Mathematics - 2019-01-09 (page 1 of 2)

Lucas Henrique

00:05

Could someone please give me advice?
I'm already familiar with the notion of rings, fields, etc and they are indeed relevant in the way I see math. I'm gonna take linear algebra - actually, I'm using @Ted's book on linear algebra. Should I read, simultaneously, another book with focus on linear algebra applied to abstract algebra? There's this specific theorem that states that any field $F$ of characteristic $p$ has order $p^n$, and this relies on linear algebra. Considering real linear algebra, I don't think I'll ever understand how (based on the formal definitions of vector space, etc, si…

Daminark

So, a lot of the linear algebra you did, even if it was in the context of $\mathbb{R}$, will transfer wholesale to other fields. For example, the theorem you quoted is a consequence of the fact that $n$-dimensional vector spaces over a field $k$ is isomorphic to $k^n$. The proof is: write down the proof for $\mathbb{R}$, and do find and replace with $k$

The main thing that's trickier in finite fields is the difference between polynomials and polynomial functions

Lucas Henrique

I know those are different. So I'll be okay if I study linear algebra on $\Bbb R$?

Daminark

For the most part, as long as you examine your proofs to see where you invoke concepts that rely on it. Polynomials vs polynomial functions, stuff over $\mathbb{C}$ is dangerous because of algebraic closure, and every now and then characteristic might kick in

Lucas Henrique

Oh, and there's also everything that cares about measure: working with measures with $\Bbb R$ is pretty easy (in 3 or less dimensions), but measures over $\Bbb Z_p$ are pretty much a problem

Daminark

When do measures kick in for linear algebra?

Lucas Henrique

00:18

Please take note that I know nothing about measure theory, so this is a loose use of the word measure. So: Cauchy-Schwarz, distance between stuff, determinants (I think? I mean, they're not dependent on $\Bbb R$, but I'm not sure if the rules there still apply everywhere).

Daminark

00:33

Oh distance is a bit different, yeah that can kick in

Yeah I guess it kinda depends on how you prove the stuff actually

I sorta learned inner products later, which I say is a bit of a good thing

Some people prefer proving facts geometrically but since people have reason to care about fields where some of these geometric ideas don't really hold as nicely I like to know which facts use algebra, which use geometry, etc

But yeah if you prove a lot of stuff using dot products it might actually be a bit dangerous for the sake of thinking about finite fields

Jacksoja

02:09

Hello

@MatheinBoulomenos Hi ! there is something a little technical i want to ask you about, found it in artin book

MatheinBoulomenos

@Jacksoja okay

Jacksoja

so , if f : G--> G' is a group hom

and we have H is a subgroup of G' , then preimage of H is a subgroup of G

Now my question is in the case of normalilty, if we start by assuming that H is normal is G' , and take its pullback, we do not require f to be surjective right ?@MatheinBoulomenos

MatheinBoulomenos

right, that's not needed

Jacksoja

it is needed if we start with normal subgroup of the domain

and then taking f of it

MatheinBoulomenos

yes

Jacksoja

02:22

artin did not asssume surjectivity on both, and I was comfused ^^

@MatheinBoulomenos thank you

MatheinBoulomenos

if we don't assume surjective for the latter, then one could show that every subgroup is normal

since if H is a subgroup of G, then H is the image of the inclusion map H->G

Jacksoja

aha yeah true

maybe that is why he did not even bother to make it in the proposition

so this correspondance thm, it sais that the number of subgroups that contain the kernel are in bijective correspondance with the number of subgroups of the target group

this seems very usefull but I do not know how much !

MatheinBoulomenos

I wouldn't say number of subgroups

just say "subgroups that contain the kernel are in correspondence with subgroups of the target group"

Jacksoja

{subgroups of G that contain K} <--> { subgroups of G' } this what i meant

I think that this also relates to , the only Subgroups of G/K , are of the form H/ K , where K is normal in H

MatheinBoulomenos

with H containing K

that's basically the same statement

Jacksoja

02:32

oh yeah true, H contains K is better ,because normality is clear

MatheinBoulomenos

since the 1. isomorphism theorem tells you that if f:G->G' is a surjective group homomorphism, then G' is isomorphic to G/K where K=ker(f)

Jacksoja

the intersection of two cosets H and K is either empty or a coset of the intersection subgr

it this possible to prove with the corres theorem ?

