Yes

@Ultradark just working on bancats

When off on tangent trying to make some coding tool

it's too difficult, just use every tool I have as is is the solution

@ShineOnYouCrazyDiamond nice I’m working on horocyclic flow across hyperbolas

@Ultradark nice :)

So, suppose we wanted to know a subring of the ring of continuous functions from $[0,1]$ to $[0,1]$ which is maximal with respect to being an integral domain. How would you try to find that subring?

When working with spectral sequences does anyone ever bother to view the bigraded modules which are the pages of the sequence as a doubly indexed direct sum of modules?

@Perturbative not really

sometimes it's reasonable to consider singly graded spectral sequences that don't come from a filtration or exact couple, which would be like collapsing it to the diagonal grading if it did come from a filtration

but beyond that I dunno that there's much to say

Ahh cool, I was silly and thought of them like that for a while

I think one can use a singly graded spectral sequence to give a proof that the Euler characteristic of a fibration is multiplicative

Sure, you only need the single grading for that application

chi(H^*(B;H^*(F))) = chi(B) chi(F) (this requires something non-trivial to check when pi_1 B acts nontrivially on the cohomology of the fiber); each successive page is the homology of a differential on the previous page, and a finite rank complex has the same Euler char as its homology. So chi(B)chi(F) = chi(E_infty), and the Euler characteristic of an associated graded agrees with the Euler characteristic of the original thing

The only proof I saw was when $\pi_1(B)$ acts trivially actually, and yeah I think the argument is the same as what you said

In general you need to prove that Euler characteristic doesn't depend on the local system, only on the generic fiber of the local system

That is, if $A$ is a pi_1-module and let's say $A_x$ the underlying abelian group, one has $\chi H^*(B;A) = \chi H^*(B;A_x)$. One needs to assume B is a finite complex

And then this follows in the end from cellular homology. Both cellular chain complexes have identical underlying abelian groups, but different differential

I am studying some introductory level undergraduate physics and I need some clarification. What is the significance of "linear" in linear vector space? All texts list the set of rules that constitute a linear vector space but no one says why it is called "linear"... are there quadratic vector spaces also?

@MikeMiller Ahh I see, I sorta follow what you've said

Do we have a general formula/steps to calculate- In how many ways we can choose k distinct balls from x blue balls y red balls and z green balls?

@MikeMiller This is quite far out of my expertise level, but I am curious, in your thesis did you intend for the homology theory you were creating to have applications to algebraic topology (specifically the part about finding invariants of rational homology spheres)

That would be considered low-dimensional topology or geometric topology rather than algebraic topology

You could say that what I did was an application of algebraic topology and analysis to the study of 3-manifolds

Oh I see, in my head, I was thinking it was the other way around

Algebraic topology is just the tool here, I don't have any neat theorems about homotopy groups of spheres or the category of spectra or whatever

The reason I asked my question is, there seems to have been a decline in the popularity of algebraic topology as a research field in the US (just from what I've seen looking at faculty in the US who do algebraic topology) and those who are doing research in it apart from the big names are doing what seems to be very specialized stuff.

I had thought (incorrectly) that you had done work in gauge theory which ended up proving some stuff in algebraic topology, which made me think that perhaps one way to do research in algebraic topology without having to do things that are too specialized would be similar to what you did (though of course, I had the wrong idea in my head)

What I do could easily be accused of being much much more specialized than algebraic topologists are

I'm not sure that algebraic topology is getting less popular, though more specialized might be true. This seems to me a function of the age of a discipline

Good morning everyone :)

The time is 11:17

11 is prime and 17 is prime...and 1117 is also prime

prime time to do mathematics

hello everyone, i have a doubt regarding expressing basis of a free abelian group as the set of row vectors of presentation matrix

Let H be a subgroup of Z^5, such that H is isomorphic to Z^3, then the book i'm studying says that the basis of H cane represented by the set of rows of a 3x5 matrix. My confusion emanates from the fact that a row of 3x5 matrix has 5 entries.

nevermind, i got it.

@TedShifrin Hi Ted

@TedShifrin I need some help making sense of the implication if then, from logic to mathematics

how is this useful

the language is a bit strange to me here also, am not sure of the meaning of P is sufficient for Q , or necessary etc.

Hi @Jacksoja. Yeah, that's confusing, although it does make sense. $P$ is necessary for $Q$ means that in order for $Q$ to hold, $P$ must hold, so $Q\implies P$. $P$ is sufficient for $Q$ means that $P\implies Q$. Whenever $P$ holds, $Q$ must hold.

