Mathematics - 2020-05-03 (page 1 of 2)

Balarka Sen

00:16

@LeakyNun The transgression thing was bugging me so I tried to work it out. Suppose $S^{n-1} \to E \to B$ is an oriented spherical bundle; $\pi_1 B$ acts trivially on $H^*(S^{n-1})$ by orientedness, so we have a cohomological SSS with $E^2_{p, q} = H^p(B; H^q(S^{n-1}))$. This is zero everywhere except the horizontal lines $q = 0$, $q = n-1$ which are filled with $H^*(B)$. The differentials go like $d_k : E^k_{p, q} \to E^k_{p+k, q-k+1}$, so they are all zero for $k < n$, which means $E^2 = E^n$, and $d_n : H^p(B; H^{n-1}(S^{n-1})) \to H^{p+n}(B; H^0(S^{n-1}))$, top-to-bottom, diagonally rig…

I suppose it shouldn't be hard to figure out why $e$ is the geometric Euler class now, how hard can the single differential $d_n : E^n_{0, n-1} \to E^n_{n, 0}$ be?

Not sure though, let me see.

Balarka Sen

00:28

Maybe I'm wrong. I have to go back to the construction of the differential and do a staircase chase. Ugh

Thorgott

how does one solve a PDE

Balarka Sen

seems like a hard question to me haha

Thorgott

same

but these people want me to solve one and I know nothing about PDEs lmao

a lot of googling lies ahead

rschwieb

02:05

I have a softball question for the commutative algebraists out there: how would you explain whether or not $\mathbb Z[x]/(x^2-1)$ is a regular ring (in this sense en.m.wikipedia.org/wiki/Regular_ring)

Balarka Sen

@rschwieb I would draw Spec of that ring as disjoint union of two copies of the "smooth curve" Spec Z. Doesn't that make it clear that it's regular?

I don't know any commutative algebra but regularity of A to me means Spec A is a smooth scheme

Balarka Sen

02:40

Hi @Ted, @Edward

rschwieb

03:14

@BalarkaSen I have zero experience with the condition. When I look at its definition in terms of generators I am already foundering. I understand all the words you said, but the reasoning is also still opaque to me. It’s a pity then: it looks like this ring provides a counterexample to a question that was recently deleted by it owner.

Ted Shifrin

Hi, a @Balarka

Balarka Sen

@rschwieb The definition of regularity of a local ring $(A, \mathfrak{m})$ that I am aware of is $\dim A = \dim_{A/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2$. If $A$ is an algebra over it's residue field $k = A/\mathfrak{m}$, I think of $\mathfrak{m}/\mathfrak{m}^2$ as the cotangent space of the $k$-scheme $\text{Spec}\, A$ at the point $\mathfrak{m}$, so the condition translates for me as "dimension of the scheme = dimension of the tangent space", which is what smoothness means to me

rschwieb

@BalarkaSen so I can see its prime ideals are of the form $(n, x-1)$ Or $(n,x+1)$ where $n$ is 0 or a prime in $\mathbb Z$. That suggests two copies of $\mathbb Z$ I guess. They intersect where $n=2$. When you say two copies of $\mathbb Z$, are you thinking rather or the quotients by (the obviously minimal) prime ideals (x-1) and (x+1)?

Balarka Sen

I agree this is not as clear if you're not already working with $k$-algebras, like the example you gave

rschwieb

@BalarkaSen fascinating... a couple parsecs ahead of what I know, but fascinating.

Balarka Sen

03:23

@rschwieb Oh, maybe you're right. I was too hasty.

The copies of Spec Z do intersect, that's a problem

So maybe one should be able to see it's not regular at the prime ideal (2, x+1)?

rschwieb

@BalarkaSen maybe. I only know some rudiments about manifolds, and none of it is on spectra, just regular smooth manifolds. Are the theories basically the same or are they radically different?

Balarka Sen

I mostly think about affine $k$-schemes (or spectrum of $k$-algebras) where the ideas are more or less the same, so there may be issues when thinking about other kinds of rings

Let me see if I can prove it's not regular at (2, x+1)

So the maximal ideal of $\Bbb Z[x]_{(2, x+1)}/(x^2 - 1)$ is $\mathfrak{m} = (2, \overline{x}+1)$, and $\mathfrak{m}/\mathfrak{m}^2$ is $(2, \overline{x}+1)/(4, 2\overline{x} + 2)$, because $(\overline{x}+1)^2 = 2\overline{x} + 2$

Funny. That seems like $\Bbb F_2$

So it seems to be regular at $(2, x+1)$, haha

Or am I doing something obviously wrong?

