Mathematics - 2020-05-10 (page 2 of 3)

Thorgott

3:04 PM

I have never seen this before and I didn't need to see it now

nbro

3:18 PM

Why or in which sense is this true "the CDF of a random variable maps the random variable to an uniform distribution"?

Leaky Nun

@Thorgott the font reminds me of old papers

which of course have different conventions

@nbro in the sense that if $F$ is the CDF of $X$ then $F(X)$ follows the uniform distribution

as long as $X$ is continuous

Thorgott

it's from 67, not that old

Leaky Nun

@Thorgott you would call someone born in 67 old

Thorgott

I'd call them middle-aged

Balarka Sen

leaky you're old

Alessandro Codenotti

3:32 PM

@LeakyNun Someone is going to be smacked by Ted

The album was really good @Balarka

I was surprised to learn that the band is German, it sounds a lot like the classic British prog scene from the 70s

Jasper

@AlessandroCodenotti Have you seen Peter Hinman's Fundamentals of Mathematical Logic? I am wondering how it compares with Joseph Shoenfield's Mathematical Logic.

Alessandro Codenotti

Nope, never heard of that book before

Despite that I'm betting that it is more readable than Shoenfield's

Jasper

Yes, since Shoenfield's book is so short.

Balarka Sen

@AlessandroCodenotti Agree.

nbro

@LeakyNun Really? I've never heard of this. Can you point me to resource that talks more in detail about the topic?

abhas_RewCie

3:46 PM

@Thorgott Yes, that the original way, we often short it as $\sum \limits_{i=1}^n$

Rather it should be written as $\sum \limits_{i=1}^{i=n}$

Leaky Nun

@nbro I think it's $F^{-1}(X)$ instead

wait no it's $F(X)$

it maps to $[0,1]$ right

so $P(F(X) \le t) = P(X \le F^{-1}(t)) = F(F^{-1}(t)) = t$ basically

and the conditions are to make sure that $F^{-1}: [0,1] \to \text{range of X}$ is well-defined

Balarka Sen

It still makes sense to do this for not necessarily continuous random variables, $X = F^{\leftarrow}(U)$ where $U = \text{Unif}((0, 1))$ and $F^{\leftarrow}$ is the "generalized inverse"

$F^{\leftarrow}(t) = \inf F^{-1}((t, 1])$

This can often be very useful, because given two completely unrelated random variables $X, Y$, you can "couple them" to be on the same probability space by replacing them with $F_X^{\leftarrow}(U)$ and $F_Y^{\leftarrow}(U)$ for some uniform variable $U$ on $([0, 1], \mathscr{B}_{[0, 1]}, \lambda)$

So that's a joint distribution whose marginals are distributed as $X$ and $Y$

abhas_RewCie

4:03 PM

@nbro Should I answer Programming Questions (which are asking RL Concept through codes)?

geocalc33

how do you compute a diffeomorphism on a lattice?

abhas_RewCie

@geocalc33 take an infinitesimal element and integrate over whole volume.

nbro

@abhas_RewCie I guess this not the right chat to ask that question

abhas_RewCie

@nbro It's okay to have informal chat here, you can see^ (unless it troubles others)

@nbro Let's talk in singularity

Rudi_Birnbaum

Hi all, just listened to a (recorded) talk from Conway on the surreal numbers. Whats that all about? Is it like "the largest" ordered field?

In the sense that it contains all the important ordered fields?

geocalc33

4:21 PM

0

Q: What is the motivation for studying this subgroup of $SL_n(\Bbb R)?$

Consider a diagonal matrix $A$ with entries $e^{s_1},e^{s_2},\cdot\cdot\cdot$ on the diagonal s.t. $\sum_{i \ge1} s_i=0.$ From what I understand this is a subgroup of $SL_n(\Bbb R).$ This is because the determinant of a diagonal matrix is the product of the diagonal entries, and when we do this c...

feel free to answer

Rudi_Birnbaum

$\sqrt{e} + \sqrt{e}^{-1} \ne 1$

oh sorry

$\sqrt{e}\sqrt{e}^{-1}=1$!

feynhat

@geocalc33 $\det$ is a homomorphism from $\text{GL}(V)$ to $F^*$. $\text{SL}(V)$ is the kernel.

Rudi_Birnbaum

@geocalc never heard of that before, but det = 1 means that the map preserves the volume.

feynhat

$\text{SL}(V)$ has an obvious action on $\Bbb P(V)$, the kernel of this action is the center of $\text{SL}(V)$. Modding out by the kernel you get $\text{PSL}(V)$.

Rudi_Birnbaum

moreover the new basis strictly orthogonal in terms of the old basis.

not only that but also diagonal

that means it describes an linear transformation that keeps the direction of the basis fixed and only stretches/compresses the axes.

but that in such a way that a volumes in the space are conserved

feynhat

4:45 PM

@Rudi_Birnbaum What do you mean by the map preserves the volume?

Do you mean if $T \in \text{SL}(n, \Bbb R)$, and $A \subset \Bbb R^n$, then $\text{vol}(TA) = \text{vol}(A)$?

Rudi_Birnbaum

No because thats wrong, take a ball, that won't work!

feynhat

right.

Rudi_Birnbaum

I mean the spats

"Späte"?