When I did the truth table for these

I did not understand that what part is the focus on

it is the implication in total (p=>Q) or the truth of Q

Q needn't be true, of course ... You only know something if you know about P.

So we dont really consider the truth value of the implication right?

when we write P ==> Q

i.sstatic.net/eQwAp.png hyperbolic flow achieved on desmos!

let me explain what I mean, so you can help me better

Whether the implication is valid depends on the P and Q.

consider the truth table for the implication

it is ofc false only when P is true and Q is false

this part I have no problem with

The point is that if the implication holds, then whenever $P$ is true, $Q$ must also be true. If $P$ is false, who cares. (This is the "vacuous" case.)

by if the implication holds

you mean true right?=

holds true*

I don't like overusing true and false, but yes.

I'd rather refer to the implication as a valid statement (like a theorem).

Okay let me see if that is what I have here

okay looking at those 3 parts when the implication holds

T T

F T

F F

Saying that P is necessary for Q, does not make sense to me

since T is false, and Q is true , also makes the implication True

You're looking at the wrong table.

If Q holds, then P must hold.

You should be looking at $Q\implies P$, not the converse.

Hi @Ted @Jacksoja et al

:)

Heya @ÍgjøgnumMeg.

@ÍgjøgnumMeg Hello !

That is strange Ted

When you move to Germany, will you acquire an easier name? :)

why would I look at the other table?

Q=> P

am doing P ==> Q

How are you both? :) Also I can if you want! hahaha I just love the Faroese language

Because I told you that necessity was $Q\implies P$.

Or else say $Q$ is necessary for $P$.

okay I see

let me look at that table and see what is what

With how many people do you converse in that language, @ÍgjøgnumMeg?

yeah but looking at that

T T

@Ted exactly none

and F T

looking at Q==> P

if really Q is necessary for P

why is the implication hold , when Q is false

Agh. You keep switching.

Read what I said carefully and let's decide which implication we're discussing.

Am not sure I understood it correctly

we take this P==> Q

I want to understand the various equivalent meaning of this

OK, so then Q should be necessary for P. In order for P to hold, Q must hold.

So you cannot have P true and Q false.

aha that is what they mean?

by necessary ?

If you read what I said 15 minutes ago, that's what I said.

sorry , am just comfused

i am not doubting that you said it before haha

Yeah, but you actually need to pay attention and think about what I say, not just ignore it.

yes, I was also comfused with the switch of premise adn conditional

@TedShifrin thank you, i should have payed more attention

You're welcome. But, yeah, it gets really frustrating in here when I have to say the same things numerous times because someone just doesn't actually read it and pay attention.

@TedShifrin and for the sufficient part , i think by saying P is sufficient for Q

that whenever P is true, Q has to be true

and nothing else been said about when P is False right?

Right. It suffices to know P in order to know Q.

If you know nothing about P, you know nothing.

okay I see, i was looking at all the combinations

that is why it did not make sense to me at first

I should have been looking only where the implication holds

I guess the point is to make sense out of the English (or whatever language) words. It should actually make sense.

Because when I saw

T F

for P=>Q

the truth of P did not seem to suffie

if you understand what I mean

since the whole implication is false

Anyway, I think that part is clear haha

Who wants to discuss c-sections? (conic sections)

Ultradark

You said something very good last time

What what what

11:17

yes I'm very proud of that

I may even try to get a paper out of it

Did you really wake up at that time or did you time it ?

I just noticed that it was prime time

It is not that good for a paper but it just funny enough

Yes, let's write a paper every time you sneeze or open your eyes.

No!

haha

Two papers every time?

Heya @Rithaniel.

On the turbulence of sneezing underwater

Sup Ted.

Ted , is there any difference between the implication symbol and this one |= in logic?

the latter is read as , it follows from in my book

like P , Q,R |= S

Sounds the same to me.

I thought so too , some strange notations in logic

Yes

formal logic

is it fun?

it is not not fun

It is normal. but something that one has to learn to be better at math I suppose

I'm taking a course in boring abstract aglebra and boring geometry

so far seems like programming

geometry ?

geometry from a transformational approach

what book you use?

so we're using matrices a lot

I can get back to you on that, I don't know off the top of my head

If you're in grad school, surely you're taking more math courses than that.

I'm only taking two

this semester

Weird.

i'll be doing three next semester

I need time to write a thesis

You can't write a thesis at the beginning.

You need to learn a ton of stuff first.

that is what am doing too

oh yeah

I did not even find a subject yet

Most masters students at UGA would take 4 courses unless some of them were Ph.D. qualifying exam courses.

