Mathematics - 2017-12-01 (page 3 of 6)

Leaky Nun

2:00 AM

$g \circ f$ is also $1$

Daminark

Also wait surjective compose injective is not bijective either

Leaky Nun

:o $f$ is invertible!!!!

Secret

hmm...

Daminark

@Leaky

Also @Balarka on general principle

MatheinBoulomenos

that's not as good as the tetrathonk

Balarka Sen

2:02 AM

Daminark

True, but we here in damiTHONK studios like variety

Leaky Nun

(diagram is commutative)

according to my book

how many exact sequences are there?

@MatheinBoulomenos

MatheinBoulomenos

Ugh, I'm not into combinatorics :P

Leaky Nun

eh

Secret

surjective o injective = injective?

Leaky Nun

2:06 AM

so all of them are exact?

Daminark

@MatheinBoulomenos >:(

MatheinBoulomenos

@Daminark graph theory is fine

Leaky Nun

@Secret lemme think it through

Daminark

@Mathein even with graph theory, not liking combinatorics is unacceptable!!

Leaky Nun

@Secret take $\{1,2\} \twoheadrightarrow \{1\} \xrightarrow \sim \{1\}$. The composition is clearly not injective. Take $\{1,2\} \xrightarrow \sim \{1,2\} \twoheadrightarrow \{1\}$. The composition is still not injective.

note that $\xrightarrow \sim \implies \hookrightarrow$

you're welcome to take my three arrows out of context

Secret

2:15 AM

I think I swap them around again: Another sanity check:
surjective o injective = surjective
injective o surjective = surjective?

Daminark

Neither is correct

Leaky Nun

I think you only have injective o injective = injective, surjective o surjective = surjective, and bijective o bijective = bijective

indeed, if $f$ and $g$ are monic, denoting $h = f \circ g$, we see that $h \circ x = h \circ y \implies f \circ (g \circ x) = f \circ (g \circ y) \implies g \circ x = g \circ y \implies x = y$, so $h$ is monic.

A dual argument shows that $h = f \circ g$ is epic if $f$ and $g$ are epic

Daminark

@LeakyNun that's pretty epic

Leaky Nun

$\Bbb Z \twoheadrightarrow \Bbb Q$ just to confuse people :P

MatheinBoulomenos

it's an epimorphism

Leaky Nun

2:19 AM

is it monic?

MatheinBoulomenos

that's perfectly fine if you're working in the category of rings

sure

Leaky Nun

@MatheinBoulomenos I know

so it's epic and monic

how do I show that in an arrow?

MatheinBoulomenos

monic + epic does not imply iso in general

not sure about the LaTeX command

Leaky Nun

I don't think there is a command

Secret

hmm...
function o relation = function
relation o function = non total relation
non total function o function = non total function
function o non total function = non total function

function o any function = function
any function o function = function
injective o surjective = function
surjective o surjective = surjective
injective o injective = injective
surjective o injective = function
bijective o injective = injective
bijective o surjective = surjective
bijective o bijective = bijective
injective o bijective = injective

MatheinBoulomenos

2:28 AM

LOL, this thread: reddit.com/r/math/comments/7gqhlc/…

Leaky Nun

@Secret the lower part looks good to me, don't want to check the upper part

Secret

hmm, bijection seemed to act like an identity under function composition...

Leaky Nun

@MatheinBoulomenos "By a simple diagonalization argument"

Daminark

@Mathein amazing

Okay so my TA in algebra, especially for the first bit of the class, was pedantic to the point of seriously pissing me off

MatheinBoulomenos

"Assuming the reader’s intellect approaches that of the writer, it should be obvious that"

Secret

2:30 AM

Is it true that bijections are identity elements under the associative algebra $(f,\circ)$?

Daminark

On our first pset I got knocked a lot because I didn't specify left vs right inverse

MatheinBoulomenos

@Secret only the identity is an identity element in that algebra

Daminark

And I remember at one point I was about to say "Clearly..." while working on something with friends and I was like

Perturbative

Didn't specify you are a human on problem set, gets marks docked

Daminark

BALEETED

2

Leaky Nun

2:32 AM

@MatheinBoulomenos what do isomorphisms correspond to in higher category?