MatheinBoulomenos

hmm, I don't think so

Jacksoja

@MatheinBoulomenos also how useful is the correpondence theorem if one knows subgroups of one group and tries to find subgroups of the unkown one

I thought so, this exercice were put on wrong section of the book ^^

famesyasd

04:47

If $a_n$ is in $\mathbb{C}$ and $\sum_0^\infty a_n$ is absolutely convergent and $b_n$ is any rearragement of $a_n$ then $b_n$ is also absolutely convergent and both $\sum_0^\infty a_n$ = $\sum_0^\infty b_n$ and $\sum_0^\infty |a_n|$ = $\sum_0^\infty |b_n|$ are equal?

Al t.

05:36

0

Q: Deterministic dynamic programming example

Consider the following example In the beginning of Solution procedure, Why the maximum of $p_3(x_3)$ is automatically achieved by allocating all $s_3$ teams? Why finding $x_2^*$ requires calculating and comparing $f_2(s_2, x_2)$ for the alternative values of $x_2$, namely, $x_2 = 0, 1, ...

Could someone help with the last question specially?

Dair

06:32

hi chat

user574848

10:02

Hello

I remember reading about some online tool that allowed you to put in a decimal and it would tell you similar approximations

Like if I put in $3.15$, it might tell me that it's around $\pi$ or $\sqrt{10}$ (I don't know how precise the software is, I only remember reading a vague description of it in a comment)

Does this ring a bell to anyone?

user574848

10:15

^ nevermind, I think I found it

Gabriel Romon

11:06

Hello, do you know if Daniel Fischer will return to the site ?

Nick

11:34

Daniel Fischer ♦

173k 16 160 282

complex-analysis real-analysis functional-analysis general-topology sequences-and-series analysis calculus

According to his profile, he was last seen 61 days ago.

Gabriel Romon

He said once he had a disease

Nick

Oct 6 '18 at 23:32, by Ted Shifrin

Sadly, various people who've become mods (like Pedro and Daniel Fischer) don't have time for us anymore.

@GabrielRomon I hope he's alright.

Jul 12 '17 at 16:04, by Just win baby

We lost Daniel Fischer :(

He's probably still very much alive, but we seem to have misplaced him somewhere.

Jul 12 '17 at 16:08, by robjohn

@Justwinbaby where did you last put him?

@38554 Where are you? The people here miss you, Dan.

Abdullah UYU

11:49

Hi chat!

I'm trying to write a condition for a set that I called $\overline{B}$. $B$ is a set that contains bijections. And I want that $\overline{B}$ contains only the bijections that cannot be generated with their compositions. So I write the condition as follows:
$\forall (f,g,h)\in B^3, f\circ g\neq h$.

Do you think it sufficies what I said?

Tobias Kildetoft

@AbdullahUYU That gives you the empty set if you want $B$ to contain all bijections from some set to itself.

Abdullah UYU

Let me elaborate a little bit. $B$ is a set of bijections between some numerically equivalent sets.

Tobias Kildetoft

if the sets are not identical, then these bijections cannot be composed.

Abdullah UYU

Let's say $S=\{X,Y,Z\}$. Then $f\circ g=h$ where $g:X\to Y, f:Y\to Z, h:X\to Z$.

@TobiasKildetoft So, I don't think that's true.

Tobias Kildetoft

@AbdullahUYU I though you said $B$ was a set of bijections

Abdullah UYU

12:04

@TobiasKildetoft Pardon, that was a typo.

Tobias Kildetoft

Which part?

Abdullah UYU

B has changed to S.

So $S$ is the set that contains the numerically equivalent sets.

Tobias Kildetoft

Ahh, I thought you wanted all the functions to have the same domain and codomain

Abdullah UYU

I had to explicitly say it though. Anyways, I think it's clear now.

Tobias Kildetoft

But your $\overline{B}$ will still probably not be uniquely determined by not containing any compositions (even if you require it to be maximal)

Nick

12:07

How is a dist. variance $E(X^2)-(E(X))^2$ ... it looks like it's going to be negative most of the time.

Abdullah UYU

And there is also a typo in what I initially wrote, it should be $\forall (f,g,h)\in \overline{B}^3, f\circ g\neq h$

We're trying to write the condition for $\overline{B}$, afterall.

Tobias Kildetoft

@AbdullahUYU What does that have to do with the previous stuff?

Abdullah UYU

@TobiasKildetoft I think these $n-1$ bijections have to satisfy the condition I wrote.

They cannot be expressed as their compositions.

Tobias Kildetoft

Sure, but there is no point in writing up the condition. Just arrange the sets along a line

Or in fact in any tree.

Abdullah UYU

Yesterday, I studied a little bit on it, tried to come up with some examples.

