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5 hours later…
04:56
Let $G$ be a finite group.

$$
\Bbb{Z}[G] \xrightarrow{\pi} G^{\text{ab}}, \\
\pi(\sum_{g \in G} x_g g) = \prod_{g \in G} \tilde{g}^{2x_g},
$$
is a group hom, where $\tilde{g} = $ the coset of $g$ in $G$'s abelianization, $G^{\text{ab}}$. $\ker \pi = \{ x \in \Bbb{Z}[G] \ | \ \prod \tilde{g}^{2x_g} = \tilde{1}\}$, but,

$$
\tilde{g}_1^{2 x_1} \cdots \tilde{g}_n^{2 x_n} = \tilde{1} \iff \\
\tilde{g}_1^{x_1} \cdots \tilde{g_n}^{x_n} = \tilde{g_1}^{-x_1} \cdots \tilde{g_n}^{-x_n} \iff \prod_{i = 1..|G|} \tilde{g_i}^{x_i} = \text{ its inverse}
05:32
@ShineOnYouCrazyDiamond what you wrote is not correct if $G$ is not abelian, remember that the inverse of $ab$ is $b^{-1}a^{-1}$
nevermind, it's correct, I'm dumb
but it's not too hard to cook up an example where the inclusion is strict: let $G$ be any finite group such that the kernel of $G \to G^{\mathrm{ab}}$ contains an element $a$ that has order at least $3$, then $a \in \ker{\pi}$, but $\mathrm{support}(a)=\{a\}$ is not a self-inverse subset of $G$
one can take for example $S_n$ with $n \geq 3$, here the abelianization has order $2$, so there are plenty of elements like that
06:06
Let $A=k[x,y,z]/(xy-z^2)$ and let $\overline{x},\overline{y},\overline{z}$ denote the images of $x,y,z$ respectively in $A$. Let $\mathfrac{p}=(\oveline{x},\overline{z})$. Is it correct to interpret $\mathfrac{p}$ as $(x,z)/(xy-z^2)$?
06:28
@MatheinBoulomenos so you suggest that that is not the kernels description?
I don't technically need a nice decsription, but it may come in handy later on
Let $x \in G^n$ for a finite non-abelian group, $x = (x_1, \dots, x_n)$. Can we somehow create a hom from $G^n$ into $\Bbb{Z}[G]$ that "has to do with" sums such as $x_1 + x_1 x_2 + \dots + x_1 \cdots x_n$?
Call such a sum $f(x), f : G^n \to \Bbb{Z}[G]$. It's definitely not a hom, but can we turn it into one say by doing $g(x) = \dfrac{1}{|G|} \sum_{g \in G} g f(x) g^{-1} - \text{ something} $ ?
If we do that exact expression, then $g(1) = 1$ so "Something" should be $1$ there.
It reminds me of fourier transform for finite groups, except I'm not working with complex numbers. I'm trying to keep this in $\Bbb{Z}[G]$ context
I mean $\dfrac{1}{n|G|}$ in that sum
I do know that when $x_1 + \dots + x_1 \cdots x_n = \sum_{g \in G} g$ then we have a hamiltonian cycle in the Cayley graph. That's why this is interesting.
So this hom $g$ would map into the kernel talked about above, and the homology at $\Bbb{Z}[G]$ module would measure something about Hamiltonian cycle problem in Cayley graphs, hopefully! :D
It's all speculation for now, but I am liking the looks of this symmetrizing sum thing
Note that $f$ preserves conjugation of its argument, so equivalently the sum becomes with $f(g x g^{-1})$
I did notice that conjugation doesn't view all the symmetries of $f(x)$. For instance if $G \simeq \Bbb{Z}_3, G = \{1, a, b\}, ab = 1$ then $f(x) = f(a,b) = a + ab$ but conjuate all you want and you don't necc get $b + ab$ in general (though you might here). So I should also through in the cycles of $(a,b,c, \dots, z) = x$
If you have a Hamilton cycle presented by $(x_1, \dots, x_n)$ then all cycle perms of that are also a Hamilton cycle presentation.
