Mathematics - 2020-12-10 (page 2 of 2)

anakhro

4:00 PM

The extent of my optimization is the second derivative test, so it's safe to say you know more than me. ;)

Rithaniel

XD fair enough

Alessandro Codenotti

@Thorgott You probably know about this, say I have some category, can I always "formally complete" it to have all limits? Does this process have a name?

Leonhard Euler

I wanna ask smth

What is your favorite result/theorem in mathematics?

Tobias Kildetoft

@AlessandroCodenotti Probably you can do that completion by considering the category of diagrams with suitable morphisms

Thorgott

the answer is "yes!" for small categories and "I believe there are subtle set-theoretic issues I never paid attention to" for moderate categories

Tobias Kildetoft

4:07 PM

@LeonhardEuler Steinberg's Tensor Product Theorem, not even close.

Leonhard Euler

Okay nice

Alessandro Codenotti

@Thorgott My categories are concrete small categories with countably many finite objects so set theoretic issues should not be a concern

Leonhard Euler

Mine is the Dirichlet theorem on arithmetic progressions

Alessandro Codenotti

@TobiasKildetoft Right it's clear I can just add an object for every diagram that's missing its limit and forcefully add morphisms to guarantee it is the limit I want, but it's not clear to me what morphisms between those new objects should be

Thorgott

so the point is that Yoneda gives you a fully faithful covariant embedding $\mathcal{C}\hookrightarrow(\mathbf{Set}^{\mathcal{C}})^{op}$ and the latter category is complete (limits of functors work "pointwise")

this embedding is universal in an appropriate sense

Alessandro Codenotti

4:10 PM

But adding those new objects also adds new diagrams, so I might need to iterate this process. I probably can stop after some infinite number of steps depending on how big the category I started with was

Thorgott

I believe the appropriate sense is that you get a right biadjoint to the inclusion of the 2-category of complete categories with continuous functors into the 2-category of categories

Alessandro Codenotti

@Thorgott Ah that's a nice trick, embed "canonically" in something complete

Tobias Kildetoft

@LeonhardEuler That is a great one too

Thorgott

the adjunction can be easily expressed as: every functor $\mathcal{C}\rightarrow\mathcal{D}$ into a complete category $\mathcal{D}$ corresponds to an up to natural isomorphism unique continuous functor $(\mathbf{Set}^{\mathcal{C}})^{op}\rightarrow\mathcal{D}$

Alessandro Codenotti

continuous?

Thorgott

4:13 PM

preserving limits

Alessandro Codenotti

I see

Leonhard Euler

@TobiasKildetoft yes sound simple but hard to prove

It is said to be the beginning of analytic NT

Balarka Sen

whats up

Astyx

Am I right that if M is a quasi-coherent sheaf on X a schmeme, a family of global sections generate M(X) iff they generate M(U) for any open U in X?

Alessandro Codenotti

Hi @Balarka

Astyx

4:14 PM

Hi Balarka and everyone else

Alessandro Codenotti

Are you familiar with solenoids by any chance?

Balarka Sen

Sort of.

Thorgott

the point should be that every object in $(\mathbf{Set}^{\mathcal{C}})^{op}$ is a limit of objects in the image of the covariant Yoneda embedding, so where that image goes determines a continuous functor up to the ambiguity of limits (but since any two limits of the same diagram are canonically isomorphic, this will give a natural isomorphism of the functors)

Balarka Sen

@Astyx Sounds right, why do you need quasicoherence?

Alessandro Codenotti

@BalarkaSen Can you convince me that they are not locally connected?

Balarka Sen

4:16 PM

Locally they feel like Cantor set x I, right?

Alessandro Codenotti

I mean it's quite clear from the pictures of the toruses wrapping inside other toruses

Astyx

@BalarkaSen You might not have that the restrictions are surjective? Looking for instance at a compact complex manifold and smooth functions

Alessandro Codenotti

I don't see it from the actual definition as an inverse limit of circles

Balarka Sen

@Astyx You have some elements of M(X) which you said by definition generate M(U) once you restrict. Then M(X) -> M(U) is automatically surjective, or no?

