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Jakobian
Jakobian
12:01 AM
I'm glad we are not discussing philosophy anymore because it made me feel like a pseudo-intellectual
(I also don't know anything about philosophy)
 
Ryder Rude
Ryder Rude
i too hav never studied philosophy (i tried but they use too many big words). i just like to question basic notions about life to organise my knowledge
 
Jakobian
Jakobian
But you do agree, that there is some sense in which we can question existence of ourselves?
So it seems like, again, this boils down to correctly interpreting the question
 
Ryder Rude
Ryder Rude
at my level of philosophical precision, i just cant question my own existence. but the discussion above made me question my previous stance a bit
because i ended up concluding that i cant be sure that anything is doing the questioning, as that itself is an assumption
i dont think philosophy is all bad. see this for a refreshing stance on philosophy
 
Jakobian
Jakobian
12:19 AM
I don't think its bad just how I don't think math is good
we are biased in the sense that there are clear ways (well, not always but you probably get what I mean) in which one can check our work
"good" is also a relative statement
 
psie
psie
@Jakobian ok, yeah, "indexed by" means a bijection I would say, i.e. instead of $s_{\alpha}$ we write $f(a)\in S$, where $a\in A$ (though this is my interpretation, but I guess the only sensible one).
Anyway, I have a basic question: say we have $S$ in bijection with an ordinal $A$. Can there be more ordinals other than $A$ in bijection with $S$? Alessandro said that the set of ordinals in bijection with $S$ is non-empty, so I'm wondering if that set simply contains $A$ or if it can contain more ordinals?
 
Xander Henderson
Xander Henderson
@psie What is your definition of an "ordinal", and how can you tell if two ordinals are distinct?
 
Jakobian
Jakobian
@psie yes there can
If $\alpha$ is an infinite ordinal, then $|\alpha| = |\alpha+1|$ but $\alpha\neq \alpha+1$
If there is just one ordinal in bijection with $S$, then $S$ is necessarily a finite set
The existence of such ordinal comes from axiom of choice
If we refute the axiom of choice, then there exists a set which cannot be well-ordered, that is a set $S$ such that there is no indexing of elements of $S$ using an ordinal
$\alpha+1 := \alpha \cup \{\alpha\}$ here, of course
The smallest ordinal $\alpha$ such that $|S| = |\alpha|$ would be called by some a cardinal number of the set $A$
That is, cardinal numbers are ordinals $\alpha$ such that if $\beta$ is an ordinal and $|\alpha| = |\beta|$, then $\alpha\leq \beta$
(I'm explaining the way in which I learnt those things and this is what a cardinal number is for me)
 
psie
psie
12:36 AM
@XanderHenderson the definition is...cumbersome, but something like this: a well-ordered set $a$ is an ordinal if for all elements $x \in a$, $x$ equals the initial segment of $a$ with respect to $x$ (initial segment of $a$ with respect to $x$ means the subset of $a$ whose elements are all "smaller" than $x$)
@Jakobian ah ok, this makes sense, I grok it I think
 
Xander Henderson
Xander Henderson
@psie The more important question was "how do you tell them apart?"
I was trying to get you to think about the problem, rather than just hand you an answer.
 
psie
psie
I see, I'll have to think
 
Jakobian
Jakobian
@XanderHenderson Xander encouraging future set theorists :P
 
robjohn
robjohn
@XanderHenderson I think he wants Mars to pay for it and keep their aliens out of here.
 
 
2 hours later…
Xander Henderson
Xander Henderson
2:36 AM
Great use of your time, legislators.
 
Jakobian
Jakobian
you either really dislike Pluto, or, tamales?
 
Xander Henderson
Xander Henderson
Pluto isn't a planet.
But also, there are other things congress should be doing.
 
 
2 hours later…
one potato two potato
one potato two potato
4:23 AM
I just read your answer post. Since I'm not very familiar with the classifying spaces, there are several questions I want to ask:
1. Why $Homeo^+(S^1)\cong Homeo^+_e(S^1)\times S^1$? 2. Why $[B,\Bbb CP^\infty] = H^2(B)$?
 
