Mathematics - 2016-10-15 (page 2 of 3)

Danu

16:00

1

In a paper I'm reading (https://arxiv.org/pdf/1009.1175.pdf), one claims the following: Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold. If $\mu$ is any $2$-form on $M$, then $$\omega^{n-1}\wedge\mu=0\Leftrightarrow \mu=0.$$ The implication $\Leftarrow$ is clear, but I have trouble ...

@BalarkaSen ^sanity check: is that comment right?

Because that'd imply that Guillemin claims something that's pretty clearly false in his paper :P

eurocoder

I am reading a paper in which it states that the terms of a series (its in Equation 1) are of the order 1/sqrt(m) for large m and hence the series is extremely slow to converge. But surely if the terms are of the order 1/sqrt(m) we have a p series with p=1/2 and hence the series shouldn't converge at all?!

Secret

is the m restricted to finite values?

eurocoder

The series I'm talking about is in Equation (1) of this paper - dii.unisi.it/~capolino/pdf%20publications/… - and the part about the slow convergence is right after Equation (3), with some more detail given after Equation (11).

@Secret It is an infinite series for m

mercio

maybe it's an alternating series ?

Mike Miller

I don't like that o someone else is now using the name studiosus.

Danu

16:11

@MikeMiller Someone else?

abenthy

Is there an example of an embedded submanifold of a smooth manifold which is not smooth?

Danu

@abenthy An embedded topological submanifold, or what?

Can't you just do a line with a cusp?

abenthy

yes

an embedded topological, but not smooth, submanifold I mean.

or is every embedded topological submanifold of a smooth manifold smooth?

Mike Miller

@Danu studiosus used to be the name of one of the best MSE contributors in topology.

Secret

The complex exp() in that series seemed to be alternating between + and -, and is symmetric wrt $\pm \infty$ thus I suspect it will converge because it is alternating at least in the imaginary direction

Mike Miller

16:19

@abenthy You can always embed a topological manifold in some large R^N

PVAL-inactive

That paper is terrifyingly bad.

Mike Miller

What paper?

Ted Shifrin

Morning, @PVAL, g'night, @MikeM. ... Wait, someone usurped studiosus's name?

DHMO

@MikeMiller dii.unisi.it/~capolino/pdf%20publications/…

PVAL-inactive

The Gulleimin paper

or more accurately the paper written by his students.

Ted Shifrin

16:21

I'm a long-time fan of Guillemin's. :)

Mike Miller

I don't know why anyone could ever care about Poisson structures.

Ted Shifrin

They show up naturally from physics, at least.

This studiosus has been around 6 months, @MikeM, so it's not a recent usurpation.

Mike Miller

Maybe that's why the old one changed his name.

PVAL-inactive

I've never seen an error that bad in a mathematical paper.

This is a paper thats primarily about the manipulation of differential forms, and they think thats true.

abenthy

@MikeMiller I just found your helpful answer here math.stackexchange.com/a/1739784/88291 Do you maybe know any reference I can cite for this result instead of copying your argument?

Mike Miller

16:26

There are classical papers on equivariant differential topology you can cite.

One by exactly that name I believe.

I wrote that answer in the generality of proper actions but in most cases you're really working with compact Lie groups anyway.

abenthy

You mean msp.org/agt/2007/7-1/agt-v7-n1-p01-p.pdf ?

PVAL-inactive

@abenthy If someone directed you to the right argument, you better say that as well as cite whatever original source you can find.

abenthy

Yes sure.

Mike Miller

I'm not sure I need to be thanked for something so classical.

But do as you please in any case.

abenthy

I couldn't find the statement that the fixed point set is a smooth submanifold if the action is smooth and proper anywhere in the literature, but maybe i just didnt find it.

and the proof you give looks nice to me

Mike Miller

16:31

Note that there are no fixed points on proper actions when the Lie group is noncompact. See the comment.

abenthy

Oh right, i read that some time ago and forgot about it... so its a statement about actions of compact Lie groups, right?

Mike Miller

Basically yeah.

Danu

@MikeMiller Oh

@PVAL-inactive So it's an actual mistake?

Is it worth it to call the attention of Guillemin to that error?

PVAL-inactive

@Danu Yes, unless there are some hidden assumptions there.

No, you should contact the corresponding author instead.

Danu

@PVAL-inactive The way it is stated seems unambiguous...