This is interesting because I haven't seen the problem broken down algebraically like this before, even though it's the most natural way to work at it
$$
$$
$\psi : G^n \to \Bbb{Z}[G], \ \ \psi(x) = \sum_{\sigma \in C_n} \sum_{g \in G} g f(\sigma x) g^{-1}$ maybe
Divided by $n|G|$ of course
 
3 hours later…
09:32
Morning all
 
1 hour later…
10:38
Is it always true that given a mapping $T:X\rightarrow X'$ (where X, X' and their respective sigma algebras are measurable spaces) we have that: $$G\subset A$$ implies $T^{-1}(G)\subset T^{-1}(A)$?
11:09
Yes, but that has nothing to do with measurable spaces, it could just be a function between sets
@JoséCarlosSantos Why did you close my question. Is there some way you can reopen it?
13:04
Let $V$ be a countable dimensional vector space over $\Bbb Q$. Is it true that dual of $V$ is uncountable?
@Silent I think the answer lies in the first few lines of kconrad.math.uconn.edu/blurbs/linmultialg/dualspaceinfinite.pdf
13:24
@Silent algebraic dual or is $V$ a topological vs and you want the topological dual?
13:38
@s.harp oh! sorry i did not know there was something like topological dual. By dual of V i meant set of all linear transformations from $V$ to $\Bbb Q$.
@ÍgjøgnumMeg Thank you!!!
@Silent let your basis be indexed by $I$, any function $f:I\to \Bbb Q$ defines an element of the dual by $f^*(\sum_{i\in I} x_i e_i) = \sum_{i\in I}x_i f(i)$ (note that in such a sum only finitely many terms are non-zero) and in turn any dual element defines such a function in that way.
if $I$ is countable infinite there exist uncountably many such functions, in particular the dual is an uncountable vectors space over $\Bbb Q$, which is a countable field. Hence any basis must be uncountable
@s.harp That is what I am not being able to figure out: We are taking linear combination of only finitely many basis at a time, right? Hence, in other words, we are looking for set of finite subsets of a countable set, which is countable. Where am i wrong?
14:04
@Silent hello
14:16
hi, @SubhasisBiswas
14:29
S.harp, please correct me! I think I have misinterpreted something you said, but can't figure out!
14:48
Hi
15:30
@Silent I don't understand why you conclude that the elements of the dual may be indexed by a the finite subsets of the basis
16:23
Can the transposition of a matrix be implemented by matrix multiplication?
@user193319 you mean whether there exist matrices $A,B$ so that for any matrix $T$ you have $T^t= ATB$?
@s.harp Yes.
I want to show that transposition is continuous, and this is the first thing that came to mind.
It seems like $A$ and $B$ would need to be permutation matrices.
There are a couple of reasons why that is false, for example $\Bbb1=\Bbb1^T= AB$ hence $A=B^{-1}$, so if $T$ is self-adjoint you'd have $T=T^t= ATB$ and $A$ and $B$ must commute with every self-adjoint, meaning $A$ is proportional to identity (so $ATB$ cannot be $T^t$ for more general $T$)
however note that transposition is a linear map $M_{n\times n}\to M_{n\times n}$. You can show that linear maps are always continuous (on finite dimensional vector spaces, which $M_{n\times n}$ is)
Ah, very nice.
16:41
Hi everyone :)
I'm not sure whether I have the imposter syndrome or whether I feel bad about being genuinely inadequate :D
What's more is that I have so many textbooks for my PhD that I don't think I can read them all within my lifetime.
I want to do more category theory but there doesn't seem to be much in combinatorial group theory, at least insofar as I have seen these last couple of years.
I'm taking time off at the moment for my mental health. I feel as if I can't pick up where I left off, even though I was behind when I did.
Stick with geometric group theory and solve the Sofic group conjecture. Then you'll be world famous.
2
In short, whilst mathematics is as fun for me as it always has been, the experience of studying it at this level is taking its toll.
I know nothing about geometric group theory.
Well, combinatorial group theory is often considered a part of geometric group theory, so you know some.
Is there a better notation for antiderivative than $\int f dx$?
That makes me feel a little better, I guess, @user193319.