Dannyu NDos
Dannyu NDos
4:38 AM
My attempt to understand intuitionistic logic: "Impossibility of rebuttal doesn't count as a proof."
 
Finn Bolton
Finn Bolton
4:48 AM
Why was my answer deleted?
 
one potato two potato
one potato two potato
@Thorgott It seems that the usual argument to classify Seifert Fibered space is using two things: Euler number and Euler characteristic. But if 2-orbifold admits CW decomposition (I don't know if this is true), then according to your argument, SFS also can be classified using Euler class only. I'm not sure if two methods are equivalent, do you know? Does Euler class know euler characteristic?
 
Finn Bolton
Finn Bolton
How exactly does it not answer the question?
 
one potato two potato
one potato two potato
5:19 AM
Ah stupid. I need to know a type of base 2-orbifold so knowing Euler characteristic should be fulfilled. If I know Euler class then I know Euler number so two arguments are equivalent.
But independently, I wonder if knowing Euler class means knowing Euler characteristic. Based on my quick google search, it seems they're related by Chern classes but I don't know any of them.
 
one potato two potato
one potato two potato
6:01 AM
(wait, does it make sense?)
 
one potato two potato
one potato two potato
6:18 AM
Ah Euler class of the tangent bundle is Euler characteristic
 
 
2 hours later…
Thomas Finley
Thomas Finley
8:07 AM
A curious question: In almost every standard real analysis book, I found that after the chapter differentiation the next chapter that follows is of Riemann integrals,
followed by improper integrals and then series and sequence of functions, and so on. But nowhere did I find a chapter on indefinite integrals. Riemann integrals only dealt with definite integrals, but what is the thing about indefinite integrals that are usually taught in a calculus course but are not mentioned in any real analysis books, I have seen so far?
 
 
1 hour later…
nickbros123
nickbros123
9:18 AM
@ThomasFinley indefinite integrals are just reapplication of FTC really
 
 
2 hours later…
one potato two potato
one potato two potato
11:21 AM
oh wait, orbifold is topologically just a manifold so of course it has a CW decomposition
But still, SFS cannot be treated as the usual $S^1$-bundle because the usual local trivialization of fiber bundle does not make sense on a nbd of a singular point
 
Ryder Rude
Ryder Rude
anyone want to play this? lichess.org/OlfTXRpJ
 
 
1 hour later…
Thorgott
Thorgott
12:44 PM
1. $f\mapsto (f(e)^{-1}\cdot f,f(e))$
2. cause $\mathbb{CP}^{\infty}$ is a $K(\mathbb{Z},2)$
@onepotatotwopotato I don't know anything about orbifolds
for compact oriented manifolds, the Euler class of the tangent bundle determine the Euler characteristic by evaluating on the fundamental class
 
X4J
X4J
1:25 PM
Suppose $f, g$ are orthogonal linear operators above $V$ and $\mathbb{F} = \mathbb{R}$. I am trying to understand if it is always the case that $f + g$ is not orthogonal
 
Thorgott
Thorgott
what if $f,g$ are multiples of another
 
X4J
X4J
Then you can prove $f + g$ is not orthogonal
intuitively I would say that if there exists a vector $v$ such that $\langle f(v), g(v) \rangle$ = 0 it will imply that $f + g$ is not orthogonal
 
one potato two potato
one potato two potato
1:41 PM
@Thorgott Ah, I see. Thanks
both questions have nothing to do with classifying space, just my lack of knowledge.
@Thorgott Yes. I think the question does not make sense. I was thinking like for given Euler class of some fiber bundle, whether I can determine the Euler char of the base space.
 
Jakobian
Jakobian
@X4J like $v=0$?
 
