Mathematics - 2014-12-12 (page 2 of 5)

Axoren

8:00 AM

Does anyone have any major experience with Presentations of groups?

I've got a quick question.

robjohn

8:35 AM

@FreeMind As soon as I come up with something, I will let you know.

Balarka Sen

@Axoren You can ask.

Axoren

Is there a fast way to find the word in the Normal Form of the presentation of a group-like object, such as a monoid, that represents a certain element of that group-like object?

Err.

I left out some words.

Balarka Sen

what's a "normal form" of a presentation?

Axoren

Every element in a group-like structure has a word of minimal "length".

Balarka Sen

ah, ok

wait no

let's replace "group-like structure" by monoids. you mean every word that represents the relations has minimal length?

Axoren

8:41 AM

Yes.

We can say the same about groups by saying that these minimal words are just the reduced words of that presentation.

Balarka Sen

makes sense.

Axoren

Which is as close a definition for minimal length as I can think of for actual groups.

Balarka Sen

@Axoren but i don't understand your question.

say you have a monoid $M$ with reduced presentation $\langle S|R\rangle$. What do you want?

Axoren

Let me be more specific, because a general presentation of S and R isn't helpful.

Balarka Sen

the presentation isn't the issue. i get what you're restricting it to. i don't know what "the word that represents a certain element of the monoid" means.

Axoren

8:44 AM

Let's say that $M$ is a generalization of $S_n$, with presentation $\langle s,\ t\ | \ s^2 = 1, t^n = 1 \rangle$.

If I say "$tts$" for $S_3$, that "$tts$" corresponds to a single element that results from $t * t * s$

But again, I need to be more specific, because those generators could be any two generators.

$s = (1\ n)$ and $t = (1\ 2\ \dots\ n)$

$tts$ would correspond to $\binom{1\ 2\ 3}{2\ 1\ 3}$ in $S_3$

Balarka Sen

you need to formalize what you mean, @Axoren. i don't understand any bit of that.

you have $M$.

where does $S_3$ come from?

Axoren

$M$ is a generalization of $S_3$, the symmetric group of 3 symbols. However, by allowing $S_3$ to be a group, I include the inverses in its presentation.

Balarka Sen

Ok, you're considering $M_3$.

wait a second @Axoren. $M_n$ is a group. it's the free product of $\Bbb Z_2$ and $\Bbb Z_n$

so i am not sure what you mean.

Axoren

My main focus is that I want to exclude inverses from the presentation of the symmetric group $S_3$, so I'm treating it like a monoid.

Balarka Sen

@Axoren OK. But then $M$ is not what you want.

Axoren

8:54 AM

So, you're using a specific $M$, not $M$ as a variable?

Balarka Sen

"variable"? it's a group.

$M_n$ is a group.

Axoren

Herein lies my confusion at this point: Did $M$ mean something specific before this conversation started? Was it a specific class of monoids/groups?

Balarka Sen

I though $M$ was defined by $<r, s | s^2 = r^n = 1>$?

there are well defined inverses of $r$ and $s$ in $M$.

it's the group $\Bbb Z_2 \star \Bbb Z_n$

look up free products.

Axoren

The only thing I'm trying to do is say that in the presentation, I will not allow terms of the form $g^{-n}$.

Balarka Sen

ok. then you have a monoid $M$ whose grothendieck groupification is $\Bbb Z_2 \star \Bbb Z_n$

then? what do you want?

Axoren

8:59 AM

I guess I want some map from the group to its presentation.

Karl Kronenfeld

Perhaps Axoren is considering any monoid that has a pair of generators satisfying the given relation, while the notation indicates a specific monoid.

Balarka Sen

that's not at all clear to me @Karl.

@Axoren which group?

Axoren

@KarlKronenfeld Could you elaborate? I lack the vocabulary and experience to explain properly to those with the vocabulary and experience.

@BalarkaSen $M$, sorry. I called it a group when we just established it was a monoid.

Balarka Sen

so you want a map from $M$ to what?

Axoren

Let's call $W$ the set of words in the presentation of $M$.

I want a map $f: M \to W$

Balarka Sen

9:02 AM

eh. so in your case, $W$ is $\{rrrrr...r, ss\}$?

