@XanderHenderson This is totally unacceptable. However, you can have it shipped in (although perhaps you have a dry state line, like Pennsylvania).

@TedShifrin I'll get more when I am in SoCal in a month.

Or at the liquor store.

A question for the experts/teachers here

How many problems should one do out of a textbook? I know it varies but is skipping some if you "know" how to do them OK? Or should one really take the time to, say, Tex it all out...

Even in the best texts there is redundancy among exercises. If you truly have done a problem perfectly, there’s no point doing essentially the same problem 5 more times.

But with proofs, carbon copies are more rare.

All proofs here, and good thing I stopped on this one because now I'm stumped!

In my algebra book there are plenty of “theory” questions that are not really proofs.

Rather, they are concrete applications of theorems ….

I am basically unravelling definitions so it's not so interesting but I think there's some property I'm missing because at the end I get stumped with not knowing how to show there is some $n \in N$ with $\varphi(n) = n''$

So you’re not using the conjugation definition of normality?

I'm using $gN = Ng$ but have enough to show that the other version is equivalent

Is that version usually easier to work with?

The invariance under conjugation is usually more efficient.

Hmm, OK I will give that a go

It should be immediate.

I suppose immediate is always relative :) I did end up managing to stitch up my way, but will think about yours now

Little bit ugly but works I guess

$\phi(gng^{-1})= ….$

@TedShifrin ok but how did 1 appear?

probability of NOT

Is that due to the first P(B|A) P(A) being the share DID raise by 5% and the second P(B|A) P(A) being the share DID NOT raise by 5%?

@TedShifrin is that correct?

Typically, Bayes computes the probability by adding the conditional possibilities … either A happened or it did not.

@TedShifrin Not quite sure I follow. We want the equality $gNg^{-1} = N$ where $N := \varphi^{-1}(N')$ and I guess you're saying there should be someway that I can immediately see that $g\varphi^{-1}(N')g^{-1} = \varphi^{-1}(N')$.

OH

I'm meant to use that conjugation is an automorphism of $G$

Hmm, but that doesn't mean its restriction to this subgroup is an automorphism does it?

It is a homomorphism. Just use that.

@TedShifrin ok, so the first P(B|A) P (A) is did raise by 5% and the second one is did NOT raise by 5% so the formula is 1 - 0.04. Is that accurate?

What formula? What is the probability of not $A$?

I don't quite see it Ted, will think about it on my walk home

I definitely want to get better at solving questions in these "high level" ways rather than always needing to unravel the definitions and get nitty gritty

Just write diwn what homomorphism gives for a product. And remember what it does to inverses.

@TedShifrin I'm new to probability so excuse my ignorance on several things. If the question said NOT raised, is that means we have to subtract 1 with the probability of the first P(A)?

@TedShifrin Maybe write it DOWN, too.

Typing on the iPad sucks, and I have to relaunch the page to get the edit button.

@Snoopy if the probability that event A occurs is $p$, then the probability that event $A$ does not occur is $1-p$, because one or the other must occur and they cannot both occur simultaneously.

I think Ted you’re saying I have to first show that the set-valued pre image function for a homorphism is itself a homéomorphism?

Autocorrect…am on phone now

homéomorphism is a different class

@EE18 in Poland we would say that its ringing ... in the wrong church

LOL

No. It’s a map of sets.

whichever church it's ringing in, I still can't follow the argument Ted is suggesting to me

i’m saying to compute $\phi(gng^{-1}$.

Missing parenthesis — add it.

Right. You just have to show $g N g^{-1}\subseteq N$. So $gng^{-1}\in N$ for $n\in N$

but $n\in N \iff \varphi(n)\in N'$

so what, $\varphi(gng^{-1}) = ?$

$\phi(gng^{-1}) = \phi(g)\phi(n)\phi(g)^{-1} = n'$ for some $n' \in N'$ because $N'$ normal

Done.

you can just write $\in N'$ without introducing a useless variable

How does this establish $g\phi^{-1}(N')g^{-1} = N'$ though? Sorry to both of you for being so dull here I just can't see how I'm done

I've established one direction $\subseteq$

didn't we already go over this

you don’t need =. Just subset. Why?

