A line OA of length $ r$ starts from its initial position OX and traces an angle AOB=$\alpha$ in the anticlockwise direction.It then traces back in the clockwise direction an angle BOC=3 $\theta $(where alpha is greater than 3 $\theta$).L is the foot of the perpendicular form C on OA.
Also $\f...
Nice, analysis is a good think to be interested in. If you want to learn it thoroughly I'd recommend you to go through the first seven chapters of Rudin, exercises included.
Given any two bounded sequences $\{a_n\}$ and $\{b_n\}$ of real numbers, there's a way to measure how far they are apart. Namely, consider $\sup_{n \geq 1} |a_n - b_n|$. Do you know how to make this idea of "distance between two sequences" precise, and put it in a formal context?
(I suppose this will be a series of questions than a single specific question, but that's ok)
I don't seem to get this, what is the reason that when I have to define the degree of a smooth map I have to take the residue class modulo 2 of the number of elements in the preimage of a regular value? What's special in mod 2. Like I get that the number of points in the preimage of a regular value is locally constant and that motivates some definition out of it but why mod 2
@Albas The number of points in the preimage of a regular value depends on the regular value, but it's residue class mod 2 doesn't, is the key reason. It's a topological invariant of the smooth map.
You can remove this mod 2 reduction at the cost of keeping track of certain signs at the preimage points coming from orientation reasons. This is the "oriented degree", and is independent of the regular value once again.
Look at Milnor's little book for some pictures and intuitions regarding this
You can easily construct a map $f : S^1 \to \Bbb R$ with regular values $x$ and $y$ such that $f^{-1}(x)$ and $f^{-1}(y)$ have different cardinalities but mod 2 they are the same. Try it out
@SubhasisBiswas Precisely. So consider the collection $X$ of all bounded real sequences and define the function $d : X \times X \to \Bbb R$ such that for any two points $\{a_n\}, \{b_n\} \in X$, $d(\{a_n\}, \{b_n\}) = \sup |a_n - b_n|$.
Consider the simplest map $f : S^1 \to \Bbb R$, just the height function on the standard circle $x^2 + y^2 = 1$ in $\Bbb R^2$. The preimage of a regular value (let's take something which is in the range of $f$) contains two points in general.
These will have opposite orientations, $+1$ and $-1$ (think of tracing the circle by your finger eg clockwise, then note that the direction you are crossing one is opposite the direction you are crossing the other)
So for my summers these people have asked me to do degree of maps and stuff from Milnors book and then do some symplectic geometry later in the project. Lots of stuff to do.
Yea, how good a book do you thing Berndt is for Symplectic topology. I am not really a fond of it but I can't McDuff's book because it is too difficult for me at the moment.
I don't know any symplectic geometry to be honest. There's a good summary of results/crash course outline in Eliashberg-Mishachev's book on h-principles that I scroll through occasionally.
@SubhasisBiswas That side question aside, let's take a look at what you have proved so far: $(X, d)$ is indeed a metric space. This is known as the space of bounded real sequences with the "sup metric".
In general your proof shows that for any two subsets $A, B \subset \Bbb R$, $\sup(A + B) = \sup(A) + \sup(B)$, which was my punchline in this digression.
Write down precisely what it means for a sequence of sequences $\{a_{n, 1}\}, \{a_{n, 2}\}, \cdots$ to converge to a sequence $\{a_n\}$ means in the sup metric.
Hmm interesting. What kind of stuff can you do with probability in things like differential geometry roughly? Whenever I hear probability and stuff in diff geometry my mind straight away goes to dynamical systems
I just took a pretty basic course in probability last semester, hope to strengthen it with concepts from measure theory in my next semester.
wait.... correct me if I am wrong. Here is my intuition :
As these sequence of sequences (all of them are convergent) approach a sequence, $\sup |a_n-x_n|$ becomes lesser and lesser. Now, this sup metric enforces that the limiting sequence must also be convergent.
Now, in order for this sequence of convergent sequences to be convergent, after a very large $n$, all of their individual limits will be almost same (Cauchy Criterion).
before getting into writing the proof, am I mistaken somewhere?
I mean think about it for a second. I have a sequence of sequences $\{\{x_{n, m}\}_n\}_m$ such that for any fixed $m$, $\{x_{n, m}\}$ is convergent, and the whole sequence $\{\{x_{n, m}\}\}$ converges in $X$ in the sup metric. Why on earth will $\{x_{n, m}\}$ for each fixed $m$ have the same limit?