16:58
@Shaun I would definitely recommend learning some more geometric group theory (GGT). There's been a surge of interest in it and a lot of funding that was given towards traditional topological research is being redirected towards GGT, because (1) it is intrinsically interesting and (2) sheds light on those traditional problems of topology.
So, in terms of being a marketable mathematician, it wouldn't hurt to learn a little GGT.
Here's a nice introduction to the subject: amazon.com.mx/Geometric-Group-Theory-Introduction-Clara/dp/…
That book is simply fantastic.
17:12
Would Collins et al.'s "Combinatorial Group Theory and Applications to Geometry" suffice? I have a copy of it. Also, as to the book you recommend, I found what I think is a free 2010 copy.
@user193319, see above, please :)
Anything written by Grigorchuk should be really good, although I haven't read that specific book.
For the most part, the free version you found follows the "printed" version, but I'd definitely recommend getting the printed version at some point.
Alternatively, you can use this site libgen.is/…
17:57
@SirCumference Not really. Many of us will define $F$ to be an antiderivative of $f$, reminding folks that this means that $F'=f$.
18:36
Hello all!
I need to find the area delimited by the curves: $y=1-x^2$, $y=1/x$, $y=x+2$
However I am not able to find a bounded region to find the area nor the intersection between these 3 curves. Any help? Thanks!
@manooooh You mean total area (so from $-\infty$ to $+\infty$)?
@Mr.Xcoder the exercise is the same as I have posted here
Ah wait, nevermind, I misread your question.
In this case, I don't think the question makes such sense. Unless you involve the axes as delimiters too, I really don't see any bounded region either.
@Mr.Xcoder ok. I guessed it too. Thanks for your time!
Well, haven't done much, so np :)
19:32
May I know why "https://math.stackexchange.com/questions/3340332/showing-g-in-g-gn-1-is-a-cyclic-subgroup-of-cyclic-g-of-order-n-for" is made duplicate? Can anyone help and check it as my points are pretty low to turn it down.
20:08
^ seems to have since been addressed
20:37
Hi @Semiclassic
ABC
ABC
Hi guys
Hi ABC
ABC
ABC
I'm calculating this integral $$\int_{0}^{+\inf}\frac{1}{(1+x^2)(x^3+1)} dx$$
Not fun, but there's a predictable method.
ABC
ABC
I'm going to use complex number
20:43
Residues, OK.
ABC
ABC
in particular I whant to use Residue theorem
yes, man
Now I'm going to explain my problem
It's not obvious how to proceed.
ABC
ABC
I'm using this curve:
I don't think that's going to work.
ABC
ABC
Why?
20:45
Why do you think it works?
ABC
ABC
Very simply because we have not studied so many cases in our course. This seemed to me what could come closer.
I think you'll find that the integrals cancel. And what do all the residues add up to?
ABC
ABC
yes infact I get 0
I also tryed to use that metod
So you KNOW that this contour won't work.
ABC
ABC
Yes but i don't understa what can I do
understand
20:48
If the function were $\dfrac 1{(x^2+1)(x^4+1)}$, what contour would you use?
ABC
ABC
I don't know
You should have seen things like this for sure in your course.
Think about symmetry of the function.
ABC
ABC
I can solve all integral like that: $$\int_{-inf}^{+inf} \frac{P(x)}{Q(x)} dx$$ with P,Q Polynomial expressions with Deg(q)-Deg(P)>1 and Q(x)!=0
Curve in that case is a semi-circumference

I can solve all integral like that: $$\int_{0}^{+inf} \frac{P(x)}{Q(x)} dx$$ with P,Q Polynomial expressions with Deg(q)-Deg(P)>1 and Q(x)!=0 and P/X not pair
Curve is the "PACMAN" (I sent you the photo above)
so here's something a bit funny
i've been looking for sources on a certain topic (Hilbert space of random variables) and there's a chapter I've seen
20:55
So, what does this tell you about my question?
while looking up the references, I noticed that the authors (in 1997) were UMN math profs :)
No, pacman doesn't always work like that.
they're now emeritus
Oh, interesting, @Semiclassic. Are they emeritus and alive?