Alessandro Codenotti

@Rudi_Birnbaum They contain all ordered fields

Thorgott

no, it will work

linear maps scale volumes by the absolute value of their determinant

Alessandro Codenotti

4:55 PM

There's some subtlety to be precise, the surreal numbers are a proper class, but if you work in a theory where this makes sense (NBG) then you can prove that every set sized ordered field embeds into the surreals

Balarka Sen

It's always true

Thorgott

volumes of all Borel-measurable sets

Lebesgue-measurable even

Balarka Sen

Since @feynhat is doing forms, $A^*(dx_1 \wedge \cdots \wedge dx_n) = \det(A) dx_1 \wedge \cdots \wedge dx_n$ :P

Thorgott

I always almost forget the very useful fact that diffeomorphisms are Lebesgue-Lebesgue-measurable

Balarka Sen

Then integrate over any subset. You get $\text{vol}_{\Bbb R^n} (AS) = \det(A) \text{vol}_{\Bbb R^n} (S)$.

feynhat

4:58 PM

@BalarkaSen How could I forget that... that's how we define orientability

Thorgott

can you integrate forms over any subset?

feynhat

k-forms with compact support on k-submanifolds

Balarka Sen

No, $dx_1 \wedge \cdots \wedge dx_n$ is exactly the Lebesgue measure on $\Bbb R^n$

You can integrate it over any measurable subset.

feynhat

That's for $\Bbb R^n$. For manifolds we have to first choose a metric to define the volume form, right?

Balarka Sen

Sure.

I should really say $|\text{vol}_g|$ is a measure, where $\text{vol}_g$ is the volume form. I don't want signed measures

Galois in the Field

5:06 PM

Hi there. Does someone who likes number theory want to tell me what the purpose of the minkowski bound is? It seems that it reduces the computation of the class number of a number field, to an analysis of the (finite) number of prime ideals laying over the primes in Z which are within this bound, is that how I should think of it?

Alessandro Codenotti

@GaloisintheField Apart from the practical usefulness in simplifying the computations, it also tells you that the class number is finite, which is not obvious from its definition

Galois in the Field

That's true.

Balarka Sen

@Alessandro Since when did you start doing number theory man

Edward Evans

who said number theory

Alessandro Codenotti

I took an ANT course in my last year of undergrad @Balarka

Balarka Sen

5:08 PM

ah

Alessandro Codenotti

I actually really liked it too

Thorgott

Edward has been summoned

Edward Evans

lol

Balarka Sen

kvlt mathematician arrives

geocalc33

number theory is the best

Edward Evans

5:09 PM

@Balarka distant blasting

Balarka Sen

amps go brrrr

geocalc33

subwoofer goes wubawuba

Balarka Sen

wtf

Edward Evans

what's bass

Alessandro Codenotti

@EdwardEvans That's a very weird way to spell downtuned 8 string guitar

Edward Evans

5:12 PM

I usually try and tune the 8th string up to a high e but the neck usually snaps

geocalc33

Bass is of fundamental importance in the theory of algebraic groups and Lie groups and in the study of the geometric objects associated to them

Edward Evans

Sanity check: Let $R$ be a PID and $x = \pi^e$ for some irreducible $\pi$ and $e \in \Bbb N$. Then the length of the $R$-Module $R/Rx$ is just $e$. If, more generally, $x = \pi_1^{e_1} \cdots \pi_g^{e_g}$ then $R/Rx \cong \bigoplus_{i=1}^g R/R\pi_i^{e_i}$ and the length of $R/Rx$ is $\sum_{i=1}^g e_i$.

Galois in the Field

In practice, one observes that $N(IJ)=N(I)N(J)$ and in particular, whenever we have $J|I$ we have $I=JK$ so that $N(I)=N(J)N(K)$ i.e. $N(J)|N(I)$, and thus when we lift a prime in $(p)\subset \Bbb Z$ to $(p)\mathcal{O}_K$, and factor it, we have a collection of primes laying over $(p)$ whose absolute norms are the degrees of their residue fields, which are divisible by $p$ (and are p if completely split, or ramified). Maybe that's wrong --

-- But regardless, the primes that lay over the primes who are less than the absolute norm generate the class group, since any ideal class for K can be represented by one of those primes in $\mathcal{O}_K$?

Balarka Sen

sounds right @Edward

R/(p^e) is local so

Edward Evans

right

geocalc33

5:17 PM

looks right to me

Thorgott

I can neither agree nor disagree, my neutrality is absolute

Balarka Sen

@Thorgott Let $X$ be a vector field on a compact manifold $M$.

Galois in the Field

Or in other words. If a prime in $\mathcal{O}_K$ lies over a prime $p\in\Bbb Z$ where $p>\lambda_K$, then necessarily all primes in the lift have absolute norm exceeding the minkowski bound too. Since every ideal class in $K$ can be represented by an integral ideal with absolute norm respecting the minkowski bound, we restrict our analysis to those primes over $p<\lambda_K$. Is that correct?

Edward Evans

yes

feynhat

Okay. This might be a good time to ask this question. Why do we need metric to construct the volume form? Metric gives us tangent-cotangent isomorphism, without it we don't have a canonical choice for dual of a frame. But suppose we force this. Like we start with a global frame and dualize this chart-wise, then wedge it all together and sum it up may be using partition of unity. What goes wrong?

