1$ indicates that $S^{n-1}$ is an $n-1$ dimensional space, even though it lives in $n$-dimensional space.) For example, $S^1 \subseteq \mathbb{R}^2$ is the unit circle, whose distance between two points is the length of the chord joining them. You can also think of it as the "boundary" of the unit ball $B^n$.
We can let $M$ be the space of continuous functions $f \colon [0,1] \to \mathbb{R}$ and define the metric by $d(f,g) = \int_0^1 \left\lvert f-g \right\rvert \; dx$. (It admittedly takes some work to check $d(f,g) = 0$ implies $f=g$, but we won't worry about that yet.)
Here is a slightly more pathological example.
Let $S$ be any set of points (either finite or infinite). We can make $S$ into a **discrete space** by declaring $$d(x,y)
=
\begin{cases}
1 & \text{if $x \neq y$} \\
0 & \text{if $x = y$}.
\end{cases}$$ If $\left\lvert S \right\rvert = 4$ you might think of this space as the vertices of a regular tetrahedron, living in $\mathbb{R}^3$. But for larger $S$ it's not so easy to visualize...
Any connected simple graph $G$ can be made into a metric space by defining the distance between two vertices to be the graph-theoretic distance between them. (The discrete metric is the special case when $G$ is the complete graph on $S$.)
Check the conditions of a metric space for the metrics on the discrete space and for the connected graph.
From now on, we will refer to $\mathbb{R}^n$ with the Euclidean metric by just $\mathbb{R}^n$. Moreover, if we wish to take the metric space for a subset $S \subseteq \mathbb{R}^n$ with the inherited metric, we will just write $S$. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Definition and examples of metric spaces | 06_metric-top.md | 1 | 1,088 |
Since we can talk about the distance between two points, we can talk about what it means for a sequence of points to converge. This is the same as the typical epsilon-delta definition, with absolute values replaced by the distance function.
Let $(x_n)_{n \ge 1}$ be a sequence of points in a metric space $M$. We say that $x_n$ **converges** to $x$ if the following condition holds: for all $\varepsilon > 0$, there is an integer $N$ (depending on $\varepsilon$) such that $d(x_n, x) < \varepsilon$ for each $n \ge N$. This is written $$x_n \to x$$ or more verbosely as $$\lim_{n \to \infty} x_n = x.$$ We say that a sequence converges in $M$ if it converges to a point in $M$.
You should check that this definition coincides with your intuitive notion of "converges".
If the parent space $M$ is understood, we will allow ourselves to abbreviate "converges in $M$" to just "converges". However, keep in mind that convergence is defined relative to the parent space; the "limit" of the space must actually be a point in $M$ for a sequence to converge.
size(9cm); Drawing(\"x_1\", (-9,0.1), dir(90)); Drawing(\"x_2\", (-6,0.8), dir(90)); Drawing(\"x_3\", (-5,-0.3), dir(90)); Drawing(\"x_4\", (-2, 0.8), dir(90)); Drawing(\"x_5\", (-1.7, -0.7), dir(-90)); Drawing(\"x_6\", (-0.6, -0.3), dir(225)); Drawing(\"x_7\", (-0.4, 0.3), dir(90)); Drawing(\"x_8\", (-0.25, -0.24), dir(-90)); Drawing(\"x_9\", (-0.12, 0.1), dir(45)); dot(\"$x$\", (0,0), dir(-45), blue); draw(CR(origin, 1.5), blue+dashed);
Consider the sequence $x_1 = 1$, $x_2 = 1.4$, $x_3 = 1.41$, $x_4 = 1.414$, ....
(a) If we view this as a sequence in $\mathbb{R}$, it converges to $\sqrt 2$.
(b) However, even though each $x_i$ is in $\mathbb{Q}$, this sequence does NOT converge when we view it as a sequence in $\mathbb{Q}$!
What are the convergent sequences in a discrete metric space?
Since we can talk about the distance between two points, we can talk about what it means for a sequence of points to converge. This is the same as the typical epsilon-delta definition, with absolute values replaced by the distance function.
Let $(x_n)_{n \ge 1}$ be a sequence of points in a metric space $M$. We say that $x_n$ **converges** to $x$ if the following condition holds: for all $\varepsilon > 0$, there is an integer $N$ (depending on $\varepsilon$) such that $d(x_n, x) < \varepsilon$ for each $n \ge N$. This is written $$x_n \to x$$ or more verbosely as $$\lim_{n \to \infty} x_n = x.$$ We say that a sequence converges in $M$ if it converges to a point in $M$.
You should check that this definition coincides with your intuitive notion of "converges".
If the parent space $M$ is understood, we will allow ourselves to abbreviate "converges in $M$" to just "converges". However, keep in mind that convergence is defined relative to the parent space; the "limit" of the space must actually be a point in $M$ for a sequence to converge.
size(9cm); Drawing(\"x_1\", (-9,0.1), dir(90)); Drawing(\"x_2\", (-6,0.8), dir(90)); Drawing(\"x_3\", (-5,-0.3), dir(90)); Drawing(\"x_4\", (-2, 0.8), dir(90)); Drawing(\"x_5\", (-1.7, -0.7), dir(-90)); Drawing(\"x_6\", (-0.6, -0.3), dir(225)); Drawing(\"x_7\", (-0.4, 0.3), dir(90)); Drawing(\"x_8\", (-0.25, -0.24), dir(-90)); Drawing(\"x_9\", (-0.12, 0.1), dir(45)); dot(\"$x$\", (0,0), dir(-45), blue); draw(CR(origin, 1.5), blue+dashed);
Consider the sequence $x_1 = 1$, $x_2 = 1.4$, $x_3 = 1.41$, $x_4 = 1.414$, ....
(a) If we view this as a sequence in $\mathbb{R}$, it converges to $\sqrt 2$.
(b) However, even though each $x_i$ is in $\mathbb{Q}$, this sequence does NOT converge when we view it as a sequence in $\mathbb{Q}$!
What are the convergent sequences in a discrete metric space? | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Convergence in metric spaces | 06_metric-top.md | 2 | 1,061 |
In calculus you were also told (or have at least heard) of what it means for a function to be continuous. Probably something like
> A function $f \colon \mathbb{R} \to \mathbb{R}$ is continuous at a point $p \in \mathbb{R}$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $\left\lvert x-p \right\rvert < \delta
> \implies
> \left\lvert f(x) - f(p) \right\rvert < \varepsilon$.
Can you guess what the corresponding definition for metric spaces is?
All we have to do is replace the absolute values with the more general distance functions: this gives us a definition of continuity for any function $M \to N$.
Let $M = (M, d_M)$ and $N = (N, d_N)$ be metric spaces. A function $f \colon M \to N$ is **continuous** at a point $p \in M$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $$d_M(x,p) < \delta \implies d_N(f(x), f(p)) < \varepsilon.$$ Moreover, the entire function $f$ is continuous if it is continuous at every point $p \in M$.
Notice that, just like in our definition of an isomorphism of a group, we use the metric of $M$ for one condition and the metric of $N$ for the other condition.
This generalization is nice because it tells us immediately how we could carry over continuity arguments in $\mathbb{R}$ to more general spaces like $\mathbb{C}$. Nonetheless, this definition is kind of cumbersome to work with, because it makes extensive use of the real numbers (epsilons and deltas). Here is an equivalent condition.
A function $f \colon M \to N$ of metric spaces is continuous at a point $p \in M$ if and only if the following property holds: if $x_1$, $x_2$, ... is a sequence in $M$ converging to $p$, then the sequence $f(x_1)$, $f(x_2)$, ... in $N$ converges to $f(p)$.
*Proof.* One direction is not too hard:
Show that $\varepsilon$-$\delta$ continuity implies sequential continuity at each point.
Conversely, we will prove if $f$ is not $\varepsilon$-$\delta$ continuous at $p$ then it does not preserve convergence.
If $f$ is not continuous at $p$, then there is a "bad" $\varepsilon > 0$, which we now consider fixed. So for each choice of $\delta = 1/n$, there should be some point $x_n$ which is within $\delta$ of $p$, but which is mapped more than $\varepsilon$ away from $f(p)$. But then the sequence $x_n$ converges to $p$, and $f(x_n)$ is always at least $\varepsilon$ away from $f(p)$, contradiction. ◻
Example application showcasing the niceness of sequential continuity:
Let $f \colon M \to N$ and $g \colon N \to L$ be continuous maps of metric spaces. Then their composition $g \circ f$ is continuous.
*Proof.* Dead simple with sequences: Let $p \in M$ be arbitrary and let $x_n \to p$ in $M$. Then $f(x_n) \to f(p)$ in $N$ and $g(f(x_n)) \to g(f(p))$ in $L$, QED. ◻
Let $M$ be any metric space and $D$ a discrete space. When is a map $f \colon D \to M$ continuous? | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Continuous maps | 06_metric-top.md | 3 | 816 |
When do we consider two groups to be the same? Answer: if there's a structure-preserving map between them which is also a bijection. For metric spaces, we do exactly the same thing, but replace "structure-preserving" with "continuous".
Let $M$ and $N$ be metric spaces. A function $f \colon M \to N$ is a **homeomorphism** if it is a bijection, and both $f \colon M \to N$ and its inverse $f^{-1} \colon N \to M$ are continuous. We say $M$ and $N$ are **homeomorphic**.
Needless to say, homeomorphism is an equivalence relation.
You might be surprised that we require $f^{-1}$ to also be continuous. Here's the reason: you can show that if $\phi$ is an isomorphism of groups, then $\phi^{-1}$ also preserves the group operation, hence $\phi^{-1}$ is itself an isomorphism. The same is not true for continuous bijections, which is why we need the new condition.
(a) There is a continuous bijection from $[0,1)$ to the circle, but it has no continuous inverse.
(b) Let $M$ be a discrete space with size $|\mathbb{R}|$. Then there is a continuous function $M \to \mathbb{R}$ which certainly has no continuous inverse.
Note that this is the topologist's definition of "same" -- homeomorphisms are "continuous deformations". Here are some examples.
(a) Any space $M$ is homeomorphic to itself through the identity map.
(b) The old saying: a doughnut (torus) is homeomorphic to a coffee cup. (Look this up if you haven't heard of it.)
(c) The unit circle $S^1$ is homeomorphic to the boundary of the unit square. Here's one bijection between them, after an appropriate scaling:
:::: center
::: asy
size(1.5cm); draw(unitcircle); pair A = (1.4, 1.4); pair B = rotate(90)\*A; pair C = rotate(90)\*B; pair D = rotate(90)\*C; draw(A--B--C--D--cycle); dot(origin); pair P = Drawing(dir(70)); pair Q = Drawing(extension(origin, P, A, B)); draw(origin--Q, dashed);
:::
::::
It may have seemed strange that our metric function on $S^1$ was the one inherited from $\mathbb{R}^2$, meaning the distance between two points on the circle was defined to be the length of the chord. Wouldn't it have made more sense to use the circumference of the smaller arc joining the two points?
In fact, it doesn't matter: if we consider $S^1$ with the "chord" metric and the "arc" metric, we get two homeomorphic spaces, as the map between them is continuous.
The same goes for $S^{n-1}$ for general $n$.
Surprisingly, the open interval $(-1,1)$ is homeomorphic to the real line $\mathbb{R}$! One bijection is given by $$x \mapsto \tan(x \pi/2)$$ with the inverse being given by $t \mapsto \frac2\pi \arctan(t)$.
This might come as a surprise, since $(-1,1)$ doesn't look that much like $\mathbb{R}$; the former is "bounded" while the latter is "unbounded". | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Homeomorphisms | 06_metric-top.md | 4 | 783 |
When do we consider two groups to be the same? Answer: if there's a structure-preserving map between them which is also a bijection. For metric spaces, we do exactly the same thing, but replace "structure-preserving" with "continuous".
Let $M$ and $N$ be metric spaces. A function $f \colon M \to N$ is a **homeomorphism** if it is a bijection, and both $f \colon M \to N$ and its inverse $f^{-1} \colon N \to M$ are continuous. We say $M$ and $N$ are **homeomorphic**.
Needless to say, homeomorphism is an equivalence relation.
You might be surprised that we require $f^{-1}$ to also be continuous. Here's the reason: you can show that if $\phi$ is an isomorphism of groups, then $\phi^{-1}$ also preserves the group operation, hence $\phi^{-1}$ is itself an isomorphism. The same is not true for continuous bijections, which is why we need the new condition.
(a) There is a continuous bijection from $[0,1)$ to the circle, but it has no continuous inverse.
(b) Let $M$ be a discrete space with size $|\mathbb{R}|$. Then there is a continuous function $M \to \mathbb{R}$ which certainly has no continuous inverse.
Note that this is the topologist's definition of "same" -- homeomorphisms are "continuous deformations". Here are some examples.
(a) Any space $M$ is homeomorphic to itself through the identity map.
(b) The old saying: a doughnut (torus) is homeomorphic to a coffee cup. (Look this up if you haven't heard of it.)
(c) The unit circle $S^1$ is homeomorphic to the boundary of the unit square. Here's one bijection between them, after an appropriate scaling:
:::: center
::: asy
size(1.5cm); draw(unitcircle); pair A = (1.4, 1.4); pair B = rotate(90)\*A; pair C = rotate(90)\*B; pair D = rotate(90)\*C; draw(A--B--C--D--cycle); dot(origin); pair P = Drawing(dir(70)); pair Q = Drawing(extension(origin, P, A, B)); draw(origin--Q, dashed);
:::
::::
It may have seemed strange that our metric function on $S^1$ was the one inherited from $\mathbb{R}^2$, meaning the distance between two points on the circle was defined to be the length of the chord. Wouldn't it have made more sense to use the circumference of the smaller arc joining the two points?
In fact, it doesn't matter: if we consider $S^1$ with the "chord" metric and the "arc" metric, we get two homeomorphic spaces, as the map between them is continuous.
The same goes for $S^{n-1}$ for general $n$.
Surprisingly, the open interval $(-1,1)$ is homeomorphic to the real line $\mathbb{R}$! One bijection is given by $$x \mapsto \tan(x \pi/2)$$ with the inverse being given by $t \mapsto \frac2\pi \arctan(t)$.
This might come as a surprise, since $(-1,1)$ doesn't look that much like $\mathbb{R}$; the former is "bounded" while the latter is "unbounded". | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Homeomorphisms | 06_metric-top.md | 5 | 783 |
Here is an extended example which will occur later on. Let $M = (M, d_M)$ and $N = (N, d_N)$ be metric spaces (say, $M = N = \mathbb{R}$). Our goal is to define a metric space on $M \times N$.
Let $p_i = (x_i,y_i) \in M \times N$ for $i=1,2$. Consider the following metrics on the set of points $M \times N$: $$\begin{align*}
d_{\text{max}} ( p_1, p_2 )
&:= \max \left\{ d_M(x_1, x_2), d_N(y_1, y_2) \right\} \\
d_{\text{Euclid}} ( p_1, p_2 )
&:= \sqrt{d_M(x_1,x_2)^2 + d_N(y_1, y_2)^2} \\
d_{\text{taxicab}} \left( p_1, p_2 \right)
&:= d_M(x_1, x_2) + d_N(y_1, y_2).
\end{align*}$$ All of these are good candidates. We are about to see it doesn't matter which one we use:
Verify that $$d_{\text{max}}(p_1,p_2)
\le d_{\text{Euclid}}(p_1, p_2)
\le d_{\text{taxicab}}(p_1, p_2)
\le 2d_{\text{max}}(p_1, p_2).$$ Use this to show that the metric spaces we obtain by imposing any of the three metrics are homeomorphic, with the homeomorphism being just the identity map.
Hence we will usually simply refer to *the* metric on $M \times N$, called the **product metric**. It will not be important which of the three metrics we select.
If $M = N = \mathbb{R}$, we get $\mathbb{R}^2$, the Euclidean plane. The metric $d_{\text{Euclid}}$ is the one we started with, but using either of the other two metric works fine as well.
The product metric plays well with convergence of sequences.
We have $(x_n, y_n) \to (x,y)$ if and only if $x_n \to x$ and $y_n \to y$.
*Proof.* We have $d_{\text{max}} \left( (x,y), (x_n, y_n) \right)
= \max\left\{ d_M(x, x_n), d_N(y, y_n) \right\}$ and the latter approaches zero as $n \to \infty$ if and only if $d_M(x,x_n) \to 0$ and $d_N(y, y_n) \to 0$. ◻
Let's see an application of this:
The addition and multiplication maps are continuous maps $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$.
*Proof.* For multiplication: for any $n$ we have $$\begin{align*}
x_n y_n &= \left( x + (x_n-x) \right)
\left( y + (y_n-y) \right) \\
&= xy + y(x_n-x) + x(y_n-y) + (x_n-x)(y_n-y) \\
\implies \left\lvert x_n y_n - xy \right\rvert
&\le \left\lvert y \right\rvert
\left\lvert x_n - x \right\rvert
+ \left\lvert x \right\rvert
\left\lvert y_n - y \right\rvert
+ \left\lvert x_n - x \right\rvert
\left\lvert y_n - y \right\rvert.
\end{align*}$$ As $n \to \infty$, all three terms on the right-hand side tend to zero. The proof that $+ \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is continuous is similar (and easier): one notes for any $n$ that $$|(x_n + y_n) - (x+y)| \le |x_n-x| + |y_n-y|$$ and both terms on the right-hand side tend to zero as $n \to \infty$. ◻
covers the other two operations, subtraction and division. The upshot of this is that, since compositions are also continuous, most of your naturally arising real-valued functions will automatically be continuous as well. For example, the function $\frac{3x}{x^2+1}$ will be a continuous function from $\mathbb{R} \to \mathbb{R}$, since it can be obtained by composing $+$, $\times$, $\div$. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Extended example/definition: product metric | 06_metric-top.md | 6 | 853 |
Continuity is really about what happens "locally": how a function behaves "close to a certain point $p$". One way to capture this notion of "closeness" is to use metrics as we've done above. In this way we can define an $r$-neighborhood of a point.
Let $M$ be a metric space. For each real number $r > 0$ and point $p \in M$, we define $$M_r(p) := \left\{ x \in M: d(x,p) < r \right\}.$$ The set $M_r(p)$ is called an **$r$-neighborhood** of $p$.
size(4cm); bigblob(\"$M$\"); pair p = Drawing(\"p\", (0.3,0.1), dir(-90)); real r = 1.8; draw(CR(p,r), dashed); label(\"$M_r(p)$\", p+r\*dir(-65), dir(-65)); draw(p--(p+r\*dir(130))); label(\"$r$\", midpoint(p--(p+r\*dir(130))), dir(40));
We can rephrase convergence more succinctly in terms of $r$-neighborhoods. Specifically, a sequence $(x_n)$ converges to $x$ if for every $r$-neighborhood of $x$, all terms of $x_n$ eventually stay within that $r$-neighborhood.
Let's try to do the same with functions.
In terms of $r$-neighborhoods, what does it mean for a function $f \colon M \to N$ to be continuous at a point $p \in M$?
Essentially, we require that the pre-image of every $\varepsilon$-neighborhood has the property that some $\delta$-neighborhood exists inside it. This motivates:
A set $U \subseteq M$ is **open** in $M$ if for each $p \in U$, some $r$-neighborhood of $p$ is contained inside $U$. In other words, there exists $r>0$ such that $M_r(p) \subseteq U$.
Note that a set being open is defined *relative to* the parent space $M$. However, if $M$ is understood we can abbreviate "open in $M$" to just "open".
<figure id="fig:example_open" data-latex-placement="ht">
<div class="asy">
<p>size(5cm); draw(unitcircle, dashed); pair P = Drawing("p", (0.6,0.2), dir(-90)); draw(CR(P, 0.3), dotted); MP("x^2+y^2<1", dir(45), dir(45));</p>
</div>
<figcaption>The set of points <span class="math inline"><em>x</em><sup>2</sup> + <em>y</em><sup>2</sup> < 1</span> in <span class="math inline">$\mathbb{R}^2$</span> is open in <span class="math inline">$\mathbb{R}^2$</span>.</figcaption>
</figure>
(a) Any $r$-neighborhood of a point is open.
(b) Open intervals of $\mathbb{R}$ are open in $\mathbb{R}$, hence the name! This is the prototypical example to keep in mind.
(c) The open unit ball $B^n$ is open in $\mathbb{R}^n$ for the same reason.
(d) In particular, the open interval $(0,1)$ is open in $\mathbb{R}$. However, if we embed it in $\mathbb{R}^2$, it is no longer open!
(e) The empty set $\varnothing$ and the whole set of points $M$ are open in $M$.
(a) The closed interval $[0,1]$ is not open in $\mathbb{R}$. There is no $\varepsilon$-neighborhood of the point $0$ which is contained in $[0,1]$.
(b) The unit circle $S^1$ is not open in $\mathbb{R}^2$.
What are the open sets of the discrete space?
Here are two quite important properties of open sets.
(a) The intersection of finitely many open sets is open.
(b) The union of open sets is open, even if there are infinitely many.
Convince yourself this is true.
Exhibit an infinite collection of open sets in $\mathbb{R}$ whose intersection is the set $\{0\}$. This implies that infinite intersections of open sets are not necessarily open.
The whole upshot of this is:
A function $f \colon M \to N$ of metric spaces is continuous if and only if the pre-image of every open set in $N$ is open in $M$.
*Proof.* I'll just do one direction...
Show that $\delta$-$\varepsilon$ continuity follows from the open set condition.
Now assume $f$ is continuous. First, suppose $V$ is an open subset of the metric space $N$; let $U = f\pre(V)$. Pick $x \in U$, so $y = f(x) \in V$; we want an open neighborhood of $x$ inside $U$.
size(12cm); bigblob(\"$N$\"); pair Y = Drawing(\"y\", origin, dir(75)); real eps = 1.5; draw(CR(Y, eps), dotted); label(\"$\varepsilon$\", Drawing(Y--(Y+eps\*dir(255)))); label(\"$V$\", Drawing(shift(-0.5,0)\*rotate(190)\*scale(3.2,2.8)\*unitcircle, dashed)); add(shift( (13,0) ) \* CC()); label(\"$f$\", Drawing( (4.5,0)--(8,0), EndArrow));
bigblob(\"$M$\"); real delta = 1.1; pair X = Drawing(\"x\", (-1.5,-0.5), dir(-45)); label(\"$\delta$\", Drawing(X--(X+delta\*dir(155)))); draw(CR(X, delta), dotted); label(\"$U = f^{\text{pre}}(V)$\", Drawing(shift(-1.5,-0.3)\*rotate(235)\*scale(2.4,1.8)\*unitcircle, dashed));
As $V$ is open, there is some small $\varepsilon$-neighborhood around $y$ which is contained inside $V$. By continuity of $f$, we can find a $\delta$ such that the $\delta$-neighborhood of $x$ gets mapped by $f$ into the $\varepsilon$-neighborhood in $N$, which in particular lives inside $V$. Thus the $\delta$-neighborhood lives in $U$, as desired. ◻ | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Open sets | 06_metric-top.md | 7 | 1,328 |
It would be criminal for me to talk about open sets without talking about closed sets. The name "closed" comes from the definition in a metric space.
Let $M$ be a metric space. A subset $S \subseteq M$ is **closed** in $M$ if the following property holds: let $x_1$, $x_2$, ... be a sequence of points in $S$ and suppose that $x_n$ converges to $x$ in $M$. Then $x \in S$ as well.
Same caveat: we abbreviate "closed in $M$" to just "closed" if the parent space $M$ is understood.
Here's another way to phrase it. The **limit points** of a subset $S \subseteq M$ are defined by $$\lim S := \left\{ p \in M :
\exists (x_n) \in S \text{ such that } x_n \to p \right\}.$$ Thus $S$ is closed if and only if $S = \lim S$.
Prove that $\lim S$ is closed even if $S$ isn't closed. (Draw a picture.)
For this reason, $\lim S$ is also called the **closure** of $S$ in $M$, and denoted $\overline S$. It is simply the smallest closed set which contains $S$.
(a) The empty set $\varnothing$ is closed in $M$ for vacuous reasons: there are no sequences of points with elements in $\varnothing$.
(b) The entire space $M$ is closed in $M$ for tautological reasons. (Verify this!)
(c) The closed interval $[0,1]$ in $\mathbb{R}$ is closed in $\mathbb{R}$, hence the name. Like with open sets, this is the prototypical example of a closed set to keep in mind!
(d) In fact, the closed interval $[0,1]$ is even closed in $\mathbb{R}^2$.
Let $S=(0,1)$ denote the open interval. Then $S$ is not closed in $\mathbb{R}$ because the sequence of points $$\frac12, \;
\frac14, \;
\frac18, \;
\dots$$ converges to $0 \in \mathbb{R}$, but $0 \notin (0,1)$.
I should now warn you about a confusing part of this terminology. Firstly, **"most" sets are neither open nor closed**.
The half-open interval $[0,1)$ is neither open nor closed in $\mathbb{R}$.
Secondly, it's **also possible for a set to be both open and closed**; this will be discussed in .
The reason for the opposing terms is the following theorem:
Let $M$ be a metric space, and $S \subseteq M$ any subset. Then the following are equivalent:
- The set $S$ is closed in $M$.
- The complement $M \setminus S$ is open in $M$.
Prove this theorem! You'll want to draw a picture to make it clear what's happening: for example, you might take $M = \mathbb{R}^2$ and $S$ to be the closed unit disk.
Let $M = (M,d)$ be a metric space. Show that $$d \colon M \times M \to \mathbb{R}$$ is itself a continuous function (where $M \times M$ is equipped with the product metric).
Consider $\mathbb{Q}$ and $\mathbb{N}$ as metric spaces (each with the obvious metric $d(x,y) = |x-y|$). Are these spaces homeomorphic?
No. There is not even a continuous injective map from $\mathbb{Q}$ to $\mathbb{N}$.
Two possible approaches, one using metric definition and one using open sets.
Metric approach: I claim there is no injective map from $\mathbb{Q}$ to $\mathbb{N}$ that is continuous. Indeed, suppose $f$ was such a map and $f(x) = n$. Then, choose $\varepsilon = 1/2$. There should be a $\delta > 0$ such that everything with $\delta$ of $x$ in $\mathbb{Q}$ should land within $\varepsilon$ of $n \in \mathbb{N}$ --- i.e., is equal to $n$. This is a blatant contradiction of injectivity.
Open set approach: In $\mathbb{Q}$, no singleton set is open, whereas in $\mathbb{N}$, they all are (in fact $\mathbb{N}$ is discrete). As you'll see at the start of , with the new and improved definition of "homeomorphism", we found out that the structure of open sets on $\mathbb{Q}$ and $\mathbb{N}$ are different, so they are not homeomorphic.
Show that subtraction is a continuous map $- \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, and division is a continuous map $\div \colon \mathbb{R} \times \mathbb{R}_{>0} \to \mathbb{R}$.
You can do this with bare hands. You can also use composition.
For subtraction, the map $x \mapsto -x$ is continuous so you can view it as a composed map
Similarly, if you are willing to believe $x \mapsto 1/x$ is a continuous function, then division is composition
If for some reason you are suspicious that $x \mapsto 1/x$ is continuous, then here is a proof using sequential continuity. Suppose $x_n \to x$ with $x_n > 0$ and $x > 0$ (since $x$ needs to be in $\mathbb{R}_{>0}$ too). Then $$\left\lvert \frac{1}{x} - \frac 1{x_n} \right\rvert
= \frac{\left\lvert x_n-x \right\rvert}{\left\lvert x x_n \right\rvert}.$$ If $n$ is large enough, then $\left\lvert x_n \right\rvert > x/2$; so the denominator is at least $x^2/2$, and hence the whole fraction is at most $\frac{2}{x^2} \left\lvert x_n-x \right\rvert$, which tends to zero as $n \to \infty$.
Exhibit a function $f \colon \mathbb{R} \to \mathbb{R}$ such that $f$ is continuous at $x \in \mathbb{R}$ if and only if $x=0$.
$\pm x$ for good choices of $\pm$.
Let $f(x) = x$ for $x \in \mathbb{Q}$ and $f(x) = -x$ for irrational $x$.
Prove that a function $f \colon \mathbb{R} \to \mathbb{R}$ which is strictly increasing must be continuous at some point.
Project gaps onto the $y$-axis. Use the fact that uncountably many positive reals cannot have finite sum. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Closed sets | 06_metric-top.md | 8 | 1,457 |
Assume for contradiction it is completely discontinuous; by scaling set $f(0) = 0$, $f(1) = 1$ and focus just on $f \colon [0,1] \to [0,1]$. Since it's discontinuous everywhere, for every $x \in [0,1]$ there's an $\varepsilon_x > 0$ such that the continuity condition fails. Since the function is strictly increasing, that can only happen if the function misses all $y$-values in the interval $(f(x)-\varepsilon_x, f(x))$ or $(f(x), f(x)+\varepsilon_x)$ (or both).
Projecting these missing intervals to the $y$-axis you find uncountably many intervals (one for each $x \in [0,1]$) all of which are disjoint. In particular, summing the $\varepsilon_x$ you get that a sum of uncountably many positive reals is $1$.
But in general it isn't possible for an uncountable family $\mathcal F$ of positive reals to have finite sum. Indeed, just classify the reals into buckets $\frac1k \le x < \frac1{k-1}$. If the sum is actually finite then each bucket is finite, so the collection $\mathcal F$ must be countable, contradiction.
Someone on the Internet posted the question "is $1/x$ a continuous function?", leading to great controversy on Twitter. How should you respond?
First answer the following question: "is $1/x$ a function?".
Like most Internet "debates" about math, the question revolves around sloppy definitions. The original posed question (which is ill-formed) is
> \(1\) Is $1/x$ a continuous function?
To make it well-formed, I want to *first* bring up the question:
> \(2\) Is $1/x$ a function?
Technically, this question is *also* ill-formed because it never specifies the domain of the function, which is part of the data needed to specify a function. One reasonable guess what the asker meant would be $\mathbb{R} \setminus \{0\}$, i.e. the set of nonzero real numbers, in which case we get the question
> (2') Does $1/x$ define a function from $\mathbb{R} \setminus \{0\}$ to $\mathbb{R}$?
which has the firm answer YES.
On the other hand, it does *not* make sense to try to define $1/x$ as a function on $\mathbb{R}$. The definition a function requires you to specify an output value for every input, so at least if you want a real-valued function[^2], there isn't any way to construe $1/x$ as a function on all of $\mathbb{R}$.
Now, returning to (1), we can now ask a well-formed question
> (1') Does $1/x$ describe a continuous function from $\mathbb{R} \setminus \{0\} \to \mathbb{R}$?
which again has the firm answer YES.
