Mathematics - 2020-01-13

Alex Wertheim

00:00

Very nice! In a parallel universe, he and I might have been fellow grad students.

Ted Shifrin

You're still Mr. Algebra :P

Alex Wertheim

It's true, but I wouldn't hold his trade against him. ;P

We all have our vices

Ted Shifrin

Well, I'm not usually so magnanimous.

Actually, @MikeM will be amused to know that, as a reward for doing a brief review of a book proposal, I now am the proud owner of Kirwan & Woolf's Introduction to Intersection Homology Theory. It brings back some memories.

Alex Wertheim

Lol. I do wish I understood more geometry! It's a defect I'm looking to fix at some point. I've had to learn some in the process of trying to do some algebraic geometry, but I'm still quite ignorant, I'm afraid.

Ted Shifrin

Well, I probably can't help, but if I can, let me know.

Alex Wertheim

00:04

Our algebra seminar this quarter is actually on perverse sheaves, coincidentally.

I'm sure you could! I'll certainly take you up on it once I'm ready to get serious, should that ever happen. Thanks!!

Ted Shifrin

Well, I've forgotten most of what I once knew.

Terran Swett

02:42

I'm curious, is there a piece of software that's "like vim but for expression rewriting and equation solving"? Like, vim is a text editor that lets you manipulate text using keystrokes, using lots of few-keystroke commands, like x to delete a single character, dd to delete a single line, and rL to replace a single character with the character "L".

It would be cool if I could type a couple of keystrokes to do a command like "move this term to the other side of the equation."

If there's no such software out there, maybe I'll have to write it myself. think emoji

Leaky Nun

04:26

@TedShifrin hi

adesh mishra

05:48

Can someone please help me in understanding how this image is an exercise?

ÍgjøgnumMeg

08:30

no wait

Is the tensor algebra of a $K$-vector space $V$ defined as $T(V) := \bigoplus_{n\geq0} V^{\otimes n}$ or $\bigoplus_{n\geq 1} V^{\otimes n}$ ?

Balarka Sen

What is $V^{\otimes 0}$

ÍgjøgnumMeg

I think it's usually defined as $K$

I'm just seeing if there's something I don't know about, cuz my exercise sheet seems to be defining it as starting from $n= 1$

Balarka Sen

I would define it as $K \oplus V \oplus (V \otimes V) \oplus \cdots$, yeah

ÍgjøgnumMeg

That's in order for the multiplication on $T(V)$ to make sense right?

Balarka Sen

Multiplication also makes sense on the truncation, no? It's graded multiplication, you have a map $V^{\otimes m} \otimes V^{\otimes n} \to V^{\otimes (m + n)}$

ÍgjøgnumMeg

08:42

Yeah you're right

I just wanna know what would break if you started from $n = 1$

Balarka Sen

I don't think anything serious would happen

ÍgjøgnumMeg

Fair

I'm "showing" that the $w$th symmetric power of $\Bbb R^n$ is isomorphic to the space of grade $w$ homogenous polynomials in $n$ variables over $\Bbb R$

which intuitively makes perfect sense

In fact I'd go as far as to say it's kind of obvious? The wth symmetric power has as a basis just $w$-tensors that you let commute (I guess, as in, identifying $v_1 \otimes \dots \otimes v_w$ with $v_{\sigma(1)} \otimes \dots \otimes v_{\sigma(w)}$ for some $\sigma \in S_w$) and you can just send these to the degree $w$ monomials

Balarka Sen

Ya

ÍgjøgnumMeg

So I just need to show that that map is $K$-linear and bijective o.O

user123

can someone take a look at my question: math.stackexchange.com/questions/3506730/…?

adesh mishra

09:16

@LeakyNun Are you around?

Leaky Nun

10:12

@AlessandroCodenotti play?

Alessandro Codenotti

I need to leave very soon

Leaky Nun

@AlessandroCodenotti we can play bullet

ÍgjøgnumMeg

Yo

Leaky Nun

@ÍgjøgnumMeg hi

@ÍgjøgnumMeg wanna play?