I don't understand what your thought process there is. You shouldn't require an example to understand this.
if $M^n$ (smooth) contains an $S^{n-1}$ (smooth, standard structure) which does not separate, can we factor an $S^1\times S^{n-1}$ (standard structure)
Right, you need orientability. You're gluing a standard $D^{n-1} \times I$ to $S^{n-1} \times [-1, 1]$ standardly by the attatching spheres, which gets you a standard embedded handlebody $D^1 \times D^{n-1}$ in $M$.
@RyanUnger No, just need to figure out how many $D^{n-1}$-bundles there are over $S^1$.
@RyanUnger cover $S^1$ by upper and lower hemicircles, over which the $D^{n-1}$-trivializes. It only remains to count the number of ways to glue it at the equator, which is determined by a map $S^0 \to \text{Diffeo}(D^{n-1})$.
for all $\epsilon>0$, $\exists k \in \mathbb{N}$ s.t $\sup_n |x_n^m -a_n| < \epsilon/2$ $\forall m\geq k$ [$(a_n)$ is the limiting sequence].
[Intuitively, the terms of the sequence of sequences gets more and more identical to that of the terms of the limiting sequence]
Therefore, $|x^m_n-a_n| <\epsilon/2$, $\forall m\geq k$ , $\forall n\in \mathbb{N}$ [since the inequality holds for the supremum, it should hold for other values too].
Again, let $\lim_{n \to \infty} x^m_n=l_m$.
Therefore, for $\epsilon>0$, $\exists k^*>0, $ s.t $|x_n^m-l_m| < \epsilon/2$, $\forall n \geq k^*$
trying to figure out what branch of math this might be classified under.
Consider the following table of numbers: 9 7 5 3 1 1 3 5 7 9 9 5 1 3 7
There's two properties here. 1) Each row contains the first five odd integers. 2) There's some way to add/subtract the numbers in each column such that the result is 1. (Doesn't need to be the same combination of add/subtract for each column)
I think there's a generic construction here. First line: Start from 2n+1 and go down in steps of 2, Second line: Start from 1 and go up in steps of 2, Third line: Start from 2n+1 and go down in steps of 4, taking absolute values as necessary
Let $X$ be a compact Hausdorff space and let $T: X\rightarrow X$ be an arbitrary continuous function, how can I prove that there exists a nonempty compact $K \subseteq X$ such that $T(K)=K$ ?
@SubhasisBiswas You should be careful about the $k, k^*$ as $k$ is in general dependent on $n$ and $k^*$ is dependent on $m$. Does it cause a problem in your proof, or easily fixable?
@RyanUnger Let's take $X$ to be a compact metric space first. Given $x \in K$, I can write $x = T(x_1) = T^2(x_2) = \cdots$. The sequence $\{T^{n-1}(x_n)\}$ has a convergent subsequence, say $\{T^{n_k - 1}(x_{n_k})\}$, which converges to some $y \in X$. So $T^{n_k - 1}(x_{n_k}) \to y$ implies $T^{n_k}(x_{n_k}) \to T(y)$. But $T^{n_k}(x_{n_k}) = x$ for all $k$, so what that means is $T(y) = x$. Just check now that $y$ is indeed from $K$.
Because what's happening at the boundary is you're gluing a standard $I \times S^{n-1}$ to a standard $S^{n-1} \times \{1, -1\}$, which is the same as taking connected sum of two $S^{n-1}$'s by a standard tube.
The standard sphere is the identity element in the group of exotic spheres :3
ok viewing this as a connected sum is probably smarter. So the boundary of the nbhd of the S^{n-1} is two spheres but then adding the S^1 makes a connected sum
@BalarkaSen Re: that description you suggested for that circle diff thing
One review article contains the following remark plus a footnote:
“In that paper [of Thurston], earthquake theory is devel- oped in the setting of the universal Teichmüller space, that is, the space parametrizing the set of complete hyperbolic metrics on the unit disk up to orientation-preserving homeomorphisms that extend continuously as the identity map on the boundary of the disk.“
Footnote (will edit): “A common trend is to call DiffC.S1/=PSL.2;R/ the physicists universal Teichmüller space and QS.S1/=PSL.2;R/theBersuniversalTeichmüllerspace. Here,DiffC.S1/denotesthegroupoforientation- preserving homeomorphisms of the circle and QS.S1/ its group of quasi-symmetric homeomorphisms, of which we talk later in this text.”
