Can I use $F \lesssim K$ to mean an isomorphic copy of $F$ is sitting inside $K$ ?

Let G be a finite group. Is the condition that for every subgroup H of G, the chain of normalizers $1 \leq H \leq N_G(H) \leq N_G(N_G(H)) \leq ...$ stablizes to the whole group G equivalent to G being a nilpotent group?

yes if I remember correctly

Hi guys can someone help me with functional and variations ?

askaway

May I link the thread

which was in the physics forum

because there I have given a good overview of what is going on

I mean I can even say the following

How would one expand $S[\vec q + \delta \vec q]$? Where $\vec q(t)=(q_1,q_2,...q_N)$?

and S is a functional

In this article: en.wikipedia.org/wiki/Functional_(mathematics)

Why isn't this a contradiction

In functional analysis, the term linear functional is a synonym of linear form;that is, it is a scalar-valued linear map.

And then the example given is

The arc length functional has as its domain the vector space of rectifiable curves – and outputs a real scalar. This is an example of a non-linear functional.

How is it a non-linear functional if the result is a scalar

and by definition a linear functional is a scalar valued linear map ?

cause it's not linear

In the answer math.stackexchange.com/questions/1806042/… in step 2, "Each component of ∂N is homeomorphic to a 2-sphere." I wonder how many component ∂N would have.

Does it only have 1 component?

as many as there are curves?

it only has one component

I try to prove component ∂N only have 1 component and is homeomorphic to S^2, given $g$ disjoint curves which represents linearly independent homology classes in ∂H_2, as characteristic curves

good luck

the euler characteristic will help :)

∂H_2 is homeomorphic to the genus $g$ orientable surface

Its first homology group is Z^{2g}

don't try to work this out live in chat, this may need some thought

I thought about it and the number doesn't match

which number

Attaching $g$ 2-handles to $H_2$, the resulting 3-manifold should have first homology group $0$.

...no?

didn't we have this a while ago? attaching a 2-handle doesn't necessarily kill a generator

ponder a genus 1 Heegaard diagram for $S^1 \times S^2$

...or, more generally, that of a lens space $L_{p, q}$

I try to prove the resulting 3-manifold has boundary $S^2$, by showing its first homology group is $0$.

oh, that

yes, do that :)

Yes

given $g$ disjoint curves which represents linearly independent homology classes in ∂H_2, as characteristic curves

But the first homology group of ∂H_2 is Z^{2g}

The numbers don't match. I must be wrong.

I see no contradiction

they match just fine

again, sit down with a piece of paper and work this out carefully

do some examples

and then think carefully about what effect attaching a 2-handle has on the euler characteristic

it is not reduction by 1

recently Hatcher uploaded a revision paper on arxiv

revision of what?

arxiv.org/abs/1312.3518

lol

that's old

why are they publishing a 12 year old expository paper

because it’s Hatcher

whos publishing it?

L'Enseignement Mathematique

French journals are not real to me

good, because it's swiss :^)

or at most half-french-half-swiss

but it seems to be owned by a swiss foundation

Let $f$ satisfy the following property. If, for some $x$, there are constants $\delta\gt0$ and $M\lt\infty$ such that $$|f(x+t)-f(x)|\leq M|t|\tag{1}$$ for all $t\in(-\delta,\delta)$, then the Fourier series of $f$ converges to $f(x)$. This is a theorem in Rudin's.

To prove this, he uses an auxilary function, namely $$g(t)=\begin{cases}\frac{f(x-t)-f(x)}{\sin(t/2)}& 0<|t|\leq\pi\\ 0&t=0.\end{cases}$$He claims at a decisive moment in the proof that $f_1(t)=g(t)\cos(t/2)$ and $f_2(t)=g(t)\sin(t/2)$ are bounded on $[-\pi,\pi]$. Is there an easy way to verify this claim? I feel like I need to use $(1)$ somehow, but I only get, in the case of e.g. $f_1$, that $$|f_1(t)|\leq M\left|\frac{t}{\sin(t/2)}\right|.$$How can I bound this further?

