Mathematics - 2018-11-02 (page 1 of 2)

user193319

00:11

Actually, I think my conjecture is slightly off. If $\mathcal{B}$ denotes the jordan basis, $\mathcal{B}_r$ its "reversion", and $[T]_\mathcal{B} = A_1 \oplus ... \oplus A_p$, then $[T]_{\mathcal{B}_r} = A_p^T \oplus ... \oplus A_1^T$.

So, I was close...Now I just need to figure out how to prove it.

Alucard

00:40

i have an idea for a bijective function between a square and a circle

Roll up and smoke Adjoint

@MatheinBoulomenos mathoverflow.net/questions/314380/…

Ted Shifrin

01:04

@MikeM: ETA not so shabby!

Rithaniel

I named a subset in a proof "I," and now I'm giggling at sentences like "Therefore, I must be countable." (I never knew this about myself)

Ted Shifrin

Are you sure I isn't giggling?

Rithaniel

I might be.

Daminark

Hey guys!

Ted Shifrin

Hi Demonark

Rithaniel

01:09

Sup Daminark.

Daminark

How's everything going?

Ted Shifrin

I'm busy scheduling alumni interviews of prospective MIT students. Never done this before. Should be interesting for the first few ...

Rithaniel

Working on a proof to show that an uncountable set with the cocountable topology is anticompact, and working on rounding off Round 5 in a TIS-100 tournament I am hosting.

Ted Shifrin

What the hell is anticompact?

I've never hoyd of this.

Rithaniel

Only finite subsets of the space are compact.

Ted Shifrin

01:11

Learn something new every year.

Rithaniel

It might be something my professor coined for this particular homework, I don't necessarily know.

Ted Shifrin

I've been around a long while and have never heard it. But I don't get that interested in point-set topology.

It does appear to appear under googling.

Rithaniel

Yeah, I just googled it too.

Ted Shifrin

But it's stilllllll obscure.

Rithaniel

Yep, there's a lot of obscure stuff in math.

Ted Shifrin

01:15

I'd like to think I prefer the less obscure.

Demonark, what are you up to?

Rithaniel

I was about to say that we all would, but actually, I can see a person getting enthused about really obscure stuff.

Ted Shifrin

There are such, yes.

Daminark

Just finished up midterms, doing a rep theory pset right now and later I'll be reading a bunch of stuff for my various classes

Ted Shifrin

I guess beginning of November makes sense for midterms. My sense of time is shot now.

UGA is 3/4 through its semester.

Daminark

Do keep in mind that we have the quarter system, this is 5th week so smack in the middle

Ted Shifrin

01:18

Right, I berember.

Daminark

Then again, not all midterms are in the middle of the term. The chemistry department has a midterm Wednesday of 10th week. Now, turns out Thursday and Friday of 10th week are reading period, and 11th week is finals

Ted Shifrin

Demonark, UGA finally passed a rule that there couldn't be exams the last two days of classes, before final exams.

Still some people love to break it.

Daminark

That's excluding reading period?

Ted Shifrin

There's like a one-day reading period.

Daminark

I see

Ted Shifrin

01:20

We didn't even have that until I made a fuss in the faculty Senate.

Daminark

Yeah here we have two days of reading period where you can't have anything due, can't hold obligatory classes, and can't give tests

Ted Shifrin

But some faculty figure they're better than rules and break them? :)

Daminark

Many

Like, you can technically get around them

Ted Shifrin

I would encourage students to complain to the Deans ...

Daminark

Homework is most common, though to the benefit of the students, since they say yeah, Wednesday is the official due date (and some professors that normally assign Friday to Friday homework actually shorten it for Wednesday), strict deadline

But we'll accept it without penalty until Friday

Hey Mathein!

MatheinBoulomenos

01:24

Hi @Daminark

Hi @Ted

Ted Shifrin

Hi, @Mathein ... whoa ... really late night for you.

Daminark

Oh I mentioned this problem the other day but you guys may like it, this was probably my favorite problem from my rep theory midterm (aside from the bonus but I haven't figured that out yet)

MatheinBoulomenos

yeah, I should try to sleep probably

Daminark

So let's say $G$ is a finite group and $V$ is an $n$-dimensional complex $G$-rep with character $\chi$ such that for any $g\in G$, $\dim(V^g) \ge \frac{n}{2}$. Show that $\text{Re}(\chi(g)) \ge 0$ for any $g$ and that $\dim(V^G) > 0$

Ted Shifrin

I guess I don't see how the character is in the hypothesis. Am I being stooopid?

