so that your initial identifications didnt even matter, if you just use the boundary circles..

you are left with $I^2 / \partial I^2$

I think that for @Koro picturing the sewing the boundary of the disk together is a good thing to see.

which is, the closed unit disk, with its boundary collapsed

Yes, although that's a different picture, actually. :)

indeed, that is trying your hardest at gift wrapping a basketball with a rubbery disk sheet

which works in topology la la land after you identify the boundary

I guess you can think of the zipper as forming a line segment on the surface of the sphere, and then you shrink the zipper to a point.

yeah, there are a ton of ways to see the answer

the reason I knew it had to be a sphere was because I knew I could choose whatever meridional circles/longitudinal circles I wanted, because up to homeomorphism, it doesn't make a difference

Spherical coordinates will also give the mapping for Koro to do his quotient topology with :)

which led me directly to just using the boundary circles, and then noting that im just collapsing the whole boundary of $I^2$ to a point (not dissimilar to how when you mod out by an ideal $(a,b)$, you can mod out by $(a)$ and then $(\overline{b})$ and you get the same ring)

Oh, not quite. That doesn't identify the top and the bottom.

Right, some fundamental homomorphism theorem :P

yeah anyway, I did preface this by saying looong story, so I've just given you the first 40%

haha

I was in on an early part of this yesterday. Sorry for causing the disruption :)

koro convinced me after a while that you get $S^2 \lor S^2$, if you take a torus, plus a meridional disk (so a closed disk sitting in the interior region the torus bounds, circumscribed by a meridional circle on the torus itself), and collapse both the meridional disk plus a longitudinal circle to a point

this is the next 60%

initially I thought huh maybe that would work, but soon realized it makes no difference

at the very least, I suspected, you need to collapse another meridional disk

in that case, I think you would get $S^2 \lor S^2$

but that wasn't what he was suggesting

Right. If you collapse the equator of the sphere to a point, you get the wedge of two spheres.

so then he claimed what he was thinking was, collapse a meridional disk, and not a longitudinal circle exactly, but a longitudinal circle that intersects the meridional disk at a circumference point, and enters the interior region of the torus afterwards,

this led to some confusion because, what he was describing was not a circle on the torus at all

he gave me this picture to explain what he was trying to convey

and I asked him if he intended the red line to be wholly in the interior region of the torus, or not

LOL. OK. You don't need to relive all the history. I do thank you for the reporting!

I then concluded that if he indeed intended for the red line to be wholly in the interior region of the torus, collapsing that plus the meridional disk (which he drew in purple/blue) , will not give $S^2 \lor S^2$, but collapsing the red line if it met the meridional disk at a circumference point , stayed for a while on the boundary, and then in the interior afterwards, should really give $S^2 \lor S^2$

no worries haha

im basically procastaworking

soo finally (end of story) I concluded from his picture, that his initial confusion, assuming that his red line actually is longitudinal circle, so it really doesnt enter the interior region, and the purple line is just a meridional circle, was that the shape after the arrow, gets disconnected after you cut along the red line

if you draw that shape in a particular way, it can appear to be disconnected

so I wagered his initial confusion stemmed from that picture, as if cutting along the red line would yield $\mathbb{D} \coprod \mathbb{D}$ with common boundary red-line, rather than just $\mathbb{D}$ with boundary red line (plus purple/blue pinch point)

and then if one were to think the former, collapsing the common boundary would yield $S^2 \lor S^2$

i feel pretty strongly this was the initial source of confusion lol, mostly because ive been here before plenty of times

because a pinched torus does resemble a sphere in someway, I think this confusion is relatable , if you cut along an equivalent red line on a sphere, you would indeed get a disjoint product of open disks with common boundary

the catch being I guess, that the genus one property gives you connectivity after the cut