ÍgjøgnumMeg

10:27

I gotta read :/

Leaky Nun

take a break

ÍgjøgnumMeg

No time for breaks rofl

Leaky Nun

there's always time for breaks

ÍgjøgnumMeg

not when I have exams in like 2 weeks

and a seminar talk next week

Leaky Nun

who cares about exams

ÍgjøgnumMeg

10:28

and 2 Psets

Leaky Nun

and seminar talks

and Psets

ÍgjøgnumMeg

I do rofl

Leaky Nun

I'm just a pile of ashes

constantly burnt out

ÍgjøgnumMeg

rofl

same

but

Leaky Nun

I stopped caring about maths a long time ago

ÍgjøgnumMeg

10:30

if I do badly then I'm kinda in trouble so

:P

Leaky Nun

I'm already dead inside

ÍgjøgnumMeg

dang

adesh mishra

11:02

@LeakyNun Are you around?

Leaky Nun

@adeshmishra why?

adesh mishra

@LeakyNun Can you help me with that attachment?

Leaky Nun

I don't understand your question

Rithaniel

(I think Leaky might need a break, to be honest)

adesh mishra

That image was supposed to be an exercise, what I have to do in that? Do I have to write it symbolically or do I have to state true and false?

Leaky Nun

11:06

@adeshmishra ask the person who gave you the exercise

@Rithaniel lets play

adesh mishra

@LeakyNun okay, I’m sorry if I disturbed you.

Lucas Henrique

11:30

@adeshmishra we can’t really know... as @Leaky said, it’s better to ask who wrote it. Nevertheless, you could do both.

Also: hi chat!

Rithaniel

Heya Lucas

adesh mishra

@LucasHenrique Thank you

Leaky Nun

12:00

@Rithaniel boy I missed a tactic to win your knight

Rithaniel

In the late game, right?

Leaky Nun

yeah

Rithaniel

It seems that we consistently go back and forth pretty hard in the early game, but then, mid and late game you destroy me. I think you have a better grasp of long term strategy than me

Leaky Nun

maybe

1010011010

13:09

Hello everybody

Very straightforward question

I am working with very very very very very very long expressions inside fractions.

What are the conventions in regards to notation for this ?

I am already at smallest point size of the font, and these very long expressions are already a short-hand of something even longer.

Something like $\times\cdots$ at the end, then start the next line with a phantom equals-to and another times?

I could repeat the capital pi's again, but that's so damn ugly...

Semiclassical

13:46

One approach would be to define some of those quantities separately, e.g., “$\Pi_{j,k}A_j B_k C_{jk}$ where $A_j=\ldots, B_k=\ldots, C{jk}=\ldots$“

That’s overkill if A,B,C are simple, but in your case they’re complex enough that this might be useful

1010011010

Yeah I get it...

I'll try...

Mathematical physics can lead to stupid expressions sometimes. I'm considering just reverting back to the more general argument that I have a polynomial of which zeroes are known, so say a $P_2$ at the top for each instance of $\ell,m$, and then just making a more general argument

1010011010

14:10

Thanks Semi

Semiclassical

Np

Akiva Weinberger

15:03

Apparently Frank P. Ramsey (of Ramsey theory) died at 26

Rithaniel

15:23

So, what're people working on?

Ultradark

"Geometry and Topology" by Bredon

Semiclassical

angle inequalities for spherical quadrilaterals

Alessandro Codenotti

Set theory, as always

Rithaniel

Any particular set theory problem or are they the sort to be too involved to easily explain?

Ultradark

Can anyone calculate 31 percent of 836,431 to within .01 percent accuracy in less than 5 seconds?

Rithaniel

15:26

(Also, spherical quadrilaterals sounds like a pain)

Semiclassical

you're not wrong

Zero Punctuation NOOOOOOOO

►

Rithaniel

I'm sure someone can, Ultradark. I'm not one of them, though

Semiclassical

to be more precise: Given two points A,B on a unit sphere, let AB be the length of the (smaller) great-circle arc which connects them. (Alternatively, AB is the central angle for A,B.)

I'm interested in what set (AB,BC,CD,DA) are possible for a sequence of four points A,B,C,D on a sphere. In the most obvious cases, those values can be interpreted as side lengths of the spherical quadrilateral ABCD

Rithaniel

This seems connected to the problem you were working on before

Semiclassical

yeah

the main problem is that it's hard to consider 4 points at once

Rithaniel

15:31

Is there a larger problem this is commected to?