I think we might have. I will review

we did

Ted giving me nightmares about reading a book, having the author hit me with a "(why?)", and not knowing why :)

i do have some (Why?js in my books.

it is meant to force students to answer.

I mostly joke...books are usually pretty good about having led you to water at that point

Now we turn in earnest (why?) to our study of the differential geometry of curves and surfaces (why? and why now?)

earnest more important than now …

Do textbook authors chuckle to themselves when they write these things

I do not.

Thank you Jakobian, you are right. Appreciate it :) I will reread what i wrote and digest

I am a follower of Socrates.

In reality you just write $N = a(a^{-1}Na)a^{-1} \subseteq aNa^{-1}$

Slick, using that the image of a subset is a subset of the range

Thank you very much

OK, so, summarizing (I made a mistake in my claim above). Ted, I was meant to argue: For arbitrary $g$, we have that for all $n \in N$, $\phi(gng^{-1}) = \phi(g)\phi(n)\phi(g)^{-1} \in N'$ so that $gNg^{-1} \subset N$, and that's enough by what Jakobian pointed out.

Not to point any fingers

Why does pointing at something feel rude?

Unsolicited attention I guess

@BalarkaSen Hmm ok I'll keep that in mind. But you know the book is too long so I kinda prefer a shorter paper. That's why I planned to read the paper.

@BalarkaSen thank you for the motivation. Now I'm motivated to read that paper.

@BalarkaSen Yeah... easy observation but important...

Is it possible in "general" to extend a covering of $n-1$ manifolds to a covering of associated mapping tori?

Should depend on a monodromy... for $n=3$ case, I can construct such covering for example by rotation but the monodromy of a covering is reducible.

For $n=3$ case, I should ask in this way: Is there an explicit example that a fibered 3-manifold $M$ can be finitely covered by another fibered 3-manifold $N$ such that $M$ and $N$ have different cyclic coverings (in geometric structure). So I rule out Balarka's example.

(all are hyperbolic)

No that condition still can't rule out that example. I can just choose a different fiber bundle representation of the same manifold.

@BalarkaSen Do you have any recommendations on a reference to complex dynamics? I remember you don't recommend Milnor's book. It seems Prof. Mahan Mj is mainly studying Sullivan dictionary these days.

or at least dynamical viewpoint of Kleinian groups.

Cech cohomology...Cech cohomology...Hmm...

I think there's no such covering I want.

Curious what others think about this solution from my child's textbook regarding the number of faces/edges on a cylinder. There were no definitions scoping "edge" or "face". Is it worth making a math.SE question?

@JoshHibschman At best, this would be a Math Ed question, but even there, I don't think it would be well received. I would still want to know what the intended definition of "face" and "edge" are.

Personally, I would either say that a cylinder is not a polyhedron, therefore it doesn't make sense to talk about edges and faces at all (the idea being that faces should be polygons, and edges should be segments; consistent with the way polygons are defined), or a cylinder has three faces, three edges, and two vertices (add a vertex to each circle, and join those vertices to make an edge) so that Euler's formula makes sense.

The image given satisfies Euler, but feels like nonsense to me. In a more general context, you are meant to build up a space out of, for example, $n$-cells (e.g. a point is a $0$-cell, a segment is a $1$-cell, a triangle is a $2$-cell, a tetrahedron is a $3$-cell, and so on). Then the Betti numbers are a topological invariant.

how does the sum become that? i didnt reliaze that this is a convergent known sum

@onepotatotwopotato There's a particular chapter on taut foliations in the book, which doesn't seem too long

@Mad The highlighting makes that difficult for me to read.

But that looks like a geometric series to me.

I thought so too, but ...

there is no but

i am absolutely a retarted potential, and i dont know how i thought "Yea thats NOT the geometric series"

it 100% is

@onepotatotwopotato Right. Presumable you need some condition on the homology to manufacture examples of hyperbolic 3-manifolds which fiber over the circle in two distinct ways.