Or did you just write it badly?

Daminark

01:31

What do you mean?

MatheinBoulomenos

okay, so the $\text{Re}(\chi(g)) \ge 0$ part is clear

Ted Shifrin

Why does the character $\chi$ get mentioned when you say $\dim(V^g)\ge n/2$?

Daminark

The first part of the problem is to show something about the character so that phrase just establishes notation

Ted Shifrin

I don't like the way the sentence is written. The character should be mentioned after. But OK.

This is true for any character, then.

MatheinBoulomenos

@Ted I think the formulation is fine

Ted Shifrin

01:33

It confuses non-experts.

I'm a non-expert.

But I'm about to leave, anyhow. :)

Daminark

And nice re Mathein

MatheinBoulomenos

okay so we have $\langle \chi,1\rangle=\frac{1}{|G|}\sum_{g \in G} \chi(g)$ and we want to show that this is non-zero

this is real anyway, so we can instead look at $\langle \chi,1\rangle = \frac{1}{|G|} \sum_{g \in G} \mathrm{Re}(\chi(g))$, this is a sum of positive numbers and $\chi(1)=\mathrm{dim} V > 0$, so this is positive

Daminark

Exactly

MatheinBoulomenos

the part about $\mathrm{Re}(\chi(g))$ follows from the fact that any $g$ acts as a diagonalizable matrix with at least $n/2$ $1$s and $n/2$ possibly other roots of unity (which have real part $\geq -1$)

nice problem!

Daminark

Yeah definitely. The others were more generic

One problem was defining the left regular rep, one problem was stating and proving Schur, one problem was proving with rep theory that finite groups of commuting matrices are simultaneously diagonalizable, one was to use the character table of S_4 to decompose the tensor product of its 2D irrep with itself

One was true-false, whether $U\oplus V \cong U\oplus W$ implies $V\cong W$, and whether the span of the orbit of a vector in a rep is an irrep

And two more I don't recall offhand

MatheinBoulomenos

01:47

okay yeah that does sound more generic

but really doable, so I assume it went well?

Daminark

Yeah, I think I got all of them right. Didn't quite the bonus, still thinking about it (so don't spoil just yet)

That the only 2D rep of a non-cyclic simple group is trivial

MatheinBoulomenos

I won't spoil anything, but I remember going through that once

Rithaniel

02:12

Is the intersection of finitely many compact sets necessarily compact?

I can't think of a counter example.

Daminark

In metric spaces?

Mike Miller

I think I see an argument that's true but I'm paranoid

Maybe it needs compact sets to be closed

Roll up and smoke Adjoint

Take an open cover $\{U_a\}$ then each is covered by a finite subfamily. Union the subfamilies.

@Rithaniel

Rithaniel

Ah, I see.

Daminark

That's for a finite union

So if your space isn't Hausdorff (more importantly the property that Mike said about compact => closed) then this isn't true. I think Alessandro mentioned a counterexample once, you take $\mathbb{N}$ union two points, put the discrete topology on $\mathbb{N}$ and let $\{x_i\} \cup \mathbb{N}$ be the only open set containing $x_i$

Then $\{x_i\} \cup \mathbb{N}$ is compact, the intersection is $\mathbb{N}$ which is not

Rithaniel

02:17

Ah, dang, that's clever.

(I am too quick to say "Ah, I see.")

Zee

02:51

I wonder if this holds for T1 spaces , probably not

Mike Miller

I think if you choose the open sets containing $\{x_i\}$ to be those with finite complement, and the discrete topology on $\Bbb N$, it's still a counterexample

Zee

03:08

Ya , hausdorrf probably nessecery

Can someone explain to me how is the curvature tensor a tensor field? You input a point on the manifold and that gives you a tensor ,then you input three vectors and that gives you a vector but aren’t tensors supposed to go to R ?

