Mathematics - 2018-01-17 (page 4 of 5)

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9:02 PM

17 mins ago, by Leaky Nun

If $f:A \to B$ is a split monomorphism of rings, does it follow that the preimage of a maximal ideal is a maximal ideal?

Let $\mathfrak b$ be a maximal ideal of $B$ such that $f^{-1}(\mathfrak b)$ is not a maximal ideal of $A$, i.e. $f^{-1}(\mathfrak b) \subsetneq \mathfrak a$ for some ideal $\mathfrak a \subsetneq A$. Then, $\mathfrak b \subsetneq \mathfrak b + \langle f(\mathfrak a) \rangle \subsetneq B$. amirite?

Captain Bohemian

I have insomnia.

Jon skeet has 1m rep.

amazing

Captain Bohemian

I am reading math meta to find out how to type math symbols, but too sleepy to concentrate on.

42

Q: Is there an "inverted" dot product?

The dot product of vectors $\mathbf{a}$ and $\mathbf{b}$ is defined as: $$\mathbf{a} \cdot \mathbf{b} =\sum_{i=1}^{n}a_{i}b_{i}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}$$ What about the quantity? $$\mathbf{a} \star \mathbf{b} = \prod_{i=1}^{n}a_{i} + b_{i}=(a_{1} +b_{1})\,(a_{2}+b_{2})\cdots \,...

linear-algebra abstract-algebra vectors terminology products

19 hours, 42 upvotes

9:13 PM

you would like the symplectic 2-form, which is like the dot product but skew-symmetric instead of symmetric

thats pretty much what a symplectic vector space is. instead of an inner product you have a "skew symmetric inner product"

tangential perhaps

Captain Bohemian

I can't decipher these LaTeX commands easily. I think I need renderer.

read room description

it's so weird to me that the symplectic boiz have no local fuckery

as opposed to riemannian boiz @Balarka

i want like a 2 sentence reason that happens

kind of a cool point

I am thinking of it like being so close to a complex structure

that it gets the rigidity

I think the "right rigidity" for Riemannian manifolds is to have a geodesic chart

as opposed to having a chart isometric to R^n

Basically the torsion = 0 is an integrability condition, which gives that you can have geodesically convex charts

wait but is there a darboux theorem for complex guys??

9:19 PM

Something something Kahler?

yeah Newlander-Nirenberg

@Semi Kahler boiz are like everything at once

Kahler is madness

symplectic + complex + Riemannian

bam bam

i dont even want to start thinking about that

@BalarkaSen I thought there was some statement like "If it's at least two of the three, then it's all three"

is newland-nirenberg the characterization of integrability

nijenhuis vanishing?

9:20 PM

but I may be full of s***

@EricSilva Right, I mean, it says if NJ = 0 then your (almost) complex structure is locally equivalent to (C^n, J)

ok sounds good

That's how we "usually" define complex manifolds - locally biholomorphic to C^n

yes yes

@Semiclassical Ah you're prolly right

9:22 PM

when you figure out darboux give me a one sentence explanation accompanied with a picture that explains it @Balarka

the punch line if you will

I will!

need to finish reading physics first

Sanity check: Let $A_0 = \Bbb Q$. Let $A_{n+1} = A_n[X] / \langle X^2 \rangle$. Let $f_n : A_n \to A_{n+1}$ be the inclusion that sends $a \in A_n$ to $a + \langle X^2 \rangle \in A_{n+1}$. Then, let $A = \varinjlim A_n$. Then, for every $n \in \Bbb N$ there is $x \in A$ such that $x^{2^n} = 0$ but $x^m \ne 0$ where $m < 2^n$.

i wanna read more non-riemannian stuff

(cc @MatheinBoulomenos)

I'd dig a Hamiltonian mechanics interpretation of Darboux's theorem

9:23 PM

where have the algebraists gone...

but im getting sucked into the riemannian rabbithole

@Leaky we have scared them off

tries to appear intimidating

@BalarkaSen then I will avenge their death

you can only do that by speaking geometry

algebra

9:24 PM

otherwise well just put you on ignore

no person, no avenge

lol

I sorta want to say that Darboux's theorem = "There's a set of canonical coordinates for any Hamiltonian"

