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21:04
I just got suspended from chat for saying "I had a great shit". Who flagged me?
I dunno.
This chat flagging is getting ridiculous.
There are some not very nice people here who likes to backstab me.
I expect to be flagged again for saying why I got suspended, and I am prepared to be suspended again.
someone who is constipated probably
@r9m So, an integral like this one is naturally approached by means of double gamma function.
r9m
r9m
@Chris'ssis I see .. :)
21:15
@r9m let me know if you don't agree ... :-)
r9m
r9m
@Chris'ssis I have to think about it for a while :) .. but +1 to the double gamma function .. I understand it'll be ultimately at the heart of any real method that we might come up with :)
@Chris'ssis What do you get then?
@Alizter I didn't continue the work on it. :-)
@Chris'ssis Is it really childs play?
@r9m :-)))))
@Alizter It seems so at first sight.
@Alizter Wait, I wanna show to you the course to the previous one ...
(I refer to your question)
21:21
@Chris'ssis Ok
@Alizter there are 3 parts (take them). However I think I can optimize this proof though.
@Chris'ssis robjohns is amazing :P
@Alizter Yeah, true.
@Chris'ssis Your's it better than mine
which I still havnt finished
@Alizter Try to get something better. I'd be glad. As I said, my proof is not optimized. :-)
I might reduce it a lot.
21:25
@Chris'ssis My new one is nice at first site. Try to get through it a bit
@Alizter I think I can put all on a single page though.
@Chris'ssis The only problem I have with these inteagrals are the series that emerge.
It is my weak point
@Alizter I think you give up too easily, you need some more practice. I saw it the last time I talked to you: you said the series you got is hard and didn't continue the work.
@Chris'ssis Who said I gave up? ;)
No i am still trying at this one
the other one i gave up
oops
21:41
$$\sum_{n=1}^{\infty} \left(\psi^{(0)}(n)+\frac{1}{2}\psi^{(1)}(n)-\log(n)\right)$$
$$\sum_{n=1}^{\infty} \left(\psi^{(0)}(n)+\frac{1}{2}\psi^{(1)}(n)-\log(n)\right)^2$$
r9m
r9m
@Chris'ssis Cool !! :D
@r9m Glad you like it. :-)
Anybody seeming broken notifications? Someone tagged me in a comment on a question, and now every page I go to says I have a new comment, even after I've read the comment.
No issue here (so far).
r9m
r9m
@ThomasAndrews happens often for me .. but idk why .. I have a slow internet connection :|
21:49
@r9m It' s happened to me before, but this has persisted for 20+ minutes or so.
r9m
r9m
@ThomasAndrews never more than a few seconds for me .. I can't imagine the same happening for 20 long mins :O
@r9m Yeah, both are great! :-)
r9m
r9m
@Chris'ssis ya .. both cool :)) do you have a smooth solution for these ? :D
don't show me the solutions though .. tomorrow is sunday :) I wanna try 'em ::)
@r9m Of course :-)
r9m
r9m
@Chris'ssis I see that you put 'let's brilliantly compute .. ' before your proofs :) .. you can try 'let's smoothly compute ..' for the smooth one .. for a change :D
21:58
@r9m hahahaha, I liked that! :D
r9m
r9m
:P
lol, you have to see what I received by e-mail now ...
hmmm, why does it look like that?
I'm not sure. The answer is 17 isn't it?
Multiply the top and bottom left number, then add the bottom right one.
22:03
That was funny, it took me 2 seconds. What kind of math puzzle is this one? :-))))
@Khallil Yeah.
I'm not entirely sure. It probably just comes under fun math problems on Google.
I remember seeing a Facebook page that posts interesting problems like these daily.
I prefer some limits, series and integrals though.
Here, what does Lang mean exactly by 'the inverse image'?
@Chris'ssis In that case, this page has quite a few interesting questions to do with those. (facebook.com/maths.monde)
@Alyosha The natural homomorphism $\phi:G\to G/H$ means that for any subgroup $N$ of $G/H$ we can find the "inverse image" $\phi^{-1}(N)=\{g\in G: \phi(g)\in N\}$, which can be proved to be a subgroup of $G$.
22:16
@Khallil I see.
More generally, if $f:X\to Y$, and if $Z\subseteq Y$ we say the inverse image of $Z$ is $$f^{-1}(Z)=\{x\in X\mid f(x)\in Z\}$$ @Alyosha
@Alyosha Did you end up making it into Cambridge? ^_^
@ThomasAndrews I thought it was that, but how does that prove the statement?
