@Thorgott oh you're just referring to the increasing pointwise limit construction?

oh wait, you also want smooth

like $\sum \mathbf{1}_[f^{-1}[\frac{i}{2^k}, \frac{i+1}{2^k}) \frac{i}{2^k}$...

yeah, that

then approximate those by smooth functins from below using bump functions

right, thats what i surmised i had to do

If you have approximation by smooth, I recon that dividing by the sup of the absolute value on the function should give approximation by bounded-by-1 smooth functions

thats what i thought as well haha, but i never put pen to paper because i figured it might not be true

I'm a bit confused why you're mentioning $L^p$ and looking at the absolute value

and yeah i wanted to go from 'abstract' approximation by smooth (no specifics like increasingly) to the result

its motivated from a problem one sec

@Astyx imgur.com/a/HPsidn4

im trying to approximate $\mathbf{1}_{y_0 \geq 0} - \mathbf{1}_{y_0 < 0}$ with smooth functions

because I want to show that $sup_{x \neq 0} |F(x)|/||x||$ is the thing in the picture

where $||x|| = max_{t \in [a,b]} |x(t)|$

What's C[a,b] ?

continuous functions on $[a,b]$

with what norm?

$||x|| = max_{t \in [a,b]} |x(t)|$

sup norm

So that doesn't explain why you're mentioning $L^p$ or why you're asking for $C^\infty$ functions

i know i can prove it if i can find a sequence of continuous functions that approximate $\mathbf{1}_{y_0 \geq 0} - \mathbf{1}_{y_0 < 0}$ in $L^1$ and are bounded by $1$ in absolute value

because then I can make $|F(s_n)|$ arbitrarily close to $||F||$ all while keeping $||s_n|| \leq 1$

and I already have $|F(x)| \leq ||F|| |x|$

that's a good idea but you're going to struggle with the details I think

yeah theres probably a slicker way :/

If you look at something like $\cos(1/x)x^2$ on [0,1]

and yes i have been

I can give you a hint at how I would do it

sure

Notice that $y_0$ is uniformly continuous, and divide [a,b] into $|y_0|>\epsilon$ and $|y_0|\le \epsilon$

Hey, @Ted!

I didn't mean to ruin your tutoring session last night. I thought just noting those angles would be useful.

I also haven't had much call to use xkcdConvert :-)

@Astyx i still fail to see :(

@robjohn: Yes, of course, but he's got to build up his basic skills. Sorta scary. Of course, you can't draw those pictures in Mathematica unless you already know the answers :)

Salut @Astyx.

@TedShifrin actually, I had a vague idea, but drawing that image made it clearer to me.

But to draw it in Mathematica, don't you have to specify all the points, etc.?

Or is there some slick command I know not about?

@TedShifrin I know the points, but those were in the original image

It was three circles around the vertices of an equilateral triangle

Right. But to draw the radii and line segments in Mathematica, don't you have to specify Line[{{,},{,}}] and know the exact points?

@Astyx Why are you making this so complicated? Or is the problem actually not the one in the figure?

o/

its the one in the figure

in particular the prove that $||F|| = ...$ part

Based on a (poorly-defined) question that was posed in chat a while back

@TedShifrin I have nice tools to draw what I drew knowing just the vertices of the triangle and radii of the circles (and the start and end angles for the arcs)

Ah, so you did need to know the angles in the first place.

Suppose I have a set of smooth functions $\{f_k\}$ on the real line. Let $A_k$ be the set of $(x,y)$ such that $y\geq f_k(x)$. I can then take the convex hull of these $A_k$. Then there should also exist a function $F$ such that $y\geq F(x)$ defines the desired hull. Is there a name for this $F$ in relation to the $f$'s?

