Mathematics - 2024-01-28

Jakobian

00:12

I'm trying to get better at linear algebra so I'm reading "Second course in linear algebra" by Brown

good choice?

I know I absolutely suck when it comes to linear algebra

leslie townes

00:39

every morning i get up at 4am and diagonalize two symmetric 10x10 matrices by hand #grindset

2

Jakobian

I guess its like sudoku for mathematician

I love how he's trying to put commutative diagrams everywhere

they actually do help

it explains why in the change of base formula there's a transpose

Ted Shifrin

01:04

@leslietownes Better than yoga

No there isn’t, unless you’re doing bilinear forms.

Jakobian

what do you mean, yes there is

Change of basis

In mathematics, an ordered basis of a vector space of finite dimension n allows representing uniquely any element of the vector space by a coordinate vector, which is a sequence of n scalars called coordinates. If two different bases are considered, the coordinate vector that represents a vector v on one basis is, in general, different from the coordinate vector that represents v on the other basis. A change of basis consists of converting every assertion expressed in terms of coordinates relative to one basis into an assertion expressed in terms of coordinates relative to the other basis.Such...

when you write $w_j$ in terms of $v_i$, the matrix is basically transposed in the definition

Ted Shifrin

Nope.

Unless you’re talking about the duality of coordinates vs basis vectors.

Jakobian

I'm talking about the matrix $A = (a_{ij})$ where $w_j = \sum_i a_{ij}v_i$

The matrix $A^t$ is the matrix of this system, we are taking transpose and call it the matrix of change of basis

maybe I'm just wrong

Soumik Mukherjee

07:36

Can someone provide some examples of such functions?

One example is $max\{\|m\|,\|n\|\}$ where $\frac{m}{n}$ is the $x$ coordinate of a rational point on an elliptic curve.

In lowest terms ofcourse

leslie townes

07:52

beats me. do you want more examples specific to rational points on elliptic curves (i.e. fixing this type of Gamma, trying to identify other height functions)? or examples for other Gammas entirely?

these conditions (and suggestive name of the theorem) make the concept look like something someone cooked up just to make an argument go through, so it seems at first glance like maybe identifying more than one h for a given gamma is not "interesting." is that right?

X4J

If I want to compute the run-time for $T(n) = aT(\frac{n}{b}) + f(n)$, $T(1) = 1$, $a, b > 0$ why can't I simply argue that $T(n) = \sum_{i=0}^{log_{b}n} a^{i}f(\frac{n}{b^i})$?

Here it seems like I'm missing something

My question is why is the second term necessary?

Soumik Mukherjee

08:16

@leslietownes Examples for other Gammas. For an arbitrary commutative group how can we make such a height function.

This proof is given as a general proof of Mordell's theorem that states the group of rational points on elliptic curve is finitely generated.

@leslietownes Yes, the height function that I wrote above does its job for group on elliptic curves. But I am asking about height functions on an arbitrary commutative group.

leslie townes

well, "arbitrary" won't quite work, right? simply because not all commutative groups are finitely generated... if you know in advance that the group is finitely generated, does the structure theorem help?

e.g. is n mapsto |n| (or some silly modification) a height function on Z, and could you piece together that with height functions on other copies of Z to get one for any direct sum of those things and a finite abelian group (where it looks like any function would be a height function)

if the height function is primarily a widget that you use to prove that an abstractly defined abelian group is finitely generated, it seems like constructing one could be potentially difficult, if only because it seems rare/hard to me [a non algebraist] to be confronted with a group where you wouldn't have that information in advance

Soumik Mukherjee

08:51

@leslietownes Condition (d) is also needed to conclude that the group is finitely generated. So can't it be that an arbitrary group may have a height function but is not finitely generated?

leslie townes

well, sure, i also just don't know of a lot of interesting examples of non-finitely-generated abelian groups (other than ones you might construct using e.g. abstract constructions and large sets)

i don't know the context, but the fact that any finitely generated abelian group satisfies (d) [e.g. from the classification theorem for such things] suggests that whether (d) holds or not might not be the focus of attention for general groups where you might be looking for height functions

might be worth a MSE post? maybe it relates to some interesting math in situations where you can't make a full descent argument work but there is a next best thing

Soumik Mukherjee

@leslietownes Yes, I will post the question.