Semiclassical

this is all in service to a QM problem, but one which basically amounts to "consider two pairs of unit vectors and the resulting angles between them"

so the underlying motivation is not helpful

I already know the answer to a simpler version of this problem, where you only take points A,B,C and consider (AB,BC,CA)

Rithaniel

You might try to find a space homeomorphic to the sphere which might be a little more cooperative when handling 4 points at once

Semiclassical

it'd be nice to be able to project this to the plane, yeah

I mean, stereographic projection is one option

Rithaniel

Turn all the circles into lines

Semiclassical

stereographic projection won't do that, unfortunately. all of the arcs are great circles, but they're not all great circles through the north pole

hence stereographic projection will map some of them to circles again

Rithaniel

15:36

You can pick the "north pole" to be the intersection of two great circles and then you get two lines and an ellipse

Semiclassical

1) two lines and a circular arc

2) that's fine for the three point case, but for the four-point case you can only pick one of the four vertices that way

say, A out of ABCD

then AB and AD will be great-circle arcs, but BC and CD won't

moreover, stereographic projection doesn't preserve lengths. one can account for this but it's not trivial

Rithaniel

Why wouldn't they be? Given any two points on a sphere, they define a great circle

Semiclassical

sure, but you can only pick one point to be the north pole while doing stereographic projection

there's no reason the great circle containing B,C should contain A as well.

user123

can someone take a look at my question : math.stackexchange.com/questions/3506730/… ?

Rithaniel

Ah, right, so two lines and two arcs

user123

15:42

@ThomasShelby lol , you saw peaky blinders i guess ^^

Semiclassical

for reference, the relevant metric in the plane after stereographic projection is $ds^2=(1+r^2)^{-1}(dx^2+dy^2)$

actually, I think that's off by a factor of 4?

oh, blah. should've been $(1+r^2)^{-2}$ as well

anyways. in the case of 3 points, one has the following IIRC:

first, the angles AB,BC,AC are between 0 and pi (inclusive) by construction

Thomas Shelby

@user123 Sure! (I accidentally deleted my previous reply, lol)

user123

@ThomasShelby Amazing show ! :)

Semiclassical

beyond that, one has $AB+BC\geq AC$, $BC+AC\geq AB$, $AC+AB\geq BC$, and $AB+BC+AC\leq 2\pi$

My understanding/suspicion is that this kind of characterization should work for the four-point case as well, namely that the set of allowed (AB,BC,CD,AD) should be determined by a finite set of linear inequalities

but figuring out those inequalities is giving me a headache. maybe working in the plane will help in that regard

Rithaniel

16:02

You never know. You could also try something where you fix two points and vary the other two, subject to some constraints

Lucas Henrique

I was wondering: is there any way to construct a group operation over $\mathbb R$ and a product $\mathbb C \times \mathbb R \to \mathbb R$ in such a way that $\mathbb R$ is a $\mathbb C$-vector space?

Since the reals and complex numbers have the same cardinality, I’ve tried to construct the group operation in such a way that $(\mathbb R, \cdot) \cong (\mathbb C, +)$

Lucas Henrique

16:23

Yup, there’s a way. :)

Kedar Lodge

16:36

If a y vs x curve is vertical then is it dependant or independant of x?What about horizontal?I have this perennial confusion.Please help me clear my doubt.

Semiclassical

I’ll assume you’ve got coordinates (x,y)

Kedar Lodge

Yes exactly

@Semiclassical

Semiclassical

If you’re only interested in y as a function of x, then a vertical line isn’t the graph of such a function

x is held fixed along that line, so you can’t obtain that line by varying x

Alessandro Codenotti

@Rithaniel forcing V=HOD

Semiclassical

By contrast, you can get a horizontal line by varying x and keeping y fixed

So that’s a plot of a constant function, ie y doesn’t depend on x

Thorgott

16:47

$(\mathbb{R}^{\times},\cdot)\not\cong(\mathbb{C},+)$ @Lucas

Also I'm not sure what the relation between those two questions is. There's a group operation of $\mathbb{C}^{\times}$ on $\mathbb{R}$, say, by multiplication with the modulus, but that's weaker than giving you a vector space structure.