@Semiclassical That's right

where have the algebraists goooone
loooong time paaassing
where have the algebraists goooone
looong time agoooo

gone for coffee every one

9:26 PM

@Semiclassical this is straight up is basically the statement of darboux

neat

death metal riff abruptly starts THEY ARE DEEEEAAAD

lol

@BalarkaSen that reminds me of nothing so much as the first three minutes of this: youtube.com/watch?v=uBzazRsbwb0

lmao

man the heat flow proof of gauss bonnet is cool

9:29 PM

That sounds neat.

@Semiclassical This is fucking good

Though tbh I think heat flow stuff is automatically neat soooo

ikr

@EricSilva How does the heat flow proof work?

@Semi heat flow has a lot of topologeometric good good

lol

i see Eric has picked up my explaining style

9:32 PM

so the idea is that you can relate $\chi(M)$ to the behavior trace of the heat operator on a riemannian manifold via the hodge decomposition of the de rham cohomology groups

tbh i'm mostly wondering if it can be translated into terms a physicist would appreciate

no clue

So in $A$ there is $0=x_0, x_1, x_2, \cdots$ such that $x_{n+1}^2 = x_n$ for every $n \in \Bbb N$.

yeah

ive thought a lot about the heat equation with 0 physics knowledge lol

9:34 PM

lol

i guess the point is that the asymptotic behavior of the heat equation on a curved dude depends on how curved the dude is

hmm

I suspect what I'm looking for is analogous to a statement in electrostatics

i would be interested in hearing something like that now that im learning baby electrostatics

every time im sitting in physics im just like "wowow so much secret ninja cohomology wowowow"

Suppose you have a metal sphere, and you put a net charge on it. Since like charges repel, that net charge will gather itself on the surface of the sphere

Moreover, from Gauss's law and spherical symmetry, the electric field outside the sphere is identical to what you'd have if all the charge was concentrated at the center

which amounts to E = kq/r^2 where q is the net charge and r is your distance from the sphere (k is Coulomb's constant)

in particular, if you look at the electric field right near the surface of the sphere, you have $E = k q/a^2$ where $a$ is the radius of the sphere.

which if you write $k=1/4\pi \epsilon_0$ (because that's how it's defined) is $E=\frac{1}{\epsilon_0} \frac{q}{4\pi a^2}=\frac{\sigma}{\epsilon_0}$

where $\sigma$ is the surface charge density (total charge / surface area of sphere)

Bernardo Meurer

@Semiclassical only tells lies

9:43 PM

You can moreover show that this is a quite generic result: The electric field at the surface of a charged conductor (not necessarily spherical) is given by $E=\sigma/\epsilon_0$ where $\sigma$ is the surface charge density at that point of the surface

I'm forgetting the next step though :/

@Balarka i just discovered my SO loves king gizzard and the lizard wizard and ive literally never been so happy

Bernardo Meurer

I love King Gizzard!

Fuck yeah

Bernardo Meurer

@EricSilva That's crazy props

I want a SO like that

I'll search on Fiverr right now

literally changed my life

ok i better go read about heat flow on curvy boiz

tchau chat

9:46 PM

cya

Bernardo Meurer

@EricSilva Inté

i hate the generic american appeal of Tame Impala as the leader of modern psychedelic rock

that band sucks ass

Bernardo Meurer

They're not the worst thing ever

The Less I Know the Better is a good track

But that's about it too

still quite terrible as pop psychedelic rock

Bernardo Meurer

Yeah, they're not psych IMHO

They're just chill pop-rock

9:48 PM

yeah

Bernardo Meurer

With those annoying trap-influenced instrumentals so they can top the charts as summer tracks

FUCK

Yeah okay they suck

lmao

Bernardo Meurer

Listening to Preoccupations (FKA. Viet Cong)

@EricSilva I was trying to find a reference to something, and this is the best I could find: somethingpositive.net/sp11102017.shtml

