i've run into that in the past, in the following way
In physics, you often deal with periodic potentials
usually you model them with trig polynomials because that's easy to write out
it's hard to come up with exact solutions to problems in that case, though. you can get approximations easily enough, but not exact
as an alternative, you may want to build up a periodic potential in another way. one way which I've had cause to use, at times, is to put down a periodic array of gaussian distributions
@Knight so, after a long calculation via lagrangian mechanics (probably way more complicated than necessary but idk) I find that hinge A actually will initially accelerate-away- from C i.e. away from the direction of pull
C moves in the expected direction, which causes B and D to move inwards. but this makes A want to move away from the center of mass
i suspect, though, that the right POV is that of the COM frame.
in that case, C moving outwards means that B,D move inwards and A moves outwards. but that means that A moves away from C
so...kinda goofy
i'm not altogether certain about my numbers, though
Why can i not find anything about the notation or the details of the Maximum function? Max ( here numbers ) = Gives you biggest number. Is this function linear? can someone give me an article to read?
Hi ! im doing (b). assuming $L^1(G)$ has a unit i want to show $G$ is discretre. i think this should use some kind of substitution with the modular function but im not sure how to do that. can someone help?
@BalarkaSen What is the argument that for the topological boundary, we can locally find neighborhoods that are diffeomorphic to the upper half plane? I can see that if the derivative is non-vanishing, then by the global rank theorem, we have slice charts, and hence the boundary is a smoothly embedded 1-dim submanifold. Is that enough for the whole thing to be a 2-manifold with boundary?
Would the general formulation be: if we have an n-manifold, and we look at a subset (giving it the subspace topology) whose topological boundary is a smoothly embedded (n-1)-manifold, then this whole subset is a smoothly embedded n-manifold?
Just pick any non-smooth function f on [-1,1] that vanishes on the boundary, and consider the graph over D^2 of f(|z|). Then the boundary of this graph is a smoothly embedded circle but the disc is not smoothly embedded. One can even rig this example so that the graph is smoothly embedded except at the boundary.
I'm not sure what the immersion theorem would be. Is it the global rank thm for immersions?
@MikeMiller thx for the counterexample. (I was just wondering if that general statement were true, then this specific case would hold by virtue of that)
in my reader for complex analysis, they talk about jordan curves
so I wanted to verify that piecewise smooth jordan curves do satisfy these conditions for Green
so basically what I want is: if we have connected bounded subset of the plane, whose boundary is given by finitely many piecewise (disjoint) smooth jordan curves
then this is an embedded 2-dim manifold of the plane
it will do for me if I can show it for a simple closed smooth jordan curve
i.e., I'd be happy enough if I understand that case
my issue was that I didn't realise that the chart was defined on an open subset of the whole manifold x) for some reason I thought it was only defined for the curve. but we have a specifik representation on the curve, that's all
If you wanted to do the corner pieces (where it's piecewise smooth), you would instead show that there is a diffeomorphism to the upper-right quadrant. That is only a little more tricky.
@ShaVuklia Yes, it's very important that this local model is for the pair $S \subset M$ of a manifold and embedded submanifold.
If it we're just of $S$, we wouldn't know anything new! We would just be seeing that S is a manifold.
This local model for immersions is what Balarka calls the immersion theorem
@BalarkaSen All of these guys should be the limit of C^1 curves if you're allowed to reparam each arc, and surely there's some limiting argument that says that if you have Greens over domains with C^1 boundary you get it with bad boundary
Sure, why not. In the time interval [0,1], trace out (a smoothing of) the unit square; in next few time intervals trace out the unit squares adjacent to this one; then for the next 9 time intervals trace out a subdivision of that unit square into smaller squares (you have more length to trace out, and so you must traverse more quickly)
Just keep doing this and for each n you have some interval [0, c_n] so that its image a curve in [-n, n] with any point say 1/n from a point in the image
Here is a completely different kind of answer to this question.
A perfectoid space is a term of type PerfectoidSpace in the Lean theorem prover.
Here's a quote from the source code:
structure perfectoid_ring (R : Type) [Huber_ring R] extends Tate_ring R : Prop :=
(complete : is_complete_hausd...
