Hm

Is there a formal framework for this homotopy limit thing?

A. K. Bousfield and Dan Kan, Homotopy limits, completions, and localizations, Springer LNM 304

That started the subject

(actually, formalized it; it was used implicitly for ages)

ooo Kan, I see.

@MarianoSuárez-Alvarez "Natural homomorphism" se traduce a "homomorfismo natural"?

Sigh I still don't understand why the two notions of colimits are different. Directly gluing and adding a cylinder in between - what causes the difference in homotopy type?

@PeterTamaroff, yup

well, it is pretty obvious in the examples that the non-homotopical constructions are not invariant under homotopy equivalence of the initial data

that is the problem one has to solve

this is really imortant if you plan to study, say, the category of spaces up to homotopy

(for otherwise things will not be well defined)

and the category of spaces up to homotopy is Something One Wants To Study™, of course

«well defined» !

hmm

So do people call $X \cup CA$ the homotopy quotient of $(X,A)$?

homotopy limit

Hm I see.

Thanks a lot! That's fun.

A very famous construction which is an homotopy-thing is the Borel construction, which is what one does when ones to quotient by groups homotopically

and what would that be?

Can it be described in a simple way?

again, if you are studying group quotients and you wwant to look at things up to homotopy, then your procedure for constructing quotients has to be homottopy invariant

Yes

that is just what the Borel construction does

in good cases, it is homotopy equivalent to the usual quotient

Oh I'm asking, what the actual construction looks like

say $\mathbb Z$ acts on a space $X$

then if the action is proper and free, the quotient and the Borel construction are homotopically equivaent

BTW, I still find this surprising - there's nothing like a "homotopy category" where all these homotopy limits/quotients etc are categorical?

if there are fixed points, then the Borel construction keeps information about what the stailizers looked like, and that sort of thing

oo, interesting

that is what Kan, Boufied, Quillen and a long et at. do

Oh, so these homotopy limits are actual limits in some category

model categories, "homotopical algebra", $\infty$-categories, etc

they are not limits

they are homotopy limits

no I'm changing the category.

it is a distinct concept

well, no

Hmm

they are not limits

ok

When I was asking whether these homotopy limits/quotients can be made categorical, I mean in this sense (that they are limits/quotients in actual categories)

Hm

just like an $\infty$-category is not a category

Oh, I didn't know that.

I don't know what $\infty$-category is anyway, although people seem to talk about them a lot these days (nLab..)

they are the rage now

haha

Thanks a lot!

This would surely sound like a contradiction to any layman looking for a physical interpretation:

the $\infty$ is like a weak

a semi-group is not a group

that one looks less surprising

Sort of like saying a unicorn is undefined as a horse, but it is horse-like.

@MarianoSuárez-Alvarez Do you think I can visit you this week?

II'm leaving on saturday for Brazil for a couple of weeks

and these two days I am writing notes for my lectures

so probably no :-/

@MarianoSuárez-Alvarez Oh, OK.

Is the trip math related?

@MarianoSuárez-Alvarez I got to the basic isomorphism theorems in Jacobson.

yup, a conference in curitiba

@MarianoSuárez-Alvarez You'll talk or listen?

( or prob. both..)

both :-)

can't say I wasn't expecting it

@MarianoSuárez-Alvarez I?m curious about the uses of the isomorphism theorems

I know that by studying it this way, alone, maybe I lose much perspective

for example, the first isomorphism theorem seems useful

the one that says $$\operatorname{im} \eta \simeq G/\ker \eta$$

Similarily, I can seem so grasp the use that of "for normal $N$ in $G$ and $\eta$ an homomorphism we have $$G/N\simeq \eta(G)/\eta(N)$$"

They are useful when they are useful, really. It is hard to say what they are useful for

there are weirder ones, like Zassenhaus's lemma, en.wikipedia.org/wiki/Zassenhaus_lemma

@MarianoSuárez-Alvarez aha

Since it relates to the fact that every homomorphic image of a group $G$ is isomorphic to $G/K$ for some normal $K$

that oone shows up notably when one proves Jordan-Holder, but also in many other places

@MarianoSuárez-Alvarez wow =P

Well, the one that looks kinda "weird" is $$\frac{HK}K\simeq \frac{H}{K\cap H}$$

but it does relate the "group theoretical" operations of product $HK$ and intersection $H\cap K$

$K$ is normal

if you have a group G, a normal subgroup H, and a random subgroup K, you cannot compute K/H

since that doesnto make sense

that isomorphism tells you pretty much all you can do

and that the two options give the same result

no it isn't

@MarianoSuárez-Alvarez I see

you'll encounter it on occasion

and it is quite useful

but naming an application is like naming uses for scotch tape

@MarianoSuárez-Alvarez haha i see

who starred that?

