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12:21 AM
@Roby5 i would modify the usual proof of Sylow by letting $H$ (instead of $G$) act on the set of Sylow subgroups by conjugation.
Let me know how that goes.
 
12:37 AM
@TedShifrin How would that help? I posted this question on the main site. This is the response I got: "You went a bit too far. A $p$-group might be contained in more than one $p$-Sylow subgroup. Sylow groups can have a nontrivial intersection."
 
12:51 AM
can I say that $(-1)^n$ has not a finite limit $l\in\mathbb{R}$ because, assuming it has limit $l$, there exists $n_0\in\mathbb{N}$ such that $n\ge n_0$ implies $|(-1)^n-l|<1$ and so it is $2=|(-1)^{n_0+1}-(-1)^{n_0}| \le |(-1)^{n_0+1}-l|+|(-1)^{n_0}-l|<2$ and this implies $2<2$, which is absurd?
 
The correct statement is that every $p$-subgroup is contained in some Sylow. I thought you were asking about that. Where did you get the unique statement?
I was doing exactly what Mark suggested there.
 
@TedShifrin Sorry. I don't really know what I was doing.
 
LOL, Ok.
 
@Sonozaki yes. What you’ve done is essentially “recognize” that the sequence is not Cauchy, and hence cannot converge.
 
Well, a very weak form of Cauchy.
 
12:58 AM
thank you very much
 
Looking just at adjacent terms is very not enough for Cauchy. Just saying …
@Sonozaki Do you know about Cauchy sequences?
 
yes, I do know, that strategy above was the first thing that I thought about and so I didn't tried using those
 
Your proof is excellent. I just wanted to elaborate on my comment. Do you know a sequence so that $a_n-a_{n+1}\to 0$ and yet the sequence does not converge (hence cannot be Cauchy)?
 
I'm stuck on this d.e, i'm given implicitly $\frac{dX}{dt} = (X-1)(1-2X), \log(\frac{2X-1}{X-1}) = t$
 
As it happens, I happen to be marking a calc test with a very similar question :)
 
1:10 AM
am I supposed to multiply out the right side*
if so I get $-2X^2 +3X-1$
 
Was that to Obliv or to me, peek?
 
@TedShifrin $a_n=\sin\sqrt{n}$ seems to do the job
 
to you, I snuck my post just before his.
 
Oh, not on my wall :)
@Sonozaki That is not trivial to prove, is it?
 
my comment was to you Ted, and it’s always funny watching their ‘creative’ answers
the question had the extra condition of making a_n unbounded.
 
1:14 AM
@Obliv … what are you doing?
 
@TedShifrin Yes, indeed now that I think better $a_n=\sqrt{n}$ is way easier haha
 
Yes, my example passes your exam.
Or even partial sums of the harmonic series :)
Good, regardless!
My challenge problem (on exams) when I taught Spivak was to give a divergent series whose partial sums are bounded. Not hard, but for an exam it’s not super obvious to most.
@Obliv First, you have the solution there? Second, what do you mean implicit? Third, what’s the most basic technique you learned in second calculus?
 
haha last year I gave my student that problem in tutorial, and some were stumped because they had just learned the week before that for $a_n\geq 0$, the series converges if and only if the partial sums are bounded
 
1:29 AM
Yes, @TedShifrin I'm given the implicit solution which can easily be written explicitly $\frac{2X-1}{X-1} = e^t$ but how do I verify this now in the original equation?
I mean, I could separate variables in the original equation but I thought I wasn't allowed to mess with it, but instead rewrite the given solution so that I can plug it in
screw it, it shouldn't matter, if the solution is correct then it's correct.
 
Solve for $X$? Or differentiate implicitly.
Yes, separating variables and using partial fractions is the way to do it.
@peek-a-boo Non-monotone sequences should be banned!
 
agreed, $[0,\infty]$ is a better place to be
 
 
3 hours later…
4:59 AM
Don't use ChatGPT for math
 
5:21 AM
I wouldn’t think of it!
 
5:36 AM
@TedShifrin I would guess that question was generated by ChatGPT...
 
Which?
 