Thorgott

5:20 PM

I don't understand. Can you phrase that in higher topos-theoretic terms?

4

Balarka Sen

Lmao

Galois in the Field

@Thorgott I can try if you want. But I'm not sure what grothendieck topology you want me to use, since I'm working with ramified covers (I'm also not sure how it would help)

Or was that not directed at me

Balarka Sen

@feynhat This is perfectly fine. You don't need a metric to construct the volume form, a volume form is just choice of a section of the top exterior power $\Lambda^n T^*M$ which is a 1-dimensional line bundle on $M$. If $M$ is orientable, there is no obstruction to constructing a nowhere vanishing section in the process you just described

Thorgott

nono, it was a joke

Balarka Sen

It's much weaker than the structure of a metric

Thorgott

5:23 PM

please don't mind me

Balarka Sen

Lmao

Galois in the Field

Ok

geocalc33

can a space be CAT(0), CAT(1), dot dot dot all at once?

Balarka Sen

@feynhat Also you cannot start with a global frame, that parallelizes the manifold. You want to choose orientation-compatible local frames chartwise

geocalc33

like in some places the space is CAT(0) but in some other places it's CAT(1) dot dot dot

Balarka Sen

5:25 PM

$S^2$ of course does not admit a global frame, it does not even have a nonvanishing vector field

Galois in the Field

I'm actually not sure the arithmetic in play even gets captured topos-theoretically. If you want to replace the associated schemes by their \'etale \infty-toposes, then you've got the correct geometric spaces associated to the (derived) deligne-\infty-stacks, but this is totally space-theoretic, and I don't know much about it yet

Balarka Sen

^ holy shit

geocalc33

what's topoi

Tobias Kildetoft

That looks like it could be taken straight out of garbology :)

feynhat

@BalarkaSen That's what I meant. I don't know why but 'local frame' doesn't feel right to say lol.

Balarka Sen

5:27 PM

To say a little more, nonvanishing volume forms on $M$ gives rise to a family of measures on $M$ which are all absolutely continuous with respect to each other

Thorgott

that message instills terror into me

Balarka Sen

haha i know what you mean yeah

i think geometers like to work inside the frame bundle $F(TM)$ to not be coordinate-dependent, but Ted can tell you more about this

Galois in the Field

Anyway I've found that replacing everything by its fancy version is a good way to completely miss the point, so I won't do that

feynhat

Galois took the meme too seriously

Balarka Sen

I can't tell if Galois is trolling or not

I feel like he is

Galois in the Field

5:28 PM

I'm serious, maybe I'm not used to the tone here

Balarka Sen

no worries, we like to leg pull infinity people

feynhat

and intergalactic believers.

Galois in the Field

I never took number theory in my undergrad, so it's lagged a little in my studies

Balarka Sen

omg imagine if one combines mochi and lurie

$\infty$-intergalactic

Galois in the Field

That would be disastrous lol

Mochi already seems to have significantly increased the weight of standard category theory, by strangely philosophically interpreting it

Balarka Sen

5:31 PM

@geocalc33 CAT(k) is in appropriate sense spaces of curvature bounded above by k. Smaller k provides tighter bounds

feynhat

@BalarkaSen I was living with the belief that you can't integrate functions if you don't have metric (because no metric -> no tangent-cotangent iso -> no volume form).

Balarka Sen

Well yeah the point is you don't have a canonical choice of the volume form

Galois in the Field

There was actually a point where Mochizuki was surprised that by “forgetting the history of a group” one still has more than a group up to isomorphism: Namely, a group. That the datum of a group is strictly more than the datum of a group up to isomorphism seemed new to Mochizuki. I believe this is the (psychological) main reason he considers these full poly-isomorphisms.

But really, forgetting “the history of the groups” you still have groups and they may still have natural commuting isomorphisms between them. Of course, for groups up to isomorphism you can’t ask for natural commuting isomorphisms between them, then you only have “full poly-isomorphisms”.)

-Scholze, April 15

feynhat

@BalarkaSen That's sounds like the final boss of a math RPG.

Tobias Kildetoft

@GaloisintheField Based on comments from people who have at least managed to understand part of the intergalactic papers, it seems that Mochizuki is not comfortable with constructions such as the product of two copies of the same object without having to first specify how one gets two copies rather than just the one you started with

Balarka Sen

5:34 PM

lol

I like how Tobias unironically used intergalactic there

Tobias Kildetoft

(this is apparently what he means with all the talk of forgetting the history of objects, since if you don't, you end up with the objects being equal, which is no good for such a construction)

@BalarkaSen You have taught me well

Galois in the Field

@TobiasKildetoft Yes, it's all very strange. I can definitely see how one can get there though. Like what does it even mean for two things not to be equal, but only isomorphic, is slightly philosophical (without digging into foundational stuff most people don't care about)

geocalc33

Polymorphic okay

Galois in the Field

His system does do away with the foundational concerns completely (by lumping an insane amount of abstraction on top - which is arguably much worse - and possibly introduces errors)

Btw, I am very much not an expert in that field, so don't quote me (if I was under my own name, I would refrain from commenting of course - as most have)

geocalc33

Consider an arbitrary TCFT over a base scheme enriched with a high-rank $O-$lattice, s.t. all commuting subgroups are all normal.

quote me on that

feynhat

5:43 PM

@Thorgott Reminded me of this.