@TedShifrin Yes. I did know that since there are three angles that sweep out $2\pi$ each must be $\frac{2\pi}3$

Yes, of course, @robjohn. My point was only that to use Mathematica as a problem-solving tool for such things you have to already know the answer. Same thing with something like drawing the intersection of two surfaces ...

as I said, I had a vague idea, the image helped to solidify the idea

Yes. I did not use it to solve the problem, just to visualize it

@porridgemathematics I see. So the only issue is that $y_0$ may change sign infinitely many times.

@TedShifrin yeah exactly

maybe I could approximate it uniformly by polynomials

Actually, I think I figured out what I want: Let $f(x):=\min \{f_k(x)\}_{k\geq 1}$. Then form the lower convex envelope of $f(x)$.

I don't think that's a wise idea at all, @porridge.

What question is yours, @Semiclassic?

@Astyx If $X$ is a scheme, would you call maps $\text{Spec} A \to X$ as geometric points if $A$ is a 0-dimensional algebra, or is that just restricted to $A$ being a field?

What evil thing has taken over Balarka's body?

Oh, sorry, @Semiclassic.

np

Meh, geometric point is only for maps from spectrum of separably closed fields

@TedShifrin Writing assignment lol

@BalarkaSen Yes, any algebra

I was trying to find the simplest description for the Maxwell construction here, e.g.

@Astyx Oh, excellent

Cuz otherwise the terminology is not closed under taking fibers lol

Oh, good grief.

@TedShifrin That's the most straightforward way I see to approximate the integrals

physics exam on monday, still haven't started learning

how do I motivation

but in retrospect it's as simple as "form the pointwise minimum of the red/blue curves, and then make it convex"

@Thorgott what's on the exam?

actually I think polynomials works

Yeah, @Astyx, I realized after I knew what the problem actually was. Obviously, we can continuize $\text{sgn}\, y_0$ without a problem unless there are infinitely many sign changes.

Exactly

the idea is that for any function on that interval which changes sign finitely many times, say $g$, there is a sequence of continuous functions $x_{n}^g$ associated with $g$, such that $|x_n| \leq 1$ and $\int x_ng \rightarrow \int |g|$ as $n \rightarrow \infty$

the interesting aspect is that, if you move the red curve up vertically, then you only need to take the lower convex envelope of the blue curve (which is mostly just a straight line)

Well not for the sup norm actually, you always want to leave little gaps around zeroes where the sign changes, but you can make these go to zero

so by moving the red curve down you end up 'splitting' that straight line into two

with the critical case being when the double-tangent of the blue curve is also tangent to the red curve

sorta obvious but also cute

then approximate $y_0$ with polynomials uniformly, say by $P_n$, and call its associated continuous functions $x_m^{P_n}$, now $\int x_m^{P_n}P_n \rightarrow \int |P_n|$ as $m \rightarrow \infty$ which can be as close as we like to $\int |y_0|$ given $n$ sufficiently large. But $x_m^{P_n}P_n \rightarrow x_m^{P_n}y_0$ uniformly in $n$. So I think these combined conclude the result?

whats wrong with this argument

@porridgemathematics Do you know how to prove that?

prove what?

classical mechanics lol

@Thorgott like, Lagrangian/Hamiltonian mechanics?

or just Newton's second law stuff

Or first course?

This : chat.stackexchange.com/transcript/message/57180092#57180092

Newtonian, yeah

yeah, it's a course for physics first-semesters

ah, then that's not too bad

Force diagrams, conservation of momentum/energy.

roller coasters

the classical-mechanics aspect I was having to pound into people's brains lately was "centripetal force is not a force that acts on objects"

@Astyx I think so, more or less, the idea is you start with a constant $+1$ on the interval its positive, minus a tiny bit at the end where you shoot to $-1$ where it starts going negative, and you decrease the length of these tiny parts so that this thing converges almost everywhere to $sgn$ of that function

I had a really good course and great professor, so I loved that course.

shoot linearly

Right, this is exactly what we want to do here

If I ask you to list the forces which act on a given object, I'd better not see centripetal force in that list.

but we want to have a continuous function that changes sign finitely often so we can actually construct this thing

which is why I was suggesting approximating $y_0$ uniformly by polynomials

My hint is meant to guide you around this problem

@Semiclassic You do not allow fictitious forces? What bigotry.