ABC
ABC
Also i can solve that integral: $$\int_{a}^{b} \frac{x}{x^3+a^3} dx$$ with the pacman curve and $f(z)=\frac{z*Ln(z)}{(z^3+a^3)}$
20:56
emeritus, yes. alive...hopefully?
@ABC ... you're just babbling.
I want you to think about my specific question.
i didn't recognize the first prof for sure, but I think I may recognize the second
@Semiclassic: I was going to say that if you have an interesting question, they might like to hear from you.
yeah, hopefully
ABC
ABC
@TedShifrin I'm not able to answer you
20:58
I told you to think about the symmetry of the function. Did you do that?
if nothing else, I would hope they could pin down the history a bit better
the book is also one I think I should try to read this semester simply to get a handle on measure-theoretic probability
Have you taken a grad course in Lebesgue integration theory, Semiclassic? You'll need to know the standard stuff.
no, and that does worry me
Yup. It's definitely a prereq.
ABC
ABC
Ok function is pair
20:59
in English, you say even.
So what is the approach?
ABC
ABC
so can I think of it from -infinite to +infinite?
Right and then do what?
ABC
ABC
Use the first type ? So semi-circumference?
Yes, and get the answer to the original question by doing what?
ABC
ABC
Making the sum of the residuals ignoring those that are not in my curve
21:02
No, no, that's not right. But how is $\int_0^\infty$ determined from $\int_{-\infty}^\infty$?
ABC
ABC
ah, divide it
@TedShifrin sup
ABC
ABC
by 2
@TedShifrin the book's prefaces states: "The prerequisite knowledge needed is an understanding of elementary linear algebra and advanced calculus. Measure theory is introduced as needed mostly within the first nine chapters." so i might still be able to get something out of it
OK.
heya @Leaky.
Oh, unusual, @Semiclassic.
ABC
ABC
21:03
@TedShifrin I took it for granted, I'm sorry
@ABC: Now let's think about what we can say about symmetry in your original question.
well, the first line of the preface is: "This book is intended as a textbook in probability for graduate students in mathematics and related areas such as statistics, economics, physics, and operations research. "
so they've evidently got an applied audience in mind as well as a theoretical one
@TedShifrin what is a point? does f(x)=x^4 and g(y)=y^2 count?
21:04
Semiclassic, Lebesgue stuff is particularly important for applied people :P
pffffffft (yeah fair)
@Leaky: I don't understand your question.
I more meant that that might explain why they chose to structure the book like that, i.e. not assuming prior exposure to measure theory
@TedShifrin I'm referencing the picture above that I sent two messages earlier
I saw that.
But the question you typed means nothing to me.
21:05
"exactly two points"
I am beginning to fall in love with the whole "Hilbert space of random variables" thing
I'm proposing a solution where the first morphism is f(x)=x^4 and the second is g(y)=y^2
I can't work it all out in my head, Leaky. What are you alleging are the two points of intersection?
I mean, it's a simple enough idea: $\langle X,Y\rangle :=E[XY]=\int X(\omega)Y(\omega)\,dP(\omega)$ is just the $L^2$ inner product w/r/t the probability measure
But it makes a lot of the formulas I know just a matter of geometry
ABC
ABC
@TedShifrin It is not even. It is not odd.
21:07
@ABC: So what stops that same approach from working with your original question?
Aha.
What contour would you use if you wanted to work JUST with $1/(1+z^3)$?
@TedShifrin (0,0) with multiplicity 2
ABC
ABC
@TedShifrin I don't know
@Leaky: My interpretation is that they intend two different points, not counting multiplicities.
the other thing I'd like to do this semester (which sorta dovetails with the above) is see if I can start something like a physics-math conversation meeting
the physics and math buildings are right next to each other, but afaik there's not much cross-talk.
@ABC: You should know the answer to this. On what curve is $z^3$ the same as it is on the positive $x$-axis?