Axoren

But it also includes things like $rrs$ and $rsrrs$.

Balarka Sen

but it's not in the presentation. if you include stuff like that, $W$ is precisely (as a set) all of $M$.

$W = M$, in short.

let $f$ be the identity map.

wait a sec

i think i see what you mean

Axoren

I'm under the assumption that words are a different object than the group elements they represent.

And that the presentation can have a many-to-one relation ship between words and elements.

Balarka Sen

@Axoren $W$ differs from $M$ in the sense that $rrrr....r$ doesn't "collapse" to $1$, right?

Axoren

Yes, but that also if there were every a word with $r^{n+k}$ in it, then it would not be in the normal form of the presentation.

But I don't think we're on the exact same page yet.

The relations I placed in the presentation weren't a closure.

I should have marked that the set $R$ in the presentation should have been the closure of the relation set $\{s^2 = 1, t^n = 1\}$

As $n$ get's arbitrarily large, the relation set $R$ contains more and more relations.

Balarka Sen

9:09 AM

whoops. i got disconnected.

Axoren

Can you see the transcript?

Balarka Sen

@Axoren from what i have inferred from your very vague stuff, you are looking at a representation of $M$ onto the free group on 2-generators.

it's nothing special. just a surjection. the image is precisely $M$.

Axoren

Again, I'm still considering the presentation and the group itself to be sets whose elements are different types.

A word in the presentation is different from its corresponding element in the Monoid.

Because I need a physical procedure to construct the word from the group element.

Balarka Sen

ok in the case $M = <r, s|r^n, s^2>$, $W$ is the free group on $s$ and $r$, right @Axoren?

if so, my interpretation is correct.

otherwise, please clarify.

Axoren

Yes, and there is some subset of that $NormForm(W)$.

That set is of lets say Strings.

Which will be fundamentally different from the group elements.

Balarka Sen

9:14 AM

yes, then $M \to W$ is just a surjection.

Axoren

Which I need a physical construction for.

Balarka Sen

in short, if you delete all the relations on the presentation of $M$, you get $W$.

Axoren

I will end up with the free group of ${s, t}$, yes.

When I say physical construction, I mean an algorithm I can follow by hand, using the description of a group element and the definitions of the generators, to form the Normal Form string of the element.

I know that there is a map from $M \to W$, but I need a procedure that gives me it, and in reasonable time.

I've been under the assumption that the presentation of a group and the group are fundamentally different, but I may have been wrong.

I'm required in my application of them to treat them as fundamentally different things.

Balarka Sen

there is no homomorphism though.

ugh this connection

if you just want a set map, it's obvious @Axoren

Axoren

It's not obvious without enumerating the whole set.

Balarka Sen

9:21 AM

why not? just use the identity map.

Axoren

Because I don't have the identity map on paper.

Balarka Sen

$f(x) = x$ for all $x \in M$.

there's your map off-hand.

Axoren

What I want is a map of $m = rrrsssrrssr$

Balarka Sen

as i said, $M \subset W$. the inclusion map is obvious.

@Axoren heh?

m is your word?

Axoren

$m$ is the element, $rrssrrssr$ is the word

Balarka Sen

9:23 AM

just map it to the word.

$rrssrrssr \mapsto rrssrrssr$

Axoren

Let me restart this discussion, may I?

Balarka Sen

sure, but i don't know what's the problem with the identity map.

you realize that $M \subset W$ don't you?

Axoren

Because the objects in the domain are fundamentally different from the objects in the range.

So I'm going to restate the problem such that that is clear.

Balarka Sen

not as sets.

there is no homomorphism as you say.

and "fundamentally different" is a vague concept.

if you're gonna formalize your question fully, i am prepared to look at it.

Axoren

I have a list of $n$ elements, $(a_1, a_2, ... a_n)$. I want to reorder them using only two operations. $t$ is an operation which takes the first element and pushes it onto the right of the list. $s$ is an element which swaps the front and back of the list.

Given an permutation of this list, I want to give a String that describes the which operations and in which order I need to perform on an ordered version of the list to get that permutation of it.

Lembik

9:29 AM

anyone interested in math.stackexchange.com/questions/1059379/… ?