I don't know of graph theory, by the way.

Why there is no point in writing up the condition? It's the only formal thing I wrote for advancing in writing the proof. @TobiasKildetoft

Tobias Kildetoft

12:22

What does writing up that condition have to do with the proof? It is just a condition that happens to hold for the set, not an important part of showing that the set exists

Abdullah UYU

Isn't that what we're proving existence of all bijections can be shown from these bijections? And to show the others' existence, we'll use the composition. So don't we have to construct it such that it satisfies that condition?

Other than that condition, what explicitly specifies the set, accordingly its existence?

Tobias Kildetoft

We are proving that we can get a path of bijections between any pair of the sets. We could do that as well with more bijections, in which case that condition would fail. So the condition is not what you are looking for.

Just construct the set of bijections explicitly.

Abdullah UYU

The point is making this by $n-1$ bijections, I think. And I speculate that it's the minimum. So we cannot waste it by taking a bijection that can be made up with the present ones.

So the set we're trying to construct will indeed contain $n-1$ bijections.

Tobias Kildetoft

Sure, but just arrange the sets in a line as I already told you. Then do induction.

Abdullah UYU

12:39

OK, I think I got the strategie.

But, can induction be made on finite sets? What if $S$ is finite? @TobiasKildetoft

Or, I didn't understand on what we're trying to do induction.

Nevertheless, I don't think induction would work there @TobiasKildetoft

If I'm understanding you correctly. Existence of the bijection $f:S_{n-1}\to S_n$ doesn't imply the existence of the bijection $g:S_n\to S_{n+1}$.

(S_k is the kth element of S, if that's what you want to say by arranging sets on a line.)

Tobias Kildetoft

13:14

You have $n$ sets. Pick one. The remaining $n-1$ sets have such a collection of bijections. Pick a bijection from the chosen set to any of them and add that. Now you have your set by induction.

Abdullah UYU

13:31

Whoa

I see, but the mean in the "by means of" in my mind was very different. But yes, this shows by induction that there can be found $n-1$ bijections that entrain the fact that the $n$ sets in $S$ is equipotant.

Where is arrangement in that proof by the way? @TobiasKildetoft

And as I side note, I realised that the condition I wrote is sufficient but not necessary.

Tobias Kildetoft

@AbdullahUYU You don't actually have to arrange them explicitly for this.

Abdullah UYU

Let me show you the rest of the exercise.

I can't really understand what is meant in (b).

I retract my words about the condition. It is indeed, sufficient and necessary.

Akiva Weinberger

15:22

Graph $\operatorname{sign}(x)\sqrt{\ln(x^2+1)}$. It's almost like a signum function

but it grows to infinity (really slowly)

*sigmoid

Abdullah UYU

16:07

@TobiasKildetoft I succeeded to prove it in my own way.

Abdullah UYU

16:25

@AkivaWeinberger Which is slower, $\ln(\ln(x))$ or the one you wrote?

Akiva Weinberger

$\sqrt x$ is a lot faster than $\ln(x)$ so $\sqrt{\ln(x)}$ will be faster than $\ln(\ln(x))$

and note that $\sqrt{\ln(x^2+1)}\approx\sqrt{\ln(x^2)}=\sqrt{2\ln x}=\sqrt2\sqrt{\ln x}$

Abdullah UYU

Yes, that seems reasonable.

Akiva Weinberger

I messed those up, give me a moment

$\sqrt{\ln(x^2+1)}$ doesn't exceed 10 until around $x=5\cdot10^{21}$, so it's still really slow

but $\ln(\ln(x))$ doesn't exceed 10 until around $x=9\cdot10^{9565}$

In general, $e^x$ has around $\frac37x$ digits, by the way

Fun fact

$e^7=1096.63316\approx10^3$

Rithaniel

16:45

Hmmm, that's useful

I never really know how to approximate exponentials of irrational numbers.

So, the $\frac{3}{7}$ is just rounding $ln(10)$ to a very roughly approximate rational number. Did you arrive at that through experimentation or was there some rule you used to find an appropriate fraction?

Akiva Weinberger

Continued fractions @Rithaniel

7/3 = 2+1/3

ln(10) = 2+1/(3+1/(3+1/(...)))

Wolfram Alpha does it if you ask it for "ln(10) continued fraction"

Rithaniel

17:03

Alright, and the precision is just how deep down the tree of fractions you want to go?

Very useful

Akiva Weinberger

Relatedly, you know how it's an open problem whether or not $\gamma$ is rational?