Mike Miller

Nope

A tensor field takes as input n vectors and spits out (a sum of tensor products of) m vectors

A section of (TM)^{otimes m} otimes (T*M)^{otimes n}

This is an (n,m) tensor field

Riemann is a (3,1)

Zee

03:59

I see , you take a section of the tensor bundle , input three vector field ls and then spits out a vector field

Buddhini Angelika

04:26

Hi, does any one know whether we can obtain the order of a semiregular automorphism alpha, of a vertex transitive graph G, if we know the order of G and sizes of orbits of <alpha>

I mean is there a relationship like the order of subgroups of a group is having with the order of the group

Order of subgroup should be a divisor of the order of the group

Something like that?

Thanks a lot in advance.

Zee

05:45

Should I give a math book to a math girl on a first date ?

Raptor

06:38

Nope, bring an old contest papers and discuss the answers of each problem over coffee. Bringing IMO papers means you will have to pause for lunch and dinner, possibly spend the night at another's house.

Isana Yashiro

08:50

hello

Leaky Nun

@loch hi

Vrouvrou

hello, if $(x_n,y_n)$ converge to $(x,y)$ then $x_n$ converge to $x$ and $y_n$ converge to $y$

always ?

@LeakyNun

hello

??????

Isana Yashiro

09:13

a little question about linear algebra: why eigenvalue/vector only defined on operator?

I suddenly realized that when talking about eigen-thing, only holds on square matrices...

is that a definition start with eigen- always means it's to solve diagonalize-problem?

Vrouvrou

09:31

someone have an answer to my question ?

Leaky Nun

@IsanaYashiro we want to make sense of $T(\vec v) = \lambda \vec v$ right

it wouldn't make sense if the codomain of $T$ is incompatible with the domain thereof

Vrouvrou

09:52

@LeakyNun hello how are you?

Isana Yashiro

@LeakyNun thanks

Bhathiya Perera

10:24

Is this a good book for someone who wants to study much further later on?
https://www.amazon.co.uk/Mathematics-Engineers-Dr-Tony-Croft/dp/1292065931/ref=pd_sbs_14_1?_encoding=UTF8&pd_rd_i=1292065931&pd_rd_r=934f9efb-de87-11e8-9a86-056d0d37c5cd&pd_rd_w=L9y3K&pd_rd_wg=EkmjN&pf_rd_i=desktop-dp-sims&pf_rd_m=A3P5ROKL5A1OLE&pf_rd_p=18edf98b-139a-41ee-bb40-d725dd59d1d3&pf_rd_r=21AEZJ6EJGNTTGSY8GSN&pf_rd_s=desktop-dp-sims&pf_rd_t=40701&psc=1&refRID=21AEZJ6EJGNTTGSY8GSN

:)

Akiva Weinberger

The existence of topology implies the existence of bottomology

@Zee Depends on the book and what areas of math she's interested in, but it could work well

(Disclaimer: I have never dated)

Actually, maybe it's more of a birthday thing

unless you pass it off as "I've finished this book and don't need it any more, do you want it?"

This is a good one

as is this one

and this one

Secret

10:40

Finally some weirdness. I thought they all died off since Slereah's intrusion a few weeks ago

@AkivaWeinberger I want some sidewayology

Akiva Weinberger

This one is definitely not an easy read but I found it enjoyable

I'm just gonna start listing every book I've ever read, aren't I

Secret

Actually... do a given family of topologies can form a tree ordering always?

Akiva Weinberger

Say again?

Secret

since the largest topology is always the discrete topology and the smallest one is the indiscrete topolpgy

Akiva Weinberger

Topologies probably form a lattice

(I need to remind myself the definition of a lattice)

> It consists of a partially ordered set in which every two elements have a unique supremum (also called a least upper bound or join) and a unique infimum (also called a greatest lower bound or meet).

Right so we just have to figure out what the join and meet of two topologies are

Secret

10:45

Hmm... let me check if there is an operator that makes topologies finer or coarser...

Akiva Weinberger

Did you see First Man, by the way?

People have had mixed feelings about it but I thought it was amazing

It's about Neil Armstrong, highly recommended

Secret

I have heard about that, but I never get a chance to see movies, cause phd talks at 13 November as we'll progress interview on 8 nivember

Akiva Weinberger

Mhm

Secret

the actor on the poster strongly reminds of The Martian though

Leaky Nun

@AkivaWeinberger yes they do, it's in Lean

instance {α : Type u} : complete_lattice (topological_space α) :=
(gi_generate_from α).lift_complete_lattice

github.com/leanprover/mathlib/blob/master/analysis/topology/…

in particular, they form a galois insertion with the powerset of the base set

which is in itself a complete lattice

Akiva Weinberger

10:52

What's a Galois insertion?

