@anon You know about the completion of topological groups $G$ by way of filtrations right? If so, I have a question about one tiny choice in the construction of the filtration of $\hat G$. I read in Qing Liu's quick overview of completions at the beginning of his text on Algebraic Geometry that $\hat G_n=\{\langle g_m\rangle\in\hat G: g_m=0\text{ for all }m\le n\}$ I believe it should be "for all $m<n$" in the description of the set. Can you suggest one over the other?
Oh, nevermind. You want the kernel of the projection $\hat G\to G/G_n$, which requires $g_n=0$ as well, obviously.
@anon I got it, but at any rate Liu defines $\hat G$ to be the inverse limit of the system $G/G_i\to G/G_j, i\ge j$, which can be realized as a quotient of the product of the groups.
So $\langle g_m\rangle $ is a representative of a class in this quotient.
@TedShifrin Are there nice moduli spaces for higher genus Riemann surfaces? For example, the moduli space for tori is nice. If so, are they controlled by some group like SL_2(Z)?
@AlexYoucis BTW, Mariano has told us a few times his computer or whatever he uses automatically logs in here, so he might not respond simply because he is not around! =)
@TedShifrin I know this is a tall order, but can you give any intuition about why there should be a unique structure on the sphere, but not one on higher genus guys? Also, I assume there are infinitely many non-biholomorphic structures on all higher genus surfaces, right?
@TedShifrin OK; that, table of contents. It allows to surf around the document, with hyperlinks and stuff. And yes, that is the new document, I did get it!
@TedShifrin The STDU Viewer makes the table of contents to the left by a "select text and I will hyperlink to it" way. It also builds up the table you see on the left. Quite neat.
@AlexYoucis This was shown by Whitney by considering an embedding of the manifold in some $R^n$, and considering for some $\epsilon>0$ the set of hyperplanes parallel to the coordinat hyperplanes and separated by $\epsilon$ units: these hyperplanes cut the maanifold in pieces, and he shows that when $\epsilon$ is sufficiently small you can in turn cut these pieces into simplices to get, in all, a triangulation.
@MarianoSuárez-Alvarez It turns out that what Spivak does is the following, in the proof of Poincaré. I worked it out today at the uni. I guess it is what you showed me the other day, but in the whole generality.
@TedShifrin No, yeah, I know why g=0 is trivial, and why that standard approach fails for g>0, but I didn't know if there was any intuition about why it should be the case that g>0 has multiple structures. Good night!
First, consider the map $F:[0,1]\times \Bbb R^n\to\Bbb R^n$ given by $F(t,{\bf x})=t\bf x$. If $\omega$ is a $k$-form on $\Bbb R^n$ then $F^\ast \omega$ is a $k$ form on $[0,1]\times \Bbb R^n$ which I computed to be $$dt\wedge (F^\ast \omega)_1(t)+(F^\ast\omega)_0(t)$$ where $$(F^\ast \omega)_1(t)=\sum_{I} \sum_{\alpha=1}^k(-1)^{\alpha-1}t^{k-1}\omega_I(t\;\cdot \;)x_{i_\alpha} dx_{i_1}\wedge\cdots\wedge\widehat{dx_{i_\alpha}}\wedge\cdots\wedge dx_{i_k}$$ and
@MarianoSuárez-Alvarez Right. Well, then when we take an arbitrary $k$ form on that cylinder we can write it as $\eta=dt\wedge \eta_1(t)+\eta_0(t)$ (this I picked up from the answer) and we may define an operator $J\eta=\int_0^1 \eta_1(t)dt$, so we integrate the coeffs wrt to $t$. Then $$d\eta=\underbrace{d\eta_0(t)}_{\text{ only w.r.t to the } x_i} +dt\wedge \left(\frac{\partial \eta_0}{\partial t}-d\eta_1\right)$$
@RamanaVenkata My favorite way to see the equality is with equivalence relations. Define $\sim$ on $HK$ via $a\sim b$ iff $aK=bK$, and $c\sim d$ on $H$ iff $c(H\cap K)=d(H\cap K)$. By abuse of notation write the quotient spaces as $HK/K$ and $H/(H\cap K)$. Then there is an isomorphism $HK/K\cong H/(H\cap K)$ given by $[hk]\leftrightarrow[h]$.
