Mathematics - 2019-07-22 (page 2 of 2)

Balarka Sen

15:00

@AlessandroCodenotti Think of a 2x2 matrix integer matrix of determinant 1 whose eigenvalues are irrational. Then, being a matrix from $\text{SL}_2(\Bbb Z)$, it acts on $\Bbb R^2$ and keeps a pair of transverse foliations on $\Bbb R^2$ by irrational slope lines fixed. This descends to a pair of transverse foliations on $T^2$ by real lines which are being fixed by the matrix

Along one, it expands, and along another, it contracts.

Ryan Unger

@AlessandroCodenotti

oops

look at the Dirichlet eigenvalue of the Laplacian over relatively compact subsets of $\tilde M$ and then take the infimum over all of them

Balarka Sen

This is an example of an Anosov diffeomorphism of $T^2$.

Ryan Unger

$\tilde M$ is the Riemannian universal covering space of $M$

Balarka Sen

A pseudo-Anosov diffeomorphism of a surface is something which keeps a pair of transverse singular foliations on the surface fixed. I think there are associated measures to the foliations in the transverse direction, and you demand you stretch one foliation by $\lambda$ and another by $1/\lambda$.

Ryan Unger

$\lambda_0$ is called the "fundamental tone" because it's the lowest frequency if you view $\tilde M$ as a big drum

Balarka Sen

15:03

this is bottom of the spectrum?

Ryan Unger

lmao I forgot to say $\lambda_0(\tilde M)>0$

that's the theorem

@BalarkaSen it's inf spec ($-\Delta$) but isn't necessarily attained unless it's zero

Tobias Kildetoft

@BalarkaSen Ahh, right. It was in connection to this qgm.au.dk/video/mc/weil-petersson that I heard about them

Balarka Sen

i see

Ryan Unger

now I believe that this shows that hyperbolic groups are amenable

MatheinBoulomenos

@ÍgjøgnumMeg which courses are you gonna take? I guess ANT and modular forms are set

Ryan Unger

15:04

by hyperbolic groups I mean those that give compact quotients of H^n

$\lambda_0(\Bbb H^n)=\frac{1}{4}$ IIRC

ÍgjøgnumMeg

@Mathein right, and I'm thinking of either taking diff top 1 or taking the seminar Quadratic Forms

Balarka Sen

Hyperbolic groups are literally non-amenable though

ÍgjøgnumMeg

or both, though I don't wanna overload myself in the first semester lol

Balarka Sen

They contains $F_2$

ÍgjøgnumMeg

@Mathein I think diff top 1 would be good to get used to some topology, since I've never really done any lol

MatheinBoulomenos

15:07

@ÍgjøgnumMeg did you see that diff top 1 overlaps with modular forms?

Balarka Sen

That's a "paradoxical group", which is what non-amenability is, IIRC (Right, @Alessandro?)

Ryan Unger

ok Balarka I'm retarded

I'm deleting my account

ÍgjøgnumMeg

@Mathein oh really

Ryan Unger

jesus christ

ÍgjøgnumMeg

Ah dienstags

Ryan Unger

15:07

the theorem is $\lambda_0(\tilde M)=0$

ÍgjøgnumMeg

oder donnerstags

Ryan Unger

so this proves they're not amenable

Balarka Sen

Ah OK

Alessandro Codenotti

@BalarkaSen yes

Ryan Unger

I am definitely right about the 1/4 at least in dimension 3

Balarka Sen

15:08

I should maybe correct my statement. Hyperbolic groups which aren't two-ended (virtually $\Bbb Z$) are non-amenable

ÍgjøgnumMeg

Could do geometric analysis lol

Ryan Unger

Heidelberg has a geometric analyst?

Alessandro Codenotti

@BalarkaSen two ended

ÍgjøgnumMeg

Jan Swoboda

Balarka Sen

@Alessandro lol whoops

corrected the correction

Ryan Unger

15:09

he does very...European geometric analysis

Balarka Sen

@TobiasKildetoft I have heard of the Weil-Petersson metric but dunno what it is exactly

ÍgjøgnumMeg

ah diff geo 1 is a requirement tho

Ryan Unger

@ÍgjøgnumMeg do you know what the course is about

Tobias Kildetoft

@BalarkaSen Same here (even though I attended most of that master class :) )

Ryan Unger

looking at what this guy does, it could be completely orthogonal to what people are doing in the states

Alessandro Codenotti

15:11

@RyanUnger I'm curious about what that means

ÍgjøgnumMeg

@Ryan differential operators on riemannian manifolds

I guess

MatheinBoulomenos

@ÍgjøgnumMeg you could take functional analysis 2. Have you had an intro functional analysis course in your undergrad?