Secret

11:36

Comparison of topologies

In topology and related areas of mathematics, the set of all possible topologies on a given set forms a partially ordered set. This order relation can be used for comparison of the topologies. == Definition == A topology on a set may be defined as the collection of subsets which are considered to be "open". An alternative definition is that it is the collection of subsets which are considered "closed". These two ways of defining the topology are essentially equivalent because the complement of an open set is closed and vice versa. In the following, it doesn't matter which definition is used. Let...

Akiva: It is a complete lattice, and the join is not quite the union

On a side note, I wish people can give a name to those "not a topology" (collection of sets such that they don't necessary become closed under unions or intersections) because I felt they can tell us more about dicontinuous notions of convergence

Might explore that more in the Star Wars room

Leaky Nun

@AkivaWeinberger a Galois insertion is some sort of "free construction", i.e. left adjoint

Leaky Nun

12:15

$$\operatorname{End}_A(A) \cong A^{op}$$

:o

Liad

LHS is $P_Mx_0$ where $P_M$ is the projection on M

i showed $||P_M x_0 || \le max \{..\}$

can someone help with the reversed inequality?

($M$ is a closed subspace of a Hilbert space $H$ )

@LeakyNun maybe you got an idea?

($x_0$ is point in $H$ )

DreamConspiracy

12:37

Hey everyone, new to the chat. I posted this question to the site a few days ago and it hasn't gotten much love despite the bounty. Just wanted to give it a shoutout in here in case someone was interested (I hope doing this isn't frowned upon too much, lmk if it is)

Secret

In case any complex people are interested. Someone have upgraded the complex plotter of David Bau so you can now see the graphs in 3D

abenthy

13:07

If $\mathbb{A}_{\mathbb{Q},f}$ is the ring of finite rational adeles, is the condition $x^2 \notin \mathbb{Q}_{<0}$ open in $\mathbb{A}_{\mathbb{Q},f}^\times$?

loch

13:23

@LeakyNun hi

Oskar Tegby

I don't get the "by compactness of $[0,M]$" argument for convergence here. Any ideas?

https://math.stackexchange.com/questions/1623027/exists-a-uniformly-convex-norm-on-banach-space-satisfying-certain-condition

CaptainAmerica16

Can someone give an example of what applied math is?

Mats Granvik

@CaptainAmerica16 Add trendline in Excel.

CaptainAmerica16

@MatsGranvik Why do you say that?

Mats Granvik

I thought adding a trend line in Excel is pretty much applied math.

To some measurement data.

Alessandro Codenotti

13:37

Computational stuff. Like approximating solutions to PDEs or routing algorithms and that kind of things

Mats Granvik

Never mind.

CaptainAmerica16

@AlessandroCodenotti Oh, so what @MatsGranvik said kind of makes sense.

Rithaniel

Perhaps you could argue that arithmetic is applied math.

CaptainAmerica16

@Rithaniel Hm...

Alucard

14:10

@Fargle thank you for teaching me the devide and conquer algorithm :)

sonicboom

How do we obtain a linear extension of a linear functional $f$ where $f$ is defined on some subspace $Z$ of $X$? I always see something like the following phrase, 'let $E$ be the set of all linear extensions $g$ of $f$'...but how do we construct such a $g$?

Alessandro Codenotti

Very often you use Hahn-Banach to extend functionals but it's hard to say without more context

sonicboom

14:25

@AlessandroCodenotti I'm asking this question as this statement is used right at the start of the proof of the Hahn-Banach theorem (Kreyszig, p.215). He starts with a functional $f$ and then in the very first step he says let '$E$ be the set of all linear extensions $g$ of $f$...' - I want to know why such $g$ exist - how can they be constructed?

Say for the case of $X = \mathbb{R}^2$ and $Z = \{(x,0) | x \in \mathbb{R}\}$. How do we define an a linear extension $g$ to a linear functional $f$ defined on $Z$?

