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04:14
@BalarkaSen turns out, I don't need to deal with that horrendous diagram at all
Dold-Thom itself is a nightmare to prove but if you don't deal with the isomorphisms being natural (which my book doesn't), the axioms just fall out like that
heloo
So I'll probably be devoting more of the paper to applications of the stuff. And maybe I'll expand the proof of Dold-Thom somewhat
04:33
@Daminark Ah ok sick
Yeah I'd like to know some applications of D-T
Same, I dunno any
Honestly I think it'd be useful in AG if anything
idk maybe
Like, I don't see too much use in AT because it's ridiculously hard to compute symmetric products
If you take a Moore space, its symmetric product will be an Eilenberg-Maclane space
And those are totally fucked
I mean I guess that's a way to postulate the existence of Eilenberg-Maclane spaces?
Well, not quite. But the point is SP^n(X) doesn't seem to have a visually clear CW structure in general
Like, doing that instead of the bar construction or the game of the geometric realization of whatever
Ah hmm
Wel
04:42
@Daminark Yeah
I guess you can take a Moore space
Take its symmetric product
Well you have to construct a Moore space first...
I mean that'll probably be a lot easier
That's not too hard
But yeah so once you take the symmetric product of the Moore space, take the geometric realization of its total singular complex
That'll give you a CW complex with the same homotopy type
04:43
why do you have to bring geometric realization to this
So that'll be an E-M space
symmetric product is a space dude
Well you want it to be a CW complex
And that's the only way I know of to get a CW complex out of a space
Ah OK. Ugh.
I think it's much easier to prove SP(X) is a CW complex if X is
Is that even true?
04:45
Pretty sure. Like, take SP^n(X). All you have to do is to produce an S_n-equivariant CW structure on X x ... x X (n times)
Give a CW complex X
I guess that's a bit of a work
It's not immediately obvious for sure
And then is CW gonna be closed under colimits?
You have to make it compatible with the inclusion I guess
SP^n(X) \subset SP^(n+1)(X)
Well, the inclusion is just stacking on a basepoint so surely that won't be the thing that kills it
If anything, you let $\bigoplus X$ be the set of sequences of elements of $X$ such that all but finitely many points are the basepoint
And quotient that out by the action of $S_{\infty}$
That'll also give you an infinite symmetric product
Wait hold on though, now that I think about it things might be sneakier
04:55
Hm, do you know how SP behaves under wedge? Like, is SP(X * Y) a mess or understandable?
See, $SP$ just gives you a $K(\mathbb{Z},n)$
Yeah you need the G coefficient analog
Wait isn't it true that a quotient of a CW is a CW?
I thought that'd just be a thing
Depends on what you mean by quotient and what you mean by CW. If you quotient a subcomplex of a CW complex, you get a CW complex
But in general if you have a CW structure on X, that need not pass to a quotient $X/ \tilde$
Consider S^2 with {pt, 2-cell} CW structure
consider ~ identifying two points on S^2
that isn't giving you a CW structure on S^2/~ fam
Oh okay
I don't know much about CW complexes so I'm just gonna roll with the business of passing through to simplicial sets and then back
05:00
okie
tbh I don't care much either
btw, if you want to build a Moore space, here
answered by yours truly
I don't understand how do they get the last inequality from the first one, I tried putting u=k*sigma/sqrt(n). I get exp(-k^2/2)
@Akiva If you didn't see earlier conversations, it turns out SP^n(2-manifold) is an n-manifold, in fact
So in particular SP^n(torus) too
I don't know what manifold it is though
OPEN QUESTIOn
I will think about it later. Gotta run now bye
Maybe if you have a (finite, or somehow nice) group acting by cellular maps on a CW complex, the quotient is a CW complex? dunno
Sick
And see you @Balarka!
05:15
@Balarka turns out I'm dumb
This totally gets you $K(G,n)$
You just ask that the integral homology groups are whatever, then pass to the symmetric product
Okay so I guess at this point the question is, do we only care about $K(G,n)$ where $G$ is finitely generated?
Do we? From a Brown representability point of view, we certainly care about cohomology in non-finitely generated coeffcients like $\Bbb Q$ or $\Bbb R$
True
Okay so Moore spaces may not be easy to construct in general
05:34
I think Moore spaces should be not that hard to contruct
you only need $n$ cells and $n+1$ cells
this is in contrast to Eilenberg-Maclane spaces which generally have cells in infinitely many dimensions
@BalarkaSen This is very clear to me now.
05:57
Hey @Alessandro!
How's it going?
I have a group theory and a Galois theory exam next week so I'm busy studying for those but quite good overall, what about you?
Doing topology for now, and after that I'll figure something out.
06:12
What kind of topology?