I am confused, but maybe you can clear my confusion

@AlessandroCodenotti Ah OK. So for example you might be working with the $p$-adic solenoid, where the maps $S^1 \to S^1$ in the inverse system are $p$-fold

Alessandro Codenotti

Ok

Balarka Sen

4:19 PM

Restrict to a fiber, that looks like the system $\cdots \to \Bbb Z/p^2\Bbb Z \to \Bbb Z/p\Bbb Z \to \{e\}$

Set-theoretically that is. The inverse limit is a Cantor set

Alessandro Codenotti

Right

Balarka Sen

So you expect (Cantor set) x I (the circle-direction) to be the local structure

Astyx

@BalarkaSen I'm asking if they do generate M(U) once you restrict, which I believe is not always the case.

Alessandro Codenotti

Fair enough, that makes sense

Astyx

I think that comes from quasi-coherentness

Balarka Sen

4:21 PM

@Astyx Ah I see now. You are right.

Alessandro Codenotti

In general a fiber looks like some inverse limit of finite discrete spaces, which is always some zero dimensional compact Hausdorff space

Balarka Sen

So you want to prove quasicoherent sheaves are flasque? Hmm.

This seems suspect.

Astyx

Yeah, but that can't be right

Balarka Sen

Do you know an example? I for sure don't!

Astyx

It's probably wrong

My knowledge in what is/isn't quasicoherent is very lacking, so no, no examples

Balarka Sen

4:26 PM

I understand these as just modules, which is what they are over affine opens

What does flasqueness mean for modules? Urgh

Flasque guys have zero sheaf cohomology, but does that help...

Thorgott

ah ok, I never learned this explicitly, but the point is that a functor $F\colon\mathcal{C}\rightarrow\mathbf{Set}$ is the limit of $\operatorname{el}(F)\rightarrow\mathcal{C}\rightarrow(\mathbf{Set}^{\mathcal{C}})^{op}$, where the first morphism is the canonical projection and the second the covariant Yoneda embedding

Astyx

I doubt it's an easy question in all generality

Balarka Sen

I think $H^1(X; F)$ vanishes for any affine scheme $X$ and a quasicoherent sheaf $F$ over it

This is what should happen

So sheaf cohomology would not detect flasqueness

Great question I think you should ask in MSE

Astyx

Will do!

nbro

1

Q: Why is second order backpropagation useful?

Raul Rojas's book on Neural Networks dedicates section 8.4.3 to explaining how to do second-order backpropagation, that is, cumputing the hessian of the error function with respect to two weights at a time. What problems are easier to solve using this approach rather than first order backpropagat...

backpropagation

Thorgott

4:38 PM

@Thorgott no, this is wrong

I'm too stupid to dualize

Balarka Sen

be wise dualize

Thorgott

Ok, the issue is that I'm in fact very stupid: the covariant Yoneda embedding takes values in $\mathbf{Set}^{\mathcal{C}^{op}}$. It is then clear that a functor $F\in\mathbf{Set}^{\mathcal{C}^{op}}$ is the *co*limit of $\operatorname{el}(F)\rightarrow\mathcal{C}\rightarrow\mathbf{Set}^{\mathcal{C}^{op}}$. This gives us the Yoneda embedding as free cocompletion.
Then we dualize and get that $(\mathbf{Set}^{(\mathcal{C}^{op})^{op}})^{op}=(\mathbf{Set}^{\mathcal{C}})^{op}$ is the opposite of the free cocompletion of $\mathcal{C}^{op}$, so the free completion of $\mathcal{C}$.

now everything works out

schn

5:05 PM

I have a Gaussian function given by 3*exp(-500*(-0.5 + x)**2-990(x-0.5)(y-0.5)-500(-0.5 + y)**2) and I'd like to get the exact same Gaussian, but instead of being symmetric around y=x, symmetric around y=0.5.

The middle term in the exponent needs to go away, but then one gets a circular Gaussian, which is not the same as the one before.

123

Hi guys..

In group theory. What is the meaning of $a^{-n} = e$

How order of the group can be negative??

$a^n = (a^{-1})^{-n}$

How order of element is negative??

Thorgott

what kind of condition is it for a chart of a Riemannian manifold to induce an orthogonal frame? it's implied by conformality of the chart, but appears to be a weaker property

so I guess I'm really asking what kind of linear maps the isomorphisms between inner product spaces mapping an orthogonal basis to an orthogonal basis are

Balarka Sen

That forces conformality.

But I am not 100% clear on what your question regarding Riemannian manifolds are. Are you asking for coordinates under which the metric is diagonal?