10 hours ago, by Koro
I see why it will be true: I take a rectangle $[0,1]\times [0,1]$ and identify the top side to a point, and the bottom to another point- we get a disk.
oops, I think it is not correct yet. I should get a disk, but I got a sphere.
 
@TedShifrin The amusing link you provided earlier.
 
But it should follow by symmetry? We map the left vertical side to upper semi-circle; right vertical side to lower semi-circle. The left half of the rectangle respecting 'symmetry' should go to upper half of the disk; the right half should go to the lower half of the disk.
that follows intuitively indeed but I'm thinking about the map which does that.
 
@copper.hat Ah. And the response was likewise?
Koro. You are starting with a cylinder now, not a Möbius strip.
Thus you get a sphere with those identifications.
 
5:49 AM
I asked ChatGPT to solve $x^2 - 4x = -4$ for $x$ and it concluded no real solution after it somehow made the -4x term completely disappear
I also asked it how many ways there were to go from (0,0) to (10,10) using only upward and rightward 1-unit moves. At first it said 1, 10 right and 10 up but I said "what about right 9, up 1, right 1, and then up 9?" and it got the correct answer
 
Ohh the diagram arrows may be misleading, I drew them for a different purpose. Please ignore them. The idea is still the same: top side to a point, bottom side to another point; left side to upper semi-circle etc.
I'm trying to map the insides now: using $(x,y)\mapsto (\frac x{\sqrt 2}, \frac y{\sqrt 2})$
 
Are the arrows opposite or not there at all? This is not cool.
 
Please ignore the arrows: they were intended (for my own consumption) to see which part of the circle f is mapping into with the orientation. Later on, I acted the cyclic group C_2 on S^1 to get the antipodal effect to take care of the opposite sides identification as is done for Mobius strip.
 
OK, so still Möbius. So what’s the issue?
 
I want to write the map which does that. f is not correct- as it maps only the boundary of the rectangle-it's not even been defined yet on the interior of the rectangle.
I'm trying to do that.
I realized this after posting the image here.
 
5:59 AM
So map the square with upper/lower edges identified to a diamond (double cone over an interval).
 
6:52 AM
whats the question? are you trying to find a homeomorphism from a $I^2$ to the closed unit disk?
at least, before your mobius shenanigans, is that the core issue here?
 
Hello! I'm wondering if someone can help me understand where the result 2.2.16 comes from... I don't really understand the notation. It seems like $\theta^a$ represents a finite set of continuous parameters $\theta^1, \theta^2$, etc. but even of that interpretation I am unsure of.
is it just saying if you compose an identity transformation with an arbitrary transformation you are just left with the arbitrary transformation?
 
@SillyGoose yeah
its just saying it in a unnecesarily obtuse way
 
what the hell XD
 
thats math for ya :P
 
this isn't even a math book XD why in the world would someone describe that in this way
 
7:06 AM
yeah I figured it wasn't one, funnily enough a lot of math books don't do so much of a better job
but for Lie groups, you should go open up Lee
he does a fantastic job at expositing virtually everything in his book (2nd edition, intro to smooth manifolds) at least in my opinion
its just that his book is pretty cumulative. you really need to invest time working through it
but depending on your background , you may be able to just open up the relevant chapters on lie groups and read on
(fyi his book isn't some treatise on lie theory, but its not a bad introduction)
also, you have to do the exercises, and go look up his errata list on his website if you get stuck on parsing a proof or exercise, there are a couple, although IMO he gives you more than enough material to fix what are mostly (non)-issues
 
ah hm. i only have exposure to a course in undergrad abstract algebra (mostly group, ring theory). i'd just like to have a good understanding of connecting lie groups with lie algebras and being able to have familiarity with how to represent such groups but perhaps that is a great deal of lie theory XD
 
oh, i've the opposite background pretty much, for me a lie group is a smooth manifold
but I know people who know this stuff way better than me and view it inversely
idk :/ im not the best person to consult on this topic
although these people seem more interested in lie algebras
(which can just be defined purely algebraically)
but to me its all nonsense without motivating the Jacobi identity with the OG lie bracket of vector fields
 
@porridgemathematics yes.
 
and connecting lie algebras to tangent spaces to lie groups at the identity... flows and the like
 
@TedShifrin I'm quite convinced that such a map can be found.
 