Right now, it's stuff around the Dold-Thom theorem
06:26
@Balarka so it turns out computing $SP(S^1)$ isn't looking too hot. For that reason I'll probably just black box the fact that $\mathbb{CP}^{\infty}$ is a $K(\mathbb{Z},2)$ and use that to prove dimension
 
2 hours later…
07:57
Morning everyone!
08:22
@MatheinBoulomenos Yes, that's what it means for the cell structure to be G-equivariant
where G acts on the thing
@Daminark SP(S^2) is easier than SP(S^1) tbh
I think SP^2(S^1) is the Moebius strip
I guess SP^n(S^1) is a nontrivial disk bundle on S^1
08:43
Yeah it's strange
I think I have a semi-slick way to think about this stuff but I need to draw some pictures to convince myself
 
6 hours later…
14:19
hey there.
I'm looking after an 'open' - free - real analysis book that I can use for self-learn. I was using wikibooks (I liked the fact that they constructed the naturals, integers and rationals first), but there's no offline version. Recommendations?
@TedShifrin I'm evading linear algebra. :P
I'm not really acquainted with the reals. Not anymore.
@LucasHenrique
but, I think it is illegal :P
you can download.
thanks @ManeeshNarayanan. It looks legal, since it has DCMA and stuff
It is ok.
I just referred two theorems fro this book b-ok.org/book/2513386/78d317. It seems simple.
14:43
Thanks. :)
14:56
If I say that $M$ is an $A$-module generated by the set $\lbrace 1, \alpha, \dots, \alpha^{n-1}\rbrace$ does that mean $M = A + A\alpha + \dots + A\alpha^{n-1}$?
15:21
yes
@LucasHenrique I found the textbook understanding analysis on library genesis.libgen.pw/download/book/5a1f047f3a044650f5fdbcf8
 
1 hour later…
16:25
Suppose the function $f(t)$ has derivative $L$ at $t_0$, is it true that $\mid f(t_0+k)-f(t_0)\mid<Lk$ for $f$ is continuous differentiable? How can I prove this? What happens if f is only differentiable?
16:52
So it appears somehow rational solutions to diophataine equations has something to do with light propagation in manifolds
17:26
Goood afternoon folks
Letssss gettt riiiight into the newssss
Thats a Youtube thing
Maybe PhillyD dunno
but I recognize a youtube thing when I see one
Keemstar actually
Oh boy. I don't know anything about that guy. But I think I hate him.
Never watched any of his videos though
Everyone hates him lol
he's a gnome
17:28
@BalarkaSen So, I know that the study of zeroes of vector fields on manifolds is a big deal. What about the study of zeros of $\mathbb{RP}$ fields on manifolds?
@KevinDriscoll What's an RP field?
Associate to each point an element in $\mathbb{R}P$ such that from point to point the variation is smooth
How the fuck do we notate this? Just $RP$? $\mathbb{RP}$? $\mathbb{R}P$?
$\Bbb{RP}^n$, you mean?
$\mathbb{R}\mathscr{P}$
So you'd be look at $\Bbb{RP}^n$-bundles on manifolds
The "fields" of interest would be sections of this bundle
17:32
Sure, although in this case we physically only care about $\mathbb{RP}^1$ and $\mathbb{RP}^2$ I think
Circle bundles are important. RP^1 = S^1
$S^1$ mod antipodal points, right?
The definition you gave is just a smooth function to $\Bbb RP$
Yeah but it's homeomorphic to S^1
@MatheinBoulomenos Depends on what he means by "associating to each point an element of $\Bbb{RP}^n$".
If he's fixing the copy of the projective plane canonically, then yes.
From context, I presume he isn't
(i.e., the bundle is not trivial)
If 2 point on the sphere have the same element of $\mathbb{RP}$ associated ot them, I don't want to think about them as going to the same point
I want to keep the fact that they're indexed byt he point on the sphere they're associated to
17:37
Not entirely sure what you're going for there
I mean I want the same thing that a vector field is.
But with $\mathbb{RP}$ instead of $\mathbb{R}$
I mean you could like take the tangent bundle, and fiberwise projectivize it
That gives you "the projectivized tangent bundle" $\Bbb P TM$
not every projective bundle arises like this (at least in alg geo, not sure about the diff geo setting)
Of course
But I don't understand what Kevin wants here
A vector field is a tangent field
Sorry I'm not being clear
17:40
If you want the tangential information to survive, better projectivize the tangent field
in alg geo there are obstructions in $H^2(X,\mathcal O ^\times)$ which describe if projective bundles can be lifted to vector bundles
I seeeeeeee. Whne I say vector field, I don't necessarily mean a section of the tangent bundle
Sorry, i forgot the difference in terminology
then what do you mean by vector field?
For me a vector field includes tangent vectors but also vector that point into the ambient space
Oh, so you're embedded somewhere?
