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12:09 AM
i cant even find the rule
Oh is it supposed to be per page?
Not one page vs other
 
12:29 AM
Is it possible to visualize $\Bbb S^3$ in a way that it is obvious that it embeds in $\Bbb C^2$?
 
is there a way to visualize so that it isn't obvious?
It's z_1\overline{z_1}+z_2\overline{z_2}=1
@Akiva what can't the computer understand?
It seems pretty easy to give a computer an algorithm to solve that problem.
There's like 4 things it needs to recognize and then theres a path between them.
 
@PVAL I interpret it as being more, most general algorithms for computer vision have a hard time teaching it that
Also I don't even know the rule
 
white disk is closer to the apostrophe, black disk is closer to the diamond thing.
 
Ah
 
on the immersed curve between the white and black disk.
 
12:43 AM
Left: White, curl, diamond, black. Right: White, diamond, curl, black.
(@io_cantor)
 
ah
 
Also @io_cantor regarding compactness, I've found Rudin to be reasonably good
 
Rudin is certainly compact.
My Rudin is bounded and for a quite a long time it has been closed.
 
Strange; mine has an open cover
 
haha good ones
Rudin is defintely closed
 
12:57 AM
I like the proof of the extreme value theorem for functions on compact sets
 
Lolol
 
Exercise: Provide two disjoint closed sets in $\Bbb R^2$ whose distance is $0$.
 
@Daminark @AkivaWeinberger any idea?
 
Also, prove that two disjoint compact sets have nonzero distance.
 
does anyone here know a way to translate $n < n_0$ to $b^n > c(a^n)$, where I can choose the value of n?
 
12:59 AM
@Leaky an idea about what?
 
@LeakyNun Well it's $\{(w,z):|w|^2+|z|^2=1\}$, isn't it?
 
I feel like its a simple solution, but i'm getting stuck
 
@AkivaWeinberger two spirals chasing each other?
@AkivaWeinberger but that is essentially 4 dimensional
 
Howdy, DogAteMy, Demonark, Leaky, et al.
@Theo, offhand they seem totally unrelated.
 
@LeakyNun So is $\Bbb C^2$
 
1:00 AM
but S^3 is not
 
It's a 3-dimensional object sitting in 4D space
 
$S^3$ is a $3$-dimensional manifold sitting in $4$-space.
 
Write $w=a+bi$ and $z=c+di$
 
@Akiva the second one should be about letting $d(x,A)$ to be a continuous function, and its restriction to $B$ is uniformly continuous
 
Why is uniform continuity relevant, Demonark?
 
1:00 AM
What I wrote is essentially be $\{(a,b,c,d)\mid a^2+b^2+c^2+d^2=1\}$
The set of things in 4-space unit distance from the origin
 
@TedShifrin Yes, but if you do some log manipulation I believe you can get somewhere
 
Oh it probably isn't actually, it's just a continuous function on a compact set so it attains the minimum
 
@TedShifrin i just keep getting stuck
 
There must be more information than just the one line I read, @Theo.
 
@Theo Do we have $a<b$ or something like that?
 
1:02 AM
And why is $n_0$ relevant?
 
ohhh yes i complete forgot to mention
 
The first might be taking a curve representing exponential decay and then flipping it across the x-axis
 
you're correct, b > a and a > 1
 
@TedShifrin We can choose the value of $n_0$ I think
 
the thing is, how to visualize the group action of S1 on S3?
 
1:03 AM
and then it should be true for all $n<n_0$
@LeakyNun It's a bunch of tori inside of each other
 
Then $b^n>a^n$ for all positive $n$?
 
@AkivaWeinberger @TedShifrin b > a and a > 1 and we can choose the value of $n$
n_0, n \in \mathbb{N}
 
@AkivaWeinberger I mean, how do you view S3 at all?
 
If $b>a\ge 1$, then all positive powers of $b$ exceed the power of $a$.
It's true for all $n$.
All you need is $b/a > 1$.
 