7:13 AM
its too awesome to bypass imo, talking out of my depth here but I think you have to see flows first
 
But it seems calculation intensive to figure out one so I am postponing my search for the map, for now.
 
i am trying to parse the math structure behind textbook quantum mechanics that is just tacitly assumed, which is the motivation to understand lie algebras but idk i am very unknowledgeable at it all XD
 
@Koro you should be. I can actually write one down even better, I can write a conformal (biholomorphism) of $I^2$ to $\text{cl}(\mathbb{D})$ extending continuously to the boundary in a 1-1 fashion
 
i shall take a look at lee !
 
look into the Schwarz-Christoffel mapping , and the Riemann mapping theorem
or if you know a little bit of complex analysis, first assume a conformal map from the interiors to each other exists. (this is fine imo, the riemann mapping theorem at this point is a staple for mathematics), and then straighten your corners
use the schwarz reflection principle a bunch of times once you've done this, then look at the logarithmic derivatives $\frac{f'}{f}$ of your continuations along all the boundary pieces of your square
 
7:15 AM
I don't yet know Riemann mapping theorem or conformal mapping. But thank you. :)
 
find their residues at the corner points, and you'll find the general form of such a map
you should though!
 
I do know Scharwz reflection though :).
 
if you know schwarz reflection, you can do this
it will take some thinking though
 
@porridgemathematics I'll get there soon :). Thanks.
 
but this is overkill, because all you care about is a homeomorphism.
there are much simpler ways of getting a homeomorphism
 
7:18 AM
But I like the idea, thank you very much. I just have to map half of rectangle to half of the disk, and then Schwarz gives me the complete thing.
 
basically yes
but you need to straighten the corners out first
and then unstraighten them
but yeah
its a really pretty picture
 
@porridgemathematics ohh to avoid loss of 'holomorphicity'. hmm.
 
precisely
 
but seems overkill as you said: we want only continuity. No need to go to differentiability. We can treat a rectangle and a sphere as the 'same' in continuous treatment.
 
the local form of such a map at four boundary points $z_1,...,z_4$ oriented say anticlockwise on the boundary of the disk, which map to the corners, is going to look like $(z-z_j)^{\frac{\pi}{2}}h(z) + c_j$, where $c_1,...,c_4$ are say, the points $(0,0), (1,0), (1,1), (0,1)$
where $h(z)$ is non-zero at $z_j$, and holomorphic in $\mathbb{D} \cap B_{\epsilon}(z_j)$ for some small $\epsilon > 0$
err, sorry, those shouldn't be corner points, those should be the midpoints between consecutive corner points, $c_1,...,c_4$, so replace those with say, $b_1,...,b_4$
nvm, ignore my previous remark, those should be corner points
but you will want your exponents to be alternating in sign, so $\frac{\pi}{2} (-1)^{j-1}$ at each $z_j$
then you will be in a position to schwarz reflect across a boundary arc, and see that when you schwarz reflect again back into the unit disk, your continuation is the same up to a rotation, so that your logarithmic derivative $\frac{f'}{f}$ remains the same, i.e, what I am demonstrating shows you can continue $\frac{f'}{f}$ to $\mathbb{C} \setminus \{z_1,...,z_4 \}$
 
7:44 AM
well sure, if you only care about continuity, then just take $[-1,1]^2$, and for all $t \in [0,1]$, map the square $[-t,t]^2$ homeomorphically onto $\text{cl}(B_{|t|}(0))$
in the 'same way' of course
'radially'
i mean the perimeters of those squares
 
yeah, it can be done. But even for a given t, it'll be involving to write such a map. Something like $(x,y)\mapsto (\frac x{\sqrt 2},\frac y{\sqrt 2})$. But I'm postponing this for some other day.
I was thinking that: identifying the set consisting of a meridianal circle and a longitudinal circle of a torus to a point should give me the quotient space as two spheres (S^2) touching exactly at one point.
But it seems that this answer is wrong.
The correct answer is $S^2$ and I don't see how.
 
those circles would just correspond to the perpendicular bisectors, vertical and horizontal, of the unit square, pre-identification
they aren't contained entirely on the boundary, besides their starting and ending points
 
First I collapse a longitudinal circle to a point a. The resulting image looks like a donut pinched at a point. Then collapsing the meridianal circle to the same point a should give me spheres touching at a point.
 