Semiclassical

17:11

ooo: math.stackexchange.com/a/177729/137524

Thorgott

Noob set theory question: does it make sense to say $\mathcal{P}(X)=\{E\colon E\subseteq X\}$? Not as in asserting existence of this "set", but is it possible to state this equality after invoking the axiom of power set or is talking about "all" $E$ (without restricting them to coming from a given set) simply not well-founded?

I guess it definitely works in NBG.

Alessandro Codenotti

17:25

What do you mean with "does it make sense"?

Thorgott

Is it well-defined, I guess

Lucas Henrique

@Thorgott I didn’t say $\mathbb R^\times$, and I’m not talking about $\cdot$ as the usual product

Alessandro Codenotti

It's definitely something that is done commonly

But if you want to work with such an object in ZFC you should justify it being a set

Any set theory book defines $H_\kappa=\{x\mid |\mathrm{trcl}(x)|<\kappa\}$, but then they justify why this is a set (by showing $H_\kappa\subseteq V_\kappa$)

Lucas Henrique

Let $f\colon \mathbb C \to \mathbb R$ be a bijection. Then $a+b := f(f^{-1}(a) + f^{-1}(b))$ defines an abelian group structure over $\mathbb R$ and $zr := f(zf^{-1}(r))$ makes $\mathbb R$ a $\mathbb C$-vector space

Thorgott

Ok, it being a set follows from the axiom of power set, so I take it that equality is fine to state with this in mind.

@Lucas ah, ok

Alessandro Codenotti

17:31

Right

Thorgott

Thanks

Alessandro Codenotti

Even in ZFC you can kind of talk about definable classes $\{x\mid \psi(x)\}$ for some formula $\psi$ in the language of set theory, but this gets tricky

Lucas Henrique

@Thorgott I guess this is right.

Thorgott

Yeah, that's what my NGB comment was getting at. $\{E\colon E\subseteq X\}$ is a class and with the axiom of power set, we can argue it's not proper.

Lucas Henrique

Test: $\circ$

Alessandro Codenotti

17:39

Right, but models of NBG that look like a model of ZFC with all of its definable classes are essentially models of ZFC, the interesting things happen when more classes are allowed

(I know very little about NBG, just what I needed to give a proper formulation of Kunen's inconsistency theorem for a talk)

Thorgott

17:49

@Lucas you can do even better and keep the standard additive structure on $\mathbb{R}$

(at least if you like Choice)

Akiva Weinberger

@BalarkaSen I don't remember. Did you ever get to the punchline of my "approximating pi with areas versus perimeters" question?

Semiclassical

Usually, the punchline of "is X bigger or less than Y" is not what you expect

so on those grounds, I'm going to guess that X=Y :)

(if it wasn't a surprise, you wouldn't be asking it)

Semiclassical

18:21

Sometimes, you look at a paper's refs and you wonder how the heck they extracted claim X from source Y

example: An unpublished REU paper from 2002 (link) states a reasonably nice inequality on spherical 4-gons

but they attribute it to a paper with this title: "On the existence of unitary flat connections over the punctured sphere with given local monodromy around the punctures" (link)

and I have utterly no idea how they get that claim from that paper

sardinsky

18:56

Customers arrive at a bank according to a Poisson process with
rate $\lambda$ per minute. Given that two customers arrived in the first 4
minutes, what is the probability that one arrived in the first minute and the other arrived in the last minute.

The setup for the problem is the following

$$P[N_1=1,N_3=1,N_3=1|N_4=2]=\frac{P[N_1=1,N_3=1,N_4=2]}{P[N_4=2]}$$

by bayes' theorem and conditionality which is equal to

$$\frac{P[N_1=1]P[N_2=0]P[N_1=1]}{P[N_4=2]}$$

I understand the inclusion of the term $P[N_2=0]$ due to the stationarity of increments of a poisson process but I don't understa…

Semiclassical

19:11

I did manage to find the advising author's current email, so I guess the simplest approach is just to ask them if they remember why that source was cited

hi @ted

Ted Shifrin

Hi @Semiclassic

Semiclassical

weird math word of the day: Triunduloids

Ted Shifrin

They undulate three times and then die?