21:10
that's not really a surprise but it's always seemed like a missed opportunity
@Semiclassic: Several of my closest friends were physicists, and we had a fair amount of interaction between the departments because of particular faculty.
yeah, doesn't seem like there's a lot of that here
ABC
ABC
@TedShifrin I can't understand where you want to get me, sorry
I do not understand
This is something very fundamental, @ABC. How do you solve $z^3 = 1$?
the place I'd start (and not only because of my interest in that probability book) is the prob/stats people
simply because so much of my experience was in statistical physics
ABC
ABC
21:14
ah Ok are you asking me the solution of that equation? Do you want me to use partial fraction decomposition?
Statistics was in a separate department at UGA.
What happens from here to here?
NO, @ABC. I'm asking you to understand basic complex numbers and arithmetic.
Type in: "-x/e^(2x)" at
https://www.integral-calculator.com
ABC
ABC
$z=e^{i\pi/3)}$
$z=e^{i\pi}$
$z=e^{i5\pi/3)}$
21:15
@Sebastian: You're showing a photo out of context.
That's why I linked to integral-calculator. It is the third last line at that integral-calculator.com
so much noise
Well, almost, @ABC. $z=1$ is one solution, but $z=e^{2\pi i/3}$ is another (and, yes, its conjugate is too).
ABC
ABC
yes ok, i don't understand why you're doing so though ..
So does that give you a clue about where the function $1/(1+z^3)$ has the same values as it does on the positive $x$-axis?
21:18
Coding is hard today
Trying to get my debug env to launch from command line
This is for BananaCats, which I forecast will be ready in 7 months
ABC
ABC
@TedShifrin I think my poor English is making me miss what you mean.
hi @TedShifrin
@Sebastian: It looks like the first integral is what you get when you do the problem with no minus sign in front. And then you take the negative of the answer.
@ABC: You told me that $z^3$ has the same value at $z=1$ and $z=e^{2\pi i/3}$.
What about at $z=2$ and ???
@TedShifrin I missed you :)
hi @Adeek
Ha ha.
21:21
@TedShifrin ?
I want to update you with information that happened to me recently
I'm in the middle of stuff now. Crazy in here.
ABC
ABC
@TedShifrin $z=2^{1/3}*e^{2πi/3}$ ??
Okay I will pm you on facebook @TedShifrin ?
Just email, Adeek. That's better.
Okay
21:23
@ABC: Well, you made it harder. I just wanted $2$ and $2\cdot e^{2\pi i/3}$. So the value at any real $a$ is the same also at $a\cdot e^{2\pi i/3}$.
What contour does this make you think of doing?
ABC
ABC
oh yes sorry 2*...
not 2^1/3
Honestly I don't have an intuitive view of everything so I honestly can't think of anything
I only know that by doing the complex integral on the curve the real integral I was looking for returns to me. This allows me to use this "theory".
Are you taking a course in complex analysis?
ABC
ABC
I don't know which is the intuitive idea that allows you to create curves on which to make complex integrals above and determine real integrals.
The whole point is to think about symmetry, as I'm trying to get you to do now. But I think this problem is more complicated than you will understand.
ABC
ABC
Yes, but nobody here explains intuitively how it came to this. They just show you that if you do the complex integral on every piece of the curve you get back the real Integral.
21:28
But it scares me that you won't answer my question. Can't you see that we should multiply every number on the positive $x$-axis by $e^{2\pi i/3}$ and get another ray (line going from $0$ to $\infty$)?
ABC
ABC
Yes I see it
It seems to me you are trying to just memorize lots of rules you've been told without understanding anything. This is a very bad idea.
ABC
ABC
I see it as a pure arithmetic construct
So what contour might you try that would take advantage of this symmetry?
ABC
ABC
I explain to you
21:29
Well, it's actually geometric, too, since we understand complex multiplication geometrically.
ABC
ABC
In our course we have seen 3-4 case studies.
We have shown them and ended. No one made us understand how we can build curves ourselves to solve integrals!
So for the simple cases I understood the demonstrations. But the intuitive idea of why it is obvious to do such a thing has none in my course.
I apply the definition of complex integral on the specific curve (which we students do not derive) and I can prove that everything returns.
Even I don't appreciate this at all, it's the course's fault!
Maybe you should ask your professor how to approach this problem.
I emailed you with my updates Dr Shifrin.