Oh, no, OK. It's $\mathfrak{m}/2\mathfrak{m}$, which as a $\Bbb F_2$-vector space is 2-dimensional, with basis $2$ and $\overline{x}+1$

@rschwieb Yeah, so $\Bbb Z[x]/(x^2 - 1)$ is not regular at $\mathfrak{m} = (2, x+1)$, because the cotangent space is a $2$-dimensional vector space over the quotient field $\Bbb F_2$ whereas the ring has Krull dimension 1. Does this sound right to you?

rschwieb

04:19

@BalarkaSen if we’re sure the maximal ideal of the localization at (2,x-1) is not principal, then I buy it. The thing I don’t understand is why (2,x-1) is different.... is the maximal ideal of the localization at (3, x-1)$ principal for some reason?

Balarka Sen

04:31

@rschwieb This is because of non-domain weirdness, I think. $x+1$ lies outside $(3, x-1)$, so since $(x + 1)(x - 1) = 0$, this forces $x - 1 = 0$ in the localization by how localization of non-domains work.

So indeed you get that the localization of the ideal is just $(3)$.

Remember that in $S^{-1} R$, $a/b = c/d$ iff there is some $s \in S$ such that $s(ad - bc) = 0$. This $s$ is usually omitted because most people work over domains

rschwieb

04:49

@BalarkaSen yes, that makes sense. Well cool, I think that puts it to rest then. You’ve helped me fill another little fact in my website... thank you!

Balarka Sen

Thanks for the discussion!

rschwieb

@BalarkaSen I think I probably got more out of it than you did, so thanks. I’m more of a noncommutative algebraist.

Balarka Sen

Oh, I see, that's cool! Nah, I am pretty bad at commutative algebra in general, these discussions help to polish up on it.

When you say noncommutative algebraist, does that fall into the genre of representation theory, or analytic stuff like C^* algebras, or more complicated things like Leavitt path algebras and rings without IBN, etc?

I am never really sure what noncommutative algebra consists of, seems like a vast arena :)

rschwieb

@BalarkaSen Module theory and structural stuff. The last thing you said is pretty close. Have you seen my site? Links in my profile. Exotic examples of rings.

Balarka Sen

Yeah I have used it once in a while.

Pretty cool

I learnt about the Leavitt path algebra story from a friend through conversations and attended a talk recently. Pretty bizarre!

northerner

05:40

In terms of notation, for modular arithmetic does the number at the end have to be in brackets? en.wikipedia.org/wiki/Modular_arithmetic

For example could $38 \equiv 14 (\mod 12)$ just as easily be written as $38 \equiv 14 \mod 12$

rschwieb

06:16

@BalarkaSen I visited Gene Abrams department to give a talk back in 2011, I think. He did a lot of work with them...

CaptainAmerica16

This is my first time with an inductive proof problem, can someone tell me if I've set this up correctly?

$1/1\cdot 2 + 1/2 \cdot 3 + ... + 1/k(k+1) + 1/(k+1)((k+1)+1) =( k/k+1) + (1/((k+1)+1))$

oof, I understand if it's a bit too messy to respond to. Any help is greatly appreciated, otherwise.

on the right-hand side, it's supposed to be $k/(k+1) + 1/(k+1)((k+1)+1)$

chat won't let me edit

Leaky Nun

06:52

@BalarkaSen it seems like you can require a lot of conditions of the chains / cochains and still end up with the same (co)homology

e.g. smoothness

Balarka Sen

@LeakyNun It's not unexpected for smooth chains because you can approximate topological simplices by smooth simplices.

That's how you would formally construct a chain homotopy equivalence, in fact. By writing down a convolution formula

Do you want to figure out the Euler class thing togather? I suspect one might just be able to do it by looking at the construction of SSS

KReiser

Let me try to give you some down-to-earth help and then later on some pie-in-the-sky stuff. The prime ideals of $Z[x]$ are well-known, for instance [here](https://math.stackexchange.com/questions/174595/classification-of-prime-ideals-of-mathbbzx). By the correspondence theorem, prime ideals of $Z[x]/(x^2-1)$ are prime ideals of $Z[x]$ containing $(x^2-1)$. This means they're of the form $(x-1)$, $(x+1)$, $(p,x-1)$, $(p,x+1)$, or $(2,x-1)=(2,x+1)$ where we take $p$ an odd prime.

Localization at any of the first four is nice: as we've inverted 2 in each of those, the ideals $(x+1)$ and $(x-1…

Leaky Nun

@BalarkaSen sure

Balarka Sen

Oh hi @KReiser

I think I got a better picture for the example we discussed earlier; essentially my map is Zariski-locally $\Bbb A^1 \setminus \{0, -1\} \to \Bbb A^1 \setminus \{0\}$, $t \mapsto t^2$. I localized at the fiber over $1$ to get the map I discussed. This is an etale map but isn't proper, and that nonproperness survives to the localization (on a little neighborhood over $1$ it's a 2:1 map everywhere except the fiber over $1$, because $-1$ is missing).

Etale maps are proper iff finite

KReiser

@BalarkaSen This is indeed a good way to see the picture. Thanks again for your help yesterday.

Balarka Sen

07:05

No, thanks for the interesting question!