Of course, you could also consider a question like "does $1/x$ describe a continuous function $\mathbb{R} \to \mathbb{R}$?". However, this feels misleading: it would be like asking "is $\sqrt{2}$ an even integer?". The question doesn't make sense to begin with because $\sqrt2$ isn't an integer, and "even" is an adjective used for integers, so trying to ask whether it applies to $\sqrt2$ is a [type-error](https://qchu.wordpress.com/2013/05/28/the-type-system-of-mathematics/). Similarly, "continuous" is an adjective used for functions; it doesn't make sense to ask whether it applies to something that isn't a function.
See <https://twitter.com/davidcpvm/status/1481024944830046209> for the Twitter post (in Spanish) and the accompanying Reddit post (one of several) at <https://www.reddit.com/r/math/comments/s82vf8>.
[^1]: Also, "metric" is a fun word to say.
[^2]: Those of you that know what $\mathbb{RP}^1$ is could consider it as a function $\mathbb{RP}^1 \to \mathbb{RP}^1$ if you insisted; but it's continuous in that case too. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Closed sets | 06_metric-top.md | 9 | 999 |
It would be criminal for me to talk about open sets without talking about closed sets. The name "closed" comes from the definition in a metric space.
Let $M$ be a metric space. A subset $S \subseteq M$ is **closed** in $M$ if the following property holds: let $x_1$, $x_2$, ... be a sequence of points in $S$ and suppose that $x_n$ converges to $x$ in $M$. Then $x \in S$ as well.
Same caveat: we abbreviate "closed in $M$" to just "closed" if the parent space $M$ is understood.
Here's another way to phrase it. The **limit points** of a subset $S \subseteq M$ are defined by $$\lim S := \left\{ p \in M :
\exists (x_n) \in S \text{ such that } x_n \to p \right\}.$$ Thus $S$ is closed if and only if $S = \lim S$.
Prove that $\lim S$ is closed even if $S$ isn't closed. (Draw a picture.)
For this reason, $\lim S$ is also called the **closure** of $S$ in $M$, and denoted $\overline S$. It is simply the smallest closed set which contains $S$.
(a) The empty set $\varnothing$ is closed in $M$ for vacuous reasons: there are no sequences of points with elements in $\varnothing$.
(b) The entire space $M$ is closed in $M$ for tautological reasons. (Verify this!)
(c) The closed interval $[0,1]$ in $\mathbb{R}$ is closed in $\mathbb{R}$, hence the name. Like with open sets, this is the prototypical example of a closed set to keep in mind!
(d) In fact, the closed interval $[0,1]$ is even closed in $\mathbb{R}^2$.
Let $S=(0,1)$ denote the open interval. Then $S$ is not closed in $\mathbb{R}$ because the sequence of points $$\frac12, \;
\frac14, \;
\frac18, \;
\dots$$ converges to $0 \in \mathbb{R}$, but $0 \notin (0,1)$.
I should now warn you about a confusing part of this terminology. Firstly, **"most" sets are neither open nor closed**.
The half-open interval $[0,1)$ is neither open nor closed in $\mathbb{R}$.
Secondly, it's **also possible for a set to be both open and closed**; this will be discussed in .
The reason for the opposing terms is the following theorem:
Let $M$ be a metric space, and $S \subseteq M$ any subset. Then the following are equivalent:
- The set $S$ is closed in $M$.
- The complement $M \setminus S$ is open in $M$.
Prove this theorem! You'll want to draw a picture to make it clear what's happening: for example, you might take $M = \mathbb{R}^2$ and $S$ to be the closed unit disk.
Let $M = (M,d)$ be a metric space. Show that $$d \colon M \times M \to \mathbb{R}$$ is itself a continuous function (where $M \times M$ is equipped with the product metric).
Consider $\mathbb{Q}$ and $\mathbb{N}$ as metric spaces (each with the obvious metric $d(x,y) = |x-y|$). Are these spaces homeomorphic?
No. There is not even a continuous injective map from $\mathbb{Q}$ to $\mathbb{N}$.
Two possible approaches, one using metric definition and one using open sets.
Metric approach: I claim there is no injective map from $\mathbb{Q}$ to $\mathbb{N}$ that is continuous. Indeed, suppose $f$ was such a map and $f(x) = n$. Then, choose $\varepsilon = 1/2$. There should be a $\delta > 0$ such that everything with $\delta$ of $x$ in $\mathbb{Q}$ should land within $\varepsilon$ of $n \in \mathbb{N}$ --- i.e., is equal to $n$. This is a blatant contradiction of injectivity.
Open set approach: In $\mathbb{Q}$, no singleton set is open, whereas in $\mathbb{N}$, they all are (in fact $\mathbb{N}$ is discrete). As you'll see at the start of , with the new and improved definition of "homeomorphism", we found out that the structure of open sets on $\mathbb{Q}$ and $\mathbb{N}$ are different, so they are not homeomorphic.
Show that subtraction is a continuous map $- \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, and division is a continuous map $\div \colon \mathbb{R} \times \mathbb{R}_{>0} \to \mathbb{R}$.
You can do this with bare hands. You can also use composition.
For subtraction, the map $x \mapsto -x$ is continuous so you can view it as a composed map
Similarly, if you are willing to believe $x \mapsto 1/x$ is a continuous function, then division is composition
If for some reason you are suspicious that $x \mapsto 1/x$ is continuous, then here is a proof using sequential continuity. Suppose $x_n \to x$ with $x_n > 0$ and $x > 0$ (since $x$ needs to be in $\mathbb{R}_{>0}$ too). Then $$\left\lvert \frac{1}{x} - \frac 1{x_n} \right\rvert
= \frac{\left\lvert x_n-x \right\rvert}{\left\lvert x x_n \right\rvert}.$$ If $n$ is large enough, then $\left\lvert x_n \right\rvert > x/2$; so the denominator is at least $x^2/2$, and hence the whole fraction is at most $\frac{2}{x^2} \left\lvert x_n-x \right\rvert$, which tends to zero as $n \to \infty$.
Exhibit a function $f \colon \mathbb{R} \to \mathbb{R}$ such that $f$ is continuous at $x \in \mathbb{R}$ if and only if $x=0$.
$\pm x$ for good choices of $\pm$.
Let $f(x) = x$ for $x \in \mathbb{Q}$ and $f(x) = -x$ for irrational $x$.
Prove that a function $f \colon \mathbb{R} \to \mathbb{R}$ which is strictly increasing must be continuous at some point.
Project gaps onto the $y$-axis. Use the fact that uncountably many positive reals cannot have finite sum. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Closed sets | 06_metric-top.md | 10 | 1,457 |
Assume for contradiction it is completely discontinuous; by scaling set $f(0) = 0$, $f(1) = 1$ and focus just on $f \colon [0,1] \to [0,1]$. Since it's discontinuous everywhere, for every $x \in [0,1]$ there's an $\varepsilon_x > 0$ such that the continuity condition fails. Since the function is strictly increasing, that can only happen if the function misses all $y$-values in the interval $(f(x)-\varepsilon_x, f(x))$ or $(f(x), f(x)+\varepsilon_x)$ (or both).
Projecting these missing intervals to the $y$-axis you find uncountably many intervals (one for each $x \in [0,1]$) all of which are disjoint. In particular, summing the $\varepsilon_x$ you get that a sum of uncountably many positive reals is $1$.
But in general it isn't possible for an uncountable family $\mathcal F$ of positive reals to have finite sum. Indeed, just classify the reals into buckets $\frac1k \le x < \frac1{k-1}$. If the sum is actually finite then each bucket is finite, so the collection $\mathcal F$ must be countable, contradiction.
Someone on the Internet posted the question "is $1/x$ a continuous function?", leading to great controversy on Twitter. How should you respond?
First answer the following question: "is $1/x$ a function?".
Like most Internet "debates" about math, the question revolves around sloppy definitions. The original posed question (which is ill-formed) is
> \(1\) Is $1/x$ a continuous function?
To make it well-formed, I want to *first* bring up the question:
> \(2\) Is $1/x$ a function?
Technically, this question is *also* ill-formed because it never specifies the domain of the function, which is part of the data needed to specify a function. One reasonable guess what the asker meant would be $\mathbb{R} \setminus \{0\}$, i.e. the set of nonzero real numbers, in which case we get the question
> (2') Does $1/x$ define a function from $\mathbb{R} \setminus \{0\}$ to $\mathbb{R}$?
which has the firm answer YES.
On the other hand, it does *not* make sense to try to define $1/x$ as a function on $\mathbb{R}$. The definition a function requires you to specify an output value for every input, so at least if you want a real-valued function[^2], there isn't any way to construe $1/x$ as a function on all of $\mathbb{R}$.
Now, returning to (1), we can now ask a well-formed question
> (1') Does $1/x$ describe a continuous function from $\mathbb{R} \setminus \{0\} \to \mathbb{R}$?
which again has the firm answer YES.
Of course, you could also consider a question like "does $1/x$ describe a continuous function $\mathbb{R} \to \mathbb{R}$?". However, this feels misleading: it would be like asking "is $\sqrt{2}$ an even integer?". The question doesn't make sense to begin with because $\sqrt2$ isn't an integer, and "even" is an adjective used for integers, so trying to ask whether it applies to $\sqrt2$ is a [type-error](https://qchu.wordpress.com/2013/05/28/the-type-system-of-mathematics/). Similarly, "continuous" is an adjective used for functions; it doesn't make sense to ask whether it applies to something that isn't a function.
See <https://twitter.com/davidcpvm/status/1481024944830046209> for the Twitter post (in Spanish) and the accompanying Reddit post (one of several) at <https://www.reddit.com/r/math/comments/s82vf8>.
[^1]: Also, "metric" is a fun word to say.
[^2]: Those of you that know what $\mathbb{RP}^1$ is could consider it as a function $\mathbb{RP}^1 \to \mathbb{RP}^1$ if you insisted; but it's continuous in that case too. | An Infinitely Large Napkin | napkin | general | advanced | Starting Out | Metric spaces | Closed sets | 06_metric-top.md | 11 | 999 |
Homomorphisms and quotient groups
ch:homomorphisms_quotient
Generators and group presentations
= < r,s r^n=s^2=1>$
Let $G$ be a group.
Recall that for some element $x G$,
we could consider the subgroup
\[ \ , x^-2, x^-1, 1, x, x^2, \ \]
of $G$.
Here's a more pictorial version of what we did:
put $x$ in a box, seal it tightly, and shake vigorously.
Using just the element $x$,
we get a pretty explosion that produces the subgroup above.
What happens if we put two elements $x$, $y$ in the box?
Among the elements that get produced are things like
\[ xyxyx, x^2y^9x^-5y^3, y^-2015, \]
Essentially, I can create any finite product of $x$, $y$, $x$, $y$.
This leads us to define:
Let $S$ be a subset of $G$.
The subgroup generated by $S$,
denoted $<S>$,
is the set of elements which can be written as a
finite product of elements in $S$ (and their inverses).
If $<S> = G$ then we say $S$ is a set of
generators for $G$,
as the elements of $S$ together create all of $G$.
Why is the condition ``and their inverses''
not necessary if $G$ is a finite group?
(As usual, assume Lagrange's theorem.)
Consider $1$ as an element of $ = (, +)$.
We see $<1> = $, meaning $\1\$ generates $$.
It's important that $-1$, the inverse of $1$ is also allowed:
we need it to write all integers as the sum of $1$ and $-1$.
This gives us an idea for a way to try and express groups compactly.
Why not just write down a list of generators for the groups?
For example, we could write
\[ < a > \]
meaning that $$ is just the group generated by one element.
There's one issue: the generators usually satisfy certain properties.
For example, consider $100$.
It's also generated by a single element $x$,
but this $x$ has the additional property that $x^100 = 1$.
This motivates us to write
\[ 100 = < x x^100 = 1 >. \]
I'm sure you can see where this is going.
All we have to do is specify a set of generators and
relations between the generators,
and say that two elements are equal if and only if
you can get from one to the other using relations.
Such an expression is appropriately called a group presentation.
The dihedral group of order $2n$ has a presentation
\[ D_2n = < r, s
r^n = s^2 = 1, rs = sr >. \]
Thus each element of $D_2n$ can be written uniquely in the form $r^$
or $sr^$, where $ = 0, 1, , n-1$.
The Klein four group,
isomorphic to $2 2$, is given by the presentation
\[ < a,b a^2=b^2=1, ab=ba >. \]
[Free group]
The free group on $n$ elements is the group
whose presentation has $n$ generators and no relations at all.
It is denoted $F_n$, so
\[
F_n = < x_1, x_2, , x_n >.
\]
In other words, $F_2 = <a,b>$ is the set of strings
formed by appending finitely many copies of $a$, $b$, $a$, $b$ together.
Notice that $F_1 $.
One might unfortunately notice that ``subgroup generated by $a$ and $b$''
has exactly the same notation as the free group $<a,b>$.
We'll try to be clear based on context which one we mean.
Presentations are nice because they provide a compact way to write down groups.
They do have some shortcomings, though.%
Actually, determining whether two elements of a presentation are equal is undecidable.
In fact, it is undecidable to even determine if a group is finite from its presentation.
[Presentations can look very different]
The same group can have very different presentations.
For instance consider
\[ D_2n = < x,y x^2=y^2=1, (xy)^n=1 >. \]
(To see why this is equivalent, set $x=s$, $y=rs$.)
Homomorphisms
$.
How can groups talk to each other?
Two groups are ``the same'' if we can write an isomorphism between them.
And as we saw, two metric spaces are ``the same''
if we can write a homeomorphism between them.
But what's the group analogy of a continuous map?
We simply drop the ``bijection'' condition.
Let $G = (G, )$ and $H = (H, )$ be groups.
A group homomorphism is a map $ G H$
such that for any $g_1, g_2 G$ we have
\[ (g_1 g_2) = (g_1) (g_2). \]
(Not to be confused with ``homeomorphism'' from
last chapter: note the spelling.)
[Examples of homomorphisms]
Let $G$ and $H$ be groups.
Any isomorphism $G H$ is a homomorphism.
In particular, the identity map $G G$ is a homomorphism.
The trivial homomorphism $G H$ sends
everything to $1_H$.
There is a homomorphism from $$ to $100$ by
sending each integer to its residue modulo $100$.
There is a homomorphism from $$ to itself by $x 10x$
which is injective but not surjective.
There is a homomorphism from $S_n$ to $S_n+1$ by ``embedding'':
every permutation on $\1,,n\$ can be thought of as a permutation
on $\1,,n+1\$ if we simply let $n+1$ be a fixed point.
A homomorphism $: D_12 D_6$
is given by $s_12 s_6$
and $r_12 r_6$.
Specifying a homomorphism $ G$ is the same as
specifying just the image of the element $1 $. Why?
The last two examples illustrate something: suppose we have a presentation of $G$.
To specify a homomorphism $G H$, we only have to specify where each generator of $G$ goes, in such a way that the relations are all satisfied.
Important remark:
the right way to think about an isomorphism is as a ``bijective homomorphism''.
To be explicit,
Show that $G H$ if and only if there exist
homomorphisms $ G H$ and $ H G$
such that $ = _H$ and $ = _G$.
So the definitions of homeomorphism of metric spaces
and isomorphism of groups are not too different.
Some obvious properties of homomorphisms follow.
Let $ G H$ be a homomorphism.
Then $(1_G) = 1_H$ and $(g) = (g)$.
Boring, and I'm sure you could do it yourself if you wanted to. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 0 | 1,559 | |
Now let me define a very important property of a homomorphism.
The kernel of a homomorphism $ G H$ is defined by
\[
\ g G : (g) = 1_H \.
\]
It is a subgroup of $G$
(in particular, $1_G $ for obvious reasons).
Verify that $$ is in fact a subgroup of $G$.
We also have the following important fact, which we also encourage the reader to verify.
[Kernel determines injectivity]
The map $$ is injective if and only if $ = \1_G\$.
To make this concrete, let's compute the kernel of each of our examples.
[Examples of kernels]
The kernel of any isomorphism $G H$ is trivial,
since an isomorphism is injective.
In particular, the kernel of the identity map $G G$ is $\1_G\$.
The kernel of the trivial homomorphism $G H$
(by $g 1_H$) is all of $G$.
The kernel of the homomorphism $ 100$ by $n n$
is precisely \[ 100 = \, -200, -100, 0, 100, 200, \. \]
The kernel of the map $ $ by $x 10x$ is trivial: $\0\$.
There is a homomorphism from $S_n$ to $S_n+1$ by ``embedding'',
but it also has trivial kernel because it is injective.
A homomorphism $ D_12 D_6$
is given by $s_12 s_6$ and $r_12 r_6$.
You can check that
\[ = \ 1, r_12^3 \ 2. \]
Exercise below.
Fix any $g G$.
Suppose we have a homomorphism $ G$ by $n g^n$.
What is the kernel?
Show that for any homomorphism $: G H$,
the image $(G)$ is a subgroup of $H$.
Hence, we'll be especially interested in the case where $$ is surjective.
Cosets and modding out
The next few sections are a bit dense.
If this exposition doesn't work for you, try ref:gowers.
Let $G$ and $Q$ be groups, and suppose there exists
a surjective homomorphism \[ G Q. \]
In other words, if $$ is injective then $ G Q$ is a bijection,
and hence an isomorphism.
But suppose we're not so lucky and $$ is bigger than just $\1_G\$.
What is the correct interpretation of a more general homomorphism?
Let's look at the special case where $ 100$ is ``modding out by $100$''.
We already saw that the kernel of this map is
\[
= 100 = \ , -200, -100, 0, 100, 200, \.
\]
Recall now that $ $ is a subgroup of $G$.
What this means is that $$ is indifferent to the subgroup $100$ of $$:
\[ (15) = (2000 + 15) = (-300 + 15) = (700 + 15) = . \]
So $100$ is what we get when we ``mod out by $100$''. Cool.
In other words, let $G$ be a group and $ G Q$
be a surjective homomorphism with kernel $N G$.
We claim that $Q$ should be thought of as the quotient of $G$ by $N$.
To formalize this, we will define a so-called
quotient group $G/N$
in terms of $G$ and $N$ only (without referencing $Q$)
which will be naturally isomorphic to $Q$.
For motivation, let's give a concrete description of $Q$ using just $$ and $G$.
Continuing our previous example, let $N = 100$ be our subgroup of $G$.
Consider the sets
N &= \ , -200, -100, 0, 100, 200, \ \\
1+N &= \ , -199, -99, 1, 101, 201, \ \\
2+N &= \ , -198, -98, 2, 102, 202, \ \\
& \\
99+N &= \ , -101, -1, 99, 199, 299, \.
The elements of each set all have the same image when we apply $$,
and moreover any two elements in different sets have different images.
Then the main idea is to notice that
We can think of $Q$ as the group
whose elements are the sets above.
Thus, given $$ we define an equivalence relation $_N$
on $G$ by saying $x _N y$ for $(x) = (y)$.
This $_N$ divides $G$ into several equivalence classes in $G$
which are in obvious bijection with $Q$, as above.
Now we claim that we can write these equivalence classes very explicitly.
% Also, for each $g G$ define $ g$ to be the equivalence class of $g$ under $_$.
Show that $x _N y$ if and only if $x = yn$ for some $n N$
(in the mod $100$ example, this means they ``differ by some multiple of $100$'').
Thus for any $g G$, the equivalence class of $_N$ which contains $g$
is given explicitly by \[ gN \ gn n N \. \]
%Note that
%\[ (x) = (y)
% 1_Q = (x)(y) = (xy)
% xy . \]
%Hence another way to describe $ x$ is
%\[ g = gN = \ gn n N \. \]
Here's the word that describes the types of sets we're running into now.
Let $H$ be any subgroup of $G$ (not necessarily the kernel of some homomorphism).
A set of the form $gH$ is called a left coset of $H$.
Although the notation might not suggest it,
keep in mind that $g_1N$ is often equal to $g_2N$ even if $g_1 g_2$.
In the ``mod $100$'' example, $3+N = 103+N$.
In other words, these cosets are sets.
This means that if I write ``let $gH$ be a coset'' without telling you what $g$ is,
you can't figure out which $g$ I chose from just the coset itself.
If you don't believe me, here's an example of what I mean:
\[ x+100 = \ , -97, 3, 103, 203, \ x = ?. \]
There's no reason to think I picked $x=3$. (I actually picked $x=-13597$.)
remark:coset_warning
Given cosets $g_1H$ and $g_2H$,
you can check that the map $x g_2g_1 x$ is a bijection between them.
So actually, all cosets have the same cardinality.
So, long story short,
Elements of the group $Q$ are naturally identified with left cosets of $N$.
In practice, people often still prefer to picture elements of $Q$ as single points
(for example it's easier to think of $2$ as $\0,1\$
rather than $\ \,-2,0,2,\, \,-1,1,3,\ \$).
If you like this picture,
then you might then draw $G$ as a bunch of equally tall fibers (the cosets),
which are then ``collapsed'' onto $Q$.
size(9cm);
for (int i=1; i<=6; ++i)
draw( (i,0)--(i,3.4) );
dot( (i,0.6) );
dot( (i,1) );
dot( (i,1.7) );
dot( (i,2.4) );
dot( (i,2.8) );
dot( (i,3) );
label(rotate(90)*scale(1.4)*"$$", (i+0.02,-0.9), dir(90));
dot( (i,-1) );
label("$G$", (0.5, 3));
label("$Q$", (0.5,-1)); | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 1 | 1,568 | |
Now that we've done this, we can give an intrinsic
definition for the quotient group we alluded to earlier.
A subgroup $N$ of $G$ is called normal if it is the
kernel of some homomorphism.
We write this as $N G$.
Let $N G$.
Then the quotient group, denoted $G/N$
(and read ``$G$ mod $N$''),
is the group defined as follows.
The elements of $G/N$ will be the left cosets of $N$.
We want to define the product of two cosets $C_1$ and $C_2$ in $G/N$.
Recall that the cosets are in bijection with elements of $Q$.
So let $q_1$ be the value associated to the coset $C_1$,
and $q_2$ the one for $C_2$.
Then we can take the product to be the coset corresponding to $q_1q_2$.
Quite importantly,
we can also do this in terms of representatives of the cosets.
Let $g_1 C_1$ and $g_2 C_2$,
so $C_1 = g_1N$ and $C_2 = g_2N$.
Then $C_1 C_2$ should be the coset which contains $g_1g_2$.
This is the same as the above definition since
$(g_1g_2) = (g_1)(g_2) = q_1q_2$;
all we've done is define the product in terms of elements of $G$,
rather than values in $H$.
Using the $gN$ notation,
and with remark:coset_warning in mind,
we can write this even more succinctly:
\[ (g_1N) (g_2N) (g_1g_2)N. \]
And now you know why the integers modulo $n$ are often written $/n$!
Take a moment to digest the above definition.
By the way we've built it, the resulting group $G/N$ is isomorphic to $Q$.
In a sense we think of $G/N$ as ``$G$ modulo the condition that $n=1$
for all $n N$''.
(Optional) Proof of Lagrange's theorem
As an aside, with the language of cosets
we can now show Lagrange's theorem in the general case.
[Lagrange's theorem]
thm:lagrange_grp
Let $G$ be a finite group, and let $H$ be any subgroup.
Then $ H $ divides $ G $.
The proof is very simple: note that the cosets of $H$
all have the same size and form a partition of $G$
(even when $H$ is not necessarily normal).
Hence if $n$ is the number of cosets,
then $n H = G $.
Conclude that $x^ G =1$
by taking $H = <x> G$.
It should be mentioned at this point that
in general, if $G$ is a finite group and $N$ is normal,
then $|G/N| = |G| / |N|$.
Eliminating the homomorphism
Let's look at the last definition of $G/N$ we provided.
The short version is:
The elements of $G/N$ are cosets $gN$, which you can think
of as equivalence classes of a relation $_N$
(where $g_1 _N g_2$ if $g_1 = g_2n$ for some $n N$).
Given cosets $g_1N$ and $g_2N$ the group operation is
\[ g_1N g_2N (g_1g_2)N. \]
Question: where do we actually use the fact that $N$ is normal?
We don't talk about $$ or $Q$ anywhere in this definition.
The answer is in remark:coset_warning.
The group operation takes in two cosets,
so it doesn't know what $g_1$ and $g_2$ are.
But behind the scenes,
the normal condition guarantees that the group operation can pick
any $g_1$ and $g_2$ it wants and still end up with the same coset.
If we didn't have this property, then it would be hard to define the
product of two cosets $C_1$ and $C_2$ because it might make a difference
which $g_1 C_1$ and $g_2 C_2$ we picked.
The fact that $N$ came from a homomorphism meant we could pick any representatives
$g_1$ and $g_2$ of the cosets we wanted, because they all had the same $$-value.
We want some conditions which force this to be true without referencing $$ at all.
Suppose $ G K$ is a homomorphism of groups with $H = $.
Aside from the fact $H$ is a group, we can get an ``obvious'' property:
Show that if $h H$, $g G$,
then $ghg H$.
(Check $(ghg) = 1_K$.)
ex:dihedral_normal_subgroup
Let $D_12 = <r,s r^6=s^2=1, rs=sr>$.
Consider the subgroup of order two $H = \1,s\$
and notice that \[ rsr = r(sr)= r(rs) = r^2s H. \]
Hence $H$ is not normal, and cannot be the kernel of any homomorphism.
Well, duh -- so what?
Amazingly it turns out that this is the sufficient condition we want.
Specifically, it makes the nice ``coset multiplication'' we wanted work out.
[For math contest enthusiasts]
This coincidence is really a lot like functional equations at the IMO.
We all know that normal subgroups $H$ satisfy $ghg H$;
the surprise is that from the latter seemingly weaker condition,
we can deduce $H$ is normal.
Thus we have a new criterion for ``normal'' subgroups which does not
make any external references to $$.
Let $H$ be a subgroup of $G$.
Then the following are equivalent:
$H G$.
For every $g G$ and $h H$, $ghg H$.
We already showed one direction.
For the other direction, we need to build a homomorphism with kernel $H$.
So we simply define the group $G/H$ as the cosets.
To put a group operation, we need to verify:
If $g_1' _H g_1$ and $g_2' _H g_2$ then $g_1'g_2' _H g_1g_2$.
Boring algebraic manipulation (again functional equation style).
Let $g_1' = g_1h_1$ and $g_2' = g_2h_2$, so we want to show that
$g_1h_1g_2h_2 _H g_1g_2$.
Since $H$ has the property, $g_2 h_1g_2$ is some element of $H$, say $h_3$.
Thus $h_1 g_2 = g_2 h_3$, and the left-hand side becomes $g_1g_2(h_3h_2)$,
which is fine since $h_3h_2 H$.
With that settled we can just define the
product of two cosets (of normal subgroups) by \[ (g_1H) (g_2H) = (g_1g_2)H. \]
Thus the claim above shows that this multiplication is well-defined
(this verification is the ``content'' of the theorem).
So $G/H$ is indeed a group!
Moreover there is an obvious ``projection'' homomorphism
$G G/H$ (with kernel $H$), by $g gH$.
%Another way to write the condition is
%\[ H = gHg \ ghg h H \. \]
%You should take a moment to check that these definitions are equivalent. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 2 | 1,563 | |
Consider again the product group $G H$.
Earlier we identified a subgroup
\[ G' = \ (g, 1_H) g G \ G. \]
You can easily see that $G' G H$.
(Easy calculation.)
Moreover, you can check that
\[ (G H) / (G') H. \]
Indeed, we have $(g, h) _G' (1_G, h)$ for all $g G$ and $h H$.
It is not necessarily true that $(G/H) H G$.
For example, consider $G = /4$
and the normal subgroup $H = \0,2\ /2$.
Then $G/H /2$,
but $/4 /2 /2$.
(Footnote: the precise condition for this kind of ``canceling'' is called the Schur-Zassenhaus lemma.)
Let $: D_8 4$ be defined by \[ r 2, s 2. \]
The kernel of this map is $N = \1,r^2,sr,sr^3\$.
We can do a quick computation of all the elements of $D_8$ to get
\[ (1) = (r^2) = (sr) = (sr^3) = 0
and
(r) = (r^3) = (s) = (sr^2) = 2. \]
The two relevant fibers are \[ ( 0) = 1N = r^2N = srN = sr^3N = \1,r^2,sr,sr^3\ \] and
\[ ( 2) = rN = r^3N = sN = sr^2N = \r,r^3,s,sr^2\. \]
So we see that $|D_8/N| = 2$ is a group of order two, or $ 2$.
Indeed, the image of $$ is \[ \ 0, 2 \ 2. \]
Suppose $G$ is abelian.
Why does it follow that any subgroup of $G$ is normal?
Finally here's some food for thought:
suppose one has a group presentation for a group $G$
that uses $n$ generators.
Can you write it as a quotient of the form $F_n / N$,
where $N$ is a normal subgroup of $F_n$?
(Digression) The first isomorphism theorem
sec:first_isomorphism_thm
One quick word about what other sources usually say.
Most textbooks actually define normal using the $ghg H$ property.
Then they define $G/H$ for normal $H$ in the way I did above,
using the coset definition
\[ (g_1H) (g_2H) = g_1g_2H. \]
Using purely algebraic manipulations (like I did) this is well-defined,
and so now you have this group $G/H$ or something.
The underlying homomorphism isn't mentioned at all,
or is just mentioned in passing.
I think this is incredibly dumb.
The normal condition looks like it gets pulled out of thin air
and no one has any clue what's going on,
because no one has any clue what a normal subgroup actually should look like.
Other sources like to also write the so-called first isomorphism theorem.There
is a second and third isomorphism theorem.
But four years after learning about them, I still don't remember what they are.
So I'm guessing they weren't very important.
It goes like this.
[First isomorphism theorem]
Let $ G H$ be a homomorphism.
Then $G / $ is isomorphic to $(G)$.
To me, this is just a clumsier way of stating the same idea.
About the only merit this claim has is that if $$ is injective,
then the image $(G)$ is an isomorphic copy
of $G$ inside the group $H$.
(Try to see this directly!)
This is a pattern we'll often see in other branches of mathematics:
whenever we have an injective structure-preserving map,
often the image of this map will be some ``copy'' of $G$.
(Here ``structure'' refers to the group multiplication,
but we'll see some more other examples of ``types of objects'' later!)
In that sense an injective homomorphism $ G H$
is an embedding of $G$ into $H$.
[18.701 at MIT]
Determine all groups $G$ for which the map $ G G$ defined by
\[ (g) = g^2 \] is a homomorphism.