@BalarkaSen, I don't see any problem with it (maybe I am missing something obvious). The final form becomes $|a_n-l_m|< \epsilon$ and $m,n ≥ g(m,n)$. Where $g(m,n)= \max\{m,n\}$.
But yeah this seems to agree with my thought that Diff+S^1/PSL2R is the space of orientation preserving diffeomorphisms of the circle which extend conformally to the hyperbolic disk
There's probably some description of this thing as the space of tuples (complex structure on upper hemisphere, smooth structure on equator, complex structure on lower hemisphere) modulo when they glue appropriately (when they glue to a complex structure on S^2?) but both me and Mike were unsure how to formulate it.
And, the keyword is "for all $m$ and $n$". We don't need to look back and watch if we are missing out any of those $m$ and $n$ s, after a certain $g(m,n)$
What perplexes me is that I’ve seen multiple assertions about the “physicists” UTS as being the set of (orientation preserving) diffeos, normalized to have three fixed points on the unit circle
But don’t Mobius transformation generically have just two fixed points?
@SubhasisBiswas This is sort of subtle from a symbolic point of view; here's what: $k^*$ might depend on $m$, but $k$ doesn't (that's what convergence in sup metric means). That's the key idea.
@Eran OK, I think the sequential proof can be slightly abstracted out. Fix some $x \in K$ and look at $A_n = T^n(X) \cap T^{-1}(x)$. $\{A_n\}$ is a decreasing sequence of compact sets again in $X$, so contains some $y \in \bigcap A_n \subset K$. That is your $y \in K$ such that $T(y) = x$ :D
@SubhasisBiswas That's not good enough, $\sup_{m, n} |a_n - l_m|$ might be large. You want $\sup_{m > k, n > k^*}$ but then again you run into dependence issues. Use what I pointed out: $k$ doesn't depend on $n$, it's a universal constant.
Okay. I am trying to write down the piecewise function corresponding to the following graph...Let $r \in [0,1]$. From $0$ to $r$, I want to the function to be $y=x$; then a vertical line connecting $(r,r)$ to $(r,0)$; and then $0$ from [r,1]$.
That middle portion is giving me trouble.
I don't know why it is giving me so much trouble...
Now, at some point (in the proof) I mentioned that the inequality holds when $m≥k$ and for all $n$. Doesn't this remove the dependency of $k$ from $n$?
@Subhasis Yes, the point is, $k$ never depended on $n$ in the first place. For all $m \geq k$, $\text{sup}_n |a_{m, n} - a_n| < \epsilon$.
That quantity in the sup doesn't depend on $n$. So when you extract out $|a_{m,n} - a_n| < \epsilon$ for all $m \geq k$ and for all $n$ from there, it's the same $k$. No dependence on $n$.
That's why convergence in sup metric is called "uniform convergence" by the way, contrary to "pointwise convergence" of sequences, where there is a dependence.
(Pointwise convergence just means $\lim_{m \to \infty} a_{m, n} = a_n$ for all $n$)
Let's write it all out precisely. Say $\{a^1_n\}, \{a^2_n\}, \cdots$ is a sequence converging to $\{a_n\}$ in sup. $\{a^m_n\}$ has limit $l_m$ for any fixed $m$. Then your technique is: $|a_n - l_m| < |a_n - a^m_n| + |a^m_n - l_m| < \sup |a_n - a^m_n| + |a^m_n - l_m|$. There is a universal constant $k$ such that $\sup |a_n - a^m_n| < \epsilon/2$ whenever $m > k$. There is a constant $k^* = k^*(m)$ depending on $m$ such that $|a^m_n - l_m| < \epsilon/2$ whenever $n > k^*(m)$.
Combining we have that for all $m > k, n > k^*(m)$, $|a_n - l_m| < \epsilon$.
How do you argue from here that $\{a_n\}$ converges, now? Remember that that was the goal.
(yeah, I do felt an ideal basically behave like a blackhole, because you can multiply something outside the ideal with something inside it, and you fell into the ideal, never to see the light again lol)
