Mathematics - 2020-12-12 (page 2 of 2)

8:02 PM

And we will still get to have the small version of the Christmas lunch in January. Obviously the full company one with 1500 people will not happen. But we will be 10 people going to a Michelin star restaurant.

Ted Shifrin

No restaurants for me until long after vaccines.

Tobias Kildetoft

Understandable

We had the restaurant planned for a short while and then they increased the restrictions which shut down all restaurants in most of the country

robjohn

@TedShifrin We have not eaten at a restaurant here since March. We've gotten take-out, but no dining, not even outdoor.

porridgemathematics

@TobiasKildetoft oh wait, maybe its not wrong? I think what I was trying to argue is that since $M$ does not contain any element of $H$, and $I = (M)$ elements of $I$ are of the form $q_1m_1 + ... + q_tm_t$ where $m_i$ are in $M$, but then if it contains a monomial of $H$, some $q_im_i$ contains a monomial of $H$, which must mean $m_i$ cannot be divisible by any $X_t$ where $t$ is not one of $i_1,...,i_k$,

so it is of the form $cX_{i_1}^{b_1}...X_{i_k}^{b_k}$ where some $b_m > a_m$, but then $m_i$ multiplied by any monomial of $q_i$ cannot be in $H$ since it is divisible by $X_{i_m}^{b_m}$ for $b_m > a_m$

does that convince you?

Ted Shifrin

Yup. I got takeout from my favorite Thai restaurant last night, so even though it's not as good as being there, I have yummy food for a few days.

BTW, was that evident parabola supposed to be a catenary?

robjohn

8:06 PM

@TedShifrin We got Indian take-out last night. Curry, and Rogan-Josh and Masala.

Ted Shifrin

Nice ;)

robjohn

@TedShifrin I commented that it was not a catenary, but got no response. I think it might have been a catenary for non-mathematicians.

Tobias Kildetoft

@porridgemathematics Yeah, that looks fine. But where do you really use that these are finitely generated?

Ted Shifrin

Oh, non-mathematicians have chains that hang like suspension bridges.

robjohn

@TedShifrin It was instead a purrabola.

@Ted: I introduced someone to envelopes this morning. You may have seen the posts.

Ted Shifrin

8:12 PM

Nope. Self-addressed, I hope.

porridgemathematics

@TobiasKildetoft ah, i guess i don't even use it ! its totally redundant by what i just said

for some reason i thought it was necessary because we recently learned about noetherian modules

Balarka Sen

Hi, all

porridgemathematics

thanks @TobiasKildetoft

Ted Shifrin

Hi, a @Balarka

Balarka Sen

@porridgemathematics Kindly fix your gravatar!

robjohn

8:16 PM

@TedShifrin: If you're interested, here is my initial response.

Balarka Sen

@TedShifrin Online exams start tomorrow.

robjohn

@BalarkaSen what is causing that?

porridgemathematics

oh is it showing up as a file for everyone?

Astyx

It's showing up as text for me

Ted Shifrin

@robjohn: I think I've answered a few questions about envelopes on main, but I don't usually go to the graphics work that you do!

Astyx

8:17 PM

And messing up the layout

Balarka Sen

I have no clue. It looks like it hasn't loaded, and instead the whole username appears

robjohn

@porridgemathematics: it looks like this

Balarka Sen

Wrecks the active participants display on the right side of the chat for me

porridgemathematics

sorry, i uploaded a picture in edit profile, is that what is needed to fix it?

Balarka Sen

Yes, working in main site now.

It'll update in chat.SE as well, in a bit

porridgemathematics

8:20 PM

sorry about thatr

Balarka Sen

Updated.

Thanks

All good now

robjohn

I refreshed their profile from math main

Now I see bowls of oatmeal ;-)

Balarka Sen

Ah, mystical mod powers.

robjohn

@BalarkaSen I've also been an owner of this room since before that.

Balarka Sen

That must be a long time back.

Tobias Kildetoft

8:24 PM

@BalarkaSen Behold the power of the ancient ones.

Balarka Sen

Hi @Tobias. Indeed

Ted Shifrin

<--- ancienter and powerless

robjohn

@BalarkaSen Over 8 years ago. I used to be as gabby as anyone here ;-p

Balarka Sen

@TedShifrin "Ancient and Powerless" sounds like it could be a good alternative to "Dazed and Confused"

Ted Shifrin

I'll wait for the script.

Balarka Sen

8:26 PM

@robjohn Ahh, I think I came here about 5 years ago

So that's a long time before I existed in SE

robjohn

@BalarkaSen: Does Allesandro's avatar mess up your 'active participants' bar, too?