As I said these conversations are very helpful for me because I suck at algebra haha

KReiser

In the words of the kids these days, all I know as an algebraic geometer is "take Spec, take the associated sheaf, and lie"

Balarka Sen

Hahah

Leaky Nun

and Lie algebra?

KReiser

@LeakyNun funny story this was actually my minor topic for my orals! I did not learn it that well, stumbled through the exam questions on it, and then never used it again, lol. I could probably still do some root system computations if you made me...

Balarka Sen

@LeakyNun So I think the Serre SS is constructed by taking the filtration of the base $B$ by the skeleta, pulling that filtration back to $E$ to get $X_p = \pi^{-1}(B^{(p)})$ and then you have a filtration of the singular cochain complex $C^*(X)$ by $C^*(X_p)$.

Leaky Nun

07:12

right

maybe use exact couple

Balarka Sen

Yeah, trying to translate that to exact couples

I guess you'd look at $A = \bigoplus_p \bigoplus_q C_q(X_p)$, the inclusions $X_p \to X_{p+1}$ gives a map $A \to A$ which shifts the $p$-degree up by $1$, and let it's cokernel be $B$

So you get an exact sequence $0 \to A \to A \to B \to 0$. $B$ is essentially the rolled up relative homology complex $\bigoplus_p \bigoplus_q C_q(X_p, X_{p-1})$

Er, I am doing homology instead of cohomology. Fuck it, I'll stick to homology.

This gives an exact couple on total homology $H(A) \to H(A) \to H(B)$ by rolling up the homology long exact sequence, and then you can iterate using exact couples

Leaky Nun

$A^1_{p,q} = H_{p+q}(X_p)$, $E^1_{p,q} = H_{p+q}(X_p, X_{p-1})$

Balarka Sen

The $E^1$ page is the total relative homology $H(X_p, X_{p-1})$

Thanks, yep

When you have Serre's monodromy construction you get that this is actually $H_p(B^{(p)}, B^{(p-1)}) \otimes H(F)$

That's some computation, but also pictorially evident. This detail is irrelevant to us

So anyway that shows $E^2_{p, q} = H_p(B; H_q(F))$ by cellular homology

Leaky Nun

$E^r_{p,q} = \dfrac{i^{r-1} A^1_{p,q}}{i^r A^1_{p-1,q+1}} = \dfrac{\operatorname{im}(H_{p+q}(X_p) \to H_{p+q}(X_{p+r-1}))}{\operatorname{im}(H_{p+q}(X_{p-1}) \to H_{p+q}(X_{p+r-1}))}$

Balarka Sen

@LeakyNun Oh, you're using a different indexing convention. $E^1_{p, q} = H_q(X_p, X_{p-1})$ to me I think

Just to get that confusion out of the way

Leaky Nun

07:20

ok

I think you write $n$ for that

$n=p+q$

Balarka Sen

Yeah

It's OK

So now to chase higher differentials. I think it's best to do it for a general double complex $\{K_{p, q}, \delta, d\}$ with horizontal differential $\delta$ and vertical differential $d$, now that we know what our double complex is.

Leaky Nun

I think if you use my formula then you know what the differential is

Balarka Sen

Uh, is it super apparent? In exact couple setup the differential is $d_r = j_r k_r$, but is it apparent what it does without unrolling the exact couple back to the double complex?

I want to read $d_r$ in the $E^0$ page itself, remember

Leaky Nun

$E^r_{p-r,q+r-1} = \dfrac{\operatorname{im}(H_{p+q-1}(X_{p-r}) \to H_{p+q-1}(X_{p-1}))}{\operatorname{im}(H_{p+q-1}(X_{p-r-1}) \to H_{p+q-1}(X_{p-1}))}$

so you want a map $H_{p+q}(X_{p+r-1}) \to H_{p+q-1}(X_{p-1})$

no

can you change all the $H$'s to $C$'s and then use the differential $d: C_n(X) \to C_{n-1}(X)$ and hope you come up with a well-defined non-trivial map

Balarka Sen

I think that's the first approximation for the differential, yeah. You have to correct for that by going down a staircase

To make it well-defined

1 sec let me do some scribbling on paper lol

Leaky Nun

07:36

another construction:
$Z^r_{s,t} = \{ x \in F_s C_{s+t} \mid \exists \varepsilon \in F_{s-1} C_{s+t}, d(x+\varepsilon) \in F_{s-r} C_{s+t-1} \}$
$B^r_{s,t} = \{ x \in F_s C_{s+t} \mid \exists y \in F_{s+r-1} C_{s+t+1}, x-dy \in F_{s-1} C_{s+t} \}$

Balarka Sen

i cant even read that mate lol

Leaky Nun

here $F_s C_{s+t} = C_{s+t}(X_s)$?