Balarka Sen

Brutal

Thorgott

^me after calling something "functorial" instead of "canonical"

Balarka Sen

@Thorgott You can define a unique family of diffeomorphisms $\Phi^X_t : M \to M$ parametrized by $t \in \Bbb R$ such that $\Phi^X_{t + s} = \Phi^X_t \circ \Phi^X_s$ and $d/dt \, \Phi^X_t(p) \, |_{t = 0} = X(p)$ for all $p \in M$

This is called flow defined by the vector field $X$

Thorgott

is that a one-parameter group thing

Mike Miller

yeah

Balarka Sen

5:52 PM

Yup. The map $t \mapsto \Phi^X_t$ defines a group homomorphism $\Bbb R \to \text{Diff}(M)$, where $\text{Diff}(M)$ is the group of self-diffeomorphisms of $M$

Why isn't a group homomorphism called a groupomorphism?

Something to think about

Thorgott

why isn't a homeomorphism called a topomorphism?

Balarka Sen

something to think about

Thorgott

How does one construct that family of diffs?

Mike Miller

Existence and uniqueness of solutions to ODEs

Balarka Sen

Plus the fact that $M$ is compact plus defining it for a small $t$ defines it everywhere because the time-additivity

@MikeMiller One should really formulate existence and uniqueness of ODEs as existence and uniqueness of germs of solutions, and because of time-linearity, linear germs extend globally

Thorgott

5:56 PM

ODEs on manifolds, scary

Balarka Sen

iamverysmart

@Thorgott Do it on charts!

Thorgott

I figured that's what it's gonna be, but why do they glue together?

Something to do with uniqueness and compatibility of charts?

Balarka Sen

Essentially. Explicitly, you would like to solve, locally, for curves $\gamma_x : (-\varepsilon, \varepsilon) \to M$ such that $\gamma_x(0) = x$, $\gamma_x'(t) = X(\gamma_x(t))$. These small curves belong to charts on $M$ because you can choose a cover by ball charts of radius some $2\varepsilon$ by picking $\varepsilon$ sufficiently small

These exist by uniqueness and existence of ODEs for R^n

So now define $\Phi^t_X(x) := \gamma_x(t)$ for $t \in (-\varepsilon, \varepsilon)$

Can you tell me why this is smooth in $x$

(Modify the construction appropriately so that smoothness on $x$ coordinate is also immediate from uniqueness and existence of ODEs)

William Sun

6:13 PM

Is there a classification for prime ideals in $k[x,y]$ for non-algebraically closed field $k$(such as $\mathbb R$)?

I am trying to compute the dimension of the ring $\mathbb R[x,y]/(x^2+y^2)$.

Tobias Kildetoft

@WilliamSun You don't need a classification for that

Modding out an integral domain by a prime ideal always decreases the dimension

(non-zero prime ideal that is)

William Sun

If there is a classification other problems like this can be solved easily...but just curious, how do we compute the dimension for this specific example?

Tobias Kildetoft

Well, you know the dimension of the original ring

William Sun

Of course $\dim \mathbb R[x,y]=2$.

Tobias Kildetoft

right, and do you see why my claim above holds?

Balarka Sen

6:22 PM

Generally: If $L/K$ is an algebraic extension then $L[x_1, \cdots, x_n]$ is integral over $K[x_1, \cdots, x_n]$, so I believe that says heights of prime ideals in the former lying above prime ideals of the latter are the same

William Sun

Yes by the correspondence of prime ideals in the original and the quotient ring

Tobias Kildetoft

exactly

So we are left with two options. But one of those corresponds to the quotient being a field

Balarka Sen

So some combination of going up/going down/lying over should let you argue for examle that $\Bbb R[x, y]/(x^2 + y^2)$ has the same dimension as $\Bbb C[x, y]/(x + iy)$

Admittedly I forgot this integral extensions trash completely

Leaky Nun

see, Balarka is an algebraist at heart

Balarka Sen

I forgot it dude

i read it with my algebro geometric roommate

Leaky Nun

6:25 PM

search your heart

Balarka Sen

never internalized it

or i can google

William Sun

@TobiasKildetoft and I wonder if we can prove $(x^2+y^2)$ is not maximal?

Balarka Sen

or i can watch a playthrough and have dinner

ciao

Leaky Nun

@BalarkaSen $\Bbb C[x,y]/(x+iy) \oplus \Bbb C[x,y]/(x-iy) = \Bbb C[x,y]/(x^2+y^2) = \Bbb R[x,y]/(x^2+y^2) \otimes_{\Bbb R} \Bbb C$

Galois in the Field

@WilliamSun Does $(x,y)$ contain it, and vice-versa?

William Sun

6:33 PM

I don't think $(x^2+y^2)$ is the whole ring.

Galois in the Field

@WilliamSun It's not, and nor is $(x,y)$ which contains it

William Sun

Oh I'm being dumb.

Thank u I understand.

np

$(x-1,y)$

@LeakyNun ?