Jakobian
Jakobian
2:08 PM
If $f, g, f+g$ are orthogonal then $1 = \|(f+g)(v)\|^2 = 2+2f(v)\cdot g(v)$ when $\|v\| = 1$ so $f(v)\cdot g(v) = -1/2$
 
X4J
X4J
@Jakobian Forgot to mention $ ||v|| > 0 $
 
Jakobian
Jakobian
I'm thinking this might never be possible
 
X4J
X4J
I was asked to find an example for $f,g$ orthogonal and $f+g$ that is not orthogonal but then I thought it might always be the case
 
Jakobian
Jakobian
This is possible, never mind
 
X4J
X4J
2:40 PM
Yeah it is possible
 
Jakobian
Jakobian
3:20 PM
@Thorgott This is a picture depicting prime ideals of $C(X)$ corresponding to various points $p\in \beta X$
As you see, $X = \mathbb{N}$ is a boring example, $X = \Sigma$ is slightly interesting, while $X = \omega+1$ is also very interesting
In the picture for $X = \omega+1$, the maximal chains of those prime ideals are only intersecting at $M_\omega$ and all of them correspond to a minimal prime ideal corresponding to an element of $\beta\mathbb{N}\setminus\mathbb{N}$ (i.e. free ultrafilter)
 
 
2 hours later…
b_jonas
b_jonas
5:02 PM
I spent way too much time trying to find this formula in Concrete Mathematics because it's in chapter 7.4 rather than in chapter 5 where I expected it would be. $ \sum_{0\le n} n^m z^n = \sum_k \genfrac\{\}{0pt}{}{m}{k} \frac{k! z^k}{(1-z)^{k+1}} $
 
nickbros123
nickbros123
" In the xy-plane, what is the length of the shortest path from
$(0, 0)$ to $(12, 16)$ that does not go inside the circle $(x - 6)^2 + (y - 8)^2 = 25$? " this was a problem in some midde school textbook, but doesnt this require variational calculus or something?
 
Xander Henderson
Xander Henderson
@nickbros123 To completely justify an answer, yes. But there is an "obvious" solution which really only requires some algebra and trig.
It still seems a bit beyond what a middle schooler would be expected to do, but you don't need super advanced tools.
Given points $A$ and $B$, and some circle which intersects the straight line path from $A$ to $B$, travel along a line through $A$ that is tangent to the circle, then travel along the circle to the line that goes through $B$ and is tangent to the circle. Take that path to $B$.
E.g. if you never hit the circle, you can find a shorter path which does intersect the circle, and if you don't intercept the circle along tangent lines, the triangle inequality can be invoked to show that the path along tangent lines is shorter.
I think that this can be made completely rigorous with only pretty elementary tools.
 
nickbros123
nickbros123
yeah, thats intuitive
i wonder if, with the math tools i currently have, I go back to coordinate geometry and its results, maybe i can show everything i took for granted
 
Koro
Koro
5:36 PM
uppose that M is simply connected, compact manifold.
Then I want to show that every closed 1 form is exact on M.
Question: is it possible to answer this without using universal coefficients theorem?
Since M is simply connected, its fundamental group pi_1 is 0.
First homology group H_1 is abeliazation of pi_1 so H_1 =0
To conclude H^1=0 I would need universal coefficients theorem.
:(
 
Koro
Koro
5:57 PM
Suppose that w is a closed 1 form on M. Consider p in M. Take a chart at p and call it h. Then (inv h)^\ast w is a 1 form on R^n hence equal to df for some Coo f.
pulling it back to M and using that fact that d anf \ast commute, I have \ast w= d(h^\ast f).
does this make sense?
 
Koro
Koro
6:16 PM
Suppose that $M$ is simply connected, compact manifold.
Then I want to show that every closed $1-$ form is exact on $M$.

This is same showing that the first de-Rham cohomology space of $M$ is $0$.

**Question**: Is it possible to answer this without using universal coefficients theorem?

(The reason for not using the universal coefficients theorem is that it contains a mystery term ext which I'm not comfortable using. Digging deeper into ext, it turns out that it is a core algebraic term which I'm not good enough at at this stage.)
 
Jakobian
Jakobian
If no one is interested then I won't share my stuff. shrug
 
Koro
Koro
except that you can never be sure about the antecedent.
 
Xander Henderson
Xander Henderson
6:35 PM
@Koro No! I refuse to uppose that!
 
Koro
Koro
Oh, I missed s there.
 