ABC
ABC
@TedShifrin I assure you that I wouldn't mind figuring out how to visualize this, do you think you can tell me in a few minutes?
I have told you the key idea, @ABC, but then it gets a bit tricky because we haven't paid attention to the $1+z^2$ factor yet.
ABC
ABC
21:36
on the $1+z^3$ i can se every $f(z)$ as $z^3*e^{2ipi/3}+1$
You forgot to take $z^3$.
No.
It might also help to write something like $\omega = e^{2\pi i/3}$ ...
@TedShifrin I hope Yau is still alive by the time I finish my school.
ABC
ABC
@TedShifrin I thank you for your availability, I won't let you waste any more time. Good evening!
I guess mathematicians tend to stay alive for long time
Take care, @ABC. Best to talk to your professor about how to proceed.
@Adeek: I'll just say this was never my favorite person in mathematics.
21:41
Really I heard good things about him in a conference that I have attended recently. One prof that works in Harvard was speaking very highly of him.
I'm not going to discuss it here. There is a huge amount of politics, back-stabbing, and game-playing with prominent Chinese mathematicians.
I see. I personally want the Chinese culture in my home country was thinking of talking to him about that in the future of sending students from my country to China. Can you send me this information over email. I am interested to know.
Hey @Ted
hi @ÍgjøgnumMeg.
How's life!
21:46
I see.
Have you moved yet, @ÍgjøgnumMeg? No, it must be still too early for the European schedule.
@Ted Right, I'll be moving on the 30th of September :)
Some places school's been going for 3 weeks here already.
I have 2 weeks left at my job and then I have to clean and empty my flat and then move to my parents' place until I go to Germany lol
brb
21:48
Wow that's early
Make sure you clean all the cigarette soot!!
nice seeing you here @MatheinBoulomenos
Indeed :(
21:49
LOL, OK, I'm done with that. :)
:D On the plus side, I have found a place to live in Mannheim which is 18km away from the university
so I can get lots of miles in on the bicycle
Probably not fun in the dark or snow, but a good plan to make you quit smoking :P
Oops, not quite done.
I cycled a lot in the dark and snow in the UK so it's fine hahaha
Well, all power to you :P
Seems I stumbled upon nerd chat once again
22:04
ha
waddup
Not too much, classes are gonna start soon so I'm super hyped
niiiice
I got 6 weeks D:
what are ya taking?
At the moment it's looking like AG, rep theory, and maybe AT or something
Niiice :D
Analytic NT is being offered but it's less my thing, topics in wombo combo sounds fun but the professor said that it'll be harder to appreciate without knowing Lie theory (even if he won't assume it)
So as a result I'm prob taking one of the intro classes as my third. There's intro algebra, analysis, and topology
Leaning toward topology since that's prob my shakiest of the three? I'll see for sure soon
22:14
wtf is wombo combo hahaha
that's sensible tho
Combinatorics
ahhh okey
I am taking an intro topology course too tbf
slash seminar
Combinatorics to combo, and then that leads to: youtube.com/watch?v=pD_imYhNoQ4
hahahaha
I wanted to take diff top too but it clashes :(
Rip
First semester AT seems to do a bit of difftop as well
Manifolds, pi_1/covering spaces, and homology first semester
Maybe I can bully the prof to cover more, we'll see
22:20
@ÍgjøgnumMeg I think you can't complain with ANT, modular forms and the minicourse on Iwasawa theory
hahaha I ain't complaining
Yeah if anything I should be jealous
But yeah what are you planning during these 6 weeks?
very excited!
Also yo Mathein whaddup
err I finish at my job in 2 weeks, then I have to move out and stuff and live at my parents' place for 2 weeks so I'm hanging with family and stuff, and then I move to Mannheim on the 30th, am in Bonn from the 1st until the 2nd of October, and then I'll be cramming some complex analysis and attending and Iwasawa theory minicourse in the 2 weeks before classes start
and hey @Mathein :)
22:25
whaddup @Daminark
and hey @ÍgjøgnumMeg
22:36
That sounds fun for sure
Just had my TA orientation stuff
nice, how was that?
It was aight
22:56
@MatheinBoulomenos hi

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