Oh, I'm fine with fictitious forces. Centrifugal force is a-ok in my book.

Centripetal force, however, is not.

sad coriolis noises

If you're thinking I'm bitching about non-inertial forces, I'm afraid you're missing my point :P

Yeah, I was missing your point.

Also, salut Ted

Forces will create centripetal acceleration but there's no centripetal force.

Right.

the course honestly isn't much more than AP high school physics

sadly I forgot all the physics my high school self knew

What about an object whirling over my head on a string? Isn't the tension in the string centripetal? :D

when students hear "centripetal force" they think "oh, like friction"

what?

when they should think "oh, like upward/downward"

@Astyx okay so $y_0$ doesn't change sign at all when its $> \eps$ in absolute value, so you are suggesting thinking about $y_0 \mathbf{1}_{|y_0| > \eps}$ and replacing $y_0$ with this?

centripetal force is a statement about a component of the net force, not about the forces which produce said net force

You didn't answer my question.

@TedShifrin It acts towards the center, so yes. Same as weight being a "downward" force.

@porridgemathematics That function is no longer continuous (it's continuous by part), but has finitely many sign changes

I rest my case.

Going to dinner, I'll read later

Bon appétit, @Astyx.

Pfft. Would you list the forces on a falling object as "weight downward & air resistence upwards & downward net force"?

(The phrase I've seen used is "centripetal force condition", i.e., it's the condition on the net force to produce circular motion)

I'm listing tension as a force and remarking to you that it is centripetal. I'm not listing "centripetal force."

Anyhow, enough.

yeah. hence why I agreed with you :P

the wording I sometimes use is that the required centripetal force is generated/provided by tension in that case

Yes, agreed.

by contrast, I can forgive centrifugal force: if you're in a non-inertial reference frame and so from your perspective you're not accelerating, then it's legit to say that there's a centrifugal force.

i still counsel students not to use it but it does produce sensible algebra.

@Clarinetist use that it evaluates to $\lim_{n} \mathbb{P}(A_1 \cap ... \cap A_n)$

Probably start by doing the simplest case

and show that for finitely many of them $\mathbb{P}(A_1 \cap ... A_n) = 1$

@Clarinetist in general if $\mu$ is a measure and $\{B_n\}$ is a decreasing sequence of sets of finite measure, then $\mu(\cap_{n} B_n) = \lim_n \mu(B_n)$

could probably also write it as $\mathbb{P}(A_1\cap B_2)$ where $B_2=\bigcap_{k=2}^\infty A_k$

and then do inductive stuff

that's inadvertently proving the result I mentioned

oh neat

its how you prove it essentially

So, as I suspected, it's just using continuity of measure from above

makes sense

actually sorry its not, lol, I'm not reading correctly

pfffft

I guess I should be saying something more like $\bigcap_{k=1}^n A_k$ and $\bigcap_{k=n+1}^\infty A_k$

splitting it up is necessary , yeah

@TedShifrin im curious why polynomials is a bad idea?

my patience for proving stuff like this has atrophied over time

gimme something to compute, rather than make me argue about why it's true :P

@porridge: Offhand, it seems to me that the sup bound on the polynomials may well go to infinity. You're going to introduce huge oscillations in the polynomials.

@TedShifrin but the polynomials aren't approximating sgn

they are approximating $y_0$

so why does there sup bound matter

Right. Just seems like you will lose control in the proof. But I'm not thinking about it.

we just want them close enough in $L^1$ to $y_0$ so we can imagine $y_0$ was that polynomial

fair enough

@Clarinetist you want that in any finite measure space, the intersection of a co-null set with any set preserves that sets measure

@TedShifrin math.stackexchange.com/q/4041138/137524

@Clarinetist because $\mu(C \cap X) + \mu(C^{c} \cap X) = \mu(C \cap X)$ oh wait, you don't even need a finite measure space

as long as its conull, and not full measure, you're good

in a finite measure space these notions coincide anyhow

so with you $A_k$, you have that $\mu(A_1 \cap A_2) = \mu(A_2) = 1$ because of this, then induct

@BalarkaSen i don't think i've ever seen people calling a map 'spec k[x]/(x^2) -> X' a geometric point?