Daminark

I don't want my name associated to that on the starboard

Balarka Sen

@Leaky Isomorphisms :P

MatheinBoulomenos

yeah

Leaky Nun

@BalarkaSen I thought morphisms become objects

Balarka Sen

No...

You just have morphisms between morphisms

and so on

Perturbative

2:33 AM

You can make a category where the objects are morhpisms though

Leaky Nun

exactly

Secret

I see. What term should I use to describe the following behaviour?
bijective o function= function
bijective o surjective = surjective
bijective o injective = injective
function o bijective = function
surjective o bijective = surjective
injective o bijective = injective
It seems in general: bijective o any function = any function o bijective = any function?

Perturbative

Then the morphisms become commutative diagrams in that case

Balarka Sen

That's not what a higher category is

Leaky Nun

@BalarkaSen I see

what is that then?

Daminark

2:34 AM

But yeah like I think at one point when we were doing something and someone was making a pedantic point, I was like yeah yeah sure, reminded me of my TA, and I was all like "Aight I'll just say "We assume that the reader has at least finished 10th grade, at which point we can confidently assume that it will be clear...""

Balarka Sen

It's not easy to define. You have n-morphisms between (n-1)-morphisms, iteratively, satisfying the various coherence conditions.

There are models to do it

Leaky Nun

> Instead of QED, I prefer "and Bob's your uncle."

Balarka Sen

For infinity categories, the easiest model is a topological space.

Leaky Nun

@BalarkaSen I mean, what's a category where morphisms become objects

Balarka Sen

idk and idc

Daminark

2:35 AM

@Balarka youtube.com/watch?v=fn9oXl9tyG0&t=75s

Secret

@MatheinBoulomenos It seems in the associative algebra $(f\circ)$ of function compositions modulo the details of each function, injections, surjections are idempotents, neither injective nor surjective functions form an absorbing element and bijection acts like a two sided identity?

MatheinBoulomenos

@LeakyNun for any two categories $C$ and $D$ you can form a category whose objects are functors from $C \to D$ and morphisms are natural transformations

Perturbative

@LeakyNun Those are slice categories

Leaky Nun

> Standard Proof that the square root of 2 is irrational:
>> Suppose for the sake of contradiction that the square root of 2 were rational. Then we could find relatively prime integers p and q such that (p/q)2 = 2. Then p2 = 2q2. This means p2 is even, therefore p is even, so we can write p = 2r. Then 4r2 = 2q2, so 2r2 = q2. This means q2 is even, therefore q is even, which contradicts p and q being relatively prime.
> Souped-up proof:
>> Suppose for the sake of contradiction that the square root of 2 were rational. Then indisputably we could find relatively prime positive integers p and q …

MatheinBoulomenos

this works in any 2-category

Perturbative

2:36 AM

@LeakyNun They are special cases of comma categories

Leaky Nun

@Perturbative I mean, where morphisms are commutative diagrams

Perturbative

Yep, that's what I'm referring to

Leaky Nun

I thought they are dual of comma

and what do isomorphisms become?

@Secret interesting

Balarka Sen

I never think about slice categories as categories where morphisms become object, but sure, that works

MatheinBoulomenos

@LeakyNun the souped-up version reads much better

Balarka Sen

2:37 AM

It's not a very good description though

Perturbative

@LeakyNun Hmmm, that's not what my textbook says

Balarka Sen

The morphisms become commutative triangles

The isomorphisms the same, but with base of the triangle being an isomorphism

Leaky Nun

@MatheinBoulomenos the last line is gold

I mean, what becomes isomorphisms

MatheinBoulomenos

isomorphisms that commute with stuff

Balarka Sen

^

it's nothing special d00d

isomorphisms don't become unicorns

2

Leaky Nun

2:39 AM

I mean, what becomes of the objects that represent isomorphisms in the original category

Secret

What is the name of the associative algebra $(f,\circ)$ that ignores the details of $f$ except whether it is injective, surjective, bijective and so on. It seemed to be some kind of quotient algebra but I am not sure what is the equivalence class?

