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4:00 PM
though in that case I guess one could factorize the summation since $M$ would be a direct sum of 1-by-1 matrices
i do feel like your sum should have some nice properties, though, since $M$ is of the form identity plus a rank-one perturbation
 
@Semiclassical right!
I may add a bounty to the MO version soon.. I feel it deserves a good answer :)
 
4:19 PM
@BalarkaSen I'm flying to Israel tonight, which means that we'll end up being only 2.5 hours apart instead of the 9.5-hour time difference we have now. I thought you'd find that interesting.
 
How do you make the connection that because $\mathbb{Z^+}$ is infinite, there must exist an infinite amount of prime factors and consequently prime numbers?
 
have you seen Euclid's proof?
 
no I'm trying to formulate my own for my textbook
0
Q: Proving the set of prime numbers in $\mathbb{Z+}$ is infinite

OblivI'm trying to prove that for any $N \in \mathbb{Z^+}$, there exists only finite many integers $n$ with $\phi(n) = N$ (i.e. finite amount of numbers that have $N$ numbers relatively prime to them) I started by addressing the lower bound of $\phi(n)$ as the set of primes $P = \{0<p\leq n~|~\phi(p)...

 
@MikeMiller thanks for your help anyway
 
4:20 PM
both the answers given to me just throw euclid's proof at me but I don't want it to be spoiled D: @Semiclassical
I should have just asked for a hint
 
$\mathbb{Z}^+$ being infinite isn't enough. after all, the set of positive even integers is also infinite but only contains one prime number.
 
that is true..
 
Hint: You essentially want to prove that if I give you any finite list of primes, you can find a prime not on that list.
 
that suggests a nice (if rather open-ended) question, actually. What are some infinite subsets of the natural numbers which unexpectedly contain an infinite number of primes?
 
Hint 2: Every number can be written as a product of primes (why?). So if I give you that finite list of primes, all you have to do is find something that's not a product of things on that list to show that my list is incomplete.
(Does that give too much away? @Semiclassical)
 
4:27 PM
well, ultimately those're just hints on Euclid's proof
 
Well, yeah
I mean, I can't hint towards Euler's proof…
 
so if that's already been seen before, then hints aren't going to be terribly relevant :)
 
well my knowledge of primes is limited. I just think of them as fundamental building blocks of numbers. Every number is a product of primes
 
It sounds like @Obliv doesn't know Euclid's proof. Unless I misread
 
I don't know it
 
4:28 PM
Euler's proof is definitely too sophisticated here.
 
@Semiclassical residue 3 modulo 4.. But that's not unexpected, just what I can think of
 
then those hints are probably fine.
Dirichlet's theorem on arithmetic progressions is a good example, yeah
 
I'll try to prove it using those hints @AkivaWeinberger thank you
 
@datalava Any arithmetic sequence (assuming it's not trivially false, i.e., residue $a$ mod $b$ where $a$ and $b$ have a common factor), in fact!
Open problem (I believe): $x^2+1$
Probably infinitely many primes but we don't know
 
i was about to suggest that, heh. it came to mind but i didn't know whether it was open or not.
 
4:30 PM
@AkivaWeinberger Nice.
 
I didn't realize that was an open question
 
$x^2$ and $x^2-1$, of course, are definitely not examples
hmm!
The Bunyakovsky conjecture (or Bouniakowsky conjecture) stated in 1857 by the Russian mathematician Viktor Bunyakovsky, asserts when a polynomial in one variable with positive degree and integer coefficients should have infinitely many prime values for positive integer inputs. Three necessary conditions are the leading coefficient of is positive, the polynomial is irreducible over the integers, and as runs over the positive integers, the numbers should be relatively prime. (Thus, the coefficients of should be relatively prime.) Bunyakovsky's conjecture is that these three conditions ar...
 
Nothing factorable over the integers I guess
 
it mentions on that page, btw: "That n^2+1 should be prime infinitely often is a problem first raised by Euler, and it is also the fifth Hardy–Littlewood conjecture and the fourth of Landau's problems." so quite a pedigree.
 
@Krijn My algebraic geometry prof let me borrow his copy of Fulton's intersection theory for a few days. Going a bit off-course shortly to read that :)
Well, read chapter 1. No way I am going to be able to read the whole book without more background, and I don't want to anyway.
 
4:37 PM
In this question, is what the asker said to prove the graph is closed enough? math.stackexchange.com/questions/77896/…
 
are there some fundamental primes (like 2,3,5) where one of them has to be in the prime factorization of all integers in $\mathbb{Z^+}$?
 