R^3 @LeakyNun
and there's an infinity as well
 
1:04 AM
Then $(b/a)^n > 1$.
 
@AkivaWeinberger and then how do you multiply by a number...
 
@LeakyNun First you need to see how $S^3$ is made up of lots of tori (plus two circles)
Well, actually
 
Hi Faust.
 
Did you learn that if you glue two solid tori together along their boundaries you can get $S^3$
 
@AkivaWeinberger :o
 
1:06 AM
That is, the complement of a solid torus in $S^3$ is another solid torus
 
is that even possible lol
oh wait, I can glue two solid circles together to form S2
ok I can believe that
 
balls/disks
 
thanks
 
hi @TedShifrin
 
oh, I see, DogAteMy
hi Karim
 
1:08 AM
It is very cool how things work out very nicely in complex analysis
 
@LeakyNun Say you have a torus. You know how you can parametrize it by $S^1\times S^1$?
 
I guess the notion of holomorphism is very very strict thing
 
@AkivaWeinberger yes
 
What are you up to now? @Adeek
 
Varying one coordinate moves you around longitudes. Varying the other moves you around meridians.
Varying them both at the same time gives you this circle that goes through the whole and around it also I guess
Just a sec
 
1:09 AM
if I draw a circle on CP1 and take its pre-image in C^2\{0}, what would I get? Would I get the same thing if I draw a line instead?
 
(Cont'd) Varying them both at the same rate gets you one of these circles
 
@Daminark many things.. So I am just making sure my complex analysis knowledge is good, also reviewing stuff in differential topology. Then I am working mainly on complexity of the higher chow group.
 
@AkivaWeinberger ok
 
@Daminark My main goal this year 2017-2018 is get as much knowledge as possible in complex geometry and strengthen my foundation in algebra and analysis (I have gaps due to my university). Also to finish my master thesis.
 
I dunno much about the last bit but complex and difftop are pretty sweet. Today our professor did the inverse function theorem and then talked a bit about whether there's a biholomorphic function between various subsets of $\mathbb{C}$
 
1:12 AM
@LeakyNun So like say we have $(z,w)\in(S^1)^2$. Then $(e^{i\theta}z,e^{i\theta}w)$ traces out a circle that goes through the hole and also around it as $\theta$ varies. Do you see how that works?
On the torus ($\!{}=(S^1)^2$).
 
yes, I see
 
yeah
 
Right. So back to the picture of the torus in $S^3$, which separates the space into two solid tori
 
For example, impossible to find one from $\mathbb{C}$ to $B(0,1)$ because the inverse would be bounded, and thus constant. Also you can't get the punctured open ball and the open annulus because you can extend the map to the whole open ball, but then by open mapping theorem, the image is still in the open annulus, which breaks it
 
Each solid torus can be filled with infinitely smaller tori plus a circle
 
1:13 AM
@AkivaWeinberger it's essentially the loop that is $ab$ right
 
yeah that is nice example
 
He mentioned a theorem that's actually really cool but I don't think he's gonna prove it
 
where $a$ and $b$ are the generators
 
@LeakyNun Yeah
 
@Daminark what is it ?
 
1:14 AM
It says that if you have a simply connected proper open subset of $\mathbb{C}$, there's a biholomorphic function from it to $B(0,1)$
 
Each solid torus can be filled with infinitely smaller tori plus a circle, as you just shrink the thickness of the boundary torus
 
i see
 
I think it's called the Riemann mapping theorem
 
At zero thickness you get the circle
Each torus itself is the union of infinitely many $ab$-type circles
(Take each point on it and multiply it by $e^{i\theta}$ like before)
 
yeah
 
1:15 AM
ok @AkivaWeinberger
 
For everything in that solid torus in $S^3$, multiplying by something in $S^1$ moves it along the circle it's on
 
@Daminark have you learned about immersions and submerssion ?
 