Alessandro Codenotti

Pick a basis $(v_i)$ of $Z$, complete it to a basis $(v_i,w_j)$ of $X$, you know $f(v_i)$ for all $i$, define $f(w_j)$ for all $j$ however you like and extend linearly to all of $X$

This shows that there are some extensions of $f$ so you can collect all of them in a set

sonicboom

@AlessandroCodenotti 'extend linearly to all of $X$'. This is the part I am not sure about. How do we extend linearly? If I have a linear functional $f$ on $Z$, how do I show that a linear extension $g$ is linear?

Rithaniel

Wolframalpha is not very good with diophantine equations.

Alessandro Codenotti

After fixing $f$ on $w_j$ you have a function defined on a whole basis. Every vector in $X$ can be written as $\sum\lambda_k z_k$ where $(z_k)=(v_i)\cup(w_j)$ (sorry for the terrible notation) and $\lambda_k$ are scalars. $f$ can be extended linearly to a function on the whole of $X$ by setting $f(\sum\lambda_k z_k)=\sum\lambda_kf(z_k)$

@Rithaniel did you solve your problem about countable space and being lindelöf?

Rithaniel

Oh, yeah, that was easier than I was thinking at the time.

Just, for every element in the space, take one cover which contains it.

Alessandro Codenotti

14:38

Yeah

It's the same proof as "finite spaces are compact", just one cardinal up

Piyush Divyanakar

I have to prove that polynomial in a ring that has zero divisors doesn't have number of roots equal to degree

is it enough to show a counter example.

Rithaniel

I had just got finished writing a page and a half on showing that the Arens-Fort space is not first or second countable. Was in a mindset where everything seemed complicated, I guess.

Alessandro Codenotti

@PiyushDivyanakar Hint: think about $\Bbb Z/15\Bbb Z$ and recall the chinese remainder theorem

Rithaniel

Though, on diophantine equations, $\sqrt{x}=(x-y)^2$ has simple solutions where $x=n^4, y=n^4\pm n, n\in\mathbb{N}$. Had to really coerce wolfram into displaying that answer

Piyush Divyanakar

@AlessandroCodenotti

I didn't get you

Alessandro Codenotti

14:47

Then think more about it

sonicboom

@AlessandroCodenotti I have posted a question just now where I explained in detail my issue, and my attempt to show linearity - math.stackexchange.com/questions/2981838/… - do you know where I'm going wrong?

Alessandro Codenotti

So your $f$ is defined on $Z$ whose basis is (for example) $\{(1,0)\}$, can you complete this to a basis of $\Bbb R^2$?

Piyush Divyanakar

@AlessandroCodenotti So taking $f(a)$ in $Z/15Z$ is like taking $f(a)$ in $Z/3Z$ and $Z/5Z$ and the isomorphism will give me the value in $Z/15Z$

Alessandro Codenotti

Pick a specific $f$. A nice one

Piyush Divyanakar

So picking a really simple one $f(x)=ax$ should have 0 as a root but if the ring has zero divisors then a $a$ can be chosen in such a way that it may have 2 roots.

Alessandro Codenotti

15:07

Hmm, what's such an $a$ in $\Bbb Z/15\Bbb Z$?

Piyush Divyanakar

I suppose $f(x)=5x$, but how does it need chinese remainder theorem

Alessandro Codenotti

It doesn't

I was thinking about $x^2-1$ but your example is easier and works as well

Leaky Nun

but your example is monic! :P

Piyush Divyanakar

:d

monic shmonic

Leaky Nun

I believe the universal polynomial is $x^2-x$

it detects whether how far a ring is from being an integral domain

Mike Miller

15:21

? Non-domains needn't have interesting idempotents. Take $\Bbb Z[x]/(x^2)$.

Idempotents detect splittings of your ring

Alessandro Codenotti

I asked that invariance of domain question from yesterday on main a little ago if you're interested in seeing if good answers come up for it

Mike Miller

Too scary

Alucard

I assume this is a known problem: What is the maximum constant speed such that you don't commit speeding, but pass all traffic lights to green? But I don't know what I should search...

Leaky Nun

@MikeMiller aha, never mind

Mike Miller

In proving that ring has no extra idempotents you'll see you need $2\neq 0$, which makes a fun observation that $(\Bbb Z/2)[x]/(x^2)$ does indeed have a splitting as $(\Bbb Z/2)^2$

In retrospect a better example was just $\Bbb Z/4$, where you have no CRT-given splitting

Leaky Nun

15:30

interesting

jsc

15:43

I just posted a question; immediately after posting, it got a downvote. Is this not the appropriate forum? math.stackexchange.com/questions/2981895/…

Alucard

@jsc maybe use codereview or stackoverflow, BUT I can't judge wether this question fits those sites either

jsc

Thanks. It's not really a coding question, but I'll check those out.