Thorgott

5:22 PM

consider like $\begin{pmatrix}1&0\\0&2\end{pmatrix}$, this maps an orthogonal basis to an orthogonal basis, but does not preserve angles

it maps the standard basis to an orthogonal basis, I should specify

Balarka Sen

@Thorgott Ah, this is what was confusing me.

If you demand any orthobasis is sent to an orthobasis you're conformal

Thorgott

right, that I agree with

Balarka Sen

So you're asking for coordinates in which the metric is diagonal. This actually forces existence of coordinates in which the metric is conformally flat in high enough dimensions, like $\geq 4$

In dimension $2$ all metrics are conformally flat so that's boring. Dimension $3$ you might find examples

Thorgott

ah, that's interesting

Balarka Sen

Look up the Weyl tensor

Astyx

5:29 PM

If I have an A-module M, why are the stalks of $\tilde M$ generated by the valuation at the stalks of generators of $M$?

Balarka Sen

Actually I think there are no examples. Weyl tensor vanishes iff conformally flat with the only exception of the Cotton tensor in dimension 3. You can check Weyl tensor dies in orthogonal coordinates, so >= 4 is done. I think this Cotton guy will vanish in orthocordinates as well

Orthogonal coordinates implies conformally flat (which is kind of surprising)

Thorgott

Yes, very surprising

Balarka Sen

I have something worked out somewhere, let me see

in random cohomology for quantum nerds, Apr 28 at 1:44, by Balarka Sen

OK. Let's do this

Mary Star

@robjohn Ahh ok!! Thank you :-)

Thorgott

Oof

123

5:44 PM

Is there any simple elements non-Abelian finite group example??? Pls share

Like $(G, \times_5)$ So, $G = \{1 , 2 , 3 , 4\}$

Astyx

6:02 PM

Is it just that if $(m_1, \dots, m_n)$ generates M as an A-module, then its restriction to $D(f)$ generates $M\otimes A_F$ as an $A_f$-module, so it generates the colimits?

Tobias Kildetoft

6:28 PM

@123 The smallest example is the group of permutations of 3 objects, which has 6 elements total

Rithaniel

7:09 PM

@123 Check out the dihedral group or the group of quaternions.

If you're asking for a simple group, as in "doesn't have any nomal subgroups," then the smallest non-abelian one you will find has 60 elements

Thorgott

7:22 PM

(this fact takes a bit of work)

Ted Shifrin

7:44 PM

I certainly do not know this. Re discussion w/@Balarka and @Thor

@Rithaniel 123 is just starting. He means simple in the naive sense.

Rithaniel

Ah, gortcha

Yeah, dihedral groups, then. Tobais's example is particularly good

123

@Rithaniel Which group has 60 elements of non-abelian group?

Alessandro Codenotti

$A_5$, the alternating group on 5 elements

Rithaniel

I'm referring to the Alternating group on 5 elements. If you get pretty deep into talking about group theory, you should start seeing it crop up

(Ninja'd by Alessandro)

123

@Rithaniel Thanks i check this.

Rithaniel

7:51 PM

No problem

Ted Shifrin

Also the group of symmetries of a regular icosahedron or dodecahedron. My preferred approach.

Alessandro Codenotti

The geometric approach strikes again

Rithaniel

Oh, geometric

Ted Shifrin

Duh.

123

@TedShifrin Oookay....

Karim Mansour

8:00 PM

@TedShifrin I am becoming lazy when doing background reading; I tend to do the problems in my head and not write them on the iPad.

Ted Shifrin

@Karim: If it's stuff you can do in your head, this is very elementary background :)

Karim Mansour

I m reading this book webusers.imj-prg.fr/~fouad.elzein/Hodge.pdf

I guess at the beginning it is elementary

the book considers the start of shimura varieties advanced

Ted Shifrin

Yes. Wow. Very cool book. I know a lot of the authors/contributors.

Karim Mansour

Yeah awesome book

Sayan Chattopadhyay

8:16 PM

@KarimMansour maybe I am thinking of the wrong person here, but weren't you interested in Mirror Symmetry sometime back?

Karim Mansour

Yes that is me

Adeek is my previous name

was *

Sayan Chattopadhyay

Right I see. Neat.

Karim Mansour

What is your interests @SayanChattopadhyay?

any topic yet or just intersection between geometry and physics ?

Levi

Hi

Sayan Chattopadhyay

@KarimMansour they vary from time to time, but I maybe slowly drifting towards symplectic geometry and mirror symmetry, more on the categorical side of it.