@AkivaWeinberger hmm
 
If you stereographically project $S^3$ onto $\Bbb R^3$, the circle at the center of the "outer" solid torus turns into just a straight line
 
In difftop yeah, the derivative being injective vs surjective
 
1:17 AM
A note:
What that is is, on the bottom-right he chooses a point on $S^2$ (which is $S^3/S^1$),
 
Ah, I know Niles. He was a postdoc at UGA. :)
 
and in the main screen you have the preimage of that point
(so the preimage of each point is a copy of $S^1$, a circle)
He has stereographically projected $S^3$ onto $\Bbb R^3$ (and then smooshed $\Bbb R^3$ into the unit ball to make it easier to see)
 
but now how do you show that S^3/S^1 is S^2?
 
(The circles should be geometrical circles, but the smooshing messed that up)
@LeakyNun Algebraically, by showing it's $\Bbb C^2\setminus\{0\}/\Bbb C=\Bbb C\rm P^1$, and then showing that the later is $\Bbb C\cup\{\infty\}$, the Riemann sphere
(The relation between $\Bbb C\cup\{\infty\}$ and a sphere is stereographic projection)
 
@AkivaWeinberger so essentially the diagram I drew yesterday
 
1:21 AM
@LeakyNun Yeah
By the way, the same thing can be done with $\Bbb H$ (the quaternions) instead of $\Bbb C$
You end up with $S^7/S^3=S^4$
 
oh
 
@Daminark how was it presented ? was it presented as the differential map is injective/surjective ?
 
Yup
 
how is $S^7/S^3$ defined?
 
cool
 
1:23 AM
Similarly, using $\Bbb O$ (the octonions) gives you $S^{15}/S^7=S^8$
 
actually when things like a looking at the figure cross i.e singularities then you start getting into algebraic geometry.
 
@LeakyNun Specifically, what I mean is there's a map from $S^7$ to $S^4$, where the preimage of every point is a copy of $S^3$
 
oh I thought it was a group action
 
But the actual map is the same as with $\Bbb C$, only with quaternions instead
 
oh ok
 
1:25 AM
@LeakyNun I think it would be a group action, yeah
$S^3$ is the unit quaternions, $S^7$ is the unit stuffs in $\Bbb H^2$
But I think it's technically not a group action for $\Bbb O$ 'cause of the lack of associativity? Not sure
In any case, you can't continue to $S^{31}/S^{15}=S^{16}$. There's no map that'll work
So it ends there
 
@Adeek I see
 
See the part of this article:
In the mathematical field of differential topology, the Hopf fibration (also known as the Hopf bundle or Hopf map) describes a 3-sphere (a hypersphere in four-dimensional space) in terms of circles and an ordinary sphere. Discovered by Heinz Hopf in 1931, it is an influential early example of a fiber bundle. Technically, Hopf found a many-to-one continuous function (or "map") from the 3-sphere onto the 2-sphere such that each distinct point of the 2-sphere comes from a distinct circle of the 3-sphere (Hopf 1931). Thus the 3-sphere is composed of fibers, where each fiber is a circle—one for each...
where they say "There are numerous generalizations of the Hopf fibration."
 
We actually had a math club talk recently where someone was talking about finding the arc length of the cubic, elliptic integrals, and algebraic geometry, though I had a hard time following
 
@LeakyNun
 
I gave undergrad talks at Berkeley and MIT on Abel's Theorem ... generalizing addition laws for sin and cos, to the cubic ... in terms of very concrete stuff. Really cool.
 
1:29 AM
Heh, $S^1/S^0=S^1$ (by folding it over)
($S^0=\{-1,1\}$)
 
or is it two of 'em, DogAteMy?
 
Oh, you're modding out by the subgroup, not the points.
 
Yeah
@LeakyNun, Balarka knows a lot about this stuff as well
He might be able to explain it better
(@BalarkaSen)
 
@Daminark very cool
 
1:32 AM
Whoops I meant arc length of the ellipse but that's pretty sick too
 
very cool @TedShifrin
 
Apparently the structure of the 120-cell is similar to the Hopf fibration
 
I wish I attended same university you taught at @TedShifrin
very cool @TedShifrin
 
Ha ha, Karim. You were spared.
 