I'm happy with basic functions, but just don't know how to change the tower starting at $G/H$ into that with $G$.
You get a downward tower in $G$ that bottoms in $H$, to which you can then add $1$.
@Khallil Yes. How did results go for you?
@ThomasAndrews Oh, so the inverse map preserves normality and so on between subgroups?
22:20
Yes, that's definitely something you'd have to prove, but it is true. @Alyosha
I thought it would turn into that tower, but didn't know how to justify my question above.
@Chris'ssis I get a divergent series :(
@Alizter Sorry :-(
or if I am naughty zeta(1/2)
@ThomasAndrews I can do that now, thanks.
22:28
@Alyosha Hallo! How're you doing with Alekseev?
I haven't touched it in a while, mainly doing other things.
I'd finished the first half.
And the second half had a few skippable things.
I am not sure what halves are you talking about.
Group theory and Complex things.
Yeah, you could skip them.
Just start with the Riemannsurfaces and do the exercises.
I've encountered RS before, so I skipped the first few RS ones.
22:32
@Alyosha I won't do that if I were you. From where have you studied it?
@Alizter Oh, by the way, your problem seems interesting. After a few hours of fiddling, I think deciding whether or not every group realizable as a galois group of some algebraic number field can be realized as a galois group of some real algebraic number field is a highly nontrivial problem.
From various places, but not a dedicated book.
However, I've just done covering spaces in AT, which has helped.
Can you draw me the Riemannsurface of $w = \sqrt[3]{z(z-1)}$?
@Alyosha You won't need much covering space theory while doing Alekseev. Just the intro.
Every Riemannsurface is more or less an "algebraic" branched covering of $\Bbb C$. In particular, this sense can be generalized, along with the concept of monodromy.
Well, that Riemann surface would behave like $z^{1/3}$ near $0$ and $1$, then glue them together?
I didn't skip quite past those sort of problems.
@Alyosha Glue what together?
Are you familiar with the sheet-and-glue way of drawing Riemannsurfaces?
I think so.
22:38
OK, let's start with something easier. Take $w = \sqrt{z}$
How'd you draw the Riemannsurface of this?
Yes, two sheets are glued along the branch cut from zero to infinity.
Three sheets for cube root.
@Alyosha How are they glued?
Why take the branch cut from 0 to infinity? Why not 0 to negative infinity?
Irrelevant, any infinity will do.
Right. And the nature of the gluing?
The right side of the $n-1$th is glued to the left of the $n$th.
Obviously $n$ is taken modulo $m$.
22:41
@Alyosha Why?
To ensure continuity.
Covering maps are continuous.
That is true, but show that there aren't any other ways of ensuring continuity.
Give the explicit calculations.
What do you mean by explicit calculations?
I'd have thought that there is a reason for gluing like that.
How is the continuity ensured?
Consider a point moving on a path?
I'm not sure what you're getting at.
22:44
@Alyosha Yes. Or more precisely, let a point traverse a loop around the neighborhood of $z = 0$.
See if it comes back.
OK fine.
What's the general method for more complicated functions?
@Alyosha Do you see that the point has winding number 2?
@ThomasAndrews That happens when they do a failover test, see here and the linked posts on Meta.SE.
And that there is a sense of cyclicity involved?
Yes.
I thought WN were associated with curves.
22:46
@Alyosha Excellent. That is precisely where the monodromy comes from. This Riemannsurface has corresponding monodromy $\Bbb Z_2$
What does monodromy mean?
I didn't get that far.
@Alyosha A rigorous definition is in Alekseev.
I'm just saying that there is a sense of group involved with the RS.
Is it the same as $\pi_1(R,0) $, where $R$ is the Riemann surface?
indeed, indeed, you'll soon come to that ;)
And $\pi_1$ is the first homotopy group.
22:49
@r9m let me know when you're done with the first series. I won't remain too long.
@Alyosha In fact this group coincides with the Galois group, but that needs some extra work.
For the Riemann surfaces of which functions does this occur?
You wouldn't want to care for the Riemannsurface, really. This sense of group can be attached to any covering map from a topological space to another topological space.
This is the so-called "geometric galois action" that I yet have to study.
By the induced homomorphism of the covering map?
@Alyosha homomorphism?
22:51
@r9m would you believe me if I told you I did it without pen and paper? :D
Indeed, it's for emphasis.