Write it out: $(ab) = (a)(b)$.
Abelian groups: $abab =a^2b^2 ab= ba$.
Consider the dihedral group $G = D_10$.
Is $H = < r >$ a normal subgroup of $G$?
If so, compute $G/H$ up to isomorphism.
Is $H = < s >$ a normal subgroup of $G$?
If so, compute $G/H$ up to isomorphism.
Yes, no.
Yes to (a): you can check this directly
from the $ghg$ definition.
For example, for (a)
it is enough to compute $(r^a s) r^n (r^a s) = r^-n H$.
The quotient group is $2$.
The answer is no for (b) by following
ex:dihedral_normal_subgroup.
Does $S_4$ have a normal subgroup of order $3$?
No.
A subgroup of order $3$ must be generated by
an element of order $3$, since $3$ is prime.
So we may assume WLOG that $H = < (1\; 2 \; 3) >$
(by renaming elements appropriately).
But then let $g = (3 \; 4)$; one can check $gHg H$.
Let $G$ and $H$ be finite groups, where $ G = 1000$
and $ H = 999$.
Show that a homomorphism $G H$ must be trivial.
$(1000,999)=1$.
$G/ G$ is isomorphic to a subgroup of $H$.
The order of the former divides $1000$;
the order of the latter divides $999$.
This can only occur if $G / G = \1\$
so $ G = G$.
Let $^$ denote the nonzero complex numbers under multiplication.
Show that there are five homomorphisms $5 ^$
but only two homomorphisms $D_10 ^$,
even though $5$ is a subgroup of $D_10$.
Find a non-abelian group $G$
such that every subgroup of $G$ is normal.
(These groups are called Hamiltonian.)
Find an example of order $8$.
Quaternion group.
[PRIMES entrance exam, 2018]
Let $G$ be a group with presentation given by
\[ G = < a,b,c
ab = c^2a^4, \; bc = ca^6, \; ac = ca^8, \;
c^2018 = b^2019
>. \]
Determine the order of $G$.
Try to show $G$ is the dihedral group of order $18$.
There is not much group theory content here --- just manipulation.
The answer is $|G| = 18$.
First, observe that by induction we have
\[ a^n c = ca^8n \]
for all $n 1$.
We then note that
a(bc) &= (ab)c \\
a ca^6 &= c^2 a^4 c \\
c a^8 a^6 &= c^2 a^4 c \\
a^14 &= c(a^4c) = c^2 a^32.
Hence we conclude $c^2 = a^-18$.
Then $ab = c^2a^4 b = a^-15$.
In that case, since $c^2018 = b^2019$,
we conclude $1 = a^- 1009 18 + 2019 15 = a^12123$.
Finally,
bc &= ca^6 \\
a^-15 c &= ca^6 \\
a^-15 c^2 &= c(a^6c) = c^2 a^48 \\
a^-33 &= a^30 \\
a^63 &= 1. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 3 | 1,537 | |
Since $(12123, 63) = 9$, we find $a^9 = 1$, hence finally $c^2 = 1$.
So the presentation above simplifies to
\[ G = < a,c a^9=c^2=1, \; ac = ca^-1 > \]
which is the presentation of the dihedral group of order $18$.
% a = r, r^9 = 1
% b = r^3
% c = s, s^2 = 2
This completes the proof.
[Homophony group]
The homophony group (of English) is the group
with $26$ generators $a$, $b$, , $z$
and one relation for every pair of English words
which sound the same.
For example $knight = night$ (and hence $k=1$).
Prove that the group is trivial.
Get yourself a list of English homophones, I guess.
Don't try too hard.
Letter $v$ is the worst; maybe $felt = veldt$?
You can find many solutions by searching ``homophone group'';
one is https://math.stackexchange.com/q/843966/229197.
% Copying math.SE content:
% F:=FreeGroup("a","b","c","d","e","f","g","h","i","j","k","l","m",
% "n","o","p","q","r","s","t","u","v","w","x","y","z");
%
% AssignGeneratorVariables(F);
%
% pairs:=[
% "bye=by", # e=1
% "lead=led", # a=1
% "maid=made", # mid=md => i=1
% "sow=sew", # o=e=1
% "buy=by", # u=1
% "sow=so", # w=1
% "lye=lie", # y=1
% "hour=our", # h=1
% "knight=night", # k=1
% "damn=dam", # n=1
% "psalter=salter", # p=1
% "plumb=plum", # b=1
% "bass=base", # s=1
% "butt=but", # t=1
% "tolled=told", # l=1
% "barred=bard", # r=1
% "dammed=damned", # m=1
% "chased=chaste", # d=1
% "sign=sine", # g=1
% "daze=days", # z=1
% "cite=sight", # c=1
% "jeans=genes", # j=1
% "queue=cue", # q=1
% "tax=tacks", # x=1
% # "ruff=rough", # f=1, we need not this relation used in the paper
% "phase=faze", # f=1
% "chivvy=chivy"]; # v=1
%
% rels:=List( pairs, r -> ParseRelators(GeneratorsOfGroup(F),r)[1]);
% G:=F/rels; | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 4 | 485 | |
Homomorphisms and quotient groups
ch:homomorphisms_quotient
Generators and group presentations
= < r,s r^n=s^2=1>$
Let $G$ be a group.
Recall that for some element $x G$,
we could consider the subgroup
\[ \ , x^-2, x^-1, 1, x, x^2, \ \]
of $G$.
Here's a more pictorial version of what we did:
put $x$ in a box, seal it tightly, and shake vigorously.
Using just the element $x$,
we get a pretty explosion that produces the subgroup above.
What happens if we put two elements $x$, $y$ in the box?
Among the elements that get produced are things like
\[ xyxyx, x^2y^9x^-5y^3, y^-2015, \]
Essentially, I can create any finite product of $x$, $y$, $x$, $y$.
This leads us to define:
Let $S$ be a subset of $G$.
The subgroup generated by $S$,
denoted $<S>$,
is the set of elements which can be written as a
finite product of elements in $S$ (and their inverses).
If $<S> = G$ then we say $S$ is a set of
generators for $G$,
as the elements of $S$ together create all of $G$.
Why is the condition ``and their inverses''
not necessary if $G$ is a finite group?
(As usual, assume Lagrange's theorem.)
Consider $1$ as an element of $ = (, +)$.
We see $<1> = $, meaning $\1\$ generates $$.
It's important that $-1$, the inverse of $1$ is also allowed:
we need it to write all integers as the sum of $1$ and $-1$.
This gives us an idea for a way to try and express groups compactly.
Why not just write down a list of generators for the groups?
For example, we could write
\[ < a > \]
meaning that $$ is just the group generated by one element.
There's one issue: the generators usually satisfy certain properties.
For example, consider $100$.
It's also generated by a single element $x$,
but this $x$ has the additional property that $x^100 = 1$.
This motivates us to write
\[ 100 = < x x^100 = 1 >. \]
I'm sure you can see where this is going.
All we have to do is specify a set of generators and
relations between the generators,
and say that two elements are equal if and only if
you can get from one to the other using relations.
Such an expression is appropriately called a group presentation.
The dihedral group of order $2n$ has a presentation
\[ D_2n = < r, s
r^n = s^2 = 1, rs = sr >. \]
Thus each element of $D_2n$ can be written uniquely in the form $r^$
or $sr^$, where $ = 0, 1, , n-1$.
The Klein four group,
isomorphic to $2 2$, is given by the presentation
\[ < a,b a^2=b^2=1, ab=ba >. \]
[Free group]
The free group on $n$ elements is the group
whose presentation has $n$ generators and no relations at all.
It is denoted $F_n$, so
\[
F_n = < x_1, x_2, , x_n >.
\]
In other words, $F_2 = <a,b>$ is the set of strings
formed by appending finitely many copies of $a$, $b$, $a$, $b$ together.
Notice that $F_1 $.
One might unfortunately notice that ``subgroup generated by $a$ and $b$''
has exactly the same notation as the free group $<a,b>$.
We'll try to be clear based on context which one we mean.
Presentations are nice because they provide a compact way to write down groups.
They do have some shortcomings, though.%
Actually, determining whether two elements of a presentation are equal is undecidable.
In fact, it is undecidable to even determine if a group is finite from its presentation.
[Presentations can look very different]
The same group can have very different presentations.
For instance consider
\[ D_2n = < x,y x^2=y^2=1, (xy)^n=1 >. \]
(To see why this is equivalent, set $x=s$, $y=rs$.)
Homomorphisms
$.
How can groups talk to each other?
Two groups are ``the same'' if we can write an isomorphism between them.
And as we saw, two metric spaces are ``the same''
if we can write a homeomorphism between them.
But what's the group analogy of a continuous map?
We simply drop the ``bijection'' condition.
Let $G = (G, )$ and $H = (H, )$ be groups.
A group homomorphism is a map $ G H$
such that for any $g_1, g_2 G$ we have
\[ (g_1 g_2) = (g_1) (g_2). \]
(Not to be confused with ``homeomorphism'' from
last chapter: note the spelling.)
[Examples of homomorphisms]
Let $G$ and $H$ be groups.
Any isomorphism $G H$ is a homomorphism.
In particular, the identity map $G G$ is a homomorphism.
The trivial homomorphism $G H$ sends
everything to $1_H$.
There is a homomorphism from $$ to $100$ by
sending each integer to its residue modulo $100$.
There is a homomorphism from $$ to itself by $x 10x$
which is injective but not surjective.
There is a homomorphism from $S_n$ to $S_n+1$ by ``embedding'':
every permutation on $\1,,n\$ can be thought of as a permutation
on $\1,,n+1\$ if we simply let $n+1$ be a fixed point.
A homomorphism $: D_12 D_6$
is given by $s_12 s_6$
and $r_12 r_6$.
Specifying a homomorphism $ G$ is the same as
specifying just the image of the element $1 $. Why?
The last two examples illustrate something: suppose we have a presentation of $G$.
To specify a homomorphism $G H$, we only have to specify where each generator of $G$ goes, in such a way that the relations are all satisfied.
Important remark:
the right way to think about an isomorphism is as a ``bijective homomorphism''.
To be explicit,
Show that $G H$ if and only if there exist
homomorphisms $ G H$ and $ H G$
such that $ = _H$ and $ = _G$.
So the definitions of homeomorphism of metric spaces
and isomorphism of groups are not too different.
Some obvious properties of homomorphisms follow.
Let $ G H$ be a homomorphism.
Then $(1_G) = 1_H$ and $(g) = (g)$.
Boring, and I'm sure you could do it yourself if you wanted to. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 5 | 1,559 | |
Now let me define a very important property of a homomorphism.
The kernel of a homomorphism $ G H$ is defined by
\[
\ g G : (g) = 1_H \.
\]
It is a subgroup of $G$
(in particular, $1_G $ for obvious reasons).
Verify that $$ is in fact a subgroup of $G$.
We also have the following important fact, which we also encourage the reader to verify.
[Kernel determines injectivity]
The map $$ is injective if and only if $ = \1_G\$.
To make this concrete, let's compute the kernel of each of our examples.
[Examples of kernels]
The kernel of any isomorphism $G H$ is trivial,
since an isomorphism is injective.
In particular, the kernel of the identity map $G G$ is $\1_G\$.
The kernel of the trivial homomorphism $G H$
(by $g 1_H$) is all of $G$.
The kernel of the homomorphism $ 100$ by $n n$
is precisely \[ 100 = \, -200, -100, 0, 100, 200, \. \]
The kernel of the map $ $ by $x 10x$ is trivial: $\0\$.
There is a homomorphism from $S_n$ to $S_n+1$ by ``embedding'',
but it also has trivial kernel because it is injective.
A homomorphism $ D_12 D_6$
is given by $s_12 s_6$ and $r_12 r_6$.
You can check that
\[ = \ 1, r_12^3 \ 2. \]
Exercise below.
Fix any $g G$.
Suppose we have a homomorphism $ G$ by $n g^n$.
What is the kernel?
Show that for any homomorphism $: G H$,
the image $(G)$ is a subgroup of $H$.
Hence, we'll be especially interested in the case where $$ is surjective.
Cosets and modding out
The next few sections are a bit dense.
If this exposition doesn't work for you, try ref:gowers.
Let $G$ and $Q$ be groups, and suppose there exists
a surjective homomorphism \[ G Q. \]
In other words, if $$ is injective then $ G Q$ is a bijection,
and hence an isomorphism.
But suppose we're not so lucky and $$ is bigger than just $\1_G\$.
What is the correct interpretation of a more general homomorphism?
Let's look at the special case where $ 100$ is ``modding out by $100$''.
We already saw that the kernel of this map is
\[
= 100 = \ , -200, -100, 0, 100, 200, \.
\]
Recall now that $ $ is a subgroup of $G$.
What this means is that $$ is indifferent to the subgroup $100$ of $$:
\[ (15) = (2000 + 15) = (-300 + 15) = (700 + 15) = . \]
So $100$ is what we get when we ``mod out by $100$''. Cool.
In other words, let $G$ be a group and $ G Q$
be a surjective homomorphism with kernel $N G$.
We claim that $Q$ should be thought of as the quotient of $G$ by $N$.
To formalize this, we will define a so-called
quotient group $G/N$
in terms of $G$ and $N$ only (without referencing $Q$)
which will be naturally isomorphic to $Q$.
For motivation, let's give a concrete description of $Q$ using just $$ and $G$.
Continuing our previous example, let $N = 100$ be our subgroup of $G$.
Consider the sets
N &= \ , -200, -100, 0, 100, 200, \ \\
1+N &= \ , -199, -99, 1, 101, 201, \ \\
2+N &= \ , -198, -98, 2, 102, 202, \ \\
& \\
99+N &= \ , -101, -1, 99, 199, 299, \.
The elements of each set all have the same image when we apply $$,
and moreover any two elements in different sets have different images.
Then the main idea is to notice that
We can think of $Q$ as the group
whose elements are the sets above.
Thus, given $$ we define an equivalence relation $_N$
on $G$ by saying $x _N y$ for $(x) = (y)$.
This $_N$ divides $G$ into several equivalence classes in $G$
which are in obvious bijection with $Q$, as above.
Now we claim that we can write these equivalence classes very explicitly.
% Also, for each $g G$ define $ g$ to be the equivalence class of $g$ under $_$.
Show that $x _N y$ if and only if $x = yn$ for some $n N$
(in the mod $100$ example, this means they ``differ by some multiple of $100$'').
Thus for any $g G$, the equivalence class of $_N$ which contains $g$
is given explicitly by \[ gN \ gn n N \. \]
%Note that
%\[ (x) = (y)
% 1_Q = (x)(y) = (xy)
% xy . \]
%Hence another way to describe $ x$ is
%\[ g = gN = \ gn n N \. \]
Here's the word that describes the types of sets we're running into now.
Let $H$ be any subgroup of $G$ (not necessarily the kernel of some homomorphism).
A set of the form $gH$ is called a left coset of $H$.
Although the notation might not suggest it,
keep in mind that $g_1N$ is often equal to $g_2N$ even if $g_1 g_2$.
In the ``mod $100$'' example, $3+N = 103+N$.
In other words, these cosets are sets.
This means that if I write ``let $gH$ be a coset'' without telling you what $g$ is,
you can't figure out which $g$ I chose from just the coset itself.
If you don't believe me, here's an example of what I mean:
\[ x+100 = \ , -97, 3, 103, 203, \ x = ?. \]
There's no reason to think I picked $x=3$. (I actually picked $x=-13597$.)
remark:coset_warning
Given cosets $g_1H$ and $g_2H$,
you can check that the map $x g_2g_1 x$ is a bijection between them.
So actually, all cosets have the same cardinality.
So, long story short,
Elements of the group $Q$ are naturally identified with left cosets of $N$.
In practice, people often still prefer to picture elements of $Q$ as single points
(for example it's easier to think of $2$ as $\0,1\$
rather than $\ \,-2,0,2,\, \,-1,1,3,\ \$).
If you like this picture,
then you might then draw $G$ as a bunch of equally tall fibers (the cosets),
which are then ``collapsed'' onto $Q$.
size(9cm);
for (int i=1; i<=6; ++i)
draw( (i,0)--(i,3.4) );
dot( (i,0.6) );
dot( (i,1) );
dot( (i,1.7) );
dot( (i,2.4) );
dot( (i,2.8) );
dot( (i,3) );
label(rotate(90)*scale(1.4)*"$$", (i+0.02,-0.9), dir(90));
dot( (i,-1) );
label("$G$", (0.5, 3));
label("$Q$", (0.5,-1)); | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 6 | 1,568 | |
Now that we've done this, we can give an intrinsic
definition for the quotient group we alluded to earlier.
A subgroup $N$ of $G$ is called normal if it is the
kernel of some homomorphism.
We write this as $N G$.
Let $N G$.
Then the quotient group, denoted $G/N$
(and read ``$G$ mod $N$''),
is the group defined as follows.
The elements of $G/N$ will be the left cosets of $N$.
We want to define the product of two cosets $C_1$ and $C_2$ in $G/N$.
Recall that the cosets are in bijection with elements of $Q$.
So let $q_1$ be the value associated to the coset $C_1$,
and $q_2$ the one for $C_2$.
Then we can take the product to be the coset corresponding to $q_1q_2$.
Quite importantly,
we can also do this in terms of representatives of the cosets.
Let $g_1 C_1$ and $g_2 C_2$,
so $C_1 = g_1N$ and $C_2 = g_2N$.
Then $C_1 C_2$ should be the coset which contains $g_1g_2$.
This is the same as the above definition since
$(g_1g_2) = (g_1)(g_2) = q_1q_2$;
all we've done is define the product in terms of elements of $G$,
rather than values in $H$.
Using the $gN$ notation,
and with remark:coset_warning in mind,
we can write this even more succinctly:
\[ (g_1N) (g_2N) (g_1g_2)N. \]
And now you know why the integers modulo $n$ are often written $/n$!
Take a moment to digest the above definition.
By the way we've built it, the resulting group $G/N$ is isomorphic to $Q$.
In a sense we think of $G/N$ as ``$G$ modulo the condition that $n=1$
for all $n N$''.
(Optional) Proof of Lagrange's theorem
As an aside, with the language of cosets
we can now show Lagrange's theorem in the general case.
[Lagrange's theorem]
thm:lagrange_grp
Let $G$ be a finite group, and let $H$ be any subgroup.
Then $ H $ divides $ G $.
The proof is very simple: note that the cosets of $H$
all have the same size and form a partition of $G$
(even when $H$ is not necessarily normal).
Hence if $n$ is the number of cosets,
then $n H = G $.
Conclude that $x^ G =1$
by taking $H = <x> G$.
It should be mentioned at this point that
in general, if $G$ is a finite group and $N$ is normal,
then $|G/N| = |G| / |N|$.
Eliminating the homomorphism
Let's look at the last definition of $G/N$ we provided.
The short version is:
The elements of $G/N$ are cosets $gN$, which you can think
of as equivalence classes of a relation $_N$
(where $g_1 _N g_2$ if $g_1 = g_2n$ for some $n N$).
Given cosets $g_1N$ and $g_2N$ the group operation is
\[ g_1N g_2N (g_1g_2)N. \]
Question: where do we actually use the fact that $N$ is normal?
We don't talk about $$ or $Q$ anywhere in this definition.
The answer is in remark:coset_warning.
The group operation takes in two cosets,
so it doesn't know what $g_1$ and $g_2$ are.
But behind the scenes,
the normal condition guarantees that the group operation can pick
any $g_1$ and $g_2$ it wants and still end up with the same coset.
If we didn't have this property, then it would be hard to define the
product of two cosets $C_1$ and $C_2$ because it might make a difference
which $g_1 C_1$ and $g_2 C_2$ we picked.
The fact that $N$ came from a homomorphism meant we could pick any representatives
$g_1$ and $g_2$ of the cosets we wanted, because they all had the same $$-value.
We want some conditions which force this to be true without referencing $$ at all.
Suppose $ G K$ is a homomorphism of groups with $H = $.
Aside from the fact $H$ is a group, we can get an ``obvious'' property:
Show that if $h H$, $g G$,
then $ghg H$.
(Check $(ghg) = 1_K$.)
ex:dihedral_normal_subgroup
Let $D_12 = <r,s r^6=s^2=1, rs=sr>$.
Consider the subgroup of order two $H = \1,s\$
and notice that \[ rsr = r(sr)= r(rs) = r^2s H. \]
Hence $H$ is not normal, and cannot be the kernel of any homomorphism.
Well, duh -- so what?
Amazingly it turns out that this is the sufficient condition we want.
Specifically, it makes the nice ``coset multiplication'' we wanted work out.
[For math contest enthusiasts]
This coincidence is really a lot like functional equations at the IMO.
We all know that normal subgroups $H$ satisfy $ghg H$;
the surprise is that from the latter seemingly weaker condition,
we can deduce $H$ is normal.
Thus we have a new criterion for ``normal'' subgroups which does not
make any external references to $$.
Let $H$ be a subgroup of $G$.
Then the following are equivalent:
$H G$.
For every $g G$ and $h H$, $ghg H$.
We already showed one direction.
For the other direction, we need to build a homomorphism with kernel $H$.
So we simply define the group $G/H$ as the cosets.
To put a group operation, we need to verify:
If $g_1' _H g_1$ and $g_2' _H g_2$ then $g_1'g_2' _H g_1g_2$.
Boring algebraic manipulation (again functional equation style).
Let $g_1' = g_1h_1$ and $g_2' = g_2h_2$, so we want to show that
$g_1h_1g_2h_2 _H g_1g_2$.
Since $H$ has the property, $g_2 h_1g_2$ is some element of $H$, say $h_3$.
Thus $h_1 g_2 = g_2 h_3$, and the left-hand side becomes $g_1g_2(h_3h_2)$,
which is fine since $h_3h_2 H$.
With that settled we can just define the
product of two cosets (of normal subgroups) by \[ (g_1H) (g_2H) = (g_1g_2)H. \]
Thus the claim above shows that this multiplication is well-defined
(this verification is the ``content'' of the theorem).
So $G/H$ is indeed a group!
Moreover there is an obvious ``projection'' homomorphism
$G G/H$ (with kernel $H$), by $g gH$.
%Another way to write the condition is
%\[ H = gHg \ ghg h H \. \]
%You should take a moment to check that these definitions are equivalent. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 7 | 1,563 | |
Consider again the product group $G H$.
Earlier we identified a subgroup
\[ G' = \ (g, 1_H) g G \ G. \]
You can easily see that $G' G H$.
(Easy calculation.)
Moreover, you can check that
\[ (G H) / (G') H. \]
Indeed, we have $(g, h) _G' (1_G, h)$ for all $g G$ and $h H$.
It is not necessarily true that $(G/H) H G$.
For example, consider $G = /4$
and the normal subgroup $H = \0,2\ /2$.
Then $G/H /2$,
but $/4 /2 /2$.
(Footnote: the precise condition for this kind of ``canceling'' is called the Schur-Zassenhaus lemma.)
Let $: D_8 4$ be defined by \[ r 2, s 2. \]
The kernel of this map is $N = \1,r^2,sr,sr^3\$.
We can do a quick computation of all the elements of $D_8$ to get
\[ (1) = (r^2) = (sr) = (sr^3) = 0
and
(r) = (r^3) = (s) = (sr^2) = 2. \]
The two relevant fibers are \[ ( 0) = 1N = r^2N = srN = sr^3N = \1,r^2,sr,sr^3\ \] and
\[ ( 2) = rN = r^3N = sN = sr^2N = \r,r^3,s,sr^2\. \]
So we see that $|D_8/N| = 2$ is a group of order two, or $ 2$.
Indeed, the image of $$ is \[ \ 0, 2 \ 2. \]
Suppose $G$ is abelian.
Why does it follow that any subgroup of $G$ is normal?
Finally here's some food for thought:
suppose one has a group presentation for a group $G$
that uses $n$ generators.
Can you write it as a quotient of the form $F_n / N$,
where $N$ is a normal subgroup of $F_n$?
(Digression) The first isomorphism theorem
sec:first_isomorphism_thm
One quick word about what other sources usually say.
Most textbooks actually define normal using the $ghg H$ property.
Then they define $G/H$ for normal $H$ in the way I did above,
using the coset definition
\[ (g_1H) (g_2H) = g_1g_2H. \]
Using purely algebraic manipulations (like I did) this is well-defined,
and so now you have this group $G/H$ or something.
The underlying homomorphism isn't mentioned at all,
or is just mentioned in passing.
I think this is incredibly dumb.
The normal condition looks like it gets pulled out of thin air
and no one has any clue what's going on,
because no one has any clue what a normal subgroup actually should look like.
Other sources like to also write the so-called first isomorphism theorem.There
is a second and third isomorphism theorem.
But four years after learning about them, I still don't remember what they are.
So I'm guessing they weren't very important.
It goes like this.
[First isomorphism theorem]
Let $ G H$ be a homomorphism.
Then $G / $ is isomorphic to $(G)$.
To me, this is just a clumsier way of stating the same idea.
About the only merit this claim has is that if $$ is injective,
then the image $(G)$ is an isomorphic copy
of $G$ inside the group $H$.
(Try to see this directly!)
This is a pattern we'll often see in other branches of mathematics:
whenever we have an injective structure-preserving map,
often the image of this map will be some ``copy'' of $G$.
(Here ``structure'' refers to the group multiplication,
but we'll see some more other examples of ``types of objects'' later!)
In that sense an injective homomorphism $ G H$
is an embedding of $G$ into $H$.
[18.701 at MIT]
Determine all groups $G$ for which the map $ G G$ defined by
\[ (g) = g^2 \] is a homomorphism.
Write it out: $(ab) = (a)(b)$.
Abelian groups: $abab =a^2b^2 ab= ba$.
Consider the dihedral group $G = D_10$.
Is $H = < r >$ a normal subgroup of $G$?
If so, compute $G/H$ up to isomorphism.
Is $H = < s >$ a normal subgroup of $G$?
If so, compute $G/H$ up to isomorphism.
Yes, no.
Yes to (a): you can check this directly
from the $ghg$ definition.
For example, for (a)
it is enough to compute $(r^a s) r^n (r^a s) = r^-n H$.
The quotient group is $2$.
The answer is no for (b) by following
ex:dihedral_normal_subgroup.
Does $S_4$ have a normal subgroup of order $3$?
No.
A subgroup of order $3$ must be generated by
an element of order $3$, since $3$ is prime.
So we may assume WLOG that $H = < (1\; 2 \; 3) >$
(by renaming elements appropriately).
But then let $g = (3 \; 4)$; one can check $gHg H$.
Let $G$ and $H$ be finite groups, where $ G = 1000$
and $ H = 999$.
Show that a homomorphism $G H$ must be trivial.
$(1000,999)=1$.
$G/ G$ is isomorphic to a subgroup of $H$.
The order of the former divides $1000$;
the order of the latter divides $999$.
This can only occur if $G / G = \1\$
so $ G = G$.
Let $^$ denote the nonzero complex numbers under multiplication.
Show that there are five homomorphisms $5 ^$
but only two homomorphisms $D_10 ^$,
even though $5$ is a subgroup of $D_10$.
Find a non-abelian group $G$
such that every subgroup of $G$ is normal.
(These groups are called Hamiltonian.)
Find an example of order $8$.
Quaternion group.
[PRIMES entrance exam, 2018]
Let $G$ be a group with presentation given by
\[ G = < a,b,c
ab = c^2a^4, \; bc = ca^6, \; ac = ca^8, \;
c^2018 = b^2019
>. \]
Determine the order of $G$.
Try to show $G$ is the dihedral group of order $18$.
There is not much group theory content here --- just manipulation.
The answer is $|G| = 18$.
First, observe that by induction we have
\[ a^n c = ca^8n \]
for all $n 1$.
We then note that
a(bc) &= (ab)c \\
a ca^6 &= c^2 a^4 c \\
c a^8 a^6 &= c^2 a^4 c \\
a^14 &= c(a^4c) = c^2 a^32.
Hence we conclude $c^2 = a^-18$.
Then $ab = c^2a^4 b = a^-15$.
In that case, since $c^2018 = b^2019$,
we conclude $1 = a^- 1009 18 + 2019 15 = a^12123$.
Finally,
bc &= ca^6 \\
a^-15 c &= ca^6 \\
a^-15 c^2 &= c(a^6c) = c^2 a^48 \\
a^-33 &= a^30 \\
a^63 &= 1. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 8 | 1,537 | |
Since $(12123, 63) = 9$, we find $a^9 = 1$, hence finally $c^2 = 1$.
So the presentation above simplifies to
\[ G = < a,c a^9=c^2=1, \; ac = ca^-1 > \]
which is the presentation of the dihedral group of order $18$.
% a = r, r^9 = 1
% b = r^3
% c = s, s^2 = 2
This completes the proof.
[Homophony group]
The homophony group (of English) is the group
with $26$ generators $a$, $b$, , $z$
and one relation for every pair of English words
which sound the same.
For example $knight = night$ (and hence $k=1$).
Prove that the group is trivial.
Get yourself a list of English homophones, I guess.
Don't try too hard.
Letter $v$ is the worst; maybe $felt = veldt$?
You can find many solutions by searching ``homophone group'';
one is https://math.stackexchange.com/q/843966/229197.
% Copying math.SE content:
% F:=FreeGroup("a","b","c","d","e","f","g","h","i","j","k","l","m",
% "n","o","p","q","r","s","t","u","v","w","x","y","z");
%
% AssignGeneratorVariables(F);
%
% pairs:=[
% "bye=by", # e=1
% "lead=led", # a=1
% "maid=made", # mid=md => i=1
% "sow=sew", # o=e=1
% "buy=by", # u=1
% "sow=so", # w=1
% "lye=lie", # y=1
% "hour=our", # h=1
% "knight=night", # k=1
% "damn=dam", # n=1
% "psalter=salter", # p=1
% "plumb=plum", # b=1
% "bass=base", # s=1
% "butt=but", # t=1
% "tolled=told", # l=1
% "barred=bard", # r=1
% "dammed=damned", # m=1
% "chased=chaste", # d=1
% "sign=sine", # g=1
% "daze=days", # z=1
% "cite=sight", # c=1
% "jeans=genes", # j=1
% "queue=cue", # q=1
% "tax=tacks", # x=1
% # "ruff=rough", # f=1, we need not this relation used in the paper
% "phase=faze", # f=1
% "chivvy=chivy"]; # v=1
%
% rels:=List( pairs, r -> ParseRelators(GeneratorsOfGroup(F),r)[1]);
% G:=F/rels; | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Homomorphisms and quotient groups | 07_quotient.md | 9 | 485 | |
In this chapter we'll introduce the notion of a **commutative ring** $R$. It is a larger structure than a group: it will have two operations addition and multiplication, rather than just one. We will then immediately define a **ring homomorphism** $R \to S$ between pairs of rings.