Axoren

These string that says corresponds to performing the first operation $t$ three times, then the second operation $s$ once is $ttts$

Balarka Sen

sorry @Axoren whatever you are saying is way too vague for me to answer. i'll let someone else look at your question.

Axoren

I understand, @BalarkaSen. But could you tell me how this statement of the problem is vague? Describing this problem is a bigger problem than solving it. If anything, I think this is the clearest description of the problem to-date

Balarka Sen

OH WAIT

You're given a permutation $\sigma'$ of $\sigma = (123...n)$. You want to determine the word in $<s, t>$ the maps $\sigma$ onto $\sigma'$, right?

Axoren

Yes. The spacing was throwing me off for a bit.

Swapnil Tripathi

9:38 AM

@Alizter: I went through the transcript. 100 rep for participating in the elections seems a good idea to me. Why don't you put it up as a feature-request on meta?

Axoren

@BalarkaSen Your sudden understanding of my problem has me curious if you've actually heard of a way to do this.

Balarka Sen

@Axoren That's a good question. I don't know of an effective algorithm to do that.

@Axoren There are some "similar looking" problems out there. Google [fifteen puzzle + mathematics]

Axoren

@BalarkaSen That's a shame. Would you be able to suggest some edits to this question then, since you know what I'm talking about? I'm afraid the language I put in the question is even more confusing than in this transcript.

And yes, this is definitely a similar case to solving Rubik's cubes and the 15 Puzzle.

But those problems have other limitations like parity constraints.

With rearranging this list, you can get to any configuration of the list from any configuration of the list.

And I'm concerned with solutions with lists of length potentially hundreds of elements in length.

Balarka Sen

This is a cute problem. Feels like an algorithm shouldn't be very hard to construct.

Axoren

It isn't. But time and space are the bounds, not the intuition.

It isn't to the "Feels like an algorithm". Not to the "This is a cute problem."

It's a very cute problem.

Tom Cruise

9:58 AM

hi

Balarka Sen

hello @TomCruise

any progress on that problem?

Tom Cruise

I am depressed because so many of my Calculus students failed the final exam :(

Balarka Sen

@TomCruise why should that make you depressed.

Tom Cruise

gonna have to fail 10/50 students

is that acceptable?

but truth be told most of them were failing before the final and should have dropped long ago

maybe I am a horrible teacher

but I also think I just got some lazy students

Fargle

I don't know about that. Calculus is one of the hardest math courses regularly taken by non-math majors.

That, too, is a possibility.

skullpatrol

10:02 AM

What was the highest mark on the final?

Tom Cruise

95

skullpatrol

Lowest?

Balarka Sen

Ted's probability classes gets worse.

Tom Cruise

like 12

someone else's student made a 4.5

they got 1 point just for signing your name

Fargle

O_o

Balarka Sen

10:04 AM

LEL

Fargle

Was it a short answer type deal? If it was MC I'm utterly mystified, but if not, I can kind of understand

Tom Cruise

all short answer

skullpatrol

Any proofs?

Tom Cruise

no this is just your standard calculus course

Fargle

I guess there is always that one guy.

Tom Cruise

10:07 AM

I wouldn't really be doing those students a favor by passing them... they're just going fail some other courses later.

skullpatrol

They need to be ready for the next coarse and it doesn't get any easier.

Balarka Sen

tests in a university in here i know of are open book. in MSc 1st sem algebra test some guy while proving a theorem also copied "see theorem 5.9.1" along the way.

LOL

Tom Cruise

yeah, I tried. Anyway, I learned a few things I need to do differently next semester.

skullpatrol

Such as?

Tom Cruise

@BalarkaSen hah, that's a pretty clear sign of cheating, unless he memorized the theorem numbers

Well for one thing I think I need to focus on the simplest examples.

Balarka Sen

10:10 AM

so the examiner asked him what was theorem 5.9.1. :P of course he couldn't answer.

Tom Cruise

And I need to somehow force them to do the damn exercises in the book.

skullpatrol

Now that^ is going to be hard to do.

user117890

Hi, I'm looking for a measurable function $f$ such that $f^+$ is integrable and $f^-$ is not. Anyone here has a nice example?