Thorgott
Thorgott
7:11 PM
@Koro I think you can prove this using the same idea as one uses to prove the FTC or the exactness of closed forms on convex domains. Namely, explicitly define such a $g$ via integration in an appropriate sense.
@Jakobian I simply don't have anything to say about it. Not trying to be rude, it's just not my subject.
 
Koro
Koro
@Thorgott I did that but go only a 'local' form.
Finding such global g seems to require some other concepts.
 
Thorgott
Thorgott
I'm not saying to use the theorem, I'm saying to imitate the proof globally
 
Koro
Koro
I can't use g(x)=integration of omega on a curve from some p to x.
 
Thorgott
Thorgott
the "patching local solutions together" approach will yield an obstruction in Cech cohomology and to show that vanishes you have to compare Cech cohomology to sheaf cohomology to singular cohomology to singular homology to get to the result
@Koro why?
 
Koro
Koro
because I know integrating form on manifold. But I don't know integrating form on a curve.
:(
 
Thorgott
Thorgott
7:16 PM
of course you do, pull back the form along the curve to get a form on $[0,1]$. you know how to integrate that.
 
Koro
Koro
$\int_{\gamma} \omega=?$
$\gamma:[0,1]\to M$ is a curve.
smooth^
 
Thorgott
Thorgott
yup
 
Koro
Koro
is it $=\int_{[0,1]}\gamma^\ast \omega$?
 
Thorgott
Thorgott
yes, over $[0,1]$
 
Jakobian
Jakobian
@Thorgott I understand those type of things. But still
 
Koro
Koro
7:25 PM
Suppose \gamma(0)= \gamma(1), then I should get 0 because of simple conncetedness.
But how to see this from the above formula?
 
Thorgott
Thorgott
use Stokes' theorem
 
Koro
Koro
@Koro $=\int_{[0,1]} d f =\int_{\{0,1\}} f= ?$
I wrote \gamma * \omega = df as LHS is a 1 form on [0,1].
 
Jakobian
Jakobian
'But still' as in, its no ones subject here really, so its pointless to share my enthusiasm with others
 
Koro
Koro
I know on R, closed form and exact form are the same.
(top forms)
 
Jakobian
Jakobian
and I came to the same conclusion with most things in mathematics
 
Koro
Koro
7:32 PM
I feel so too at my college.
so I don't ask them anymore.
I either ask online or wait till time helps me learn it on my own.
 
Xander Henderson
Xander Henderson
@Jakobian You can share your enthusiasm, but you have to recognize that mathematics becomes very specialized, and if there are 20 people in the world who understand what you are doing, you are reaching a large audience.
I know of about 15-20 people who are actively studying properties of the Assouad dimension (most of them in Japan and Norway; one person in Scotland, one in England, and my masters advisor); and another 15-20 people working on complex dimensions (my phd advisor, his students, and a couple of others he's picked up on the way).
The intersection of those groups consists of exactly two people, and I think that Sean has maybe left academia.
 
Jakobian
Jakobian
I mentioned a law of math.stackexchange before, that the only people answering my questions are me and Eric Wofsey. It seems like this is still true on mathoverflow, its just not Eric Wofsey anymore but K. P. Hart
Which is kinda funny
 
Koro
Koro
Is it true that continuous image of a topological manifold also a manifold?
what about smooth image of a smooth manifold?
 
Jakobian
Jakobian
Of course not
 
Xander Henderson
Xander Henderson
@Koro I feel like the answer should be "no".
Consider one of those fancy flowers in polar coordinates.
 
Thorgott
Thorgott
7:42 PM
to be fair, Eric Wofsey answers every question
 
Xander Henderson
Xander Henderson
They are the continuous image of an interval, but are self-intersecting, thus not manifolds.
 
Thorgott
Thorgott
what Xander said, and that's still a nice image
 
Xander Henderson
Xander Henderson
Did I do a topology right?!
 
Thorgott
Thorgott
continuous images can get really horrible
@XanderHenderson yes!
 
Koro
Koro
I suppose the answer is no even for smooth images?
 