This time, instead of having normal subgroups $H \trianglelefteq G$, rings will instead have subsets $I \subseteq R$ called **ideals**, which are not themselves rings but satisfy some niceness conditions. We will then show you how to define $R/I$, in analogy to $G/H$ as before. Finally, like with groups, we will talk a bit about how to generate ideals.
Here is a possibly helpful table of analogies to help you keep track:
Group Ring
--------------- -------------------- ---------------------
Notation $G$ $R$
Operations $\cdot$ $+$, $\times$
Commutativity only if abelian for us, always
Sub-structure subgroup (not discussed)
Homomorphism grp hom. $G \to H$ ring hom. $R \to S$
Kernel normal subgroup ideal
Quotient $G/H$ $R/I$
In this chapter we'll introduce the notion of a **commutative ring** $R$. It is a larger structure than a group: it will have two operations addition and multiplication, rather than just one. We will then immediately define a **ring homomorphism** $R \to S$ between pairs of rings.
This time, instead of having normal subgroups $H \trianglelefteq G$, rings will instead have subsets $I \subseteq R$ called **ideals**, which are not themselves rings but satisfy some niceness conditions. We will then show you how to define $R/I$, in analogy to $G/H$ as before. Finally, like with groups, we will talk a bit about how to generate ideals.
Here is a possibly helpful table of analogies to help you keep track:
Group Ring
--------------- -------------------- ---------------------
Notation $G$ $R$
Operations $\cdot$ $+$, $\times$
Commutativity only if abelian for us, always
Sub-structure subgroup (not discussed)
Homomorphism grp hom. $G \to H$ ring hom. $R \to S$
Kernel normal subgroup ideal
Quotient $G/H$ $R/I$ | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Some motivational metaphors about rings vs groups | 08_ring-intro.md | 0 | 580 |
I wrote most of these examples with a number theoretic eye in mind; thus if you liked elementary number theory, a lot of your intuition will carry over. Basically, we'll try to generalize properties of the ring $\mathbb{Z}$ to any abelian structure in which we can also multiply. That's why, for example, you can talk about "irreducible polynomials in $\mathbb{Q}[x]$" in the same way you can talk about "primes in $\mathbb{Z}$", or about "factoring polynomials modulo $p$" in the same way we can talk "unique factorization in $\mathbb{Z}$". Even if you only care about $\mathbb{Z}$ (say, you're a number theorist), this has a lot of value: I assure you that trying to solve $x^n+y^n = z^n$ (for $n > 2$) requires going into a ring other than $\mathbb{Z}$!
Thus for all the sections that follow, keep $\mathbb{Z}$ in mind as your prototype.
I mention this here because commutative algebra is *also* closely tied to algebraic geometry. Lots of the ideas in commutative algebra have nice "geometric" interpretations that motivate the definitions, and these connections are explored in the corresponding part later. So, I want to admit outright that this is not the only good way (perhaps not even the most natural one) of motivating what is to follow.
I wrote most of these examples with a number theoretic eye in mind; thus if you liked elementary number theory, a lot of your intuition will carry over. Basically, we'll try to generalize properties of the ring $\mathbb{Z}$ to any abelian structure in which we can also multiply. That's why, for example, you can talk about "irreducible polynomials in $\mathbb{Q}[x]$" in the same way you can talk about "primes in $\mathbb{Z}$", or about "factoring polynomials modulo $p$" in the same way we can talk "unique factorization in $\mathbb{Z}$". Even if you only care about $\mathbb{Z}$ (say, you're a number theorist), this has a lot of value: I assure you that trying to solve $x^n+y^n = z^n$ (for $n > 2$) requires going into a ring other than $\mathbb{Z}$!
Thus for all the sections that follow, keep $\mathbb{Z}$ in mind as your prototype.
I mention this here because commutative algebra is *also* closely tied to algebraic geometry. Lots of the ideas in commutative algebra have nice "geometric" interpretations that motivate the definitions, and these connections are explored in the corresponding part later. So, I want to admit outright that this is not the only good way (perhaps not even the most natural one) of motivating what is to follow. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | (Optional) Pedagogical notes on motivation | 08_ring-intro.md | 1 | 715 |
Well, I guess I'll define a ring[^1].
A **ring** is a triple $(R, +, \times)$, the two operations usually called addition and multiplication, such that
(i) $(R,+)$ is an abelian group, with identity $0_R$, or just $0$.
(ii) $\times$ is an associative, binary operation on $R$ with some identity, written $1_R$ or just $1$.
(iii) Multiplication distributes over addition.
The ring $R$ is **commutative** if $\times$ is commutative.
As usual, we will abbreviate $(R, +, \times)$ to $R$.
For simplicity, assume all rings are commutative for the rest of this chapter. We'll run into some noncommutative rings eventually, but for such rings we won't need the full theory of this chapter anyways.
These definitions are just here for completeness. The examples are much more important.
(a) The sets $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are all rings with the usual addition and multiplication.
(b) The integers modulo $n$ are also a ring with the usual addition and multiplication. We also denote it by $\mathbb{Z}/n\mathbb{Z}$.
Here is also a trivial example.
The **zero ring** is the ring $R$ with a single element. We denote the zero ring by $0$. A ring is **nontrivial** if it is not the zero ring.
Show that a ring is nontrivial if and only if $0_R \neq 1_R$.
Since I've defined this structure, I may as well state the obligatory facts about it.
For any ring $R$ and $r \in R$, $r \cdot 0_R = 0_R$. Moreover, $r \cdot (-1_R) = -r$.
Here are some more examples of rings.
Given two rings $R$ and $S$ the **product ring**, denoted $R \times S$, is defined as ordered pairs $(r,s)$ with both operations done component-wise. For example, the Chinese remainder theorem says that $$\mathbb{Z}/15\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$$ with the isomorphism $n \bmod{15} \mapsto (n \bmod 3, n \bmod 5)$.
Equivalently, we can define $R \times S$ as the abelian group $R \oplus S$, and endow it with the multiplication where $r \cdot s = 0$ for $r \in R$, $s \in S$.
Which $(r,s)$ is the identity element of the product ring $R \times S$?
Given any ring $R$, the **polynomial ring** $R[x]$ is defined as the set of polynomials with coefficients in $R$: $$R[x] = \left\{ a_n x^n+a_{n-1}x^{n-1}+\dots+a_0
\mid a_0, \dots, a_n \in R \right\}.$$ This is pronounced "$R$ adjoin $x$". Addition and multiplication are done exactly in the way you would expect.
Happily, polynomial division also does what we expect: if $p \in R[x]$ is a polynomial, and $p(a) = 0$, then $(x-a)q(x) = p(x)$ for some polynomial $q$. Proof: do polynomial long division.
With that, note the caveat that $$x^2-1 \equiv (x-1)(x+1) \pmod 8$$ has *four* roots $1$, $3$, $5$, $7$ in $\mathbb{Z}/8\mathbb{Z}$.
The problem is that $2 \cdot 4 = 0$ even though $2$ and $4$ are not zero; we call $2$ and $4$ *zero divisors* for that reason. In an *integral domain* (a ring without zero divisors), this pathology goes away, and just about everything you know about polynomials carries over. (I'll say this all again next section.)
We can consider polynomials in $n$ variables with coefficients in $R$, denoted $R[x_1, \dots, x_n]$. (We can even adjoin infinitely many $x$'s if we like!)
The **Gaussian integers** are the set of complex numbers with integer real and imaginary parts, that is $$\mathbb{Z}[i] = \left\{ a+bi \mid a,b \in \mathbb{Z} \right\}.$$
Careful readers will detect some abuse in notation here. $\mathbb{Z}[i]$ should officially be "integer-coefficient polynomials in a variable $i$". However, it is understood from context that $i^2=-1$; and a polynomial in $i = \sqrt{-1}$ "is" a Gaussian integer.
As another example (using the same abuse of notation): $$\mathbb{Z}[\sqrt[3]{2}] = \left\{ a + b\sqrt[3]{2} + c\sqrt[3]4
\mid a,b,c \in \mathbb{Z} \right\}.$$ | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Definition and examples of rings | 08_ring-intro.md | 2 | 1,087 |
Well, I guess I'll define a ring[^1].
A **ring** is a triple $(R, +, \times)$, the two operations usually called addition and multiplication, such that
(i) $(R,+)$ is an abelian group, with identity $0_R$, or just $0$.
(ii) $\times$ is an associative, binary operation on $R$ with some identity, written $1_R$ or just $1$.
(iii) Multiplication distributes over addition.
The ring $R$ is **commutative** if $\times$ is commutative.
As usual, we will abbreviate $(R, +, \times)$ to $R$.
For simplicity, assume all rings are commutative for the rest of this chapter. We'll run into some noncommutative rings eventually, but for such rings we won't need the full theory of this chapter anyways.
These definitions are just here for completeness. The examples are much more important.
(a) The sets $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are all rings with the usual addition and multiplication.
(b) The integers modulo $n$ are also a ring with the usual addition and multiplication. We also denote it by $\mathbb{Z}/n\mathbb{Z}$.
Here is also a trivial example.
The **zero ring** is the ring $R$ with a single element. We denote the zero ring by $0$. A ring is **nontrivial** if it is not the zero ring.
Show that a ring is nontrivial if and only if $0_R \neq 1_R$.
Since I've defined this structure, I may as well state the obligatory facts about it.
For any ring $R$ and $r \in R$, $r \cdot 0_R = 0_R$. Moreover, $r \cdot (-1_R) = -r$.
Here are some more examples of rings.
Given two rings $R$ and $S$ the **product ring**, denoted $R \times S$, is defined as ordered pairs $(r,s)$ with both operations done component-wise. For example, the Chinese remainder theorem says that $$\mathbb{Z}/15\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$$ with the isomorphism $n \bmod{15} \mapsto (n \bmod 3, n \bmod 5)$.
Equivalently, we can define $R \times S$ as the abelian group $R \oplus S$, and endow it with the multiplication where $r \cdot s = 0$ for $r \in R$, $s \in S$.
Which $(r,s)$ is the identity element of the product ring $R \times S$?
Given any ring $R$, the **polynomial ring** $R[x]$ is defined as the set of polynomials with coefficients in $R$: $$R[x] = \left\{ a_n x^n+a_{n-1}x^{n-1}+\dots+a_0
\mid a_0, \dots, a_n \in R \right\}.$$ This is pronounced "$R$ adjoin $x$". Addition and multiplication are done exactly in the way you would expect.
Happily, polynomial division also does what we expect: if $p \in R[x]$ is a polynomial, and $p(a) = 0$, then $(x-a)q(x) = p(x)$ for some polynomial $q$. Proof: do polynomial long division.
With that, note the caveat that $$x^2-1 \equiv (x-1)(x+1) \pmod 8$$ has *four* roots $1$, $3$, $5$, $7$ in $\mathbb{Z}/8\mathbb{Z}$.
The problem is that $2 \cdot 4 = 0$ even though $2$ and $4$ are not zero; we call $2$ and $4$ *zero divisors* for that reason. In an *integral domain* (a ring without zero divisors), this pathology goes away, and just about everything you know about polynomials carries over. (I'll say this all again next section.)
We can consider polynomials in $n$ variables with coefficients in $R$, denoted $R[x_1, \dots, x_n]$. (We can even adjoin infinitely many $x$'s if we like!)
The **Gaussian integers** are the set of complex numbers with integer real and imaginary parts, that is $$\mathbb{Z}[i] = \left\{ a+bi \mid a,b \in \mathbb{Z} \right\}.$$
Careful readers will detect some abuse in notation here. $\mathbb{Z}[i]$ should officially be "integer-coefficient polynomials in a variable $i$". However, it is understood from context that $i^2=-1$; and a polynomial in $i = \sqrt{-1}$ "is" a Gaussian integer.
As another example (using the same abuse of notation): $$\mathbb{Z}[\sqrt[3]{2}] = \left\{ a + b\sqrt[3]{2} + c\sqrt[3]4
\mid a,b,c \in \mathbb{Z} \right\}.$$ | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Definition and examples of rings | 08_ring-intro.md | 3 | 1,087 |
Although we won't need to know what a field is until next chapter, they're so convenient for examples I will go ahead and introduce them now.
As you might already know, if the multiplication is invertible, then we call the ring a field. To be explicit, let me write the relevant definitions.
A **unit** of a ring $R$ is an element $u \in R$ which is invertible: for some $x \in R$ we have $ux = 1_R$.
(a) The units of $\mathbb{Z}$ are $\pm 1$, because these are the only things which "divide $1$" (which is the reason for the name "unit").
(b) On the other hand, in $\mathbb{Q}$ everything is a unit (except $0$). For example, $\frac 35$ is a unit since $\frac 35 \cdot \frac 53 = 1$.
(c) The Gaussian integers $\mathbb{Z}[i]$ have four units: $\pm 1$ and $\pm i$.
A nontrivial (commutative) ring is a **field** when all its nonzero elements are units.
Colloquially, we say that
A field is a structure where you can add, subtract, multiply, and divide.
Depending on context, they are often denoted either $k$, $K$, $F$.
(a) $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ are fields, since the notion $\frac 1c$ makes sense in them.
(b) If $p$ is a prime, then $\mathbb{Z}/p\mathbb{Z}$ is a field, which we usually denote by $\mathbb{F}_p$.
The trivial ring $0$ is *not* considered a field, since we require fields to be nontrivial.
Although we won't need to know what a field is until next chapter, they're so convenient for examples I will go ahead and introduce them now.
As you might already know, if the multiplication is invertible, then we call the ring a field. To be explicit, let me write the relevant definitions.
A **unit** of a ring $R$ is an element $u \in R$ which is invertible: for some $x \in R$ we have $ux = 1_R$.
(a) The units of $\mathbb{Z}$ are $\pm 1$, because these are the only things which "divide $1$" (which is the reason for the name "unit").
(b) On the other hand, in $\mathbb{Q}$ everything is a unit (except $0$). For example, $\frac 35$ is a unit since $\frac 35 \cdot \frac 53 = 1$.
(c) The Gaussian integers $\mathbb{Z}[i]$ have four units: $\pm 1$ and $\pm i$.
A nontrivial (commutative) ring is a **field** when all its nonzero elements are units.
Colloquially, we say that
A field is a structure where you can add, subtract, multiply, and divide.
Depending on context, they are often denoted either $k$, $K$, $F$.
(a) $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ are fields, since the notion $\frac 1c$ makes sense in them.
(b) If $p$ is a prime, then $\mathbb{Z}/p\mathbb{Z}$ is a field, which we usually denote by $\mathbb{F}_p$.
The trivial ring $0$ is *not* considered a field, since we require fields to be nontrivial. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Fields | 08_ring-intro.md | 4 | 764 |
This section is going to go briskly -- it's the obvious generalization of all the stuff we did with quotient groups.[^2]
First, we define a homomorphism and isomorphism.
Let $R = (R, +_R, \times_R)$ and $S = (S, +_S, \times_S)$ be rings. A **ring homomorphism** is a map $\phi \colon R \to S$ such that
(i) $\phi(x +_R y) = \phi(x) +_S \phi(y)$ for each $x,y \in R$.
(ii) $\phi(x \times_R y) = \phi(x) \times_S \phi(y)$ for each $x,y \in R$.
(iii) $\phi(1_R) = 1_S$.
If $\phi$ is a bijection then $\phi$ is an **isomorphism** and we say that rings $R$ and $S$ are **isomorphic**.
Just what you would expect. The only surprise is that we also demand $\phi(1_R)$ to go to $1_S$. This condition is not extraneous: consider the map $\mathbb{Z} \to \mathbb{Z}$ called "multiply by zero".
(a) The identity map, as always.
(b) The map $\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$ modding out by $5$.
(c) The map $\mathbb{R}[x] \to \mathbb{R}$ by $p(x) \mapsto p(0)$ by taking the constant term.
(d) For any ring $R$, there is a trivial ring homomorphism $R \to 0$.
Because we require $1_R$ to $1_S$, some maps that you might have thought were homomorphisms will fail.
(a) The map $\mathbb{Z} \to \mathbb{Z}$ by $x \mapsto 2x$ is not a ring homomorphism. Aside from the fact it sends $1$ to $2$, it also does not preserve multiplication.
(b) If $S$ is a nontrivial ring, the map $R \to S$ by $x \mapsto 0$ is not a ring homomorphism, even though it preserves multiplication.
(c) There is no ring homomorphism $\mathbb{Z}/2016\mathbb{Z} \to \mathbb{Z}$ at all.
In particular, whereas for groups $G$ and $H$ there was always a trivial group homomorphism sending everything in $G$ to $1_H$, this is not the case for rings.
This section is going to go briskly -- it's the obvious generalization of all the stuff we did with quotient groups.[^2]
First, we define a homomorphism and isomorphism.
Let $R = (R, +_R, \times_R)$ and $S = (S, +_S, \times_S)$ be rings. A **ring homomorphism** is a map $\phi \colon R \to S$ such that
(i) $\phi(x +_R y) = \phi(x) +_S \phi(y)$ for each $x,y \in R$.
(ii) $\phi(x \times_R y) = \phi(x) \times_S \phi(y)$ for each $x,y \in R$.
(iii) $\phi(1_R) = 1_S$.
If $\phi$ is a bijection then $\phi$ is an **isomorphism** and we say that rings $R$ and $S$ are **isomorphic**.
Just what you would expect. The only surprise is that we also demand $\phi(1_R)$ to go to $1_S$. This condition is not extraneous: consider the map $\mathbb{Z} \to \mathbb{Z}$ called "multiply by zero".
(a) The identity map, as always.
(b) The map $\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$ modding out by $5$.
(c) The map $\mathbb{R}[x] \to \mathbb{R}$ by $p(x) \mapsto p(0)$ by taking the constant term.
(d) For any ring $R$, there is a trivial ring homomorphism $R \to 0$.
Because we require $1_R$ to $1_S$, some maps that you might have thought were homomorphisms will fail.
(a) The map $\mathbb{Z} \to \mathbb{Z}$ by $x \mapsto 2x$ is not a ring homomorphism. Aside from the fact it sends $1$ to $2$, it also does not preserve multiplication.
(b) If $S$ is a nontrivial ring, the map $R \to S$ by $x \mapsto 0$ is not a ring homomorphism, even though it preserves multiplication.
(c) There is no ring homomorphism $\mathbb{Z}/2016\mathbb{Z} \to \mathbb{Z}$ at all.
In particular, whereas for groups $G$ and $H$ there was always a trivial group homomorphism sending everything in $G$ to $1_H$, this is not the case for rings. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Homomorphisms | 08_ring-intro.md | 5 | 984 |
Now, just like we were able to mod out by groups, we'd also like to define quotient rings. So once again,
The **kernel** of a ring homomorphism $\phi \colon R \to S$, denoted $\ker \phi$, is the set of $r \in R$ such that $\phi(r) = 0$.
In group theory, we were able to characterize the "normal" subgroups by a few obviously necessary conditions (namely, $gHg^{-1} = H$). We can do the same thing for rings, and it's in fact easier because our operations are commutative.
First, note two obvious facts:
- If $\phi(x) = \phi(y) = 0$, then $\phi(x+y) = 0$ as well. So $\ker \phi$ should be closed under addition.
- If $\phi(x) = 0$, then for any $r \in R$ we have $\phi(rx) = \phi(r)\phi(x) = 0$ too. So for $x \in \ker \phi$ and *any* $r \in R$, we have $rx \in \ker\phi$.
A (nonempty) subset $I \subseteq R$ is called an ideal if it satisfies these properties. That is,
A nonempty subset $I \subseteq R$ is an **ideal** if it is closed under addition, and for each $x \in I$, $rx \in I$ for all $r \in R$. It is **proper** if $I \neq R$.
Note that in the second condition, $r$ need not be in $I$! So this is stronger than merely saying $I$ is closed under multiplication.
If $R$ is not commutative, we also need the condition $xr \in I$. That is, the ideal is *two-sided*: it absorbs multiplication from both the left and the right. But since rings in Napkin are commutative we needn't worry with this distinction.
Consider the set $I = 5\mathbb{Z} = \{\dots,-10,-5,0,5,10,\dots\}$ as an ideal in $\mathbb{Z}$. We indeed see $I$ is the kernel of the "take mod $5$" homomorphism: $$\mathbb{Z} \twoheadrightarrow \mathbb{Z}/5\mathbb{Z}.$$ It's clearly closed under addition, but it absorbs multiplication from *all* elements of $\mathbb{Z}$: given $15 \in I$, $999 \in \mathbb{Z}$, we get $15 \cdot 999 \in I$.
If $K$ is a field, show that $K$ has exactly two ideals. What are they?
Now we claim that these conditions are sufficient. More explicitly,
Let $R$ be a ring and $I \subsetneq R$. Then $I$ is the kernel of some homomorphism if and only if it's an ideal.
*Proof.* It's quite similar to the proof for the normal subgroup thing, and you might try it yourself as an exercise.
Obviously the conditions are necessary. To see they're sufficient, we *define* a ring by "cosets" $$S = \left\{ r + I \mid r \in R \right\}.$$ These are the equivalence classes under $r_1 \sim r_2$ if and only if $r_1 - r_2 \in I$ (think of this as taking "mod $I$"). To see that these form a ring, we have to check that the addition and multiplication we put on them is well-defined. Specifically, we want to check that if $r_1 \sim s_1$ and $r_2 \sim s_2$, then $r_1 + r_2 \sim s_1 + s_2$ and $r_1r_2 \sim s_1s_2$. We actually already did the first part -- just think of $R$ and $S$ as abelian groups, forgetting for the moment that we can multiply. The multiplication is more interesting.
Show that if $r_1 \sim s_1$ and $r_2 \sim s_2$, then $r_1r_2 \sim s_1s_2$. You will need to use the fact that $I$ absorbs multiplication from *any* elements of $R$, not just those in $I$.
Anyways, since this addition and multiplication is well-defined there is now a surjective homomorphism $R \to S$ with kernel exactly $I$. ◻
Given an ideal $I$, we define as above the **quotient ring** $$R/I := \left\{ r+I \mid r \in R \right\}.$$ It's the ring of these equivalence classes. This ring is pronounced "$R$ mod $I$".
The integers modulo $5$ formed by "modding out additively by $5$" are the $\mathbb{Z}/5\mathbb{Z}$ we have already met.
But here's an important point: just as we don't actually think of $\mathbb{Z}/5\mathbb{Z}$ as consisting of $k + 5\mathbb{Z}$ for $k=0,\dots,4$, we also don't really want to think about $R/I$ as elements $r+I$. The better way to think about it is
$R/I$ is the result when we declare that elements of $I$ are all zero; that is, we "mod out by elements of $I$".
For example, modding out by $5\mathbb{Z}$ means that we consider all elements in $\mathbb{Z}$ divisible by $5$ to be zero. This gives you the usual modular arithmetic!
Earlier, we wrote $\mathbb{Z}[i]$ for the Gaussian integers, which was a slight abuse of notation. Convince yourself that this ring could instead be written as $\mathbb{Z}[x] / (x^2+1)$, if we wanted to be perfectly formal. (We will stick with $\mathbb{Z}[i]$ though --- it's more natural.) Here the shorthand $(x^2+1) := (x^2+1) \mathbb{Z}[x] = \{ (x^2+1) f \mid f \in \mathbb{Z}[x] \}$ denotes the ideal of multiples of $x^2+1$ within $\mathbb{Z}[x]$.
Figure out the analogous formalization of $\mathbb{Z}[\sqrt[3]{2}]$. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Ideals | 08_ring-intro.md | 6 | 1,309 |
Let's give you some practice with ideals.
An important piece of intuition is that once an ideal contains a unit, it contains $1$, and thus must contain the entire ring. That's why the notion of "proper ideal" is useful language. To expand on that:
Let $R$ be a ring and $I \subseteq R$ an ideal. Then $I$ is proper (i.e. $I \neq R$) if and only if it contains no units of $R$.
*Proof.* Suppose $I$ contains a unit $u$, i.e. an element $u$ with an inverse $u^{-1}$. Then it contains $u \cdot u^{-1} = 1$, and thus $I = R$. Conversely, if $I$ contains no units, it is obviously proper. ◻
As a consequence, if $K$ is a field, then its only ideals are $(0)$ and $K$ (this was ). So for our practice purposes, we'll be working with rings that aren't fields.
First practice: $\mathbb{Z}$.
Show that the only ideals of $\mathbb{Z}$ are precisely those sets of the form $n\mathbb{Z}$, where $n$ is a nonnegative integer.
Thus, while ideals of fields are not terribly interesting, ideals of $\mathbb{Z}$ look eerily like elements of $\mathbb{Z}$. Let's make this more precise.
Let $R$ be a ring. The **ideal generated** by a set of elements $x_1, \dots, x_n \in R$ is denoted by $I = (x_1, x_2, \dots, x_n)$ and given by $$I = \left\{ r_1 x_1 + \dots + r_n x_n \mid r_i \in R \right\}.$$ One can think of this as "the smallest ideal containing all the $x_i$".
The analogy of putting the $\{x_i\}$ in a sealed box and shaking vigorously kind of works here too.
If you know linear algebra, you can summarize this as: an ideal is an $R$-module. The ideal $(x_1, \dots, x_n)$ is the submodule spanned by $x_1, \dots, x_n$.
In particular, if $I = (x)$ then $I$ consists of exactly the "multiples of $x$", i.e. numbers of the form $rx$ for $r \in R$.
We can also apply this definition to infinite generating sets, as long as only finitely many of the $r_i$ are not zero (since infinite sums don't make sense in general).
(a) As $(n) = n\mathbb{Z}$ for all $n \in \mathbb{Z}$, every ideal in $\mathbb{Z}$ is of the form $(n)$.
(b) In $\mathbb{Z}[i]$, we have $(5) = \left\{ 5a + 5b i \mid a,b \in \mathbb{Z} \right\}$.
(c) In $\mathbb{Z}[x]$, the ideal $(x)$ consists of polynomials with zero constant terms.
(d) In $\mathbb{Z}[x,y]$, the ideal $(x,y)$ again consists of polynomials with zero constant terms.
(e) In $\mathbb{Z}[x]$, the ideal $(x,5)$ consists of polynomials whose constant term is divisible by $5$.
Please check that the set $I = \left\{ r_1 x_1 + \dots + r_n x_n \mid r_i \in R \right\}$ is indeed always an ideal (closed under addition, and absorbs multiplication).
Now suppose $I = (x_1, \dots, x_n)$. What does $R/I$ look like? According to what I said at the end of the last section, it's what happens when we "mod out" by each of the elements $x_i$. For example...
(a) Let $R = \mathbb{Z}$ and $I = (5)$. Then $R/I$ is literally $\mathbb{Z}/5\mathbb{Z}$, or the "integers modulo $5$": it is the result of declaring $5 = 0$.
(b) Let $R = \mathbb{Z}[x]$ and $I = (x)$. Then $R/I$ means we send $x$ to zero; hence $R/I \cong \mathbb{Z}$ as given any polynomial $p(x) \in R$, we simply get its constant term.
(c) Let $R = \mathbb{Z}[x]$ again and now let $I = (x-3)$. Then $R/I$ should be thought of as the quotient when $x-3 \equiv 0$, that is, $x \equiv 3$. So given a polynomial $p(x)$ its image after we mod out should be thought of as $p(3)$. Again $R/I \cong \mathbb{Z}$, but in a different way.
(d) Finally, let $I = (x-3,5)$. Then $R/I$ not only sends $x$ to three, but also $5$ to zero. So given $p \in R$, we get $p(3) \pmod 5$. Then $R/I \cong \mathbb{Z}/5\mathbb{Z}$.
By the way, given an ideal $I$ of a ring $R$, it's totally legit to write $$x \equiv y \pmod I$$ to mean that $x-y \in I$. Everything you learned about modular arithmetic carries over. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Generating ideals | 08_ring-intro.md | 7 | 1,083 |
What happens if we put multiple generators in an ideal, like $(10,15) \subseteq \mathbb{Z}$? Well, we have by definition that $(10,15)$ is given as a set by $$(10,15) := \left\{ 10x + 15y \mid x,y \in \mathbb{Z} \right\}.$$ If you're good at number theory you'll instantly recognize this as $5\mathbb{Z} = (5)$. Surprise! In $\mathbb{Z}$, the ideal $(a,b)$ is exactly $\gcd(a,b) \mathbb{Z}$. And that's exactly the reason you often see the GCD of two numbers denoted $(a,b)$.
We call such an ideal (one generated by a single element) a **principal ideal**. So, in $\mathbb{Z}$, every ideal is principal. But the same is not true in more general rings.
In $\mathbb{Z}[x]$, $I = (x,2015)$ is *not* a principal ideal.
For if $I = (f)$ for some polynomial $f \in I$ then $f$ divides $x$ and $2015$. This can only occur if $f = \pm 1$, but then $I$ contains $\pm1$, which it does not.
A ring with the property that all its ideals are principal is called a **principal ideal ring**. We like this property because they effectively let us take the "greatest common factor" in a similar way as the GCD in $\mathbb{Z}$.
In practice, we actually usually care about so-called **principal ideal domains (PID's)**. But we haven't defined what a domain is yet. Nonetheless, all the examples below are actually PID's, so we will go ahead and use this word for now, and tell you what the additional condition is in the next chapter.
To reiterate, for now you should just verify that these are principal ideal rings, even though we are using the word PID.
(a) As we saw, $\mathbb{Z}$ is a PID.
(b) As we also saw, $\mathbb{Z}[x]$ is not a PID, since $I = (x,2015)$ for example is not principal.
(c) It turns out that for a field $k$ the ring $k[x]$ is always a PID. For example, $\mathbb{Q}[x]$, $\mathbb{R}[x]$, $\mathbb{C}[x]$ are PID's.
If you want to try and prove this, first prove an analog of Bézout's lemma, which implies the result.
(d) $\mathbb{C}[x,y]$ is not a PID, because $(x,y)$ is not principal.
What happens if we put multiple generators in an ideal, like $(10,15) \subseteq \mathbb{Z}$? Well, we have by definition that $(10,15)$ is given as a set by $$(10,15) := \left\{ 10x + 15y \mid x,y \in \mathbb{Z} \right\}.$$ If you're good at number theory you'll instantly recognize this as $5\mathbb{Z} = (5)$. Surprise! In $\mathbb{Z}$, the ideal $(a,b)$ is exactly $\gcd(a,b) \mathbb{Z}$. And that's exactly the reason you often see the GCD of two numbers denoted $(a,b)$.