I work too though

OK

But I'm going to write my thesis on the dynamics of horocyclic flow, knots, lattices and the riemann hypothesis

if one can give an accurate gauge of the speed at which a certain curve fills space, then one can solve the riemann hypothesis

twitter.com/74WTungsteno/status/1099000016482590720

yo

yo

how goes it

me?

It's going pretty well. Need to do homework today on finding the order of a group and filling out a group table

How's Princeton? How's your work going with Fernando Marques?

or was that Ryan, i forget

both of us

move in was this week so i’ve yet to do anything

howdy @Eric ... welcome to the next chapter of your life!

hey there :)

@TedShifrin hopefully it’s decent lol

first time using the chat

I have a questions. I made a post which has already been answered. Now I found new material to it which questions the answer I received. Do I make a new post about this? Or how do I handle such cases?

This is hard to answer without seeing specifics.

@ÉricoMeloSilva sup

@TedShifrin you can do like Yau and learn it all in one semester and write the thesis over winter break

Somehow, I don't think that principle applies in this case.

we just started discussing it in the comments. but its not that good. have trouble understanding the one line comments. maybe a new post would be better

@Mr.Sh4nnon: You need link to the actual post.

It's a bad idea to create multiple posts about the same question.

hmm I actually just pasted the link. how do i paste links correctly?

href: math.stackexchange.com/questions/3344081/…

nevermind^^

@RyanUnger sup

i finally made the trip up to nyc, kind of sucks but i can see myself doing this like p often

nice

I was there yesterday

rainy as hell

yeah i heard

today is niiiice tho

@Mr.Sh4nnon: I'm not reading it carefully, but it sounds like you don't know some basic terminology from linear algebra, like span, basis, etc.

You should read up on the basic terminology rather than pestering the answerer or posting more questions.

@ÉricoMeloSilva oh you're there now?

@Eric: Going to the US Open finals? :)

yeah i’m in brooklyn rn

@TedShifrin is that tennis?

LOL, yes.

idk sports for europeans

smack

lol we had this conversation the other day

lol

Two Colombians won the men's doubles (having already won at Wimbledon). South America ≠ Europe.

@ÉricoMeloSilva have you done laundry yet

it seems like we have to pay in the NGC

I won't even mention Serena.

BBIAB.

@RyanUnger i got a laundry card i assume i have to load w money, i was gonna launder today

wait what where did you get the card

I haven't tried doing it yet I just went to the room to look around

@TedShifrin i wasn’t serious but also i was 100% serious and they’re all honorary euros

actually though i just don’t watch sports unless it’s soccer and only at the big events

Rick Schoen is big on tennis and he's 100% American

@RyanUnger idk if it’s an ogc specific thing but i got a mailbox key and a laundry card from the porters lodge

Oh I see

Welp I won't be doing laundry this weekend

I haven't gotten my mail key yet

Thankfully I'm in no rush...

Please!!! please someone answer me that!!

did u find out if ngc is like separate from ogc for this @Ryan

cuz i had to sign up for a mailbox like online before i got one assigned to me

and i think u have to do that by next week or u can’t receive mail at the gc

maybe i could be wrong on that don’t quote me

uhhhh the paper they gave me during check in says go to the porters lodge during business hours

where did you go online for that

and where did it say this deadline

in the gc survival guide thing

we got an email about it i think

they sent me an email i think about mailbox sign up

i presume u can do it in person too

I didn't get an email like that

just double checked

it was titled “Graduate Move In 2019”

if u got that one

@TedShifrin I have problems understanding your notation. I am not a mathematician but an engineer. okay my liner algebra skills are not the best, but I assume I could understand this one. Just don't see where those indices come frome

oh from a while ago

thx

no problem

removed for stupidity...

lol

I was reading it out of order somehow

if x is a Boolean variable, what does x* mean?

y0

Hi IgjognumMeg

I'm reading an expository paper on Thompson's groups and came across the phrase "universal conjugacy idempotent" What could that possibly mean? The author's say that Thompson group $F$ has a universal conjugacy idempotent.

@ÉricoMeloSilva wholesome group chat :)

@RyanUnger unfortunate that ppl are finally gonna socialize and i’m not in town lol

Hi

@ÉricoMeloSilva there was a group of us at the d bar the other night too

let me know when you get back and I'll tell u if we're still out

@RyanUnger i was also not around then lol

I understand its just the sum of all possible gcd of all possible k combinations divided by nCk

But how do I find the sum of all possible gcds

Are well-founded relations reflexive?