Balarka Sen

i dont how people write formulas for specseqs

I guess so @LeakyNun

Leaky Nun

$Z^r_{s,t} = \{ x \in C_{s+t}(X_s) \mid \exists \varepsilon \in C_{s+t}(X_{s-1}), d(x+\varepsilon) \in C_{s+t-1}(X_{s-r}) \}$
$B^r_{s,t} = \{ x \in C_{s+t}(X_s) \mid \exists y \in C_{s+t+1}(X_{s+r-1}), x-dy \in C_{s+t}(X_{s-1}) \}$

is this more legible

Balarka Sen

I have to try something myself first haha, I'm slow with these things

So say you have the double complex $\{K, \delta, d\}$, $\delta$ is horizontal, $d$ is vertical, with total differential of $\text{Tot}(K)$ being $D = \delta + (-1)^p d$. The natural filtration is by $\bigoplus_{p \geq k, q} K_{p, q}$, the stuff on the right of $p \geq k$. The $E^1$ page is homology of the associated graded $GK = \bigoplus_{p, q} K_{p, q}/K_{p+1, q}$, which has total differential $D = (-1)^p d$.

Leaky Nun

07:43

@loch why is $D$ the Euler class of $\mathcal O(D)$?

and what does it mean?

Balarka Sen

What is $H_{(-1)^p d}(GK)$? I think it's the same as $H_d(K)$

Leaky Nun

how come?

northerner

In terms of notation, for modular arithmetic does the number at the end have to be in brackets? https://en.wikipedia.org/wiki/Modular_arithmetic
For example could 38≡14 (mod 12) just as easily be written as 38≡14 mod 12

Balarka Sen

Oh, I wrote $GK$ wrong. $(GK)_p$ is $\bigoplus_{i \geq p, q} K_{i, q}/\bigoplus_{i \geq p+1, q} K_{i, q} = \bigoplus_q K_{p, q}$.

So the $d$-cohomology (vertically up in $q$ index) is indeed $H_d(K)$

So the $E^1$ page is $H_d(K)$ and $E^2$ page is $H_\delta H_d (K)$

So an element of $E^1_{p, q}$ is represented by an element $\alpha \in E^0_{p, q}$ such that $d\alpha = 0$, and an element of $E^2_{p, q}$ is represented by an element $\alpha \in E^0_{p, q}$ such that $d\alpha = 0$ and $\delta \alpha \in \ker d$, right?

That's the staircase diagram. If you push $\alpha$ up by an index to $(p+1, q)$, it becomes zero. If you push $\alpha$ left by an index to $(p, q+1)$, it's image of something from $(p-1, q+1)$, one step down, by $d$

That guy, let's call it $\beta$ so that $\delta \alpha = d\beta$, might be the one representing $d_1[\alpha] \in E^1_{p-1, q+1}$

loch

i dont know how to see this from the sss definition lol but if you use the thom classes stuff then it should be doable

the upshot being that the euler class is really the intersection between a generic section and the zero section of O(D)

then if you remember, a general section cuts out something that is linearly equivalent to D (in particular homologous)

Balarka Sen

07:56

^ That's how I think about Euler class

Leaky Nun

why does it generalize Euler characteristic?

Balarka Sen

OK, I think I see $d_r$

Euler characteristic of M is Euler class of TM, @LeakyNun

Leaky Nun

why?

Balarka Sen

That's the famous Poincare-Hopf theorem; if you take a generic vector field the signed count of the zeroes computes $\chi(M)$

Proof-idea: The number does not depend on the vector field you choose (why?). Pick a triangulation of $M$ and on each face $\Delta^n$ construct a vector field which vanishes at the center of $\Delta^n$ and is a source at that point, vanishes at the center of each $k$-face and is a saddle of order $k$ at those points, and vanishes at the vertices of $\Delta^n$ and are sinks at those points

The signed count is $V - F_1 + F_2 - F_3 + \cdots \pm F_n$

Euler characteristic famio

@Leaky So I think $d_r : E^r_{p, q} \to E^r_{p - r, q + r - 1}$ is defined as follows. Let's call $E^0_{p, q}$ be the double complex you started off with. Pick an element $\alpha \in E^0_{p, q}$ representing $[\alpha] \in E^r_{p, q}$, by your formula. Then $d_{r-1} [\alpha] = 0$ translates to $d\alpha = 0, \delta \alpha = d\alpha_1, \delta \alpha_1 = d \alpha_2, \cdots, \delta \alpha_{r-2} = d\alpha_{r-1}$. Then $d_r [\alpha] = [\delta \alpha_{r-1}]$

It's an elongated snake map

Leaky Nun

hmm

what does this mean in terms of H

Balarka Sen

08:08

Translating lol hol up

Lol hype news that heartburn tablets can cure corona

Is medical science retarded?

Ban this subject

2

Leaky Nun

lol

Balarka Sen

It's literally alchemy at this point

Just mixing random chemicals to see if you get a cure for Corona

What the fuck are we paying pharmaceutical chemists for

user434058

proofwiki.org/wiki/Bibhorr_Formula

user434058

Is this formula true?☝️

user434058

I think it's an approximation.

user434058

08:18

How can we sidestep trigonometry and do the stuff which was supposed to be done using trigonometry. Also, there aren't really any factors in the formula which might make up for values of sines and cosines of any random angle.