ÍgjøgnumMeg

from the description

@Mathein yeah I have

MatheinBoulomenos

functional analysis 2 counts as applied, lol

ÍgjøgnumMeg

rofl

MatheinBoulomenos

15:12

"applied" by Heidelberg standards

ÍgjøgnumMeg

there's gonna be a certain amount of catch up required by me anyway, since my undergrad was so poor

Ryan Unger

@AlessandroCodenotti if you look at his papers, half of them are gauge theory

few geometric analysts do gauge theory here, or at least they don't call themselves geometric analysts

ÍgjøgnumMeg

@Mathein I could just take a seminar and 2 lectures lol

Alessandro Codenotti

@RyanUnger I see

What's in functional analysis 2?

Ryan Unger

@MatheinBoulomenos I'm going to take that Hodge-Arakelov theory course

going in with no background

wish me luck

Balarka Sen

15:16

lol

Ryan Unger

>what's a scheme

MatheinBoulomenos

@Ryan good luck! I'm praying for you

Alessandro Codenotti

@RyanUnger A thing that locally looks like an affine scheme

ÍgjøgnumMeg

rofl

Alessandro Codenotti

(I'm not even joking that's the usual definition)

Ryan Unger

15:16

I actually know what a scheme is

but not more than that

Tobias Kildetoft

@AlessandroCodenotti With an affine scheme being a type of functor of course

Ryan Unger

> Description
For a smooth and projective variety X over a global field of dimension n with an adelic polarization, we propose canonical local and global height pairings for two cycles Y, Z of pure dimension p, q satisfying p + q = n*1. We give some explicit arichmedean local pairings by writing down explicit formula for the diagonal green current for some Shimura varieties.

Tobias Kildetoft

(and a scheme as well)

Ryan Unger

this is the only number theory course offered

that means it has to be accessible right

Balarka Sen

diagonal green current eh

MatheinBoulomenos

15:19

the definition of a Shimura variety is quite ... technical

Ryan Unger

actually if I have time to take a meme course it will be homological algebra

Leaky Nun

19 hours ago, by Balarka Sen

I never understood this obscene fascination with injective resolutions

@MatheinBoulomenos do you have anything to say about this?

:P

Balarka Sen

u pullin mathein on me eh

im sure he'll have some functorial reason to debunk me

Tobias Kildetoft

Well, if you have no projective objects, then injective resolutions are hard to avoid

Balarka Sen

@Tobias I work with $\Bbb Z\mathsf{-Mod}$

Tobias Kildetoft

15:21

@BalarkaSen Well stop that

Balarka Sen

No

ÍgjøgnumMeg

rofl

Leaky Nun

I work with $(\Bbb Z\mathsf{-Mod})(X)$

Balarka Sen

What if you don't have enough injectives

but enough projectives

Tobias Kildetoft

You want $\mathbb{Z}$-comod

Alessandro Codenotti

15:22

If you have no projective objects then you're working in the wrong category, simple as that

Balarka Sen

What exactly goes wrong with that definition of abelian category

WHY INJECTIVE

MatheinBoulomenos

it's just a cofibrant resolution for the standard model category structure on the category of chain complexes over an abelian category

Leaky Nun

oh right, of course

Balarka Sen

Well yeah I do fibrant resolutions

user1

I have a differential geometry question, anyone that might be able to answer that here?

Tobias Kildetoft

15:24

@AlessandroCodenotti There are some pretty important categories that have no projective objects. Such as those of rational representations of algebraic groups (at least, the category will be either pretty boring or have no projectives)

where "pretty boring" is of course due to my own warped tastes

MatheinBoulomenos

@Balarka enough injectives is more common than enough projectives. For example, all Grothendieck categories have enough injectives and for examples all category of sheaves of modules are Grothendieck categories

Balarka Sen

The only resolution of sheaves I had ever taken were acyclic

Which is good enough for me to take derived functors with :)

MatheinBoulomenos

even if you don't use injectives object for computation, it's nice to have them for proofs. How do you get your derived functor to act on morphisms? Injective resolutions have the nice lifting property which makes this work, which general acyclic resolutions lack

Balarka Sen

And it's a fact that every sheaf on a topological space admits an acyclic resolution, the Godement resolution. I don't have to appeal to your Grothendieck disease!!

CowperKettle

Ryan Unger

15:32

Wtf are you people talking about

Balarka Sen

@MatheinBoulomenos That's true.