Balarka Sen

Hmm, nope.

Alessandro Codenotti

What's wrong with my avatar?

Balarka Sen

I see the album cover

Alessandro Codenotti

@BalarkaSen I just searched "@Alessandro" in the chat, apparently I've been here at least 5 years. Also that search string gives 128 pages of results, maybe I spend too much time here

Tobias Kildetoft

8:28 PM

Not sure how long I have used the chat here. I was one of the fairly early users of the main site, coming from MO

robjohn

Oops, now it's back. I was getting the default name display

Balarka Sen

@Alessandro Haha. I think I saw you pretty early in here, but we didn't interact too much.

robjohn

@AlessandroCodenotti perhaps something got refreshed when you spoke.

Balarka Sen

You had a different gravatar then, a bunch of people staring down at the well.

Alessandro Codenotti

I had that one for a long time

Balarka Sen

8:29 PM

@TobiasKildetoft Yeah

I still frequently find MO comments of yours :P

Ted Shifrin

I retired over 5 1/2 years ago, and I think I remember most of you from before that.

Balarka Sen

Oh, ok.

Alright, I have been active in main site for 7 years.

Tobias Kildetoft

@BalarkaSen What are you doing browsing those parts of MO :)

Astyx

Apparently I arrived here in 2016

Ted Shifrin

Balarka 1.0 was super-into algebra.

Astyx

8:31 PM

Feels like a century ago

Ted Shifrin

Thanks a lot, Astyx.

I guess we all had crêpes in Paris in summer 2017.

Astyx

haha I didn't mean it like that

We did!

Balarka Sen

I remember everyone despised me when I was here in the beginning lol

Ted Shifrin

wonders wistfully if he'll get to go back to Paris before being dead

That's too strong a word, Balarka.

There are numerous people who despise me now :D

Balarka Sen

Lol

Astyx

8:34 PM

I remember you learning about the spectral theorem

Ted Shifrin

@Balarka @robjohn Have you seen anything like this before?

Balarka Sen

I learnt linear algebra after I learnt algebraic topology

Ted Shifrin

You learned basic analysis/calculus after algebraic topology, as well.

Balarka Sen

I would learn the definition of an inner product and go "Ah, so this is like the intersection pairing, got it".

Ted Shifrin

So you knew how to reduce modules to standard form before you knew basic linear maps. :D

Balarka Sen

8:35 PM

No way haha, it was all kind of an intuitive mess.

I learnt JNF and primary decomposition way later

robjohn

@TedShifrin I assume that $S$ is the area of the triangle?

Ted Shifrin

yes

Balarka Sen

I have not seen this before.

robjohn

@TedShifrin I've computed the areas of spherical triangles in a couple of different forms, but I haven't seen that one in particular before.

Balarka Sen

I am having a hard time. Curvature is constant here?

Ted Shifrin

8:38 PM

Yeah, this is not the usual spherical/hyperbolic law of cosines at all.

Balarka Sen

Yeah.

Ted Shifrin

Yes, assume constant curvature (because "infinitesimal" triangle).

Balarka Sen

If it's true it must fall out of understanding what $\cos_K(c) = \sqrt{|K|} [\cos_K(a) \cos_K(b) + \text{sgn}(K) \sin_K(a)\sin_K(b)\cos(C)]$ means upto first order.

Ted Shifrin

or maybe third order ...

Balarka Sen

Let's assume $K > 0$. Then $\cos_K(x) = \cos(x\sqrt{K})/\sqrt{K}$ is like $1/\sqrt{K}(1 - K x^2/2 + O(x^3))$.

Ted Shifrin

8:43 PM

What's strange, though, is that with $K<0$ we expect hyperbolic trig.

Balarka Sen

Right.

Ted Shifrin

I suspect that your $\cos_K$ should be $\cosh$ when $K<0$.

Balarka Sen

Ya, that's all

Ted Shifrin

But does this make it look more like what I expect?

porridgemathematics

why does the hilbert basis theorem imply that a non-noetherian subring of a noetherian ring must be finitely generated?

*infinitely generated

Ted Shifrin

8:45 PM

It would be noetherian if it were finitely-generated?

Balarka Sen

I am thinking maybe the second order term gives some insight.

Tobias Kildetoft

@porridgemathematics infinitely generated as what?

Ted Shifrin

As I posted in my comment, @Balarka, what's strange is that the $K$ term shows up in the cubic term of the metric in geodesic normal coordinates.

porridgemathematics

@TedShifrin is that finitely generated over R ? as in like R[a_1,...,a_n]?