Balarka Sen

Looks fucking suspicious lol

Bibhorr, what a name

user434058

@BalarkaSen BTW, he's Indian :P

Balarka Sen

Horrible name

user434058

And to be honest, the proof really looks like dark magic to me, I mean how can we even get something like this!!!

user434058

It's sorcery.

skullpatrol

08:28

@BalarkaSen corona is the virus, the disease is Covid which they can't cure

Balarka Sen

potato pohtato

skullpatrol

tomato tohmato

~~tohmato~~ tomahto

Balarka Sen

tomaeto you mean

Mike Miller

@BalarkaSen Sounds like a proof of the 4-color theorem

2

Balarka Sen

Lol

skullpatrol

08:43

This virus knows no colour boundaries.

(not that any others do :)

mostly native American Indians are dying

in the US

skullpatrol

09:25

@r9m :-D

maths student

09:59

Show that the function $f$ given by $f(x)=|x|$ is continuous on $\mathbb{R}^{n}$.

First I Consider $|x-a|^{2}=(x-a) \cdot(x-a)$ then try to do some epsilon delta stuff but I am not able to get any conclusion somebody can help?

Tobias Kildetoft

@mathsstudent Well, what epsilon-delta stuff did you try?

maths student

@TobiasKildetoft I tried it for $\mathbb{R}$ but for R^{n} I didn't know ?

Tobias Kildetoft

That does not actually answer my question

Simone

10:41

@mathsstudent remember the results from the triangle inequality.

then it's easy to see what $|x-a|$ should be smaller than.

Secret

11:01

Fun fact: I have maths dreams like Ramanujan, but the maths I saw in my dreams are often nonsensical in the real life lens, unlike Ramanujan's which lead to profound breakthroughs

Simone

Well sometimes I have epiphanies whilst I'm running XD

But I'm not going to run marathons anytime soon

Balarka Sen

I get visions when I am having a vision

Leaky Nun

ah yes, the floor here is made of floor

Balarka Sen

No, but like, one of the visions I had

consisted of a vision

Not even joking

rschwieb

11:21

@KReiser thank you for that extended intuition! From regular smooth manifolds, I'm aware of the "badness" of an intersection like that since it means the manifold can't be expressed as the graph of a function there. It sounds like spectra considered as manifolds share much of the same theory.

HokieFan7

Anyone here good with complex line integrals? And can veryify my work

verify*

Simone

I have to turn the equation of an ellipse with the origin at one focus from its cartesian form to polar form but I'm having issues with the algebra.
Can someone provide a link of someone working through it?

Remember its with the origin at one focus.

Simone

11:57

Maybe you can help?
I need to turn $\frac{(x-c)^2 ì}{a^2}+\frac{y^2}{b^2}=1$ into $r(1-\frac{c}{a} cos(\theta))=\frac{b^2}{a}$, with $a^2 =b^2 +c^2$

I have pages of aimless calculations now :(

HokieFan7

Are you talking to me?

Simone

I'm not talking to anyone in particular.

HokieFan7

mathopenref.com/coordgeneralellipse.html I found this link idk if you've seen similar pages but it might help

Simone

alas they don't derive the formula in polar coordinates :/

Secret

12:28

@BalarkaSen vison-ception, otherwise, I have no clue

Balarka Sen

12:51

@Secret Haha I meant it tautologically though

@MikeMiller Hey PDE man tell me how functions $f$ satisfying $f' \leq a - f^2$ behave

Alessandro Codenotti

PDE man lol

Mike Miller

Take $a = c^2$, then $f' + f^2 = c^2$ is given by the family $f_t(x) = c \tanh(ct + cx)$

Alessandro Codenotti

What is $\Bbb Z_{p^\infty}$?

Balarka Sen

The Sylow p-subgroup of $\Bbb Q/\Bbb Z$, I suppose.

$\Bbb Z[1/p]/\Bbb Z$ equivalently

Thorgott

Prüfer group?

Balarka Sen

12:56

@MikeMiller The inequality should force $f$ to grow slower than any guy in your family I suppose

Mike Miller

Yeah, I'm trying to remember the precise statement of that

Balarka Sen

Sturm's theorem, or something

Alessandro Codenotti

@Thorgott I think that's the same as what Balarka said

Thanks to both

Balarka Sen

Oh no Sturm is linear ODEs

Gross

Sturm–Picone comparison theorem

In mathematics, in the field of ordinary differential equations, the Sturm–Picone comparison theorem, named after Jacques Charles François Sturm and Mauro Picone, is a classical theorem which provides criteria for the oscillation and non-oscillation of solutions of certain linear differential equations in the real domain. Let pi, qi i = 1, 2, be real-valued continuous functions on the interval [a, b] and let ( p 1 ( x ) y ′...