Semiclassical

so it would seem

the abstract probably means more to you than it does to me:
"In 1841, Delaunay constructed the embedded surfaces of revolution with constant mean curvature (CMC); these unduloids have genus zero and are now known to be the only embedded CMC surfaces with two ends and finite genus. Here, we construct the complete family of embedded CMC surfaces with three ends and genus zero; they are classified using their asymptotic necksizes. We work in a class slightly more general than embedded surfaces, namely immersed surfaces which bound an immersed three-manifold, as introduced by Alexandrov."

(and somehow this connects back with the spherical polygon inequalities I've been after. I have no idea how)

Ted Shifrin

19:36

Yes, I know about these unduloids and constant mean curvature. I even have a little about them in the penultimate section of my diff geo notes.

Semiclassical

"The spherical polygon inequalities also appear in the context of holomorphic vector bundles[1]." Now if only I could figure out how on earth they get that from [1]

(I have utterly no idea what holomorphic vector bundles are supposed to do with spherical polygons)

Ted Shifrin

Well, that could fit with monodromy if they're flat bundles.

They're doing bundles on the sphere, presumably, but I don't know.

Semiclassical

the cited paper is "On the existence of unitary flat connections over the punctured sphere with given local monodromy around the punctures", so that seems legit

Ted Shifrin

Yes, I read what you put above.

Semiclassical

kk

Akiva Weinberger

20:45

●●●●/■ = ○○○○/□ = π

Ted Shifrin

say what, DogAteMy?

Akiva Weinberger

Interpret ● and ■ as the area of a circle and square

(the circle inscribed in the square)

and interpret ○ and □ as their perimeters

They're Unicode stuff so they might not show up in your browser

Ted Shifrin

That's the most horrid looking $\pi$ I've ever seen.

So you're taking the fourth powers for the circle?

Akiva Weinberger

No, adding them

Area of four circles next to each other

Basically just saying $4(\pi r^2)/(2r)^2=4(2\pi r)/(8r)=\pi$

Slereah

Hello people of math

I am looking for

Mixed geometry

Some paper claims that, given a manifold with a "metric" that is asymmetric

such a field is called mixed geometry

but I am having troubles finding anything on the topic

is that what it is actually called?

Akiva Weinberger

20:52

That's interesting. I wonder how one would visualize such a thing

I guess the length of a curve would be the same as the symmetric version (replacing the metric $g$ with $(g+g^\top)/2$)

Slereah

It was in vogue in the 30's because it was part of some effort to generalize general relativity to a unified field theory

The basic idea was that since the gravity metric was symmetric, and the tensor of electromagnetism was antisymmetric

Heck let's put them together

Thorgott

I've seen these referred to as "quasimetrics" before, but I don't know anything about what geometry they result in.

Slereah

There's a few papers on the topic, but they're mostly physicsy

and fairly focused on 4 dimensions

Akiva Weinberger

Hold on, there's two things this could mean

One is, $d(p,q)\ne d(q,p)$

Ted Shifrin

I'm not sure I know what to do with a non-symmetric metric tensor. So $u\cdot v \ne v\cdot u$?

Akiva Weinberger

20:56

The other is $\langle u,v\rangle\ne\langle v,u\rangle$

I'm assuming the second

Slereah

$g(X, Y) \neq g(Y, X)$

Ted Shifrin

I don't know what this even means, though.

Akiva Weinberger

Aight

Slereah

I guess it's just a generic tensor field, really

Of a certain signature

Ted Shifrin

So we work with a non-symmetric bilinear form. But what does it mean geometrically?

How do I define anything using it?

Lengths of curves don't care, as Akiva pointed out. I can symmetrize anyhow.

Alessandro Codenotti

20:58

Hi everyone

Akiva Weinberger

Geodesics are the same as in $(g+g^\top)/2$ I assume

Ted Shifrin

But I don't know what a non-symmetric "inner product" even means.

Slereah

It changes the connection, though

Akiva Weinberger

if they're defined as "locally minimizing length"

Ted Shifrin

Hi, demonic @Alessandro.