8:05 AM
it would give you a sphere.
think about it this way, lets cut and paste to simplify things
 
8:17 AM
Hey, I couldn't help but notice your conversation
I found a couple links that might help
0
Q: Collapsing longitudinal and meridian circles of a torus.

Anil Bagchi. What will be the quotient space obtained from torus by identifying the longitudinal and the meridian circles of it? The problem is that I can't visualize the space but in Hatcher's book it is claimed that the quotient space is homeomorphic to the $2$-sphere $\Bbb S^2.$ Is there any easier way ...

In the answer to that question, they also link this site, which has some informative graphics
https://www.math3ma.com/blog/clever-homotopy-equivalences
 
oh..
i just spent some time drawing a picture
to elaborate on what i meant by cut and paste
anyway ill upload it in case its different to those links and may be of use to Koro
that red cross is the outcome of identifying your meridonial/longitudinal circles
 
Thanks @porridgemathematics.
 
so... I think this is correct, if I understand what you mean exactly @Koro, you start with a torus, i.e. with $I^2$ with the usual identifications, and then you want to collapse the set equal to the union of the meridonial and longitudinal circle to a single point.
 
Thanks for the links @JadeVanadium. I haven't yet looked into the links.
 
if this is what you are trying to do, I think what I've illustrated shows why you get a sphere
 
8:30 AM
My confusion arose because I mistakenly assumed the torus to be a solid torus.
serves me right to regard 'torus' as a donut with filling inside. Haha
 
funnily enough, before krispy kreme abominations, donuts were better approximations to mathematical torus
they didn't really half a filling
bagels on the other hand, were never good approximations to mathematical torus
unless you painstakingly kept saying, "like the surface of a bagel, is what I mean by 'torus'!"
*have
anyway.. even if the torus was just the boundary of what you thought it was, i don't see how you wouldn't end up with a filled in sphere at the end
unless by 'identify longitudnal/meridonial circles' to a a point, you meant something very different..
the interior regions of the hemispheres bounded them aren't touching anyway
to lose an entire dimension, you will need to collapse something of comdimension 1
your circles have codimension 2 in the filled in torus
(codimension 1 or less)
im not able to think of a way to collapse something of dimension 3 to something of dimension 2 in the realm of nice spaces, lets stick to CW complexes or smooth manifolds
without just collapsing it all to a point
like if we just think about things in terms of hausdorff dimension, it would look like to do that you need to collapse some codimension 0 subset... but really what you would have to do is find a way to hollow out via collapsing.. is that possible?
my intuition says no, because if you collapse a codimension 0 subset that is not of full measure.. it seems like you would never lose a dimension
by the way, if you're taking an AT course that follows a book like hatcher, and come from something like a hardcore analysis background, you can justify all this cut and pasting business formally , look into adjunction spaces and see that you can basically view gluing as quotienting by an ideal
in the sense that, when quotienting by $(a,b)$ you can always quotient by $(b)$ and then quotient out $(\overline{a})$ posthoc
thats essentially what you need to justify to make all this business concrete, but the proofs are basically routine
(on a set-theoretic level at least, it should be clear everything checks out)
and the quotient construction is not much more than the set theoretic quotient, theres no other topology you can put on quotients than the final one, basically by definition of colimits
hatchers book imo is not great for these types of minds, and to make do with only using it for a first course you'd have to basically justify by drawing a lot of pictures, and hope that your manipulations are legitimate (but in the context of the spaces he works with they always are)
that being said, i haven't read other AT books, funnily enough, I learned most of my AT using books about riemann surfaces / smooth manifolds
(except a bit of May's , but that is basically antipodal to hatcher in style)
ahlfors-sario is a classic
the first section gives a ground up proof of the classification of surfaces
using triangulations etc, and its written by ahlfors, so if you're used to his style from say his complex analysis classic, then it might be worth looking into
okay.. ive procastinated enough, ciao
 
10:04 AM
suppose I have a group homomorphism $\Phi:\mathrm{SU)(2)}\rightarrow\mathrm{O}(3)$. Now, $\mathrm{SU}(2)$ is connected, while $\mathrm{O}(3)$ is not. Without proving continuity first, I shouldn't be able (in general) to conclude that $\mathrm{Im}\Phi\subset\mathrm{SO}(3)$
I mean, I need continuity to conclude that the image of a connected set is connected, unless there is something that makes thing easier for groups (?)
 
