Hey @TedShifrin this below is represented mathematically as follows right ? $\bigcup \Delta_{\alpha}^n\right)\big/{\sim}$ w$here $x \in \Delta^{n_{\alpha}} ~ y \in \Delta^{n_{\beta}} iff \sigma_{\alpha}(x) = \sigma_{\beta}(y)$ right ?
It's like triangulations, but weaker requirements. (For example, for triangulations, two faces can't share two vertices unless they share the edge joining the two vertices. Or they can't share two edges.)
I see that $a$, $b$, and $c$ are the latitude circle, the longitude circle, and this spiraling curve that cuts the torus in half. This information is coming from the 1-dimensional simples after identifying them to a point right ?
Maybe it would be good for you to see the $\Delta$-complex structure if you wanted to cut the (surface of the) bagel horizontally into two half-bagels (like you would eat).
Or warm up by doing the sphere. Suppose you cut it into two hemispheres. Or eight eighth-spheres. (Not just one 2-cell and one 0-cell.)
@Semiclassical Alright I proved it for n>50 (and checked it up to n=50). I made use of Brieskorn's formula for the signature as a signed count of elements in an open tetrahedron.
Hey guys! looking from the context of Gronwalls inequality, I am trying to answer "What can I say about a continuous function $f \colon \mathbb{R} \rightarrow [0, \infty]$
For example to build the sphere as CW complex structure we start with 1 disk then identify its boundary to 0-disk then we take 2 disk and identify its boundary with the circle we constructured.
the first version in which the derivative is included has the $f(0)$ constant, but the second version dosnt. (in particular, the textbook that's asking the question is in the context of the second version)
An interesting question might be: What are the criteria that a mathematical object $M$ has to satisfy in order for the question "Proof some properties in $M$ without using some properties $A$" to have a nonempty solution
I actually needed help asking a question, rather than having it answered. It involves a vector space and two metrics, and it's driving me bonkers. I've been trying to find a new way to ask this question here: math.stackexchange.com/q/1637996/187120
Intuitively, I already know a sufficient answer, but I'm not happy with the fact that I don't really know how to ask the question.
nope I am wondering whether these's a systematic way for multiple methods of solving a given problem to exist, that is ,treat the problem itself as some abstract object and trying to work out what criteria it has to obey in order to have multiple ways of solving it
I'm having trouble figuring out how to prove the convergence of the $p$-series, that is,
$$\sum_{n=1}^{\infty}{\frac{1}{n^p}}$$
where $p > 1$.
I'm in a real analysis course and I have a midterm coming up. I think I might need to prove this on the midterm, but without using the integral test. ...
I don't know that there's an abstract way to decide if there are truly different proofs for a theorem, @Secret. But your question has a well-known alternative argument with the comparison test (see Rudin, for example).
if we treat the question linked above as an abstract object, one question that we can ask is how to show that this "solution without integral test" method exists
so in general, generalising all possible problems as some abstract objects, the question is an existence proof of alternate solution methods and the number of them
@Secret This would be feasible if you already had the totality of provable statements. Otherwise, iterating over the set of known truths and combining them in a brute-force manner to generate a Knowledge Tree would be seriously infeasible.
You're essentially asking for a way to search the Tree of Knowledge.
Where a proof would be a path through the tree.
Potentially, with a heuristic, it's possible. But the best heuristic for that would be intuition, and that's hard to develop an algorithmic system for.
@TedShifrin @MikeMiller It's literally the # of lattice points inside a rectangle divided by the # of lattice points in a right triangle dividing it in halves. No idea how Brieskorn showed this.
@TedShifrin Technically, you could argue that there is a specific truth statement which cannot be used in a proof. However, this doesn't stop you from taking all that precedes it and simply building a proof that subverts that one statement. It's very muddy watter, @Secret
Karim: I'm going to go cook dinner, but the cheapest way I can do the torus with the two bagel halves is with 2 0-cells, 4 1-cells, and 2 2-cells. See if you can do that.