6:43 PM

(1,0) is a point on the circle

so $(x-1,y) \supset (x^2+y^2)$

Galois in the Field

No

It's a point on the circle of radius 1, not the circle of radius 0

Leaky Nun

oops

looks like I need a break

Thorgott

I've been trying to convince myself of just the existence and uniqueness part, but I do not see why this ought to be locally Lipschitz (or is this based on another uniqueness result?) @Balarka

Tobias Kildetoft

good old circle of radius $0$ (with more than one point)

Galois in the Field

Don't we all

Leaky Nun

6:44 PM

oh yeah

Thorgott

Why am I even thinking about ODEs, I was planning to do algebra

Leaky Nun

(0,0) is a point

so $(x,y) \supset (x^2+y^2)$

Tobias Kildetoft

And good old AG being as counterintuitive as ever

@Thorgott Then do algebra

Galois in the Field

@LeakyNun Well, I was expecting that to be clear without AG motivation :P

Tobias Kildetoft

always do algebra in fact

2

Galois in the Field

6:45 PM

Since $(x,y)$ is additively closed, and clearly $x^2,y^2\in (x,y)$

Balarka Sen

7:25 PM

@LeakyNun Yes but what about a general irreducible variety in $\Bbb R^n$

@Thorgott The idea is, for a chart $U \subset M$, solve for $F : U \times (-\varepsilon, \varepsilon) \to M$, $d/dt \, F(x, t)|_{t = 0} = X(x)$

The curves $\gamma_x$ are just $F(x, -)$, and smoothness in $x$ follows from existence/uniqueness

These solutions patch over charts by uniqueness, like you said

Thorgott

don't we need to post-compose with a chart or something

that's what I was trying to do

cause I know nothing about ODEs with codomain $M$

Balarka Sen

7:43 PM

Right, you need to set it up appropriately. Choose a chart $(U, \varphi : U \stackrel{\cong}{\to} \Bbb R^n)$ in $M$ and pushforward the vector field $X$ on $U$ to $\Bbb R^n$ by $\varphi$. Let us call $V = \varphi_* X$ to be this vector field on $\Bbb R^n$.

By existence uniqueness, there exists $\delta, \delta' > 0$ and $\varepsilon > 0$ such that there exists a unique $F : B_\delta(0) \times (-\varepsilon, \varepsilon) \to B_{\delta'}(0)$ solving the initial value problem $dF(x, t)/dt |_{t = 0} = V(x)$, $F(x, 0) = x$.

Pull this back to $M$ by applying $\varphi^{-1}$. For every point $x \in M$ you can prepare this solution, and then $\{\varphi_x^{-1}(B_{\delta_x}(0)) : x \in M\}$ has a finite subcover, indexed by $x_1, \cdots, x_n \in M$

Let $\delta = \min\{\delta_{x_1}, \cdots, \delta_{x_n}\}$ and $\varepsilon = \min\{\varepsilon_{x_1}, \cdots, \varepsilon_{x_n}\}$

That $\delta$ was unnecessary. Anyway, then you have a finite cover, which I will denote by $\{U_1, \cdots, U_n\}$, on $M$ such that there exists a unique $\Phi^i(x, t) : U_i \times (-\varepsilon, \varepsilon) \to M$ satisfying $d\Phi^i(x, t)/dt|_{t = 0} = X(x)$ for all $i \in \{1, \cdots, n\}$.

$\Phi^i(-, t)$ and $\Phi^j(-, t)$ agree on the overlap $U_i \cap U_j$ by uniqueness, so this defines a global map $\Phi(x, t) : M \times (-\varepsilon, \varepsilon) \to M$

Define $\Phi^X_t(x) := \Phi(x, t)$. What is needed to be checked now is that whenever $s, t, s + t \in (-\varepsilon, \varepsilon)$, $\Phi^X_{t + s} = \Phi^X_t \circ \Phi^X_s$. Setting $t = -s$ gives you diffeomorphism.

Thorgott

sorry, gotta go for a bit

I'll come back to this later

Balarka Sen

Sure. You can look up the details in Milnor, "Morse theory"

He has a nice section on flows

user10478

Is there a rule for computing $e^tA$ given a Jordan normal form, or do the superdiagonal 1's throw it off completely?

hchar

8:10 PM

Let $F$ be a field, and $K$ a finite extension of $F$. If $b$ is algebraic over $K$, then $[K(b):F] \le [F(b):F]\implies [K(b):K][K:F] \le [F(b):F]$. Hence, $[K(b):K] \le [F(b):F]$. Is this a reasonable chain of arguments?

Leaky Nun

@user10478 yes. try to do it by hand first and find the pattern

@hchar no, because the assumption $[K(b):F] \le [F(b):F]$ has not been justified

hint: argue using minimal polynomials

hchar

The minimal polynomial of $K(b)$ cannot have a higher degree than that of $F(b)$, since $F\subseteq K$. I mean at worst the same minimal polynomial $p(x)\in F[x]$ can be used in $K[x]$ such that $b$ is a root.

Leaky Nun

one can be more precise here

if $p \in F[x]$ and $q \in K[x]$ are the minimal polynomials of $b$ over $F$ and $K$ resp., then $q \mid p \in K[x]$, since $p(b) = 0$.

so $[K(b):K] = \deg q \le \deg p = [F(b):F]$

minimal polynomial forms the following "adjunction": for any polynomial $r$, $r(b) = 0 \iff p \mid r$

hchar

8:30 PM

ok, thanks.