 
2 hours later…
11:43 AM
Sometimes I forget I'm more than 12 hours ahead of the US in timezone
 
 
1 hour later…
1:09 PM
@porridgemathematics thanks a lot for the insights. :-)
@porridgemathematics ohh, I meant two spheres touching at a point is what we'll get after collapsing the set {a meridianal circle, longitudinal circle} to a point (suppose that the meridianal circle that we want to collapse is inside the donut.)
but of course since by torus we mean only the surface, we get the sphere.
 
1:49 PM
huh, in algebraic topology class tutorial here, TA says to exercises-this question is very easy and writes (copy pastes) the solution from Greg grant website. I say why bother arranging the tutorial if this has to happen.
 
Btw, @XanderHenderson, you recommended Spivak's Calculus, but there seem to be multiple editions, it has a 3rd edition, an international student's edition, apparently a 4th edition, ..., does it matter which one you use?
Or should it be pretty much the same regardless of the edition
 
2:06 PM
Can anyone help me with this:Find the nth derivative of $\frac{x+1}{(x^2+x+2)^2}$ ?
 
2:17 PM
Anyone ?
 
I've forgotten most of this, all I can remember from calculus is how to form the integration by parts formula. How about you try WolframAlpha or ChatGPT?
@ILikeMathematics I think your main objective when you get a textbook like that is to first start working on the question bank and then throwing them randomly at people until they figure out for themselves which questions to challenge themselves with.
 
2:47 PM
@Koro you need to be really careful with terminology here, I know what you want to say, but you are not collapsing a circle inside the donut (even in common parlance). What you want to do is collapse a meridional disk circumscribed along a meridional circle of a torus, and collapse it, as well as a longitudinal circle on the suface of the torus intersecting the disc along its circumeference only at one point.
then indeed you will get two spheres that are tangent to each other at a single point
here you only need a torus (the surface, not a filled torus), union a meridional disk.
you dont need the entire torus to be filled in
so there isn't any strange change in dimension, you aren't starting with a solid torus, your object you started with is still very much only two dimensional
and after collapsing, you dont lose a dimension
disk and circle are unfortunately used pretty interchangeably so its understandable, but its going to confuse when speaking in a technical setting
the thing is, you don't get a 'pinched torus' after collapsing a meridional disk. You get one after collapsing a meridional circle. What you get after collapsing a meridional disk is homeomorphic, to a sphere. A pinched torus is not.
 
Oh right, I was collapsing a meridional disk :(.
thanks a lot. :)
 
3:02 PM
a pinched torus is homeomorphic to two napped cones [surfaces!] (i.e. the with their boundary circles) identified
(identified along their boundary circles)
sorry, correction not quite
its homeomorphic to a sphere with two points identified
(i forgot to identify the vertices of the two napped cones)
once you identify the vertices of the two napped cones, then yeah you get a sphere with two points identified
the two napped cones with their boundary circles identified is indeed homeomorphic to a sphere
again this can be figured out using the cut and paste technique
or when you collapse the meridional *circle + the longitudinal circle, what you have just done is collapse the boundary of a closed disk to a point
and thats another maybe easier way to see that leaves you with a sphere, because this is a well known fact, you can basically think of it as wrapping the sphere up with a disk and closing into the north pole by shrinking the boundary to it
 
3:48 PM
Repost:Can anyone please help me with this:Find the nth derivative of $\frac{x+1}{(x^2+x+2)^2}$ ?
 