Ted Shifrin

Better have this conversation with a serious algebraist, like Tobias, @porridge. :)

Tobias Kildetoft

8:47 PM

@porridgemathematics A subring does not have a structure that we usually care about being finitely generated

porridgemathematics

@TobiasKildetoft that is my issue, this was the ambiguous wording in my notes, if it means finitely generated like as in the smallest subring of R containing a finite set, then it seems false

Ted Shifrin

@Balarka, Having posted that comment, I thought about writing down the length integral and the area integral and didn't see how anything would remotely give me the "classical" law of cosines.

Tobias Kildetoft

@porridgemathematics As I said, there is not really any useful notion of a ring being finitely generated on its own

Balarka Sen

@TedShifrin Ah OK yes I remember that.

This is because Ricci is cubic deviations of the geod. ball from Euclidean ball.

porridgemathematics

my lecturer is a ring theorist and the wording of this hint to a question was 'the hilbert basis theorem shows that such a subring must be generated by infinitely many elements', the question is given an example of a non noetherian subring of $\mathbb{C}[X,Y]$

thats why i asked

Tobias Kildetoft

8:49 PM

@porridgemathematics I think he means as a complex vector space

Ted Shifrin

Yeah, basically that's right, @Balarka.

Tobias Kildetoft

Sorry, that is not strong enough clearly

Ted Shifrin

you mean volume?

Balarka Sen

Yeah, volume.

Tobias Kildetoft

He probably means as a $\mathbb{C}$-algebra

Thorgott

8:51 PM

well, it can't be f.g. as a C-algebra by Hilbert, but that has little to do with C[X,Y]

Balarka Sen

@BalarkaSen Computing without writing anything, but in here, second order term LHS is $-1/\sqrt{K} \cdot K c^2/2$. Second order term RHS is $\sqrt{K} \cdot (1/K \cdot -K a^2/2 + 1/K \cdot -K b^2/2 + 1/K \cdot K ab \cos(C))$. And that doesn't help because it gives me $c^2 = a^2 + b^2 - 2ab \cos(C)$, the Euclidean stuff.

So it is indeed third order contributions that we must look for @Ted

porridgemathematics

must every subring be a $\mathbb{C}$ algebra?

Balarka Sen

Your hint makes sense now.

Ted Shifrin

LOL, but I don't see how to throw it into a blender and get anything resembling that to come out.

I can approximate by a straight-sided triangle in the tangent plane, but even if I do the integral for length $c$ and integral for area $S$, it's a total mess.

Taylor polynomials aren't the right approach for the trig, because these are definitely angles well away from $0$.

Thorgott

not necessarily

at least not in a natural manner

porridgemathematics

8:55 PM

hmm

Balarka Sen

Yeah I am not sure how to get what he wants.

porridgemathematics

perhaps he means it cannot be equal to $(\mathbb{C}[X,Y])[a_1,...,a_n]$ or else as a $\mathbb{C}[X,Y]$ module it is finitely generated and so noetherian?

Ted Shifrin

Should we even believe it? Does it work on a sphere? I suppose I should try that.

porridgemathematics

but then again not all subrings are of this form surely

Thorgott

for example, R[X,Y] is a subring of C[X,Y], but not a C-algebra

Balarka Sen

8:58 PM

I mean I guess I am a little weirded out because what does this question mean anyway? What does it mean for infinitisimal quantities to be equal? He demands a very small triangle, so all $a, b, c, S$ are small quantities, but he doesn't specify any error.

Surely it doesn't hold on the nose for constant curvature surfaces?

Seems totally suspect.

robjohn

@TedShifrin It seems to be saying that the spherical excess in this particular usage, is divided up equally among the three angles.

@TedShifrin I'm looking at the sphere.

Ted Shifrin

That was the first comment, @robjohn, which struck me as nonsense. I mean, fix the vertex $C$ and let the sides $a$ and $b$ grow, keeping angle $C$ fixed.

I worked out the classical versions of the spherical law of cosines in my algebra book, so I have those, at least :P I wonder what Taylorizing those will give.

Thorgott

I'm really not show what to say about subrings here

Balarka Sen

That's what I tried above. Second order gives nothing, just the Euclidean version.

Thorgott

the point should be that f.g. algebras over a noetherian ring are noetherian

Ted Shifrin

9:02 PM

You had $\cos A = \dfrac{\cos a-\cos b\cos c}{\sin b\sin c}$?

Balarka Sen

Yeah

Ted Shifrin

Ah.