Picone what a name

Mike Miller

Those are second order

Balarka Sen

12:59

Ah yeah

Mike Miller

I suspect I can't find the statement of the first order one because it's just an exercise lol

Thorgott

right, it is

Mike Miller

Probably you just need that $f' \leq G(x,f)$ for $G$ Lipschitz

Balarka Sen

Seems believable

Mike Miller

You were interested in asymptotics to infty?

Balarka Sen

13:03

I think something is true pointwise here, not just asymptotically.

Mike Miller

Do you have an idea the kind of statement you're looking for

Balarka Sen

Let $f' = c - f^2$ and $g' \leq c - g^2$. Then I think $g \leq f$ pointwise.

Lol

I should work it out on my own

Mike Miller

Yes that's true

Balarka Sen

But then you made me do a calculus exercise

Mike Miller

Well

Yeah that's true too

I mean that if $f' \leq G(x,f)$ and $g' = G(x,g)$ with $f(x_0) \leq g(x_0)$ then $f(x) \leq g(x)$ pointwise

I thought you were using this to get asymptotics on $g$

I swapped f and g from you sorry

Balarka Sen

13:07

Ah OK I see it yeah

Differentiate $(g - f) e^{\int (g + f)}$.

@MikeMiller Why is that true

I should know these comparison theorems for ODEs lolol

Oh, because $(f - g)' \leq G(x, f) - G(x, g) \leq C |f - g|$ by Lipschitzness

Thumbs

Mike Miller

14:02

@BalarkaSen I'm surprised this doesn't have a name though

Balarka Sen

@MikeMiller I will call it the Ur comparison theorem

All comparison theorems in Riemannian geometry should boil down to this

Mike Miller

Well sure

You should lightly imply that there's an ancient mathematician named Ur

Balarka Sen

Lool

Thorgott

I'm trying to find all smooth functions $p,q\colon\mathbb{R}^2\rightarrow\mathbb{R}$ satisfying $\frac{\partial q}{\partial x}-\frac{\partial p}{\partial y}=y$. Can I just fix an arbitrary $q$ and then get $p(x,y)=\int_0^y\frac{\partial q}{\partial x}(x,t)dt-\frac{y^2}{2}+h(x)$, where $h$ is an arbitrary smooth function $\mathbb{R}\rightarrow\mathbb{R}$, as all $p$ such that $(p,q)$ satisfies the given by FTC? For some reason this feels very wrong.

Balarka Sen

That's correct Thorgott

Mike Miller

14:07

It's a very unconstrained equation

Balarka Sen

Super underdetermined stuff

Mike Miller

Write $V$ for the space of solutions to your equation. Of course, there is a map $V \to C^\infty(\Bbb R^2)$ given by sending $(p,q)$ to $q$. What you've observed is that this is surjective (you have given an appropriate $p$). You identified the fibers as being affine over $C^\infty(\Bbb R)$.

Not much more to say than that

Thorgott

thanks!

Balarka Sen

These are known in literature as partial differential relations; if $\Phi$ is a differential operator aka a polynomial on $\partial^m/\partial x_{i_1} \cdots \partial_{i_m}$ for indices running over subtuples of $\{1, \cdots, n\}$, a differential relation is something of the form $\Phi(f_k) = g$ where $\Phi(f_k)$ is to be interpreted as feeding the function $f_k$ to the $k$-th differential operator term in $\Phi$, all these functions different

Highly underdetermined, in the sense that the space of solutions $V$ of such a relation over $\Bbb R^n$ usually maps surjectively into $C^\infty(\Bbb R^n, \Bbb R^m)$, space of smooth functions from $\Bbb R^n \to \Bbb R^m$ for some $m \leq n$, like Mike said

The fibers won't be affine in general, of course.

The tuple of functions $(f_k)$ is known as a "formal solution" to the partial differential relation "$\Phi = g$". An actual solution would be when all the functions $f_k$ are equal, $\Phi(f) = g$ in the normal sense, solving the PDE.

If $\Phi$ is a differential operator such that whenever you have a formal solution you have an actual solution it is said to satisfy $h$-principle. This seems like a strong restriction but surprisingly happens often.

Mike Miller

@BalarkaSen That's just a certain kind of differential equation

Thorgott

14:19

did you just use this to segue into Gromov..

Mike Miller

A differential relation would be $\Phi(f) \subset C$

for some C, maybe defined by an inequality or whatever

Balarka Sen

@MikeMiller Yes, but I didn't want to talk about general PDR's

Mike Miller

I think they're only worth mentioning if you're talking about a feature that distinguishes them from PDEs though

Balarka Sen

Shrug. I can do that, but that'd be a long detour :)

Balarka Sen

14:33

@MikeMiller Actually there aren't too many examples I know of where these kind of $h$-principles hold for a PDR of the form $\Phi = 0$. I suppose the isometric immersion theorem is one.

Ample differential relations are like this I believe; $\text{im}\,\Phi \subset C$ where $C$ is a closed set.

Usually you approximate from the interior of the convex hull of $C$, so the formal solutions are not quite cooked up like the way I suggested.