@loch @Mathein I heard y'all define $BG$ as $pt/G$ for a group scheme $G$ lol

What the hell does that mean

Insider information tells me it's a stack

loch

I mostly just think of it as some 'thing' where if I want to give a map from S to BG, it's the same as giving a principal G bundle on S (as in topology)

Balarka Sen

Ah

loch

and in general if you have [X/G], then a map from S to [X/G] is the same as a principal G-bundle P on S, with a G-equivariant map P to X

Balarka Sen

Gotcha. That makes perfect sense.

Leaky Nun

15:47

oh no

that's just cheating

Balarka Sen

If $G$ acts freely on $X$, then $[X/G]$ should be realizable as a scheme, yes?

loch

you need to be more careful i think - see mathoverflow.net/questions/1558/…

Balarka Sen

oh dear

loch

"In general, the quotient of a scheme by a free action of a group is an algebraic space."

Balarka Sen

lol aren't those like

sheaves on the category of schemes

loch

15:54

yeah

well

so are schemes :p

Balarka Sen

lol

so representable algebraic spaces are schemes

i feel like algebraic geometry is just a big meme

its just really hard to get into the meme

loch

that's what i think about homotopy theory

lmao

Balarka Sen

lolol

loch

but i guess it's true for AG as well
so BG is the classifying space of principal G-bundles - in particular you have a universal family given by pt-> G

dim(pt) = 0, so dim(BG) = -dim(G)
(it's negative)

Ryan Unger

What

Balarka Sen

16:08

lmao

Leaky Nun

yeah it's just a big meme

Ryan Unger

I don’t think geometric analysis has many memes. Feels bad

Maximum principle and integration tricks and you’re good to go

loch

16:25

Here’s another meme - gromov witten theory is a curve counting theory. Youre supposed to count the number of curves in some space.

you can get fractional/negative answers

Leaky Nun

@loch do you have any examples?

loch

in the context of GW theory - i dont know an example with negative answer off the top of my head

but actually it only looks like a meme because i phrased it for clickbait. negative answers show up fairly naturally in topology because you are meant to count things with a sign

Ryan Unger

Boo

Balarka Sen

16:41

fraction numbers sounds like rational divisors

anakhro

17:05

Hi hot cats

Balarka Sen

17:18

I have a question

anakhro

Nice question.

Alessandro Codenotti

Just ask, don't ask to ask, that's how the chat works if you're new here

Balarka Sen

Suppose $E$ and $E'$ are fiber bundles (with fiber space $F$ and $F'$ respectively) over $B$, and there is a map $f : E \to E'$ which commutes with the bundle projections.

lol

I can speak of this as a collection $\{f_x : F \to F' \}_{x \in B}$. This is not a continuous family of collections - that's the right thing to say when both are trivial bundles! It's locally continuous, and satisfies some cocycle conditions

Would that be correct?

Balarka Sen

17:33

I'd like to say this would be a section of a $\text{Map}(F, F')$-bundle over $B$, maybe. But that sounds really annoying!

anakhro

So $\pi\colon E\to B$ and $\pi'\colon E'\to B$, with $f\colon E\to E'$ such that $f\circ \pi = \pi'\circ f$.

Balarka Sen

Right, $f$ is fiber-preserving

anakhro

And then the $f_x$ is just restriction to the fibres at $x$?

Balarka Sen

Ya

Leaky Nun

@AlessandroCodenotti b... but he isn't new h... here

@BalarkaSen are bundles like sheaves?

anakhro

17:42

They are more like manifolds, I would say.

@LeakyNun are you familiar with covering spaces?

Leaky Nun

oh the sections form a sheaf

a bit

anakhro

A covering space is a fibre bundle with discrete fibres.

Balarka Sen

@LeakyNun In a sense, yes.

Leaky Nun

I see

Balarka Sen

Covering spaces are locally constant sheaves

Leaky Nun

17:43

cool

and locally constant sheaves are covering spaces...

Alek Murt

hey guys someone could help me trying to find values $A$ and $B$, for this question math.stackexchange.com/questions/3300566/…

Balarka Sen

The correct analogue of that is that locally free sheaves correspond to vector bundles.

anakhro

@BalarkaSen what do you mean by "locally continuous"?

Leaky Nun

@BalarkaSen how many separation axioms do you need if you want $\varinjlim\limits_{U \ni p} U = \{p\}$

@anakhro read it again

anakhro

@LeakyNun I am referencing what he said way above.

MatheinBoulomenos

17:45

@LeakyNun $T_1$ I think

anakhro

@BalarkaSen here.