Got it, thanks @porridge

@loch yeah lol

Maybe "geometric point" is a more specific notion, but in general, if you have an A-scheme X, and an A-algebra B, you define a B point of X to be a morphisme B -> X over A

yeah i just wanted to call it a point

doesnt matter

I'm guessing geometric requires an (algebraically closed?) field instead of just an algebra

@Astyx i still don't know how to make use of your hint , I only managed to prove it via replacing $y_0$ with a polynomial uniformly close enough to it

forget about polynomials

i guess my issue is I don't know how to really visualize $\mathbf{1}_{|y_0| > \epsilon} y_0$, why does it only have finitely many sign changes? And even if it does, how do we construct a companion function that approximates sgn of this?

I don't have time right now but I'll type something later

ive forgotten about them, but the proof using them does work as far as I can tell

sure thing

@Semiclassical Thanks. Answered.

@porridge I don't see how you're going to get a limiting function $x(t)$ with $F(x) = \|F\|$ with your approach.

BTW, what's a quick proof that the trace form $\text{tr}_{L/K} : L \times L \to K$ is nondegenerate for a separable extension $L/K$?

Or are you just arguing that $\|F\|$ is an upper bound?

that its $\sup_{||x|| \leq 1} |F(x)|

I don't need a limiting function, I just need to be able to get as close as I want to $\int_{a}^b |y_0(t)|dt$ via computing $|\int_{a}^b y_0(t)x(t) dt|$ for some $||x|| \leq 1$

Correct.

Is the issue fundamentally not about trace and more about $L \otimes_K \overline{K}$ not being a reduced algebra perhaps?

(after showing its an upper bound)

and polynomials works for this, although its a bit annoying notationally to write down

@BalarkaSen non-degeneracy of the trace form is equivalent to the trace map not being the zero map by conjugation invariance, the latter is true since trace is inseparability degree times sum over embeddings into an algebraic closure and those are linearly independent by Dedekind

Ahh, nice. I always forget that Dedekind lemma or whatever

Independence of characters

its a weird lemma

I always forget why it's true

yeah haha

it's a manipulation proof

you write the n equations and subtract lol

I bet there's a more conceptual explanation

but I don't know it

questions like this are tiresome: math.stackexchange.com/questions/4041178/…

"Help, I don't have enough information to solve!" "What information are you given?" "Oh, I know this stuff." Then why didn't you say that in the first place

@Thorgott It must be true that the trace map on $L$, basechanged to $\text{tr}_L \otimes 1 : L \otimes_K \overline{K} \to \overline{K}$, decomposes as sum of traces on the various $K$-algebra factors on $L \otimes_K K$.

But if there are nilpotent factors the trace map is zero

Doing it the etale way

@Semiclassical It's so exhausting that so many people ask questions without specifying the conditions and assumptions

Yeah, that decomposition stuff is true

@BalarkaSen only compact stuff on the exam

crab rave

@Semiclassical I agree, but with all the people wanting us to do their exams/homework, it's nice that someone might be trying to think for himself a little bit more. Little.

@user2103480 Cool! How was it?