I did :-(

@skullpatrol I was just trying to see who else was around =P

It's all good pal

What's the difference between $\sigma$-ring and $\sigma$-algebra?

complements

a sigma-algebra is just a sigma-ring which contains the whole set, in other words

Well, I see.

Since the collection could be uncountable, it's different.

huh?

$\{\emptyset,\{1\}\}$ is a finite sigma-ring on $\mathbb R$ which is not a sigma-algebra

and a sigma-algebra on the set $\{1\}$.

At first, I thought $\bigcup_{A\in\mathcal R}A\in\mathcal R$.

If $\mathcal R$ is countable, we can say that $\mathcal R$ is a $\sigma$-algebra, where $A^c=\left(\bigcup_{X\in\mathcal R}X\right)\mathinner\backslash A$.

the cardinality of $R$ has nothing to do with the distinction

something is a sigma-algebra on a set

I exactly mean, could we always consider a $\sigma$-ring a $\sigma$-algebra on some set? Then the cardinality of $\mathcal R$ plays some role in that.

@MarianoSuárez-Alvarez I have a problem. =/

well, if you are willing to change the set, sure

Take $G$, $K$ normal in $G$, $H$ a subgroup of $G$

@PeterTamaroff, ?

Consdier the mapping $v$ with $g\mapsto gK$

And let $v'=v\mid H$.

The book claims $\operatorname{im}{v'}=HK/K$

@MarianoSuárez-Alvarez That's wrong. For example, $\mathcal R$ is the collection of all at-most-countable subsets of $\mathbb R$.

But as far as I can see it is $H/K$.

$H/K$ does not make sense

@MarianoSuárez-Alvarez That's wrong. For example, $\mathcal R$ is the collection of all at-most-countable subsets of $\mathbb R$.

the collection of all at most countable subsets of $\mathbb R$ is surely not countable

@MarianoSuárez-Alvarez I mean $\{hK:h\in H\}$

Sorry.

@MarianoSuárez-Alvarez Yes, so $\mathcal R$ is not a $\sigma$-algebra.

Frank, you wrote earlier: «If $R$ is countable then...» and I said yes, that is true

you are now giving me an example which has $R$ not countable

I dunno what your point is, really :-)

@PeterTamaroff, check that that set is the same as $HK/K$

The book says "Since any coset of the form $hkK$ coincides with $hK$, it is clear that ..."

I can follow that.

I don't know, something made me hesitate.

simply because $kK=K$.

@MarianoSuárez-Alvarez Yes, yes.

That was not my problem.

@MarianoSuárez-Alvarez I mean that the difference between $\sigma$-algebra and $\sigma$-ring is essential, not a business of changing the universe set.

you are arguing against a point I did not make

@MarianoSuárez-Alvarez I mean that the difference between $\sigma$-algebra and $\sigma$-ring is essential, not just a business of changing the universe set.

@MarianoSuárez-Alvarez I mean that the difference between $\sigma$-algebra and $\sigma$-ring is essential, not a business of changing the universe set.

I only said that yes, it is true that if R is a countable sigma-ring then it can be viewed as a sigma-algebra over the set which is its union

which was something you said

Okay, now it's solved.

since countable sigma algebras are rather rare, the point is not very interestng

@MarianoSuárez-Alvarez OK, assuming that $K$ was normal in $H$, how would $H/K$ and $HK/K$ differ? That is what I fail to see.

if $K$ is contained in $H$, then $HK=H$

so in no way

Oh, good. So I wasn't that derailed after all.

I'll think harder next time.

heh

It is kinda late! =P

@MarianoSuárez-Alvarez do you have to get up early?

nope

I'm in writing mode

I work much much more in the night

@MarianoSuárez-Alvarez I think the peace of the night is great to work

indeed

and then they invented the internet

@MarianoSuárez-Alvarez HAHAHAHAAH, oh well...

hi

lol i suggested the same edit about half an hour ago but my one wasn'T accepted ...

is odd = 2n

or is that one even `?

ah odd = 2n+1 and even =2n

"The «abstract meaning» of things is peanuts. Peanuts. "

If there are any mods in the room, could they do something about Silent man posts? He keeps defacing his question and "rollback war is going on. @MarianoSuárez-Alvarez @robjohn

Thanks!