@BalarkaSen After I'm done with 2.3 of Hartshorne, I'm diving in Milne's notes on étale cohomology too, so we will be off-course both!
 
@Obliv No.
 
Also, no library?
 
oh wait yeah that's trivial. just multiple two primes with each other :p
 
4:42 PM
@Krijn Hah. I'll only be off-course for a short while though.
 
eh, even more trivial: what about the powers of $7$? @obliv
 
yes that is even more trivial
 
I wonder if PVAL saw the counterexample to his question I came up with.
It's not even trivial. If you have a purported finite set of primes which appear in the prime factorization of every integer, just take a prime outside that finite set. (primes are infinite, so you can always do that).
This contradicts existence of any such "fundamental primes".
 
I meant to say composite integers I don't know why I didn't specify
 
@datalava It's not phrased very well, but it's enough, at least if the OP invokes the theorems that I assume they invoke. Pointwise convergence of the ($C^1$) functions $(f_n)$ plus uniform convergence of the derivatives $(f_n')$ implies that the limit function is continuously differentiable, and its derivative is the limit of $f_n'$.
 
4:46 PM
right. and if you don't want to take the "primes are infinite" step, just take the product of all those primes and add 1.
 
OK. Then either of your or SemiC's solutions work.
@Krijn Not sure what you mean by no libraries.
 
@DanielFischer I see; thanks!
 
@BalarkaSen Borrowing from the prof instead of getting it from the library?
 
@user1618033 Sorry, I was out pretty much all day Monday and Tuesday.
Has anyone else noticed a slow down when editing answers on main? As the answer gets longer, it gets unbearably slow for me.
 
@Krijn Oh sure there are libraries. But prof gave it to me from his own good will, so I took it :) After all library books don't compare with owned books, especially is the owner is an expert on the topic it's about.
*especially if
 
4:58 PM
@BalarkaSen Yeah, they are great if they have those little notes on the sides
 
@robjohn Haven't noticed anything. But I haven't posted much in the last month, so I wouldn't notice. Could it be related to this?
 
ah, not sure if mine has notes though.
 
@BalarkaSen Ah yes, the classic bit where they imbue the book with their essence.
 
@DanielFischer Heh... I need to change my window size so that I can click on that link (the flag, star, link icons block it for me normally). I was thinking that it might be that, but if others did not notice anything, it might just be something with my system.
 
Hah, indeed.
 
5:02 PM
"imbue" is a beautiful world I never heard before
 
Apparently my bomb message was flagged. I tried to say that it was a valid flag, but the system wouldn't let me.
 
Good night @MikeM. Hi @robjohn, @Balarka, @Krijn.
 
I guess the military-industrial complex was offended by the implication that they are of fundamentally no value to the world; I think that's a flaggable offense.
 
Imbue is a great word.
Hi @DanielF.
 
@Krijn and adam would be chased off heaven one other time!
 
5:05 PM
@MikeMiller lol
 
Ah, I see anon is in here with multiplicity two. :)
 
@TedShifrin Did you see the question PVAL posed yesterday? It's fun.
 
@robjohn i think i pinged u one time but someone removed my ping !!!!
 
No, @Balarka. At the moment, I'm stuck on a stupid point-set question that's bothering me.
 
Of course, Balarka has given up on what I asked him, and going further back, given up on the very concept of calculus.
 
5:06 PM
@TedShifrin Hey, Ted! what's up? Are you still in SD?
@Agawa001 when was that?
 
or is this somewhat a kinda rude autoresponder ?
 
Yup, @robjohn. I'm here permanently except for travel. :)
 
@MikeMiller I have not given up on calculus ;) in fact I just asked prof to explain what differential forms are about today. To get started.
 
not long time ago
 
You going to enlighten me, @Balarka?
 
5:07 PM
I don't have a reasonable idea how to approach the torus problem.
 
@TedShifrin travel is what I meant. Every time I have asked you how things in SD were, you said that you were somewhere else :-)
 
@TedShifrin Sure.
 
i asked u if removing my account can influence on my chat history
 
@ForeverMozart: Do you know an example of a normal (Hausdorff) space $X$ for which $X\times [0,1]$ is not normal?
 
@Agawa001 I think I answered that...
 
5:08 PM
LOL, hardly every time, @robjohn. I wish I traveled that much. Leaving in a month for two weeks in Chicago/Atlanta, however.
 
So suppose I have a smooth manifold $M$. 0-forms on $M$ are just real valued smooth functions on $M$. $1$-forms are smooth sections of the cotangent bundle, i.e., maps $M \to T^*M$ which sends a point to a covector of the contagent space $T_xM$. $k$-forms in general are smooth sections of the antisymmetric tensor product $\bigwedge T^*M$ of the cotangent bundle.
 