Karim Mansour

8:20 PM

very interesting

I am interested in Hodge theory, mirror symmetry, and symplectic geometry.

Sayan Chattopadhyay

But it largely depends on how my master thesis pans out. If I get to do things like deformation quantisation, Kontseivich stuff, maybe I go more into the categorical aspects, if not I might end up with more of symplectic geometry, Hodge theory

Though I have just very recently started studying Hodge theory.

Karim Mansour

That is cool. My master thesis was mainly Hodge theory.

it is freely available I cover a lot of stuff

Invariants Associated to K-theoretic Methods and Complexity of Algebraic. Cycle Groups

Sayan Chattopadhyay

Oh I will check it out :)

Ted Shifrin

You two can seminar together :)

Karim Mansour

I don't mind

that would be great actually

let me know when you have it

Sayan Chattopadhyay

8:27 PM

I have it

Ted Shifrin

I'm feeling older and older. I remember Sayan 1.0, like Balarka 1.0. And Adeek 1.2.

Sayan Chattopadhyay

Lol I know literally nothing @Ted, maybe sometime when I can say something about anything

Karim Mansour

time flies

Ted Shifrin

I wish it had flown right from 2019 to 2022.

Karim Mansour

hahaha

Sayan Chattopadhyay

8:30 PM

The satisfaction of kicking the orange man would have been lost then :p

Ted Shifrin

Oh, it's still going on. He's dragging the US Constitution and democracy into hell with him.

A third of the country will not accept this election.

Karim Mansour

I asked my uncle why is he doing this he lives in U.S in orange county he mentioned he probably want to secure a future after he is out of the white house

I agree with him it is probably for publicity

Ted Shifrin

Well, this is the most un-American thing that's ever happened in this country. I think he beats Joe McCarthy by a long shot.

Anyhow, go back to math. I'm just justifying my wish.

Karim Mansour

yeah I agree about Joe McCarthy. Back to work.

Sayan Chattopadhyay

Yeah I should sleep. I shouldn't mess up my sleep cycle.

Ted Shifrin

8:35 PM

Balarka has permanently messed up half the room's sleep cycle. I think he takes pride. :P

Sayan Chattopadhyay

Lol, it's like Balarka is running an experiment to see how sleep cycles react to math and we are the rats

Ted Shifrin

Well, he's still the main rat.

@LucasHenrique FYI, I spent my career being a beetlehead professor. :P

Balarka Sen

@TedShifrin You're putting a lot of faith in 2022

Ted Shifrin

9:26 PM

@BalarkaSen Apparently so.

Astyx

What's wrong with 2021?

Thorgott

odd number

Lelouch

9:40 PM

If $X$ is compact metric space, then for subsets of $C(X)$, uniformly equicontinious = equicontinious and uniformly bounded = pointwise bounded, right? In particular, we don't need the "uniformly" hypothesis in Arzela-Ascoli

Alessandro Codenotti

the family of constant functions $f_n(x)=n$ is uniformly equicontinuous, but not very bounded

Lelouch

it's not pointwise bounded

Alessandro Codenotti

The first thing in your chain of equalities says "uniformly equicontinuous", it does not mention boundedness

Lelouch

I was askign whether I can replace "uniformly bounded, uniformly equicontinious" with "pointwise bounded and pointwise equicontinious"

Thorgott

pointwise bounded + equicontinuous implies uniformly bounded on a compact space, that's for sure

equicontinuity means that a bound at a point is in fact a bound in a neighborhood of that point, then you cover the space with those neighborhoods, pass to finite subcover by compactness and thus obtain a uniform bound

Lelouch

9:49 PM

yeah so for arzela ascoli if a family is "pointwise bounded and pointwise equicontinious" then its closure is compact, right?

Astyx

What does "pointwise equicontinuous" mean?

Thorgott

yeah

Lelouch

For each $x$ and $\epsilon > 0$, you can find a $\delta$ such that if $f \in F$, then $|x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon$

"uniform equicontinious" means you can find such a $\delta$ working for every $x$, though if I'm not mistaken for compact sets both are same

Astyx

yes ok

Thorgott

yeah, you can get this from a contradiction

Lelouch

9:54 PM

lol ok, so I wonder why this is not spelled out explicitly (that we don't acutally need the uniform hypothesis) on most textbooks on Arzela Ascoli

Thorgott

if it weren't uniformly equicontinuous, you can find sequences $x_n,y_n$ of points and $f_n$ of functions such that $|x_n-y_n|<1/n$, but $|f_n(x_n)-f_n(y_n)|\ge\varepsilon$ for some $\varepsilon$ and all $n$. By compactness, WLOG assume $x_n,y_n$ converge and they necessarily converge to the same value by the first condition. Then the second condition contradicts equicontinuity at that point.