I love to work hard if the professor can explain what he is doing :D
 
1:37 AM
Well, I'm heading out. Keep up the lessons, DogAteMy :) You're doing great.
See ya, Karim, Demonark, Leaky.
 
See you!
 
cya
Hey @Daminark
have you seen the proof of Lioville theorem ?
 
Is that the one what says bounded stuff can uh
not be uh
not constant
I have good words today
 
Yeah an entire bounded function is constant
the proof is easy using Cauchy integral formula
but I came up with proof different than the book
 
Remind me how the usual one goes
Oh wait so we have the integral formula
 
1:52 AM
So you know you can compute the derivative using integral formula right ?
 
and the integrand gets really small away from the point you want
 
Yeah
 
if it's bounded
 
r can be very close to zero
 
so you do a really wide contour?
 
1:53 AM
i.e $|f`(z)| <= \frac{1}{2\pi} * 2\pi *r * C$ for every r.
this is for every r though
 
Oh. You integrate $\frac{f(z)}{(z-w)^2}$ rather than $\frac{f(z)}{z-w}$, is what you're saying?
Which gets you $f'(w)$
 
yeah
using the generalized cauchy integral formula
 
and by doing a wide contour, one which the integrand is really small, you show it's zero
 
yeah
 
Yeah I've seen two proofs of Liouville. One using what you're talking about called the Cauchy estimates, and the other being just directly finding f(x)-f(y) = 0 by Cauchy
 
1:54 AM
So what's your proof?
 
Proof in the book is weird though
It is similar to mine but
he divides C/r^2
I don't know where he got the r^2 from
 
@Daminark Oh, like $\int\frac{f(z)}{z-x}-\frac{f(z)}{z-y}dz$?
Cool I guess I see it
 
$|f`(z)| \leq \frac{1}{2\pi} * \int_{\alpha} * C$
 
Yeah, you integrate both over a large circle
 
I don't know how he got $1/r^2$
ohh
oh yeah sure
 
1:56 AM
Well |z-s|^2 = r^2
 
Denominator of the integrand @Adeek
 
oh ok
 
With this you can either do it directly for the derivative to get that it's 0 or try to get a similar bound for all the terms in the power series
 
nvm
 
Lol rip
 
2:10 AM
@AkivaWeinberger Well, I just use the snipplet tool in windows on a blank space, and then use the doodling tools within
 
Ah. I don't have Windows
 
2:21 AM
@LeakyNun If I have a group $G$ where $|G|=n$, would it be the case that $|\operatorname{Inn}(G)| = n$ too? I am thinking this because one can construct one inner automorphism $\phi_g$ for every $g \in G$.
 
hi handsome folks
so i want to find how many 3 sylow subgroups they are in a Group of order 90
i see that 1 and 10 satifies the criterion
my question is , if we have more than one number that satifies the crterion, how do we determine what number it is?
 
Handsome, huh? Laying it on a little thick, there.
 
On a mac, I open pixlr.com, doodle stuff there, then use the mac screen cap, and then post the picture
If no pen strokes were needed, I use powerpoint to assemble images, shapes and then screencap and post it
 
@ALannister unnecessary
 
2:58 AM
@Argon Makes sense to me
 
@AkivaWeinberger Thanks (:
 
3:19 AM
@AkivaWeinberger But wait, isn't $$|\operatorname{Inn}(G)| = |G/Z(G)| = |G|/|Z(G)| \le |G|$$
 
3:41 AM
Question
actually no one's here
so nvm
 
3:55 AM
@Argon Oh wow you're right
Sorry
 
 
2 hours later…
5:28 AM
@Argon: Did you sort out the symmetries of the cube last night?
 