We call such an ideal (one generated by a single element) a **principal ideal**. So, in $\mathbb{Z}$, every ideal is principal. But the same is not true in more general rings.
In $\mathbb{Z}[x]$, $I = (x,2015)$ is *not* a principal ideal.
For if $I = (f)$ for some polynomial $f \in I$ then $f$ divides $x$ and $2015$. This can only occur if $f = \pm 1$, but then $I$ contains $\pm1$, which it does not.
A ring with the property that all its ideals are principal is called a **principal ideal ring**. We like this property because they effectively let us take the "greatest common factor" in a similar way as the GCD in $\mathbb{Z}$.
In practice, we actually usually care about so-called **principal ideal domains (PID's)**. But we haven't defined what a domain is yet. Nonetheless, all the examples below are actually PID's, so we will go ahead and use this word for now, and tell you what the additional condition is in the next chapter.
To reiterate, for now you should just verify that these are principal ideal rings, even though we are using the word PID.
(a) As we saw, $\mathbb{Z}$ is a PID.
(b) As we also saw, $\mathbb{Z}[x]$ is not a PID, since $I = (x,2015)$ for example is not principal.
(c) It turns out that for a field $k$ the ring $k[x]$ is always a PID. For example, $\mathbb{Q}[x]$, $\mathbb{R}[x]$, $\mathbb{C}[x]$ are PID's.
If you want to try and prove this, first prove an analog of Bézout's lemma, which implies the result.
(d) $\mathbb{C}[x,y]$ is not a PID, because $(x,y)$ is not principal. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Principal ideal domains | 08_ring-intro.md | 8 | 1,145 |
If it's too much to ask that an ideal is generated by *one* element, perhaps we can at least ask that our ideals are generated by *finitely many* elements. Unfortunately, in certain weird rings this is also not the case.
Consider the ring $R = \mathbb{Z}[x_1, x_2, x_3, \dots]$ which has *infinitely* many free variables. Then the ideal $I = (x_1, x_2, \dots) \subseteq R$ cannot be written with a finite generating set.
Nonetheless, most "sane" rings we work in *do* have the property that their ideals are finitely generated. We now name such rings and give two equivalent definitions:
For a ring $R$, the following are equivalent:
(a) Every ideal $I$ of $R$ is finitely generated (i.e. can be written with a finite generating set).
(b) There does *not* exist an infinite ascending chain of ideals $$I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \dots.$$ The absence of such chains is often called the **ascending chain condition**.
Such rings are called **Noetherian**.
In the ring $R = \mathbb{Z}[x_1, x_2, x_3, \dots]$ we have an infinite ascending chain $$(x_1) \subsetneq (x_1, x_2) \subsetneq (x_1,x_2,x_3) \subsetneq \dots.$$
From the example, you can kind of see why the proposition is true: from an infinitely generated ideal you can extract an ascending chain by throwing elements in one at a time. I'll leave the proof to you if you want to do it.[^3]
Why are fields Noetherian? Why are PID's (such as $\mathbb{Z}$) Noetherian?
This leaves the question: is our prototypical non-example of a PID, $\mathbb{Z}[x]$, a Noetherian ring? The answer is a glorious yes, according to the celebrated Hilbert basis theorem.
Given a Noetherian ring $R$, the ring $R[x]$ is also Noetherian. Thus by induction, $R[x_1, x_2, \dots, x_n]$ is Noetherian for any integer $n$.
The proof of this theorem is really olympiad flavored, so I couldn't possibly spoil it -- I've left it as a problem at the end of this chapter.
Noetherian rings really shine in algebraic geometry, and it's a bit hard for me to motivate them right now, other than to say "most rings you'll encounter are Noetherian". Please bear with me!
The ring $R = \mathbb{R}[x] / (x^2+1)$ is one that you've seen before. What is its name?
$R = \mathbb{R}[i]$.
This is just $\mathbb{R}[i] = \mathbb{C}$. The isomorphism is given by $x \mapsto i$, which has kernel $(x^2+1)$.
Show that $\mathbb{C}[x] / (x^2-x) \cong \mathbb{C} \times \mathbb{C}$.
Show that the map $$\begin{align*}
\mathbb{C}[x] &\to \mathbb{C} \times \mathbb{C} \\
p &\mapsto \left( p(0), p(1) \right)
\end{align*}$$ is surjective and calculate its kernel.
Note that the map $$\begin{align*}
\mathbb{C}[x] &\to \mathbb{C} \times \mathbb{C} \\
p &\mapsto \left( p(0), p(1) \right)
\end{align*}$$ is indeed a surjective ring homomorphism. Its kernel consists of those polynomials $p$ such that $p(0) = p(1) = 0$; this is the set of polynomials divisible by both $x$ and $x-1$, so it is $x(x-1)$.
In the ring $\mathbb{Z}$, let $I = (2016)$ and $J = (30)$. Show that $I \cap J$ is an ideal of $\mathbb{Z}$ and compute its elements.
Let $R$ be a ring and $I$ an ideal. Find an inclusion-preserving bijection between
- ideals of $R/I$, and
- ideals of $R$ which contain $I$.
Let $R$ be a ring.
(a) Prove that there is exactly one ring homomorphism $\mathbb{Z} \to R$.
(b) Prove that the number of ring homomorphisms $\mathbb{Z}[x] \to R$ is equal to the number of elements of $R$.
For (b) homomorphism is uniquely determined by the choice of $\psi(x) \in R$
Prove the Hilbert basis theorem, .
Let $\mathbb{F}_p$ denote the integers modulo a fixed prime number $p$. Define $\Psi \colon \mathbb{F}_p[x] \to \mathbb{F}_p[x]$ by $$\Psi\left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n a_i x^{p^i}.$$ Let $S$ denote the image of $\Psi$.
(a) Show that $S$ is a ring with addition given by polynomial addition, and multiplication given by *function composition*.
(b) Prove that $\Psi \colon \mathbb{F}_p[x] \to S$ is then a ring isomorphism.
We say a ring $R$ is **Artinian** if it satisfies the **descending chain condition**: there does *not* exist an infinite descending chain of ideals $I_1 \supsetneq I_2 \supsetneq I_3 \supsetneq \dots$. Show that if a ring $R$ is Artinian, then it's Noetherian.
(Artinian rings are better understood in the context of algebraic geometry, even more so than Noetherian ones --- that is, shows a much better definition of Artinian ring. This problem is more here as a teaser --- it looks like it should be easy, because the ascending chain condition and descending chain condition are both simple and similar-looking, but turns out to be difficult.)
[^1]: Or, according to some authors, a "ring with identity"; some authors don't require rings to have multiplicative identity. For us, "ring" always means "ring with $1$".
[^2]: I once found an abstract algebra textbook which teaches rings before groups. At the time I didn't understand why, but now I think I get it -- modding out by things in commutative rings is far more natural, and you can start talking about all the various flavors of rings and fields. You also have (in my opinion) more vivid first examples for rings than for groups. I actually sympathize a lot with this approach --- maybe I'll convert Napkin to follow it one day.
[^3]: On the other hand, every undergraduate class in this topic I've seen makes you do it as homework. Admittedly I haven't gone to that many such classes. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Noetherian rings | 08_ring-intro.md | 9 | 1,552 |
If it's too much to ask that an ideal is generated by *one* element, perhaps we can at least ask that our ideals are generated by *finitely many* elements. Unfortunately, in certain weird rings this is also not the case.
Consider the ring $R = \mathbb{Z}[x_1, x_2, x_3, \dots]$ which has *infinitely* many free variables. Then the ideal $I = (x_1, x_2, \dots) \subseteq R$ cannot be written with a finite generating set.
Nonetheless, most "sane" rings we work in *do* have the property that their ideals are finitely generated. We now name such rings and give two equivalent definitions:
For a ring $R$, the following are equivalent:
(a) Every ideal $I$ of $R$ is finitely generated (i.e. can be written with a finite generating set).
(b) There does *not* exist an infinite ascending chain of ideals $$I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \dots.$$ The absence of such chains is often called the **ascending chain condition**.
Such rings are called **Noetherian**.
In the ring $R = \mathbb{Z}[x_1, x_2, x_3, \dots]$ we have an infinite ascending chain $$(x_1) \subsetneq (x_1, x_2) \subsetneq (x_1,x_2,x_3) \subsetneq \dots.$$
From the example, you can kind of see why the proposition is true: from an infinitely generated ideal you can extract an ascending chain by throwing elements in one at a time. I'll leave the proof to you if you want to do it.[^3]
Why are fields Noetherian? Why are PID's (such as $\mathbb{Z}$) Noetherian?
This leaves the question: is our prototypical non-example of a PID, $\mathbb{Z}[x]$, a Noetherian ring? The answer is a glorious yes, according to the celebrated Hilbert basis theorem.
Given a Noetherian ring $R$, the ring $R[x]$ is also Noetherian. Thus by induction, $R[x_1, x_2, \dots, x_n]$ is Noetherian for any integer $n$.
The proof of this theorem is really olympiad flavored, so I couldn't possibly spoil it -- I've left it as a problem at the end of this chapter.
Noetherian rings really shine in algebraic geometry, and it's a bit hard for me to motivate them right now, other than to say "most rings you'll encounter are Noetherian". Please bear with me!
The ring $R = \mathbb{R}[x] / (x^2+1)$ is one that you've seen before. What is its name?
$R = \mathbb{R}[i]$.
This is just $\mathbb{R}[i] = \mathbb{C}$. The isomorphism is given by $x \mapsto i$, which has kernel $(x^2+1)$.
Show that $\mathbb{C}[x] / (x^2-x) \cong \mathbb{C} \times \mathbb{C}$.
Show that the map $$\begin{align*}
\mathbb{C}[x] &\to \mathbb{C} \times \mathbb{C} \\
p &\mapsto \left( p(0), p(1) \right)
\end{align*}$$ is surjective and calculate its kernel.
Note that the map $$\begin{align*}
\mathbb{C}[x] &\to \mathbb{C} \times \mathbb{C} \\
p &\mapsto \left( p(0), p(1) \right)
\end{align*}$$ is indeed a surjective ring homomorphism. Its kernel consists of those polynomials $p$ such that $p(0) = p(1) = 0$; this is the set of polynomials divisible by both $x$ and $x-1$, so it is $x(x-1)$.
In the ring $\mathbb{Z}$, let $I = (2016)$ and $J = (30)$. Show that $I \cap J$ is an ideal of $\mathbb{Z}$ and compute its elements.
Let $R$ be a ring and $I$ an ideal. Find an inclusion-preserving bijection between
- ideals of $R/I$, and
- ideals of $R$ which contain $I$.
Let $R$ be a ring.
(a) Prove that there is exactly one ring homomorphism $\mathbb{Z} \to R$.
(b) Prove that the number of ring homomorphisms $\mathbb{Z}[x] \to R$ is equal to the number of elements of $R$.
For (b) homomorphism is uniquely determined by the choice of $\psi(x) \in R$
Prove the Hilbert basis theorem, .
Let $\mathbb{F}_p$ denote the integers modulo a fixed prime number $p$. Define $\Psi \colon \mathbb{F}_p[x] \to \mathbb{F}_p[x]$ by $$\Psi\left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n a_i x^{p^i}.$$ Let $S$ denote the image of $\Psi$.
(a) Show that $S$ is a ring with addition given by polynomial addition, and multiplication given by *function composition*.
(b) Prove that $\Psi \colon \mathbb{F}_p[x] \to S$ is then a ring isomorphism.
We say a ring $R$ is **Artinian** if it satisfies the **descending chain condition**: there does *not* exist an infinite descending chain of ideals $I_1 \supsetneq I_2 \supsetneq I_3 \supsetneq \dots$. Show that if a ring $R$ is Artinian, then it's Noetherian.
(Artinian rings are better understood in the context of algebraic geometry, even more so than Noetherian ones --- that is, shows a much better definition of Artinian ring. This problem is more here as a teaser --- it looks like it should be easy, because the ascending chain condition and descending chain condition are both simple and similar-looking, but turns out to be difficult.)
[^1]: Or, according to some authors, a "ring with identity"; some authors don't require rings to have multiplicative identity. For us, "ring" always means "ring with $1$".
[^2]: I once found an abstract algebra textbook which teaches rings before groups. At the time I didn't understand why, but now I think I get it -- modding out by things in commutative rings is far more natural, and you can start talking about all the various flavors of rings and fields. You also have (in my opinion) more vivid first examples for rings than for groups. I actually sympathize a lot with this approach --- maybe I'll convert Napkin to follow it one day.
[^3]: On the other hand, every undergraduate class in this topic I've seen makes you do it as homework. Admittedly I haven't gone to that many such classes. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Rings and ideals | Noetherian rings | 08_ring-intro.md | 10 | 1,552 |
We continue our exploration of rings by considering some nice-ness properties that rings or ideals can satisfy, which will be valuable later on. As before, number theory is interlaced as motivation. I guess I can tell you at the outset what the completed table is going to look like, so you know what to expect.
Ring noun Ideal adjective Relation
----------------- -------------------- -------------------------------------------------
PID principal $R$ is a PID $\iff$ $R$ is an integral domain,
and every $I$ is principal
Noetherian ring finitely generated $R$ is Noetherian $\iff$ every $I$ is fin. gen.
field maximal $R/I$ is a field $\iff$ $I$ is maximal
integral domain prime $R/I$ is an integral domain $\iff$ $I$ is prime
We continue our exploration of rings by considering some nice-ness properties that rings or ideals can satisfy, which will be valuable later on. As before, number theory is interlaced as motivation. I guess I can tell you at the outset what the completed table is going to look like, so you know what to expect.
Ring noun Ideal adjective Relation
----------------- -------------------- -------------------------------------------------
PID principal $R$ is a PID $\iff$ $R$ is an integral domain,
and every $I$ is principal
Noetherian ring finitely generated $R$ is Noetherian $\iff$ every $I$ is fin. gen.
field maximal $R/I$ is a field $\iff$ $I$ is maximal
integral domain prime $R/I$ is an integral domain $\iff$ $I$ is prime | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | 09_ring-flavors.md | 0 | 417 | |
We already saw this definition last chapter: a field $K$ is a nontrivial ring for which every nonzero element is a unit.
In particular, there are only two ideals in a field: the ideal $(0)$, which is maximal, and the entire field $K$.
We already saw this definition last chapter: a field $K$ is a nontrivial ring for which every nonzero element is a unit.
In particular, there are only two ideals in a field: the ideal $(0)$, which is maximal, and the entire field $K$.
In practice, we are often not so lucky that we have a full-fledged field. Now it would be nice if we could still conclude the zero product property: if $ab = 0$ then either $a = 0$ or $b = 0$. If our ring is a field, this is true: if $b \neq 0$, then we can multiply by $b^{-1}$ to get $a = 0$. But many other rings we consider like $\mathbb{Z}$ and $\mathbb{Z}[x]$ also have this property, despite not having division.
Not all rings though: in $\mathbb{Z}/15\mathbb{Z}$, $$3 \cdot 5 \equiv 0 \pmod{15}.$$ If $a, b \neq 0$ but $ab=0$ then we say $a$ and $b$ are **zero divisors** of the ring $R$. So we give a name to such rings.
A nontrivial ring with no zero divisors is called an **integral domain**.[^1]
Show that a field is an integral domain.
Suppose $ac = bc$ in an integral domain, and $c \neq 0$. Show that $a = b$. (There is no $c^{-1}$ to multiply by, so you have to use the definition.)
Every field is an integral domain, so all the previous examples apply. In addition:
(a) $\mathbb{Z}$ is an integral domain, but it is not a field.
(b) $\mathbb{R}[x]$ is not a field, since there is no polynomial $P(x)$ with $xP(x) = 1$. However, $\mathbb{R}[x]$ is an integral domain, because if $P(x) Q(x) = 0$ then one of $P$ or $Q$ is zero.
(c) $\mathbb{Z}[x]$ is also an example of an integral domain. In fact, $R[x]$ is an integral domain for any integral domain $R$ (why?).
(d) $\mathbb{Z}/n\mathbb{Z}$ is a field (hence integral domain) exactly when $n$ is prime. When $n$ is not prime, it is a ring but not an integral domain.
The trivial ring $0$ is *not* considered an integral domain.
At this point, we go ahead and say:
An integral domain where all ideals are principal is called a **principal ideal domain (PID)**.
Recall that the ideal $(a, b)$ is the ring-analog of the $gcd$ operation, so essentially what this definition is saying is that: If any family of elements $\{ a_i \}$ is taken, then the ideal generated by all of the $a_i$ is in fact generated by a single element $a$.
In other words,
In a PID, you can take the $gcd$ of any collection of elements.
The ring $\mathbb{Z}/6\mathbb{Z}$ is an example of a ring which is a principal ideal ring, but not an integral domain. As we alluded to earlier, we will never really use "principal ideal ring" in any real way: we typically will want to strengthen it to PID.
In practice, we are often not so lucky that we have a full-fledged field. Now it would be nice if we could still conclude the zero product property: if $ab = 0$ then either $a = 0$ or $b = 0$. If our ring is a field, this is true: if $b \neq 0$, then we can multiply by $b^{-1}$ to get $a = 0$. But many other rings we consider like $\mathbb{Z}$ and $\mathbb{Z}[x]$ also have this property, despite not having division.
Not all rings though: in $\mathbb{Z}/15\mathbb{Z}$, $$3 \cdot 5 \equiv 0 \pmod{15}.$$ If $a, b \neq 0$ but $ab=0$ then we say $a$ and $b$ are **zero divisors** of the ring $R$. So we give a name to such rings.
A nontrivial ring with no zero divisors is called an **integral domain**.[^1]
Show that a field is an integral domain.
Suppose $ac = bc$ in an integral domain, and $c \neq 0$. Show that $a = b$. (There is no $c^{-1}$ to multiply by, so you have to use the definition.)
Every field is an integral domain, so all the previous examples apply. In addition:
(a) $\mathbb{Z}$ is an integral domain, but it is not a field.
(b) $\mathbb{R}[x]$ is not a field, since there is no polynomial $P(x)$ with $xP(x) = 1$. However, $\mathbb{R}[x]$ is an integral domain, because if $P(x) Q(x) = 0$ then one of $P$ or $Q$ is zero.
(c) $\mathbb{Z}[x]$ is also an example of an integral domain. In fact, $R[x]$ is an integral domain for any integral domain $R$ (why?).
(d) $\mathbb{Z}/n\mathbb{Z}$ is a field (hence integral domain) exactly when $n$ is prime. When $n$ is not prime, it is a ring but not an integral domain.
The trivial ring $0$ is *not* considered an integral domain.
At this point, we go ahead and say:
An integral domain where all ideals are principal is called a **principal ideal domain (PID)**.
Recall that the ideal $(a, b)$ is the ring-analog of the $gcd$ operation, so essentially what this definition is saying is that: If any family of elements $\{ a_i \}$ is taken, then the ideal generated by all of the $a_i$ is in fact generated by a single element $a$.
In other words,
In a PID, you can take the $gcd$ of any collection of elements.
The ring $\mathbb{Z}/6\mathbb{Z}$ is an example of a ring which is a principal ideal ring, but not an integral domain. As we alluded to earlier, we will never really use "principal ideal ring" in any real way: we typically will want to strengthen it to PID. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Fields | 09_ring-flavors.md | 1 | 1,477 |
We know that every integer can be factored (up to sign) as a unique product of primes; for example $15 = 3 \cdot 5$ and $-10 = -2 \cdot 5$. You might remember the proof involves the so-called Bézout's lemma, which essentially says that $(a,b) = (\gcd(a,b))$; in other words we've carefully used the fact that $\mathbb{Z}$ is a PID.
It turns out that for general rings, the situation is not as nice as factoring elements because most rings are not PID's. The classic example of something going wrong is $$6 = 2 \cdot 3 = \left( 1-\sqrt{-5} \right)\left( 1+\sqrt{-5} \right)$$ in $\mathbb{Z}[\sqrt{-5}]$. Nonetheless, we can sidestep the issue and talk about factoring *ideals*: somehow the example $10 = 2 \cdot 5$ should be $(10) = (2) \cdot (5)$, which says "every multiple of $10$ is the product of a multiple of $2$ and a multiple of $5$". I'd have to tell you then how to multiply two ideals, which I do in the chapter on unique factorization.
Let's at least figure out what primes are. In $\mathbb{Z}$, we have that $p \neq 1$ is prime if whenever $p \mid xy$, either $p \mid x$ or $p \mid y$. We port over this definition to our world of ideals.
A *proper* ideal $I \subsetneq R$ is a **prime ideal** if whenever $xy \in I$, either $x \in I$ or $y \in I$.
The condition that $I$ is proper is analogous to the fact that we don't consider $1$ to be a prime number.
(a) The ideal $(7)$ of $\mathbb{Z}$ is prime.
(b) The ideal $(8)$ of $\mathbb{Z}$ is not prime, since $2 \cdot 4 = 8$.
(c) The ideal $(x)$ of $\mathbb{Z}[x]$ is prime.
(d) The ideal $(x^2)$ of $\mathbb{Z}[x]$ is not prime, since $x \cdot x = x^2$.
(e) The ideal $(3,x)$ of $\mathbb{Z}[x]$ is prime. This is actually easiest to see using below.
(f) The ideal $(5) = 5\mathbb{Z} + 5i\mathbb{Z}$ of $\mathbb{Z}[i]$ is not prime, since the elements $3+i$ and $3-i$ have product $10 \in (5)$, yet neither is itself in $(5)$.
Ideals have the nice property that they get rid of "sign issues". For example, in $\mathbb{Z}$, do we consider $-3$ to be a prime? When phrased with ideals, this annoyance goes away: $(-3) = (3)$. More generally, for a ring $R$, talking about ideals lets us ignore multiplication by a unit. (Note that $-1$ is a unit in $\mathbb{Z}$.)
What do you call a ring $R$ for which the zero ideal $(0)$ is prime?
We also have:
An ideal $I$ is prime if and only if $R/I$ is an integral domain.
Convince yourself the theorem is true; it is just definition chasing. (A possible start is to consider $R = \mathbb{Z}$ and $I = (15)$.)
I now must regrettably inform you that unique factorization is still not true even with the notion of a "prime" ideal (though again I haven't told you how to multiply two ideals yet). But it will become true with some additional assumptions that will arise in algebraic number theory (relevant buzzword: Dedekind domain). | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Prime ideals | 09_ring-flavors.md | 2 | 813 |
We know that every integer can be factored (up to sign) as a unique product of primes; for example $15 = 3 \cdot 5$ and $-10 = -2 \cdot 5$. You might remember the proof involves the so-called Bézout's lemma, which essentially says that $(a,b) = (\gcd(a,b))$; in other words we've carefully used the fact that $\mathbb{Z}$ is a PID.
It turns out that for general rings, the situation is not as nice as factoring elements because most rings are not PID's. The classic example of something going wrong is $$6 = 2 \cdot 3 = \left( 1-\sqrt{-5} \right)\left( 1+\sqrt{-5} \right)$$ in $\mathbb{Z}[\sqrt{-5}]$. Nonetheless, we can sidestep the issue and talk about factoring *ideals*: somehow the example $10 = 2 \cdot 5$ should be $(10) = (2) \cdot (5)$, which says "every multiple of $10$ is the product of a multiple of $2$ and a multiple of $5$". I'd have to tell you then how to multiply two ideals, which I do in the chapter on unique factorization.
Let's at least figure out what primes are. In $\mathbb{Z}$, we have that $p \neq 1$ is prime if whenever $p \mid xy$, either $p \mid x$ or $p \mid y$. We port over this definition to our world of ideals.
A *proper* ideal $I \subsetneq R$ is a **prime ideal** if whenever $xy \in I$, either $x \in I$ or $y \in I$.
The condition that $I$ is proper is analogous to the fact that we don't consider $1$ to be a prime number.
(a) The ideal $(7)$ of $\mathbb{Z}$ is prime.
(b) The ideal $(8)$ of $\mathbb{Z}$ is not prime, since $2 \cdot 4 = 8$.
(c) The ideal $(x)$ of $\mathbb{Z}[x]$ is prime.
(d) The ideal $(x^2)$ of $\mathbb{Z}[x]$ is not prime, since $x \cdot x = x^2$.
(e) The ideal $(3,x)$ of $\mathbb{Z}[x]$ is prime. This is actually easiest to see using below.
(f) The ideal $(5) = 5\mathbb{Z} + 5i\mathbb{Z}$ of $\mathbb{Z}[i]$ is not prime, since the elements $3+i$ and $3-i$ have product $10 \in (5)$, yet neither is itself in $(5)$.
Ideals have the nice property that they get rid of "sign issues". For example, in $\mathbb{Z}$, do we consider $-3$ to be a prime? When phrased with ideals, this annoyance goes away: $(-3) = (3)$. More generally, for a ring $R$, talking about ideals lets us ignore multiplication by a unit. (Note that $-1$ is a unit in $\mathbb{Z}$.)
What do you call a ring $R$ for which the zero ideal $(0)$ is prime?
We also have:
An ideal $I$ is prime if and only if $R/I$ is an integral domain.
Convince yourself the theorem is true; it is just definition chasing. (A possible start is to consider $R = \mathbb{Z}$ and $I = (15)$.)
I now must regrettably inform you that unique factorization is still not true even with the notion of a "prime" ideal (though again I haven't told you how to multiply two ideals yet). But it will become true with some additional assumptions that will arise in algebraic number theory (relevant buzzword: Dedekind domain). | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Prime ideals | 09_ring-flavors.md | 3 | 813 |
Here's another flavor of an ideal.
A proper ideal $I$ of a ring $R$ is **maximal** if it is not contained in any other proper ideal.
(a) The ideal $I = (7)$ of $\mathbb{Z}$ is maximal, because if an ideal $J$ contains $7$ and an element $n$ not in $I$ it must contain $\gcd(7,n) = 1$, and hence $J = \mathbb{Z}$.
(b) The ideal $(x)$ is *not* maximal in $\mathbb{Z}[x]$, because it's contained in $(x,5)$ (among others).
(c) On the other hand, $(x,5)$ is indeed maximal in $\mathbb{Z}[x]$. This is actually easiest to verify using below.
(d) Also, $(x)$ is maximal in $\mathbb{C}[x]$, again appealing to below.
What do you call a ring $R$ for which the zero ideal $(0)$ is maximal?
There's an analogous theorem to the one for prime ideals.
An ideal $I$ is maximal if and only if $R/I$ is a field.
*Proof.* A ring is a field if and only if $(0)$ is the only maximal ideal. So this follows by . ◻
If $I$ is a maximal ideal of a ring $R$, then $I$ is prime.
*Proof.* If $I$ is maximal, then $R/I$ is a field, hence an integral domain, so $I$ is prime. ◻
In practice, because modding out by generated ideals is pretty convenient, this is a very efficient way to check whether an ideal is maximal.
(a) This instantly implies that $(x,5)$ is a maximal ideal in $\mathbb{Z}[x]$, because if we mod out by $x$ and $5$ in $\mathbb{Z}[x]$, we just get $\mathbb{F}_5$, which is a field.
(b) On the other hand, modding out by just $x$ gives $\mathbb{Z}$, which is an integral domain but not a field; that's why $(x)$ is prime but not maximal.
As we saw, any maximal ideal is prime. But now note that $\mathbb{Z}$ has the special property that all of its nonzero prime ideals are also maximal. It's with this condition and a few other minor conditions that you get a so-called *Dedekind domain* where prime factorization of ideals *does* work. More on that later. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Maximal ideals | 09_ring-flavors.md | 4 | 532 |
As long as we are here, we take the time to introduce a useful construction that turns any integral domain into a field.
Given an integral domain $R$, we define its **field of fractions** or **fraction field** $\Frac(R)$ as follows: it consists of elements $a / b$, where $a,b \in R$ and $b \neq 0$. We set $a / b \sim c / d$ if and only if $bc = ad$. Addition and multiplication is defined by $$\begin{align*}
\frac ab + \frac cd &= \frac{ad+bc}{bd} \\
\frac ab \cdot \frac cd &= \frac{ac}{bd}.
\end{align*}$$
In fact everything you know about $\mathbb{Q}$ basically carries over by analogy. You can prove if you want that this indeed a field, but considering how comfortable we are that $\mathbb{Q}$ is well-defined, I wouldn't worry about it...
Let $k$ be a field. We define $k(x) = \Frac(k[x])$ (read "$k$ of $x$"), and call it the **field of rational functions**.
(a) By *definition*, $\Frac(\mathbb{Z}) = \mathbb{Q}$.
(b) The field $\mathbb{R}(x)$ consists of rational functions in $x$: $$\mathbb{R}(x) = \left\{ \frac{f(x)}{g(x)} \mid f,g \in \mathbb{R}[x] \right\}.$$ For example, $\frac{2x}{x^2-3}$ might be a typical element.
Just like we defined $\mathbb{Z}[i]$ by abusing notation, we can also write $\mathbb{Q}(i) = \Frac(\mathbb{Z}[i])$. Officially, it should consist of $$\mathbb{Q}(i) = \left\{ \frac{f(i)}{g(i)} \mid g(i) \neq 0 \right\}$$ for polynomials $f$ and $g$ with rational coefficients. But since $i^2=-1$ this just leads to $$\mathbb{Q}(i) = \left\{ \frac{a+bi}{c+di} \mid a,b,c,d \in \mathbb{Q},
(c,d) \neq (0,0) \right\}.$$ And since $\frac{1}{c+di} = \frac{c-di}{c^2+d^2}$ we end up with $$\mathbb{Q}(i) = \left\{ a+bi \mid a,b \in \mathbb{Q} \right\}.$$
As long as we are here, we take the time to introduce a useful construction that turns any integral domain into a field.
Given an integral domain $R$, we define its **field of fractions** or **fraction field** $\Frac(R)$ as follows: it consists of elements $a / b$, where $a,b \in R$ and $b \neq 0$. We set $a / b \sim c / d$ if and only if $bc = ad$. Addition and multiplication is defined by $$\begin{align*}
\frac ab + \frac cd &= \frac{ad+bc}{bd} \\
\frac ab \cdot \frac cd &= \frac{ac}{bd}.
\end{align*}$$
In fact everything you know about $\mathbb{Q}$ basically carries over by analogy. You can prove if you want that this indeed a field, but considering how comfortable we are that $\mathbb{Q}$ is well-defined, I wouldn't worry about it...