You can use continued fractions to get a lower bound on what its denominator would be if it were rational

and I forget exactly what it is but it's something massive

so it most likely isn't rational

Rithaniel

That's a number associated with a particular infinite sum, correct?

I might be confusing it with another constant, actually

Akiva Weinberger

It's $\displaystyle\lim_{n\to\infty} \left(\sum_{k=1}^n\frac1k\right)-\ln n$

Basically, $H_n:=1+\dotsb+\frac1n$ is almost $\ln n$ as $n$ gets large, and the difference is around 0.577

Rithaniel

Ah, yes, this is in fact what I was thinking about

Akiva Weinberger

You can also see it as the area of the blue region (between the step function and the curve)

in that image

because the step function gives you $H_n$ and the curve gives you $\ln n$

If you translate all the pieces to the left you see that it's less than 1

Rithaniel

17:12

Hmmm, a little bit of disconnect somewhere. $ln(n)$ should be sloping upward as $n$ increases, right?

Semiclassical

hi chat

Rithaniel

Sup Semiclassical

Semiclassical

here's a question I don't know how to answer.

consider the set of 4-by-4 symmetric matrices with unit diagonal elements. that set is parametrized by the six elements above the diagonal.

If I restrict to the subset of such matrices which are positive semidefinite, I get what's known as the 4-elliptope. Furthermore, I'm only interested in the subset of such matrices that are singular.

That amounts to the principal minors all being nonnegative and the determinant in particular being zero.

What I want to know is the hypervolume of that singular set, understood as a region in 6D space

No idea where to start. I suppose what I want to do is parametrize that set but...not so obvious how

Rithaniel

You've used many terms I'm not familiar with, so I imagine that I won't be much help.

Semiclassical

heh

Rithaniel

17:21

Though, I could act as a rubber duck, to whom you explain the issue and thus gain some realization through the act of talking about it.

Semiclassical

The principal minors are what you get when you deleting rows and corresponding columns from a matrix and then take the determinant

If you don't delete any rows/columns, then you just get the determinant itself

If you delete all but one row and its corresponding column, you're just left with the corresponding diagonal element

The reason this matters is that there's a result due to Sylvester: a matrix is positive semidefinite---i.e. has nonnegative eigenvalues---iff all principal minors are nonnegative

Rithaniel

The term "semidefinite" makes me wonder what "definite" refers to.

Semiclassical

that's when all eigenvalues are strictly positive

it's positive definite in that case because $x^\top A x >0$ for all $x\neq 0$ when $A$ has positive eigenvalues

Rithaniel

Ah, gotcha, and "negative definite" would be all negative eigenvalues?

Semiclassical

yep

Rithaniel

17:28

and "negative semidefinite" is the same as "positive semidefinite?"

Semiclassical

no. negative semidefinite is "nonpositive eigenvalues"

if it's positive semidefinite, then you're forbidding there to be any negative eigenvalues. in my case, I specifically want the set where all the eigenvalues are nonnegative but at least one is zero, i.e. I'm only interested in those matrices which are PSD but not PD

Rithaniel

Ah, so a positive semidefinite matrix could also be positive definite, but a negative semidefinite matrix could not?

Semiclassical

Right. All PD matrices are PSD, and no PD matrix is also ND.

Rithaniel

So, you could say PD = "not NSD" and ND = "not PSD."

I've now associated these concepts with logical AND and OR, respectively.

Semiclassical

If it's "not NSD", that means that some eigenvalues are positive. That doesn't mean all of them are

You can have a matrix which has both positive and negative eigenvalues. That won't be PD, ND, PSD, or NSD

Rithaniel

17:33

Okay, then I've misunderstood something.

Semiclassical

Think of the case of 2-by-2 matrices with real eigenvalues.

If both eigenvalues $\lambda_1,\lambda_2$ are positive, that's a PD matrix. In the plane of $(\lambda_1,\lambda_2)$, that's the first quadrant with the axes not included

Rithaniel

Ah, is the distinction between PD and PSD the distinction between strictly positive and "just positive?"

Semiclassical

If we weaken that to both being nonnegative, that's a PSD matrix. In the plane, that's just the first quadrant with its boundary included.

Well, in English anyways, strictly positive means the same thing as positive i.e. $x>0$

nonnegative means $x\geq 0$ i.e. $x$ is not $<0$

so it's the distinction between "all eigenvalues are positive" and "no eigenvalues are negative".

Rithaniel

Okay, so PSD can include eigenvalues of 0, but PD cannot.