Alessandro Codenotti

$\star\star\star$ is spaced well and $\star\star\star\star$ is not, are you drunk LaTeX?

6

Leaky Nun

@AlessandroCodenotti can't blame him, it's friday

Mike Miller

16:00

@AlessandroCodenotti In retrospect invariance of domain for simplicial complexes follows from the version for $\Bbb R^n$ that says there's no injection into a lower dimensional Euclidean space

Even CW complexes of finite dimension

Every neighborhood has a chunk of a top dim cell

Ahh that's not necessarily true

But if you assume that there are no lower-dimensional simplices that are not a face :p

Alessandro Codenotti

That's interesting

I'm looking more at the other direction though, assuming invariance of domain holds

Mike Miller

Yeah, I am skeptical there is much to say

Alessandro Codenotti

(further investigation reveals that an odd number of stars is well spaced and an even number of stars is not, this typesetting system is literally unusable!)

You're probably right @Mike. Do you happen to have an example of a connected $X$ for which invariance of domain fails?

Mike Miller

2 is a counterexample to that assertion

As there is only one gap

Oh sure take a 2-simplex with a whisker sticking out of it

Alessandro Codenotti

Oh of course

user10354138

16:15

@AlessandroCodenotti you need something after the last star for even >2, e.g. $\star\star\star\star{}$

Alessandro Codenotti

@user10354138 nice

Mike Miller

@AlessandroCodenotti That was what prompted me to add the "always a face" condition

Alessandro Codenotti

I don't need to change the example I'm using for stars and bars to only have odd numbers of stars now @user10354138 :P

tatan

16:43

How can I find the number of continuous function(s) $f[0,1]\rightarrow\mathbb{R}$ such that $$\int_0^1xf(x)dx=\frac13+\frac14\int_0^1(f(x))^2dx$$ ?

Isana Yashiro

is this statement about orthogonal projection matrix correct?: it's always similar to $$\pmatrix{I_r & O\\O & O_{n-r}},$$ assumed $\textrm{rank }A=r$?

MatheinBoulomenos

that's true for any matrix that satisfies $A^2=A$ with rank r

Isana Yashiro

@MatheinBoulomenos got it, thanks

And another question is: why we draw two vectors which are orthogonal as perpendicular on the plane $\mathbb R^2$

MatheinBoulomenos

@MikeMiller I don't believe that. $(\Bbb Z/2)^2$ is reduced

Isana Yashiro

I always thought orthogonal means perpendicular but then it seems like it's just for the case of standard inner product for $\mathbb R^n$?

Mike Miller

17:01

The two generators are 1 and 1+x

Oh maybe not

Ok I give

I see my failure

MatheinBoulomenos

@Mike do you have any advice on learning about spectra?

Mike Miller

what do you want to know?

Isana Yashiro

I encounter a terminology called perspective projection but is there any book specific about this?

MatheinBoulomenos

idk, just understanding some basics like what the sphere spectrum and the spectrum associated to K-theories etc. are

and how homotopical algebra relates to homological algebra

it seems like a kind of topology that fits my algebraic/categorical taste

but I'm not sure where to start

it also seems important to follow what Lurie et al. are doing

Leaky Nun

17:19

@MatheinBoulomenos I relearnt Artin-Wedderburn today

it's good stuff

MatheinBoulomenos

I told ya it's good stuff

Leaky Nun

like wtf, End_A(A) = A^op

I should prove this in Lean

after they fix modules

MatheinBoulomenos

that's not so wtf-ish

Leaky Nun

also

modules over algebras over fields

MatheinBoulomenos

there was an algebra course here once where Artin-Wedderburn was an exercise

but that's okay since you can prove it by just Schur's lemma basically

@LeakyNun you can use Artin-Wedderburn and Galois descent to show that tensor products of central simple algebras over a field K are again central simple over K

which means that the multiplication in the Brauer group is well-defined

@LeakyNun I can link you a relevant exercise sheet if you want

Math geek

17:41

please help me.