Tom Cruise

sure

Sawarnik

@TomCruise Which university?

Tom Cruise

10:15 AM

we can construct $f$ on $[0,\infty)$ to be borel measurable

on the even intervals construct $f$ to be positive step functions. We need the values of these step functions to to go 0 sufficiently fast as the intervals to to $\infty$.

on the odd intervals take step functions with height $-1$

then $f^+$ is integrable but $f^-$ is not

and clearly $f$ is measurable

there is a much easier way

I am too tired to think clearly

I'm going to dunkin donuts, back in a jiffy!

daOnlyBG

10:47 AM

@KarlKronenfeld I don't know if you're still around, but if you are, I could use some help

user117890

11:34 AM

Let $f_n(x)=nx^{n-1}-(n+1)x^n$,$0<x<1$. I want to prove that $\sum_{n=1}^\infty(\int_0^1|f_n(x)|dx)=\infty$. How can I expand $|f_n(x)|$ here?

Integrator

11:44 AM

@robjohn I edited this question but someone objected so I kind of rolled back the edit. could you please tell me if I was wrong?

Integrator

11:55 AM

@skullpatrol Maybe, but I didn't wanted to make any (immodest) statement from my side!

skullpatrol

you did the right thing pal, just make a note of that user

Integrator

@skullpatrol I don't have any problem with anybody, It's just that I wanted make sure If it was really a mistake.

skullpatrol

ok

user117890

Now I used triangle inequality and proved it.

Chris's sis

"A side note: Being a moderator ⇎ being a competent mathematician. That said, you are currently one of two serious applicants. What's wrong with these others? – AlexR"

One of the professors in one of my uni (well, one of the uni where it's not that easy to enter) used to tell that a mathematician can excel in everything.

I think he was totally right although there might be some exceptions.

One of the guys on MSE I consider the be very good

David Moews

11k 1 13 29

18

A: Study the convergence of $\sum_{n=1}^{\infty}\Bigl( \sqrt[n]{1+\frac{1}{n}}-1\Bigr)$

By the binomial theorem, $$1\le 1+\frac 1n\le (1+\frac 1 {n^2})^n=1+n\frac 1 {n^2} +\binom{n}{2} (\frac 1 {n^2})^2+\dots,$$ so, taking $n$th roots, $$1\le \sqrt[n]{1+\frac{1}{n}}\le 1 + \frac1 {n^2},$$ and the sum converges by the comparison test.

Excellent!

evinda

12:19 PM

Hey @DanielFischer!!! How could we show that we can sort $m$ integers with values between $0$ and $m^2-1$ in $O(m)$ time?

Daniel Fischer

@evinda I don't know whether that is possible.

evinda

@DanielFischer Could we use maybe an other version of radix sort, since it isn't based on comparisons?

Daniel Fischer

12:36 PM

@evinda We have $O(\log m)$ bits, so I think that would also give an $O(m\log m)$ algorithm. Haven't thought much about sorting algorithms recently, though, so I can't say for sure that won't work.

evinda

@DanielFischer How do we conclude that we have $O(\log m)$ bits?
Also, could we somehow change the base of the logarithm to $m$ ?

@DanielFischer so that the time complexity is $O(m)$ ?

Daniel Fischer

@evinda An integer $n > 0$ needs $1 + \lfloor \log_2 n\rfloor$ bits. We have an upper bound of $m^2-1$, so by and large, $2\log_2 m$ bits.

John Jack

Hi @DanielFischer could you please check my proposed solution for the question, it is in the comments. Do you think that we can choose one $\epsilon$ such that the result holds?

evinda

@DanielFischer So, do we need $1+\lfloor \log_2{(m^2-1)} \rfloor \leq 1+\log_2{m^2}=1+2 \log_2 m=O(\log m)$ bits?

Hey @KonradVoelkel!!! Could you tell me what's the property that should be satisfied so that a point $\in \mathbb{P}^2(\mathbb{C})$ ?