You're just giving me a definition, @Balarka. What insight do you have?
Why are they important?
 
Well, as far as I understood, a 1-form eats a vector field (section of $TM$) on $M$ and spits out a continuous function.
 
In the large scale, @Ted, nothing we do is important
@BalarkaSen Who cares?
 
LOL, thanks, @MikeM.
 
5:11 PM
hiya
 
Hi @Semiclassic
 
Actually, you don't even need to make the scale too large.
 
hi
 
@MikeM: In semi-seriousness, your response is why I spent more energy on teaching (and book writing) and less on research over the last half of my career.
Hi Karim.
 
I am not sure what answer you want to the "important" bit. I can form a cochain complex out of the vector space of all $k$-forms and compute it's cohomology. That's an invariant of the smooth manifold, so seems like we shouldn't ignore that.
 
5:13 PM
@Ted: That's a noble goal. My sol'n is to do non-math things that are good for people/
 
Would that classify as important?
 
@BalarkaSen What a boring reason to care about differential forms.
 
Well, I'd like to think I've been a good person, too, @MikeM, but that is clearly open for debate.
No, @Balarka. It's too much algebraic topology and too fancy.
 
what i like most out of research are the conversations that come out of it.
 
Unfortunately the union meeting tomorrow is right during the section I TA, so I'm forced to teach algebraic topology instead.
 
5:15 PM
beyond that, it's interesting in the same way that a crossword or a jigsaw puzzle is interesting. it's a satisfying intellectual challenge, but i don't hold any illusions about it actually mattering
 
@Balarka: But there is a good question. Your first answer was very abstract vector-bundle-ish. What is special about the exterior powers of $T^*M$ is that you do have $d$. Why do we have $d$ in this case and not in general?
@MikeM: You'd think they'd put those meetings in the evening and not in the middle of the day.
 
$d$ being the boundary operator $\Omega^k(M) \to \Omega^{k+1}(M)$?
 
one of the main motivations for me to care about differential forms: to have Maxwell's equations look nicer :)
 
Yes, @Balarka. Of course, Stokes's Theorem is the answer to a lot of these questions. But to me a lot of the motivation comes from differential geometry (and I mean geometry, not manifolds).
Sure, @Semiclassic, but isn't that just window-dressing?
 
@Ted: No, this is a good decision. Usually they are held in the union office at 5PM on a weekday. Nobody comes but leadership. We've moved them to mid-day (when people are still on campus) to the sculpture garden. This means some people have conflicts, but it also means more people will probably come.
 
5:17 PM
i didn't say it was a good reason
 
Of course, physicists are still in the 19th century with their zillions of indices on everything ...
 
hahaha
 
I see @MikeM.
 
yes, well, having access to coordinate systems comes in handy at times :P
 
By the way, that point-set question I left for ForeverMozart is available to all: Any idea of a normal space $X$ for which $X\times [0,1]$ is not normal? Surely that can't be.
(In general, the product of normal spaces needn't be normal, of course. But $[0,1]$?)
 
5:18 PM
i'm the kind've person who appreciates diagramatic notation and abstract-index nonsense, mind
 
@Semiclassical You say this as if it's obviously necessary to doing anything, but there are a number of mathematicians who do alright by working more-or-less invariantly.
 
@TedShifrin Can you elaborate without revealing what you mean when you say "why do we have $d$ in this case and not in general"? In particular, can you explain what you mean by "general"? Just trying to understand the question a bit better.
 
General vector bundles, @Balarka.
 
Ah, ok, I see.
 
i don't see how that conflicts with "comes in handy at times"
 
5:19 PM
@TedShifrin He's asking what the second bundle is.
 
Your original answer was very vector bundle-ish. So what's special about $\bigwedge^k T^*M$?
 
I see, yup. Let me see if I can find a fundamental reason.
 
Or maybe he's not.
 
@Balarka: I still say you need to do computations to appreciate this at something other than a formal level.
 
i think i missed what the specific question was
 
5:20 PM
Scrolling is always allowed, @Semiclassic :D
 
Unless you have carpal tunnel in your mouse hand. :)
 
Sorry to interrupt your discussion but can someone help me with this please.This has taken a lot of my time math.stackexchange.com/questions/1750854/…
 
@TedShifrin Oh of course I agree. I am going back home 23rd, and going to study forms then. I was merely getting a general overview before plunging it :)
 
well, i'm on a laptop. so i don't really have a mouse hand at the moment
 
5:21 PM
@john smith: That is not something I'm qualified to help with.
 