@PVAL-inactive Ah, of course.

Maybe I will

There is no indicated corresponding author, however

Mike Miller

16:38

Contact Pires. She posted it.

PVAL-inactive

arxiv.org/abs/1009.1175 The corresponding author is indicated on the arxiv papge

Danu

@PVAL-inactive I didn't know how that stuff works. Thanks, also @Mike

Time to fly back to Munich today...

Ted Shifrin

Safe travels @Danu

@Danu: It seems that the correct conclusion is that the 2-form $\mu$ must be orthogonal to the symplectic form. Certainly not 0.

Adeek

17:02

Hey @TedShifrin could you help me understand this example ? I just want a sanity check

I don't understand how is this sequence defined ?

is it from right to left ?

Ted Shifrin

What do you mean by right to left? He drew the graph of the $n$th function.

He's defining a continuous function that has value $1$ at $a$ and is $0$ on $[a+1/n,b]$.

Adeek

so this graph represent each nth function right ?

Ted Shifrin

Yes, that's precisely what your notes say.

Adeek

oh ok nvm I understand it now thank you

yeah

Anubhav

@MikeMiller I wonder if $f$ is a no-zero degree map from S^n\to M , where $M$ is a closed oriented $n_$ manifold, then how close the map $f$ goes to become an covering map?

In other word, is it true, that $f$ is homotopic to some covering map?

Ted Shifrin

17:07

What if $M$ is simply connected, @Anubhav?

PVAL-inactive

No just precompose with some self map of $S^n$ of degree $2$.

Mike Miller

For nontrivial examples there are degree 2 maps $S^5 \to SU(3)/SO(3)$.

Ted Shifrin

How about the squaring map $\Bbb CP^1\to\Bbb CP^1$?

Mike Miller

That's exactly what PVAL said

Ted Shifrin

Well, yes, he was precomposing. I'm just letting $M=S^2$. :P

Anyhow ... :)

Anubhav

17:15

Yes, I actually wanted to see some non-trivial example.

Mike Miller

I forget how you construct that map. But it can be done.

Ted Shifrin

What qualifies as a trivial vs. non-trivial example? Maps of degree $k$ from $S^n$ to $S^n$ are trivial?

Anubhav

@TedShifrin I didn't get your question?

Ted Shifrin

You said you wanted to see some non-trivial example. What does non-trivial mean?

Anubhav

Ohh, I mean where $M$ is not-simply connected.

Sorry for not speaking out in the first time.

Ted Shifrin

17:21

OK.

Mike Miller

When M is not simply connected you just factor through the universal cover.

The interesting thing is when the codomain is not S^n. The example I gave is simply connected.

Ted Shifrin

Mike, isn't $SU(3)/SO(3)$ simply-connected?

Oops. Too slow.

I'm trying to think of unitary frames on $\Bbb C^3$, but I don't yet see the map.

Anubhav

@MikeMiller yes, because in that case we have Hopf Degree theorem

Mike Miller

Ok, use the Hurewicz theorem mod C, where C is finite groups. That says the image of pi_5 SU(3)/SO(3) in H_5 has finite index. In particular there is a map of nonzero degree.

Harder to see that it's 2. But in particular any simply connected rational homology sphere has a map of nonzero degree from a sphere.

Now i need to come up with an argument for Georges.

Anubhav

@MikeMiller what argument?

Mike Miller

17:31

You can prove this using G&P degree.

Ted Shifrin

Interesting: $SU(3)/SO(3)$ double covers $SU(3)/SU(2) \cong S^5$, so I don't get a map the right way.

Or maybe I'm being stoooopid again.

Mike Miller

@TedShifrin That's nonsense. :)

Double covers of $S^5$ are disconnected. And the two groups $SO(3)$ and $SU(2)$ inside $SU(3)$ are unrelated: one is real matrices, and one is block diagonal matrices.

Andrew Thompson

Our Lie group course definitely did not mention that.

Anubhav

@MikeMiller I didn;t get this, in what context you are saying this?

Mike Miller

@Anubhav He claims you need to invoke homology. I'm sure that's not true. I'm trying to write a degree proof.

Ted Shifrin

17:35

Yup, stooopid.

There's got to be a geometric way to see this

Anubhav

Ohh, yes, even there I wrote a comment... for finite pre image of a regular point+ the degree concept of G&P will implies the multiplicity.

