@yCalleecharan there's an with a picture in the bar above the editing frame. If you hover with your mouse over it it says "image <img>" or something of that kind. Click it and follow the instructions.
@BrianMScott Einstein once said time is relative. If a beautiful girl sits on your lap for an hour it feels like an minute; but if an ugly girl sits on your lap for a minute it feels like an hour...
“Put your hand on a hot stove for a minute, and it seems like an hour. Sit with a pretty girl for an hour, and it seems like a minute. THAT'S relativity.”
@JonasTeuwen True, but as I said, I don't know enough to reduce it to an incomplete Bessel right now. Maybe later today I will have time to look deeper.
I wonder why they keep writing "additive group" instead of just "group". To me it makes no difference whether the group op is written $\cdot$ or $+$. This makes me think I'm missing something important.
I think a further reason is that he has applications in number theory in mind, and there is the additive and the multiplicative brand of number theory (among other things)...
But not even Ben Green can give a good definition of "additive" combinatorics. See here
im sure its all the same, i just can't get it to integrate a function. so i probably can't get it to differentiate. Couldn't get it to graph the other day either
Can you confirm this? $Z$ with the cofinite topology is a topological group. I need to check that $f_- : Z \to Z, x \mapsto -x$ and $f_{+y} : Z \to Z, x \mapsto x + y$ are continuous. If $S \subset Z$ is open, i.e. $S^c$ is finite then $f_-^{-1}(S)^c = f_-^{-1}(S^c)= -S^c$ is finite. Similarly $f_{+y}^{-1}(S)^c = f_{+y}^{-1}(S^c) = S^c - y$ is also finite so this is a topological group.
And every set in $Z$ is compact, right? For that let $S_i$ be an open cover of $S \subset Z$ with $S_i^c = \{z_{i1}, \dots, z_{in} \}$. Then construct a finite open subcover by taking $S_0$ and then for each $z_{k0} \in S_0^c$ one set $S_i$ with $z_{k0}$ in $S_i$?
@tb What is? I just made up an example because I don't really know what a topological group is. And since Tao mentions Pontryagin dual without giving a clue what it is and its definition on Wikipedia requires me to know what that is first I decided to look at examples.
The way you wrote the axioms. Those are good for abelian groups but not for non-commutative ones. And you do want to look at Hausdorff groups if you want to talk about Pontryagin duality.
But apparently the dual group is just all the characters of the group. Why does the group have to be commutative? I guess I'll know the answer to this in a few hours...
Yes, it is locally compact in every possible sense for spaces. But when people mention locally compact groups they usually assume that it is Hausdorff (or equivalently for groups $T_0$).
Since a character is a continuous homomorphism $\chi: G \to S^1$ to an abelian group, it factors through the abelianization of $G$, which is again a locally compact groups. The characters only see the "abelianization" $G / \overline{[G,G]}$ of $G$.
Good examples to have in mind are $\mathbb{R}$, $\mathbb{Z}$ and $S^1$ for locally compact groups.
Others would $p$-adic stuff like $\mathbb{Z}_p$ or $\mathbb{Q}_p$.
@MattN well, you can take the multiplicative group of $\mathbb{C}$ for the values of a character... But for Pontryagin duality people usually take its subgroup $S^1$. Why would it be important for $S^1$ to be a multiplicative group of a field?
@tb Because according to Wikipedia "A multiplicative character (or linear character, or simply character) on a group G is a group homomorphism from G to the multiplicative group of a field."
Should that say subgroup of the multiplicative group of a field?
@MattN that's what people dealing with algebraic groups do (because $F^\times$ is an algebraic group while $S^1$ isn't). I don't know Tao's conventions.
But for Pontryagin duality people usually consider $\chi: G \to S^1$ (and they write $\mathbb{T}$ for $S^1$). See here
@MattN well, the image of a homomorphism is always a subgroup...
@robjohn That may be a new feature of the software. Ilya most definitely edited and I asked him immediately what was wrong. He seems to have rolled the edit back (probably within the usual five-minutes time window) and all traces of his edit disappeared.
@tb I'm not sure if I should approve that. People that know special functions will probably open it with this title, and people that don't will probably not be able to solve it anyway.
The title is just some thing to attract readers, is my opinion.
I also didn't know what to make of this. To me, adding $3$ to itself a bunch of times is more prone to error than just knowing that $3^2 = 9$; I'm bad at mental arithmetic but it seems crazy to not have that one down.
@JonasTeuwen I think your question title was fine. Gerry was probably talking about titles of the type "induction proof help." Or the vast majority here
I don't know how many Sunday mornings I spent in my old flat, lying in my bed and waiting for the church to finally stop ringing its bells, while having the most colorful dreams of how I could go and burn that darn noisy thing down....
Yes, but Sundays were the worst. I could get used to the five minutes four times a day and the time bells didn't particularly disturb me, but the entire hour spread over the whole Sunday morning was almost too much for me to take...
hello, I'm having a bit of a brain fart moment here... can someone point out (and smack my head) as to why $\sum_{n=k}^\infty 1/n^2<\int_{k-1/2}^\infty 1/x^2\ dx$ is true, when the same isn't if the lower integration limit were simply $k$?
E.g. you have for $\xi \in \mathbb{R}$ the homomorphism $\chi_{\xi} : x \mapsto e^{i\pi \xi \cdot x}$ and write $\hat{f}(\chi_{\xi}) = \int_{\mathbb{R}} f(x) \overline{\chi_{\xi}(x)}\,dx$ (in slightly more classical notation this would be $\hat{f}(\xi)$)slightly
From Tao's blog: "...given by Fourier coefficients $\hat{f}(\xi)$, which take values in some dual object such as the Pontryagin dual $\hat{G}$ of $G$." Does this not mean that $\hat{f}$ is a homomorphism $? \to \hat{G}$?
@yoda I believe that the expression on the left hand side of the < symbol uses integer values of n (and hence k is also an integer) while the integral on the right hand side uses all real numbers. So, the inscribed rectangles will be less in the summation. Does that make sense Yoda?
The homomorphisms $\mathbb{R} \to S^1$ are precisely the functions $x \mapsto e^{i\xi x}$, where $\xi \in \mathbb{R} \cong \widehat{\mathbb{R}}$. That's why you take the Fourier "coefficients" $\hat{f}(\xi) = \int f(x) \overline{\langle x,\xi\rangle} \,dx = \int f(x) e^{-i\xi x}\,dx$ to be indexed by $\xi \in \mathbb{R}$
@MattN Oh, just something frugal, a bit of salad, bread and cheese.
@DylanMoreland well, I think since it's constantly decreasing, if you shift it by half, then the excess area to the left of the integer point is greater than the void (undercount) to the right of the integer, so you'll always end up with an excess, providing a sufficient bound