I'm in the process

Thanks a lot, Mariano!

people can be so so so obnoxious...

grrrrrrr

@MarianoSuárez-Alvarez GUSFRAVA

UNAGI

hm?

Whatever fits you!

I'd had suspended him for 5 years....

Gusfraba

Unagi

@MarianoSuárez-Alvarez I think those links are self-explanatory.

i wonder if i ever get the 25th upvote for my badge :D

I have a question about Baby Rudin's proof about $\mathfrak M_F(\mu)$.

Suppose $A_n\in\mathfrak M_F(\mu)$ and $B\in\mathfrak M(\mu)$, and $A_n\cap B\in\mathfrak M(\mu)$, why can we conclude that $\mu^*(A_n\cap B)\le\mu^*(A_n)<+\infty$?

What do your notations mean?

It's proving that $\mathfrak M(\mu)$ is a $\sigma$-ring.

I don't have rudin with me, and I don't know what your $\mu$, $\mathfrak{m}$, $\mathfrak{m}_F$, $\mu^*$ are..

anyway, maybe someone more familiar with baby rudin can help you here.

$\mathcal E$

Let me see.

Well, got it.

@DominicMichaelis Good answer?

@robjohn What time is it over there?

@PeterTamaroff Almost 11:30 PM

@PeterTamaroff what's about this recursive restatement?

@robjohn Which one?

Oh, the "over there" part?

@PeterTamaroff this one

@robjohn Well, Julian restated my statement and Byron restated Julian's restatement and you restated Byron's restament of Julian's restatement!

Well, it is not really a restamente, but a generalization.

So change restated by generalized.

Well, I said Byron restated $\lim\limits_{n\to\infty}f_n(x_n)=\lim\limits_{n\to\infty}f_n\left(\lim\limits_{‌​n\to\infty}x_n\right)$ when $\{f_n\}$ are equicontinuous at $\lim\limits_{n\to\infty}x_n$ since mine was first :-)

I didn't check the others.

@robjohn I was simply fooling around

@PeterTamaroff I know, I was just confused

but I think that theorem about equicontinuity is what you were asking.

@robjohn I had never heard about equicontinuity, really.

@PeterTamaroff Rudin talks a lot about equicontinuous families... I don't remember which: baby or big.

@robjohn Baby.

@robjohn Ah. Dunno if I'll ever read it. I guess so. Now I'm trying to learn some algebra.

@robjohn But Rudin's is exactly uniformly equicontinuous.

@FrankScience You're studying from Rudin, I guess?

@FrankScience well, the definition is similar, simply localized

Suppose $\mu^*$ is an outer measure of a regular, nonnegative, bounded function $\mu$, and $A_1\supset A_2\supset\dotsb$, $\bigcap_n A_n=0$. Is it true that $\mu^*(A_n)\to0$?

@PeterTamaroff I regard it as a reference book.

@FrankScience I see.

$\bigcap_nA_n=\emptyset$, for convention.

@FrankScience Actually, the Wikipedia link I give above says that Big Rudin on p.249 talks about equicontinuity at a point.

@robjohn I see. There's another reference: Munkres's Topology.

and my axe?

I doubt the conclusion is false.

@PeterTamaroff Is there a site that produces those?

$\Huge\text{One does not simply fool around with robjohn}$

@robjohn memegenerator.net might be an option.

I was looking here

@robjohn That's another option.

@robjohn Today I skipped many chapters of Baby Rudin and go into the last chapter about The Lebesgue Theory.

@robjohn Your enlarged gravatar looks great there >8(

very...intimidating

Ha, good one!

@PeterTamaroff Dude, the guy looks cool. Could be me. :-( Bad choice of background. :-(

Here is all I have to say.

@skullpatrol Haha.

@robjohn i don't think such a good one math.stackexchange.com/questions/305559/…

but that thread got a lot of attention, don't unterstand why at all

but i don't have the 25 upvotes badge if you mean that

oh thanks :)

It was me. :P

thanks :)

btw how is the hotness of a question calculated ?

The upvotes seem random.

@DominicMichaelis Depends on the gender, obviously.

@PeterTamaroff THIS IS NOT FACEBOOK!

@DominicMichaelis That is covered by this answer

@OrangeHarvester Ah? Why are you screaming at me?

