Mathematics - 2018-06-16 (page 3 of 3)

MatheinBoulomenos

20:01

one direction is always true: if $f$ is measurable, then $e^f$ is measurable

because $e^x$ is continuous, thus Borel-measurable

Nicholas Roberts

Yes, that is the trivial direction. Im more interested in the other direction and it seems as though you need additional hypotheses

Mike Miller

@BalarkaSen It's unfortunate that topologists might really end up being n-category theorists

But what can be done :(

Balarka Sen

only backstabbers like u

the real ones will avoid categories like the plague

Eric Silva

@MikeMiller y do u say this

MatheinBoulomenos

@BalarkaSen classical "no true scotsman"

Balarka Sen

20:03

He probably had in mind how category theory is just everywhere now

Eric Silva

sp00ked

MatheinBoulomenos

I think Mike was talking about modern homotopy theory

Balarka Sen

Gromov talks about microflexible sheaves in this book I am reading on h principles

@MatheinBoulomenos I think the plague has consumed much more than modern homotopy theory

Mike Miller

I was thinking cobordism hypothesis stuff, computating invariants by codimension n cut and paste techniques

At worst we'll have to be 3-category theorists though

Leaky Nun

@MatheinBoulomenos you here?

MatheinBoulomenos

20:06

@LeakyNun yes

Leaky Nun

how should I explain what my poster is doing

MatheinBoulomenos

there's a counterexample for the topological field thing on nlab by the way

@LeakyNun "Die 9. Symphonie kann man nicht auf dem Kamm blasen." - Ernst Bloch

Mike Miller

I'm coming to believe there's real deep content to all the TQFT stuff, but it's all just a feeling

Balarka Sen

I don't know enough examples of TQFTs to like it beyond a formal perspective admittedly

You probably work with those on a daily basis

MatheinBoulomenos

I like commutative Frobenius algebras

Balarka Sen

20:09

Mike Miller

I like complex surfaces

Leaky Nun

@MatheinBoulomenos seriously

MatheinBoulomenos

you mean finitely generated field extensions of $\Bbb C$ of transcendence degree $1$?

@LeakyNun yes, do you understand what this quote means?

Leaky Nun

one cannot play the 9th symphony on a comb?

MatheinBoulomenos

yeah

Leaky Nun

20:11

was soll es bedeuten

Balarka Sen

@LeakyNun hold my bear breaks the comb in half by repeatedly throwing it in the floor

(Intentional typo)

Fargle

oh god I really don't like the font in Spivak's Calculus on Manifolds

it's icky

Balarka Sen

Do not read that book

Fargle

you're not my dad

Balarka Sen

correct, I am your mom

Fargle

20:13

:O

MatheinBoulomenos

it's supposed to mean that you can't always break down high-level stuff to a simple description that is understandable without knowing some background (Bloch was talking about the complicated language of philosophers in the original context)

It's hard to get the motivation behind Langlands if you don't know at least a bit of CFT, so I don't know if you can say much other than "yeah, we relate representations of this group to representations of another group"

sorry if I'm not helpful

Leaky Nun

@MatheinBoulomenos go on about the "yeah, we relate representations of this group to representations of another group"

@Fargle [insert joke about your mom]

Fargle

:O

MatheinBoulomenos

@LeakyNun I don't know Langlands for tori which is what you do (right?) I know some local langlands for GL2, but I assume the result of the papers you're studying is some kind of relation between representations of the $K$-rational points of tori and representations of the Langlands group?

Leaky Nun

yes

MatheinBoulomenos

20:21

Maybe talk about why those groups are interesting

Leaky Nun

why are they interesting?

MatheinBoulomenos

and maybe mention local CFT. If you're defining the Weil group, it's not hard to state that its abelianization is isomorphic to the multiplicative group

and all one-dimensional representations of a group are determined by it's abelianization

Leaky Nun

@MatheinBoulomenos what does this mean?