Daniel Fischer

12:56 PM

@JohnJack Not sure. If you have $u_k \to u$ in $L^{p-\epsilon}$ and $f(u_k)$ bounded in $L^{p'+\epsilon}$ for all $\epsilon > 0$, then it works, but if only for a specific $\epsilon > 0$, I'm doubtful.

evinda

@DanielFischer Or am I wrong? :/

Daniel Fischer

@evinda Yes, we need approximately $2\log_2 m$ bits.

evinda

@DanielFischer Why is this: $1+ \lfloor \log_2{(m^2-1)} \rfloor$ equal to $2 \log_2 m$ ?

John Jack

@DanielFischer Is it not possible to say that if we know that there is an $\epsilon > 0$ such that $u_{k} \rightarrow u$ in $L^{p-\epsilon}$ and $f(u_{k})$ is bounded in $L^{p'+\epsilon}$ then there exists a possibly different $\epsilon$ such that $$\int_{\Omega}f(u_{k})(u_{k}-u)dx \rightarrow 0$$?

Daniel Fischer

It's not equal, it's approximately equal (if you take the same bases for the logarithm, when talking about bits, base $2$ is the convenient one).

robjohn

1:06 PM

@Integrator By adding the LaTeX, it kind of makes the question look trivial. What you did was done with good intentions, but the intent was to identify an image, so I can see why someone might object.

Daniel Fischer

@JohnJack Possible. But I have no inclination to check it, I don't like that sort of computations.

John Jack

I don't like it either! @DanielFischer

Chris's sis

@AlexanderGruber what would you recommend me for this one? $$\frac{1}{1!}+\frac{\displaystyle 1+\frac{1}{2^2}}{2!}+\frac{\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}}{3!}+\frac{\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}}{4!}+\cdots$$

@Anastasiya-Romanova秀 ^^^

evinda

@DanielFischer Could you explain me further how you found it? :/

Chris's sis

@AlexanderGruber Oh, I wanted to add @DanielFischer ...

Integrator

1:16 PM

@robjohn Few minutes before I flagged this answer as not an answer but the flag was disputed, and I cannot flag it again.

evinda

@DanielFischer Also, I found at a site the following:

Let there be d digits in input integers. Radix Sort takes $O(d \cdot (n+b))$ time where $b$ is the base for representing numbers, for example, for decimal system,$ b$ is $10$. Since $n^2-1$ is the maximum possible value, the value of $d$ would be $O(\log_b(n))$. So overall time complexity is $O((n+b) \cdot \log_b(n))$. Which looks more than the time complexity of comparison based sorting algorithms for a large $k$. The idea is to change base $b$. If we set $b$ as $n$, the value of $O(\log_b(n)) $ becomes $O(1)$ and overall time complexi…

robjohn

@Integrator You can always leave a comment...

Integrator

@robjohn The answer is deleted now. Problem solved.

robjohn

@Integrator No, I converted it to a comment to the answer it was intended for

Integrator

@robjohn Ah! Thanks by the way!

robjohn

1:21 PM

@Integrator no sweat... it's part of the job ;-) (tips may be left in the hat)

@Chris'ssis Hey... I was just discovered while playing around with an answer that $$\sum_{n=1}^\infty\frac{H_n}{n!}x^n=e^x\int_0^x\frac{1-e^{-t}}{t}\mathrm{d}t$$

That is sort of related to your series

Chris's sis

@robjohn Did you use the fact that $$\sum_{n=1}^{\infty} x^n H_n=-\frac{\log(1-x)}{1-x}$$?

skullpatrol

hat?

:-)

Dec 15

robjohn

@Chris'ssis no... I was trying to find a non-alternating series for $\operatorname{Ein}(x)$

Chris's sis

@robjohn I see. Interesting.

Daniel Fischer

@evinda I don't even know whether that actually works or is an error. If it works, the description at that site should tell you how it works.

Konrad Voelkel

1:29 PM

@evinda To me that question doesn't make sense as it stands. What is "a point"? For being in CP² to be a property of a point, that point should already be in some superset of CP² (and I don't work with mathematical foundations which allow for anything else). By definition, points of CP² are precisely lines in C². So an answer would be "the condition is to be a line in C²".

teadawg1337

Morning, everyone

Chris's sis

@robjohn That series can be tackled by Ramanujan's master formula.

skullpatrol

hi

N3buchadnezzar

Need more close votes math.stackexchange.com/questions/1064896/…

Anastasiya-Romanova 秀

@Chris'ssis I have no idea about this one $$\sum_{n=1}^\infty\frac{H_{n^2}}{n!}$$I have lost my interest in advance math at the moment because I'm studying a ton of social science stuff. Sorry...