Leaky Nun

oh

@MatheinBoulomenos is limit just intersection

Balarka Sen

@anakhro Since the bundles are locally trivial, that means over a chart $U$ that trivializes both bundles, I can say $\{f_x\}_{x \in U}$ is a continuous family

Leaky Nun

yeah of course it is

MatheinBoulomenos

but don't you mean $\varprojlim$ instead of $\varinjlim$?

Leaky Nun

17:46

perhaps

MatheinBoulomenos

colimit would be a union here

anakhro

@BalarkaSen that makes sense. Though "locally continuous" sounds like a weird name for it.

Balarka Sen

I guess, I didn't have a better terminology for it.

Sayan Chattopadhyay

@BalarkaSen Yea I guess. Since each bundle has its corresponding cocycle conditions you can get cocycle conditions since it commutes with bundle projections.

Balarka Sen

That's my thought.

anakhro

18:00

continuous when indexed by trivializing charts , maybe

a worse one would be "trivially continuous". :P

Balarka Sen

Lol

Alek Murt

hey guys someone could help me trying to find values $A$ and $B$, for this question https://math.stackexchange.com/questions/3300566/how-to-find-the-values-of-some-constants-of-a-transformed-function?noredirect=1#comment6790335_3300566

anakhro

@AlekMurt most people here can only help someone upon the 3rd time they ask.

Ryan Unger

19:29

@BalarkaSen hah, "conformorphism" = conformal diffeomorphism

that's a new one

Akiva Weinberger

https://www.flickr.com/photos/yoshinobu_miyamoto/sets/

Neat stuff

Tanner Swett

20:30

So the "picture-hanging puzzle" asks how to hang a picture on a wall using string and two nails, such that if either nail is removed, the picture falls.

66

Q: Hanging a picture on the wall using two nails in such a way that removing any nail makes the picture fall down

A friend of mine told me that it's possible to hang a picture on the wall from a string using two nails in such a way that removing either of the two nails will make both the string and picture fall down. My friend also told me that I need to be acquainted with the concept of fundamental groups t...

manooooh

Hello!

Tanner Swett

There aren't very many obvious practical applications of this.

Hey there!

But I did think of a potential application, although probably not a practical one.

I'm a student glider pilot. A glider is essentially a small airplane with no engine. The way you get a glider up into the air in the first place is by towing it behind an airplane.

One of the hazards of this is that the tow rope may fail to release.

So, while I was at the gliderport the other day, I had a thought: you could have two separate release mechanisms, and thread the rope through them in such a way that the rope will be released if either mechanism is activated.

I'm sure that wouldn't actually be practical, but I thought it was an interesting application of a seemingly useless discovery.

manooooh

Suppose we have a binary relation $R$ defined in $A\times A$. And we have a formula that defines this relation. My question is, in which set is defined the result of the expression (formula)?

Suppose that the relation is $R$ defined on $\Bbb{N}\times\Bbb{N}$, where $(a,b)R(c,d)$ if and only if $2(a-c)=d-b$. In which set is defined $2(a-c)=d-b$?

I think this question has no sense, but I am not sure

Tanner Swett

If I understand you're right, you're looking for the set of truth values, which is $\mathbf{2} = \{T, F\}$.

Or maybe I'm not understanding you.

manooooh

@TannerSwett thanks for the answer! I do not understand your notation, so I cannot tell you if you understand me or not

Tanner Swett

20:39

Well, the name of the set is $\mathbf{2}$ (or just 2), and the set is the set with two elements, $T$ and $F$.

$T$ for "true" and $F$ for "false".

manooooh

@TannerSwett but why "2" as the name of the set?

Tanner Swett

Just because it has 2 elements.

It's common to use "0" as the name of a set with 0 elements, "3" as the name of a set with 3 elements, and so on.

manooooh

@TannerSwett oh, ok. Perhaps you are right

MatheinBoulomenos

@manooooh in this case, the integers $\Bbb Z$, but you can't answer it abstractly as you posed it, it will depend on the formula

I think you're asking because $a-c$ and $d-b$ are not in $\Bbb N$, right?

manooooh

@MatheinBoulomenos thanks! Suppose that the formula is given like the example. Yes, I think you are right with the integers set

@MatheinBoulomenos no, that's why I think you are right about integers. But a relation is a set, and we could replace $R$ by all the (infinite) pairs of numbers that verify the formula