actually, im still not convinced, I think collapsing one meridional disk and then a longitudinal circle, will still be at least homotopy equivalent to just .. one sphere
if you want something homotopy equivalent two two spheres tangent to each other, you need to collapse two meridional disks.
not one meridional disk and one longitudinal circle
in fact it should even still be a homeomorphism
if you collapse two meridional disks on the other hand.. , you will get two spheres but touching at two points
if you collapse two meridional disks, as well as a longitudinal circle touching each of those disks only once, you will get two spheres touching only at one point
but i dont know how you are getting two spheres touching at a single point by collapsing one meridional disk and a longitudinal circle
collapsing one meridional disk should still give you a pinched torus, so a sphere with two points identified, then collapsing the longitudinal circle is just going to give you a single sphere, because its the same as collapsing the boundary of a closed disk..
yeah.. so im not sure how you're getting two spheres touching at a single point even if you collapse one meridional disk and a longitudinal circle..
at the very least you will need to collapse another meridional disk for that to happen
maybe im just crazy.. i dont see how any version of what you said is possible
 
4:14 PM
@porridgemathematics: the red line denotes meridianal disk.
 
yeah okay, so after you collapse it, you still get a pinched torus?
whether you collapse the disk, or the circumference, wouldn't you still get a pinched torus?
as long as you are just collapsing a single disk , that is
 
I think that after collapsing the disk, we get two spheres touching.
 
okay, so how does your picture demonstrate that
 
because we are collapsing to the same point where we collapsed the longitudinal circle to.
 
sure
but you're also doing that when you just collapse a meridional circle
after all, the longitudinal circle only touches the meridional disk on the circumference
 
4:18 PM
I don't know how to draw that. But I thought that: hold the red line and crash it to the collapsed point.
 
because its a longitudinal circle on a surface
its not going to touch the disk at an inner point or something?
okay, so im guessing your red line is really 'inside' the torus
its not a longitudinal circle at all
 
@porridgemathematics indeed
 
urghhh
that is something you omitted..
even after clarifying your meridional circle was inside
 
inside the torus. You see, that's why I was getting confused: should I take hollow torus or the solid torus?
 
(turned out to be a disk)
you didnt say your longitudinal circle is just dangling inside the torus
a torus is hollow period
 
4:21 PM
@porridgemathematics yes.
 
a solid torus is an entirely different beast
they aren't homeomorphic at all
or even homotopy equivalent
a solid torus is the one point compactification of $\mathbb{R}^3$
 
@porridgemathematics actually I somehow thought that it won't matter because I thought that meridional circles could lie inside torus.
but meridional refers to circles on the boundary so that was wrong.
 
right..
you can say meridional disk circumscribed by a meridional circle
but a longitudinal circle inside a hollow torus.. okay, lets say you start with a solid torus, now I think I see where you are going with this
if you start with a solid torus, i think what you are saying does work
err, hold on.... it still doesnt seem to work
 
Can anyone please help me with this:Find the nth derivative of $\frac{x+1}{(x^2+x+2)^2}$ ? ...pleaaaaaaaaaase I am really feeling helpless here 🥲🥲🥲🥲🥲
 
@Franklin stop spamming in the chat :(
 
4:39 PM
@SineoftheTime Suggest me something except: asking me to leave the site and try it more 😂😂😂😂😂. And help me pleaaaase if you can. One thing more:dont get me wrong, I am not ordering you, but imploring you no wait, implore is a wrong word, requesting you, that's it
 
err, sorry, a solid torus should be one 'component' of the one point compactification of $\mathbb{R}^3$, the other one will be orthogonal to it
its the one point compactification of $\mathbb{R} \times \mathbb{D}$
 
franklin, you could use complex numbers and partial fraction decomposition as someone suggested in math.stackexchange.com/questions/4632214/…, which you asked and had answered three hours ago.
generally speaking, it isn't reasonable to expect immediate answers to anything. a few hours is not a long time.
 
hmm, okay @Koro, I'm willing to believe if you do what I think you are suggesting with a filled in torus, you will get two filled in spheres touching at one point.
but you will need your "longitudinal circle" to actually be both inside the solid torus, intersecting your meridional disk at the center, and then on the opposite side you need it to eventually become half an actual longitudinal circle
as in, it needs to touch the boundary of the solid torus on the opposite side and stay on it after it intersects the meridional disk at the center
so basically draw two rays stemming from the center of the meridional disk that eventually touch opposite points of the boundary of the solid torus , exactly at the half way point, from then on, it needs to be half an actual longitudinal circle lying on the boundary of the solid torus
then what you are suggesting should give you a $S^2 \lor S^2$, when you start with a regular torus (not solid)
 