I'll go back to finish relistening the second half of Trout Mask Replica

Bernardo Meurer

9:50 PM

And I'll go back to reading type theory

Teach me later

Bernardo Meurer

@BalarkaSen I'm still going to finish that Spinoza article one day

I shall look forward to it

Bernardo Meurer

I just need time alone in the ocean to think

I'm renting a kayak this weekend to go do that

Can someone check my answer here? He is saying his TA has told him that the answer is "weaker" instead of "not comparable". I want to make sure I'm not leading him in the wrong direction:

https://math.stackexchange.com/questions/2609657/another-statement-using-the-epsilon-delta-definition/2609669#2609669

9:51 PM

~~pickle rick~~ corn spongebob

Bernardo Meurer

Lol, yeah

I like the sea dawg

There's no one there

@EricSilva so, the result I was (dimly) recalling is an exercise in Purcell: books.google.com/…

MatheinBoulomenos

10:09 PM

@LeakyNun what you have written down is just $\Bbb Q[X_1, X_2, \dots]/(X_1^2, X_2^2, \dots)$

Hi,

$f\in C^2([0,1])$ with $f''$ convex and $f(0)=f'(0)=0$ is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?

@MatheinBoulomenos $(X_1-X_2^2)$...

can a prime ideal lie inside another?

Bernardo Meurer

I believe that's incest

MatheinBoulomenos

I agree with "for every $n \in \Bbb N$ there is $x \in A$ such that $x^{2^n} = 0$ but $x^m \ne 0$ where $m < 2^n$", I don't see why it holds that "in $A$ there is $0=x_0, x_1, x_2, \cdots$ such that $x_{n+1}^2 = x_n$ for every $n \in \Bbb N$"

@LeakyNun sure. Easy example $(0) \subset (p)$ in $\Bbb Z$

@Bernardo Why are you still here

Just to suffer?

Every night

10:19 PM

@MatheinBoulomenos well if $x^{2^n}=0$ then $(x^{2^{n-1}})^2 = x^{2^n}$

Bernardo Meurer

@BalarkaSen I'm bored

I can't feel my leg

Bernardo Meurer

My bike got stolen

I have nothing to do

Usually I'd be at the beach now

But I can't get there without my bike

You're happy, though, aren't you

MatheinBoulomenos

@LeakyNun that doesn't imply the statement

10:20 PM

The total happiness in the world increased

Bernardo Meurer

No

No it didn't

@MatheinBoulomenos because my ring is constructed that way

Bernardo Meurer

It was probably just some crackhead

And I had dreams about it too

I dreamed my bike was left alone in the side of a hill

With many other lonely bikes

Waiting to be sold on the internet

for crack cocaine

Bernardo Meurer

Yes

10:21 PM

such is fate

Bernardo Meurer

Now I need a new bike

But I was saving money for furniture

oof

Bernardo Meurer

So I guess I need to sell some vinyl records

Which I really didn't want to

yup

oh i thought you said need some vinyl records

that sux

MatheinBoulomenos

Are you sure? I think you mean a different ring than you actually wrote down. What you wrote down is $\Bbb Q[X_1, X_2, \dots]/(X_1^2, X_2^2, \dots)$. I think you wanted to write down $\Bbb Q[X_1, X_2, \dots](X_1^2,X_1-X_2^2, X_2-X_3^3, \dots)$

Bernardo Meurer

10:23 PM

@MikeMiller Yeah, I'll sell a couple to buy a bed frame and stuff

@MatheinBoulomenos I'm not following why what I wrote down is what you said what I wrote down

1 hour ago, by Leaky Nun

Sanity check: Let $A_0 = \Bbb Q$. Let $A_{n+1} = A_n[X] / \langle X^2 \rangle$. Let $f_n : A_n \to A_{n+1}$ be the inclusion that sends $a \in A_n$ to $a + \langle X^2 \rangle \in A_{n+1}$. Then, let $A = \varinjlim A_n$. Then, for every $n \in \Bbb N$ there is $x \in A$ such that $x^{2^n} = 0$ but $x^m \ne 0$ where $m < 2^n$.

should I have done an inverse limit instead?