@leslietownes Z can be the group of rational points on some elliptic curve. So n mapsto |n| will be a height function.

Frieren

10:02

I must prove that $f(X) \setminus f(A)\subseteq f(X\setminus A)$ for $A\subseteq X$ and $f:X\to Y$. Let $y \in f(X) \setminus f(A)$, this means $y \in f(X)$ and $y \notin f(A)$. Hence, there exists $x \in X$ such that $y=f(x)$ and there is no $a\in A$ such that $y=f(a)$. Hence there exists $x \in X$ such that $y=f(x)$ and $y\ne f(a)$ for each $a\in A$. Since $A \subseteq X$, this means that there exists $x \in X \setminus A$ such that $y=f(x)$ and finally $y\in f(X\setminus A)$. Is this right?

Soumik Mukherjee

@Frieren Yes, also you can remove the 2nd last line.

Sahaj

10:56

hi

Anacardium

11:12

Do anyone have any idea on math.stackexchange.com/q/4852580/512080

@Jakobian Your proof is great. Fully convinced with what you have argued there.

ephe

12:35

Let R be a unital ring, M be an R-module and K be a free module over R. What is the form of R-module homomorphisms from M to K?

I have been thinking about this problem for a while and I couldn't come up with anything at all. The inverse situation is a lot simpler since the homomorphism will be completely determined by the image of the basis elements but in this case since M is somewhat arbitrary, the only homomorphism that I know to work is just the trivial one.

Thorgott

13:03

if rank(K) is finite, the module of such homomorphisms is isomorphic to rank(K) copies of the dual module of M

and the structure of the dual module can vary greatly depending on the module

ephe

If it's infinite dimensional with basis X then do we have that the homomorphism module is isomorphic to R^(X)?

Thorgott

if which one is free with basis X?

Jakobian

13:19

@leslietownes rational numbers

ephe

13:48

@Thorgott if the codomain is free with basis X

Oh I understand your confusion, sorry. So I suppose then I should expect this to be M^(X) instead not R^(X) right?

What would be the isomorphism between $\mathrm{Hom}_{R}(M, R^{(X)})$ and $M^{*^{(X)}}$?

Thorgott

14:02

only if $X$ is finite

the point is that $R^n$ is both a direct sum and a direct product, a homomorphism with codomain $R^n$ is the same data as $n$ homomorphisms with codomain $R$ (the components)

ephe

Okay I think I got it, thanks. I'll try to figure out an example where this isomorphism doesn't hold for an infinite basis.

Thorgott

14:19

that could be somewhat annoying

Sahaj

14:39

Have I made any error in this answer? math.stackexchange.com/questions/4852605/… I don't understand why it was downvoted..

Soumik Mukherjee

15:41

@leslietownes posted

1

Q: Existence and examples of height functions

Let $\Gamma$ be a commutative group. Let there be a function $h:\Gamma\to[0,\infty)$ with the following properties. For every real number $M$, the set $\{P\in\Gamma : h(P)\leq M\}$ is finite. For every $P_0\in\Gamma$, there is a constant $k_0$ so that $h(P+P_0)\leq2h(P)+k_0$ for all $P\in\Gamma...

Thomas Finley

16:05

0

Q: Need some clarifications regarding a computational example in ring theory provided in the book Topics in Algebra by IN Herstein in Section-3.9

I was reading about polynomial rings from the book "Topics in Algebra" by IN Herstein. Particularly, I was going through a computation given on Pg-157, Section-3.9. I am going to quote the portion that I am talking about. Here it is: Let $F$ be the field of rational numbers and consider the polyn...

abstract-algebra ring-theory

I need some clarifications with this issue.

Jakobian

16:18

@Sahaj asking about answers/questions is better in CURED, but let me see

I think the downvote was because you answered a poor quality question

this is discouraged on this site

or at least, people try to discourage it

Sahaj

I see. Thanks for letting me know.