Oh wasnt yet

next week

Oh ok so he announced no noncompact crap

That's good yeah

I dunno the etale viewpoint

yeh exactly, after someone voiced their concerns about homology-orientations and "limits of cap products with orientations"

For any $K$-algebra $A$, there's $\text{tr}_{A/K} : A \to K$ by sending $a$ to $\text{tr}_K(\times a)$.

but collar and tubular neighborhood are relevant

So one should prove this is nonzero if $A$ is reduced

I could imagine that this gets somehow combined with our introductory homotopy theory stuff at the start of the course

The point I think is if you tensor with $\overline{K}$ you end up with $\text{tr}_{A\otimes\overline{K}/\overline{K}}$

But $A \otimes \overline{K} \cong \overline{K}^n$ as algebras

For which the trace map is obviously nonzero

@Semiclassic Good luck with integrating $\theta'' = \pm\sin\theta$. :P

Sine Gordon equation lol

@TedShifrin well, it's not actually as bad as that. you have the first integral $\theta'^2/2 \pm g\cos\theta=$const

so that gives you the angular velocity as a function of angle, which may be all that's desired here

if you do want the angle as a function of time, of course, then have fun learning about elliptic functions

hey guys! what's an example of an incomplete orthonormal system in $\ell_2$?

i would appreciate any hint on how to find it

@user153330 What have you tried?

@TedShifrin i tried making a non-closed orthonormal system $e_1=(1/\sqrt{2},1/\sqrt{2},0,...), e_2=(0,1,0,0,...), ...$

What do you mean by non-closed?

@TedShifrin an orthonormal system A is closed in V iff $\sum_{n=1}^\infty |\langle v,e_n\rangle|^2=\|v\|^2$ for all $v\in V$

closed implies complete, that's why i need it to be non-closed

So I don't understand your list. What does ... mean there?

BTW, the first two vectors aren't orthogonal, are they?

@TedShifrin oh that's true

@TedShifrin like it just goes on $e_3=(0,0,1,0,...)$ but this is useless as you showed

Sounds like the standard orthonormal system to me, once you fix the first one.

@TedShifrin yeah standard basis but e_1 was different

Well, as we've agreed, that won't work.

You could change $e_2$ to make it orthogonal to $e_1$. But what good does that do you?

@TedShifrin i can't think of any other simple basis for $\ell_2$ unfortunately

Well, the whole point is that you do not want a basis.

$e_1=(1/\sqrt{2},1/\sqrt{2},0,0,...), e_2=(0,0,1/\sqrt{2},1/\sqrt{2},...)$ would this construction work?

it's not a basis

So that's like listing $e_1,e_3,e_5,\dots$

@TedShifrin essentially yeah

Seems fine to me. There is an easier answer, too.

@TedShifrin easier than your $e_1,e_3,e_5,\ldots$ : ) ?

Yup.

$e_2,e_3,...$ ?

Well, that'll work. But there's still an easier solution, I think.

i'm confused. wouldn't just dropping e1 from the basis give you such a system?

oh

That's what he/she just did. ;)

@TedShifrin lol can't think of anything easier unfortunately :P

yeah, was scrolled up when i typed that out

What about any finite orthonormal system?

@TedShifrin ohh that's true :)

thanks a ton @TedShifrin

Sometimes we miss the most obvious things :)

You're welcome.

@porridgemathematics I'm back if you want to discuss your problem

Ah, Astyx est revenu ... ayant bien mangé :)

J'ai aussi regardé un film avec ma famille

Un bon film?

Une série plus précisemment. Pas du grand cinéma mais quelque chose de divertissant :)

Eh bien. Quelque chose de divertissant ces jours-ci nous convient!

how do you guys survive working without getting back pain?

Oui il faut quelquechose pour survivre le covid

My back and neck are total messes. I do all sorts of exercises.

back pain ? you mean from sitting ?

Absolument d'acc, @Astyx.

Walk around a lot, work on your posture, @user153330.

Do some exercise

Maybe a stand up desk could be useful too

At least walk a few miles every day.

@TedShifrin i'm especially interested in your case since you're the oldest among us :) do you ever experience back pains?

I just said so :) I've had degenerating disks for quite a while, thanks to my relatives. I have gone to chiropractors for 15 years now.

Now going to physical therapy and massage, too. Lower back and neck have been bad for years.

But my posture is terrible.