17:44
Ya, the specific case that should be relevant for nematic liquid crystals should be embedding $S^2$ or $T^2$ in $\mathbb{R}^3$
@MatheinBoulomenos So by projective bundles you mean fiber bundles with projective spaces as fibers?
You're basically asking if a map $M \to B\text{Diffeo}(\Bbb{RP}^n)$ can be homotoped inside $B\text{PGL}(n)$, in that case
Can you write what BDiffeo(RPn)and BPGL(n) mean in english?
Ah, so these are rather complicated objects known as "classifying spaces". If you have a fiber bundle on a manifold with the transition functions taking values in a group $G$, there is a homotopy class of a map $M \to BG$ that classifies that fiber bundle.
So "$BG$" is more or less that object
I think I brought it up momentarily during our conversations about the clutching construction?
Ok, yea you probably did. I'll file it under 'things to think about after exams are done'
Good file
17:56
Thanks, and sorry again for my confusing language
Hey everyone!
Suppose $M$ is a manifold of dimension $n \geq 1$ and $B \subseteq M$ is a regular co-ordinate ball. Show that $M \setminus B$ is a $n$-manifold with boundary, whose boundary is homeomorphic to $\mathbb{S}^{n-1}$
I'm trying to prove the above, I've already proved that $\text{Bd}(B)$ is homeomorphic to $\mathbb{S}^{n-1}$
And that $M \setminus \overline{B}$ is a $n$-manifold
Well then you're done!
Oh wait sorry I misread
@KevinDriscoll How so? Ohhh by $\text{Bd}$ I mean topological boundary, not manifold boundary
18:07
The boundary of $M-B$ is the same (except for orientation issues) as the boundary of $\bar B$.
You need to check the charts near $\partial B$
$\partial\bar B$.
Thanks, yes
Keep topological boundary out of this.
Also, good morning @Ted
18:10
@Kevin: Re your discussion with Balarka, this is a fancy generalization of the fact that if you have a compact manifold $M$ and a vector bundle $E$ of rank $k$ on $M$, there is a (continuous, smooth, whatever) map $M\to G(k,N)$ (for some large $N$) so that $E$ is the pullback of the tautological rank $k$ bundle $\tilde E\to G(k,N)$.
@TedShifrin Yep I was just about to say that, I picked $x \in \text{Bd}(B)$, and I tried to construct a neighbourhood $W$ of $x$ in $M \setminus B$ homeomorphic to an openset of $\mathbb{H}^n$
Think of picking a chart on $M$ at a point of $S^{n-1}$ that flattens out the submanifold $S^{n-1}$, Perturbative.
You need to use standard facts about submanifolds (e.g., from Chapter 1 of G&P).
@TedShifrin Wait I'm not working out of G&P though, the manifolds I'm working with are purely topological (not smooth)
Well, this is not the most expedient use of your time if you're trying to get to Riemannian stuff.
You need to know that $S^{n-1}$ is tamely embedded as a topological manifold. Then the same is true.
If you're actually trying to get to Riemannian geometry, don't be worrying about topological manifold stuff.
@TedShifrin Ohh I dropped the Riemannian stuff for now, sorry if I didn't tell you
18:13
rolls 6 1/2 eyes
/roll 1d6.5
My own personal taste is to concentrate on smooth stuff.
Kevin, you saw my Grassmannian comment?
I told him the Grassmannian thing before
No I don't think I did
my grammar suck
18:15
Oh jsut now, ya
Thats what G() means, Grassmannian. I was oging to ask.
Didn't John do Grassmannians?
Not in G&P, but definitely belongs in a grad manifolds course.
We had 1 homework on Grassmannians
Hrumph.
It was to show that using the inner product on $\mathbb{R}^n$ $G(k,n) \cong G(n-k,n)$
@TedShifrin At the moment I'm just trying to complete Lee's book on Topological Manifolds (because there's also a decent intro to Algebraic Topology there), and there are a bunch of cool exercises and problems I haven't seen anywhere else. And I'm also trying to complete G&P and Milnor
18:17
Well, he must have introduced Grassmannians in class, then.
Let me check my notes
@Perturbative: Whatever.
Not the place to learn algebraic topology, I imagine.
I don't know the book, personally.
No, no mention of Grassmannian in the notes from that region of the course. There's a line 3 page discussion in Lee though where he talks about how you construct the charts and stuff that we used though
Interesting how different mathematicians have such different tastes for a first course. :)
Because of my algebraic geometry slant, I think Grassmannians and general vector bundles are important.
And I also talk about them as examples of homogeneous spaces/symmetric spaces.
Real/complex projective spaces are special cases, of course.
I <3 Grassmannians
18:21
That said, I never mentioned them at all in the undergraduate course, because they don't "obviously" get defined as subsets of some $\Bbb R^n$.
My thesis was pages and pages of flag manifolds defined in terms of Grassmannians, with corresponding "pointed" Gauss mappings.