MatheinBoulomenos

20:44

I think it should be clear from context in which set the formula is defined and it shouldn't make a difference. In the example, you could also say that we interpret the formula in $\Bbb Q$ or $\Bbb R$ or $\Bbb C$ or even quaternions $\Bbb H$

manooooh

@MatheinBoulomenos sorry but I am asking using this particular relation. It "takes" 4 natural numbers, and what returns? That's my question

MatheinBoulomenos

@manooooh that's true, but we only care for which natural numbers $a,b,c,d$ we have $2(a-c)=d-b$, so even if we use $\Bbb Z$ (or something else) to define the subset of $\Bbb N\times \Bbb N \times \Bbb N \times \Bbb N$

@manooooh ah, if you want to think of it as a function of four variables, then you can think of it as returning a truth value, either true ($T$) or false ($F$), as @TannerSwett mentioned

manooooh

@MatheinBoulomenos but $\Bbb Z\not\subseteq\Bbb N\times \Bbb N$, $\Bbb Z\not\subseteq\Bbb Z\times \Bbb Z$ etc.

@MatheinBoulomenos yeah, probably! Ok, so let's define $f(a,b,c,d)=2(a-c)-d-b$. Now what?

MatheinBoulomenos

yes, that's right. Maybe I haven't been clear enough. The idea is this: the relation $R$ is a subset of $(\Bbb N\times \Bbb N) \times (\Bbb N \times \Bbb N)$, consisting of all $((a,b),(c,d)) \in (\Bbb N\times \Bbb N) \times (\Bbb N \times \Bbb N)$ such that $2(a-c)=d-b$. This means that all elements in the relation $R$ are pairs of (pairs of) natural numbers. We only need the integers to understand _which_ pairs of natural numbers are in relation to other pairs of natural numbers.
But it's still a relation only on pairs of natural numbers

manooooh

Hm, I think we are not going in a proper way... What happens if we have a relation $S$ on $\Bbb Z$ defined by $aSb\iff a\leq b$? We can't define a function since $\leq$ is not an equal sign :S

MatheinBoulomenos

20:52

@manooooh but that's a fine relation. Why do you think that we need an equal sign in the definition?

you can also rephrase $a \leq b$ as $\exists c \in \Bbb Z:a+c=b$, if you absolutely want an equal sign

BAYMAX

Heyo chat

I am thinking about this

I have 2 by 2 matrix A which has expressions consisting of three parameters

manooooh

@MatheinBoulomenos I could understood all of the first part of your message, thanks!! And yes, since we have for example 2(2-3)=4-6$ then $((2,6),(3,4))\in R$, and $4-6=-2\in\Bbb Z$

BAYMAX

I am thinking about the values of those parameters which can give the positive eigenvalues of the matrix A

I am thinking of doing it computationally but how to ?

manooooh

@MatheinBoulomenos well, because we have only take an example which included an equal sign :P, but it seems that any math formula can be written using equal sign, as you have shown

@MatheinBoulomenos so the final answer would be $\Bbb Z$, right?

MatheinBoulomenos

it should be $\exists c \in \Bbb N_0: a+c=b$ by the way

@BAYMAX the characteristic polynomial will be quadratic, you can solve the quadratic in terms of the parameters and then you have to consider when the parameters give you positive roots. This could lead to some difficult inequalities, depending on how your entries depend on the on the parameters

@manooooh for the example you gave, yes

manooooh

21:00

Ok, so finally the question had sense.. Thanks!

Ryan Unger

@BalarkaSen

BAYMAX

yes@MatheinBoulomenos but the equation is bit complicated, hence I was thinking of using computation

manooooh

@MatheinBoulomenos please help me finding the set where the inequality $a\leq b$ is defined, knowing that $a,b\in\Bbb Z$ and $aSb\iff a\leq b$

I suspect that it is also $\Bbb Z$, but I am not sure

MatheinBoulomenos

it's defined on $\Bbb Z$, since $a,b \in \Bbb Z$ as you said

Leaky Nun

lol

manooooh

21:09

@LeakyNun lol what? :)

@MatheinBoulomenos thanks!!

Leaky Nun

21:30

quadruple complex...

loch

Lol what are you reading

Leaky Nun

@loch Cartan Eilenberg homological algebra

hypercohomology

loch

Oh

uhoh

23:42

1

Q: Crystallographic terminology associated with the honeycomb structure

I'm trying to quickly learn basic crystallographic principles in 2D under pressure (it's me whose under pressure, not the lattice) so I'm checking many sources to find those that "speak to me" the best (including this excellent answer). Many sources (books, papers, researcher's websites) cite the...

crystal-structure

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$Mathematics$ Mathematics