Stupid Questions Inc

quick analysis question: I'm trying to show that given any open Jordan region $J$ on $\mathbb{R}^n$ we can approximate it using a finite union of aligned rectangles $R$ such that $\text{vol}(J\setminus R)<\epsilon$, I'd appreciate any hint (I tried to use the definition, which is that the characteristic function of $J$ is Riemann integrable to no avail, also tried to play with the $\iff \text{vol}(\partial J)=0 \wedge J\text{ is bounded}$ theorem)

Balarka Sen

8:46 PM

Unsure where you got stuck in using $\chi_J$ is Riemann integrable. That says you can cover $J$ by a finite union of rectangles $R_1, \cdots, R_n$ intersecting only at the edges such that $|\text{vol}(J) - \text{vol}(R_1 \cup \cdots \cup R_n)| < \varepsilon$, no?

Maybe I am missing something super subtle you have in mind

Stupid Questions Inc

@BalarkaSen let me recheck all my definition to see if that's workeable

Thorgott

ok, time to do this ODE stuff

Balarka Sen

I mean, of course, one has to know $\chi_J$ is Riemann integrable in the first place!

Which is essentially because it's discontinuous exactly on $\partial J$ which is a volume zero set in $\Bbb R^n$ by Jordan-ness (Jordanity?)

Thorgott

@Balarka I realize it's possible to choose as such WLOG, but is there a particular purpose behind your choice of having $U$ map onto all of $\mathbb{R}^n$?

Balarka Sen

@Thorgott Only psychological purpose, really.

Hi @Ted

Ted Shifrin

8:51 PM

Hi, a.

Hi, @Thorgott, @StupidQ.

Stupid Questions Inc

@TedShifrin Hi Prof, hope you're safe and confined :)

Ted Shifrin

Locked up, yup.

Stupid Questions Inc

@BalarkaSen Yes, everything flows smoothly, thanks a ton! :) Sometimes a small mosquito disturbs your view :)

Balarka Sen

Glad to be of help

I get confused a lot personally so that is very relatable

Thorgott

Hi Ted

Ted Shifrin

8:54 PM

Now you've got everyone thinking about flows, a.

Balarka Sen

Haha

Leaky Nun

@BalarkaSen please approve my use of the word "adjunction"

Balarka Sen

@Ted: I'm still confused why saying $f, g : M \to N$ are $r$-jet equivalent at $x \in M$ is not the same as saying $T^{(r)} f, T^{(r)} g : T^{(r)} M \to T^{(r)} N$ agree for all points $\tilde{x} \in T^{(r)} M$ lying over $x$ under $\pi^{(r)} : T^{(r)} M \to M$ :) To be clear, $T^{(r)} M = T \cdots T M$ is the $r$-iterated tangent bundle, not your $r$-th order tangent bundle

It seems true to me if I write what $T^{(r)} f$ is out explicitly for a function $f : \Bbb R^m \to \Bbb R^n$. I get a bunch of stuff involving all the derivatives of $f$ upto $r$ orders.

Ted Shifrin

Well, as Pig and I were saying, dimensions are just way off because there are $\binom{n+k-1}k$ partial derivatives of order $k$. But I don't see intuitively explicitly how to get all higher order partials from the iterated tangent bundle.

Balarka Sen

(Regardless, your comments yesterday were enlightening and got me thinking - thanks for that!)

Ted Shifrin

8:59 PM

I don't claim that anything is enlightening.

But I don't see how you get the usual contact structures going on your bundle that one has with the jet bundle.

Balarka Sen

I can trust that the bundles aren't the same.

I am reformulating what $r$-jet equivalency means, not claiming anything at a bundle level

Ted Shifrin

So if your fiber has too small a dimension, how can I recover all the info?

By the way, an official reference for what I was doing with $T^{(2)}$ is W. Pohl's paper from the 1960's. I'll get you the reference.

"Differential Geometry of higher order," first issue of Topology (!), 1962, p. 169-211.

Another mathematical brother :P

Thorgott

What exactly is the pushforward $\varphi_{\ast}X$? It should send $x\in\mathbb{R}^n$ to $d\varphi\vert_{\varphi^{-1}(x)}(X\vert_U(\varphi^{-1}(x)))\in T_x\mathbb{R}^n$, right? Can't really think of any other natural choice.

Ted Shifrin

I had all sorts of fun doing Chern classes with these various bundles for the projective tangent bundle for complex surfaces in $\Bbb P^5$.

@Thorgott, you have a diffeo $\phi$?

Balarka Sen

@TedShifrin Well, OK, first of all I don't have a bundle to compute dimension of fiber of - but I see your point. A map $f : \Bbb R^n \to \Bbb R$, for example, gives rise to $T^{(r)} f : T^{(r)} \Bbb R^n \to T^{(r)} \Bbb R$ and $T^{(r)} \Bbb R^n$ is $2^r n$-dimensional whereas $T^{(r)} \Bbb R$ is $2^r$ dimensional. This is slightly confusing to me.

Thorgott

9:03 PM

a chart from an open $U\subseteq M$ to $\mathbb{R}^n$

Ted Shifrin

Yeah, you can push forward vector fields by diffeos in general, but not by arbitrary maps. One major reason differential forms have so much more power.