@TedShifrin Did you teach a general grad manifolds course?
Yes, Kevin. Numerous times. But I included a bit of diff geo in it, to motivate things.
That was one final I sent you.
Sometimes flows and Frobenius were in the second quarter/semester, though. Can't fit everything into one course.
Oh okay, I thought that was a mostly diff geo final. I'm saving that one for tomorrow, final test so-tos-peak
Or maybe later tonight
Do you mind if I look at your syllabus? I'm just curious to compare.
Well, it had curves and surfaces using moving frames (I wanted to actually use the differential forms for something geometric).
I never had an official syllabus. I just listed topics in the first lecture.
sup dudes
18:25
Hmm, I don't think I did, anyhow. I'll look.
sdown, Eric.
@EricSilva sup, breh
Nope, Kevin. The courses were too small for me to bother with official syllabi.
I discussed stuff in first class.
Cool, thanks for checking
holy jesus griffiths and harris is expensive
18:34
If i ahd to summarize the topics for our course I would say we covered:basics (smooth manifold definitions, derivations/tangent spaces, smooth functions, etc.), full-rank maps, submanifolds, whitney embedding, bundles and flows, homotopy (including approximations and stable properties), transversality (mod2 intersection and degree), tensors (exterior algebra, k-forms), orientations, integration, Lie derivatives, De Rham cohomology, M-V, and cup-product/Poincare duality
Most books are expensive. I'm ashamed by how much the publishers charge for my books. That is the main reason I left my diff geo notes as free .pdf.
amazon seems to be selling the new paperbacks at like 135 dollars
i aint got that kinda scratch lying around
Kevin: Because we had a separate undergraduate G&P course, I totally skipped all the G&P transversality stuff and did things like flows and Frobenius (which are super important).
I also left Poincaré duality discussion for the algebraic topology course, although at some point late in the course (doing characteristic classes) I would talk about it.
My two-semester outline was something like: basics on manifolds, tangent spaces, mappings; vector bundles and tensors; differential forms and Stokes's Thm.; curves and surfaces (including minimal surfaces); flows, flowbox thm, Lie derivative, Frobenius; Riemannian connections, curvature, geodesics, exponential, Cartan-Hadamard Thm.; connections on vector bundles, Grassmannians, symmetric spaces, Gauss-Bonnet and characteristic classes.
One time I did Bochner techniques and some complex geometry (up to Kodaira embedding theorem).
hi Tobias
Hi chat
@TedShifrin Apparently I have vastly overestimated what the students are expected to be able to do for the exam. Fortunately, I had someone more experienced look over it before I finalized it
on the phone ... back shortly
yeah ... able to reserve the right to reschedule plane tickets
@Tobias: It takes many years of experience to gauge these things. I also make it a practice to write out complete solutions to the exam and I should be able to do so in approximately 1/3 the allotted time.
18:49
@TedShifrin In this case I could probably do it in 1/4 if not less
complete solutions (like for students to understand?)
Yeah, perhaps for this level course, 1/3 is too generous.
Well, I won't offer to give my opinion since you already have plenty :P
I have gotten a lot of practice doing that this term, as I did so for one old exam problem per week for them to practice on (with the solution for after they had spent some time solving it themselves)
Of course, we know exactly how to approach problems, and they often struggle with that. In lower level courses, that part isn't so nebulous.
18:50
@TedShifrin Yeah, good thing I have coworkers to ask this sort of thing (in this case a guy who both lectured the course last year and also happens to be undergraduate coordinator, so he knows what is expected)
excellent
Quick question if $M$ is a topological space, and we show that $\text{Int}(M)$ is a $n$-manifold without boundary and $\text{Bd}(M)$ is a $n-1$-manifold without boundary can we conclude that $M$ is a $n$-dimensional topological manifold with boundary?
You need to know something about how nicely the boundary is embedded, don't you?
Suppose that $x_n$ a convergent sequence of real numbers. I am trying to show that $\sup \{x_k \mid k \ge n \} - x_n$ converges to zero. I could use a hint.
@Perturbative But $Int(M)$ is just $M$ unless you consider $M$ as a subset of some larger space
18:56
@user193319: Where are you stuck?
@TobiasKildetoft Ahh you're right
heya @Antonios
Nearly prepped for the finals... the struggle continues.
@TedShifrin I haven't really gotten anywhere with the problem. I said let $\epsilon > 0$. Then there exists a natural number $N_\epsilon$ such that $\sup \{a_k \mid k \ge n \} - \epsilon < a_{N_\epsilon}$, but this doesn't seem very helpful.
...where $N_\epsilon \ge n$.
18:59
@user193319: You need to write down and use the fact that the sequence converges!
Okay. I'll give it another go. Thanks.
Otherwise this is definitely false.

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