Leaky Nun

the Secret algebra

Perturbative

The isomorphisms could be anything in the original category satisfying the conditions for it to be an isomorphism

The construction works for any category $C$

Leaky Nun

@Secret take $X$ a set, $Y$ the set of endomorphisms of $X$, then $\operatorname{Inj}_X, \operatorname{Sur}_X, \operatorname{Bi}_X \subseteq Y$, and the sets are preserved under composition

MatheinBoulomenos

@BalarkaSen do you ever think of compact Hausdorff spaces as algebras over the ultrafilter monad?

Balarka Sen

2:42 AM

oh g0d no

what is that

MatheinBoulomenos

do you know monads?

Leaky Nun

@Secret and $\operatorname{Inj}_X, \operatorname{Sur}_X \subseteq \operatorname{Bi}_X$

Secret

@LeakyNun Interesting, I see

Leaky Nun

That generalizes to $\operatorname{Mon}_X, \operatorname{Epi}_X \subseteq \operatorname{Iso}_X \subseteq \operatorname{End}_X$

Perturbative

Does anyone think of groups as groupids with a single object?

Leaky Nun

2:43 AM

@Perturbative me, I do that all the time

Balarka Sen

@MatheinBoulomenos Nope

Secret

I think I am going to like category theory, I am sometimes so annoyed by details and embraces minimalism

Balarka Sen

there is a difference between minimalism and vulgarism

Leaky Nun

@BalarkaSen how dare you insult category theory

MatheinBoulomenos

@BalarkaSen okay a monad on a category $C$ is an endofunctor $T$ of $C$ together with natural transformations $\mu: T^2 \to T$ and $\eta: \operatorname{id} \to T$ satisfying certain obvious identities, i.e. associativity for $\mu$ and $\eta$ behaves like a left and right unit wrt to $\mu$. If you have a monad, you can define algebras over that monad, which is an object $A$ in $C$ together with a morphism $TA \to A$ such that certain diagramms comute.

Balarka Sen

2:47 AM

@LeakyNun category theory insults itself

MatheinBoulomenos

There's also a notion of morphisms for algebras over a monad

Leaky Nun

no blasphemy is allowed in this christian chat

we worship category theory our lord and saviour

2

MatheinBoulomenos

the cool thing is that there are monads on $\mathbf{Set}$ such that all the algebraic structures arise as algebras over these monads

Balarka Sen

The Old Testament does not approve of category theory

You can go look it up

MatheinBoulomenos

but here's the thing. The functor $\mathbf{Set} \to \mathbf{Set}$ that sends a set to the set of ultrafilters on it, can be turned into a monad such that algebras over that are exactly compact hausdorff spaces

Leaky Nun

2:49 AM

ultrafilters

MatheinBoulomenos

This means that the category of compact hausdorff spaces is an algebraic category

Perturbative

The Old Testament does not have the final say in the approval of Category Theory

Leaky Nun

I mean, ultrafilter is used to prove compactness theorem :P

Balarka Sen

@MatheinBoulomenos Ah I see interesting

MatheinBoulomenos

That monad comes in a certain sense (which is quite involved, it's a "codensity monad") from the inclusion functor from the category of finite sets to the category of sets

So we can go from "we forget that a set is finite" to the category of compact spaces just by abstract nonsense

Leaky Nun

2:52 AM

@MatheinBoulomenos what is an algebraic category?