Q: $Sp(n,\mathbb R)$ vs. $Sp(2n,\mathbb R)$

@usukidoll what do you mean by "well-founded"?

:S... This is like set theory stuff but I'm trying to understand well-founded :/

I know for total order, reflexive, antisymmetric, transitive, and Trichotomy laws are satisfied. If we have something that's well-founded but not total order, then something from the total order definition does not hold

Is $(\mathbb N, <)$ "well-founded" for you?

It's not reflexive.

I know but ... Ok wait suppose I have something like the relation $xRy: \rightarrow x+3 =y$ for $x=1,2,3$ then my $y$ would be $4,5,6$ and $4$ is the minimal in $y$ so I think it's not a total order because reflexive holds due to $4 \leq 4$

Your relation is $xRy \iff x + 3 = y$?

That's certainly not reflexive.

Yeah

Since $xRx \iff x + 3 = x \iff 3 = 0 \iff \bot$

That's what I came up with.

If only mobile can produce latex. I could copy and paste this in Overleaf

is a path a one dimensional manifold

Isn't $\{ 0 \}$ not a total order too? There's only one element which is zero and that's the minimal element... But aren't well-founded relations not reflexive?

@usukidoll What do you mean when you ask if {0} is a total order? What is {0}? Definitely not the total order itself, since a total order is a relation, and a relation is a subset of a cartesian product.

@Ultradark topological manifold, possibly with boundary.

Depends on if you include end points or not.

$\{0 \}$ is a Singleton set with only one element which is zero. That's the minimum as well because there isn't a number that is less than zero

@anakhro okay sure. let's say with boundary and the endpoints are undefined

Oh dayyum ok

.

Wait.. I think it's not a total order because we can't use the definitions of antisymmetric, transitive, and one of the Trichotomy laws. There's no second element int he Singleton set to compare

s.t. the manifold is defined on $(0,1)^2$

@usukidoll I am not sure what you are saying with {0}. If that's your set that you are defining the relation on, then you only have two possible relations: {} or {(0,0)}.

It's an entirely diff

Oops

@anakhro any thoughts on this?

@anakhro bookofproofs.org/branches/examples-of-well-founded-relations claims that the Singleton set $\{0 \}$ is not a total orsee

I may be wrong to assume that the manifolds are analytic, I would have to prove a lot of the stuff I say in this question

but think the assumptions are reasonable and allow for a natural and interesting expansion of the model in several different directions

@usukidoll where?

On example 1a on the link I gave

It does not say anything about {0} being or not being a total order.

@anakhro I uploaded a screenshot

As I said, it's not a relation.

?!?!?! Then why would they use the Singleton as an example?

You aren't reading it.

You need to be more precise about what you say and read.

"{0} is not a total order" is not a good statement, because "{0}" is not even a relation.

:C

And the screenshotted part doesn't even say that.

you need to be more precise usukidoll

@Ultradark I cannot help you with that, sorry.

it's so easy though

Oh :/ so $(\mathbb N, <)$ is not a total order because reflexive fails to hold... How so? They did the $\{0 \}$ on there.

Then why don't you answer it?

I guess I will

@usukidoll Example 1a says "$(N,\leq)$" is a total order, a well-order, but not well-founded.

I don't want to be weird and figure out my own question

@Ultradark why do you ask it if you already know how to find the answer and insist it is "easy"?

Or do I even want to know?

@anakhro I don't know I guess it seems too easy

but I may be missing something

Ohhhh. I'm looking for the other way around. Well founded but not total order. Maybe I should just come up with another relation like... $xRy: \leftrightarrow 2x=y+6$

@usukidoll try $xRy \iff x|y$.

x divides y

2 divides 2 would be reflexive

Sorry, let's fix that.

How would you fix it?

If you don't want it to be reflexive anymore?

We can test transitivity

If x divides y and y divides z then x divides z

Yeah

So just make it $xRy \iff x|y$ and $x\neq y$.

Is this well-founded?

Yeah but not total order because $x \neq x$ ?

No, that's not why it is not a total order.

Wait

scratch that. :P

Oh what if

Sorry, the reason I originally suggested x|y was because of an example like $2\not|3$

Like take the reflexive definition

$xRx: \leftrightarrow x \mid x$ and $ x \neq x$ which is a false statement because if $x=2$ then $2 \mid 2$ like 2 does divide 2 but we also need 2 to not equal 2 which doesn't make sense, so reflexive fails to hold because of $2 \neq 2$?