MatheinBoulomenos

every one-dimensional representation is a homomorphism into an abelian group, so it factors over the abelianization

Leaky Nun

oh right

MatheinBoulomenos

20:30

so local CFT describes one-dimensional representations of the Weil group

basically local Langlands for GL2 describes some kinds of two-dimensional representations of the Weil group

I don't know the specifics for tori

Leaky Nun

ok

MatheinBoulomenos

but I think the idea is still an extension of LCFT: of course you can consider the multiplicative group as a torus, just like you can consider it as GL1

Leaky Nun

yes, local langlands for torus is just a slightly strengthened form of LCFT

MatheinBoulomenos

okay, then that's what you should tell people

Leaky Nun

yeah, like they know what LCFT is

Xander Henderson

20:35

What about LGBT?

Leaky Nun

what about it

MatheinBoulomenos

Do they know what a local field is?

Leaky Nun

I don't expect them to

I'll just say locally compact non-discrete topological field

as if that means something

MatheinBoulomenos

How are you defining the Frobenius element, then?

Leaky Nun

why should I

MatheinBoulomenos

20:36

because you want to define the Weil group?

Leaky Nun

$\Bbb R$ doesn't have Frobenius

MatheinBoulomenos

nobody cares about that archimdean stuff

Leaky Nun

I did write Frob on my poster

what i write is different from what I will explain

let's give you more background

my poster will be marked by two processors (both of which know what I'm doing) as well as some people from year 2

God have mercy on the souls of the latter

and I basically don't care about them

MatheinBoulomenos

do they know what a representation is?

Leaky Nun

but I still need to get marks from them

no, that's in year 3

MatheinBoulomenos

20:38

lol

why do you get marks from unqualified people?

Leaky Nun

we aren't supposed to do local langlands correspondence for our poster

MatheinBoulomenos

I don't mean it in a derogative way

Leaky Nun

@MatheinBoulomenos why would it offend me lol

MatheinBoulomenos

you can't get from not knowing what a representation or a local fields is to undestanding anything about local langlands in a reasonable amount of time, I'd say

Leaky Nun

2 mins ago, by Leaky Nun

and I basically don't care about them

MatheinBoulomenos

20:41

So you're asking me what you should tell to the people who know what you're doing?

Leaky Nun

no

I'm saying, I don't really care if they can learn much from my poster

but I still need to ~~deal with them~~ cater for their needs

Poline Sandra

hello

someone help me please math.stackexchange.com/questions/2821953/…

Leaky Nun

@MatheinBoulomenos what would you do waerst ich du

waerst du ich?

Luigi M

Can someone provide me an example of a non-invertible non-zero element of the Clifford algebra CL(R^n,<,>) ? where clearly n has to be at least 3 and <,> is the usuale inner product? I'm struggling to find one just to have an idea of what goes "wrong" when it comes to multiplication

Poline Sandra

someone can help me on topology?

MatheinBoulomenos

20:49

@LuigiM ignore the guys who don't have the background and live with bad marks if they might give you some (maybe they'll just give you good marks even if they don't understand anything)

Luigi M

@MatheinBoulomenos sorry I don't get what do you mean

Leaky Nun

@LuigiM he meant to ping me

@MatheinBoulomenos i wouldn't ignore them...

MatheinBoulomenos

@LuigiM yeah, wrong ping sorry

Luigi M

@LeakyNun I see.

Leaky Nun

@MatheinBoulomenos was wurdest du sie sprechen

MatheinBoulomenos

21:08

@LeakyNun I can't help you with that

Leaky Nun

ok thanks

@MatheinBoulomenos is $A \mapsto A[X]$ a functor?

MatheinBoulomenos

yes

apply the morphism on the coefficients

Leaky Nun

@MatheinBoulomenos what are the source and target categories though

MatheinBoulomenos

Rings to Rings?

or commutative rings, if you want

also works for algebras over a commutative ring

Leaky Nun

hmm

MatheinBoulomenos

21:12

If you allow arbitrary sets of variables, then it's a bifunctor Sets x Rings -> Rings

Leaky Nun

nice

I think I have a bifunctor Mon x CRing -> CRing

by sending (M, A) to A[M]

MatheinBoulomenos

true

Leaky Nun

and then your bifunctor is a special case of mine

since you just need to compose it with the functor F x 1 : Set x CRing -> Mon x CRing

where F is the free functor

MatheinBoulomenos

yeah

you want commutative monoids though

or else you get giant non-commutative free algebras

$k\langle X, Y \rangle$ is not even Noetherian

with $k$ a field

Jake S

Thanks!