MatheinBoulomenos

Which is pretty mindblowing if you ask me

@LeakyNun the definition I know is that it is a category which is equivalent to the category of algebras over some monad on the category of sets

Leaky Nun

ok

Balarka Sen

@MatheinBoulomenos I am not an algebraist, so it's not totally clear to me, but that does sound quite nice

MatheinBoulomenos

The steps to go from "forgetting that a set is finite" to the ultrafilter monad and to prove that algebras over the ultrafilter monad are compact hausdorff spaces are both quite nontrivial

but it's purely abstract nonsense

I'm just amazed by the fact that compact Hausdorff spaces arise this naturally (no pun intended) from purely formal reasoning

Leaky Nun

$f(x) = \displaystyle \sum_{n=1}^\infty \frac{\sin\left(n\pi x\right)}{n}$

is $f$ continuous at $0$?

Balarka Sen

2:58 AM

noooobody knooows

MatheinBoulomenos

@BalarkaSen if you want to know the details, this paper is quite good: math.leidenuniv.nl/scripties/BachStekelenburg.pdf

Leaky Nun

@BalarkaSen i'm serious

Balarka Sen

@MatheinBoulomenos Thanks, I'll check it out!

@Leaky I am not

Leaky Nun

i can tell

Balarka Sen

(obligatory supa hot fire here)

Secret

3:05 AM

typo: Isoooooooooooooooomorphism, lol

Leaky Nun

"has no name", fair enough

@MatheinBoulomenos what is the name?

Secret

Is there really no names for surjective, or injective endomorphisms?

MatheinBoulomenos

No name I am aware of

Secret

Also image-po

Leaky Nun

@Secret this is so awesome

3:07 AM

rip automorphism

~~ooops~~ fixed

@MatheinBoulomenos @BalarkaSen how to find the residue of $\sin(\cot(z))$ around $z=0$? I couldn't do it when I was in a competition lol

@Secret fair enough

MatheinBoulomenos

who would want to find a residue of such a horrible function?

Leaky Nun

apparently the designers of the competition

MatheinBoulomenos

3:10 AM

I suck at competition math

Balarka Sen

same

Secret

but yeah, since only injection, surjection and bijection are idempotent, while the neither function absorbs everything, there is no straightforward composition rule for the glyphs

Leaky Nun

I should have just found the coefficient of $z$ of $\sin(\cot(z^{-1}))$ in hindsight

Secret

Normal morphism

In category theory and its applications to mathematics, a normal monomorphism or conormal epimorphism is a particularly well-behaved type of morphism. A normal category is a category in which every monomorphism is normal. A conormal category is one in which every epimorphism is conormal. == Definition == A monomorphism is normal if it is the kernel of some morphism, and an epimorphism is conormal if it is the cokernel of some morphism. A category C is binormal if it's both normal and conormal. But note that some authors will use the word "normal" only to indicate that C is binormal. == ...

There's also this thing

sounds like a generalisation of exact sequences

Daminark

Hold your breath for one second for each time the word "normal" appears in mathematics

You'll die of old age before you finish counting and start the challenge

Leaky Nun

3:18 AM

that's normal @Daminark

Balarka Sen

This book does every problem in the most convoluted way possible like jeebuz

Suppose $x_1, \cdots, x_n$ are roots of $x^n + x^{n-1} + \cdots + 1 = 0$. Compute $\sum_{i = 1}^n 1/(x_i - 1)$

This book spends a million words on this

MatheinBoulomenos

"p-adics are weird man, I mean just look at their topology" "idk what you mean, they're perfectly normal"

Perfectly normal is actually a thing

Leaky Nun

@BalarkaSen unfortunate choice of symbols

Secret

it also does not help that all "normals" don't mean the same notion. e.g. normal subgroups N are those which $gN=Ng$ holds for all g, normal distribution is a particular pdf that is symmetric at the centre, and a vector is normal to another if the inner product is zero

John Ma

@LeakyNun Hi!

Balarka Sen

3:21 AM

@Leaky blame the book, not me

Leaky Nun

@JohnMa hi

Daminark

Nah we blame you

Leaky Nun

@BalarkaSen how to do it though

Balarka Sen

I used symmetry

it's a one-liner

Do you want me to explain, or try it yourself?