@TedShifrin So what does $d$ really do? Suppose let's just look at $\Omega^0(M) \to \Omega^1(M)$. Then $d$ eats a smooth function $f$ on $M$ and spits out {1-form which eats a vector field $X$ on $M$ and spits out directional derivative $X(f) : M \to \Bbb R$}.
 
My old mouse died and so I had to spend $70 on a new bluetooth mouse that would only work with the latest OS, and so I had to update that ... Ugh.
 
So it seems the tangent information is really important. I can differentiate smooth functions along tangent directions.
 
@Ted In thinking about this while I'm more awake, I'm a little worried that my answer here is too undetailed to be helpful... thoughts?
 
@TedShifrin ok thanks for atleast replying
 
5:23 PM
Yes, @Balarka. And if you have a general vector bundle $E$, you could ask about directional derivatives of sections of $E$ (and you could multiply sections by smooth functions, too).
@MikeM: Seems the key point (as it often is) is that $(\text{im}\, T)^\perp = \ker(T^*)$ and the companion formula.
 
Right. Do you think anything in the answer needs to be clarified?
 
oh. i think i asked @mike this already, but not sure about you @ted. is the Hamilton-Jacobi equation something you've ever run into? i'm curious as to how obscure that topic is to non-physicists.
 
Oh yeah. I can still take a vector field on $E/M$, which is a smooth section $X : M \to E$ and take a continuous function $f$, directional derivate along each $X(p)$. Hmm.
So what does really go wrong? Interesting. I'll ponder a bit more.
 
I didn't read it carefully, @MikeM. But long paragraphs with tons of formulas are hard to read. So better to emphasize the key point and then put in details.
 
@BalarkaSen can you explain to me why is $\bar{f}_{*} : H_0(RP^n,Z_2) \rightarrow H_0(RP^n,Z_2)$ is an isomorphism where $\bar{f} : RP^n \rightarrow RP^n$ defined as $[x] \mapsto [f(x)]$ where f is a odd map between $S^n$ and itself.
 
5:28 PM
@Balarka: Replace "continuous" in your brain with "smooth." Everywhere :P
Karim: What is $H_0$?
 
@Adeek Think about what $H_0$ is.
 
@BalarkaSen In your original case, you take a function and spit out a 1-form. In your case, what's the input and what's the output? (At least, what should it be.)
 
Copycat @Balarka :D
 
And what a map on $H_0$ means.
@TedShifrin I don't know of a better or more elaborate answer :)
 
Anyways, guys, I'm off to a party in Antwerp
 
5:29 PM
Granted: He asked you and I replied out of order.
 
Have a good night
 
Have fun, @Krijn!
 
well it a cycle $\sigma : \Delta^0 \rightarrow RP^n$
 
@TedShifrin Right, oops.
 
so that means
 
5:30 PM
@Adeek Cycles are not maps from simplices, but linear combinations of those. Anyway, what is a 0-simplex?
 
yeah I meant to say the basis elements are of that form
the 0 simplex are just points in $RP^n$
 
@Ted: OK, I'll try to do so later.
 
Also you're telling me representative of elements of $H_0$, not what $H_0$ is itself.
 
yes
ohhh
just a sec
 
what I'm curious about at the moment is what Hamilton-Jacobi corresponds to at the level of symplectic manifolds
 
5:32 PM
that is a representative yes but then the $H_0$ is the equivalence class of the representative
 
What is that equivalence relation?
Think about it explicitly. Without just writing down the definitions.
 
Without words?
Hmm, nobody's answering my point-set query.
 
$\sigma_1 = \sigma_2(Im \partial_1)$
 
@TedShifrin I meant to say without big words, but edited.
 
so if they differ by boundary
 
5:33 PM
Huh? Karim
 
@TedShifrin That went from "Oh, no!" to "Oh." really quickly
 
?
 
I thought you meant a real mouse for a second.
 
LOL, DogAteMy ... Always a pleasure to see you.
 
What does it mean two say two 0-simplices differ by a boundary?
 
5:34 PM
@TedShifrin Yup.
 
@TedShifrin ?
 
@TedShifrin "See"
You too!
 
Ah, here's my savior.
 
@TedShifrin There is a space like that. I think they are called Dowker spaces.
 
Wow. Never hoyd of those.
 