@porridgemathematics Except that the first homology/fundamental group are a bit different. :)

right, its homemorphic to a sphere with antipodal points identified

anyway, i should try and get back to this analysis crap lol, ive been procrastaworking a bit too long ( not that this wasnt enjoyable, it was definitely a bit too attractive considering how much time ive spent not working on what im supposed to be working on )

i was kinda proud during procastaworking I managed to answer this question math.stackexchange.com/questions/4471447/…

funnily enough im working on stochastic integration related stuff atm

and stumbled upon it

@porridgemathematics Why is $|PSL_2(q)|= \frac{q^3-q)}{\gcd(2, q-1)}$? Since $Z(PSL_2(q))=\{\pm I_2\}$, we should have a denominator of $2$.

sorry i cant help atm, but Im sure someone in here eventually will

@porridgemathematics Thank you.

@Roby You're forgetting the $P$.

When you projectivize, $\pm a$ become the same element.

@TedShifrin I don't get it. Give me an example.

And please stop spamming. You do not need to ping individuals and repost your question. If people know how to do it, they will respond.

What is the definition of $PSL_2(q)$?

gcd(2,q-1) seems to be just a symbolic way of capturing "2 if q is an odd prime, and 1 if q is even"

If $q=2$, then $\pm 1$ are the same thing.

@TedShifrin $PSL_2(\FF_p):=SL_2(\FF_p)/Z(SL_2(\FF_p))$

But Roby is forgetting what projectivization means in the first place.

if q is even, i should say

Yes, precisely. So you've modded out by $\pm 1$ when you make $P$.

This is a special case of projectivization, where all nonzero multiples of an element are declared equivalent.

@TedShifrin I have not studied projectivization. This appeared in my abstract algebra course.

So what. I'm teaching you something.

You did not pay attention to the definition — is the simple answer to your question. Instead of spamming, you should look carefully at the definition and make sure you understand it.

You kept posting the center of $PSL_2$, but in fact you were posting the center of $SL_2$.

you absolute distracted banana

when q is even, the elements 1 and -1 are the same thing in the underlying field. so the matrices I_2 and -I_2 are also the same thing.

@TedShifrin I think I understand now. Thank you for your help.

they are distinct when q is odd.

But, more confusingly, there is a denominator of $2$ (unless $q=2$), so I'm not sure what the complaint is all about.

@leslie Is Munchkin in detention yet?

she's at ice skating class right now.

nvm i was meant to differentiate that

Time off for good behavior!

@shintuku hahaha

(looking at porridge's comments)

I can't tell if i did this wrong or if this is indeed not family of solutions to the function. $\frac{dP}{dt} = P(1-P), P = \frac{C_1 e^t}{1+C_1 e^t}$ I did quotient rule $\frac{(1+C_1e^t)(C_1e^t)-(C_1e^t)(C_1e^t)}{(1+C_1e^t)^2} = \frac{C_1e^t}{(1+C_1e^t)^2}$ doesn't work when I plug it in to verify though

Your algebra probably sucks. Note that $P = 1-\frac1{1+C_1e^t}$.

yeah wolfram confirms the derivative is right

I guess it isn't a solution, i'm just so used to confirming them

It is a solution.

As I said, your algebra sucks.

It's strange how the discussion with @porridgemathematics turned to 'why I thought the answer was $S^2$ touching another $S^2$' from 'what space I get by collapsing {meridianal circle, longitudinal circle} to a point'.

Do you mean $\land$ or $\vee$?

neither of these :D

$\frac{C_1e^t}{(1+C_1e^t)^2}=(\frac{C_1e^t}{1+C_1e^t})(1-\frac{C_1e^t}{1+C_1e^t})$

i don't see how that's equal

No, that's $\vee$, Koro.

Well, work on it, @Obliv. Seriously.

Did you check my "note that ..." above?

I think it was like the following situation: Suppose I am grading someone's answer script, and there is a question whose answer is say 3.1. So I would know what type of mistake student will make to get an answer say 4.3 or say 9 or say 11.

