@anilorap Ok, try this then. You know that if G is our group it has an element x of order 3 and an element y of order 2. Convince yourself that what group you get is entirely dependent on what yxy is. Now, if yxy=x then we have an abelian group which must necessarily be Z_6, right? If yxy !=x then we must clearly have that yxy=x^2 for clearly yxy!=1, yxy!=y, and yxy!=xy, yxy!=yx. Thus, we see there are only two choices for what yxy is and thus only two possible nonisomorphic groups.
@BrianMScott I must have a huge misconception then: at each step I remove $2^{n-1}$ intervals of length $\frac{a}{3^n}$, so the complement has measure $\sum_{n=1}^\infty \frac{2^{n-1}a}{3^n} = a$.
@JM Unsolicited opinion: I support the protection of the question in, uhurm, question. The bar is virtually non-existent for anyone whom I'd like to see contributing to the thread.
@AntonioVargas I would say it's your call entirely. I rarely make my answers CW except if the question is really "too soft" (e.g. how do you call that in English). On the other hand Martin Sleziak tends to write very helpful and extensive answers full of links and references and turns them into CW for whatever reason.
@AlexYoucis I think it's a relic from the olden days where only users with 2k rep could edit other posts (as it still is the case on MO). The CW mode lowers the bar to 100 points (or something like that).
@AlexYoucis In one recent case someone had a half recollection of a relevant example and posted it as CW, inviting anyone who could fill it in to do so. He was right, and it was an interesting example that I'd not seen before, but I had to fill in all of the details, and the answer as it now stands is entirely my writing (or was when I last looked). That was a very sensible use of CW, since any worthwhile answer was going to end up being a collaboration of some kind.
@Cowbell, just wanted to say sorry if I misinterpreted your question earlier. I thought you were asking how to finish the proof after successfully showing the limit was finite. I see now you were asking more about that first part!
@MarianoSuárezAlvarez How many examples of abelian groups do you know such that the tensor algebra (thought of as a Z-module, of course) is calculable? Things like Q/Z, Z/nZ, Z, etc. are. Are there any other interesting ones that you know of?
Also, being into Hoschild, so I assume you deal with algebras a lot, how important is the tensor algebra? I assuming being the free algebra it's got to be HUGE, but I don't see it show up very often in my mathematical life
I guess I just don't know see why if a function $F(z)$ has a Laurent expansion of form $(\sum_{n=-\infty}^{-1}a_nz^n+\sum_{n=0}^k a_nz^n-P_k(z))/z^k$ is bounded near $0$.
@tb So the Taylor expansion and the Laurent expansion around $0$ are the same? So subtracting $P_k(z)$ only leaves terms with $z^m$ for $m>k$, and is that why it's bounded?
@tb Oh, thanks! Sorry if this is a dumb question, but does this mean Laurent expansions for entire functions always have $0$ for the coefficients of negative powers of $z$?
@tb Can you tell me things about Riemann surfaces? Any perspectives on them you'd like to share. Approaches to learning them, cool facts (sans Uniformization), etc.?
Hmm. Is it often just enough to know it's a cokernel of some map, which should give you some amount of information?
How do you view tensor products? Do you literally just view them as corresponders for biadditive and linear maps? Or, do you view them in terms of their uses in group algebras, Kunneth and Universal Coefficients, coordinate rings, etc.
I apologize if these questions are annoying you--I just feel like you could give me some good insight :P
@AlexYoucis Oy. I don't think I'm the right person to ask. I only know a few simple facts and Miranda's book is somewhere hidden in the depths of my "to read pile mountain"
@kahen That is often times the best way to think about them in terms of problem solving, but as I get more into math I get this weird desire to understand things "metaphysically"--make them feel intuitive, understand why and where they came from, and understand why their existence should not only be natural, but should be expected.
@kahen P.S. The only reason I am aware of your existence is your fantastic response to that ordered field question.
@AlexYoucis I never read any book on distribution theory, so I can't recommend anything. I had quite a few very good courses on mathematical physics and PDEs and I draw the little I know about them from there.
@tb Ok, plan three. Can you explain to me how you think about harmonic and subharmonic functions? Do you think about them in terms of the Dirichlet problem?
it is nice to know that tensor products also show up in analysis, like $L^2(R^2)=L^2(R)\otimes L^2(R)$, and the $L^2(R)$-values distributions are elements of $D'\otimes L^2(R)$, and so on
it V and W are Banach spaces, for example, @AlexYoucis, you can put on V\times W, the usual tensor product, a norm, and then complete with respect to that
doing this for general topological vector spaces is considerably suble and in fact essentially the general solution of how to do this is Grothendieck's phd thesis
@AlexYoucis This place reminds me of the lounge in Van Vleck Hall when I was a grad student at Madison. Same enormous range of conversation, mathematical and otherwise.
