I have a group $G$ defined inductively $G^0$ is the group of upper triangular matrices, with $1$'s on the diagonal, and $G^1 = [G^0,G^0], G^{i+1}=[G^0,G^i]$
$G=G^0$ and I want to prove that $G^n =\{1\}$ at some $n$
I have played around, and the inverse is long and doesn't seem to have an obvious simplification(from being an ugly set of sums and products produced by the augmented identity method)
I have done some examples on 4x4 matrices with mathematica and I can see it works
We introduce a $0$ superdiagonal when we obtain $G^1$
Perhaps a proof could restrict attention to removing the superdiagonal and then the diagonal above that, while treating the remaining upper triangular elements as irrelevant
So @TedShifrin why is Rudin going to be challenging for me? I mean it is. I am not understanding the implicit function theorem. But I can read John Lee's books on manifolds so I don't get why I am having more trouble with Rudin. Rudin has some weird notations tho, or at least ones I haven't seen
I have a group $G$ defined inductively $G^0$ is the group of upper triangular matrices, with $1$'s on the diagonal, and $G^1 = [G^0,G^0], G^{i+1}=[G^0,G^i]$
$G=G^0$
I can see that when I do each iteration of $G^i$ I lose the next diagonal up
Its given that the $span(a_1,a_2.......,a_n)=V$ where V is a vector space so ${a_1,a_2,a_3........,a_n}$ is a basis for V , so since the basis is finite,V is finite dimensional right?
@r9m There were months when I didn't know how to do it elementarily. Finally my research opened amazing simple doors to crack them all. :-) Indeed, without having at hand a proper tool, that is proving to be horrible. I initially did that on 6 pages or so, and using CAS.
@r9m then, bang-bang, the proper gun killed it easily. This is definitely a good example where without having at hand all the proper tools, things become horrible.:-)
I am thinking of some other way.......Tell me is this right or not.....Let me take a subset W of V and lets say that all of it basis have the same cardinality n-1 then since it is a subset all these basis will be in V but by the theorem we know that dimW<dimV now if i take the basis whose cardinality is n-1 as the one which tells the dimension the inequality does not hold but if i take the basis whose cardinality is n then the inequality holds.@BalarkaSen
So by this inequality V cannot have Basis of Cardinality n-1 and since any other cardinality will also disprove the inequality the only cardinality will be n
I might be talking nonsense so please don't get angry
@Rememberme you're using later theorems to prove previous ones. dim V isn't even defined if you can't prove that cardinality of basis is an invariant of vector spaces. so you're proof is ambiguous.
just read the theory carefully, @remember. study the book thoroughly.
@BalarkaSen IF i can prove this Let V be a vector space which is spanned by a finite set of vectors P1, . . . , Pm. Then any independent set of vectors in V is finite and contains no more than m elements then i can prove what you said right?
@Balarka i think i have got an idea i think it will be enough to show that every subset of V which contains more than m elements is linearly dependent right?
@iwriteonbananas That group is an example of a Baumslag-Solitar group, and they have a pretty nice linear representation, although normally not isomorphisms, so if you are lucky you can just calculate the matrix product and see if it is trivial. Sadly if it ends up being trivial you still won't know if it is inside the group.
Can you explain the sentence: The natural embedding $H\hookrightarrow G$ composed with the natural projection $G\twoheadrightarrow H$ is an isomorphism?
you're learning about splitting of exact sequence right now, @incurrence. the statement above is equivalent to the embedding being a section of the short exact sequence $0 \to N \to G \to H \to 0$.
i'm just wrapping everything you are doing with abstract-nonsensical language. a short exact sequence is a bunch of homomorphisms with 0s at the ends and kernel of some map equal to image of the next map.
@Remember your proof goes above my head. you say $\alpha_j$ as scalars and then the next moment you express $\alpha_j$ as linear combination of $\beta_i$s.
@Incurrence the $N \to G$ map is just the embedding.
i.e., has kernel $0$. that's why there is a $0$ at the start.
it brings up surprisingly few google results... probably just a scrapper(?)
@ADG I don't understand the point of that proposed site, competitive math problems are welcomed on math.se (as long as the competition is not an ongoing one of course), isn't it the same for other disciplines that already have a site?
> The Indian Institutes of Technology Joint Entrance Examination (IIT-JEE) was an annual engineering college entrance examination in India. It was used as the sole admission test by the 16 Indian Institutes of Technology (IITs) and Indian School of Mines Dhanbad (now getting converted into IIT).[1] The examination was organised each year by one of the various IITs, using a round robin rotation pattern. It had a very low admission rate (about 10,000 in 500,000 in 2011).