Leaky Nun

21:19

yes, I mean CMon x CRing -> CRing

everything is commutative

Jake S

equiv to vector space*

Leaky Nun

@JakeS yes

Jake S

Oops, i edited my question by mistake, haha

Leaky Nun

they are also equivalent to $\prod_{x \in \Bbb R} \Bbb R$

Jake S

Thanks!

Poline Sandra

21:20

@LeakyNun hello

Leaky Nun

@PolineSandra I don't know about your topology question

Jake S

A question for the masses here. I didn't do many math intensive courses in high school and this year I changed majors from Biology to Math as I decided I would like to go towards a different field. I've always passively enjoyed math but I never dedicated myself to it.

Now, I find myself enjoying my courses, but they're difficult for me. Particularly Linear Algebra.

I've read questions on here asking for strategies to build mathematical maturity, but I'm interested in hearing people's thoughts

I've recently picked up a copy of Velleman's How to Prove It, which I'm hoping will help me on my way

Poline Sandra

no it is not about that

@LeakyNun

Leaky Nun

@PolineSandra what is it

@JakeS youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

Jake S

Haha, I have that bookmarked as well

Poline Sandra

21:26

is it possible to fide $a>0$ such that $y=ax-1$ is tangent to the circle $y^2+x^2=1$

@LeakyNun

Leaky Nun

@PolineSandra desmos.com/calculator/hxemztxmdw

Poline Sandra

how i found it mathematicaly that a=10?

@LeakyNun

Leaky Nun

show me

Poline Sandra

desmos.com/calculator/hxemztxmdw

@LeakyNun

Leaky Nun

mathematically?

Poline Sandra

21:40

yes

the line i $y=-ax+1$

it passe from (0,1) and (1/a, 0)

i have to find a=?

famesyasd

22:04

if f : X \to Y a function. If I write f(3)=5 what does it mean?

Xander Henderson

It means that the image of 3 under the map $f$ is 5.

famesyasd

alright and what if 3 was not in my domain?

geocalc33

Hey guys

famesyasd

well anyway what was my originally intended question is if we have a group (G, \circ), does associativity alone imply that G is closed under \circ?

Leaky Nun

@famesyasd no

geocalc33

22:29

is there a way to solve for $s$?

for $ y= \zeta(s) $

where $ \zeta(s) $ is the riemann zeta function

famesyasd

@LeakyNun does G being closed under \circ mean that \forall x,y \in G, \circ(x,y) \in G also?

geocalc33

@LeakyNun

Leaky Nun

@famesyasd yes

@geocalc33 no idea

famesyasd

22:55

@LeakyNun @LeakyNun Alright, suppose that we have a function $f:G \times G \to Y$ and we have a functional equation $$ \forall x,y,z \in G \; f[f(x,y),z]=f[x,f(y,z)]$$ and we want to prove that $$\forall x,y \in G \; f(x,y) \in G.$$

Let's take arbitrary $x,y \in G$ and $z=x$ then by associativity we have
$$f[f(x,y),x] = f(x,f(y,x)]$$

Now the notation $f(x)=y$ means that $(x,y) \in f$

thus we conclude that

$$ [(f(x,y),x),\; f(x,f(y,x))] \in f \subset (G\times G)\times Y$$

that is, $\exists u \in (G\times G)$, $\exists v \in Y$ such that

loch

@LeakyNun your best bet is to be able to do some simple examples - but I don’t know anything about langlands to know what simple examples would be (I guess as @MatheinBoulomenos suggested probably cft or something - also something which the average y2 wouldn’t know lol)

rep theory itself is easy to motivate though so there’s that..

Leaky Nun

@loch hmm

how would you motivate rep theory

famesyasd

23:44

yes, associativity implies closedness

Secret

hmm... I need to think later how the reverse implication failed to hold, and here a countaexample might be hard to found due to the abstract setting here

famesyasd

@Secret do you mean me?

user441848

$z^tHz$ .To multiply this z vector with Hessian matrix, the order doesn't matter right ? I mean it could be $z^tH$ first and then Hz ?

Leaky Nun

@user441848 correct

famesyasd

@user441848 if a matrix is symmetric then yes, (z^tH)^t=H^tz=Hz

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