Semiclassical

yeah

With ND/NSD, you just flip the signs so that you're looking at the third quadrant in $(\lambda_1,\lambda_2)$

but of course this only addresses the first and third quadrants. it says nothing about the second and tfourth

Rithaniel

17:38

Okay, I understand now. Didn't mean to distract from your explanation of your initial problem.

Semiclassical

so if I have an eigenvalue pair in the second/fourth quadrant, then the matrix won't be definite period

I mean, the problem at the end of the day is that I've got a certain set in 6D space, as defined by a set of polynomial inequalities and one polynomial constraint

and I want to know the measure of that set

typically if one wants the volume of a set, you parametrize it and perform the integral that way

So the challenge is to find a useful parametrization

Akiva Weinberger

@Rithaniel Oh, sorry, the curve is $y=1/x$, but the area under that curve between $x=1$ and $x=n$ is $\ln n$

Semiclassical

I have something like a parametrization, but it's a many-to-one mapping from one space to my set of interest

Akiva Weinberger

(Exercise: see why the area under $y=1/x$ between $x=1$ and $x=2$ is the same as the area under that curve between $x=2$ and $x=4$, without appealing to the formula I just gave)

Rithaniel

@AkivaWeinberger Ah, that clarifies that. It's just the integral of $\frac{1}{x}$ (Didn't think about that)

Semiclassical

17:46

That may yet be enough, but I haven't figured it out yet

Akiva Weinberger

Right yeah

but think about the exercise without appealing to integrals @Rithaniel

Rithaniel

Alright, that's an interesting brainteaser.

Well, just looking at rectangles, we have an area bounded by $1$ above and $\frac{1}{2}$ below between 1 and 2, and an area bounded by $2(\frac{1}{2})=1$ above and $2(\frac{1}{4})=\frac{1}{2}$ below between 2 and 4.

I suppose there is an argument to be made that the area under the curve from $n$ to $2n$ is also equal to the area under the curve from $1$ to $2$, right?

Alright, yeah, because the area involves some measure of horizontal length multiplied by height, which is given by $\frac{1}{n}$, and the length from $n$ to $2n$ is $n$, so, we end up with $n(\frac{1}{n})=1$ and $n(\frac{1}{2n})=\frac{1}{2}$, which is really just the area under the curve from 1 to 2.

Ted Shifrin

18:26

@AkivaWeinberger Just assigned that to my calculus students.

@Rithaniel: You really do need to think about Riemann sums, though, since the function is non-constant.

Rithaniel

So, the bit that I haven't fleshed out is writing it out as the limit of the sum of areas of rectangles as the rectangles become thinner and thinner and the number of rectangles goes towards infinity, and then showing that this limit is invariant of $n$ if the region is from $n$ to $2n$, correct?

Ted Shifrin

You don't need to worry about the limit, actually, @Rithaniel. Think about how one particular sum of areas of rectangles corresponds to something analogous for the integral from 1 to 2.

Akiva Weinberger

Here's a picture of our situation

Think of it in terms of transformations of the plane

Ted Shifrin

@Semiclassic: Of course, that's not what I call a region. It's a compact 5-dimensional hypersurface.

Akiva Weinberger

With a hypersurface hyperarea?

Ted Shifrin

18:38

Yup.

Akiva Weinberger

Hyperarright.

Ted Shifrin

The induced "hyperarea" form on a hypersurface in $\Bbb R^n$ is the given by the $(n-1)$-form $\omega = \sum (-1)^{i-1}a_i dx_1\wedge\dots\wedge\widehat{dx_i}\wedge\dots\wedge dx_n$, where $a$ is the unit normal.

Rithaniel

Well, I believe I can visualize what's happening. As you shrink/grow the interval and increase/decrease the height of each point along that interval, it's equivalent to moving the interval to other regions along the curve.

However, direct numeric justifications is another issue.

Ted Shifrin

What's important, @Rithaniel, is the precise way in which the growth of one is related to the shrinking of the other.

Akiva Weinberger

I think you got it; stretching the plane by a factor of n (either horizontally or vertically) multiplies areas by n, and the inverse transformation of squishing by a factor of n divides areas by n

Rithaniel

18:43

Well it's $\frac{1}{n}$ versus a width of $n$, so they're direct reciprocals of each other.

Akiva Weinberger

You can get from the blue area to the orange area by stretching horizontally by a factor of 2 and squishing vertically by a factor of 2

So like the point $(x,y)$ maps to the point $(2x,y/2)$

Rithaniel

Ah, excellent, I thought you were looking for an equation to justify it. :P

Akiva Weinberger

And points on the curve $y=1/x$ stay on that curve under that transformation, because if $y=\frac1x$ then $y/2=\frac1{2x}$.