It's "obvious" that $f_2$ is bounded. But with $f_1$ I struggle.

I am looking for a recommendation for a numerical analysis book. What book do you guys recommend?

@psie actually, it is not so obvious that $f_2$ is bounded either, since $(1)$ only holds on $(-\delta,\delta)$. I feel stuck. :(

Take a break.

Is there a consensus on what a "category with products" is? I can think of three possible meanings: (i) binary products always exist, (ii) finite products always exist (so that in particular a terminal object exists), (iii) arbitrary products always exist.

If you presented the term to me without context I would assume it means "has all small products"

Okay, thanks. The context I am actually interested in is to do with a quite general definition of a sheaf that appears in the Stacks Project. It defines what it means for a functor to be a "sheaf on $X$ with values in $\mathcal C$" for any topological space $X$ and category $\mathcal C$ with products.

see the last paragraph here stacks.math.columbia.edu/tag/001R

I think C, but that looks weird

...that's why you don't put a comma :)

I know this is somewhat controversial, but I always put punctuation in my diagrams

@Joe I would call a category satisfying (ii) "cartesian", but I admit this terminology is not universal either

Mac Lane puts the punctuation in the lowest line, rightmost column

rightmost column of the whole diagram, that is

so in a diagram of your shape the puncutation is "orphaned"

see e.g. the diagram on page 63 of CFTWM

it's hideous

ok I definitely won't do that

:)

otherwise the natural reading direction suggests putting it after C, but I feel like either B or C are about equally permissible

it's not going to feel quite right either way

can't you justify reorienting the diagram?

well, the horizontal arrow is an equivalence between two models of the infty-category of (infty,infty)-categories and the object at the bottom is the category of gaunt omega-categories fully faithfully embedding into either, so it feels a bit weird putting it at the top

but I'll probably do that cause there's no good alternative

can't you just move C to the right?

that sounds like the chaotic good option

Suppose we have some operation on a set $Y$, $\circledast$, and consider the induced operation on $Y^X$ defined by $(f \odot g)(x) = f(x) \circledast g(x)$ for all $x$. It's correct that $\odot$ is commutative/associative if and only if $\circledast$ is commutative/associative, right?

@EE18 Nope :)

but that's almost correct

it seems correct to me

ah ok nevermind

it's a French type of issue

something something universal counterexample something

the operation on $Y$ has the property if and only if the operation on $Y^X$ has the property for all sets $X$, by the Yoneda lemma :)

(this is not a serious remark)

roaring applause

Why in the defintiion for twice differentiable at multivariable calculus it makes sense to require the existence of the limit for any sequence $(v_n)$ and not just for unit vectors sticking to one direction?

I have no clue how I did this accidentally

that arrow has no business being this long

@Thorgott Arrow envy?

Does anybody know how to write multiple things under the limit sign in latex, as above?

\substack

$\displaystyle\varprojlim_{\substack{U \in \mathcal{B} \\ U \subseteq V}} \mathscr{F}(U)$

hmm that makes me wonder, what if

$\displaystyle\varprojlim_{\substack{A \\ B \\ C \\ D \\ E \\ F \\ G \\ H \\ I \\ J \\ K \\ L \\ M}}$

splendid

finally, real math

finally, real typesetting

this is no ordinary typesetting. we're quantum-typesetting through the 5th dimension

@X4J i am not sure if i understand this question but an example to keep in mind might be something like f given by f(x,y) = 1 if y = x^2 and 0 otherwise, at the origin. if you only approach (0,0) via paths or sequences that lie on a fixed straight line (i.e. fixing a "direction" and only approaching from that direction) you will get 0 in the limit because a line through the origin is going to cut the parabola in at most two places, one of which doesn't matter to the value of the limit.

you can still get a nonzero limit however if you approach the origin along the curve y = x^2.

so for examples like that you would not want to generally define "the limit of f as you approach (0,0)" in terms of paths to/through the origin that do not change direction. even if you might be able to interpret the limit that way a lot of the time for "nice" functions, the definitions of "nice" that you want to formulate probably also implicate some general definition of limit.