@DominicMichaelis (fraction of people who can understand that question)*(fraction of people who can understand the solutions)*(a score for posting the question/answer on a good time when most of the users are active)*(a score of wittiness/humor in the answer)*(how good is the question/answer)

@DominicMichaelis at least the irrationality of $\sqrt3$

@PeterTamaroff did not mean to offend. just wanted to make a point. :-)

@OrangeHarvester Bleh.

I am baffled by your username.

@PeterTamaroff Yeah.

Can't make heads or tails out of it

is meh<bleh

@PeterTamaroff It goes back to my name tomorrow. My region is somewhat famous for exporting oranges from the country. So, it was for all the farmers in my region who produce oranges.

@skullpatrol That is open for debate.

@robjohn What da?

Lacrosse?

It's an orange harvester.

That is an orange harvester (fruit picker) and it's orange, to boot :-)

So the orange square can harvest round oranges.

@BenW. the mean orange square

@robjohn Never forget that!

That should be your gravatar @OrangeHarvester

i will try to be productive (estimated time where i sucess ~ 5 minutes) see you later :)

@robjohn I wouldn't want to cross a mean orange square armed with an orange picker. Looks like it could do some damage.

@skullpatrol Ahaa.

@DominicMichaelis later. :-)

Everybody (deep) refresh!!

And so it is done.

Perfect match

@OrangeHarvester Do you know this guy ;-D

@skullpatrol I don't think so.

18k NERDIZZLES

Hello! Do you know in which program this picture is made? upload.wikimedia.org/wikipedia/commons/thumb/b/b9/…

i do not know in which program it was made, but it can be done in vtk.

@OrangeHarvester I'm looking vtk gallery... hm, it isn't beautiful I think, at least in comparison with matplotlib

@Nimza VTK can run on 65536 cores.

@OrangeHarvester heh, that's the argument)

But I have only 4096 :D

I guess matplotlib runs only on 1 of them. (Though I guess IPython might make it better.)

@OrangeHarvester hm, didn't hear about it. Now I'm thinking what to do, it's difficult to write large programs in Matlab for me (I don't like Matlab objects), so I'm looking where can I go without loss in easyness of programming mathematical stuff. What do you think?

I was thinking about python

@Nimza Try python/cython.

@Nimza Yes. Python should be the way. You can use cython whenever you want to write some performance intensive code (that is not already implemented.)

@OrangeHarvester thanks, but is there some argument why not just C++ with its libraries?

Cython interfaces naturally with python, so no problems there.

@Nimza There is no technical argument as such. C++ is definitely more "industrious" though. Writing in C++ is somewhat more industrious than writing in python. If you want a parallel/distributed environment and your major focus is computation, then C++ should be your choice. If you major focus is not computation, and you are also interested in post-processing the results a lot, then python/cython development model is much more easier.

@OrangeHarvester nice, thank you

:-)

If you do decide to go with python, there is an Ohio Supercomputer Tutorial on Parallel Python. (Python with parallel is not as widely tested as c++, so there might be pitfalls.) In case, you decide to go with C++ and you have not yet picked up any specific library to use, I suggest a combination of DUNE/PetSc/Slepc for your computations. All three are excellent libraries.

@OrangeHarvester aha, good

I think, that C# better for such tasks. First - beautiful IDE support. Second - good memory management (not an ARC, but pretty good). Three - many libraries/wrappers for it. Forth - your code-debugging time tremendous decreased. Fifth - expressiveness more better than Java and readability more better than any dynamic-typed language. Sixth - it can be more faster sometimes than C++ (my friend have a problem with big data and pointers)

choke

The Art of Mathematics

0# is the real number (read: set of integers) encoding all the truth of the inner model L. Since the truth is not definable (internally) the existence of this real number implies that L is not the whole universe of ZFC (i.e. V≠L). But it implies more actually, it means that there are some large cardinals in the background.

It tells us that L itself is small enough that we know almost everything possible, like we do if we have a set-model of ZFC inside the universe (we can say things, externally, about that model). 0† is similar, but for a generalized case. – Asaf Karagila 17 mins ago

gahwd

That is a bit over my head >_<

Plagarized ansers indicate that the question is duplicate and should be marked as such.

@JacobBlack Okay. Have you noticed my new gravatar?

@Tobias In particular, every abelian group is sudokable, and the direct sum of two sudokable groups is again sudokable.

And I think sudokability behaves well under exact sequences: if this is true, then all solvable groups are sudokable!