Akiva Weinberger

20:58

Hm. If you define geodesics by $\nabla_\gamma\gamma=0$ then is it different?

Slereah

Well the other part of the idea was that it used general affine connections

Akiva Weinberger

Where $\nabla$ is the connection

Alessandro Codenotti

I have a question for you Ted

Ted Shifrin

But in what sense is the connection compatible with the "metric"? Just $\nabla g = 0$? I guess that makes sense, but there will be strange goings-on.

Yes, @Alessandro?

Slereah

There were plenty of strange goings-on, certainly

Alessandro Codenotti

21:00

Do I need to define an infinitesimal generator in the context of flows or is that standard terminology?

Slereah

And the theories never really worked out

But apparently Einstein really wanted this to work?

So plenty of people worked on it

Ted Shifrin

It's usually used in general for a group action, @Alessandro. For a flow, it's just the associated vector field.

Alessandro Codenotti

The flow gives an Abelian group of $\mathrm{Diff}(M)$

One homemorphic to the additive group of the reals to be precise

Ted Shifrin

Why semigroup? You can flow backwards.

Alessandro Codenotti

(my flow is defined everywhere for simplicity)

Ted Shifrin

21:01

LOL

Alessandro Codenotti

Is there a natural Banach space structure on $\mathrm{Diff}(M)$ such that this group is strongly continuous and the notion of infinitesimal generator from differential geometry agrees with the one from functional analysis?

Ted Shifrin

@Slereah: There certainly is plenty of stuff done on Finsler geometries, which is more general.

Slereah

I'll look into it

Thanks

Ted Shifrin

I don't know the answer to that, @Alessandro. I never think about manifold structure on the diffeomorphism group, but Mike knows the answer.

Alessandro Codenotti

I'll ask him the next time I see him online, thanks

Semiclassical

21:08

I feel like this is 'obvious' but I'm not sure: Suppose I've got a convex set $D\subseteq \mathbb{R}^n$. Now let $f:\mathbb{R}\to \mathbb{R}$ be some monotonic function. If I act $f$ on $D$ elementwise, do I get another convex set?

Ted Shifrin

What do you mean elementwise?

You mean component by component?

Semiclassical

Right.

Ted Shifrin

Convexity is checked component by component.

Semiclassical

Right.

So the R^n nature of this is irrelevant.

Ted Shifrin

I'm saying that $f=(f_1,\dots,f_n)$ is convex iff $f_j$ is convex for all $j$. It sounds right.

Semiclassical

21:11

yeah

except, wait.

I'm not talking about whether $f$ is convex, I'm asking about whether it sends convex sets to convex sets.

Ted Shifrin

So the question is: Is a set convex iff it's convex coordinate by coordinate?

Semiclassical

Hrm.

Ted Shifrin

Clearly $\implies$ is true.

Oh, it doesn't make sense.

The region may not be polygonally closed (where I take my polygons along coordinate axes).

Semiclassical

The example I'm working with, for specificity: I start with some convex subset of $[0,\pi]^3$ and then I map that to $[-1,1]^3$ using $\theta\mapsto \cos\theta$

Though I guess I should be able to drop that down to 2D without losing anything essential

Ted Shifrin

Oh, I see the issue. Line segments won't map to line segments.

Paths work fine, but they're not linear paths.

So, yeah, hopeless.

Semiclassical

21:19

hrm

Ted Shifrin

Probably stay in the convex hull by monotonicity.

Semiclassical

yeah

Ted Shifrin

Although convex hull of what I'm not sure.

Akiva Weinberger

22:36

Test

Lucas Henrique

23:02

Should it be easy to read a math textbook? By easy I mean paced, and by paced I mean without being stuck in the same page for 3 days or so.

Ted Shifrin

Depends on the book and the subject matter and the reader, of course. In general, no, it's not "easy" to read math textbooks.

But one should almost never read passively. One should read actively, with paper and pencil (and brain), working things out in detail unless they're totally obvious.

Michael Albanese

23:56

Hi @TedShifrin

How did you deal with the pressure to publish early on in your career?

Akiva Weinberger

@TedShifrin Someone I follow on Twitter got anonymous feedback from their class:

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$Mathematics$ Mathematics