 
1 hour later…
6:43 AM
@TobiasKildetoft Tobias:D
 
:D
can you please tell me what an embedding is `?
I tried to google it and did not make much sense of it
 
embedding in what context?
 
well
i was trying to understand the proof of A_5 is simple
 
Hey @Kasmir, @Alessandro, and @Tobias!
 
6:44 AM
@Daminark Sup dami
 
@Daminark Hi
 
well
my teacher said something about
 
@KasmirKhaan In the context of groups an embedding is just a injective homomorphism
 
if we have a subgroup of index n
[G : H] = n , we let G act on G/H
this gives rise to permutation representation
G --> G _ G/H
@TobiasKildetoft thanks, what i wrote there also did not get it well
 
Hi chat
 
6:47 AM
G _ G/H since index of H is n, this is isomorphic to S_n
why from there also is the order of G must divide n!
many comfusing things :(
 
Hi @Dami
And everyone else
 
Now that you guys are here I may as well get your suggestions: so we've started Sylow today and will finish on Friday, leaving 4 classes left after that. I asked my professor what he had in mind, and he said "I was thinking some subset of mobius transformations, presentations (done properly this time), finite subgroups of SO(3), the (lack of) simple groups of small order. But open to requests."
Alessandro and Tobias, you two probably know a good bit more group theory than I do, are there any topics you think would be worth requesting over these?
 
7:03 AM
@Daminark Hmm, good question
 
7:19 AM
Yeah. I considered asking more about stuff like primitive actions and the like, though given this choice of topics and having heard him talk a bit, I'm getting the vibe that he does want to include geometry
Also because that's just the main other topic in group theory I have heard of
 
It also depends on how deep he wants to go. There is stuff like transfer and fusion which are pretty interesting
Especially fusion seems to be a pretty big topic currently, with some generalizations being written about fairly frequently
 
He may not be able to go too deep since we only have 4 classes left (much as we probably could've done what we did 2 weeks faster but oh well)
 
how much is a class?
 
50 minutes
 
ahh, not much time for anything then
how often do you have classes?
 
7:27 AM
The standard is either twice a week for 80 minutes or thrice a week for 50 minutes
The thing is, we're missing the next two Fridays (one for Thanksgiving and one for reading period)
And we've only got 2 weeks left in the quarter before exam week
 
7:47 AM
[Random]
I am specialise on what is known as dual trolling:
Given a chat room with one crank, one mainstream and one homework person, I ask question that give a chance for the crank to demonstrate they are not crank, but at the same time, twisting some content so that the mainstreamer will be confused and the homework person paranoid
If the crank failed to answer the question, then discussion ends
The underlying motive is that I am affiliated with the maintremer mostly, but to display an environment that I am not side with them to expose a crank
 
 
1 hour later…
 
1 hour later…
10:07 AM
can anyone please help me with: math.stackexchange.com/questions/2521236/… ?
i am stuck and i don't know how to show Hv=v
 
0
Q: choose correct statement ?10

kalom legoconsider the three subsets of $\mathbb{R}^2$ namely $$ \begin{split} A_1 &= \left\{(x,y)\left| x^2 +y^2 \le 1 \right. \right\}\\ A_2 &= \left\{(1,y) | y \in \mathbb{R}\right\}\\ A_3 &= \{(0,2)\} \end{split} $$ then there always exist a continuous real valued function $f$ on $\mathbb{R}^2$ such t...

I got $A_1$ and $A_2$ are connected.
so $a_1=a_2$
so 4 is true .Right?
How to judge the other answer?
Please help me
2 also true.
3. is false.
1. is true
Am I right?
 
 
1 hour later…
11:20 AM
@Argon $\le$
 
11:40 AM
0
Q: Reducing a non regular graph to a regular one.

user8469759I'm not a big expert in graph theory, I know basic algorithms, but I'm not quite familiar with proofs, here. I do wonder about the following (I came up with this problem, and I don't know whether or not is true). Let $G$ be a connected graph such that $\delta(G) = 5$ and $\Delta(G) = 6$, then the...