Bernardo Meurer

Like my 45RPM 200g "Getz Gilberto", and my 180g 45RPM "Take Five"

@Bernardo :( Not good

oh!

MatheinBoulomenos

@LeakyNun in each step, you're just adjoining an extra variable and modding out by the square, you don't adjoin any square root. Note that the colimit is just a union in this case

10:24 PM

$A_{n+1} = A_n[X] \langle X^2 - X\rangle$

right?

no, that isn't right

Bernardo Meurer

@BalarkaSen Yeah, I'm not particularly happy with it

MatheinBoulomenos

That doesn't do what you want either, I think. $A_n[X]/(X^2-X)$ is just $A_n[X] \times A_n[X]$ by Chinese remainder theorem.

I want $A_1 = \Bbb Q[X_1]/\langle X_1^2 \rangle$, so $(X_1 + \langle X_1^2 \rangle)^2 = 0$. Then, I want $A_2 = A_1[X_2]/\langle X_2^2 - X_1 \rangle$ so that $X_2^2 = X_1$

Also, I don't see why in $\Bbb Q[X_1,X_2,\cdots]/\langle X_1^2,X_2^2,\cdots \rangle$ there is $x$ such that $x^4 = 0$ but $x^2 \ne 0$

is that $X_1+X_2$ you're talking about?

MatheinBoulomenos

$(X_1+X_2)$

sniped :P

MatheinBoulomenos

10:28 PM

yes

but that's cube

$(X_1+X_2)^3 = X_1^3 + 3X_1^2 X_2 + 3X_2 X_1^2 + X_2^3 = 0+0+0+0 = 0$

hm

MatheinBoulomenos

what about $X_1+X_2+X_3$?

that should be fifth

oh wait, that's fourth

nice discovery

so actually for every $n \in \Bbb N_{>0}$ there is a nilpotent element with "nil-order" $n$

MatheinBoulomenos

the correct term is "order of nilpotency"

10:30 PM

very interesting

MatheinBoulomenos

iirc

oh thanks

hey @MatheinBoulomenos

want to discuss an important conceptual point

MatheinBoulomenos

Hi @Adeek

how are you

10:31 PM

@MatheinBoulomenos now using whichever construction, let $Y_n$ be a sequence of elements of strictly increasing order of nilpotency

user image

and consider the element in $A[[X]]$ that is $f = \sum Y_n X^n$

So here I am trying to understand the second important conceptual point

is it true that $f$ is not nilpotent?

MatheinBoulomenos

Fine how are you?

10:32 PM

so here the second point what do they mean by parameter t plays the role of coordinate on the given line.

@MatheinBoulomenos good I am learning some classical algebraic geometry in order to improve my intuition

MatheinBoulomenos

@LeakyNun yes. Good obversation, it's actually an exercise in Atiyah-Macdonald to come up with a power series all whose coefficients are nilpotent, but which is not nilpotent itself

@Adeek $t$ is like a coordinate function on a local chart on a manifold

@MatheinBoulomenos yes, I'm referring to the same thing :P

ohh

that is what I was guesing

It's Ch.1 Ex.5 ii), but it refers me to Ch.7 Ex.2, but I'm thinking, what I constructed works :P

so I advanced 6 chapters :P

10:35 PM

@BalarkaSen yeah so we trying to build coordinate charts for varieties

MatheinBoulomenos

@LeakyNun I think if you want to have increasing orders of nilpotency, it's easier to come up with $\Bbb Q[X_1, X_2, X_3, \dots]/(X_1,X_2^2,X_3^3, \dots)$ at least that's what I did

@MatheinBoulomenos well every road leads to rome

MatheinBoulomenos

sure, that other thing works fine as well

> that over thing

MatheinBoulomenos

@Adeek I'd like to help you, but I'm afraid I don't know much classical algebraic geometry at all

:P I never claimed my English was good

10:37 PM

very interesting typo

th-v merger?