Jakobian

or the downvote was random, it happens

@ThomasFinley Interpret $2$ as $2+A$

Thomas Finley

@Jakobian Ok, that really makes sense. But I have another issue:

"Also, if $a_0 + a_1t + a_2t^2 = b_0 + b_1t + b_2 t^2,$ then $$(a_0 - b_0) + (a_1 - b_1 )t +(a_2 - b_2 )t^2=0,$$ whence $$(a_0 - b_0 ) + (a_1 - b_1 )x + (a_2 - b_2 )x^2$$ is in $A = (x^3 - 2).$"

The issue is with this above part

Jakobian

@ThomasFinley alright, whats the issue

Thomas Finley

@Jakobian I interpret the 0 in $(a_0 - b_0) + (a_1 - b_1 )t +(a_2 - b_2 )t^2=0$ as $A.$ But how do they assert," whence $$(a_0 - b_0 ) + (a_1 - b_1 )x + (a_2 - b_2 )x^2$$ is in $A = (x^3 - 2)$" ?

Jakobian

16:32

Left side evaluates to $(a_0-b_0)+...+(a_2-b_2)x^2+A$

and what it means for $z+A = y+A$ is $x-y\in A$

Thomas Finley

Do you mean simplifying the equality, $(a_0 - b_0) + (a_1 - b_1 )t +(a_2 - b_2 )t^2=A=0$ by substituting $t=x+A$ ?

Jakobian

sure

Thomas Finley

@Jakobian Ok, so that is how they obtain $(a_0 - b_0 ) + (a_1 - b_1 )x + (a_2 - b_2 )x^2\in A$. That makes sense again! Thanks! I get it.

psie

In this answer, they prove by contradiction that if a closed set has a supporting hyperplane at every point of its boundary, then it is convex. I do not really understand the contradiction in the answer.

We assume $C$ is not convex, so there exists points $p,q$ in $C$ such that some of the line segment is outside of $C$. In particular, the line segment must go through a point where there's a hyperplane separating $p$ and $q$, and apparently that's a contradiction. Contradicting what?

Thomas Finley

I think I should delete the OP now, since the problem is resolved. But I think, it was not quite right for the author to write $2+0=2$ as $2+A=2+0=2$ seems absurd. Would you agree @Jakobian ?

psie

16:37

I have never really studied convex analysis, by the way.

Jakobian

people doing those things are so sloppy as to not distinguish between $x+A$ and $x$

I think the author was very formal about this anyway

Thomas Finley

@Jakobian very formal or very "informal" ?

Jakobian

formal

Thomas Finley

Maybe I don't have these sort of exposure to the way things are worded till now. But indeed it did create some confusion.

Jakobian

I might have just written $(a_0-b_0)+...+(a_2-b_2)x^2 = 0$ in $F[x]/A$ and call it a day

not even caring that this is some kind of quotient construction

Thomas Finley

16:39

@Jakobian That makes sense a lot more. And seems more decipherable imho.

Ok, all in all the problem's resolved and thanks again!

Jakobian

well the formal treatment is there so you don't get confused because of the hidden meaning of things when treated informally

Thomas Finley

@Jakobian that might be one explanation.

Or maybe the only one.

Jakobian

but when you're already more experienced you stop caring about this as much and write such informal equations

Thomas Finley

@Jakobian I understand.

Jakobian

@psie yeah your shift from real analysis to convex analysis surprised me a bit

Thomas Finley

16:43

All that's needed is more exposure and experience as you mention.

Jakobian

if you get confused about those concepts, then probably yes

psie

@Jakobian haha it's just this one proof of a theorem in a real analysis book that I'm currently reading that is completely absurd :) and I've been forced to look at alternative proofs and have stumbled upon convex analysis

Jakobian

@psie Suppose that $C$ is such set, and take half-spaces $H_x$ at every $x\in \partial C$ and consider $\bigcap_{x\in\partial C} H_x$

then the intersection contains $C$, is convex

psie

alright, that explains it I guess

Jakobian

Now take $z\in \bigcap_{x\in \partial C} H_x$. I don't think it should be hard to show $z\in C$

Yeah I think you should pick some $x_0\in C$ and consider the segment from $x_0$ to $z$ like in what Robert Israel is suggesting