Rithaniel

Okay, I can see how you can make a point-by-point comparison, too.

Akiva Weinberger

It's like you're "rotating" the hyperbola around

Ted Shifrin

18:44

I was thinking about individual rectangles for a Riemann sum. The base stretches by $n$, and the height shrinks by $1/n$.

Akiva Weinberger

Right

Rithaniel

Yeah, that's what I was trying to argue, but I thought I would need to write out the actual equation to make the argument as clear as I could make it.

Akiva Weinberger

Just like a rotation preserves areas and turns points on a circle into other points on a circle, this stretch-squeeze transformation preserves areas and transforms points on the hyperbola $y=1/x$ into other points on the hyperbola

@Rithaniel Well, to make it rigorous, you would say that the stretch-squeeze transformation preserves areas of rectangles, and that these shapes can be approximated arbitrarily well by collections of rectangles

Rithaniel

Ah, I like that.

Akiva Weinberger

Or actually, you could take $\int_1^2\frac1xdx$, substitute $2x=u$, and get $\int_2^4\frac1{u/2}\frac12 du=\int_2^4\frac1udu$

That's essentially the same proof!

Ted Shifrin

18:49

No, we didn't want to do that. We were being geometric with the definition of the integral :)

Akiva Weinberger

It is geometric - the $\frac12du$ says that stretching horizontally multiplies areas by $2$

That is, $du=2dx$

Ted Shifrin

Yeah, yeah, I know. I wanted upper/lower or Riemann sums.

You can also prove it by the First Fund. Thm. of Calculus (that was an exercise I gave my kidlets a few weeks ago).

Rithaniel

Also, unrelated question, but is this abuse of notation? I want to denote that I will be referring to points in this set with just the symbol "$p$" -- $\{p=(a,A)\in X\times P(X)|a\in A\}$

Ted Shifrin

Prove that $\int_1^a dt/t = \int_b^{ab} dt/t$ by considering the function of $b$.

Akiva Weinberger

$\sum_{x=1,1+\Delta x,\dots,2}\frac1x\Delta x$

Ted Shifrin

18:51

It holds for any partition, any Riemann sum, DogAteMy.

Akiva Weinberger

Right

Ted Shifrin

@Rithaniel: This is totally unrelated to what we're doing? What you're talking about is called an incidence correspondence.

Akiva Weinberger

$\sum_n\frac1{x_n}(x_{n+1}-x_n)$, $x_0=1$, $x_N=2$?

And then let $u_n=2x_n$

Ted Shifrin

DogAteMy: More generally, $1/c_n$, where $x_n\le c_n\le x_{n+1}$.

Rithaniel

Really? This is something that's been studied?

Akiva Weinberger

18:53

OK right sure fine but you let $u=2x$ and the same stuff happens

Rithaniel

Incidence correspondence is something I want to look up, now.

Ted Shifrin

A classic example would be this, @Rithaniel: Let $X$ be the space of lines $\ell$ in the plane. Consider $\{(x,\ell): x\in\ell\}\subset\Bbb R^2\times X$.

Akiva Weinberger

You're basically looking at pointed subsets of $X$, yeah?

Ted Shifrin

I changed notation from yours, sorry.

Akiva Weinberger

Like, subsets of $X$ with one distinguished point

Ted Shifrin

18:55

You can look at generalizations of this, called flag spaces.

This is super important in algebraic geometry and related places.

Akiva Weinberger

That would be if your subsets are specifically sub vector spaces

(lines and planes etc)

Ted Shifrin

Or affine spaces.

Akiva Weinberger

Right

Daminark

Hey guys!

Ted Shifrin

Heya Demonark.

Happy back to school.

Perturbative

19:01

Hey everyone!

Hey @TedShifrin and @Daminark :)

Daminark

Yup, it's looking pretty fun. Think I've decided on 3 classes, algorithms, AG, and difftop. If I take a 4th it'd prob be a non-math/CS class but I might take it a bit easy after last quarter

Ted Shifrin

Hi Perturb.

For sure, non-math/CS, Demonark.