 
12:08 PM
0
Q: Commutative ring conjecture?

mickLet $A$ be a commutative ring where every element has a multiplicative norm $N$. So if $a,b,c$ are in $A$, such that $ab = c$ Then $ N(a) N(b) = N(c)$ ( So $N$ is a multiplicative homomorphism ) Let $B$ be a commutative ring. Usually we take $N$ to be integers. But When $N$ is defined as ele...

 
1:06 PM
1
Q: Prove $\exp: \Bbb C \to \Bbb C^\times$ is surjective

brotWe define the complex exponential function: $$\begin{array}{rcl} \exp:\Bbb C &\to& \Bbb C^\times \\ z &\mapsto& \exp(z)=\displaystyle{\sum_{n=0}^\infty \frac{z^n}{n!}.} \end{array}$$ I wan't to show that this map is surjective. My idea is to show that the real exponential $\exp|_\Bbb R$ maps...

 
1:38 PM
@Daminark As you mentioned that you're interested in Algebraic Number Theory, there's a book by Frazer Jarvis in the Springer Undergraduate Mathematics Series "Algebraic Number Theory" that develops almost all algebra needed from scratch (there is little that is just assumed, mostly the definition of a group, ring, field, vector space and their homomorphisms). I think you would enjoy it
Due to its elementary and hands-on approach, it doesn't treat some topics in full generality and some techniques appear ad-hoc from a higher point of view, but there's even some stuff missing from some of the standard references, like class groups of quadratic fields in terms of quadratic forms (that's how Gauss thought about class groups!) and factoring algorithms that use ANT. Plus there's solutions/hints for exercises, making it great for self-study
 
what would you understand as a "sampling property" of a function?
what does it mean to sample a function?
 
2:06 PM
snp
 
sup
 
svp
 
hmm
the chat froze
amazon.com
when I typed that
 
2:27 PM
When one writes $\displaystyle \prod_{n \ne 0}a_n$ does that mean $n \ge 1$ or $n \in \mathbb{Z}\setminus \left\{0\right\}$?
I have never seen a product that ranges all over the integers, but this notation confuses me.
 
if you randomly picked one instance of that notation out of every time it was used in math, but weren't allowed to read any context, it'd be best to guess it's $n>0$, but that doesn't mean you can't use it the other way, and context is always king
for instance all nonzero integers are used for sine's wallis product
actually I guess it's called Euler's infinite product, and the Wallis product is a corollary
in any case, even in that product, they group the terms with $n$ and $-n$ together to make sense of convergence
 
I think in this case it's a type of Euler product, I think.
$\displaystyle \sin \pi z = \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n} = \pi z\prod_{n=1}^\infty \left(1-\left(\frac{z}{n}\right)^2\right)$
Oh, I see.
I think what you said about grouping the $n$ and $-n$ together is happening here.
 
yes
 
2:43 PM
$\displaystyle \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n} = \sin \pi z = \pi z \prod_{n \ge 1} \left(1-\frac{z}{n}\right)e^{z/n} \prod_{-n \ge 1} \left(1+\frac{z}{n}\right)e^{-z/n} = \pi z \prod_{n \ge 1} \left(1-\frac{z^2}{n^2}\right).$
Works out nicely.
Thanks @anon
 
2:54 PM
Hey people, can you give me some hints! Weierstrass' substitution doesn't work1
 
may I ask a question about planar geometry?
 
3:29 PM
@Mockingbird hint: look at the bounds
in order to add them together, what must you do?
@user8469759 yes
 
3:44 PM
@user8469759 You can ask anything you want :)
 
Just wondering if
say I have any planar figure for which $V - E + F = 2$ (euler formula) is it true that the related graph is connected?
It can be actually a solid as well
I have some cases where I'd say yes trivially
like square
or triangle
or some figure where I use euler operators when I add a new vertex it also hold
say I have a square and I add a new vertex inside the square, and I add also arcs between the new vertex and all the vertices in the square
it's still a connected graph
 

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