I think you can do it pretty easily

Leaky Nun

the former :P

Balarka Sen

3:22 AM

Ok. Say that sum right there, that little piece of shit is $S$

5

Look at $S + n$

add $1$ termwise to get $\sum_i x_i/(x_i - 1)$

that's $\sum_i 1/(1 - 1/x_i)$

$1/x_i$ are all roots of that equation so this is just a rearrangement

$S+n = -S$

solve

Semiclassical

fun new word I learned today: elliptope

Leaky Nun

interesting

ermergerd that's brilliant

conjugate is an involution / automorphism of Q[z_n]

Balarka Sen

@Semiclassical How does that look like

Semiclassical

well, an n-elliptope is the set of n-by-n correlation matrices

Daminark

I dunno what it looks like but it sounds pretty tope

2

Balarka Sen

3:25 AM

lmao

Semiclassical

so the 3-elliptope is another name for my volume from earlier

Balarka Sen

I am going to end up like Chryssippus if this continues

@Semiclassical ah

MatheinBoulomenos

@BalarkaSen that's how I'm going to introduce variables from now on

Semiclassical

@BalarkaSen www4.ncsu.edu/~clvinzan/slides/…

Balarka Sen

Let me catch my breath first

Nope, this isn't working. I'll come back later

Semiclassical

3:29 AM

kk

anon

proof is more down-to-earth if you switch x_i with its inverse to begin with

Balarka Sen

That's true

I was more exasperated at my book's approach to it, which is by constructing a polynomial with roots x_1 -1, ... x_n - 1....

it's not good

Leaky Nun

@BalarkaSen newton sums and geometric series :P

Balarka Sen

right

@Semiclassical aha, nice

I like how they mention secant varieties is the algebraic analogue of convex hull in the semialgebraic context

Semiclassical

ooo, nice

Leaky Nun

3:37 AM

what's a variety but for power series?

Semiclassical

the more important thing is that learning that word lead me to this: perimeterinstitute.ca/videos/…

Balarka Sen

It's a nice way to think about it

@Semiclassical whistles

Semiclassical

yeah

that's pretty much the kind of thing I was aiming for

Balarka Sen

so it seems

Semiclassical

so...on the one hand, I was definitely onto something

on the other hand, it looks like others have had that same thought

Balarka Sen

3:38 AM

Maybe you should think more about it

and write a paper

Semiclassical

maybe

Balarka Sen

and add me in the acknowledgements :P:P:P

I mean it definitely seems like you have some genuine ideas about this, given all the materials flying around

Semiclassical

well, I probably will get a paper out of it

Balarka Sen

Coolio

Semiclassical

not sure much beyond that

Balarka Sen

3:40 AM

Maybe you'll become a semialgebraic geometer

the semi-adjective is going to haunt you forever

Semiclassical

ha

semi-academic :/

Balarka Sen

lmao

MatheinBoulomenos

sounds almost as good as an algebraic geometer

Balarka Sen

I wish I knew more about semialgebraic geometry. It's really a beautiful branch

Also a lot more analytical in nature than algebraic geometry

MatheinBoulomenos

Is semialgebraic geometry the same as real algebraic geometry?

Leaky Nun

3:45 AM

@MatheinBoulomenos semi-same

Balarka Sen

Well, they are closely related. You study polynomial inequalities along with equalities

Leaky Nun

Semiprofessor @Semiclassical

MatheinBoulomenos

@BalarkaSen that's the description I heard of real algebraic geometry as well

Balarka Sen

Hm

Well the only actual theorem in real algebraic geometry I know of is Nash's theorem

MatheinBoulomenos

The thing is, over $\Bbb R$, even the image of an algebraic set under a polynomial map need not be algebraic

Balarka Sen

3:46 AM

which says any compact manifold can be realized as a real algebraic variety in some Euclidean space

MatheinBoulomenos

consider the projection of the unit circle $x^2+y^2=1$ in $\Bbb R^2$ onto the first coordinate

you get an interval

which is not an algebraic set

Balarka Sen

mhm, that's why you take into account the semialgebraic sets

That's a good point

It's like how in affine algebraic geometry you want to look at quasi-affine sets too

because those arise as image of affine varieties by regular maps

(xy = 1 in A^2 projects to A^1 - 0, which ain't affine bruh)

Secret

I need a semiprojection

but it appears to only exist in computational science

Leaky Nun

@MatheinBoulomenos what is an algebraic set?