Akiva Weinberger

My favorite sorting algorithm is the one that sets a timer for each number in the list

Ted Shifrin

BBIAB

Akiva Weinberger

for the number's number of seconds

and then they get returned to you in the right order

Perturbative

19:03

@TedShifrin I have a basic differential geometry question if you have some time to look at : math.stackexchange.com/questions/3066977/…

Akiva Weinberger

"Sleep sort"

Semiclassical

@TedShifrin yeah, "region" is a careless word there

The annoying thing is that, if I focus on rank 3 psd matrices (which is all I think I need, since the rank<3 stuff is the boundary of the rank 3 stuff?) then I can characterize any matrix of interest as of the form $M=LL^\top$ where $L$ is 3-by-4 with unit columns

Ted Shifrin

@Perturb: Think of this picture. Draw on your manifold the curves you get by fixing $x^2,\dots,x^n$ and varying $x^1$. You do this, in turn, for each $x^i$. This gives you an $n$-dimensional grid. (Start with surfaces. You'll see plenty of these pictures in my diff geo notes.) $\partial/\partial x^i$ is the tangent vector (of appropriate length, of course) to the $x^i$-curve.

@Semiclassic: Boundary in a topological sense, but not in the manifold sense. You have a stratification by rank. @Balarka likes these things :)

Semiclassical

yeah, that sounds right

Ted Shifrin

@Perturb: You have to understand that what I mean by all that is $\phi^{-1}$ of the grid in $\Bbb R^n$.

Perturbative

19:14

@TedShifrin I'll do that now

Semiclassical

Anyways. In that sense, I have a 'parametrization' (that's the wrong word but I'm being silly) of all possible L by 4-tuples of points on the unit 3-sphere

but...that's not a one-to-one map, so it's not actually a parametrization

there's lots of different L which give the same M=LL^T

Ted Shifrin

Yeah, Semiclassic, you want a local unique way of doing it.

Semiclassical

Right.

Érico Melo Silva

sup

Ted Shifrin

Heya Eric.

Semiclassical

19:18

I think I can assume WLOG that the first point is just the north pole

Érico Melo Silva

how go things

Daminark

Yo Eric

Érico Melo Silva

@Daminark sup nerd

Semiclassical

hmm, maybe not

Ted Shifrin

If you do the canonical $LDL^\top$ decomposition of a symmetric matrix by the usual algorithm, @Semiclassic, don't you get a unique answer since your $D$ is the identity?

Daminark

19:19

Not too much, waiting to get some forms signed to take AG and difftop

You?

Ted Shifrin

Oh, my $L$ is lower-triangular with $1$s on the diagonal.

Maybe $D$ will have other things on the diagonal.

Semiclassical

yeah, they in general will

Daminark

Also finally my last recommender finished all his schools lmao

Ted Shifrin

Better late than never, Demonark.

Érico Melo Silva

@Daminark in civ and bored

Daminark

19:20

Rip

Semiclassical

I mean, they should all be nonnegative since my matrix is psd

Érico Melo Silva

big yikes my guys all finished in november

Semiclassical

so sqrt(D) will always make sense

Érico Melo Silva

@Daminark who is teaching AG, nori?

Semiclassical

so I'll always be able to get a factorization into matrix * matrix transpose

Ted Shifrin

19:21

Right, the product of the diagonals is the determinant, of course, @Semiclassic.

Daminark

Most of them were done in December, but one guy hadn't done 3 schools which I hadn't finished the apps for until after finals, and another guy vanished until three days ago, so that was nerve wracking. But yeah now the waiting game. And yeah Nori

Ted Shifrin

Right, but the usual algorithm for this decomposition produces a unique answer (like completing the square by a certain algorithm).

Semiclassical

Is it unique when you have lower rank, though? Can't remember

Ted Shifrin

Eric, are you continuing with Sid? Not that you and I discussed stuff much at all ...

Érico Melo Silva

i ended up not being able to bc of a scheduling quagmire im in trying to finish my degree

Semiclassical

19:23

But this does seem like the right idea

Ted Shifrin

Sure, @Semiclassic. I mean, you follow an algorithm.

Semiclassical

Hmm, true

Ted Shifrin

Oh ugh, @Eric. For independent study I figured you wouldn't have conflicts :P

Semiclassical

Not a unique factorization, but a unique result of the factorization algorithm

Érico Melo Silva

so no math classes at all this quarter

@TedShifrin we tried to make it work but we just couldnt find a space to meet regularly :.

Ted Shifrin

19:24

Rats.

Érico Melo Silva

it's highly unfortunate

Ted Shifrin

Well, I'm still happy to chat about G/H if you forge onward.

Érico Melo Silva

im still gonna try to go on ahead yeah, now that i got in somewhere im so much less stressed and worried

Ted Shifrin

I was never worried about you, but I understand.