0

Q: If $d(x,y)$ is a metric on $X$, then $d'(x,y)=\frac{d(x,y)}{1 + d(x,y)}$ and $d(x,y)$ generates the same topology.

If $d(x,y)$ and $d'(x,y)$ are a metrics on $X$, then $d'(x,y)=\frac{d(x,y)}{1 + d(x,y)}$ and $d(x,y)$ generates the same topology. My attempt:- Let $\mathscr T$ be the topology generated by $d$ and $\mathscr T'$ be the topology generated by $d'$. Let $U\in \mathscr T\implies U=\bigcup_{\alpha\...

Buddhini Angelika

For a g

Sorry

For a cayley graph G of a group, if we obtain the quotient graph G/H (H is a subgroup of the group), is there a way we can say that G/H is also a cayley graph

?

Mike Miller

@MatheinBoulomenos The thing is that you're starting from a very different point than I did

@MatheinBoulomenos Do you already know model categories etc?

MatheinBoulomenos

@Mike no, should I learn that first?

Mike Miller

probably not! i thought you might specifically know that theory and could base my suggestions from that as a beginning point

but i think it's not good to go that way

MatheinBoulomenos

I only know some basic homology and homotopy theory

Mike Miller

17:56

I think historically spectra came motivated by the observation that Alexander duality in homology was actually something that could be observed at the space level as long as you suspended enough times, called Spanier-Whitehead dualiyy

MatheinBoulomenos

interesting, we covered Alexander duality, but not Spanier-Whitehead duality

Mike Miller

Try, appropriately, Spanier

Actually that is probably a good place to learn spectra

MatheinBoulomenos

Thanks!

Mike Miller

Fundamentally they come from the observation that all homology theories have a suspension isomorphism, so only depend on spaces up to stable homotopy equivalence instead of just homotopy equivalence

MatheinBoulomenos

I guess Brown representability is related to this spectra stuff?

loch

17:59

speaking of model categories, I remember when I was briefly trying to read about it my impression is that we're axiomatising all the things that we can do w/ fibrations and cofibrations in algebraic topology - but i don't have a very good sense of what makes this powerful

Mike Miller

Absolutely

MatheinBoulomenos

@loch that sounds reasonable

loch

ah i intended that to be more of a question! was wondering if you guys might have some perspective that might make someone who just learnt about homotopy theory to think "hey this might actually be very useful if i stick this into algebra"!

MatheinBoulomenos

@loch maybe one can motivate it (partially) by the Dold-Kan correspondence. If we have a category equivalence of simplicial abelian groups and (positively graded) chain complexes of abelian groups, then it seems natural to ask what fibrations correspond to since we have a notion of fibrations for simplicial sets

though "fibrations" corresponding to Kan fibrations are just surjections in this case

probably the motivation is to give a framework to compare different homotopy theories, like the standard homotopy theory and simplicial homotopy theory

I'm just guessing

loch

18:21

Yeah I've heard of the Dold-Kan stuff - which I guess I think of it as OK now I have this machinery which recovers homological algebra, but now this works for other categories too so I can compute some interesting things there

I guess it's just not obvious how all these lifting properties and fibrant/cofibrant replacements are good - but I guess that's probably a matter of actually looking at examples

Zee

@MikeMiller does the base point of a pointed connected topological space and the space form a “good pair” in homology theory ?

Mike Miller

If the space is a CW complex yes

@loch I meant to answer but started talking to a postdoc here

Fundamentally the ideas are things you've seen in homological algebra - if you want to take a quotient of a G-action on a chain complex in a way that plays well with Homology (is quasi iso invariant) you need to take a projective resolution first

And once you do that q.i.s are actually homotopy equivalences

The general idea is the same. You have some nice operations or functors you'd really like to respect weak equivalences but usually don't. The model category setup says you can just as well work with the homotopy category of sufficiently good objects/maps

In the settlement of of spectra the most obvious important functor is "taking homotopy groups", and the notion of Omega-spectrum comes to be good for that; same thing more generally for making good mapping spectra or a good product of spectra

One can say that one may do all that by defining a symmetric monoidal model category of spectra

There are also good strict models for these things nowadays, eg symmetric spectra or orthogonal spectra