If $p$ and $q$ are distinct odd primes, how could I approach showing that $x^{\varphi(pq)/\gcd(p-1,q-1)}\equiv 1\mod pq$ for all $x\in (\mathbb Z/pq\mathbb Z)^\times$?
(I realize the numerator of the exponent is $(p-1)(q-1)$)
A quick question if someone has time. I was watching a video and the lecturer said: "This is a group mod 2" What does modular 2 mean in this case? Thx.
Another question...are we experiencing a bug today...i dont get any reputation points for the last few hours
@GBeau But what does it mean. Can you dumb it down a little, to my level of group theory knowledge, and give a simple explaination if its possible to do that in a few sentences?
He was talking about the fundamental group of the projective plane. Which has two elements {e, a}. I dont think you need to know its about the projective plane tho but just a simple group. Then he says "the only non-trivial multiplication" is a^2=e. In terms of group theory its called Z_2, its the same as the additive group {0, 1} mod 2 or {-1, 1} multiplicitive mod 2, depending on how you want to think about it"
@JKnecht do you understand why it's the same as the additive group $\{0,1\}$, or are you still lost at groups?
the additive group $\{0,1\}$ mod $2$ describes the list of all possible elements: 1 and 0, the group operation: adding, and the way items are determined to be equivalent: taking the remainder modulo 2
so in this case $a\cdot a=e$ is the same as $1+1=0$
($1+1=2$, and $2$ has a remainder of $0$ modulo $2$)
@BalarkaSen: Mike once told me that "$\pi_1(S)$ acts discretely on $\mathbb{H}^2$" is very likely meant as "properly discontinuously". properly discontinuously can be stated as for all $x, y \in \mathbb{H}^2$ there exist neighbourhoods $U_x, U_y$ such that the set $\{g \in \pi_1(S): g U_x \cap U_y \neq \emptyset\}$ is finite. is that correct? if yes, can I also choose $x = y$ and then it follows that any stabilizer must be finite?
Your definition is correct. And yes, it does mean that there exists nbhds $U, V$ of $x$ such that only finitely many translates of $U$ hits $V$. In particular, the stabilizer of $x$ is finite.
@BalarkaSen: that's odd though, because I don't think it makes any sense for them to be finite in this case, but maybe I'm wrong. is there any other reasonable interpretation of "discrete action" that doesn't imply finite stabilizer?
Hold on, let me check if what I said is correct, since I am prone to saying wrong things.
OK, assume stabilizer is not finite. There are infinitely many $g$'s such that $gx = x$.
Pick any nbhds $U, V$ of $x$. $gU \cap V$ for precisely those $g$'s must be nonempty, as they'd contain $gx = x$, yeah? So that's a contradiction, because you get infinitely many such things.
So there is no such nbhds $U, V$. Proper discontinuity fails.
@Huy: Really it should just mean that the group is a discrete (and closed?) subgroup of the group of diffeomorphisms. For whatever reason I thought this was equivalent to a properly discontinuous action.
@BalarkaSen: proof that for any hyperbolic $S$, the centralizer of nontrivial elements of $\pi_1(S)$ is cyclic. identify $\pi_1(S) \cong \operatorname{Deck}(p)$, $p: \mathbb{H} \to S$. the deck transformation are nothing else than so called *parabolic* and *hyperbolic* isometries. those are classified by their fixed points on $\partial \mathbb{H}^2$: one fixed point means parabolic, two means hyperbolic. you can prove that two such isometries commute iff they have the same fixed points.
so take for example just one fixed point in $\partial \mathbb{H}^2$. take some parabolic isometry corres…
@MikeMiller: just to be sure: from the bigon criterion (transverse scc are in minimal position iff they don't form a bigon) there's a corollary that distinct simple closed geodesics in a hyperbolic surface are in minimal position, the argument being that if such geodesic curves bounded a bigon, since this bigon is simply connected, one can lift it to $\mathbb{H}^2$, but there, geodesics between two distinct points are unique, a contradiction.
can't I just argue that lifting a geodesic yields a geodesic and having two intersections in $S$ I also need at least two in $\mathbb{H}^2$, so I get a bigon? or am I using simply connectedness implicitly here?