Érico Melo Silva

honestly it's just the confirmation of "grats u have a future kiddo"

Buk Lau

19:26

Hello, I have a question... does anyone know where I can find good "exam level" questions for calculus? like, questions about limits and derivatives.. continuity of functions, sequences.. (first semester of maths degree)

Semiclassical

@ted I think, at the very least, that that way of thinking justifies taking the first column of L to be (1,0,0)^T

Ted Shifrin

Calculus exams are all over the internet, @BukLau.

Semiclassical

and I think you can also use it to justify taking the second column to be $(\cos\theta,\sin\theta,0)^T$

Érico Melo Silva

maybe try schaum's outlines or something those books have billionso f problems

Ted Shifrin

@Semiclassic: If you think of it as $LDL^\top$ and adjust later, with $L$ having $1$'s on the diagonal, I think the algorithm is clearer.

Buk Lau

19:27

@TedShifrin yes but I couldn't find some that are suitable for me :( they're too advanced

Semiclassical

Hm.

Ted Shifrin

Wait, I want $L$ to be lower-triangular, @Semiclassic.

Semiclassical

Lower?

Oh

Ted Shifrin

Yup, lower.

Semiclassical

Yeah.

got myself mixed up with L^T

Ted Shifrin

19:27

@BukLau: Only your teacher and you can know what's the right level for you. We certainly can't.

Buk Lau

True, I'll keep looking :P thank you @TedShifrin

Semiclassical

I should probably start with a simpler version of this problem

Which I guess would be 3-by-3 singular symmetric psd matrices with unit diagonal

Ted Shifrin

Maybe work out the $3\times 3$ case, Semiclassic?

Ha :)

Semiclassical

yeah

if I ignore the 'singular' part, then I know the answer

but I haven't done the calculation in that variation

Perturbative

@TedShifrin Okay so I'm vaguely starting to get the picture of what's going on, that visualization really helped. Then I'm sure that I was wrong in the question I asked on main since when taking the partial derivatives of the coordinate representation of smooth functions in local coordinates, the local coordinates need not be the standard coordinates on $\mathbb{R}^n$

Ted Shifrin

19:32

They are the standard coordinates on $\Bbb R^n$.

They may represent something else on a different $\Bbb R^n$.

Understand the polar coordinate chart. You use $r,\theta$ as coordinates on a suitable part of $\Bbb R^2$.

But $r$ and $\theta$ become the axes in that $\Bbb R^2$.

They of course look different on your manifold (which is regular $\Bbb R^2$).

Perturbative

Okay let me just think about that for a while

N.T.

@Semiclassical You probably have seen it already: mathoverflow.net/questions/225152/…

Semiclassical

huh, no i hadn't

Ted Shifrin

That's a cool post.

Robert Bryant (who knows everything, my good buddy) points out what I was pointing out less clearly: You can only get local parametrizations.

Semiclassical

hmm

Érico Melo Silva

19:46

what can't this man do

the legend

Semiclassical

that bodes ill

Ted Shifrin

LOL, I've been impressed by him since we were colleague grad students.

There was a time I knew stuff he didn't ... but that time ended decades ago.

Did you mention him in your app, Eric?

Érico Melo Silva

of course

Perturbative

@TedShifrin Okay so all local coordinate representations of smooth functions take place in standard coordinates on $\mathbb{R}^n$?

Ted Shifrin

Yes, @Perturb.

N.T.

19:49

Indeed, it does not look very good for rank-deficient matrices, which is also an inconvenience for covariance matrices. (But check out the linked "nice paper" in the question, it is very nice.)

Semiclassical

yeah, that looks a lot like the kind of thing I want to know about

Perturbative

In fact, are all local coordinates the standard coordinates on $\mathbb{R}^n$? So say I have a smooth manifold $M$ and a smooth chart $(U, \phi)$ on $M$, then does the local coordinate grid (i.e $\phi[U]$) look like the standard coordinates on $\mathbb{R}^n$ which would be just a square $n$-dimensional grid?

Ted Shifrin

Right, but what it looks like on $M$ depends on $\phi$, of course.

Semiclassical

glancing at that paper, I think what I'm after could be summed up like this. I have two optimization problems; one is combinatorial, the other is its convex relaxation.

I'm pretty sure I know how to find the hypervolume of the feasible set in the former case; I'd like to compare that with the same for the latter

And see how different they are

Perturbative

Ahh that clears up a lot! @TedShifrin

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$Mathematics$ Mathematics