Integrator

1:36 PM

@N3buchadnezzar Now it needs only one :)

Balarka Sen

@MikeMiller It's been a week.

Chris's sis

@Anastasiya-Romanova秀 social science stuff? Now while in the election period??? :-)

N3buchadnezzar

@Integrator =) I always get a gut feeling when I have seen the problems before. Then it takes me two bazillion years to find the duplicates.

robjohn

@Chris'ssis I was not starting with the series, but the integral. I knew that there was a better series for $$e^{x^2}\int_0^xe^{-t^2}\mathrm{d}t$$ that doesn't alternate, so I tried a similar trick here

Integrator

@Anastasiya-Romanova秀 Why not use your powers on this

Balarka Sen

1:37 PM

sigh. Can anyone tell Mike that it's been a week?

3

Chris's sis

@robjohn I see.

teadawg1337

@N3buchadnezzar I'd cast the final vote if I could, but I unfortunately don't have enough reputation :(

N3buchadnezzar

@teadawg1337 You have to start racking cred in da hood to earn some reputation then homie.

robjohn

@Integrator the answers there are about all you can say. Ah, you are closing it.

Anastasiya-Romanova 秀

@Chris'ssis History, Economics, etc. I'm visiting this site just to review & to see election news.

Balarka Sen

1:38 PM

Thank you, whoever starred.

Chris's sis

@Anastasiya-Romanova秀 No math anymore? That's terribly sad ...

skullpatrol

@BalarkaSen ;-)

N3buchadnezzar

@BalarkaSen Is this a weekly thing? Has mike some kind of disease leaving him unable to compute modulo seven?

Integrator

@robjohn I've closed it :)

Anastasiya-Romanova 秀

@Chris'ssis Not now, maybe next year. Right now just the easy one to pass my exams

Balarka Sen

1:41 PM

@N3buchadnezzar Heh, no. Mike and I wagered on something (related to the fundamental group of a very very very very pathological space). The bet was that if he won, he'd get to ignore me for a week and if I won, he'd persuade Ted to unignore me. He won, and you know...

I need an algebraic topologist, so he needs to realize that it's been a week :P

Chris's sis

@Anastasiya-Romanova秀 Well, it's NOT that much until the next year :D

Integrator

@N3buchadnezzar @robjohn Is it okay to post a CW-ed answer to questions that are answered in comment?

N3buchadnezzar

@Integrator CollegeWeed-ed answer?

ClasifiedWrong-ed answer?

Daniel Fischer

@Integrator It is. You don't even need to make it CW. You could also consider pinging the comment-answerer asking to convert the comment to an answer. Sometimes that works.

Integrator

@N3buchadnezzar :P

Balarka Sen

1:43 PM

CW complex <--- much better pun could be made with this

N3buchadnezzar

@BalarkaSen CountWeek-ed answer? Double slam :p

robjohn

@Integrator I would comment to the answerer that they should convert their comment to a full answer. If they don't do that, then you can post a non-CW answer.

teadawg1337

BTW guys, I bought Baby Rudin, Rudin's Real and Complex Analysis, and Rudin's Functional Analysis! They should all arrive within the next two weeks :D

Integrator

@robjohn @DanielFischer Okay! Now I need to see if MSE's list of comment templates have anything for me.

Balarka Sen

Heya @Alyosha. Watcha been upto?

Alyosha

1:46 PM

Sleeping a lot after term has ended.

Now I've started looking into more AT.

Balarka Sen

Ah, cool.

I am studying that stuff.

Alyosha

Oh, excellent.

Integrator

@DanielFischer I'll start with you math.stackexchange.com/questions/486539/…

Alyosha

It's particularly nice.

Balarka Sen

Indeed. I have changed my perspectives about it.

Alyosha

1:47 PM

Anyone: is it obvious that homology is a functor?

Balarka Sen

... haven't studied homology.

Alyosha

Directed to anyone...