But it would shock me if someone gets the answer as $\pi$. That would put me to thinking -how he might have done it. Haha

Yeah, when I wrote math competition questions that were multiple choice, I tried to include answers that one might get by making some obvious common mistakes.

forgot the +1

because it was not in my list of 'pitfalls'. How did he do it? What mistake did he make? I want to know that point of mistake!!

thanks Ted.

Sure thing. :)

You have to learn to double-check your work efficiently.

I have this problem in other areas too, like in chess, I just can't seem to step away from a problem and let my thoughts diffuse

I just tunnel vision and try to brute force

@TedShifrin ohh I'm yet to learn wedge product in algebraic topology :).

and then there is smash product.

Yeah, that's one-point union, @Koro.

will come with experience and proper training though.

Smash is a bit more like the problem you were working on :P

@TedShifrin Oops. I didn't know. I just have some reminiscence of the symbol :D.

I forgot to thank you ted I understood the priori bound.

Great!

It was so simple and obvious why didn't I thought of that before

that's why I changed my comment to include words to show connection between two 2 spheres.

@Koro: $X\wedge Y = X\times Y/X\vee Y$.

Looks like $S^1\times S^1/(S^1\times p \cup p\times S^1)$.

also today I actually understood why 'product topology' is like the 'opposite' of quetient topology.

@CuriousMind As I suggested yesterday, go back to all those limit proofs with quadratics or reciprocals. You have to do this a priori bound every time. When I taught this material, I always talked about "stipulate" [in case there were potential lawyers in the class] :D

@Koro Is that really true? I mean, if you mod out by a whole factor, sure, but not in general.

@Koro: Actually, look at the example here :P

@TedShifrin ohh, it seems to be same as identifying something in a torus-somewhat like the what I was discussing with porridge earlier. Sorry, I don't immediately see how to visualize $S^1\times p$. I'm still at rectangle description :).

I was doing it with the rectangle earlier. See my remarks about the zipper to make the sphere.

I say opposite in the following sense: We want to define product topology on cartesian product to be the smallest (coarsest) topology so that all projection maps become continuous but in quotient topology we want the largest (finest) topology on the codomain to make the onto map continuous.

@TedShifrin okay.

Oh, but in one case you have the cartesian product and not in the other.

@TedShifrin Ohh, so what I was doing earlier was smashing 1 sphere with another 1-sphere!!

Right.

oh thanks. I'll remember this example now when I study smash product soon.

Indeed!

@TedShifrin So I think the opposite is meant in the following sense: If we see both these concepts (product topology and quotient topology) as ways to construct new topological spaces, then the ways of constructing these are opposite (in the sense that in one case we want the coarsest such some map(s) become continuous, in the other finest).

It's always different whether you're talking maps out of a space or into a space. You've observed the difference before in looking at continuous maps $\Bbb R^2\to \Bbb R$ versus $X\to\Bbb R^2$.

how would I devise a D.E. that doesn't have any real solutions? Does that mean one with complex solutions?

@Obliv: look at $(dy/dx)^2+1=0$

if it's a consistent equation, shouldn't every D.E. have a solution

What about $y^2 + (y')^2 = 0$, $y(0)=1$?

Similar ideas.

What do you mean by "consistent equation"?

quaternions or octonions if we want ot get exotic with the number systems....

To me that means it has a solution. So that's begging the question.

You're not helping, DC.

Ted is the real ChatGPT.

back on track. Obliv asked for help...no distractions

@ペガサスSeiya: won't it be amazing if Klaus from the Originals went to the walking dead universe and help the people fight the virus?

Yeah those examples involve $\sqrt{-1}$ which makes sense, they have no real solutions.

@Koro No because that means I wouldn't get an awesome zombie game to play on PC

@ペガサスSeiya but you see zombie vs vampire is rare. I have never seen them in the same show.

yes, as Ted says, some topologies are understood in terms of maps into them and some topologies are understood in terms of maps out of them

now, Ted won't like this next step, but this is just an instance of understanding the dual roles played by limits and colimits in arbitrary categories

@Koro would be interesting. Zombies vs Vampires at night. If one bites the others what happens?

the hint was that it is not all reals

but I don't see why.

for $\sqrt{1-y^2}=\frac{dy}{dx}, y=\sin x$ we have $\sqrt{1-\sin^2(x)}=\cos x$ isn't the interval of definition all reals?