Balarka Sen

@TedShifrin Thanks!

Ted Shifrin

I wouldn't write $d\varphi|_{\varphi^{-1}(x)}$. I would just evaluate, not restrict.

Thorgott

$d\varphi\vert_{\varphi^{-1}(x)}$ is just supposed to be the differential of $\varphi$ at $\varphi^{-1}(x)$

Ted Shifrin

If $\varphi(u)=p$, I would write $(\varphi_*X)(p) = d\varphi_u(X(u))$.

Thorgott

9:06 PM

the double use of vertical bars may be suboptimal notationally

Ted Shifrin

Well, evaluating a function at a point and restricting domain to the point are technically the same, but I don't like it.

Thorgott

I've never even thought about $d\varphi$ on it's own, but I guess it makes sense as function defined on the tangent bundle

Ted Shifrin

Yes. Given a smooth map $f\colon M\to N$ (manifolds), you get an induced map $df\colon TM\to TN$.

Alessandro Codenotti

(which is also smooth)

Ted Shifrin

and a bundle map

Balarka Sen

9:09 PM

@Ted Will you be willing to look at an example of what $T^{(3)} f$ looks like for a map $f : \Bbb R^3 \to \Bbb R$, or nah

Ted Shifrin

Although some would write it as a bundle map $TM\to f^*TN$ of bundles over $M$.

I don't care about $\Bbb R^3$. I'll be satisfied with $\Bbb R^2$ and $k=3$, I think.

Balarka Sen

OK. Let me write it carefully.

Thorgott

ok, and if $X$ is smooth, then $\varphi_{\ast}X$ is also smooth, of course

Ted Shifrin

Sure.

Balarka Sen

To get this out of the way, I will identify $T\Bbb R^2 = \Bbb R^2 \times T_0 \Bbb R^2$, $T^{(2)} \Bbb R^2 = T(\Bbb R^2 \times T_0 \Bbb R^2) = \Bbb R^2 \times T_0 \Bbb R^2 \times T_0 \Bbb R^2 \times T_0T_0\Bbb R^2$, $T^{(3)} \Bbb R^3 = \cdots = (\Bbb R^2 \times T_0\Bbb R^2) \times (T_0 \Bbb R^2 \times T_0T_0\Bbb R^2) \times (T_0 \Bbb R^2 \times T_0 T_0 \Bbb R^2) \times (T_0T_0\Bbb R^2 \times T_0T_0T_0 \Bbb R^2)$.

As a convention, I will denote elements of $\Bbb R^2$ by $x$, elements of $T_0 \Bbb R^2$ by $u$, elements of $T_0 T_0 \Bbb R^2$ by $v$ and elements of $T_0 T_0 T_0 \Bbb R^2$ by $w$…

Let me know if you agree with notation. I have not computed anything so far

Ted Shifrin

9:22 PM

How can you blithely plug in different kinds (orders) of tangent vectors to the same multilinear form?

That doesn't make sense to me.

Balarka Sen

We are in Euclidean space, where $T_0 \Bbb R^n$ is canonically $\Bbb R^n$. So if $Df : \Bbb R^2 \to \text{Hom}(\Bbb R^2, \Bbb R)$ is the Jacobian, it's derivative $D_x(Df)$ at $x \in \Bbb R^2$ will eat a tangent vector $u \in \Bbb R^2$ and spit out $D_x(Df)(u)$, an element in $T_{Df_x} \text{Hom}(\Bbb R^2, \Bbb R) = \text{Hom}(\Bbb R^2, \Bbb R)$

Therefore, $D_x (Df)$ is an element of $\text{Hom}(\Bbb R^2, \text{Hom}(\Bbb R^2, \Bbb R))$

Hence $D(Df)$ is a map $\Bbb R^2 \to \text{Hom}(\Bbb R^2, \text{Hom}(\Bbb R^2, \Bbb R))$, making the $x$-variable free.

Ted Shifrin

I'm complaining that $w$, $v$, and $u$ are not naturally parallel objects.

Balarka Sen

I am sorry, I am unsure what is the precise issue in what I said. I could be dumb

I understand your complaint, but I do not see the symbolic problem in these identifications

Ted Shifrin

I'm bothered by plugging in "different order" tangent vectors to the hessian and third derivative. To me, all the slots of those multilinear maps come from the same vector space.

So it makes me suspicious when you just say $T_0(T_0(\dots V)\dots) = V$.

It feels like something is not intrinsically meaningful.

Balarka Sen

OK, I see what you mean. Does it help if I write $Df : \Bbb R^2 \to \text{Hom}(T_0 \Bbb R^2, \Bbb R)$, and $D^2 f : \Bbb R^2 \to \text{Hom}(T_0 T_0 \Bbb R^2, \text{Hom}(T_0 \Bbb R^2, \Bbb R))$?

Ted Shifrin

9:30 PM

But then in what sense can we say that these are symmetric multilinear maps?

Something feels suspicious to me.

Balarka Sen

Right, we can't. I don't think that quite matters in the computation I am about to do.

Ted Shifrin

Hmmm ...

So, if we try to track $\partial^2/\partial x\partial y$ and $\partial^2/\partial y\partial x$ into your story, we find what ... ?