5:35 PM
is that if their difference is inside of the image of $\partial_1$
 
In the mathematical field of general topology, a Dowker space is a topological space that is T4 but not countably paracompact. They are named after Clifford Hugh Dowker. The non-trivial task of providing an example of a Dowker space (and therefore also proving their existence as mathematical objects) helped mathematicians better understand the nature and variety of topological spaces. Topological spaces are sets together with some subsets (designated as "open sets") satisfying certain properties. Topological spaces arose as generalization of the open sets of spaces studied in elementary mathematics...
see the second equivalence
 
@MikeMiller My laptop's nearly out of charge, so I'll ponder on this on my way to the guest house and get back to you afterwards.
 
Wow, @ForeverMozart. Thank you. Very esoteric examples.
 
It's not even my question!
 
yes very strange
 
5:36 PM
I'm just telling you to make the first step at answering...
 
@TedShifrin tell me the answer :D
 
I think we constructed one in my set theoretic topology class
 
Right. I'll think about it.
 
but I'd have to look through my notes
 
I have to present this after tomorrow
so I want to make sure I explain every detail correctly
 
5:36 PM
@Adeek this does not classify as "without big words" or "explicit"
 
I was wondering why Munkres phrased his homotopy extension stuff in terms of $X\times I$ normal. Very cool, @ForeverMozart. Thanks again.
 
Be back in a few minutes.
 
Karim: Balarka is asking you the right thing and you need to answer it. What is the meaning of $H_0$ and the meaning of $f_*\colon H_0(X)\to H_0(Y)$.
 
@Adeek What are you presenting?
 
5:38 PM
Ulam-Borsuk theorem
 
(I seem to have come in in the middle of a conversation or several)
 
it is right at the end of the proof
 
Aha.
Yeah, that was the one where you could construct a map that's odd degree on the equator, right?
 
@AkivaWeinberger You usually do :P
 
But moving the equator down and keeping the function gives a nullhomotopy
@TedShifrin This is very true
 
5:40 PM
no @AkivaWeinberger that is in the proof of the any odd function $f : S^n \rightarrow S^n$ will have odd degree
 
@AkivaWeinberger You should answer this.
 
Karim: Here's a specific question. Consider the 2-fold covering map $f\colon S^n\to \Bbb RP^n$. What is $f_*\colon H_0(S^n,\Bbb Z/2) \to H_0(\Bbb RP^n,\Bbb Z/2)$?
 
so it will be
 
@MikeMiller Something something horned spheres?
 
5:43 PM
$[\gamma] \mapsto [f_{*}(\gamma)]$
 
It's your answer, not mine!
 
Karim: I want the answer in this concrete case, not the abstract definition. What are those groups, for starters?
I mean as concrete groups, not definition.
 
$H_0(X,A) = Ker(\partial_0) / Im(\partial_1)$
 
There's no $A$ here.
 
where A is an abelian group
 
5:45 PM
Oh.
 
I am just answering in general.
 
Do you know what $H_0(X)$ is when $X$ is path connected?
I don't want definitions. I want concrete answers.
 
yes when $H_0(X)$ is path concented then it is just $\mathbb{Z}$
 
And with $\Bbb Z/2$ coefficients ...
 
it will be $Z_2$
 
5:46 PM
OK. Now answer my sphere/projective space question.
 
$f_{*} : \mathbb{Z}_2 \rightarrow \mathbb{Z}_2$
 
And what is that map?
 
it is either multiplication by zero or is an isomorphism as the image is subgroup of $\mathbb{Z}_2$
 
Right. How do you decide?
 
I am not sure why it wouldn't be a zero map ?
why
 
5:50 PM
@Adeek How do you check if any map is zero or not? You check where it sends, say, $1 \in \Bbb Z/2$.
 
What is the image of the generator?
 
So it sends the generator $[\sigma] \mapsto [f\circ\sigma]$
 
How do I geometrically think of a generator?
 
oh and that isn't the zero map
as f isn't a zero map.
but how do we know that $\bar{f} : H_0(RP^n,Z_2) \rightarrow H_0(RP^n,Z_2)$ is an isomorphism then ?
how @TedShifrin
by the same logic as above !
 
What do you mean f isn't a zero map?
 
5:56 PM
nvm
I am just little confused
There is only one element basis element of $H_0(RP^n,\mathbb{Z}_2)$ which is just $[\sigma]$ where $\sigma$ is a cycle.
right ?
 
@Adeek Isn't the generator of that group the homology class of a point? Unless I'm missing something, the fact that $\bar f$ sends a point to a point means that it sends the generator to itself
so it's an isomorphism
 
yeah
yeah it is only one point $[\sigma]$
nvm I dunno why I am confusing myself.
it is just I have bunch of definitions stacked on top of each other some time it can confuse me
It is good to go back and think about what those groups actually represent.
before saying anything
@AkivaWeinberger I have good excerise for you
 

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