@Thorgott Also yes, that was pure bait

Shiranai

Hello, I have a really silly question that doesn't deserve a post:

I am trying to prove that the commutator subgroup is a subgroup, so $x^{-1}y^{-1}xy\alpha^{-1}\beta^{-1}\alpha\beta$ should be a commutator

but I don't see how that is of the form $a^{-1}b^{-1}ab$

Balarka Sen

It need not be

Commutator subgroup is generated by the commutators, the commutators itself need not form a group

Shiranai

ahhhhhh I undestood the question wrong, thanks!

half an hour looking at that expresion :)

Balarka Sen

It's not exactly easy to come up with explicit examples

Alessandro Codenotti

What about $[a,b][c,d]$ in the free group generated by $a,b,c,d$

Balarka Sen

14:41

That works, but needs a proof :)

Alessandro Codenotti

Fair enough

Mike Miller

@BalarkaSen If C is closed it's still a PDE I think

It's d(Phi, C)^2 = 0

Balarka Sen

@MikeMiller Sure, I am asking which PDEs satisfy h-principles

Don't really care too much about terminology

Oh, Whitney's theorem on singular maps is one. If $f : (\Bbb R^2, 0) \to (\Bbb R^2, 0)$ is a map germ such that $\det(df) = 0$, then $f$ is either a fold or a cusp catastrophe

Eh

I guess?

Balarka Sen

15:19

Oh, there's also: If $X$ is a Stein manifold, continuous maps $X \to \Bbb C$ can be approximated by holomorphic maps $X \to \Bbb C$. That's an $h$-principle for the Cauchy-Riemann relation

Hm, that can't be right. Take $X = \Bbb C$ lol, uniform limit of holomorphic functions are holomorphic

What is the theorem

"Uniform approximation on holomorphically convex subsets of $X$ with interpolation on closed complex subvarieties of $X$"

Whatever that means

Thorgott

I'm relieved you don't know what it means either

JamalS

15:56

Best class field theory book (local and global) that DOES use Galois cohomology?

Pole_Star

Bro

What's the best book for vector calculus

For introduction to divergence,curl and gradient

Archer

@Pole_Star Thomas Calculus is good

Mike Miller

16:20

@JamalS I think someone else already suggested Cassels and Frohlich

Mad Spaces

Is there some kind of a book or a page or some kind of collection calculating using integration the moment of inertia of all possible different bodies?

Astyx

17:29

hi chat

Ted Shifrin

17:40

@BalarkaSen You know what a closed complex submanifold/subvariety is. Closed complex hypersurface is the simplest case.

Salut, @Astyx.

Knight

@TedShifrin Sir do you help with physics also? I’m asking because once you mentioned your high school physics teacher who needed your help in teaching physics to the whole class

Ted Shifrin

LOL, that was 50 years ago.

Semiclassical is an actual physicist.

Knight

Well I suspect if that Ted (fifty years ago Ted) is no more :-)

Edward Evans

17:59

Wat

Ted Shifrin

howdy @Edward

Edward Evans

Hiya @TedShifrin

How's it going?

Ted Shifrin

To answer your question from a few days ago, I haven't heard from or seen Lukas.

Edward Evans

Ohhh okay

Thanks :)

Ted Shifrin

I assume you folks don't go anywhere near campus these days.

Alessandro Codenotti

18:03

Hi @Ted @Edward

Edward Evans

No not at the moment, my lecture courses are held asynchronously online and seminars via a conferencing tool

Hey @Alessandro

Ted Shifrin

Yup, I figured. Hi, demonic @Alessandro

Thorgott

if a function is continuous on $\mathbb{C}$ and holomorphic on $\mathbb{C}\setminus\gamma$, where $\gamma$ is (the image of) a Jordan curve, is it necessarily entire?

this should be true heuristically

Ted Shifrin

This should be a Morera's Theorem proof.

Thorgott

yeah, that's my heuristic

Ted Shifrin

18:06

Continuity along $\gamma$ tells you integrals over triangles will still all be $0$.

LOL, I don't think of a proof as a heuristic.

Thorgott

it's not that easy though, right

you need to approximate the boundary

while guaranteeing convergence

that didn't quite make sense

Ted Shifrin

No, you just need to make sure your triangles can be chosen not to overlap the curve too seriously.

And you can get away with very "thin" families of triangles or rectangles or whatever.

But I will retract my criticism of your use of "heuristic." Jordan curves can be a headache.

Thorgott

you want to split the integral along the triangle up according to how the triangle is split by the curve, no?

then you need to know that Cauchy's theorem still works when you only have holomorphicity on the interior and continuity on the boundary

Ted Shifrin

Oh, I'm not thinking of it that way.

Thorgott

is holomorphicity even a word?

Ted Shifrin

18:12

I'm letting the triangle cut across the curve.

@Thorgott Sure.

I think the approach you're thinking of, splitting into triangles on either side of the curve, shows up when you do Schwarz reflection arguments, but you don't need it here, methinks.