Balarka Sen

I believe you need $\partial/\partial x \in T_0 \Bbb R^2$ and $\partial/\partial y \in T_{\partial/\partial x} T_0\Bbb R^2$ to write $D^2 f_0(\partial/\partial y, \partial/\partial x)$. This happens to be the same as the mixed partial $\partial^2/\partial x \partial y$

Symmetry completely goes out of the window in this interpretation, as you said.

Thorgott

@Balarka Actually, I have no clue what theorem you're relying on to get existence and uniqueness there. Is it some higher-dimensional version of Picard-Lindelöf I'm unaware of?

Ted Shifrin

Somehow, I should be seeing some second-order partials in your second-order tangent space directly.

@Thorgott: That should be an $n$-dimensional theorem.

Balarka Sen

9:35 PM

Ted's right, @Thorgott. I am using the $n$-dimensional version.

Thorgott

damn, I don't know the $n$-dimensional version

time to look this up

Ted Shifrin

It's no different :P

Oh, OK, yeah, so the thing you wrote down is $\partial^2 f/\partial y\partial x(0)$, presumably.

Balarka Sen

Yeah

The problem is exactly as you said, to see the mixed partials, one needs to think about a tuple (tangent vector, tangent vector at a tangent vector)

Ted Shifrin

But now my dimension count complaint still keeps bothering me.

The lack of perceived symmetry makes the dimension discrepancy even WORSE.

But I think that proofs in Dieudonné or Lang of symmetry may apply to your viewpoint.

Balarka Sen

Hm, I felt that this lack of symmetry makes me carry more redundant information.

Ah I see.

Ted Shifrin

9:42 PM

But you can't afford redundancy. You don't have enough dimensions to start with.

Balarka Sen

Yeah, which is the confusing part. Maybe I will show you the calculation if you're convinced by notation, and you can tell me what I am doing wrong.

Ted Shifrin

For $n=2$ and $k=3$, my $T^{(3)}$ has fiber dimension $2+3+4 = 9$ and you have $8$ only?

Of course, my $T^{(k)}$ naturally has $T^{(k-1)}$ inside it. Is that true of yours?

Oh, your fiber dimension is only 6, not 8, I think.

Naturality (change of coordinates) requires lower order terms when you transform $k$th derivatives.

Balarka Sen

That makes sense

@TedShifrin $TTM$ naturally has $TM$ sitting inside of it, but most certainly not as a subbundle.

Ted Shifrin

Oh, I guess it does hold for yours, too. You can be vertical for $T^{j}$ but then "horizontal" at the $j+1$ stage.

Yeah, only at $0$ do you get a canonical subspace.

Otherwise you need connections. Damn, this is confuzling.

Balarka Sen

Yeah!

Ted Shifrin

9:46 PM

I like mine way better.

Balarka Sen

F'd up. I like yours too!

Ted Shifrin

LOL

So, if I count only my highest derivatives, I still should have too many for the dimension of your thingy. Right? We're comparing $\binom{n+k-1}k$ with $(2^k-1)n$, generically, I guess. Oh, so who does win that race?

If $n$ is big, you lose, no matter how big $k$ is.

Balarka Sen

Exactly, that's confusing to me! I guess there's some discrepancy in $n$

Ted Shifrin

But we won't see this with $n=2$.

(I think.)

Yeah, with $n=2$ you are right about having plenty of room (hence redundancy).

Oh, if you fix an $n$ and vary $k$, your dimension always beats mine.

It's only if we fix $k$ and let $n$ get bigger that we see problems.

Something's wrong with my math, evidently.

Maybe "always" means for sufficiently large $k$.

Balarka Sen

I got this too, which is what is so confusing to me. But let me write down the calculation of $T^{(2)} f : T^{(2)} \Bbb R^n \to T^{(2)} \Bbb R$ for general $n$, it looks weird.

Stupid Questions Inc

9:56 PM

@BalarkaSen sorry for the interruption but apparently I hit some difficulties. So in my definitions $J$ is a Jordan region iff $\chi_J$ is Riemann integrable on a rectangle $R$ that contains $J$, so in the definition we have to work with partitions of $R$ and not $J$, and I'm trying to guarantee that these rectangles must all lie inside $J$

Balarka Sen

Here is the confusing calculation, if you care (I will do this with general $n$): We use all the identifications and notations from earlier. The map $f : \Bbb R^n \to \Bbb R$ gives derivative $Tf : T\Bbb R^n \to T\Bbb R$, $Tf(x, u) = (x, Df(x)u)$. Taking derivative again, and using $T^{(2)} \Bbb R^n = \Bbb R^n \times T_0 \Bbb R^n \times T_0 \Bbb R^n \times T_0T_0 \Bbb R^n$, I get $Tf(x, u_1, u_2, v) = (x, Df_x(u_1), Df_x(u_2), D^2f_x(v, u_2))$. Now, using

$$T^{(3)} \Bbb R^n = (\Bbb R^n \times T_0\Bbb R^n) \times (T_0 \Bbb R^n \times T_0T_0\Bbb R^n) \times (T_0 \Bbb R^n \times T_0 T_0 \Bbb …

So I seem to keep track of all derivatives! What's wrong?

William Sun

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