Balarka Sen

For $K = 1$ that's what my formula reads out

Ted Shifrin

I guess we'd need 4th order for numerator and 3rd order for denominator, which suggests an error coming into $A$.

porridgemathematics

@Thorgott i agree, ill use that as the hint instead, thanks

Balarka Sen

9:04 PM

Yeah, probably. But how in the world does that relate to area?

Seems like esoteric trigonometry

Ted Shifrin

Well, via Gauss-Bonnet or lunes, of course.

Except we're ignoring all angles but one here.

Yeah, totally bizarre.

Balarka Sen

Right

Ted Shifrin

My geodesic normal coordinates approach seemed logical until I thought about it and then it seemed hopeless.

Balarka Sen

That has to be the origin of the appearance of $K/3$

But that's a numerological reason and not a mathematical one :P

Ted Shifrin

Well, also the $6$ in the sin expansion :P

But, yes, that's why I thought of that comment immediately.

Balarka Sen

9:08 PM

Hahah. Touch\'e.

Ted Shifrin

I will play with this more after lunch. I didn't intend to get sucked in ... but ...

Balarka Sen

Let me know if you get something! I will get back to preparing for my finals...

Ted Shifrin

Yes, sorry to have bamboozled you on it.

love_sodam

If $M$ is R-module, then $aM = 0$ for some $a\in R$ implies $M/aM$ is $R/aR$-module?

Oh change the question. $M$ is $R/aR$-module?

Astyx

what is $aM$ in this case ?

love_sodam

9:15 PM

{am:m\in M}

Astyx

I mean explicitely

love_sodam

like how?

Thorgott

you can answer this question

Astyx

Sry i'm tired, you already said aM is zero

I meant what is M/aM

love_sodam

M?

Astyx

9:17 PM

right

love_sodam

hmm. that was really stupid question

traducerad

what the fck

Thorgott

the general fact is that $R/I$-modules are in natural bijection with $R$-modules that get annihilated by $I$

traducerad

so apparently some guy became a multi-millionaire just by saying "guys, your estimations are incorrect. Your sample may not be representative of the entire population. Let s take multiple samples and average those samples instead"

He even pattented this

Resampled efficient frontier

In investment portfolio construction, an investor or analyst is faced with determining which asset classes, such as domestic fixed income, domestic equity, foreign fixed income, and foreign equity, to invest in and what proportion of the total portfolio should be of each asset class. Harry Markowitz (1959) first described a method for constructing a portfolio with optimal risk/return characteristics. His portfolio optimization method finds the minimum risk portfolio with a given expected return. Because the Markowitz or Mean-Variance Efficient Portfolio is calculated from the sample mean an...

copper.hat

10:30 PM

@traducerad many things are obvious in retrospective.

robjohn

@TedShifrin I believe it's true! I will finalize my computations.

traducerad

10:47 PM

@copper.hat yes but this is extreme. It s as if one guy invents a fridge a second guy invents the wheel and then a third guy invests fridges on wheels and pattents this

I absolutely hate it when things look easy/obvious in retrospective and I was too stupid to not see/notice it.

Ted Shifrin

11:00 PM

@robjohn Wow! Can't wait to see!

robjohn

@TedShifrin just posted

Ted Shifrin

11:15 PM

@robjohn: Ah, I wasn't persistent enough with the higher-order terms. But why is the Euclidean formula $\Delta = \frac12 ab \sin C$ valid?

robjohn

@TedShifrin for a small triangle, it is within the given errors

Ted Shifrin

Hmm ... I need to think about that. Very clever, though — I couldn't see how to get the sin term in there for the addition formula for cos.

robjohn

I changed the '$=$' in that step to a '$\approx$'

Ted Shifrin

You could say (9) is the addition formula for cos instead of using more differential calculus :P

Nice solution, @robjohn. I guess the moral of the story is that "infinitesimally" there is no Gauss-Bonnet because geodesic triangles are flat triangles.

schn

11:32 PM

Maybe a quite simple question, but how many planes are there in a Euclidean space, as a function of the number of dimensions $n\geq2$, given one adds up the planes from the previous dimension?

Thorgott

what does that last clause mean

schn

Maybe I'm confusing myself. I need to double check how many planes there as a function of the dimension $n$.

Ted Shifrin

What does "how many planes" mean? Infinitely many always?

Or are you talking about dimensions of Grassmannians?

schn

I mean coordinate planes.

Like $xy$-plane.

Ted Shifrin

Well, you certainly need to say so. And always $2$-dimensional coordinate planes?