There exists least $t \geq 0$ such that $x_0+t(z-x_0)\in \partial C$

3

A: Boyd & Vandenberghe, problem 2.27 — converse supporting hyperplane theorem

This is Robert Israel's answer https://math.stackexchange.com/a/27290/27978 marginally expanded. It is clear that $C \subset H$. Suppose $ p \notin C$, choose $x \in C^\circ$ and let $t = \sup \{ s \in [0,1] | x+\lambda(p-x) \in C \ \forall \lambda \in [0,s]\}$. Since $C^c$ and $C^\circ$ are o...

psie

16:57

I will have to read that

Jakobian

yeah you can just @copper.hat summon copper in chat

copper.hat is knowledgeable in convex analysis from what I recall

I mean... I'm sure I can get into the proof and handle it myself. But why would I, I'm sure copper would be excited to talk about convex analysis

Thomas Finley

18:13

0

Q: Need some clarifications regarding a computational example in ring theory provided in the book Topics in Algebra by IN Herstein in Section-3.9

I was reading about polynomial rings from the book "Topics in Algebra" by IN Herstein. Particularly, I was going through a computation given on Pg-157, Section-3.9. I am going to quote the portion that I am talking about. Here it is: Let $F$ be the field of rational numbers and consider the polyn...

abstract-algebra ring-theory

@Jakobian The clarifications seemed to be a fix temporarily. It seems that I got into a lot more trouble than expected when I proceeded. I edited the OP and made the necessary changes.

Jakobian

@TedShifrin does there exist $A\in\text{SO}(3)$, $A\neq 1$ which sends the rotations by angle $\pi$ to rotations by angle $\pi$?

@Thorgott

by left multiplication $B\mapsto AB$

Jakobian

18:44

elements of $\text{SO}(3)$ correspond to unit quaternions by $v\mapsto qv\overline{q}$

which ones correspond to rotation by angle pi

oh, the ones for which real part is $0$

I see how this is impossible now

If such $A$ existed, then it would correspond to unit quaternion $q_0$ such that $q_0q$ has real part $0$ when $q$ does. But then setting $q = i, j, k$ we have that $q_0$ has to be real. But then $q_0 = \pm 1$ which corresponds to the identity rotation

1

Q: Brouwer's fixed point theorem for quotient spaces of the unit ball

Consider the closed unit ball $\overline{\mathbb B ^n}$, and let $\sim$ be some equivalence relation on $\overline{\mathbb B ^n}$ that only identifies points on the boundary $\mathbb S ^{n-1}$ of $\overline{\mathbb B ^n}$, that is, if $x,y\in \overline{\mathbb B ^n}$ are such that $x\sim y$ and $...

general-topology algebraic-topology differential-topology fixed-point-theorems

psie

19:04

Jakobian, do you understand why in Robert Israel's hint we can choose a $0<t<1$ such that $tp$ is on the boundary of $C$? I guess one has to choose a $p\not\in C$ sufficiently close to $C$ for that to be possible.

Jakobian

The segment $[0, p]$ from $0$ to $p$ is just like the interval on the real line

and $C\cap [0, p]$ is closed

if you consder $\sup\{t\in [0, 1] : tp\in C\}$ for example, then this should correspond to $t$ such that $tp\in \partial C$

such $t$ won't be non-zero in general, but it will certainly be smaller than $1$

psie

ok

psie

19:20

So if we consider the supporting hyperplane at $tp$, i.e. $a^T(x-tp)=0$, why if $0$ is in $C$, is $a^T(-tp)<0$?

This implies $a^Tp>0$. Then $a^T(p-tp)=(1-t)a^Tp>0$, but if $0<t<1$, this isn't contradicting anything, right?

I'm commenting on the comment by syeh_106 under the answer.

Jakobian

The convention seems to be that $C$ is contained in $a^T(x-tp)\leq 0$

I don't think it matters, you can pick sign however you like

@psie can you link the post again

psie

oops

here

my suspicion is that they are referring to another theorem...

Jakobian

@psie alright, the commenter had some proof by contradiction in mind, I'd pay it no heed

They're not saying $0\in C$ but $0$ is the interior of $C$

It shows that the point $p$ doesn't belong to the hyperplane $a^T(x-tp)\leq 0$

So they're just trying to prove the other inclusion, that the intersection of hyperplanes is $C$

ah yeah. The condition that it has a non-empty interior sounds pretty important

I believe it doesn't have to be convex otherwise

The point is that all points in hyperplane $a^T(x-tp) = 0$ which are in $C$ must lie on the boundary of $C$

If $0$ is in the interior then the inequality must be strict

eh... I've ended up explaining it anyway

psie

19:35

@Jakobian you mean what the commenter wrote? I don't see how $a^T(p-tp)=(1-t)a^Tp>0$ is saying anything different than $a^Tp>0$ (provided $0<t<1$)?