Let $k$ be a field. We define $k(x) = \Frac(k[x])$ (read "$k$ of $x$"), and call it the **field of rational functions**.
(a) By *definition*, $\Frac(\mathbb{Z}) = \mathbb{Q}$.
(b) The field $\mathbb{R}(x)$ consists of rational functions in $x$: $$\mathbb{R}(x) = \left\{ \frac{f(x)}{g(x)} \mid f,g \in \mathbb{R}[x] \right\}.$$ For example, $\frac{2x}{x^2-3}$ might be a typical element.
Just like we defined $\mathbb{Z}[i]$ by abusing notation, we can also write $\mathbb{Q}(i) = \Frac(\mathbb{Z}[i])$. Officially, it should consist of $$\mathbb{Q}(i) = \left\{ \frac{f(i)}{g(i)} \mid g(i) \neq 0 \right\}$$ for polynomials $f$ and $g$ with rational coefficients. But since $i^2=-1$ this just leads to $$\mathbb{Q}(i) = \left\{ \frac{a+bi}{c+di} \mid a,b,c,d \in \mathbb{Q},
(c,d) \neq (0,0) \right\}.$$ And since $\frac{1}{c+di} = \frac{c-di}{c^2+d^2}$ we end up with $$\mathbb{Q}(i) = \left\{ a+bi \mid a,b \in \mathbb{Q} \right\}.$$ | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Field of fractions | 09_ring-flavors.md | 5 | 966 |
Here is one stray definition that will be important for those with a number-theoretic inclination. Over the positive integers, we have a fundamental theorem of arithmetic, stating that every integer is uniquely the product of prime numbers.
We can even make an analogous statement in $\mathbb{Z}$ or $\mathbb{Z}[i]$, if we allow representations like $6 = (-2)(-3)$ and so on. The trick is that we only consider everything *up to units*; so $6 = (-2)(-3) = 2 \cdot 3$ are considered the same.
The general definition goes as follows.
A nonzero non-unit of an integral domain $R$ is **irreducible** if it cannot be written as the product of two non-units.
An integral domain $R$ is a **unique factorization domain** if every nonzero non-unit of $R$ can be written as the product of irreducible elements, which is unique up to multiplication by units.
Verify that $\mathbb{Z}$ is a UFD.
(a) Fields are a "degenerate" example of UFD's: every nonzero element is a unit, so there is nothing to check.
(b) $\mathbb{Z}$ is a UFD. The irreducible elements are $p$ and $-p$, for example $5$ or $-17$.
(c) $\mathbb{Q}[x]$ is a UFD: polynomials with rational coefficients can be uniquely factored, up to scaling by constants (as the units of $\mathbb{Q}[x]$ are just the rational numbers).
(d) $\mathbb{Z}[x]$ is a UFD.
(e) The Gaussian integers $\mathbb{Z}[i]$ turns out to be a UFD too (and this will be proved in the chapters on algebraic number theory).
(f) $\mathbb{Z}[\sqrt{-5}]$ is the classic non-example of a UFD: one may write $$6 = 2 \cdot 3 = \left( 1-\sqrt{-5} \right)
\left( 1+\sqrt{-5} \right)$$ but each of $2$, $3$, $1 \pm \sqrt{-5}$ is irreducible. (It turns out the right way to fix this is by considering prime *ideals* instead, and this is one big motivation for .)
(g) Theorem we won't prove: if $R$ is a UFD, so is $R[x]$ (and hence by induction so is $R[x,y]$, $R[x,y,z]$, ...).
We have the following theorem:
Let $R$ be a PID. Then $R$ is a UFD.
If we look at the non-example above: $$6 = 2 \cdot 3 = \left( 1-\sqrt{-5} \right) \left( 1+\sqrt{-5} \right)$$ The failure of $\mathbb{Z}[\sqrt{-5}]$ to be a UFD here is reflected by the fact that we cannot decompose any factor into further irreducible factor, indeed, the ideal $$\left( 2, 1 + \sqrt{-5} \right)$$ is not principal --- there is no element that is the $\gcd$ of $2$ and $1 + \sqrt{-5}$.
In a similar manner, we can prove that a PID is an UFD, assuming a decomposition into irreducible factors exist.
Here is one stray definition that will be important for those with a number-theoretic inclination. Over the positive integers, we have a fundamental theorem of arithmetic, stating that every integer is uniquely the product of prime numbers.
We can even make an analogous statement in $\mathbb{Z}$ or $\mathbb{Z}[i]$, if we allow representations like $6 = (-2)(-3)$ and so on. The trick is that we only consider everything *up to units*; so $6 = (-2)(-3) = 2 \cdot 3$ are considered the same.
The general definition goes as follows.
A nonzero non-unit of an integral domain $R$ is **irreducible** if it cannot be written as the product of two non-units.
An integral domain $R$ is a **unique factorization domain** if every nonzero non-unit of $R$ can be written as the product of irreducible elements, which is unique up to multiplication by units.
Verify that $\mathbb{Z}$ is a UFD.
(a) Fields are a "degenerate" example of UFD's: every nonzero element is a unit, so there is nothing to check.
(b) $\mathbb{Z}$ is a UFD. The irreducible elements are $p$ and $-p$, for example $5$ or $-17$.
(c) $\mathbb{Q}[x]$ is a UFD: polynomials with rational coefficients can be uniquely factored, up to scaling by constants (as the units of $\mathbb{Q}[x]$ are just the rational numbers).
(d) $\mathbb{Z}[x]$ is a UFD.
(e) The Gaussian integers $\mathbb{Z}[i]$ turns out to be a UFD too (and this will be proved in the chapters on algebraic number theory).
(f) $\mathbb{Z}[\sqrt{-5}]$ is the classic non-example of a UFD: one may write $$6 = 2 \cdot 3 = \left( 1-\sqrt{-5} \right)
\left( 1+\sqrt{-5} \right)$$ but each of $2$, $3$, $1 \pm \sqrt{-5}$ is irreducible. (It turns out the right way to fix this is by considering prime *ideals* instead, and this is one big motivation for .)
(g) Theorem we won't prove: if $R$ is a UFD, so is $R[x]$ (and hence by induction so is $R[x,y]$, $R[x,y,z]$, ...).
We have the following theorem:
Let $R$ be a PID. Then $R$ is a UFD.
If we look at the non-example above: $$6 = 2 \cdot 3 = \left( 1-\sqrt{-5} \right) \left( 1+\sqrt{-5} \right)$$ The failure of $\mathbb{Z}[\sqrt{-5}]$ to be a UFD here is reflected by the fact that we cannot decompose any factor into further irreducible factor, indeed, the ideal $$\left( 2, 1 + \sqrt{-5} \right)$$ is not principal --- there is no element that is the $\gcd$ of $2$ and $1 + \sqrt{-5}$.
In a similar manner, we can prove that a PID is an UFD, assuming a decomposition into irreducible factors exist. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Unique factorization domains (UFD's) | 09_ring-flavors.md | 6 | 1,422 |
This chapter will not be used later on, but it is of historical interest.
Recall that a PID is a ring where you can take the $\gcd$ of any family of elements.
We all know that the most popular algorithm to compute the $\gcd$ of two elements in $\mathbb{Z}$ is the Euclidean algorithm:
- Start with two integers $a$ and $b$.
- If either of $a$ and $b$ is $0$, we're done. The $\gcd$ is the nonzero element.
- Otherwise, assume $|a| \geq |b|$, divide $a$ by $b$ to get some remainder $r$ such that $|r| < |b|$, replace $a$ with $r$, and continue the algorithm.
This algorithm is very efficient --- it can be proven that the algorithm only takes logarithmically many steps in the size of $a$ and $b$ --- for instance, if $a$ and $b$ are on the order of $10^{100}$, at worst $500$ steps are needed.
Naturally, the following questions are raised:
> On which rings can we perform the same algorithm?
We will see that we can in fact do it on several rings! For example, the ring of Gaussian integers $\mathbb{Z}[i]$, the Eisenstein integers $\mathbb{Z}[\omega]$, and so on.
If we look at the algorithm description above, what makes the algorithm work? It's the absolute value $|\cdot|$ which is used to compare the magnitude of two numbers, and this absolute value satisfies two conditions:
- It outputs nonnegative integer values --- that way, the algorithm will eventually terminate.
- For any two ring elements $a$ and $b$, where $b \neq 0$, there exist some $q$ such that $r = a-qb$ has smaller absolute value than $b$.
So, naturally, for any ring with a similar integer-valued function, we can perform the algorithm. We call a function $\Norm \colon R \to \mathbb{Z}_{\geq 0}$ that satisfies the two conditions above an **Euclidean norm**, and an integral domain $R$ that has a norm an **Euclidean domain**.
On $\mathbb{Z}[i]$, the usual norm $$|a + bi| = a^2 + b^2$$ is a Euclidean norm.
Indeed, for any elements $a$ and $b$ with $b \neq 0$, we can compute the remainder $r$ by dividing $a$ by $b$, let $q$ be the Gaussian integer that is closest to $\frac{a}{b}$ (that is, $|\frac{a}{b}-q|$ is minimized) and let $r = a-b q$, then it can be proven that $|r| < |b|$.
The proof is done by showing $|\frac{a}{b}-q| < 1$ --- if we look at the lattice of points contained in $\mathbb{Z}[i]$ embedded in the complex plane, then for any value of $\frac{a}{b} \in \mathbb{C}$, rounding it to the nearest integer will move it by at most $\frac{\sqrt 2}{2} < 1$.
size(4cm); for(int x=-2; x\<=2;++x) for(int y=-2; y\<=2;++y) dot((x, y), blue); for(var p: new pair\[\] (0.3, 0.4), (-1.1, 1.6), (-1.5, -1.5), ) dot(p, red); draw(p--(round(p.x), round(p.y)), Arrow); graph.xaxis(xmin=-2.4, xmax=2.4); graph.yaxis(ymin=-2.4, ymax=2.4);
Similarly, let $\omega = \frac{1 + \sqrt 3 i}{2}$ (that is $\omega^3 = -1$), then $\mathbb{Z}[\omega]$ is a Euclidean domain with the usual norm $$|a + bi| = a^2 + b^2$$ or equivalently $$|a + b\omega| = a^2 + ab + b^2.$$
As before. This time, the natural norm[^2] will be: $$\Norm_{\mathbb{Q}(\sqrt{11})/\mathbb{Q}}(a + b \sqrt{11}) = (a + b \sqrt{11}) (a - b \sqrt{11})
= a^2 - 11 b^2.$$ Since we need a Euclidean norm, we will take $\Norm(a + b \sqrt{11}) = |a^2 - 11 b^2|.$
Given two elements $a$ and $b$ in $\mathbb{Z}[\sqrt{11}]$ with $b \neq 0$, we will try to compute $r$ such that $\Norm(r) < \Norm(b)$ as $r = a - q b$ as before.
This time around, we cannot draw $\mathbb{Z}[\sqrt{11}]$ as a lattice of points --- it is dense in $\mathbb{R}$ --- so, each point $a + b \sqrt{11}$ will be drawn at the coordinate $(a + b \sqrt{11}, a - b \sqrt{11})$.
The set of points with norm $< 1$ will be drawn below.
size(6cm); pair pointof(real a, real b) return (a+b\*sqrt(11), a-b\*sqrt(11)); real threshold=6.5; bool valid(real x, real y) return -threshold\<=x && x\<=threshold && -threshold\<=y && y\<=threshold; bool valid(pair p) return valid(p.x, p.y);
for(int a=-8; a\<=8;++a) for(int b=-2; b\<=2;++b) var p=pointof(a, b); if(valid(p)) dot(p, blue);
var axisthreshold=threshold+0.2; graph.xaxis(xmin=-axisthreshold, xmax=axisthreshold); graph.yaxis(ymin=-axisthreshold, ymax=axisthreshold);
var t=graph.graph(new real(real a) return 1/a; , -axisthreshold, -1/axisthreshold); var p=red; draw(t, p); draw(scale(-1)\*t, p); draw(scale(1, -1)\*t, p); draw(scale(-1, 1)\*t, p);
real norm(pair p) return abs(p.x\*p.y);
pair\[\] tryround(pair q) pair\[\] l; for(int a=-8; a\<=8;++a) for(int b=-2; b\<=2;++b) var p=pointof(a, b); if(valid(p) && norm(p-q)\<1) l.push(p); l=sort(l, new bool(pair a, pair b) return norm(a-q)\<norm(b-q); ); return l;
for(pair p: new pair\[\] (-2, 1.4), (-1.7, 1.8), (3.1, 3.7), (4, -5), ) var l=tryround(p); draw(p--l\[0\], Arrow); for(var q: l\[1:\]) draw(p--q, Arrow, p=gray); dot(p, red);
for(int a=-8; a\<=8;++a) for(int b=-2; b\<=2;++b) var p=pointof(a, b); if(valid(p)) dot(p, blue);
Instead of a ball (as in imaginary quadratic fields, that is, $\mathbb{Q}(\sqrt{-d})$ for integer $d$), the set of points with norm $1$ forms a hyperbola.
As such, rounding to the nearest point is not always the best way --- nevertheless, it can be proven (by exhaustive case checking, similar to the case of $\mathbb{Z}[i]$) that for every value of $\frac{a}{b} \in \mathbb{Q}(\sqrt{11})$, there is some $q \in \mathbb{Z}[\sqrt{11}]$ such that $\Norm(\frac{a}{b}-q) < 1$. Thus $\Norm$ is a Euclidean norm. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Extra: Euclidean domains | 09_ring-flavors.md | 7 | 1,535 |
That having said, sometimes the natural norm of a Euclidean domain need not be Euclidean. $\mathbb{Z}[\frac{1 + \sqrt{69}}{2}]$ is the first example.
Similarly, in $\mathbb{Q}[x]$ we can let the norm be the degree of a polynomial --- the polynomial division with remainder algorithm will take care of computing the $\gcd$.
Back to the topic of PID. In a Euclidean domain, you can compute the $\gcd$ of any two elements. What about an infinite family of elements?
Turns out the situation is very nice:
A Euclidean domain is a PID.
Actually, we don't need to provide an explicit algorithm to compute the $\gcd$ of an infinite family of elements --- of course any such algorithm cannot terminate in a finite amount of time! --- but we only need to show the $\gcd$ exists, we can cheat our way out.
Note that in the Euclidean algorithm, the norm of the elements *keep decreasing* until one of the elements become $0$. So, if we're given an arbitrary family of elements, we take the ideal generated by these elements --- certainly the $\gcd$ is inside that ideal --- and we *take the nonzero element with the smallest norm*. This is the $\gcd$.
With that intuition in mind, we formalize our proof:
*Proof.* Let $I$ be any ideal. We need to show $I$ is principal.
Let $a$ be a nonzero element with smallest norm in $I$ --- such an element exists because $\mathbb{Z}_{\geq 0}$ is well-ordered.
Show that, for every other elements $b \in I$, then $a \mid b$.
Thus, $I = (a)$, we're done. ◻
Let $R = \mathbb{Z}[\frac{1 + \sqrt{-19}}{2}]$.
With the above example in mind, what can we say about this ring?
This is in fact a principal ideal domain (we will not prove it here), but there is in fact no Euclidean norm on this ring.
We will prove the above claim. The general plan is:
- Show that the existence of a Euclidean norm implies the existence of something that we calls an **universal side divisor**.
- Show that $R$ has no universal side divisor.
- Thus, $R$ cannot have a Euclidean norm.
First, look at the examples above of $\mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$. We see that the units are the elements with the smallest norm.
So far, nothing useful --- every ring has the unit $1$. Next, we look at the elements with the next-smallest norm:
- In $\mathbb{Z}[i]$, the element $1 + i$ has norm $2$, and is an element with smallest norm that is not $0$ or an unit. (The other elements are $\pm 1 \pm i$.)
- In $\mathbb{Z}[\omega]$, the units are $\{ \pm 1, \pm \omega, \pm (\omega-1) \}$ with norm $1$. An elements with next-smallest norms are $1 + \omega$ with norm $3$.
Now, in order to proceed with the proof, we have to define a *side divisor*. Recall that $b$ is a divisor of $a$ if there is some $q$ such that $a = bq$. We say $b$ is a **side divisor** (read: "almost divisor") of $a$ if there is some $q$ such that the remainder $a-bq$ is either $0$ or an unit.
In $\mathbb{Z}[i]$, consider $b = 3 + i$. The set of numbers $a$ for which $b \mid a$ is drawn in red below.
size(8cm); for(int x=-7; x\<=7; ++x) for(int y=-7; y\<=7; ++y) var q=(x+I\*y)/(3+I); dot((x, y), q==(round(q.x), round(q.y)) ? red: mediumgray); graph.xaxis(xmin=-7.4, xmax=7.4, \"$x$\"); graph.yaxis(ymin=-7.4, ymax=7.4, \"$y$\");
If we add an unit to these values of $a$, we get the set of numbers $a$ for which $b$ is a side divisor of $a$, thus give a picture to the concept of "almost divisor".
The points $a$ where $b$ is a side divisor of $a$ is marked in red and blue below.
size(8cm); for(int x=-7; x\<=7; ++x) for(int y=-7; y\<=7; ++y) var q=(x+I\*y)/(3+I); var r=(round(q.x), round(q.y)); dot((x, y), q==r ? red: abs((x+I\*y)-r\*(3+I))\<=1 ? blue: mediumgray); graph.xaxis(xmin=-7.9, xmax=7.9, \"$x$\"); graph.yaxis(ymin=-7.9, ymax=7.9, \"$y$\");
Finally, we define a **universal side divisor** to be a number $b$ such that $b$ is a side divisor of every element $a \in R$.
Drawing a picture similar to the above, in $\mathbb{Z}[i]$, then $2 + i$ and $1 + i$ are side divisors.
size(8cm); for(int x=-7; x\<=7; ++x) for(int y=-7; y\<=7; ++y) var p=x+I\*y, q=p/(2+I), r=(round(q.x), round(q.y)); if(abs(p-r\*(2+I))==1) dot(p, blue); var x=(r\*(2+I)).x, y=(r\*(2+I)).y; if(-7\<=x && x\<=7 && -7\<=y && y\<=7) draw((r\*(2+I))--p, Arrow, DotMargin); for(int x=-7; x\<=7; ++x) for(int y=-7; y\<=7; ++y) var p=x+I\*y, q=p/(2+I), r=(round(q.x), round(q.y)); if(q==r) dot(p, red); graph.xaxis(xmin=-7.9, xmax=7.9, \"$x$\"); graph.yaxis(ymin=-7.9, ymax=7.9, \"$y$\");
Now, the connection between the two concepts considered above.
In a Euclidean domain, the smallest-norm nonzero element $b$ that is not an unit is a universal side divisor.
*Proof.* Just run the Euclidean algorithm between any number and $b$ for one step, the remainder must be $0$ or an unit. ◻
And finally,
There is no universal side divisor in $R$. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Extra: Euclidean domains | 09_ring-flavors.md | 8 | 1,379 |
*Proof.* All numbers are of the form, $\frac{a}{2} + \frac{b\sqrt{-19}}{2}$ where $a$ and $b$ have the same parity. The absolute value of a complex number, defined as $\frac{a^2}{4} + \frac{19b^2}{4}$, is multiplicative and is greater than $1$ for all numbers in $\mathbb Z [\frac{1+\sqrt{-19}}{2}]$ except for $-1, 0, 1$, which have absolute values of $1, 0, 1$, respectively. Since there are no numbers in the ring with absolute values greater than $0$ and less than $1$, all numbers in the ring with absolute values greater than $1$ do not have multiplicative inverses in the ring. Hence, the only units are $1$ and $-1$.
When we multiply all the elements in the ring by $x$, the distances between pairs of elements in the complex plane multiplies by $|x|$. Initially, the distances between pairs of elements were at least $1$.
- If $x$ is real and $|x| \geq 2$, then $xR$ doesn't contain any elements of the form $a + \frac{\sqrt{-19}}{2}$, so $x$ is not a universal side divisor.
- If $x$ is non-real and $|x| \geq 2$, then an element in $xR$ can only be real if it was previously non-real. This means the elements of $xR$ on the real axis have absolute value of at least $2|\frac{1+\sqrt{-19}}{2}| = 2\sqrt{5} \geq 4$. Hence, $x$ is not a side divisor of $2$, so $x$ is not a universal side divisor.
Since $R$ does not have a universal side divisor, $R$ is not an Euclidean domain. ◻
Not olympiad problems, but again the spirit is very close to what you might see in an olympiad.
Consider the ring $$\mathbb{Q}[\sqrt2] = \left\{ a + b\sqrt2 \mid a,b \in \mathbb{Q} \right\}.$$ Is it a field?
Yes.
Let $K$ be a field and $R$ a nontrivial ring. Prove that any homomorphism $\psi \colon K \to R$ is injective.[^3]
The kernel is an ideal of $K$!
Suppose $\phi \colon R \to S$ is a ring homomorphism, and $I \subseteq S$ is a prime ideal. Prove that $\phi\pre(I)$ is prime as well.
This is just a definition chase.
Consider $ab \in \phi\pre(I)$, meaning $\phi(ab) = \phi(a) \phi(b) \in I$. Since $I$ is prime, either $\phi(a) \in I$ or $\phi(b) \in I$. In the former case we get $a \in \phi\pre(I)$ as needed; the latter case we get $b \in \phi\pre(I)$.
Let $R$ be an integral domain with finitely many elements. Prove that $R$ is a field.
Fermat's little theorem type argument; cancellation holds in integral domains.
Let $x \in R$ with $x \neq 0$. Look at the powers $x$, $x^2$, .... By pigeonhole, eventually two of them coincide. So assume $x^m = x^n$ where $m < n$, or equivalently $$0 = x \cdot x \cdot \dots \cdot x
\cdot \left( x^{n-m} - 1 \right).$$ Since $x \neq 0$, we get $x^{n-m} - 1 = 0$, or $x^{n-m} = 1$. So $x^{n-m-1}$ is an inverse for $x$.
This means every nonzero element has an inverse, ergo $R$ is a field.
Let $R$ be a ring and $J$ a proper ideal.
(a) Prove that if $R$ is Noetherian, then $J$ is contained in a maximal ideal $I$.
(b) Use Zorn's lemma () to prove the result even if $R$ isn't Noetherian.
Just keep on adding in elements to get an ascending chain.
For part (b), look at the poset of *proper* ideals. Apply Zorn's lemma (again using a union trick to verify the condition; be sure to verify that the union is proper!). In part (a) we are given no ascending infinite chains, so no need to use Zorn's lemma.
Describe the prime ideals of $\mathbb{C}[x]$ and $\mathbb{R}[x]$.
Use the fact that both are PID's.
The ideal $(0)$ is of course prime in both. Also, both rings are PID's.
For $\mathbb{C}[x]$ we get a prime ideal $(x-z)$ for each $z \in \mathbb{C}$.
For $\mathbb{R}[x]$ a prime ideal $(x-a)$ for each $a \in \mathbb{R}$ and a prime ideal $(x^2 - ax + b)$ for each quadratic with two conjugate non-real roots.
How many prime ideals of $\mathbb{Z}[\sqrt{2017}]$ are *not* maximal ideals?
Show that the quotient $\mathbb{Z}[\sqrt{2017}]/I$ has finitely many elements for any nonzero prime ideal $I$. Therefore, the quotient is an integral domain, it is also a field, and thus $I$ was a maximal ideal.
Only one; the ideal $(0)$ which is not maximal. We contend every other prime ideal is maximal.
Indeed, let $I$ be any ideal (not necessarily prime), and let $a + b \sqrt{2017}$ be a nonzero element of it. Then $I$ also contains $(a^2-2017b^2)$. That means when taking modulo $I$ we may take modulo the integer $n \coloneqq |a^2-2017b^2| \neq 0$.
So every element in $R$ is equivalent modulo $I$ to an element of the form $x + y \sqrt{2017}$, where $x,y \in \{0, 1, \dots, n-1\}$. In other words, the quotient $R/I$ has at most finitely many elements.
When $I$ is prime, it follows $R/I$ is an integral domain, too. An integral domain with finitely many elements must be a field. Hence, from $R/I$ being a field, we conclude $I$ is maximal.
Let $R$ denote the set of rational numbers $q$ such that, when $q$ is written in lowest terms, the denominator is not a multiple of $5$. Then $R$ is a ring (under the usual addition and multiplication). Classify all the ideals of $R$. Which of these ideals are prime / maximal?
The ideals are $(0)$, $(1) = R$, and $(5^n) = 5^n R$ for each $n \ge 1$. The ideal $(0)$ is prime and the ideal $(5)$ is maximal (because the quotient $R/(5) \cong \mathbb{F}_5$ is a field).
Recall from that a ring $R$ is **Artinian** if it satisfies the **descending chain condition**: there does *not* exist an infinite descending chain of ideals $I_1 \supsetneq I_2 \supsetneq I_3 \supsetneq \dots$.
(a) Show that if $R$ is Artinian and an integral domain, then it's a field. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Extra: Euclidean domains | 09_ring-flavors.md | 9 | 1,563 |
(b) More generally, show that every prime ideal in an Artinian ring is maximal.
[^1]: Some authors abbreviate this to "domain", notably Artin.
[^2]: See for the explanation why this norm is the natural one.
[^3]: Note that $\psi$ cannot be the zero map for us, since we require $\psi(1_K) = 1_R$. You sometimes find different statements in the literature. | An Infinitely Large Napkin | napkin | general | advanced | Basic Abstract Algebra | Flavors of rings | Extra: Euclidean domains | 09_ring-flavors.md | 10 | 102 |
At the end of the last chapter on metric spaces, we introduced two adjectives "open" and "closed". These are important because they'll grow up to be the *definition* for a general topological space, once we graduate from metric spaces.
To move forward, we provide a couple niceness adjectives that applies to *entire metric spaces*, rather than just a set relative to a parent space. They are "(totally) bounded" and "complete". These adjectives are specific to metric spaces, but will grow up to become the notion of *compactness*, which is, in the words of , "the single most important concept in real analysis". At the end of the chapter, we will know enough to realize that something is amiss with our definition of homeomorphism, and this will serve as the starting point for the next chapter, when we define fully general topological spaces.
At the end of the last chapter on metric spaces, we introduced two adjectives "open" and "closed". These are important because they'll grow up to be the *definition* for a general topological space, once we graduate from metric spaces.
To move forward, we provide a couple niceness adjectives that applies to *entire metric spaces*, rather than just a set relative to a parent space. They are "(totally) bounded" and "complete". These adjectives are specific to metric spaces, but will grow up to become the notion of *compactness*, which is, in the words of , "the single most important concept in real analysis". At the end of the chapter, we will know enough to realize that something is amiss with our definition of homeomorphism, and this will serve as the starting point for the next chapter, when we define fully general topological spaces. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Properties of metric spaces | 10_metric-prop.md | 0 | 485 | |
Here is one notion of how to prevent a metric space from being a bit too large.
A metric space $M$ is **bounded** if there is a constant $D$ such that $d(p,q) \le D$ for all $p,q \in M$.
You can change the order of the quantifiers:
A metric space $M$ is bounded if and only if for every point $p \in M$, there is a radius $R$ (possibly depending on $p$) such that $d(p,q) \le R$ for all $q \in M$.
Use the triangle inequality to show these are equivalent. (The names "radius" and "diameter" are a big hint!)
(a) Finite intervals like $[0,1]$ and $(a,b)$ are bounded.
(b) The unit square $[0,1]^2$ is bounded.
(c) $\mathbb{R}^n$ is not bounded for any $n \ge 1$.
(d) A discrete space on an infinite set is bounded.
(e) $\mathbb{N}$ is not bounded, despite being homeomorphic to the discrete space!
The fact that a discrete space on an infinite set is "bounded" might be upsetting to you, so here is a somewhat stronger condition you can use:
A metric space is **totally bounded** if for any $\varepsilon > 0$, we can cover $M$ with finitely many $\varepsilon$-neighborhoods.
For example, if $\varepsilon = 1/2$, you can cover $[0,1]^2$ by $\varepsilon$-neighborhoods.
size(4cm); draw(shift( (-0.5,-0.5) )\*unitsquare, black+1); real d = 0.4; real r = 0.5; draw(CR(dir( 45)\*d, r), dotted); draw(CR(dir(135)\*d, r), dotted); draw(CR(dir(225)\*d, r), dotted); draw(CR(dir(315)\*d, r), dotted);
Show that "totally bounded" implies "bounded".
(a) A subset of $\mathbb{R}^n$ is bounded if and only if it is totally bounded.
This is for Euclidean geometry reasons: for example in $\mathbb{R}^2$ if I can cover a set by a single disk of radius $2$, then I can certainly cover it by finitely many disks of radius $1/2$. (We won't prove this rigorously.)
(b) So for example $[0,1]$ or $[0,2] \times [0,3]$ is totally bounded.
(c) In contrast, a discrete space on an infinite set is not totally bounded.
Here is one notion of how to prevent a metric space from being a bit too large.
A metric space $M$ is **bounded** if there is a constant $D$ such that $d(p,q) \le D$ for all $p,q \in M$.
You can change the order of the quantifiers:
A metric space $M$ is bounded if and only if for every point $p \in M$, there is a radius $R$ (possibly depending on $p$) such that $d(p,q) \le R$ for all $q \in M$.
Use the triangle inequality to show these are equivalent. (The names "radius" and "diameter" are a big hint!)
(a) Finite intervals like $[0,1]$ and $(a,b)$ are bounded.
(b) The unit square $[0,1]^2$ is bounded.
(c) $\mathbb{R}^n$ is not bounded for any $n \ge 1$.
(d) A discrete space on an infinite set is bounded.
(e) $\mathbb{N}$ is not bounded, despite being homeomorphic to the discrete space!
The fact that a discrete space on an infinite set is "bounded" might be upsetting to you, so here is a somewhat stronger condition you can use:
A metric space is **totally bounded** if for any $\varepsilon > 0$, we can cover $M$ with finitely many $\varepsilon$-neighborhoods.
For example, if $\varepsilon = 1/2$, you can cover $[0,1]^2$ by $\varepsilon$-neighborhoods.
size(4cm); draw(shift( (-0.5,-0.5) )\*unitsquare, black+1); real d = 0.4; real r = 0.5; draw(CR(dir( 45)\*d, r), dotted); draw(CR(dir(135)\*d, r), dotted); draw(CR(dir(225)\*d, r), dotted); draw(CR(dir(315)\*d, r), dotted);
Show that "totally bounded" implies "bounded".
(a) A subset of $\mathbb{R}^n$ is bounded if and only if it is totally bounded.