@TedShifrin sorry i didn't notice ur first comment

Yes, I'm antique.

But I did mess my back up doing some stupid things when I had my house redone 13 years ago or so.

Now I no longer play tennis because of the neck pain :(

@TedShifrin sorry to hear that

and thanks for giving me some of your experience on this topic

i will do my best as a youngster to take this matter seriously

Yes, at least exercise regularly, and try to get up every half hour and do some stretching when you're at your desk.

And get a desk that's adapted to you!

Chair more important, maybe.

it's very hard to keep a good posture when your desk is too low/too high

But Astyx is right, too.

is there a categorical property characterizing the ring $R$ as object in the category of $R$-modules

here's a cool observation: the connecting map in the snake lemma is the unique natural such map up to natural automorphism of the identity functor

If $I$ is an invertible fractional ideal of a domain $A$, I can just write the inverse as the denominator ideal $(A : I)$, right?

$A \subseteq I (A : I) \subseteq A$ yeah ok

Uniqueness of inverse

If $E/\Bbb Q$ is a number field the different ideal is $\delta_E = \{x \in \mathcal{O}_E | xy \in \mathcal{O}_E \forall y\in E : \text{tr}_E(y\mathcal{O}_E) \subseteq \Bbb Z\}$

$a_1, \cdots, a_n$ be an integral basis of $\mathcal{O}_E$, tr-dualize to a $\Bbb Q$-basis $b_1, \cdots, b_n$ of $E$ so everything in $E$ which satisfies $\text{tr}_E(-\mathcal{O}_E) \subseteq \Bbb Z$ is a $\Bbb Z$-linear combo of $b_1, \cdots, b_n$

Is it literally an equal to? Yeah, $\delta_E^{-1} = \bigoplus_{i = 1}^n \Bbb Zb_i$.

Ah ok this is just taking dual of a lattice wrt some form

One second

Covolume of the lattice is discriminant, so volume of the dual lattice should be discriminant, or something like this.

$[\mathcal{O}_E : \delta_E] = [\delta_E \delta_E^{-1} : \delta_E \mathcal{O}_E] = [\delta_E^{-1} : \mathcal{O}_E]$

I guess I can write $a_i$ as linear combo of $b_i$, $(a_i) = M(b_i)$... $a_i = \sum m_{ij} b_j$

$\text{tr}(a_i b_j) = \delta_{ij}$ so I get $m_{ij} = \text{tr}(a_i a_j)$

det(M) is discriminant, so that's the index

@Thorgott Hom(R,M) = M

ich habe seinen Name vergessen

yeah, it represents the forgetful functor

or, it's free on one generator

lol

@BalarkaSen do you know what such an object is called?

what object

Hom(R,M) = M

the object R

lol no

why would i know that man

as I said, it's a representing object for the forgetful functor

that's the name

I guess it's also a representing object for the identity functor with respect to internal hom

this is giving me an overwhelming sense of impending doom [credits: youtube.com/watch?v=sPof3taFRto ]

how did you like my snake lemma fact

cool i guess lol

@BalarkaSen wtf

@BalarkaSen Isn't covol of the lattice the square root of the discriminant?

yeah

that was a flippant comment, but the point is you can switcharoonie

if $I$ is a fractional ideal in a Dedekind domain $R$, $R/I \cong I^{-1}/R$

as $R$-modules

If you see the discriminant $d_E$ as the norm of the different $\delta_E$, then you have $1=N(\mathcal O_E)=N(\delta_E\delta_E^{-1}) = N(\delta_E)N(\delta_E^{-1})=(d_E)N(\delta_E^{-1})$

You have to set up the theory of norm of a fractional ideal to do stuff like that

ye

@Astyx me 2. Let us discuss algebraic geometry sometimes

I am doing my PhD in it.

Hi @TedShifrin

@TedShifrin I just bought a gaming chair haha

Cool, on what topic ?

variant of the Hodge conjecture

Something I know nothing about. Have you already started?

Yeah just started research this semester.