MatheinBoulomenos

@LeakyNun the set of common zeroes of a bunch of polynomials

Balarka Sen

3:51 AM

same as algebraic variety, really

Leaky Nun

so a variety?

dem

MatheinBoulomenos

some authors require varieties to be irreducible

Balarka Sen

meh

But yeah I always found Nash's theorem very surprising

MatheinBoulomenos

it's pretty surprising

Antonios-Alexandros Robotis

theproofistrivial.com

lul

Balarka Sen

3:55 AM

You can't actually realize every compact manifold as $F = 0$ inside some Euclidean space where $0$ is a regular value of $F$. A necessary restriction is that you have to have your manifold to be stably parallelizable

cuz being cut out by $F$ immediately means the normal bundle is trivial

The key point is Nash doesn't require transversality. You'll cut out your thing non-transversely at the expense :/

Leaky Nun

var intros = [
    "Just biject it to a",
    "Just view the problem as a",
];

var adjectives = [
    "abelian",
    "associative",
    "computable",
    "Lebesgue-measurable",
    "semi-decidable",
    "simple",
    "combinatorial",
    "structure-preserving",
    "diagonalizable",
    "nonsingular",
    "orientable",
    "twice-differentiable",
    "thrice-differentiable",
    "countable",
    "prime",
    "complete",
    "continuous",
    "trivial",
    "3-connected",
    "bipartite",
    "planar",

Semiclassical

"The proof is trivial! Just view the problem as a context-free orbit whose elements are context-free Betti numbers"

lol, that's a pretty good one

Leaky Nun

> Mathematical objects that can't contain elements (unless you're a set theorist)

Balarka Sen

I feel like learning intersection homology and semialgebraic geometry together could be a fun reading project at some point

MatheinBoulomenos

intersection homology? I'm not really into such perversities

Balarka Sen

4:10 AM

is that an intentional pun?

MatheinBoulomenos

I'd say it's an attempt at a pun

Balarka Sen

Close enough

I don't know what's up with perverse sheaves though

MatheinBoulomenos

One of my profs actually wrote a book on perverse sheaves

but I don't know what they're about either

still have to figure out that derived business

Adeek

hi @BalarkaSen

Balarka Sen

hi

Adeek

4:21 AM

I was wondering if we look at the family of all vector bundles over a complex manifolds does this give something ?

Balarka Sen

That seems like an immensely vague question

Adeek

I.e let us fix a complex manifold and look at all possible Vector bundle over a complex manifold X. Does let us say new complex structure on X ?

or all possible complex structure on X ?

as we vary through the family ?

Balarka Sen

This is too vague for me to give any intelligent comments

MatheinBoulomenos

If your space is compact Hausdorff, then topological K-theory does involve looking at all possible vector bundles on X

but that gives you a ring

not a topological space

Balarka Sen

There is also nothing special about having an underlying complex structure

Adeek

4:25 AM

oh

MatheinBoulomenos

Right, it's just the only thing that I could think of that involves looking at all possible vector bundles on a space

Balarka Sen

Besides, the K-groups are not literally all possible v.b.s on the space; it's v.b.s upto an equivalence relation (stable equivalence? I don't even remember anymore)

@MatheinBoulomenos Fair

Adeek

I mean I was thinking if each $\pi$ is holomorphic we get induced complex manifold on E. We can also think the other way. I was thinking maybe we can somehow associate to each topological space all possible complex structure by looking at all possible twists of such a space?

MatheinBoulomenos

@BalarkaSen now that I think about it, there's nothing that stops you from constructing a K-theory by considering only holomorphic bundles

It's not something I've seen before, but I would be surprised if it's not out there

Balarka Sen

KU theory something something

I mean you'll get the cohomology theory dual to the spectrum BSU(n) that's all

'cuz holomorphic v.b. means the structure group is SU(n) (right?)