@Mathein Email me and I will give another less public reference

Balarka Sen

Hey

loch

So you're finding a resolution of your chain complex such that this new thing behaves well when you quotient by $G$, and this thing is q.i.s. to your original chain complex

(Perhaps something which is philosophically similar (but possibly irrelevant here) imo is e.g. when I take a $G$- space $X$ and quotient by $G$, I can also consider the Borel construction and then quotient by $G$)

And I guess you're saying that in general there are functors we'd like to behave well (in some sense), so as in homological algebra where we take proj/inj resolution, we take new resolutions (so in homotopy t…

Hi @BalarkaSen

Rithaniel

18:37

Hmmm, given any three natural numbers in $(0, 1000)$, chosen at random, what is the probability that they will have a greatest common divisor greater than three?

MatheinBoulomenos

Hi @Balarka!

Mike Miller

@loch Yes, the Borel construction is very very familiar to me and essentially the same spirit

Daminark

Henlo

Balarka Sen

Borel construction is $(X \times EG)/G$, right?

Mike Miller

Your summary is quite accurate

Yes

Alessandro Codenotti

18:42

Hi @Balarka

Balarka Sen

Basically if $x \in X$ is a $G$-fixed point then you get a $BG$ sticking out of it in the quotient, which contributes to a group cohomology in $H^*((X \times EG)/G)$. In general for every $G$-orbit you get a contribution of $H^*(BG)$, right?

Mike Miller

For a $G$-orbit with stabilizer $H$ you get a $BH$ sticking above that point

Balarka Sen

Ah yes

Mike Miller

You can do a sort of equivariant Morse theory which reduces a $G$-manifold to a finite number of $G$-orbits, and then try to understand how the flowlines go between the various orbits; so you have a complex (spectral sequence) which starts from a bunch of $H^*(BH)$ things and with differentials flying around between them

Balarka Sen

sp00ky

Sir Cumference

19:07

Howdy

Semiclassical

hi chat

Érico Melo Silva

sup nerdlets

loch

@MikeMiller great - what are some typical examples to keep in mind? I guess theres the usual one for (nice) topological spaces, and i know a tiny bit about andre quillen homology (for rings). I supppse the ‘general examples’ are things like simplicial objects - but i cant say im that familiar with them (although maybe i should try to be)

CaptainAmerica16

@TedShifrin I'm going to be coming into some free time within the next few days. I could use some chapter 1 recommendations now.

Mike Miller

19:25

@loch I am njot terribly familiar with them

There is a good paper "6 model structures on dgas" that everyone should read

I like the computational stuff later in the paper which is a modern rewriting of a papery by Gugenheim and May from 50 years ago

My impression is that the people really familiar with this stuff tend to just also be familiar with $\infty$-categorical setup and that seems to be "stronger" but I don't know it either

I just try to understand the philosophy and use it in my own work in clear and down-to-earth ways when necessary

Li357

I've proved $f$ is bijective given $f : \mathcal P(\mathbb N) \rightarrow \mathcal P(\mathbb N)$ and $f(X) = \mathbb N - X$ but how am I supposed to find an inverse? How could you invert a set difference?

Tobias Kildetoft

@Li357 What happens if you do the same thing again?

Mike Miller

I mean a bijection by definition has an inverse

Li357

Exactly

@TobiasKildetoft What do you mean? Apply difference again?

Tobias Kildetoft

yes, apply the same function once more

Li357

19:34

$\mathbb N - (\mathbb N - X) = \mathbb N \cap X$

Mike Miller

$X$ is by definition a subset of $\Bbb N$...

Li357

Oh so it's just $X$? But what about the inverse? I used that to prove surjection

Wait a sec, how is it guaranteed that X is a subset of N?

Semiclassical

it's a mapping into $\mathcal{P}(\mathbb{N})$, isn't it?

(I actually don't remember what the $\mathcal{P}$ is supposed to mean)

Alessandro Codenotti

Powerset

Semiclassical

ah

then yeah

Mike Miller

19:53

@Li357 I think you should start by understanding the objects you're working with... this should be clear if you know what $X \in \mathcal P(\Bbb N)$ means

MatheinBoulomenos

@Mike I sent you a mail

Mike Miller

My email takes a while to refresh so I'll see it in a bit

Thanks for the heads up

Transcript for

$Mathematics$ Mathematics