This is for Euclidean geometry reasons: for example in $\mathbb{R}^2$ if I can cover a set by a single disk of radius $2$, then I can certainly cover it by finitely many disks of radius $1/2$. (We won't prove this rigorously.)
(b) So for example $[0,1]$ or $[0,2] \times [0,3]$ is totally bounded.
(c) In contrast, a discrete space on an infinite set is not totally bounded. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Properties of metric spaces | Boundedness | 10_metric-prop.md | 1 | 1,092 |
So far we can only talk about sequences converging if they have a limit. But consider the sequence $$x_1 = 1, \; x_2 = 1.4, \; x_3 = 1.41, \; x_4 = 1.414, \dots.$$ It converges to $\sqrt 2$ in $\mathbb{R}$, of course. But it fails to converge in $\mathbb{Q}$; there is no *rational* number this sequence converges to. And so somehow, if we didn't know about the existence of $\mathbb{R}$, we would have *no idea* that the sequence $(x_n)$ is "approaching" something.
That seems to be a shame. Let's set up a new definition to describe these sequences whose terms **get close to each other**, even if they don't approach any particular point in the space. Thus, we only want to mention the given points in the definition.
Let $x_1, x_2, \dots$ be a sequence which lives in a metric space $M = (M,d_M)$. We say the sequence is **Cauchy** if for any $\varepsilon > 0$, we have $$d_M(x_m, x_n) < \varepsilon$$ for all sufficiently large $m$ and $n$.
Show that a sequence which converges is automatically Cauchy. (Draw a picture.)
Now we can define:
A metric space $M$ is **complete** if every Cauchy sequence converges.
(a) $\mathbb{R}$ is complete. (Depending on your definition of $\mathbb{R}$, this either follows by definition, or requires some work. We won't go through this here.)
(b) The discrete space is complete, as the only Cauchy sequences are eventually constant.
(c) The closed interval $[0,1]$ is complete.
(d) $\mathbb{R}^n$ is complete as well. (You're welcome to prove this by induction on $n$.)
(a) The rationals $\mathbb{Q}$ are not complete.
(b) The open interval $(0,1)$ is not complete, as the sequence $0.9$, $0.99$, $0.999$, $0.9999$, ... is Cauchy but does not converge.
So, metric spaces need not be complete, like $\mathbb{Q}$. But we certainly would like them to be complete, and in light of the following theorem this is not unreasonable.
Every metric space can be "completed", i.e. made into a complete space by adding in some points.
We won't need this construction at all, so it's left as .
The completion of $\mathbb{Q}$ is $\mathbb{R}$.
(In fact, by using a modified definition of completion not depending on the real numbers, other authors often use this as the definition of $\mathbb{R}$.)
So far we can only talk about sequences converging if they have a limit. But consider the sequence $$x_1 = 1, \; x_2 = 1.4, \; x_3 = 1.41, \; x_4 = 1.414, \dots.$$ It converges to $\sqrt 2$ in $\mathbb{R}$, of course. But it fails to converge in $\mathbb{Q}$; there is no *rational* number this sequence converges to. And so somehow, if we didn't know about the existence of $\mathbb{R}$, we would have *no idea* that the sequence $(x_n)$ is "approaching" something.
That seems to be a shame. Let's set up a new definition to describe these sequences whose terms **get close to each other**, even if they don't approach any particular point in the space. Thus, we only want to mention the given points in the definition.
Let $x_1, x_2, \dots$ be a sequence which lives in a metric space $M = (M,d_M)$. We say the sequence is **Cauchy** if for any $\varepsilon > 0$, we have $$d_M(x_m, x_n) < \varepsilon$$ for all sufficiently large $m$ and $n$.
Show that a sequence which converges is automatically Cauchy. (Draw a picture.)
Now we can define:
A metric space $M$ is **complete** if every Cauchy sequence converges.
(a) $\mathbb{R}$ is complete. (Depending on your definition of $\mathbb{R}$, this either follows by definition, or requires some work. We won't go through this here.)
(b) The discrete space is complete, as the only Cauchy sequences are eventually constant.
(c) The closed interval $[0,1]$ is complete.
(d) $\mathbb{R}^n$ is complete as well. (You're welcome to prove this by induction on $n$.)
(a) The rationals $\mathbb{Q}$ are not complete.
(b) The open interval $(0,1)$ is not complete, as the sequence $0.9$, $0.99$, $0.999$, $0.9999$, ... is Cauchy but does not converge.
So, metric spaces need not be complete, like $\mathbb{Q}$. But we certainly would like them to be complete, and in light of the following theorem this is not unreasonable.
Every metric space can be "completed", i.e. made into a complete space by adding in some points.
We won't need this construction at all, so it's left as .
The completion of $\mathbb{Q}$ is $\mathbb{R}$.
(In fact, by using a modified definition of completion not depending on the real numbers, other authors often use this as the definition of $\mathbb{R}$.) | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Properties of metric spaces | Completeness | 10_metric-prop.md | 2 | 1,279 |
There is something suspicious about both these notions: neither are preserved under homeomorphism!
Let $M = (0,1)$ and $N = \mathbb{R}$. As we saw much earlier $M$ and $N$ are homeomorphic. However:
- $(0,1)$ is totally bounded, but not complete.
- $\mathbb{R}$ is complete, but not bounded.
This is the first hint of something going awry with the metric. As we progress further into our study of topology, we will see that in fact *open sets and closed sets* (which we motivated by using the metric) are the notion that will really shine later on. I insist on introducing the metric first so that the standard pictures of open sets and closed sets make sense, but eventually it becomes time to remove the training wheels.
There is something suspicious about both these notions: neither are preserved under homeomorphism!
Let $M = (0,1)$ and $N = \mathbb{R}$. As we saw much earlier $M$ and $N$ are homeomorphic. However:
- $(0,1)$ is totally bounded, but not complete.
- $\mathbb{R}$ is complete, but not bounded.
This is the first hint of something going awry with the metric. As we progress further into our study of topology, we will see that in fact *open sets and closed sets* (which we motivated by using the metric) are the notion that will really shine later on. I insist on introducing the metric first so that the standard pictures of open sets and closed sets make sense, but eventually it becomes time to remove the training wheels. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Properties of metric spaces | Let the buyer beware | 10_metric-prop.md | 3 | 414 |
As we've already been doing implicitly in examples, we'll now say:
Every subset $S \subseteq M$ is a metric space in its own right, by reusing the distance function on $M$. We say that $S$ is a **subspace** of $M$.
For example, we saw that the circle $S^1$ is just a subspace of $\mathbb{R}^2$.
It thus becomes important to distinguish between
(i) **"absolute" adjectives** like "complete" or "bounded", which can be applied to both spaces, and hence even to subsets of spaces (by taking a subspace), and
(ii) **"relative" adjectives** like "open (in $M$)" and "closed (in $M$)", which make sense only relative to a space, even though people are often sloppy and omit them.
So "$[0,1]$ is complete" makes sense, as does "$[0,1]$ is a complete subset of $\mathbb{R}$", which we take to mean "$[0,1]$ is a complete subspace of $\mathbb{R}$". This is since "complete" is an absolute adjective.
But here are some examples of ways in which relative adjectives require a little more care:
- Consider the sequence $1$, $1.4$, $1.41$, $1.414$, .... Viewed as a sequence in $\mathbb{R}$, it converges to $\sqrt 2$. But if viewed as a sequence in $\mathbb{Q}$, this sequence does *not* converge! Similarly, the sequence $0.9$, $0.99$, $0.999$, $0.9999$ does not converge in the space $(0,1)$, although it does converge in $[0,1]$.
The fact that these sequences fail to converge even though they "ought to" is weird and bad, and was why we defined complete spaces to begin with.
- In general, it makes no sense to ask a question like "is $[0,1]$ open?". The questions "is $[0,1]$ open in $\mathbb{R}$?" and "is $[0,1]$ open in $[0,1]$?" do make sense, however. The answer to the first question is "no" but the answer to the second question is "yes"; indeed, every space is open in itself. Similarly, $[0, \frac{1}{2})$ is an open set in the space $M = [0,1]$ because it is the ball *in $M$* of radius $\frac{1}{2}$ centered at $0$.
- Dually, it doesn't make sense to ask "is $[0,1]$ closed"? It is closed *in $\mathbb{R}$* and *in itself* (but every space is closed in itself, anyways).
To make sure you understand the above, here are two exercises to help you practice relative adjectives.
Let $M$ be a complete metric space and let $S \subseteq M$. Prove that $S$ is complete if and only if it is closed in $M$. In particular, $[0,1]$ is complete.
Let $M = [0,1] \cup (2,3)$. Show that $[0,1]$ and $(2,3)$ are both open and closed in $M$.
This illustrates a third point: a nontrivial set can be both open and closed.[^1] As we'll see in , this implies the space is disconnected; i.e. the only examples look quite like the one we've given above.
Let $M = (M,d)$ be a complete metric space. Suppose $T \colon M \to M$ is a continuous map such that for any $p, q \in M$, $$d\left( T(p), T(q) \right) \le 0.999 d(p,q).$$ (We call $T$ a **contraction**.) Show that $T$ has a unique fixed point.
The main task is to show there exists some fixed point. Start at some point $x_0$ and consider the sequence $x_1 = T(x_0)$, $x_2 = T(x_1)$, $x_3 = T(x_2)$, ..., and so on.
Uniqueness of the fixed point follows from noting that if $T(p) = p$ and $T(q) = q$ and $p \neq q$ then we get a direct contradiction by plugging this into the given statement. Hence the main task is to show there exists some fixed point.
Start with any point $x_0$. Let $x_1 = T(x_0)$, $x_2 = T(x_1)$, $x_3 = T(x_2)$, ..., and so on. We contend that $(x_0, x_1, x_2, \dots)$ is a Cauchy sequence. Indeed, if we let $r \coloneqq 0.999 < 1$ and $c \coloneqq d(x_0, x_1)$, then $$\begin{align*}
d(x_1, x_2) &< r \cdot c \\
d(x_2, x_3) &< r^2 \cdot c \\
d(x_3, x_4) &< r^3 \cdot c \\
&\vdotswithin< \\
\end{align*}$$ and so for large $M < N$ we have $$d(x_M, x_N) < \left( r^M + r^{M+1} + \dots + r^N \right) \cdot c
< \frac{r^M}{1-r} \cdot c$$ which tends to zero once $M$ is large enough.
Hence, because $M$ is complete, the sequence must converge to some limit $x$. Because $T$ is continuous, we get $$T(x) = T\left( \lim_{n \to \infty} x_n \right)
= \lim_{n \to \infty} T(x_n)
= \lim_{n \to \infty} x_{n+1} = x$$ as desired.
We let $M$ and $N$ denote the metric spaces obtained by equipping $\mathbb{R}$ with the following two metrics: $$\begin{align*}
d_M(x,y) &= \min \left\{ 1, \left\lvert x-y \right\rvert \right\} \\
d_N(x,y) &= \left\lvert e^x - e^y \right\rvert.
\end{align*}$$
(a) Fill in the following $2 \times 3$ table with "yes" or "no" for each cell.
::: center
Complete? Bounded? Totally bounded?
----- ----------- ---------- ------------------
$M$
$N$
:::
(b) Are $M$ and $N$ homeomorphic?
(a): $M$ is complete and bounded but not totally bounded. $N$ is all no. For (b) show that $M \cong \mathbb{R} \cong N$.
Part (a) is essentially by definition. The space $M$ is bounded since no distances exceed $1$, but not totally bounded since we can't cover $M$ with finitely many $\frac{1}{2}$-neighborhoods. The space $M$ is complete since a sequence of real numbers converges in $M$ if it converges in the usual sense. As for $N$, the sequence $-1$, $-2$, ... is Cauchy but fails to converge; and it is obviously not bounded.
To show (b), the identity map (!) is an homeomorphism $M \cong \mathbb{R}$ and $\mathbb{R} \cong N$, since it is continuous. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Properties of metric spaces | Subspaces, and (inb4) a confusing linguistic point | 10_metric-prop.md | 4 | 1,497 |
This illustrates that $M \cong N$ despite the fact that $M$ is both complete and bounded but $N$ is neither complete nor bounded. On the other hand, we will later see that complete and totally bounded implies *compact*, which is a very strong property preserved under homeomorphism.
Let $M$ be a metric space. Construct a complete metric space $\overline M$ such that $M$ is a subspace of $\overline M$, and every nonempty open set of $\overline M$ contains a point of $M$ (meaning $M$ is **dense** in $\overline M$).
As a set, we let $\overline M$ be the set of Cauchy sequences $(x_n)$ in $M$, modulo the relation that $(x_n) \sim (y_n)$ if $\lim_n d(x_n, y_n) = 0$.
Show that a metric space is totally bounded if and only if any sequence has a Cauchy subsequence.
See <https://math.stackexchange.com/q/556150/229197>.
Prove that $\mathbb{Q}$ is not homeomorphic to any complete metric space.
The standard solution seems to be via the so-called "Baire category theorem".
[^1]: Which always gets made fun of. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Properties of metric spaces | Subspaces, and (inb4) a confusing linguistic point | 10_metric-prop.md | 5 | 290 |
As we've already been doing implicitly in examples, we'll now say:
Every subset $S \subseteq M$ is a metric space in its own right, by reusing the distance function on $M$. We say that $S$ is a **subspace** of $M$.
For example, we saw that the circle $S^1$ is just a subspace of $\mathbb{R}^2$.
It thus becomes important to distinguish between
(i) **"absolute" adjectives** like "complete" or "bounded", which can be applied to both spaces, and hence even to subsets of spaces (by taking a subspace), and
(ii) **"relative" adjectives** like "open (in $M$)" and "closed (in $M$)", which make sense only relative to a space, even though people are often sloppy and omit them.
So "$[0,1]$ is complete" makes sense, as does "$[0,1]$ is a complete subset of $\mathbb{R}$", which we take to mean "$[0,1]$ is a complete subspace of $\mathbb{R}$". This is since "complete" is an absolute adjective.
But here are some examples of ways in which relative adjectives require a little more care:
- Consider the sequence $1$, $1.4$, $1.41$, $1.414$, .... Viewed as a sequence in $\mathbb{R}$, it converges to $\sqrt 2$. But if viewed as a sequence in $\mathbb{Q}$, this sequence does *not* converge! Similarly, the sequence $0.9$, $0.99$, $0.999$, $0.9999$ does not converge in the space $(0,1)$, although it does converge in $[0,1]$.
The fact that these sequences fail to converge even though they "ought to" is weird and bad, and was why we defined complete spaces to begin with.
- In general, it makes no sense to ask a question like "is $[0,1]$ open?". The questions "is $[0,1]$ open in $\mathbb{R}$?" and "is $[0,1]$ open in $[0,1]$?" do make sense, however. The answer to the first question is "no" but the answer to the second question is "yes"; indeed, every space is open in itself. Similarly, $[0, \frac{1}{2})$ is an open set in the space $M = [0,1]$ because it is the ball *in $M$* of radius $\frac{1}{2}$ centered at $0$.
- Dually, it doesn't make sense to ask "is $[0,1]$ closed"? It is closed *in $\mathbb{R}$* and *in itself* (but every space is closed in itself, anyways).
To make sure you understand the above, here are two exercises to help you practice relative adjectives.
Let $M$ be a complete metric space and let $S \subseteq M$. Prove that $S$ is complete if and only if it is closed in $M$. In particular, $[0,1]$ is complete.
Let $M = [0,1] \cup (2,3)$. Show that $[0,1]$ and $(2,3)$ are both open and closed in $M$.
This illustrates a third point: a nontrivial set can be both open and closed.[^1] As we'll see in , this implies the space is disconnected; i.e. the only examples look quite like the one we've given above.
Let $M = (M,d)$ be a complete metric space. Suppose $T \colon M \to M$ is a continuous map such that for any $p, q \in M$, $$d\left( T(p), T(q) \right) \le 0.999 d(p,q).$$ (We call $T$ a **contraction**.) Show that $T$ has a unique fixed point.
The main task is to show there exists some fixed point. Start at some point $x_0$ and consider the sequence $x_1 = T(x_0)$, $x_2 = T(x_1)$, $x_3 = T(x_2)$, ..., and so on.
Uniqueness of the fixed point follows from noting that if $T(p) = p$ and $T(q) = q$ and $p \neq q$ then we get a direct contradiction by plugging this into the given statement. Hence the main task is to show there exists some fixed point.
Start with any point $x_0$. Let $x_1 = T(x_0)$, $x_2 = T(x_1)$, $x_3 = T(x_2)$, ..., and so on. We contend that $(x_0, x_1, x_2, \dots)$ is a Cauchy sequence. Indeed, if we let $r \coloneqq 0.999 < 1$ and $c \coloneqq d(x_0, x_1)$, then $$\begin{align*}
d(x_1, x_2) &< r \cdot c \\
d(x_2, x_3) &< r^2 \cdot c \\
d(x_3, x_4) &< r^3 \cdot c \\
&\vdotswithin< \\
\end{align*}$$ and so for large $M < N$ we have $$d(x_M, x_N) < \left( r^M + r^{M+1} + \dots + r^N \right) \cdot c
< \frac{r^M}{1-r} \cdot c$$ which tends to zero once $M$ is large enough.
Hence, because $M$ is complete, the sequence must converge to some limit $x$. Because $T$ is continuous, we get $$T(x) = T\left( \lim_{n \to \infty} x_n \right)
= \lim_{n \to \infty} T(x_n)
= \lim_{n \to \infty} x_{n+1} = x$$ as desired.
We let $M$ and $N$ denote the metric spaces obtained by equipping $\mathbb{R}$ with the following two metrics: $$\begin{align*}
d_M(x,y) &= \min \left\{ 1, \left\lvert x-y \right\rvert \right\} \\
d_N(x,y) &= \left\lvert e^x - e^y \right\rvert.
\end{align*}$$
(a) Fill in the following $2 \times 3$ table with "yes" or "no" for each cell.
::: center
Complete? Bounded? Totally bounded?
----- ----------- ---------- ------------------
$M$
$N$
:::
(b) Are $M$ and $N$ homeomorphic?
(a): $M$ is complete and bounded but not totally bounded. $N$ is all no. For (b) show that $M \cong \mathbb{R} \cong N$.
Part (a) is essentially by definition. The space $M$ is bounded since no distances exceed $1$, but not totally bounded since we can't cover $M$ with finitely many $\frac{1}{2}$-neighborhoods. The space $M$ is complete since a sequence of real numbers converges in $M$ if it converges in the usual sense. As for $N$, the sequence $-1$, $-2$, ... is Cauchy but fails to converge; and it is obviously not bounded.
To show (b), the identity map (!) is an homeomorphism $M \cong \mathbb{R}$ and $\mathbb{R} \cong N$, since it is continuous. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Properties of metric spaces | Subspaces, and (inb4) a confusing linguistic point | 10_metric-prop.md | 6 | 1,497 |
This illustrates that $M \cong N$ despite the fact that $M$ is both complete and bounded but $N$ is neither complete nor bounded. On the other hand, we will later see that complete and totally bounded implies *compact*, which is a very strong property preserved under homeomorphism.
Let $M$ be a metric space. Construct a complete metric space $\overline M$ such that $M$ is a subspace of $\overline M$, and every nonempty open set of $\overline M$ contains a point of $M$ (meaning $M$ is **dense** in $\overline M$).
As a set, we let $\overline M$ be the set of Cauchy sequences $(x_n)$ in $M$, modulo the relation that $(x_n) \sim (y_n)$ if $\lim_n d(x_n, y_n) = 0$.
Show that a metric space is totally bounded if and only if any sequence has a Cauchy subsequence.
See <https://math.stackexchange.com/q/556150/229197>.
Prove that $\mathbb{Q}$ is not homeomorphic to any complete metric space.
The standard solution seems to be via the so-called "Baire category theorem".
[^1]: Which always gets made fun of. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Properties of metric spaces | Subspaces, and (inb4) a confusing linguistic point | 10_metric-prop.md | 7 | 290 |
Topological spaces
ch:top_more
In ch:metric_space we introduced the notion
of space by describing metrics on them.
This gives you a lot of examples, and nice intuition,
and tells you how you should draw pictures of open and closed sets.
However, moving forward, it will be useful to begin
thinking about topological spaces in terms of just their open sets.
(One motivation is that our fishy ex:fishy shows that
in some ways the notion of homeomorphism really wants
to be phrased in terms of open sets, not in terms of the metric.)
As we are going to see, the open sets manage to actually retain
nearly all the information we need, but are simpler.The
reason I adamantly introduce metric spaces first
is because I think otherwise the examples make much less sense.
This will be done in just a few sections,
and after that we will start describing more adjectives
that we can apply to topological (and hence metric) spaces.
The most important topological notion is missing from this chapter:
that of a compact space.
It is so important that I have dedicated a separate chapter just for it.
Quick note for those who care:
the adjectives ``Hausdorff'', ``connected'',
and later ``compact'' are all absolute adjectives.
%Note that in contrast to the warning on open/closed sets I gave earlier,
%
% The adjectives in this chapter will be used to describe spaces.
%
%As I alluded to earlier, sequences in metric spaces are super nice,
%but sequences in general topological spaces suck (to the point where
%I didn't bother to define convergence of general sequences).
Forgetting the metric
Recall thm:open_set:
A function $f M N$ of metric spaces is continuous
if and only if the pre-image of every open set in $N$ is open in $M$.
Despite us having defined this in the context of metric spaces,
this nicely doesn't refer to the metric at all,
only the open sets.
As alluded to at the start of this chapter,
this is a great motivation for how we can forget
about the fact that we had a metric to begin with,
and rather start with the open sets instead.
A topological space is a pair $(X, T)$,
where $X$ is a set of points,
and $ T$ is the topology,
which consists of several subsets of $X$, called the open sets of $X$.
The topology must obey the following axioms.
$$ and $X$ are both in $ T$.
Finite intersections of open sets are also in $ T$.
Arbitrary unions (possibly infinite) of open sets are also in $ T$.
So this time, the open sets are given.
Rather than defining a metric and getting open sets from the metric,
we instead start from just the open sets.
We abbreviate $(X, T)$ by just $X$,
leaving the topology $ T$ implicit.
(Do you see a pattern here?)
Given a metric space $M$, we can let $ T$ be
the open sets in the metric sense.
The point is that the axioms are satisfied.
In particular, discrete space
is a topological space in which every set is open. (Why?)
Given $X$, we can let $ T = \ , X \$,
the opposite extreme of the discrete space.
Now we can port over our metric definitions.
An open neighborhoodIn literature,
a ``neighborhood'' refers to a set
which contains some open set around $x$.
We will not use this term,
and exclusively refer to ``open neighborhoods''.
of a point $x X$ is an open set $U$ which contains $x$ (see figure).
size(4cm);
bigblob("$X$");
pair p = Drawing("x", (0.3,0.1), dir(-90));
real r = 1.55;
draw(shift(p) * scale(1.6,1.2)*unitcircle, dashed);
label("$U$", p+r*dir(45), dir(45));
Just to be perfectly clear:
by an ``open neighborhood'' I mean any open set containing $x$.
But by an ``$r$-neighborhood'' I always mean the
points with distance less than $r$ from $x$,
and so I can only use this term if my space is a metric space.
Re-definitions
Now that we've defined a topological space,
for nearly all of our metric notions we can write down
as the definition the one that required only open sets
(which will of course agree with our old definitions
when we have a metric space).
Continuity
Here was our motivating example, continuity:
We say function $f X Y$ of topological spaces
is continuous at a point $p X$ if the pre-image of any
open neighborhood of $f(p)$ contains an open neighborhood of $p$.
The function is continuous if it is continuous at every point.
Thus homeomorphism carries over:
a bijection which is continuous in both directions.
A homeomorphism of topological spaces
$(X, _X)$ and $(Y, _Y)$
is a bijection $f X Y$
which induces a bijection from $_X$ to $_Y$:
i.e.\ the bijection preserves open sets.
Show that this is equivalent to $f$ and its inverse
both being continuous.
Therefore, any property defined only in
terms of open sets is preserved by homeomorphism.
Such a property is called a topological property.
The later adjectives we define
(``connected'', ``Hausdorff'', ``compact'') will all be defined
only in terms of open sets, so they will be topological properties. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 0 | 1,391 | |
%
% A sequence $(x_n)$ of points in a topological space $X$ is said to converge to $x X$ if for every open neighborhood of $x$,
% eventually all terms of the sequence lie in that open neighborhood.
%
%
% Unfortunately, for general topological spaces we no longer have the nice property
% that any function which preserves sequential limits is automatically continuous.
%
%There's one other property of open sets that we have in a metric space that isn't implied by the above: for any two points of $X$, we can find an open set containing one but not the other.
%A space which also has this property is called a Kolmogorov space.
%This property is a good property to have, because if $x,y X$ are in the same open sets, the topology can't tell them apart.
Closed sets
We saw last time there were two equivalent definitions
for closed sets, but one of them relies only on open sets, and we use it:
def:closure
In a general topological space $X$, we say that $S X$ is
closed in $X$ if the complement $X S$ is open in $X$.
If $S X$ is any set, the closure of $S$,
denoted $ S$,
is defined as the smallest closed set containing $S$.
Thus for general topological spaces,
open and closed sets carry the same information,
and it is entirely a matter of taste whether we define everything in terms
of open sets or closed sets.
In particular, you can translate axioms and properties of open sets to closed ones:
Show that the (possibly infinite) intersection of closed sets is closed
while the union of finitely many closed sets is closed.
(Look at complements.)
Show that a function is continuous if and only if the pre-image
of every closed set is closed.
Mathematicians seem to have agreed that they like open sets better.
Properties that don't carry over
Not everything works:
[Complete and (totally) bounded are metric properties]
The two metric properties we have seen,
``complete'' and ``(totally) bounded'',
are not topological properties.
They rely on a metric,
so as written we cannot apply them to topological spaces.
One might hope that maybe,
there is some alternate definition (like we saw for ``continuous function'')
that is just open-set based.
But ex:fishy showing $(0,1) $
tells us that it is hopeless.
[Sequences don't work well]
You could also try to port over the notion
of sequences and convergent sequences.
However, this turns out to break a lot of desirable properties.
Therefore I won't bother to do so,
and thus if we are discussing sequences you should
assume that we are working with a metric space.
Hausdorff spaces
As you might have guessed,
there exist topological spaces which cannot be realized
as metric spaces (in other words, are not metrizable).
One example is just to take $X = \a,b,c\$ and the topology
$_X = \ , \a,b,c\ \$.
This topology is fairly ``stupid'':
it can't tell apart any of the points $a$, $b$, $c$!
But any metric space can tell its points apart (because $d(x,y) > 0$ when $x y$).
We'll see less trivial examples later,
but for now we want to add a little more sanity condition onto our spaces.
There is a whole hierarchy of such axioms, labelled $T_n$ for
integers $n$ (with $n=0$ being the weakest and $n=6$ the strongest);
these axioms are called separation axioms.
By far the most common hypothesis is the $T_2$ axiom,
which bears a special name.
A topological space $X$ is Hausdorff if
for any two distinct points $p$ and $q$ in $X$,
there exists an open neighborhood $U$ of $p$
and an open neighborhood $V$ of $q$ such that
\[ U V = . \]
In other words, around any two distinct points we should be
able to draw disjoint open neighborhoods.
Here's a picture to go with above,
but not much going on.
size(5cm);
pair p = (-1.5,0);
pair q = ( 1.5,0);
filldraw(CR(p,1), opacity(0.1)+lightcyan, blue+dashed);
filldraw(CR(q,1), opacity(0.1)+lightred, red+dashed);
dot("$p$", p, dir(-90), blue);
dot("$q$", q, dir(-90), red);
Show that all metric spaces are Hausdorff.
I just want to define this here so that I can use this word later.
In any case, basically any space we will encounter other than
the Zariski topology is Hausdorff.
Subspaces
One can also take subspaces of general topological spaces.
Given a topological space $X$, and a subset $S X$,
we can make $S$ into a topological space
by declaring that the open subsets of $S$ are $U S$ for open $U X$.
This is called the subspace topology.
So for example, if we view $S^1$ as a subspace of $^2$,
then any open arc is an open set,
because you can view it as the intersection of an open disk with $S^1$.
size(3cm);
draw(unitcircle, black+1);
MP("S^1", dir(60), dir(60));
MP(" R^2", dir(-45)*1.2, dir(-45));
pair A = dir(-30);
pair B = dir(50);
draw(CP(dir(10), A), dotted);
draw(arc(origin,A,B), blue+2);
dotfactor *= 2;
opendot(A, blue);
opendot(B, blue);
Needless to say, for metric spaces it doesn't matter
which of these definitions I choose.
(Proving this turns out to be surprisingly annoying, so I won't do so.)
Connected spaces
Even in metric spaces, it is possible for a set to be both open and closed.
A subset $S$ of a topological space $X$
is clopen if it is both closed and open in $X$.
(Equivalently, both $S$ and its complement are open.)
For example $$ and the entire space are examples of clopen sets.
In fact, the presence of a nontrivial clopen set other
than these two leads to a so-called disconnected space. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 1 | 1,531 | |
Show that a space $X$ has a nontrivial clopen set (one other than $$ and $X$)
if and only if $X$ can be written as a disjoint union of two nonempty open sets.
We say $X$ is disconnected if there are nontrivial clopen sets,
and connected otherwise.
To see why this should be a reasonable definition,
it might help to solve prob:disconnected_better_def.
The metric space
\[ \ (x,y) x^2+y^2 1 \
\ (x,y) (x-4)^2+y^21\ ^2 \]
is disconnected (it consists of two disks).
The space $[0,1] [2,3]$ is disconnected:
it consists of two segments,
each of which is a clopen set.
A discrete space on more than one point is disconnected,
since every set is clopen in the discrete space.
Convince yourself that the set
\[ \ x x^2 < 2014 \ \]
is a clopen subset of $$.
Hence $$ is disconnected too -- it has gaps.
$[0,1]$ is connected.
%
% For general topological spaces $X$, it is still true that if $S$ is closed
% (meaning $X S$ is open), then it contains the limits of all its sequences.
% But the converse of this statement no longer holds.
%
Path-connected spaces
A stronger and perhaps more intuitive notion
of a connected space is a path-connected space.
The short description: ``walk around in the space''.
% We have general topological spaces, but so far they've just been sitting there.
% Let's start walking in them.
% Throughout this section, we let $I = [0,1]$ with the usual topology
% inherited from $$.
%Also, recall that $S^n$ is the surface of the sphere living
%in $n+1$ dimensional space; hence $S^1$ is a unit circle, say.