Thorgott

the corresponding statement for when $\gamma$ is a line can be proven explicitly without much effort and be used to prove the Schwarz reflection principle (at least that's how I learned it)

how would you show the integrals vanish then?

Ted Shifrin

If the triangle in question cuts $\gamma$ in only a set of measure $0$, then the integral is still $0$.

I was thinking a finite number of points, I confess, but ...

Thorgott

hmm, why is that?

Mike Miller

The thing you're integrating is 0 except on a set of measure 0, by hypothesis

But being able to do this sort of cutting is unclear because there are Jordan curves of positive measure

Ted Shifrin

Well, no, it's holomorphic.

I hate Jordan curves.

Mike Miller

18:19

Nonsense. I'll see myself out

Ted Shifrin

LOL, well, my original thought is likely to be nonsense, too, thanks to your Jordan tempest in a teacup.

Mike Miller

Mine?

Ted Shifrin

"Jordan curves of positive measure"

Thorgott

I don't even see why finitely many points make no issue without the other argument

Stupid Questions Inc

One question concerning math.stackexchange.com/q/1370328 the answer says "Any prime that divides both $b$ and $n$ will also divide $b^{n-1}$, making it impossible to have $b^{n-1} \equiv 1 \mod n$." I don't see how $b^{n-1} \equiv 1 \mod n$ would be impossible in that case

Mike Miller

18:20

I was trying to pass to an integral over the disc.

Where you're integrating f' dA or something.

Ted Shifrin

@Thorgott: So my lemma was: If $f$ is continuous and holomorphic except at $z=a$, then $\int_\gamma f(z)\,dz = 0$ even if $a\in\gamma$.

$\gamma$ closed blah blah blah ...

Thorgott

how would you prove this? let the curve make a small bump towards the inside to avoid $a$ and let that go to $0$?

Ted Shifrin

@StupidQuestionsInc If $p|c$ and $p|n$, then can $c\equiv 1\pmod n$?

@Thorgott: Yeah.

Using uniform continuity.

Stupid Questions Inc

@TedShifrin so we would get that $p|c$ and $p|c-1$ which is impossible since $\gcd(c,c-1)=1$, right?

Ted Shifrin

You don't need to go that route. You deduce $p|1$ directly.

Stupid Questions Inc

18:26

@TedShifrin thanks a ton Ted, this was bugging me for 20 min lol

i still don't see how we get $p|1$, you mean we don't need to mention gcd when we can just say that then $p|c\wedge p|c-1 \implies p| c - (c-1)=1$??

Thorgott

I don't see if that generalizes nicely to worse intersections though (without going the entire non-trivial way to prove that nice approximation is always possible)

Ted Shifrin

@Stupid, precisely.

Stupid Questions Inc

@TedShifrin oh ok, thanks again :)

Thorgott

ah, this has a name apparently

encyclopediaofmath.org/index.php/Painlev%C3%A9_theorem

you need rectifiability, it seems

Ted Shifrin

Yeah, I just read that there too.

Mike Miller

18:33

Knew I should have checked Garnett first

I was trying to hack something together with Caratheodary, which is one of the few complex analysis theorems I know about general Jordan domains

Ted Shifrin

OK, I'm convinced. This is way too hard for me.

Thorgott

I'm looking at a paper where Caratheodory appears to be the method of proof

but this is way too hard for me as well

good to know that this is true though

Mike Miller

Jordan domains can be some pretty pathological stuff

Ted Shifrin

This was a routine homework exercise? :D

Mike Miller

@Thorgott Now find out if it's true for an arbitrary 1-dimensional rectifiable set in place of $\gamma$!

Ted Shifrin

18:57

What is a rectifiable non-Jordan set with Hausdorff dimension 1? Yikes.

Oh, self-intersections allowed.

Mike Miller

Self-intersections allowed, and I'm sure there's horrible pathologies that are worse than that.

Ted Shifrin

Yeah, I'm sure :(

Mike Miller

It was not a serious suggestion

Simone

19:14

Hello.

Thorgott

oh god

please spare me

Simone

Tomorrow the Italian government will ease some of its restrictions. We'll be able to travel from one municipality to another again... I've been in lockdown for 2 months now ugh

Anonymous

Hi. I'm trying to find a group $G$ with composition factors $A_5$ and $C_2$. In that context, could someone help me interpret this comment: "If there is a normal subgroup isomorphic to $A_5$, then an element outside of $A_5$ either induces an inner automorphism of $A_5$, in which case the group $G$ is $A_5 \times C_2$, or not, in which case $G \leq \mathrm{Aut}(A_5)=S_5$, so $G \cong S_5$."

Anonymous

If an element $g \in G\setminus A_5$ induces an inner automorphism on $A_5$ I can understand why $G$ may be written as a semi-direct product $A_5 \rtimes_\varphi C_2$ but I'm not sure why the semi-direct product should necessarily degenerate into a direct product or rather why is the homomorphism $\varphi_g: A_5 \to A_5$ necessarily trivial. Could someone explain?

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