@MarianoSuárez-Alvarez Can you try to explain this: "More specifically, we associate to $X$ the cylinder $X×[0,1]$. Identify the top and bottom of the cylinder with the maps $j_1(x) = (x, 1)$ and $j_0(x) = (x, 0)$ respectively. On the differential forms, the induced maps $j_1^*$ and $j_0^*$ are related by a cochain homotopy $K$: $$K d + d K = j_1^* - j_0 ^*$$
@RamanaVenkata I'm not. In general the coset-space G/H is well-defined even when H isn't normal, it just doesn't have a group structure (it does come equipped with a natuarl G-action though, left multiplication, and indeed all orbits of actions arise in this way, as stabilizer coset spaces, up to isomorphism)
yes, but going that route is actually waaaaay too long
it requires that you rpove homotopy invariance of de Rham cohomology, and doing that before proving Poincaré's lemma is either absurdly convoluted or circular
Bott and Tu in their book on differential forms give a recursive construction of (a slightly different) K
@MarianoSuárez-Alvarez I won't worry too much then. =)
@MarianoSuárez-Alvarez I have this problem from Polya and Szego I couldn't solve. Maybe I should give more thinking to it. I showed it to Fava today.
Suppose $t_n$ is a bounded (Fava noted unboundedness provides counterexamples) sequence of reals such that $$t_{n+1}>t_n-\varepsilon_n$$ for some $\varepsilon_n\to 0$ of nonnegative entries.
Then $t_n$ is everywhere dense between its limsup and liminf.
They say it is related to this problem (two before) "If the general term of a series which is neither convergent nor properly divergent tends to $0$, the partial sums are everywhere dense between their highest and lowest limit points".
Let $b_n$ be a sequence of positive real numbers such that $b_0=1$ and $$b_n=2+\sqrt{b_{n-1}}-2 \sqrt{1+\sqrt{b_{n-1}}}$$ Compute $$\sum_{n=1}^{\infty} b_n 2^n$$
Let $b_n$ be a sequence of positive real numbers such that $b_0=1$ and $$b_n=2+\sqrt{b_{n-1}}-2 \sqrt{1+\sqrt{b_{n-1}}}$$ Compute $$\sum_{n=1}^{\infty} b_n 2^n$$
@Ethan still, it's $b_0=1$!!!. It's ok my first statement!
@Ethan I just computed it again and this is the right value.
hmmm, I should create a pack with such questions ... (very nice these ones)
@Ethan I'm preparing to write the 2nd solution to the problem mentioned above. Actually I've largely outlined the main ideas and now try to see on paper if everything matches there.
(however I miss some justifications in my 2nd proof - - hope to improve that)
There is a simple way to graphically represent positive numbers $x$ and $y$ multiplied using only a ruler and a compass: Just draw the rectangle with height $y$ in top of it side $x$ (or vice versa), like this
But is there a way to draw the number $xy$ directly on the real line (i.e. not as an...
the key seems to be $$ \begin{align} b_n &=2+\sqrt{b_{n-1}}-2 \sqrt{1+\sqrt{b_{n-1}}}\\ &=(\sqrt{1+\sqrt{b_{n-1}}}-1)^2\\ 1+\sqrt{b_n} &=\sqrt{1+\sqrt{b_{n-1}}} \end{align} $$
I see. For me till now, reference-request always provided me with books, rather than such questions, so I was not sure it is used actually to find a reference that much as much as it is used to find books.
I'm trying to solve this: Let $f$ a complex function defined on $[0,2 \pi]$. Let $f_m:[a,b] → ℂ$ be the function with $f_m(x) = e^{imx}$. Solve $$ \tfrac {1}{2 \pi } \sum_{m=-n}^{n} \left [ \int_{0}^{2 \pi} f(t) e^{-imt} dt f_m \right ]$$
@DanielFischer The actual question was, calculate $P_{W_n}$ the orthogonal projection of a function $f$ on the space $｛f_{-m},...,f_0,...,f_m｝$. After evaluating I get this expression.
@Kasper Yes, it is. (I assume that both spaces are over the same field.) Try to google for vector space "L(V,W)" or vector space "Hom(V,W)" to find some basic information about this space.
There is a simple way to graphically represent positive numbers $x$ and $y$ multiplied using only a ruler and a compass: Just draw the rectangle with height $y$ in top of it side $x$ (or vice versa), like this
But is there a way to draw the number $xy$ directly on the real line (i.e. not as an...