@Huy: There is no reason the paths lift to paths that intersect in more than one point up above.
Dumb example: Take the latitude and longitude and the torus. They intersect infinitely many times (when considered as maps from $\Bbb R$)! But they only intersect once in the lift.
@MikeMiller: ah, wait. nullhomotopic closed curves are lifted to closed curves, so that guarantees that the bigon (bounded by a nullhomotopic closed curve) is lifted to a closed curve. is that it?
@Huy Here's a half-baked argument. Start with a not necessarily compact closed orientable manifold. Then top homology vanishes by noncompact Poincare duality. So there's no top dimensional cell upto homotopy (I am assuming the surface can be given a finite cell complex structure - finiteness is necessary for well definedness of $\chi$ and that 2-folds admit cell complex structure should be a standard yet not so easy to prove fact). So my surface is homotopy equivalent to a 1-complex.
Any 1-complex is wedge of a bunch of circles, $\chi$ of which is $1 - n$ where $n$ is the number of 1-cells. This is positive iff $n = 1$, in which case my surface has homotopy type of a point, hence is simply connected.
And yes, cell complex structure comes from the more general fact that 2-folds admit triangulation. You should find a proof by googling: this is something I took as granted because the proof is involved.
So I guess that should constitute a proof.
(Typo, meant $n = 0$ up there)
@Huy Note that this actually shows that any closed connected noncompact surface with $\chi > 0$ is contractible, not just simply connected.
btw, I keep reading about the "homotopy lifting property". does it mean that if $\alpha$ is given and some homotopy $H$ starting with $\alpha$, there is a (unique?) homotopy corresponding to the lift of $\alpha$ projecting down to the original homotopy? @BalarkaSen
@Huy yes. if $H : I^2 \to X$ is a homotopy of paths starting at $\alpha$, you can lift it to a unique homotopy $\tilde{H} : I^2 \to \tilde{X}$ starting at $\tilde{\alpha}$ upstairs.
@BalarkaSen: and what kind of covering space do I need to have this "homotopy lifting property"? universal should be sufficient, right? anything less strict?
@GPhys Take any compact metric space $X$ with cardinality $|\Bbb R|$. Set up a bijection with $\Bbb R$ and label the reals by the elements of $X$ by that bijection. Then transfer the metric.
Pick a bijection $f : \Bbb R \to X$. Define $d(x, y) := d_X(f(x), f(y))$. That's all.
I mean, there's nothing special about $\Bbb R$. Without a metric, or any kind of extra structure, it's a bare uncountable set. There are millions of metrics on a uncountable set which makes it into a compact metric space.
I want to maximize say $x + y + z$ with the constraint $xy^2z^3 = 108$, $x, y, z$ positive real. Using Lagrange, this takes a few seconds. But I want to do it the plain vanilla way.
@Huy Because I need to learn calculus, not just keep catchwords at the back of my mind.
:)
OK, differentiation tells me the only critical point of $F$ is at $(2, 3)$. There, $F(2, 3) = 2 + 3 + 1 = 6$. This is clearly a local minimum. But I have to prove it's a global minimum.
Now, I realize I need to construct a compact set $X$ of $\Bbb R^2$ containing $(2, 3)$ such that $F$ is more than $6$ outside the interior of $X$. Then max value theorem tells me $(2, 3)$ is global minimum.
I remember using Lagrange for all those typical calculus problem and then in my physics class, we used Lagrange to derive some formula that was of actual physical significance, which was rather nice for a change
I think it was some sort of (probability) distribution for some isolated system with distinguishable particles under the constraint of a given energy and number of atoms.
that was probably something involving the single particle partition function, but I don't remember the very details
I'm having trouble figuring out how to prove the convergence of the $p$-series, that is,
$$\sum_{n=1}^{\infty}{\frac{1}{n^p}}$$
where $p > 1$.
I'm in a real analysis course and I have a midterm coming up. I think I might need to prove this on the midterm, but without using the integral test. ...