I'm using a foreign keyboard.
So how are more complex Riemann surfaces calculated, in general (not just those in the Abel book)?
Thanks @Daniel. Oddly, the comment notification stopped when I got some rep points, but the rep points notification won't go away now. :( - You must login to post
@Alyosha There is no general algorithm! For example, I saw something a few days ago that has an infinite-genus Riemannsurface. Devil knows how to draw them.
Hello @TedShifrin
22:59
@BalarkaSen So what is the rigorous definition of a Riemann surface?
It's not in the Abel book, and I wonder whether there's a non covering-map definition.
Not that I know of.
Though I suppose this is manifolds so it must be topology.
The only rigorous definition I know comes from covering maps.
@Alyosha I haven't studied manifolds, sorry.
@Alyosha Have you ever thought of mimicking Galois theory in the sense of groups? We have pretty much all that we want for mimicking : a structure, a sense of inclusion, automorphisms.
@BalarkaSen What do you mean by that?
Well, define Aut_A(B) to be automorphisms of B fixing A pointwise.
23:11
How would you mimick it?
I think I can prove that if N is a characteristic subgroup of H a characteristic subgroup of G then there is a short exact sequence 1 --> Aut_H(G) --> Aut_N(G) --> Aut_N(H) --> 1
^ That is what precisely happens with fields.
Oh, I see what you mean.
The galois extensions are probably mimicked by charecterisitc subgroups in this case.
Why is this used less, then?
I have never seen it used, let along less used.
The exactness probably follows by noting that since H char G, Aut(G) --> Aut(H) induced from restriction automatically gives the map Aut_N(G) --> Aut_N(H) which is clearly of kernel Aut_H(G), as Aut_H(G) is sitting inside Aut_N(G)
I am hesitating with the shortness. I think it's just left-exact.
23:16
Proving exactness is equivalent to proving
Aut_N(G)/Aut_H(G)= Aut_N(H) ?
Forgive me, I've still not seen the light of exact sequences.
@Alyosha More or less.
If you deduce that you'll deduce short exactness, a much stronger version.
Aren't all short exact sequences isomorphic to 1--> H --> G --> G/H --> 1?
It's pretty much obvious that it's left exact (i.e., Aut_N(G)/Aut_H(G) = f(Aut_N(H))), what I am not seeing is whether or not he equality holds.
What is f here?
@Alyosha What do you mean by isomorphism of sequences?
@Alyosha The map.
It's just the image there. I am not sure of the surjectivity. That's what I am telling you.
23:19
Yes.
OK.
lol, my dogs have just found a hedgehog and are making a lot of noise. I barely can focus on problems. :-)
@Chris'ssis make $\mapsto$ are making
@BalarkaSen Yeah, that works better. :-)
That looks better. You'll get the hang of it. =)
Anyway, I'm finally off to read a proof of the fundamental theorem.
23:21
@Alyosha of GT?
Yes.
I've put it off for too long.
I thought you'd done reading it.
I hadn't started yet.
gah, I am too sleepy to check that the short exact sequence 1 --> Aut_H(G) --> Aut_N(G) --> Aut_N(H) --> 1 even works with some practical examples.
Do you have a favourite proof? It may be useful for me to read that one first.
23:25
@Alyosha Favorite proof? Artin has the one.
Does he use much linear algebra in it?
He used too much in the classification of finitely generated modules.
LA? In GT? Are you sleepy too?
@Alyosha He uses linear algebra in pretty much everything, but galois theory is way too algebraic.
Artin uses LA in seemingly unnecessary places.
Yes.
It's 1:30 am here.
@Alyosha I liked the one he used in proving simplicity of $A_5$ however.
Hah, and it's 5:00 already
@Alyosha Well I guess I have to go now. Byes. Have fun with FTGT.
Yes. I preferred Lang's proof, it was shorter.
23:39
@BalarkaSen to better understand my thoughts on "beauty of mathematics", you look at $$\sum_{n=1}^{\infty} \left(\psi^{(0)}(n)+\frac{1}{2}\psi^{(1)}(n)-\log(n)\right)$$ and know the answer. No need of pen and paper.
I didn't invent a special way here, but I learned a lot, and learning a lot I'm able now to produce such beautiful connections that lead to an amazing proof, you know.
This is the beauty of mathematics, not the stuff for kids. The stuff for kids is beautiful for kids.
23:56
Hello beautiful mathematicians
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