@Mad All those extra constants are dumb. Ignore them.

Something like, take $b_2 \geq 2$ and choose a fibered face of the Thurston norm ball

Pick two guys from there

@XanderHenderson Yes i just did it too

Xander which software did you use to write that?

thanks

Looks pretty cool

@onepotatotwopotato No clue really, I don't understand complex dynamics

@Sahaj Notability.

And a Wacom Cintiq.

That is my genuine, terrible handwriting.

It's fine compared to most mathematicians I know.

@XanderHenderson It's confusing but makes sense from a high level point of view. The cylinder has a CW structure with two 0-cells, two 1-cells and one 2-cell.

The Euler characteristic is then computed as alternating sum of the number of cells of a CW complex, which (by a theorem) is indeed alternating sum of betti numbers of any triangulation you choose.

Should not be in a kid's textbook. Maybe as an "intrigue" (ah, this works, even if it doesn't make sense!).

@BalarkaSen Yes... I feel like I said that already. :P

Oh, wait... no.

That isn't how I constructed the cylinder.

Yup, you didn't

Yeah, I don't like your construction. It is less "obvious" to me.

I don't see your one $2$-cell.

Oh, wait, no, I do mean three 2-cells.

Top, bottom and side

Yes, now we are saying the same thing.

No but what I don't agree with is it feels like nonsense. :)

Actually there should be three 1-cells, also. There's a 1-cell joining the top circle and the bottom circle.

The construction is two $0$-cells, three $1$-cells, and three $2$-cells.

Yes.

Which is exactly what I said, only in more elementary language. Two vertices, three edges, three faces. :P

LOOK, MA! I CAN DO ALGEBRAIC TOPOLGY!

;)

say a continuous map between top. spaces is proper if it is closed and has compact fibers. is a proper image of a LCH space LCH?

The point is the exact decomposition doesn't matter.

the H part is easy

It is not entirely invalid to say that the cylinder has no edges or vertices.

But this should be an intrigue than actual math writing

@BalarkaSen Sure, I made that argument, too---"edges" should be segments (i.e. no wibbly-wobbly curvy bits), "faces" should be polygons surrounded by edges, and "vertices" should be points where three or more faces meet.

But then we are restricting our attention only to polyhedra, not more general solid figures.

But, in that case, I would argue that it doesn't even make sense to talk about the "faces", "edges", or "vertices" of a cylinder, since it isn't a polyhedron.

@Thorgott yes

@Thorgott that's called perfect map not proper

Engelking calling it that is why I initially didn't find the result there

neither terminology, however, is definite

The usual terminology I've seen is that on wikipedia i.e. closed + compact fibers = perfect, pre-image of compact set is compact = proper

Engelking is not really to be trusted every time when it comes to standard definitions

See his definitions of dimension for example

it's a mess, Bourbaki call a map proper if its product with the identity map of any space is a closed map

this turns out to be equivalent to closed + compact fiber

and also equivalent to closed + "pre-image of compact is compact"

the "pre-image of compact is compact" condition I have also seen referred to as "quasi-proper" (mostly by algebraists who replace compact by "quasi-compact" following the Bourbaki convention that "quasi-compact" = "every open cover has a finite subcover" and "compact" = quasi-compact + Hausdorff)

Yeah because perfect implies proper. In a lot of contexts they are equivalent

curiously, iirc Engelking does follow the Bourbaki convention of requiring compact spaces to be Hausdorff

right, they're equivalent for LCH spaces, potentially in more general scenarios too

For compactly generated Hausdorff spaces they are equivalent (this includes metrizable spaces)

ah right

I'm thinking about those right now, even though I don't really want to

do you happen to know if the weak infinite product of countably many pointed LCH spaces is LCH?

(weak infinite product is the subspace of the product where all but finitely many entries are basepoints)

It shouldn't be in general for the same reason that infinite products of LCH spaces aren't LCH

If you take weak product of countably many copies of R, take a compact neighbourhood of 0, then its projection is whole of R for infinitely many indexes

yeah, turns out it isn't :\

@XanderHenderson All fair points

how can i show that $\langle g \rangle = \{ g^n | n \in \mathbf{Z} \}$ ?