Jakobian

@psie sure. What Robert Israel wrote is inaccurate

@psie You're getting it wrong

$a^T(p-tp) > 0$ says that $p$ is outside the half-plane $a^T(x-tp)\leq 0$

this much should be obvious

$a^Tp > 0$ is something we're using to be able to say that

and this follows from $0$ being in the interior of $C$

psie

ok, just trying to understand the commenter's try at a contradiction, but there seems to be none

contradiction seems not necessary anyway

Jakobian

12 mins ago, by Jakobian

@psie alright, the commenter had some proof by contradiction in mind, I'd pay it no heed

I've said this at the beginning

Ted Shifrin

20:19

@Jakobian Sure. Just do a change of basis: Change the axis of rotation. Oh, just left multiplication. I will need to think more.

copper.hat

20:33

@Jakobian did someone say convex?

Sahaj

hi

Jakobian

21:13

@copper.hat yeah psie was struggling with the proof that a closed set with non-empty interior which has supporting hyperplane at every point of the boundary is convex

so I've linked your proof

@TedShifrin If you look at homomorphism from unit quaternions to $SO(3)$, then I figured it boils down to some unit quaternion which maps unit quaternions with zero real part to themselves by left multiplication. But that makes it $\pm 1$ which corresponds to the identity matrix

so no such element of $SO(3)$ exist

copper.hat

21:35

Thanks!

Ted Shifrin

21:47

@copper I answered a lame convexity question.

copper.hat

:-).

psie

22:07

@copper.hat I have a question. Regarding this answer and the comments below, if you can show that the point $p$ is on the other side of hyperplane, have you then shown convexity of the set $C$? This seems to be the goal of the OP.

copper.hat

@psie If $H$ is the intersection of the supporting hyperplanes and $C$ is the original set, you have $C \subset H$ by construction. So, to finish, we need to show equality. If they are equal, then since $H$ is convex you are finished. To show equality we suppose $p \notin C$ and show that $p \notin H$.

psie

I see

basically showing $H\subset C$

copper.hat

Yup.

psie

22:23

@copper.hat Ok, could you explain why, in your answer, you define $t = \sup \{ s \in [0,1] | x+\lambda(p-x) \in C \ \forall \lambda \in [0,s]\}$ and why is it in $(0,1)$? You write this is because $C^c$ and $C^\circ$ are open. I don't understand that.

I have a couple of questions about your answer :)

Daniel Donnelly

22:58

@Shaun. Available to study 📚 group homology / cohomology? I like Romyar Sharifis lecture notes 📝 or weibel. In part way through weibel

Shaun

Hi @DanielDonnelly :)

Daniel Donnelly

Hey

So ur going for phd?

Shaun

My schedule is a little busy. I'm going to have to decline at the moment. If you get back to me about it over Easter, I'd love to study it with you then :)

Daniel Donnelly

When’s that?

Shaun

Yep :)

Daniel Donnelly

23:00

Lol

Oh end of march

Let’s do it then. In the meantime I’ll be doing Pereq study on abelian cats

Weibel has that data in appendix more so than cftwm

It’s odd though. You learn about abelian cats but the book proceeds in an element wise fashion. Which of course is most natural for set based mathematicians (nearly every1)

There do exist element-free diagram chases but they’re not as popular

Xander Henderson

@DanielDonnelly I prefer non-commutative dogs.

Daniel Donnelly

@XanderHenderson lol 😂 cheeky fellow

Shaun

I'm over a year into this UK PhD and all I have is some GAP computations :/

(I mentioned UK because an American PhD would have me do the research later on.)

I have ideas though, plenty of them.

I look at the maximal diameter of the Cayley graph of SL(n,q) (for now) w.r.t. normally generating subsets.

My supervisor's previous student looked at, in particular, SL(n,C) and SL(n,R).

I also look at inverse semigroups; semilattices, in particular.