Integrator

@Alyosha Are you interested in converting your comment on this question to answer ?

Balarka Sen

I am interpreting all of covering space theory using Galois theory @Alyosha :)

Alyosha

Maybe I should also do Galois over the holidays also.

@Integrator Are you interested in seeing the answer?

Integrator

1:49 PM

@Alyosha Why not? :)

Alyosha

That is, would you like me to post an answer that is basically the wiki one, but with more explanation of the details?

If so, which wiki step(s) are particularly convoluted to you?

So I can focus more on them.

Mike Miller

@Alyosha It should be obvious. A map of chain complexes induces a map of their homology, and a continuous map between spaces induces a chain map between the singular complexes.

Balarka Sen

@Alyosha I am thinking a lot about $\mathbf{Gal}(\overline{\Bbb Q}/\Bbb Q)$ these days :P

Mike Miller

The induced homomorphism is the desired one.

Balarka Sen

@MikeMiller test.

Daniel Fischer

1:50 PM

@Integrator Robjohn pinged André, so I'll wait a bit whether he makes it an answer.

Balarka Sen

gah.

Daniel Fischer

@MikeMiller Balarka says the week is over.

Mike Miller

If you mean "is it obvious it's a functor from hTop", then no, I don't consider that obvious. But just from Top the above works.

Chris's sis

@robjohn The complexity of my question is pretty high, and one can easily see this by looking at that series as if it was a generating function we can denote by f(x) and then differentiate it once and try to put all into an differential equation.

Alyosha

@BalarkaSen I need to do some basic Galois exercises.

Mike Miller

1:51 PM

@DanielFischer I have an alarm set for when it actually is, and it's not. He just got an extra day.

Alyosha

@MikeMiller One second.

Balarka Sen

@Alyosha Dummit-Foote has a lot of those.

Alyosha

Maybe too many...

Daniel Fischer

@MikeMiller Poor him. When is it over (including the additional day)?

Alyosha

I almost prefer the look of Jacobson.

teadawg1337

1:52 PM

Are Rudin's texts late-undergrad/early-grad level? Should I expect a challenge?

Mike Miller

Sometime Sunday, @DanielF.

Balarka Sen

Oh noes @Mike.

:(

Alyosha

@MikeMiller Oh, so it's a functor from chain complexes to Ab?

@Integrator if you don't reply I won't be able to write as good an answer!

Integrator

@Alyosha You don't need to post CW answer.

Khallil

Morning all!

Hakim

1:57 PM

Hi @Khallil

teadawg1337

Hello @Khallil

Mike Miller

@Alyosha That's the most general formulation: a functor from the category of chain complexes on Ab to Ab itself, where Ab is some Abelian category. For topological spaces, we're sending a topological space to its singular chain complex and then taking its homology.

Khallil

Long time no see, @Hakim!
How've you been?

How's the day treating you, @teadawg1337?

^_^

Hakim

@Khallil Yeah, as always, how about you?

Mike Miller

Looking at it like this is what I consider the easiest way to view functoriality. Many probably disagree.

Alyosha

1:58 PM

Right, OK. Thank you.

@MikeMiller Aha, whilst I have your most useful attention: how do I view adjunctions in cat thoery?

teadawg1337

@Khallil I've been fighting a nasty cold for the past week, I'm struggling a bit

Hakim

@Chris'ssis Do you know whether your series diverges or converges? (numerical check seems to suggest the latter)

Balarka Sen

I am struggling to determine the Cech nerve, or at least the 1-skeleton of the Cech nerve of some "nice" open cover of the standard p-adic solenoid.

Chris's sis

@Hakim Which series?

Balarka Sen

:(

Khallil

1:59 PM

The same, really. Nothing much has changed! I'm back at home during the holiday between semesters which is pretty nice, @Hakim. Have you been doing any interesting math lately?

Hakim

@Chris'ssis chat.stackexchange.com/transcript/message/19031016#19031016

Chris's sis

@Hakim $$\frac{1}{1!}+\frac{\displaystyle 1+\frac{1}{2^2}}{2!}+\frac{\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}}{3!}+\frac{\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}}{4!}+\cdots$$?

Transcript for

$Mathematics$ Mathematics