$\langle g \rangle$ is defined to be the intersection of all subgroups that contain $g$.

Show that every subgroup containing g contains the RHS, and that the RHS is a subgroup containing g

ah, I got sniped

You also got a and b in the wrong order, clearly

Its a very common procedure

If you are thinking in more general settings, you'll see it everywhere

I showed that it's a subgroup.

stuck at inclusion

oh wait, i am reading this wrong

Well suppose H is a subgroup containing g. Can you tell me other elements of H?

Define closure of a subset $A\subseteq X$ as intersection of things containing $A$, and then give an internal characterization of that object

Then by doing what you are doing right now, prove they are equal

this works like that very often

Hmm

@AlessandroCodenotti this sounds very natural to me, ok

will try this now

@AlessandroCodenotti This is a very witty reformulation of the statement, but it is what it is I guess :D

You implicitly embed a set theoretical fact there! :D

Alright, I wrote it up. No hassle.

its all very standard

Yeah I think I overthought it a little bit

Hi

On MSE, is it OK to reproduce proofs from research papers? Of course with citations

Why would you want to do that?

I'm just asking

Well, that is far from an answer.

I saw someone do that in an answer to a question on MSE. I'm wondering if it's allowed or not.

I don't know the rules. I personally do not think people should copy walls of text (whether from a book or a paper) unless it's absolutely essential, and it rarely is. Ordinarily, one would cite a reference.

Admittedly, I couldn't find the result anywhere else online (the research paper doesn't appear to be available and is quite old) and I wouldn't have known of it without the answer being posted.

Maybe I just didn't look enough though. I'll try to find the paper

hello

@Sahaj why not?

you're not stealing someone's work, since you're citing the result, and this makes it readily available for the ones reading

True. Is there some rule of MSE about it though?

I can't find it on Meta

I did find a post on it but it seems to be regarding a single answer and not a definitive policy on whether it is allowed or not.

...

its allowed

Alright. Thanks.

people on mse are constantly copying arguments from other sources, whetever consciously or not

why would it be not allowed when you add a citation?

thats even better

novelty does exist, and for a lot of things there seems to be a limit of ways a person can possibly think of to reach an argument

fact is, people often learn things by copying others, you might not even be aware of when you are just repeating an argument you learnt

@TedShifrin Above is a point for you specifically

What is that supposed to mean?

Notice I said "walls of text."

I don't think "learning from others" and "copying walls of text from a book/paper" are synonymous lol

You have vast knowledge and I imagine it happens often that you repeat an argument you read somewhere

Writing something in your own words is different from pasting in the words of someone else. Of course, most of what you write here is stuff you've learned from books or from someone else. But you're writing it in your own way.

Can someone help me to do the proof of the second De Morgan relation : $A - (B \cap C) = (A - B) \cup (A - C)$

I don't recall ever cutting and pasting from my own textbooks here.

Where are you stuck, Pizza?

my brain is somehow rewired in a weird way and every time I hear "cut-paste" I think of topology

To me, the English meaning of "reproduce" suggests literal copying.

Not surprising, @Balarka :)

@Pizza Are you not writing words and interpreting what union, intersection, and difference mean?

> produce something very similar to (something else) in a different medium or context.

there is this meaning of the word reproduce

I'm only a native English speaker. What do I know?

You're obviously in one of your argumentative moods, Jakobian. I'm done with it.

But Sahaj isn't, as far as I can tell

so its not a valid argument

Jakobian should become a debatebro and open a twitch stream

i thinked i need to start from these 2

bet he'll have a million followers

Not start from those two, @Pizza. You need to show both of those.

yes

I agree.

So what does it mean for $x$ to be an element of $A-(B\cap C)$?

@Balarka Out of curiosity, do you consider yourself a native English speaker?

I would consider you so.

@Sahaj If you mean "copy and paste" another's proof, no. Don't do that. It is plagiarism, even if you give a citation.