I didn't mean to delete it woops

@Thorgott Well, I have said that $\times I$ and integration over the fiber are adjoint functors, so even Ted waxes categorical from time to time.

@Obliv Stop and think, man. What if $x=\pi$?

that's weird, @TedShifrin you have $\sqrt{1} = -1$ which is false if you do the sqrt on the left side but true if you square both sides

I thought performing an operation on both sides of an equation left the equation unchanged

or rather doesn't affect its consistency

negative numbers are confusing

In just found out. ln(1+2+3)=ln(1)+ln(2)+ln(3)

@ペガサスSeiya you're on fire

@Koro oh I sincerely apologize if it conveyed any disrespect, I just found it useful to contemplate , its a thing I tend to do in general, because there's often valuable meaning in trying to figure out what thought process lead to a technically incorrect answer (I make no claims that what I postulated before was what led to anything, its just an error I and a lot of people I know comitted before learning AT so I was fairly confident it could be why)

before and after of course, at least I tend to recommit errors if I dont keep in touch with certain areas of math

also sometimes some thought process that leads to a technically incorrect answers can be tweaked here and there to be correct under some additional hypothesis, idk, i feel like this is the flora and fawna of 'mathematical thinking' if such a thing were to be defined

yeah, its spelled fauna :(

and commit

so actually it wasn't because the answers shocked me or anything, i think its fairly rare that any math that broaches this sort of stuff can ever really be called 'intuitive' in common parlance, it may be to some folks, but it seems more likely to be the case that people confuse high familiarity with something based on experience with conflating some problem that they can see the answer for quickly with being "easy", and then 'intuitive' kind of loses meaning

this definitely is a thing for professors, a lot of them may end up teaching some undergrad courses and carry out certain steps that involve invoking theorems, and its often the case they wouldn't be able to give you more than an intuitive explanation for why those theorems are true (as in something very far from a sketch leading to a complete proof) - normally of course this happens if the course is pretty far removed from their research specialization

and you'll often here phrases like 'this result is trivial because its just an immediate consequence of insert theorem that is a staple of mathematics here'

@ペガサスSeiya ln(4+5+6)=ln(1)+ln(3)+ln(5)

More generally, $\ln((n-1)+n+(n+1))=\ln(1)+\ln(3)+\ln(n)$

ln(190+191+192+193+194+195) = ln(1)+ln(3)+ln(5)+ln(7)+ln(11)

ln(4287+4288+4289+4290+4291+4292+4293)=ln(1)+ln(2)+ln(3)+ln(5)+ln(7)+ln(11)+ln(13)

Try $\ln(\sum_{n=26316}^{33744}n)$

I've been using prime factorizations because they work nice with sums of logs. I don't know what Gamma do

$\ln(\sum_{n=26316}^{33744}n)=\ln(1)+\ln(2)+\ln(3)+\ln(5)+\ln(7)+\ln(11)+\ln(13)+\ln(17)+\ln(23)$

I might have messed that one up

Fixed: $\ln(\sum_{n=26316}^{33744}n)=\ln(1)+\ln(2)+\ln(3)+\ln(5)+\ln(7)+\ln(11)+\ln(13)+\ln(17)+\ln(19)+\ln(23)$

Yeah, you get the prime factorization of the average of the numbers summed plus prime factorization of the number of numbers summed

Also, $\ln(1)=0$, and $\Gamma$ is a generalization of the factorial function

Gamma must explain why graphing $x!$ on Desmos doesn't make sense when the thing seems continuously increasing for $x\gt 0.463$. That never made sense. In fact, it still doesn't, why is there a local minimum at $(0.462, 0.886)$? Are these rational numbers?