franklin, if you find the roots b and c of the denominator, you can even get something like wolfram alpha to do most of the computation for you. see wolframalpha.com/…
 
but this is far from a longitudinal circle... its a completely different construction to what you drew even
from what you drew, your thing wont work
your red line can't literally be contained in the interior region the torus bounds
because if it were, and you collapsed that to a point first, you would end up with a (non-solid) torus , plus a meridional disk circumscribed ..
since besides touching the center of the meridional disk, its completely away from the torus.. so collapsing it just leaves you with the torus plus the disk
the only 'identification' that happens at this stage is the red line gets identified with the center of your meridional disk
then when you collapse your meridional disk, you just get a pinched torus
so the way you drew it, it can't work
 
5:01 PM
@Franklin please don't put in my mouth words I've never said. I've never told you to leave the chat, just stop spamming. This is not the first time ;)
 
unless you drew it so that it intersects the meridional disk at a circumference point, and is then inside the hollow torus (which I guess could be a possible interpretation of your drawing), but if you meant it to be wholly in the interior region of the torus, it shouldn't work
anyway, if you meant the former then I am now in agreement with you lol.. what a conversation
i swear 99% of difficulties I see on SE chat and irc come down to semantics or terminology not being used adequately
yeesh, in a weird way im glad i dont do algebraic topology lol.. after only thinking of things being the same up to things like local isometry or at least local diffeo for a while.. homeomorphisms are just so weird..
and this isnt even mentioning homotopy equivalence
 
5:24 PM
I think I know what your confusion was, im assuming when you answered with $S^2 \lor S^2$ the first time, for $S^1 \times S^1$ with $S^1 \times \{0 \} \cup \{0 \} \times S^1$ collapsed, the problem was you didn't see that after collapsing your meridional circle/disk (doesn't matter), if you were to make a cut along the red line in your most recent image (assuming it does lie entirely on the circumference), you get something disconnected, like $\mathbb{D} \coprod \mathbb{D}$
where the red line is a common boundary
then indeed you would get $S^2 \lor S^2$
but if you look carefully you will see that no matter how you draw your red line, as long as it intersects the meridional circle/disk once, and is truly a longitudinal circle, so really lies strictly on the torus, not at all in the interior region, you will get $\mathbb{D}$ after cutting along it
not two connected components
if I were to bet on it, Im guessing that was what stemmed your confusion
 
As room owner, I'm politely asking you, @Franklin, not to repeatedly post your pleas. If someone wants to help, one request will be enough. Whining for attention is not good and is very annoying. Besides, it's a horrible question.
As leslie pointed out, you got the best possible answer on the main site. And you accepted that answer hours ago.
 
I think I know this because ive definitely convinced myself of the same before, and i remember a lot of students getting tripped up on this in class back in an AT course I took at uni a while back
 
@SineoftheTime No no you got me wrong. I never said you said them. I forbade u to say those words becoz I wouldn't have accepted it...
 
@porridge I've lost track in these long discussions. What are you and koro working on now?
 
@TedShifrin Just a thing: that question in main site is a different one and the one I posted here although are on the same topic but different
 
5:34 PM
oh super long story, but sorta interesting (at least for me), so last you were here I drew Koro a picture trying to explain why collapsing $S^1 \times \{0 \} \cup \{0 \} \times S^1$ in $S^1 \times S^1$ will indeed yield $S^2$, not a wedge sum of two spheres
 
Well, good grief. You cannot post every single question in your homework. The solution you accepted shows you how to do all such problems.
 
after that Koro remarked that he thinks he knows what tripped him up, namely that what he wanted to really do, was collapse a meridional disk , as well as a longitudinal circle
 
Oh, this is why I suggested a day ago to @koro that he turn the square with top edge identified to a point and bottom edge identified to a point into a diamond.
 
so I looked into that and initially bought it, but realized that also doesn't work
 
More geometrically, we take a cylinder, squeeze the top circle to a point, squeeze the bottom circle to a point, and that is a double cone, which is homeomorphic to a sphere. shrug
 
5:37 PM
@TedShifrin May I ask what's your opinion about Frankel's book "The Geometry of Physics"?
 