I cannot officially declare it because I still mess up apostrophe's

Indeed you do :P

But most Americans mess up so many more things than that.

True.

A better approach would be to highlight the key ideas of the proof, and give a citation. (Note that answers on Math SE should be as self-contained as possible, so you should give enough discussion of the proof to ensure that the asker can put the pieces together without necessarily needing to go look at the original source).

Thanks for a thoughtful response, @Xander.

I find myself not "thinking in English" often enough though, so I'm hesitant to say I'm a native English speaker. Especially while counting numbers, I'd revert to my native tongue.

I think in our modern society we have just accepted de facto plagiarism as a fact of life.

@BalarkaSen so which language is your native one?

Balarka, yes, that's natural. That's how they caught spies, I'm told.

bengali

It is also fine to paraphrase a proof given by someone else, in your own language. Generally speaking, a copy-paste won't really work on Math SE, because (if nothing else) the notation may not align perfectly. But rewriting a proof in your own words and citing the original (again, to keep things self-contained, when possible) is fine, and is common in the mathematics literature.

@TedShifrin Oh wow makes sense

My impression is that educated people in India learn English from the very beginning of school and all the Indians I've encountered are completely fluent and "native speakers."

Indeed, a large chunk of the first chapter of my PhD thesis is, essentially, a restatement of the major results in two chapters of one of my advisor's books, written in notation that I was going to use for the rest of the thesis, with a pile of citations, and some commentary.

Interestingly, I think most people in France and Germany may also learn English from the beginning of school, but they are not necessarily as fluent — although most around here sure are.

@BalarkaSen Man, counting is maybe the one place where I still often find myself thinking in Russian (I am a native English speaker, and learned Russian in high school; but I often count in Russian, and have a good "feeling" of numbers in that language).

@BalarkaSen I see, I'm very ignorant of indian languages, how closely related are they? Can you understand, say, Hindi?

@Xander I can do numbers fluently in French, but if I have to do an addition problem, I'm not going to think through it in French.

Xander outing himself as a Russian spy was an unexpected plot twist

True for increasingly many nowadays, @Ted. In South, people can't communicate in Hindi (unofficially the common language in India), but they all know English.

@XanderHenderson Damn haha

@AlessandroCodenotti Yeah I understand/speak Hindi.

@TedShifrin can i draw a venn diagram?

Americans are notoriously horrible about learning any foreign language.

@Pizza You can draw one for you to understand it intuitively, but that is not considered a proof.

@TedShifrin I'm pretty comfortable adding and subtracting in Russian. I never got good at multiplication in Russian, and I don't actually know that I ever knew how to talk about rationals, so more complicated ideas are right out. But basic natural number stuff works just as well for me in either language.

Hindi and Bengali are close, but the former has more Persian-origin words and the latter has more Sanskrit-origin words

Interesting, thanks

@Balarka Are the written alphabets the same? (I know I'm showing my ignorance.)

Alphabet is the wrong word, I'm sure, too.

Did you mean the script?

You're in good company in your ignorance @Ted

@AlessandroCodenotti My father's mother's mother was an ethnically Russian Jew who was born in Ukraine. Her family got chased out of Ukraine in 1900. Ukrainians have always hated Russians, Russians have always hated the Jews, and Ukrainians have always hated the Jews. It was not a healthy place to be a Russian Jew. :D

Depending on whether you consider my company good I guess. Maybe you're in demonic company

The scripts are different, @Ted. Hindi script is Devnagari, Bengali script is... well, I don't know if it has a name

@Xander My grandparents all fled around 1917.

@TedShifrin No, but similar

Thanks, Sahaj, Soumik, Balarka.

@BalarkaSen I’ve seen it called the eastern nagari

Oh interesting @Sahaj

If anyone here can "write" in fraktur, raise your hand :)

So wikipedia tells me that there are 22 officially recognized languages in India, do most of them also have their own script?

@TedShifrin That's about the time my grandmother got arrested during a labor action. :D

I'm still working on my "g"

Can't get a consistent fraktur g

@AlessandroCodenotti Yeah many do