@iwriteonbananas I am in the process of learning some small cancellation theory, there are some algorithms that work on some groups (they satisfy some "small cacellation" condition), so if the group ends up being suseptable to the algorithm, and is not awful I will attempt to figure out if that element is trivial. (after I learn the relevant stuff I will look at it)
@Incurrence thanks :) would you vote to reopen? I think it's a fun and interesting problem, expected it to be welcome. But I understand this site would vote to close Fermat's last theorem "show your working"
@iwriteonbananas A quick calculation, if I didn't screw up, shows that the group satisfies the condition $C'(1/6)$, so the algorithm should work. Now all you got do is wait long enough for my scattered brain to figure some of this small cancellation theory stuff.... :D
@r9m $$\int_0^{\infty} \frac{\cos(x)}{x} \left(\int_0^x \frac{\sin(t)}{t} \ dt \right)^2 \ dx$$ that you said once you didn't see a solution without the use of CAS. It was posted on I&S.
hi. I have used the FFT analysis tool in audacity to tune my guitar, since I can observe the peaks in the graph, and the other component frequencies, but how does FFT actually work?
My attempt. This is by no means closer to the answer, but I want to address several equivalent forms that might be helpful for future calculations.
First, from Landen's identity of the following form
$$ \mathrm{Li}_2(z) = -\mathrm{Li}_2\left(-\frac{z}{1-z}\right) - \frac{1}{2}\log^{2}(1-z), \qu...
Curious, how long is the book going to be or, maybe a better measure, how many problems and will you be including solutions or will it just be a problem book @Chris'ssis
@Chris'ssis I mean knowing that you are going to release a book at one-point is not better than the state of ignorance (where one does not know you are going to release a book sometime)
@r9m It's hard because I need to do the whole thing alone, even the English part until some point when I'll contact an English professor from my country and ask for support.
@r9m To be honest, I'm very glad for any such book that is out. Inside Interesting Integrals was a very good book to me too. I mean one doesn't need to show in a book how academic one may be, but to send the message to the readers in a very enjoyable way.
I have a group $G$ defined inductively $G^0$ is the group of upper triangular matrices, with $1$'s on the diagonal, and $G^1 = [G^0,G^0], G^{i+1}=[G^0,G^i]$
@r9m Keep in mind this (it remains recorded on MSE): you would love a lot my way to that integral I mentioned above. :-) (I wanted to have this message here and at the proper time to remind it)
@Chris'ssis just one request I have as an Indian reader ... please don't leave the non-trivial problems as an exercise :P @Balarka might get the pun :P
@r9m I also include this one $$\int_0^{\pi/2} \operatorname{Li}_2 (\sin(x)) \ dx$$ and depending on the space, I might also add the trilogarithm version. Hope to do it to have them both there!
@Chris'ssis okay :) ... I have loads of exams this week .. :| (last set of exams in my final semester) .. after that I'll try to pick up the axe that I dropped in the river at one point in time :)
@Chris'ssis yea !! that is one dragon to slay :) $\displaystyle \int_0^1 \frac{\operatorname{Li}_2 (x) \log^2 (1-x)}{x}\,dx$ .. and rest of the stuff on RHS are manageable by MZV algebra :)
@Incurrence I guess then you will end using some clever stuff with the inverse of $I+N$, $N$ nilpotent and $I$ the identity, but I don't really feel like working through the problem and don't have a clear idea of wear to go
@BalarkaSen Ah, yeah, Riemann never to forget. The same about Gauss.
@r9m in middle school already we were pushed so much to learn with hard math stuff ... I don't say that was a wrong thing, but the way things were done I never lked. The hours were held with much much fear you're going to be smashed by bad marks.
@r9m The professor used to say: "you see this, I doubt any high scool student would do that or an undergraduated". It's good to teach hard math but not with so much pressure. One had to work like hell.
@Chris'ssis I don't like pressure .. it's an useless baggage :( .. the only way people survive pressure is possibly by forgetting about it and focusing on work at hand :)
@iwriteonbananas here is the wikipedia page, also there is the book Combinatorial Group Theory by Lyndon and Schupp. But basically the idea is that groups that satisfy this condition have a small amount of cancellation when relations are multiplied, so you get a bunch of control over what identity elements can look like. The small cancellation is one of the things that makes studying certain problems in free groups easy.
@r9m In one of the days, the professor asked at the blackboard a pupil and asked him to says some theorems, and since he didn't know, he pushed him with the head to the blackboard. Then asked the best girl in the class about other questions, and since she didn't know, he gave her a hand over nose until she began to bleed. It was much terror in that class.
@r9m I'm not the type of person able to learn in such an environment. People must learn with pleasure, not for fear. It was sad. That story continued for years.
@iwriteonbananas I am actually pretty sure this small cancellation stuff will be able to help me solve a problem that I have been thinking about on and off for around 2 years, so I might be able to get myself to focus long enough so that (hopefully) I will have the knowledge to do your problem within the next couple of days.