@Mr.Feynman I only looked at briefly one time. It seemed OK, but I don't have an opinion.
 
Oh, I understand
 
He has great stuff in the book.
 
right, but wouldn't you need to identify the vertices of the double cone after that? also, wouldn't you need to then collapse the circle you get touching the identified vertices to a point?
wtf
(not to you ted)
 
@Franklin. Enough. You need to get a grip.
 
5:41 PM
calm down, you are on the internet
 
calm down. i linked to a wolfram alpha calculation that would do the partial fraction composition for you.
 
Please do not yell.
 
cool people get mad in lowercase
 
now im more curious about what the deal with the all caps stuff was about lol
but its definitely besides the point
 
There is someone in my uni's offices writing mails in caps lock lmao
 
5:44 PM
@porridge Oh. I missed the point that we're identifying the top and bottom to the same point. That had changed from yesterday. So it's a torus with one meridian squeezed to a point. It actually is neither a sphere nor two spheres attached at a point.
 
I guess they just don't know but it's really funny to read
 
@TedShifrin its a torus with one meridian, and one longitudinal both collapsed to a point
i.e. the union of that set, collapsed to a point
 
It's a single sausage link with the ends of the sausage identified.
@porridgemathematics Why
 
uh, not quite as far as I understand?
that would be if you only collapse one meridian
as you say
 
Maybe I'm screwing up the words. This always confuses me. I collapse one of the small circles on the torus. Nothing else.
 
5:46 PM
i mean the question as I understood it was, take a torus, draw a meridian loop, and a longitudinal loop, orthogonal to each other on the surface
and collapse the set comprising both of them, to a point
what is the resulting space
*and these loops intersect each other at precisely one point
 
Oh, sorry. I still didn't read Koro's question correctly. I was stuck on the cylinder.
 
and the resulting space should be a sphere
 
Right, right. I have been completely sloppy.
 
although as you say, if you just collapse one, you get a pinched torus
if you collapse the other, you get a disk with its boundary identified
since the other (longitudinal, say) line plus the pinch point now form the boundary of a homeomorph of the open unit disk
so when you collapse that to a point, you are collapsing $\partial \mathbb{D}$ to a point, and then you get $S^2$
anyway, koro thought initially, you should get $S^2 \lor S^2$, two spheres kissing at a single point/tangent
 
So in my pinched torus picture, we now collapse the "inner" longitude to a point. Right.
 
5:50 PM
and after I drew a picture trying to convey why you should get $S^2$ he reopened the dialogue with an explanation for why he was under the impression you get $S^2 \lor S^2$, which was to me initially convincing, so he said he was thinking of a solid torus number 1
@TedShifrin yes, thats right
 
So you squeeze that inner circle to a point and you get a sphere.
 
right, it actually doesn't matter what longitude you collapse
thats the confusing part
you could have also collapsed the 'outer' longitude
and you will still get a sphere
 
It's best perhaps to think of a movie homotoping that inner circle to a point. So the sausage gets fatter and fatter and the hole disappears.
 
(a quick way to see this has to be the case is that you can homeomorph them onto each other)
yes thats probably the best way to intuit the result
 
There should be a clean way to get this from the rectangle picture of the torus.
Oh, of course.
 
5:52 PM
@TedShifrin I like "to homotope" as a verb.
 
i think i have a pretty clean one, let me find it, its the one I gave koro
 
It is the disk with $aa^{-1}$ on the boundary circle.
 
That's too hard.
 
i just felt I should show how it can be done concrete
but i didnt come up with $S^2$ through that picture
apparently it convinced koro though
 
5:54 PM
Take the rectangle with opposite edges identified (in the correct directions). Compress the left and right to a point. Now attach the top edge to the bottom edge.
 
pretty topology
 
If I sew the top semicircle to the bottom (matching orientations), I'm zipping up the disk along that edge and that gives me a sphere. I can even write that down in coordinates, I think.
 
its actually even simpler than that if you want to get super technical, I purposely used different longitudinal /meridonial circles in my drawing than the standard ones
the most straightforward way imo is this. you want to compute $(I^2 / \sim) / (\partial I^2 / \sim)$
because you are collapsing the set (a set, but it can be any such set) that is the union of a longitudinal and meridonial circle
 
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