My supervisor, Ben, said I probably know more about inverse semigroups than he does, given my undergraduate dissertation.

Does SAGE have a package for inverse semigroups? I saw something about them ages ago but I haven't found it since . . .

But I'm rusty re inverse semigroups.

I'm just talking aloud here. Feel free to say something though.

Xander Henderson

23:19

@Shaun I almost did my PhD in the UK---I had an offer at Warwick, but my then-wife couldn't work, so the money didn't quite work out.

Shaun

I got a book recently: "Semilattice Structures". Hopefully, it'll give me the skills to articulate my ideas.

@XanderHenderson That's cool. Warwick is damn good. Where did you end up?

Xander Henderson

University on California, Riverside.

Shaun

Nice :)

Xander Henderson

Which is not on the same level, but the logistics were better. I also had an offer at University of Arizona.

Which had tight connections to NASA and JPL, but I eventually decided that I didn't want to go applied.

Shaun

I think American PhDs are better than UK ones. The duration, the taught aspects; it's just superior.

Xander Henderson

23:24

Meh. It really depends.

Shaun

How so?

Xander Henderson

I think it is largely down to your advisor.

And what your long term plans are.

Shaun

Well, I'm set there. Prof. Ben Martin is world renowned in the field. He and his collaborators came up with the function I study, and it's foundational to a major conjecture in the area.

I'd like to be an academic.

That's where I might fall short.

copper.hat

@psie The map $l(\lambda) = x + \lambda(p-x)$ is continuous, so $l^{-1}(C^C)$ and $l^{-1}(C^\circ)$ are open. A little diagram might help here.

Shaun

If success on MSE is anything to go by, I could be okay; I don't know. I asked on Academia SE about whether it's an indicator and not many people think it is.

Xander Henderson

23:29

Success on SE is irrelevant for most things.

Shaun

I'm 33. That's still young but ideally I think not having a PhD by now is not great.

@XanderHenderson I agree :)

I just spend too much time here

Xander Henderson

Meh. I was 40 when I finished. And there was a guy a year ahead of me in his mid 50s.

3

Academia is generally not too concerned with age.

Shaun

Oh. That's cool. You seem to be doing pretty well.

I guess I have preconceived notions.

Xander Henderson

I am happy enough where I am.

Shaun

Awesome :)

Xander Henderson

23:34

Though I wish I had more time/remit to spend on research. :/

Shaun

I think most academics would say that.

Daniel Donnelly

@Shaun zettlr.com

I searched for stackedit.io alternatives and that popped up

looks great. It's FOSS

So probably let's you insert either Tikz or at least images from Quiver

Xander Henderson

@Shaun yeah, but I have a teaching gig at the very edges of academia.

I don't get paid to produce papers. :)

Daniel Donnelly

@Shaun rather than just GAP alone or SAGE, I recommend HomAlg project, probably has those inverse semigroups

I think it includes GAP codes

Shaun

Or inductive groupoids, which are categorically equivalent.

Daniel Donnelly

23:40

@Shaun what do you normally use to write your papers?

Software I mean

Shaun

TeXnicCenter.

Daniel Donnelly

Nice

I personally need a drag-n-drop wysiwyg editor

Shaun

@XanderHenderson I see. I couldn't do that. I prefer research over teaching.

@DanielDonnelly I don't know what that is.

The jargon . . .

Daniel Donnelly

WYSIWYG = what you see is what you get - usually instantly rendered LaTeX preview.

drag-n-drop, visual way to layout a page

BaKoMaTeX had wysiwyg

😎😎

Shaun

That makes sense, @DanielDonnelly

Overleaf is like that.

I'm nor sure about drag-n-drop for Overleaf though.

It's nearly midnight here and I have something in the morning. See you later!

Xander Henderson

23:55

A text editor and a makefile are my tools of choice.

copper.hat

vi hopefully

Xander Henderson

Extra bonus points if the text editor does syntax highlighting.

Though I am not quite old enough / self hating enough to use vi or emacs.

I use BBEdit on a Mac.

Semiclassical

@XanderHenderson i've had to get some comfort with using makefiles due to inheriting some files from the previous lecturer

copper.hat

uggh

emacs was developed by control perverts

Transcript for

$Mathematics$ Mathematics