Adeek

4:29 AM

cool I want to study topological K-theory at some point.

Balarka Sen

@Adeek I am not sure how easy it is to talk about the space of all complex structures. You could look at the space of all almost complex structures, which are just endomorphism $J : TM \to TM$ such that $J^2 = -I$. That's a subbundle of $End(TM)$, I guess?

I don't know how to construct the moduli space of all complex structures, which should be a subspace of the space of all almost complex structures

Adeek

oh I see

MatheinBoulomenos

For surfaces, there is Teichmüller theory

Balarka Sen

For surfaces there's the Teichmuller spaces, because complex structure = hyperbolic structure

great minds, great minds, @Mathein

Adeek

cool

MatheinBoulomenos

4:34 AM

But that only classifies complex structures modulo something (I can't remember the exact thing, something with isotopy)

Daminark

@BalarkaSen greater minds think faster OOOOHHHHHH

Balarka Sen

@Daminark fucking hell

@MatheinBoulomenos Ah, yes, upto the action of MCG I think

Adeek

brb

Kevin Driscoll

4:46 AM

I have an ill-formed question: Supposed you have a fiber-bundle. Locally the space is just $M \times F$ where $M$ is some base manifold and $F$ the fiber space. What are the possibilities for the global structure though? I know we can have twisted-products. And there could be lots of twists maybe. But are there other fucked-up kind of structures that we can have?

Balarka Sen

It's really twists of various sorts. Let me tell you a story

What's the simplest manifold in all dimensions?

Kevin Driscoll

$\mathbb{R}^n$?

Balarka Sen

Right. If I demand compactness?

Spheres, right? $S^n$

Kevin Driscoll

Thatd be my guess

Balarka Sen

Ok, so a natural question is to classify vector bundles on the simplest compact manifold

aka, classify vector bundles on the sphere

Kevin Driscoll

4:50 AM

Ok ya good call

Its the simplest compact manifold, and each fiber is the simplest manifold

Balarka Sen

Yup

Kevin Driscoll

or at least the ones we think we knwo the most about

Balarka Sen

Small note is that fiber bundles with R^n fibers are not quite vector bundles though; it's the transition functions which are important - in R^n-bundles you are allowed to have transition functions in Homeo(R^n), whereas in vector bundles they have to be strictly linear, i.e., in GL_n(R)

But yeah whatever

@KevinDriscoll What are the vector bundles on the circle that you know of? This is the simplest non-dumb dimension, $n = 1$.

Kevin Driscoll

So, the trivial bundle (AKA a cylinder). The mobius band. And then..... I was thinking about thing like a mobius band but with 2 twists the other day, and one part of my brain was saying this is somehow like just being a cylinder again. But anotther part of my brain wasnt so sure that there not some way to see the 2 twists

And then I imagine one can just keep twisting maybe

orbit-stabilizer

Question

anon

4:57 AM

the question isn't whether or not you can see the 2 twists (you can, in 3-space), it's whether or not that's equivalent to a band with 0 twists

Balarka Sen

@KevinDriscoll Your brain is quite right. The Moebius strip with "two half-twists" is indeed the same, upto vector bundle equivalence, as the trivial rank 1 bundle on the circle.

You can't see this in $\Bbb R^3$, though

anon

is it not ambient isotopic in R^3 or whatever to a trivial band?

Balarka Sen

Any rank > 1 bundle on $S^1$ that you know of?

@anon Oh, no, it can't be. The boundary is a nontrivial link

anon

ah

orbit-stabilizer

If I'm looking at a group of order 24, why are the only possibilities for a 2-sylow subgroup either 1 or 3. I thought that the number of 2-sylow subgroups has to be congruent to 1 mod 2, ie odd and it has to be a multiple of 3. So it should be possible to have {1,3,9,15,21} 2-sylow subgroups.

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