MatheinBoulomenos

Somehow I make a lot more typos when chatting, no idea why

do you merge voiced th with v?

MatheinBoulomenos

apparently

@MatheinBoulomenos I see. Yeah I thought it would be nice to develop intuition for classical algebraic geometry while doing modern as well.

I mean in your speech

MatheinBoulomenos

10:39 PM

I think I can differentiate them when I speak

@EricSilva if I think a bit about the formula I linked in Purcell

MatheinBoulomenos

@Adeek actually I'm going to take a seminar on classical alg geo starting in April :)

@MatheinBoulomenos Better than the usual German "asa" instead of other

@MatheinBoulomenos Nice :D. I prefer learning things on my own though.

@MatheinBoulomenos aber in deine(n/m/r) Denkweise sind sie gleich?

10:41 PM

I learn things on my own much faster than classes and more efficient

MatheinBoulomenos

@LeakyNun Nein, ich habe keine Ahnung, wieso ich den Fehler gemacht habe. (deine Denkweise heißt es übrigens)

suppose I have a family of surfaces $f(x,y,z)=c$

@MatheinBoulomenos very interesting indeed

Then I should (usually) have enough info to deduce the mean curvature of each such surface

sehr interessant tatsächlich

die Welt der Algebra

10:46 PM

andersrum lol

ja, lol

deutsche sprache, schwere sprache

Hehe

That’s what I think Purcell’s formula boils down to

In fact, it might just be a version of this diff-geo formula: en.m.wikipedia.org/wiki/First_variation_of_area_formula

Not 100 on that though

Akiva Weinberger

Gauss–Bonay tells us that the total curvature of closed surfaces is just the genus (times $4\pi$ or something). What about surfaces with boundary?

I feel like it's also invariant under topological meddling, as long as you keep stuff near the boundary the same

Don’t you usually need to include terms for the corners?

Akiva Weinberger

10:51 PM

I dunno, do you?

All I know about Gauss–Bonnet is what I wrote above

And by "genus" I mean Euler characteristic, whoops

@Akiva The boundary term comes from the geodesic curvature of the boundary inside the surface

Right

That’s what I was vaguely remembering

Akiva Weinberger

Like, the 1D curvature of the boundary in the 2D surface?

$$\int_{\text{int} M} K dA + \int_{\partial M} k_g d\ell = 2\pi \chi(M) $$

Akiva Weinberger

So, like, a flat disk has 0 total curvature

10:53 PM

@Akiva The geodesic curvature determines how far your curve is from being a geodesic

Akiva Weinberger

Its Euler characteristic is... 1, I think? And the boundary has curvature $1/r$, over a length of $2\pi r$

Boundary circle has curvature tho

I have forgotten the actual definition but it's in Ted's notes

Akiva Weinberger

which makes $0+2\pi=2\pi$

which is nice and good and what it should be

How about for a hemisphere?

Akiva Weinberger

10:54 PM

Hm. Curvature of the boundary is $0$ now I think,

The equator is a geodesic

k_g = 0

Kk

Akiva Weinberger

curvature of the surface is $1/r^2$, over an area of $2\pi r^2$ so $2\pi+0=2\pi$

Arright, that checks out

That's cool, I didn't know that

I think it's okay :) I like closed surfaces

But yeah

Akiva Weinberger

10:56 PM

Ninja no henshū

I’m sorta tempted to work out the case of a spherical sector

@Akiva You should see my proof of G-B using Poincare-Hopf

Just the outer surface I mean

(That says sum of indices of zeroes of a vector field is chi(M))

Akiva Weinberger

@Semiclassical Like a lune?

How do you do it if the boundary isn't smooth?

10:57 PM

here ho

Akiva Weinberger

Is the curvature at a corner like a Dirac delta?

I can’t remember the terminology

@Akiva It suffices for the boundary to be piecewise differentiable.

I think you take account to angles at the corners then

But I don’t want the cone on the inside

It's all worked out in Ted's notes

10:58 PM

Just the spherical surface bounded by a line of latitude

Akiva Weinberger

mlerp

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