> Theorem: Let $(R,R^+)$ be an affinoid perfectoid $K$-algebra, with tilt $(R^\prime,R^{\prime +})$. Let $X=\mathrm{Spa}(R,R^+)$, with $\mathcal{O}_X$ etc., and $X^\prime = \mathrm{Spa}(R^\prime,R^{\prime +})$, etc. . i) We have a canonical homeomorphism $X\cong X^\prime$, given by mapping $x$ to $x^\prime$ defined via $|f^\prime(x^\prime)| = |[f^\prime] (x)|$. Rational subsets are identified under this homeomorphism. ii) For any rational subset $U\subset X$, the pair $(\mathcal{O}_X(U),\mathcal{O}_X^+(U))$ is affinoid perfectoid with tilt $(\mathcal{O}_{X^\prime}(U),\mathcal{O}_{X^\prime}^…
@BalarkaSen "a few years back"
I feel old
I was witnessing this whole thing in the Lean chat
> Kevin Buzzard (May 30 2018 at 13:22): Ok so here is the perfectoid spaces thread. As many people here know, I've long been mulling over the idea of formalising the notion of a perfectoid space in Lean. To the CS people -- it's just some structure, like a group, just a few more axioms and things.
IF $X$ is a topological space and every continuous function $f\colon X\rightarrow\mathbb{R}$ attains a maximum, is $X$ necessarily compact? It's true for metric spaces.
Take the deleted Tychonoff plank. Every continuous function extends to a continuous function on the one-point compactification, since the Stone-Cech compactification and the one-point compactification of the Tychonoff plank agree.
Okay, so I have a question about Poincare duality's proof.
I've seen a couple of proofs of Poincare duality. One is Mayer-Vietorisy proof, and the other is Hodge theoretic one. But neither give any insight into what is actually going on. Like how, Hatcher discusses the notion of dual cell structures, but abandons it for the cap-product formalism when he gets into the actual proof. I am sure none of this nonsense existed during OG Poincare's time. Has some one read the proof by Poincare? Or some other proof that has more geometric intuition?
As far as I know, the only textbook reference for this approach, which is Poincare's original approach, is Seifert and Threlfall's text "A textbook of topology". It's available in English translation but the original was in German. Moreover, Seifert and Threlfall's proof isn't as efficient as i...
I can see that $\omega_1=\bigcup_{\alpha<\omega_1}\alpha$ is an open cover without finite subcover since $\omega_1$ is a limit ordinal, but I don't see why every continuous function $f\colon\omega_1\rightarrow\mathbb{R}$ attains a maximum.
I thought I had an idea, but just now I realized that initial segments aren't connected :/
Thanks, I'll think about this for a bit. Meanwhile I'll ask the follow-up: is there a nice condition on $X$ (weaker than metrizability) such that all continuous functions $f\colon X\rightarrow\mathbb{R}$ attaining a maximum implies $X$ is compact?
A k-facet of an n-simplex can be thought of as a k+1-element subset of {0,...,n}. An (n-1)-facet containing the k-facet, an n-element subset containing your k+1 elements.
To find that n-element set you just need to know which element you don't include.
So that's picking an element out if the remaining (n+1)-(k+1) = n-k elements.
In the example given in the post, he considers an edge of $\Delta^3$, this edge is contained in that edge itself, two adjacent copies of $\Delta^2$ and the $\Delta^3$.
So, I think when he writes facet undecorated, he means any dimensional facet.
Also, ignore what I wrote about the facet being contained in (n-k) many facets.
@loch the only assumption about $G$ is that $G$ is locally compact. i thought maybe taking $U \subset G\setminus \{e\}$ open and look at $\delta X_U$. but i got stuck with this approach.
Write $x_0, \cdots, x_n$ for the vertices in $\Delta^n$. We say that $[a_0, \cdots, a_n]$ are barycentric coordinates for a point $P$ in $\Delta^n$ if we have $$P = \frac{a_0 x_0 + \cdots + a_n x_n}{a_0 + \cdots + a_n},$$ where $a_i \geq 0$ and not all $a_i$ are zero. We consider these equivalent if they are positive scalar multiples of each other.
Here is a completely different kind of answer to this question.
A perfectoid space is a term of type PerfectoidSpace in the Lean theorem prover.
Here's a quote from the source code:
structure perfectoid_ring (R : Type) [Huber_ring R] extends Tate_ring R : Prop :=
(complete : is_complete_hausd...