Xander Henderson
Xander Henderson
7:44 PM
Oh, the von Koch curve, too.
@Koro Dem flowers is pretty smoove.
 
Koro
Koro
Then why is image of \gamma a 1 dimensional manifold?
 
Jakobian
Jakobian
Every Peano continuum is an image of an interval
but Peano continua can get not what would a normal person call a nice space
 
Koro
Koro
@tychonovs-scholar: So if $\gamma:[0,1]\to M$ is a smooth map, then image ($\gamma)$ is a $1-$ manifold? — Koro 6 mins ago
 
Thorgott
Thorgott
@Koro who said it is?
 
Jakobian
Jakobian
what if you collapse interval onto itself
 
Thorgott
Thorgott
7:46 PM
what Xander described can easily be realized as a smooth curve
 
Jakobian
Jakobian
like with that picture you posted once
then the image is not a manifold
the map will be an immersion though
 
Koro
Koro
@Thorgott then why does the integration of \omega make sense on \gamma?
 
Thorgott
Thorgott
another instructive example is a shar map like $(x,y)\mapsto(xy,y)$ on $\mathbb{R}^2$
@Koro cause it makes sense to integrate forms on $[0,1]$
 
Xander Henderson
Xander Henderson
@Koro $1$-manifold, not $1-$manifold. The horizontal line there is a hyphen, not a subtraction. :(
 
Thorgott
Thorgott
$\int_{\gamma}\omega$ is defined to be $\int_{[0,1]}\gamma^{\ast}\omega$
 
Koro
Koro
7:48 PM
@XanderHenderson let's take this matter to the UN.
 
Jakobian
Jakobian
look at what mathematicians make us subtract!
 
Xander Henderson
Xander Henderson
@Koro Or you could just be better.
 
Koro
Koro
my chatjax is not installed so both looked same to me :(
@Thorgott Can you please direct me a source where it's defined like this?
Tu doesn't do it in his book.
 
Thorgott
Thorgott
why would you need this
 
Koro
Koro
He does say something about parametrisation though.
for my question on 1 forms.
 
Derivative
Derivative
8:11 PM
What are the advantages of homology compared to cohomology?
 
Koro
Koro
easier to compute
But I'm no expert at this.
 
Xander Henderson
Xander Henderson
@Derivative Fewer letters.
 
Thorgott
Thorgott
more directly related to homotopy groups
tends to be better behaved in the absence of finiteness hypotheses when it comes to various algebraic constructions like the Künneth LES/SS
 
Derivative
Derivative
hmmm I admit I'm not convinced
 
Thorgott
Thorgott
well, there's not a lot of advantages, cohomology is preferred most of the time
 
Derivative
Derivative
8:28 PM
so I'm thinking about this because last semester I learned about deRham cohomology from a class that followed the Tu/Lee approach and I liked it somewhat. Next semester I'm going to try to place out of an algebraic topology class so I'm reading Hatcher. I like Hatcher, but it feels like he's spending a lot of time and effort to build a weaker and less general theory of homology before doing cohomology when the way backwards is smoother
 
Thorgott
Thorgott
There is an important application of what I suggest above, though, namely the improved Whitehead theorem, which states that a homology equivalence of simply connected (more generally, simple or even nilpotent) CW-complexes is a homotopy-equivalence. This does not hold true for cohomology equivalences without any finiteness hypotheses.
@Derivative smoother how?
you have to define singular chains before you can define singular cochains, it makes no sense the other way round
and a lot of the basic properties are proved more easily for singular homology and then simply dualized (as Hatcher does)
plus, pedagogically, singular homology is a lot more intuitive/accessible than singular cohomology
 
Derivative
Derivative
it feels like you're just doing analysis (I'm good at analysis, or at least better than I am at topology) and then you somehow get the deRham cohomology ring for free
Hatcher feels like you're grinding away in the topology mines for hundreds of pages to get something which is just marginally better
 