@MarianoSuárezAlvarez I heard that theory, too. But there certainly is also the more mundane aspect that he was trying to come to grips with the approximation problem which wouldn't succumb to his efforts. For a good reason, though: it's got a negative solution in general.
@BrianMScott My favorite hall named after a mathematician is in Göttingen: Der Hilbertraum :)
Let $\nu(n)$ be the number of distinct prime factors of $n$. I can't figure out why $$\sum_{d^2|n} 2^{\nu(n/d^2)}$$ gives the number of positive divisors of $n$.
I feel like there's a simple counting argument.
$2^{\nu(n/d^2)}$ is the number of squarefree divisors of $n/d^2$...
@tb But I also consider Patrick's comment relatively harmless so I don't think any upvotes do any damage (especially given Brian's comment which I also upvoted).
If $d^2|p^k$ then $d = p^j$ with $0 \leq 2j \leq k$, so that $j \leq \lfloor k/2 \rfloor$. Then $$\sum_{d^2|p^k} 2^{\nu(p^k/d^2)} = \sum_{j=0}^{\lfloor k/2 \rfloor} 2^{\nu(p^{k-2j})}.$$ Suppose $k$ is even. Since $\nu(1) = 0$, the sum is $2 \cdot \frac{k}{2}$, since there are $k/2$ nonzero terms and each term is $2$. If $k$ is odd, then no terms are zero and so the sum is $2 \cdot \frac{k+1}{2} = k+1$.
I think that if $h$ is multiplicative and its restriction to prime powers injective, and $f$ satisfies $f(ab,cd)=f(a,c)f(b,d)$ when $a,b$ and $c,d$ are each coprime pairs, then $$\sum_{h(d)|n}f(d,n)$$ is multiplicative.
No, there needs a little more restriction on $h$, $h(p^r)$ and $h(q^s)$ must be coprime for any distinct primes $p,q$, I think..
Dear Emperor of Notation. Could you please decree that $x_n \rightharpoonup x$ for weak convergence and $x_n \rightharpoondown x$ for weak$^\ast$ convergence are no longer viable options and will henceforth be punished with 20 years of incarceration?
@BenjaminLim Sheaf cohomology is certainly one of the things you're going to have to look at. But instead of preparing all the machines, why don't you go straight to the subject, by like reading Reid or Shafarevich?
@tb Atm I have finished looking at localisation and going into noetherian rings, I am trying to build up my commutative algebra base before tackling AG....
@BenjaminLim that, too. I'm basically saying it is a huge mistake by many people who think that you need ten years of preparation before you can understand a darn complex polynomial...
@BenjaminLim I understand that, and yes, commutative algebra certainly is attractive, neat and clean, but it also is very easy and soft. Get your hands dirty, look at examples, and then see what the machines provide you with.
@tb I don't understand what you mean by easy and soft.......
@tb I guess what you're saying now is that I am learning the technical machinery but don't know what it's for, I guess that can be said of my understanding of "flatness" and " localisation"
easy: just a bit of abstraction, nothing hard, everything's very tidy. soft: the machines do the work for you.
But what work exactly? To really understand and appreciate that, you need to know what's simple and what's hard and that you can only learn by getting your hands dirty, i.e., by looking at explicit examples and working through them thoroughly.
I'm not saying you're not doing that but I fear by going after the highlights of the abstract theory too quickly you're going to miss out on a lot.
Do I remember correctly that some weak form of choice is true under Axiom of Determinacy. IIRC it was something like choice for countable systems of subsets of $\mathbb R$ or something similar.
Anyway, it seemed to me when I saw that result that at least some of the basic results in analysis would work under ZF+AD, too.
$\mathbf{CC}(\Bbb R)$ holds under $\mathbf{AD}$: the product of a sequence $\langle X_n:n\in\omega\rangle$ of non-empty subsets of $\Bbb R$ is non-empty.
The Weak Axiom of Choice wAC says that for any countable family of non-empty subsets of a given set of cardinality $2^{\aleph_0}$ there exists a choice function.
I find it quite impressive that when I asked, two people from this room gave the correct statement of that result from the top of their heads within a minute. (If Asaf were here, it would probably bee three people.)
BTW is from the top of their heads correct way to put this idiom in plural? (If grammar questions are allowed here....)