Rudin only has pointwise bounded and equicontinuous in his hypotheses

Ted Shifrin

10:42 PM

I think equicontinuity is ordinarily defined on a set in the first place, not at a point.

Rithaniel

Points are singleton sets, though. Right?

Ted Shifrin

I meant a nontrivial compact set or an interval.

I no longer have a whole shelf of books to look in to see ...

Thorgott

well, equicontinuous means equicontinuous at each point

Ted Shifrin

No, I don't think so. If so, you do need to say uniformly equicontinuous, and never in my life have I heard that.

Thorgott

no, uniformly equicontinuous is a different notion

I believe there's some miscommunication going on

Ted Shifrin

10:54 PM

OK, I just looked at Baby Rudin. He talks about equicontinuity on all of the domain $E$.

The $\delta$ works for all $x,y\in E$ and all $f\in\mathscr F$.

I dunno what I'm miscommunicating.

Alessandro Codenotti

The "for all $x,y\in E$" is why I'd call it uniformly equicontinuous

Ted Shifrin

I've never heard that before.

Thorgott

oh, you're right, I hadn't checked the definition in Rudin

Ted Shifrin

But it's interesting. Dieudonné has your definition, and he says the family is equicontinuous (no point mentioned) if it's equicontinuous at each point of the domain.

Which is very different.

Thorgott

but that's what's usually called uniformly equicontinuous, as Alessandro says

Ted Shifrin

10:56 PM

Maybe in Europe. Not in the US.

Thorgott

it's the same on compact spaces

Ted Shifrin

Ahlfors agrees with Rudin. I swear I have never seen the European notion here.

Thorgott

the thing is that equicontinuity at a point generalizes to topological spaces, whereas uniform equicontinuity doesn't

Ted Shifrin

Yes, so you can talk about uniformities when you're not in a metric space, but that to me is NOT a convincing argument.

Thorgott

and Arzelà-Ascoli remains true on compact Hausdorff spaces

Ted Shifrin

10:58 PM

Anyhow, I will accept your apology for accusing me of miscommunicating.

Munkres in his topology book agrees with you. He says equicontinuous at a point.

So it is a weaker notion unless the metric space is compact.

Ascoli is for compact spaces, so meh.

Alessandro Codenotti

@Thorgott uniform spaces have entered the chat

Ted Shifrin

Pedantic point-set people to the back of the room.

Alessandro Codenotti

uniform spaces are not that bad

(but I'm kind of forced to work with them so that's my almost unbiased opinion)

Thorgott

I was trying to sweep them under the rug

Ted Shifrin

Sweep Alessandro under the rug while you're at it :D

Rithaniel

11:08 PM

Waiting for emails sucks

Ted Shifrin

A watched mailbox never receives.

Rithaniel

True enough

Alessandro Codenotti

11:23 PM

@Thorgott they're still kind of annoying for noncompact spaces because there are many possible uniformities to be fair

Thorgott

watching a recorded lecture and the prof mentions the famous Newton-Euler calculus controversy.....

@Alessandro all spaces are compact

Leaky Nun

11:36 PM

what controversy

you mean Newton Leibniz?

Thorgott

yes, but the prof said it was between Newton and Euler

hence my shock

Ted Shifrin

Weren't they rather far apart in centuries?

Thorgott

Newton was in his 60s when Euler was born

Ted Shifrin

Wow, I thought Newton was way earlier.

Thorgott

he was early enough to invent calculus before Euler was born, still

Rithaniel

11:42 PM

Euler was born in 1707. Newton died in 1727

Ted Shifrin

OK, so my mistake is that I thought Euler was 19th century. Silly me.

Rithaniel

So, if they were gonna have a debate, it'd be a very old Newton and a very young Euler

(I just googled it. I also thought they were a few centuries apart)

Thorgott

hmm, who was prolific during the 19th century

Gauß, Riemann, I think

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$Mathematics$ Mathematics