A path in the space $X$ is a continuous function
\[ [0,1] X. \]
Its endpoints are the two points $(0)$ and $(1)$.
You can think of $[0,1]$ as measuring ``time'', and so we'll often write $(t)$
for $t [0,1]$ (with $t$ standing for ``time'').
Here's a picture of a path.
bigblob("$X$");
pair A = Drawing("(0)", (-3,-1));
pair B = Drawing("(1)", (2,1), dir(90));
path p = A..(-2,0)..(0,2)..(1,0)..B;
draw(p, EndArrow);
MP("", midpoint(p), dir(90));
Why does this agree with your intuitive notion of what a ``path'' is?
A space $X$ is path-connected if
any two points in it are connected by some path.
[Path-connected implies connected]
Let $X = U V$ be a disconnected space.
Show that there is no path
from a point of $U$ to point $V$.
(If $ [0,1] X$, then we get
$[0,1] = (U) (V)$,
but $[0,1]$ is connected.)
$^2$ is path-connected,
since we can ``connect'' any two points with a straight line.
The unit circle $S^1$ is path-connected, since
we can just draw the major or minor arc to connect two points.
% Easy, right?
Homotopy and simply connected spaces
$.
sec:meteor
Now let's motivate the idea of homotopy.
Consider the example of the complex plane $$ (which you can
think of just as $^2$) with two points $p$ and $q$.
There's a whole bunch of paths from $p$ to $q$ but somehow
they're not very different from one another.
If I told you ``walk from $p$ to $q$'' you wouldn't have too many questions.
unitsize(0.8cm);
bigbox("$ C$");
pair A = Drawing("p", (-3,0), dir(180));
pair B = Drawing("q", (3,0), dir(0));
draw(A..(-2,0.5)..(0,2)..(1,1.2)..B, red, EndArrow(TeXHead));
draw(A--B, mediumgreen, EndArrow(TeXHead));
draw(A--(-1,-1)--(2,-1)--B, blue, EndArrow(TeXHead));
// draw(A..(1.5,-2)..(-1.5,-2)..B, orange, EndArrow(TeXHead));
So we're living happily in $$ until a meteor strikes the origin,
blowing it out of existence.
Then suddenly to get from $p$ to $q$, people might tell you two different things:
``go left around the meteor'' or ``go right around the meteor''.
unitsize(0.8cm);
bigbox("$ C \0\$");
pair A = Drawing("p", (-3,0), dir(180));
pair B = Drawing("q", (3,0), dir(0));
draw(A..(-2,0.5)..(0,2)..(1,1.2)..B, red, EndArrow(TeXHead));
draw(A--(-1,-1)--(2,-1)--B, blue, EndArrow(TeXHead));
filldraw(scale(0.5)*unitcircle, grey, black);
So what's happening?
In the first picture, the red, green, and blue paths somehow all looked
the same: if you imagine them as pieces of elastic string pinned down
at $p$ and $q$, you can stretch each one to any other one.
But in the second picture,
you can't move the red string to match with the blue string:
there's a meteor in the way.
The paths are actually different.If you know about winding numbers,
you might feel this is familiar.
We'll talk more about this in the chapter on the fundamental group.
The formal notion we'll use to capture this is homotopy equivalence.
We want to write a definition such that in the first picture,
the three paths are all homotopic, but the two paths in the
second picture are somehow not homotopic.
And the idea is just continuous deformation.
Let $$ and $$ be paths in $X$ whose endpoints coincide.
A (path) homotopy from $$ to $$ is a continuous function
$F [0,1]^2 X$, which we'll write $F_s(t)$ for $s,t [0,1]$,
such that
\[ F_0(t) = (t) and F_1(t) = (t)
for all $t [0,1]$ \]
and moreover
\[ (0) = (0) = F_s(0)
and
(1) = (1) = F_s(1)
for all $s [0,1]$. \]
If a path homotopy exists, we say $$ and $$
are path homotopic and write $ $.
While I strictly should say ``path homotopy'' to describe this relation
between two paths, I will shorten this to just ``homotopy'' instead.
Similarly I will shorten ``path homotopic'' to ``homotopic''.
Animated picture: https://commons.wikimedia.org/wiki/File:HomotopySmall.gif.
Needless to say, $$ is an equivalence relation. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 2 | 1,520 | |
What this definition is doing is taking $$ and ``continuously deforming'' it to $$, while keeping the endpoints fixed.
Note that for each particular $s$, $F_s$ is itself a function.
So $s$ represents time as we deform $$ to $$:
it goes from $0$ to $1$, starting at $$ and ending at $$.
size(9cm);
bigbox("$ C$");
pair A = Drawing("p", (-3,0), dir(180));
pair B = Drawing("q", (3,0), dir(0));
draw(A..MP("F_0 = ", (0,2), dir(45))..B, heavygreen);
draw(A..MP("F_0.25", (0,1), dir(45))..B, mediumgreen);
draw(A..MP("F_0.5", (0,0), dir(90))..B, palecyan);
draw(A..MP("F_0.75", (0,-1), dir(-45))..B, mediumcyan);
draw(A..MP("F_1 = ", (0,-2), dir(-45))..B, heavycyan);
// draw(A..(1.5,-2)..(-1.5,-2)..B, orange, EndArrow);
Convince yourself the above definition is right.
What goes wrong when the meteor strikes?
So now I can tell you what makes $$ special:
A space $X$ is simply connected if it's path-connected and
for any points $p$ and $q$, all paths from $p$ to $q$ are homotopic.
That's why you don't ask questions when walking from $p$ to $q$ in $$:
there's really only one way to walk. Hence the term ``simply'' connected.
Convince yourself that $^n$ is simply connected for all $n$.
Bases of spaces
You might have noticed that the open sets of $$ are a little annoying to describe:
the prototypical example of an open set is $(0,1)$,
but there are other open sets like
\[
(0,1)
( 1, 32 )
( 2, 73 )
(2014, 2015). \]
Check this is an open set.
But okay, this isn't that different.
All I've done is taken a bunch of my prototypes and threw a bunch of $$ signs at it.
And that's the idea behind a basis.
A basis for a topological space $X$
is a subset $ B$ of the open sets such that every open set in $X$
is a union of some (possibly infinite) number of elements in $ B$.
And all we're doing is saying:
The open intervals form a basis of $$.
In fact, more generally we have:
The $r$-neighborhoods form a basis of any metric space $M$.
Kind of silly -- given an open set $U$, for every point $p$ inside $U$,
draw an $r_p$-neighborhood $U_p$ contained entirely inside $U$.
Then $_p U_p$ is contained in $U$ and covers every point inside it.
Hence, an open set in $^2$ is nothing more than a union
of a bunch of open disks, and so on.
The point is that in a metric space, the only open sets you really
ever have to worry too much about are the $r$-neighborhoods.
Let $X$ be a topological space.
Show that there exists a nonconstant continuous function $X \0,1\$ if and
only if $X$ is disconnected (here $\0,1\$ is given the discrete topology).
prob:disconnected_better_def
Let $X$ and $Y$ be topological spaces
and let $f X Y$ be a continuous function.
Show that if $X$ is connected then so is $f(X)$.
Show that if $X$ is path-connected then so is $f(X)$.
[Hausdorff implies $T_1$ axiom]
Let $X$ be a Hausdorff topological space.
Prove that for any point $p X$ the set $\p\$ is closed.
[ref:pugh, Exercise 2.56]
Let $M$ be a metric space with more than one point
but at most countably infinitely many points.
Show that $M$ is disconnected.
Let $p$ be any point.
If there is a real number $r$ such that $d(p,q) r$ for any $q M$,
then the $r$-neighborhood of $p$ is clopen.
Let $X$ be a topological space.
The connected component of a point $p X$
is the union of all subspaces $S X$ which are connected and contain $p$.
Does the connected component of a point have to be itself connected?
Does the connected component of a point have to be an open subset of $X$?
(a) is yes, and (b) is no even for metric spaces.
In fact, a totally disconnected space is one for which
every connected component consists of only a single point,
and there are examples of totally disconnected metric spaces with
non-discrete topologies.
Part (a) is straightforward:
assume for contradiction that the connected component of $p$ is
a disjoint union $U V$ of two nonempty sets open in $X$.
WLOG, assume $x U$
Let $S$ be one of the subspaces containing $X$ that intersects $V$.
Then $S = (S U) (S V)$ rewrites $S$ as the disjoint union
of two sets which are open in $S$, contradicting the connectedness of $S$.
(Though note that as $S$ is not necessarily open in $X$,
the sets $S U$ and $S V$ are not necessarily open in $X$ either.)
For (b), a counterexample is to take any totally disconnected space
like the Cantor set or the $p$-adic numbers.
We declare a subset of $$ to be open if it's the union (possibly empty or infinite)
of arithmetic sequences $\ a + nd n \$,
where $a$ and $d$ are positive integers.
Verify this forms a topology on $$,
called the evenly spaced integer topology.
Prove there are infinitely many primes by considering $_p p$
for primes $p$.
Note that $p$ is closed for each $p$.
If there were finitely many primes, then
$ p = \-1,1\$ would have to be closed;
i.e.\ $\-1,1\$ would be open, but all open sets here are infinite.
Prove that the evenly spaced integer topology on $$ is metrizable.
In other words, show that one can impose a metric $d ^2 $
which makes $$ into a metric space whose open sets are those described above.
% https://teratologicmuseum.wordpress.com/2009/05/05/a-metric-for-the-evenly-spaced-integer-topology/
The balls at $0$ should be of the form $n! $.
Let $d(x,y) = 2017^-n$, where $n$ is the largest integer
such that $n!$ divides $ x-y $. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 3 | 1,514 | |
%
% A topological space $X$ is called locally path-connected
% if for every point $x X$ and open neighborhood $U$ of $x$,
% some open neighborhood $V$ of $x$ contained in $U$ is path-connected.
% Prove that $X$ is path-connected if and only if it is connected
% and locally path-connected.
% prob:local_path_connected
%
We know that any open set $U $
is a union of open intervals (allowing $$ as endpoints).
One can show that it's actually possible to write $U$ as the
union of pairwise disjoint open intervals.You are invited to try
and prove this, but I personally found the proof quite boring.
Prove that there exists such a disjoint union with at most countably many
intervals in it.
Appeal to $$.
You can pick a rational number in each interval and
there are only countably many rational numbers. Done! | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 4 | 231 | |
Topological spaces
ch:top_more
In ch:metric_space we introduced the notion
of space by describing metrics on them.
This gives you a lot of examples, and nice intuition,
and tells you how you should draw pictures of open and closed sets.
However, moving forward, it will be useful to begin
thinking about topological spaces in terms of just their open sets.
(One motivation is that our fishy ex:fishy shows that
in some ways the notion of homeomorphism really wants
to be phrased in terms of open sets, not in terms of the metric.)
As we are going to see, the open sets manage to actually retain
nearly all the information we need, but are simpler.The
reason I adamantly introduce metric spaces first
is because I think otherwise the examples make much less sense.
This will be done in just a few sections,
and after that we will start describing more adjectives
that we can apply to topological (and hence metric) spaces.
The most important topological notion is missing from this chapter:
that of a compact space.
It is so important that I have dedicated a separate chapter just for it.
Quick note for those who care:
the adjectives ``Hausdorff'', ``connected'',
and later ``compact'' are all absolute adjectives.
%Note that in contrast to the warning on open/closed sets I gave earlier,
%
% The adjectives in this chapter will be used to describe spaces.
%
%As I alluded to earlier, sequences in metric spaces are super nice,
%but sequences in general topological spaces suck (to the point where
%I didn't bother to define convergence of general sequences).
Forgetting the metric
Recall thm:open_set:
A function $f M N$ of metric spaces is continuous
if and only if the pre-image of every open set in $N$ is open in $M$.
Despite us having defined this in the context of metric spaces,
this nicely doesn't refer to the metric at all,
only the open sets.
As alluded to at the start of this chapter,
this is a great motivation for how we can forget
about the fact that we had a metric to begin with,
and rather start with the open sets instead.
A topological space is a pair $(X, T)$,
where $X$ is a set of points,
and $ T$ is the topology,
which consists of several subsets of $X$, called the open sets of $X$.
The topology must obey the following axioms.
$$ and $X$ are both in $ T$.
Finite intersections of open sets are also in $ T$.
Arbitrary unions (possibly infinite) of open sets are also in $ T$.
So this time, the open sets are given.
Rather than defining a metric and getting open sets from the metric,
we instead start from just the open sets.
We abbreviate $(X, T)$ by just $X$,
leaving the topology $ T$ implicit.
(Do you see a pattern here?)
Given a metric space $M$, we can let $ T$ be
the open sets in the metric sense.
The point is that the axioms are satisfied.
In particular, discrete space
is a topological space in which every set is open. (Why?)
Given $X$, we can let $ T = \ , X \$,
the opposite extreme of the discrete space.
Now we can port over our metric definitions.
An open neighborhoodIn literature,
a ``neighborhood'' refers to a set
which contains some open set around $x$.
We will not use this term,
and exclusively refer to ``open neighborhoods''.
of a point $x X$ is an open set $U$ which contains $x$ (see figure).
size(4cm);
bigblob("$X$");
pair p = Drawing("x", (0.3,0.1), dir(-90));
real r = 1.55;
draw(shift(p) * scale(1.6,1.2)*unitcircle, dashed);
label("$U$", p+r*dir(45), dir(45));
Just to be perfectly clear:
by an ``open neighborhood'' I mean any open set containing $x$.
But by an ``$r$-neighborhood'' I always mean the
points with distance less than $r$ from $x$,
and so I can only use this term if my space is a metric space.
Re-definitions
Now that we've defined a topological space,
for nearly all of our metric notions we can write down
as the definition the one that required only open sets
(which will of course agree with our old definitions
when we have a metric space).
Continuity
Here was our motivating example, continuity:
We say function $f X Y$ of topological spaces
is continuous at a point $p X$ if the pre-image of any
open neighborhood of $f(p)$ contains an open neighborhood of $p$.
The function is continuous if it is continuous at every point.
Thus homeomorphism carries over:
a bijection which is continuous in both directions.
A homeomorphism of topological spaces
$(X, _X)$ and $(Y, _Y)$
is a bijection $f X Y$
which induces a bijection from $_X$ to $_Y$:
i.e.\ the bijection preserves open sets.
Show that this is equivalent to $f$ and its inverse
both being continuous.
Therefore, any property defined only in
terms of open sets is preserved by homeomorphism.
Such a property is called a topological property.
The later adjectives we define
(``connected'', ``Hausdorff'', ``compact'') will all be defined
only in terms of open sets, so they will be topological properties. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 5 | 1,391 | |
%
% A sequence $(x_n)$ of points in a topological space $X$ is said to converge to $x X$ if for every open neighborhood of $x$,
% eventually all terms of the sequence lie in that open neighborhood.
%
%
% Unfortunately, for general topological spaces we no longer have the nice property
% that any function which preserves sequential limits is automatically continuous.
%
%There's one other property of open sets that we have in a metric space that isn't implied by the above: for any two points of $X$, we can find an open set containing one but not the other.
%A space which also has this property is called a Kolmogorov space.
%This property is a good property to have, because if $x,y X$ are in the same open sets, the topology can't tell them apart.
Closed sets
We saw last time there were two equivalent definitions
for closed sets, but one of them relies only on open sets, and we use it:
def:closure
In a general topological space $X$, we say that $S X$ is
closed in $X$ if the complement $X S$ is open in $X$.
If $S X$ is any set, the closure of $S$,
denoted $ S$,
is defined as the smallest closed set containing $S$.
Thus for general topological spaces,
open and closed sets carry the same information,
and it is entirely a matter of taste whether we define everything in terms
of open sets or closed sets.
In particular, you can translate axioms and properties of open sets to closed ones:
Show that the (possibly infinite) intersection of closed sets is closed
while the union of finitely many closed sets is closed.
(Look at complements.)
Show that a function is continuous if and only if the pre-image
of every closed set is closed.
Mathematicians seem to have agreed that they like open sets better.
Properties that don't carry over
Not everything works:
[Complete and (totally) bounded are metric properties]
The two metric properties we have seen,
``complete'' and ``(totally) bounded'',
are not topological properties.
They rely on a metric,
so as written we cannot apply them to topological spaces.
One might hope that maybe,
there is some alternate definition (like we saw for ``continuous function'')
that is just open-set based.
But ex:fishy showing $(0,1) $
tells us that it is hopeless.
[Sequences don't work well]
You could also try to port over the notion
of sequences and convergent sequences.
However, this turns out to break a lot of desirable properties.
Therefore I won't bother to do so,
and thus if we are discussing sequences you should
assume that we are working with a metric space.
Hausdorff spaces
As you might have guessed,
there exist topological spaces which cannot be realized
as metric spaces (in other words, are not metrizable).
One example is just to take $X = \a,b,c\$ and the topology
$_X = \ , \a,b,c\ \$.
This topology is fairly ``stupid'':
it can't tell apart any of the points $a$, $b$, $c$!
But any metric space can tell its points apart (because $d(x,y) > 0$ when $x y$).
We'll see less trivial examples later,
but for now we want to add a little more sanity condition onto our spaces.
There is a whole hierarchy of such axioms, labelled $T_n$ for
integers $n$ (with $n=0$ being the weakest and $n=6$ the strongest);
these axioms are called separation axioms.
By far the most common hypothesis is the $T_2$ axiom,
which bears a special name.
A topological space $X$ is Hausdorff if
for any two distinct points $p$ and $q$ in $X$,
there exists an open neighborhood $U$ of $p$
and an open neighborhood $V$ of $q$ such that
\[ U V = . \]
In other words, around any two distinct points we should be
able to draw disjoint open neighborhoods.
Here's a picture to go with above,
but not much going on.
size(5cm);
pair p = (-1.5,0);
pair q = ( 1.5,0);
filldraw(CR(p,1), opacity(0.1)+lightcyan, blue+dashed);
filldraw(CR(q,1), opacity(0.1)+lightred, red+dashed);
dot("$p$", p, dir(-90), blue);
dot("$q$", q, dir(-90), red);
Show that all metric spaces are Hausdorff.
I just want to define this here so that I can use this word later.
In any case, basically any space we will encounter other than
the Zariski topology is Hausdorff.
Subspaces
One can also take subspaces of general topological spaces.
Given a topological space $X$, and a subset $S X$,
we can make $S$ into a topological space
by declaring that the open subsets of $S$ are $U S$ for open $U X$.
This is called the subspace topology.
So for example, if we view $S^1$ as a subspace of $^2$,
then any open arc is an open set,
because you can view it as the intersection of an open disk with $S^1$.
size(3cm);
draw(unitcircle, black+1);
MP("S^1", dir(60), dir(60));
MP(" R^2", dir(-45)*1.2, dir(-45));
pair A = dir(-30);
pair B = dir(50);
draw(CP(dir(10), A), dotted);
draw(arc(origin,A,B), blue+2);
dotfactor *= 2;
opendot(A, blue);
opendot(B, blue);
Needless to say, for metric spaces it doesn't matter
which of these definitions I choose.
(Proving this turns out to be surprisingly annoying, so I won't do so.)
Connected spaces
Even in metric spaces, it is possible for a set to be both open and closed.
A subset $S$ of a topological space $X$
is clopen if it is both closed and open in $X$.
(Equivalently, both $S$ and its complement are open.)
For example $$ and the entire space are examples of clopen sets.
In fact, the presence of a nontrivial clopen set other
than these two leads to a so-called disconnected space. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 6 | 1,531 | |
Show that a space $X$ has a nontrivial clopen set (one other than $$ and $X$)
if and only if $X$ can be written as a disjoint union of two nonempty open sets.
We say $X$ is disconnected if there are nontrivial clopen sets,
and connected otherwise.
To see why this should be a reasonable definition,
it might help to solve prob:disconnected_better_def.
The metric space
\[ \ (x,y) x^2+y^2 1 \
\ (x,y) (x-4)^2+y^21\ ^2 \]
is disconnected (it consists of two disks).
The space $[0,1] [2,3]$ is disconnected:
it consists of two segments,
each of which is a clopen set.
A discrete space on more than one point is disconnected,
since every set is clopen in the discrete space.
Convince yourself that the set
\[ \ x x^2 < 2014 \ \]
is a clopen subset of $$.
Hence $$ is disconnected too -- it has gaps.
$[0,1]$ is connected.
%
% For general topological spaces $X$, it is still true that if $S$ is closed
% (meaning $X S$ is open), then it contains the limits of all its sequences.
% But the converse of this statement no longer holds.
%
Path-connected spaces
A stronger and perhaps more intuitive notion
of a connected space is a path-connected space.
The short description: ``walk around in the space''.
% We have general topological spaces, but so far they've just been sitting there.
% Let's start walking in them.
% Throughout this section, we let $I = [0,1]$ with the usual topology
% inherited from $$.
%Also, recall that $S^n$ is the surface of the sphere living
%in $n+1$ dimensional space; hence $S^1$ is a unit circle, say.
A path in the space $X$ is a continuous function
\[ [0,1] X. \]
Its endpoints are the two points $(0)$ and $(1)$.
You can think of $[0,1]$ as measuring ``time'', and so we'll often write $(t)$
for $t [0,1]$ (with $t$ standing for ``time'').
Here's a picture of a path.
bigblob("$X$");
pair A = Drawing("(0)", (-3,-1));
pair B = Drawing("(1)", (2,1), dir(90));
path p = A..(-2,0)..(0,2)..(1,0)..B;
draw(p, EndArrow);
MP("", midpoint(p), dir(90));
Why does this agree with your intuitive notion of what a ``path'' is?
A space $X$ is path-connected if
any two points in it are connected by some path.
[Path-connected implies connected]
Let $X = U V$ be a disconnected space.
Show that there is no path
from a point of $U$ to point $V$.
(If $ [0,1] X$, then we get
$[0,1] = (U) (V)$,
but $[0,1]$ is connected.)
$^2$ is path-connected,
since we can ``connect'' any two points with a straight line.
The unit circle $S^1$ is path-connected, since
we can just draw the major or minor arc to connect two points.
% Easy, right?
Homotopy and simply connected spaces
$.
sec:meteor
Now let's motivate the idea of homotopy.
Consider the example of the complex plane $$ (which you can
think of just as $^2$) with two points $p$ and $q$.
There's a whole bunch of paths from $p$ to $q$ but somehow
they're not very different from one another.
If I told you ``walk from $p$ to $q$'' you wouldn't have too many questions.
unitsize(0.8cm);
bigbox("$ C$");
pair A = Drawing("p", (-3,0), dir(180));
pair B = Drawing("q", (3,0), dir(0));
draw(A..(-2,0.5)..(0,2)..(1,1.2)..B, red, EndArrow(TeXHead));
draw(A--B, mediumgreen, EndArrow(TeXHead));
draw(A--(-1,-1)--(2,-1)--B, blue, EndArrow(TeXHead));
// draw(A..(1.5,-2)..(-1.5,-2)..B, orange, EndArrow(TeXHead));
So we're living happily in $$ until a meteor strikes the origin,
blowing it out of existence.
Then suddenly to get from $p$ to $q$, people might tell you two different things:
``go left around the meteor'' or ``go right around the meteor''.
unitsize(0.8cm);
bigbox("$ C \0\$");
pair A = Drawing("p", (-3,0), dir(180));
pair B = Drawing("q", (3,0), dir(0));
draw(A..(-2,0.5)..(0,2)..(1,1.2)..B, red, EndArrow(TeXHead));
draw(A--(-1,-1)--(2,-1)--B, blue, EndArrow(TeXHead));
filldraw(scale(0.5)*unitcircle, grey, black);
So what's happening?
In the first picture, the red, green, and blue paths somehow all looked
the same: if you imagine them as pieces of elastic string pinned down
at $p$ and $q$, you can stretch each one to any other one.
But in the second picture,
you can't move the red string to match with the blue string:
there's a meteor in the way.
The paths are actually different.If you know about winding numbers,
you might feel this is familiar.
We'll talk more about this in the chapter on the fundamental group.
The formal notion we'll use to capture this is homotopy equivalence.
We want to write a definition such that in the first picture,
the three paths are all homotopic, but the two paths in the
second picture are somehow not homotopic.
And the idea is just continuous deformation.
Let $$ and $$ be paths in $X$ whose endpoints coincide.
A (path) homotopy from $$ to $$ is a continuous function
$F [0,1]^2 X$, which we'll write $F_s(t)$ for $s,t [0,1]$,
such that
\[ F_0(t) = (t) and F_1(t) = (t)
for all $t [0,1]$ \]
and moreover
\[ (0) = (0) = F_s(0)
and
(1) = (1) = F_s(1)
for all $s [0,1]$. \]
If a path homotopy exists, we say $$ and $$
are path homotopic and write $ $.
While I strictly should say ``path homotopy'' to describe this relation
between two paths, I will shorten this to just ``homotopy'' instead.
Similarly I will shorten ``path homotopic'' to ``homotopic''.
Animated picture: https://commons.wikimedia.org/wiki/File:HomotopySmall.gif.
Needless to say, $$ is an equivalence relation. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 7 | 1,520 | |
What this definition is doing is taking $$ and ``continuously deforming'' it to $$, while keeping the endpoints fixed.
Note that for each particular $s$, $F_s$ is itself a function.
So $s$ represents time as we deform $$ to $$:
it goes from $0$ to $1$, starting at $$ and ending at $$.
size(9cm);
bigbox("$ C$");
pair A = Drawing("p", (-3,0), dir(180));
pair B = Drawing("q", (3,0), dir(0));
draw(A..MP("F_0 = ", (0,2), dir(45))..B, heavygreen);
draw(A..MP("F_0.25", (0,1), dir(45))..B, mediumgreen);
draw(A..MP("F_0.5", (0,0), dir(90))..B, palecyan);
draw(A..MP("F_0.75", (0,-1), dir(-45))..B, mediumcyan);
draw(A..MP("F_1 = ", (0,-2), dir(-45))..B, heavycyan);
// draw(A..(1.5,-2)..(-1.5,-2)..B, orange, EndArrow);
Convince yourself the above definition is right.
What goes wrong when the meteor strikes?
So now I can tell you what makes $$ special:
A space $X$ is simply connected if it's path-connected and
for any points $p$ and $q$, all paths from $p$ to $q$ are homotopic.
That's why you don't ask questions when walking from $p$ to $q$ in $$:
there's really only one way to walk. Hence the term ``simply'' connected.
Convince yourself that $^n$ is simply connected for all $n$.
Bases of spaces
You might have noticed that the open sets of $$ are a little annoying to describe:
the prototypical example of an open set is $(0,1)$,
but there are other open sets like
\[
(0,1)
( 1, 32 )
( 2, 73 )
(2014, 2015). \]
Check this is an open set.
But okay, this isn't that different.
All I've done is taken a bunch of my prototypes and threw a bunch of $$ signs at it.
And that's the idea behind a basis.
A basis for a topological space $X$
is a subset $ B$ of the open sets such that every open set in $X$
is a union of some (possibly infinite) number of elements in $ B$.
And all we're doing is saying:
The open intervals form a basis of $$.
In fact, more generally we have:
The $r$-neighborhoods form a basis of any metric space $M$.
Kind of silly -- given an open set $U$, for every point $p$ inside $U$,
draw an $r_p$-neighborhood $U_p$ contained entirely inside $U$.
Then $_p U_p$ is contained in $U$ and covers every point inside it.
Hence, an open set in $^2$ is nothing more than a union
of a bunch of open disks, and so on.
The point is that in a metric space, the only open sets you really
ever have to worry too much about are the $r$-neighborhoods.
Let $X$ be a topological space.
Show that there exists a nonconstant continuous function $X \0,1\$ if and
only if $X$ is disconnected (here $\0,1\$ is given the discrete topology).
prob:disconnected_better_def
Let $X$ and $Y$ be topological spaces
and let $f X Y$ be a continuous function.
Show that if $X$ is connected then so is $f(X)$.
Show that if $X$ is path-connected then so is $f(X)$.
[Hausdorff implies $T_1$ axiom]
Let $X$ be a Hausdorff topological space.
Prove that for any point $p X$ the set $\p\$ is closed.
[ref:pugh, Exercise 2.56]
Let $M$ be a metric space with more than one point
but at most countably infinitely many points.
Show that $M$ is disconnected.
Let $p$ be any point.
If there is a real number $r$ such that $d(p,q) r$ for any $q M$,
then the $r$-neighborhood of $p$ is clopen.
Let $X$ be a topological space.
The connected component of a point $p X$
is the union of all subspaces $S X$ which are connected and contain $p$.
Does the connected component of a point have to be itself connected?
Does the connected component of a point have to be an open subset of $X$?
(a) is yes, and (b) is no even for metric spaces.
In fact, a totally disconnected space is one for which
every connected component consists of only a single point,
and there are examples of totally disconnected metric spaces with
non-discrete topologies.
Part (a) is straightforward:
assume for contradiction that the connected component of $p$ is
a disjoint union $U V$ of two nonempty sets open in $X$.
WLOG, assume $x U$
Let $S$ be one of the subspaces containing $X$ that intersects $V$.
Then $S = (S U) (S V)$ rewrites $S$ as the disjoint union
of two sets which are open in $S$, contradicting the connectedness of $S$.
(Though note that as $S$ is not necessarily open in $X$,
the sets $S U$ and $S V$ are not necessarily open in $X$ either.)
For (b), a counterexample is to take any totally disconnected space
like the Cantor set or the $p$-adic numbers.
We declare a subset of $$ to be open if it's the union (possibly empty or infinite)
of arithmetic sequences $\ a + nd n \$,
where $a$ and $d$ are positive integers.
Verify this forms a topology on $$,
called the evenly spaced integer topology.
Prove there are infinitely many primes by considering $_p p$
for primes $p$.
Note that $p$ is closed for each $p$.
If there were finitely many primes, then
$ p = \-1,1\$ would have to be closed;
i.e.\ $\-1,1\$ would be open, but all open sets here are infinite.
Prove that the evenly spaced integer topology on $$ is metrizable.
In other words, show that one can impose a metric $d ^2 $
which makes $$ into a metric space whose open sets are those described above.
% https://teratologicmuseum.wordpress.com/2009/05/05/a-metric-for-the-evenly-spaced-integer-topology/
The balls at $0$ should be of the form $n! $.
Let $d(x,y) = 2017^-n$, where $n$ is the largest integer
such that $n!$ divides $ x-y $. | An Infinitely Large Napkin | napkin | general | advanced | Basic Topology | Topological spaces | 11_top-more.md | 8 | 1,514 |
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