Thorgott
Thorgott
singular homology is already much better than de Rham cohomology in some (but not all) ways, singular cohomology is better than de Rham cohomology in every way
de Rham cohomology is not all that useful from the perspective of algebraic topology, its a tool for smooth theory and computations
they're things made for different reasons, is what I'm trying to say
 
psie
psie
I'm reading about the construction of the Borel sigma-algebra on Wikipedia. They provide a short proof by transfinite induction that it is obtained by iterating a certain operation up to the first uncountable ordinal. If someone has the energy to look at this, I have some questions.
> To prove this claim, any open set in a metric space is the union of an increasing sequence of closed sets. In particular, complementation of sets maps $G^m$ into itself for any limit ordinal $m$; moreover, if $m$ is an uncountable limit ordinal, $G^m$ is closed under countable unions.
Two claims are being made here I think. One; what is meant by and why is it that complementation of sets maps $G^m$ into itself for any limit ordinal? Two; why is $G^m$ closed under countable unions?
 
Thorgott
Thorgott
8:48 PM
for 2., if $m$ is an uncountable ordinal and you have a countable collection of sets in $G^m=\bigcup_{i<m}G^i$, then there is an ordinal $j<m$ s.t. the countable collection is contained $G^j$, hence its union is contained $G^{j+1}\subseteq G^m$
 
psie
psie
9:19 PM
why does there exist a $j<m$ such that the countable collection is contained in $G^j$? The picture I have in my head: couldn't the countable collection of sets be scattered in a countable collection of subsets of $G^m$?
maybe my last question doesn't make sense, since the countable collection of sets in $G^m$ are already (sub)sets in $G^m$
 
Jakobian
Jakobian
@psie 1. $A\in G^m \implies A^c\in G^m$
 
psie
psie
why does this implication hold?
 
Jakobian
Jakobian
you have to prove by transfinite induction that $A\in G^i \implies A^c\in G^{i+1}$
 
psie
psie
9:39 PM
ok, this is trickier than I thought :(
 
Thorgott
Thorgott
9:50 PM
@psie sure, but the the $G^i$ are totally ordered
and the point is that an upper bound for those ordinals appearing is still $<m$
because $m$ is an uncountable limit ordinal
 
psie
psie
makes sense 👍
 
Jakobian
Jakobian
10:03 PM
@Thorgott this is not true by the way
let $m = \omega_1+\omega_0$ then $m$ has countable cofinality
I mean it is true in the sense that $G^\alpha = G^{\omega_1}$ for $\alpha \geq \omega_1$, but its not the way to take an arbitrary uncountable limit ordinal
 
Thorgott
Thorgott
oh, you caught me, yeah
I did $m=\omega_1$ in my head
 
Xander Henderson
Xander Henderson
@Thorgott You just wrote it upside down. And left off a subscript.
 
Jakobian
Jakobian
@psie yea its though
 
psie
psie
10:52 PM
ok, so Thorgott's argument above is correct even if $m$ does not equal $\omega_1$?
 
Jakobian
Jakobian
@psie no
you do this argument for $m = \omega_1$ and case of $\alpha \geq \omega_1$ separately
 
psie
psie
ok 👍
 
Jakobian
Jakobian
$\text{cf}(\omega_1) = \omega_1$, which means that there is no countable set $A\subseteq \omega_1$ such that for all $\alpha < \omega_1$ there exists $a\in A$ with $\alpha \leq a$
In particular, if you take any countable (non-empty) subset $A\subseteq \omega_1$, then $\sup A < \omega_1$
this is what you need but not all uncountable ordinals have uncountable cofinality
so you need a separate argument for why $G^m$ stabilizes when $m$ is uncountable, in the sense that its equal for any two uncountable ordinals
it would be nice if you picked just one thing you are uncertain about instead of asking about a bunch of them, too
those are all intuitively obvious but it takes some energy to write up each one
 
Thorgott
Thorgott
yeah, my argument only works for $m=\omega_1$. the crux is that any ordinal smaller than $\omega_1$ is countable and the supremum of a countable collection of countable ordinals is countable, so my argument works in that case
once you grant that, though, $G^{\omega_1}$ is closed under complementation and countable unions, so that implies the transfinite sequence stabilizes from there onwards
 

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