Ted Shifrin

This is coming from the Hopf bundle on $\Bbb CP^2$.

Andrew Thompson

I'm currently taking a reading course with curriculum "whatever you want in chapter 4 in Hatcher." Any recommendations on the additional topics?

Mike Miller

All of them.

I have found 4A-4G all useful at various times. 4H-4K I haven't found useful but I do found interesting.

Anubhav

@AndrewThompson you should read as much as you can digest from that chapter... you may skip a few proof.

Mike Miller

17:37

4L is eminently useful.

Andrew Thompson

There are other people in the course I have to consider as well. We will do Brown representability at least.

Mike Miller

Why do you have to consider them?

Ted Shifrin

LOL

@AndrewT: Follow in Mike's footsteps and be selfish. :)

shredalert

Hello

Ted Shifrin

Hello @shred

Mike Miller

17:40

I'm having trouble making the argument work. maybe someone else can.

Andrew Thompson

Well, they may have their own interests and/or needs. I laughed a bit when I saw Hatcher introducing pushouts in 4H. He should've done that in Chapter 0.

Mike Miller

@GeorgesElencwajg I disagree. Consider a map $f: M \to X \to N$, where $M, N$ are oriented, $X$ noncompact; call the first map $f_1$, the second $f_2$. $f_2^{-1}(p) \cap f_1(M)$ is a finite set for generic $p$. Because $X$ is noncompact, you can find a proper map $[0,\infty) \to X$, where $0$ maps to one of the points and $X$.

Anubhav

2

Q: Is every vector bundle over an open subset of $\mathbb{R}^n$ flat?

Can every smooth, real vector bundle over an open subset $U \subset \mathbb{R}^n$ be equipped with a flat connection? Intuitively I feel the answer is "yes", but I don't know how to prove this.

differential-geometry algebraic-topology

Mike Miller

Somehow this should lead to a proof that $f_2f_1$ is degree zero.

shredalert

Could someone please help me with a composite relation proof? Been stuck on it for a good few hours now.

Mike Miller

17:41

@Anubhav The answer to that is no.

Anubhav

@TedShifrin can you explain me the answer.

Ted Shifrin

Huh? @Anubhav What answer?

Anubhav

I didn;t understand the answer, why $TS^2$ is not flat?

Ted Shifrin

Because the Euler characteristic comes from integrating the curvature.

Just ask your question, @shred.

Anubhav

ohk, and since $TS^2$ is deformation retract onto $S^2$.

Now I get it.

Ted Shifrin

17:44

Not really. We're not thinking of a metric on $TS^2$. We're computing the Euler class of $S^2$ from the curvature of the connection on that bundle over $S^2$.

shredalert

http://math.stackexchange.com/questions/1969608/problem-with-composite-relations
That's the link to the question. My main problem is I don't see how I can show that each relation is a function when the composition of $R_1\circ R_2=diagonal of X$ and $R_2\circ R_1= diagonal of Y$

PVAL-inactive

@MikeMiller There aren't any non-trivial examples. Any degree k map factors through the universal cover.

Ted Shifrin

@shred: Ugh, I hate math filled with so many symbols.

Anubhav

@TedShifrin I never encountered any of these kind of relation.

shredalert

@ted I'm sorry!

Ted Shifrin

17:47

@shred: OK, so let's stay calm and think. What do we need to know $R_i$ is a function? We need the vertical line test for its graph as a relation, right?

shredalert

@ted I'm studying from Zorich's Analysis book and it's pretty heavy on the symbols.

Anubhav

@TedShifrin Maybe I should read your note, becasuse do Carmo didn't do much with vector bundle in his Riemannian geometry book

Ted Shifrin

I was trained to write in English, fewer symbols :)

@Anubhav: My notes are undergraduate diff geo, nothing to help you.

But, yeah, you need to discuss connections on vector bundles or principal bundles, not just Riemannian geometry.

Anubhav

@TedShifrin then reffer me something.

I've completed the book of Do Carmo

and solved almost all problems

shredalert

@ted Umm, let's see. For a relation to be a function every element in the domain needs to have a unique element in the range.

Ted Shifrin

17:50

Milnor/Stasheff discuss this stuff briefly in an appendix in Characteristic Classes. There's Kobayashi-Nomizu. Books on complex geometry discuss it, because complex geometry uses this heavily.

Wells, Griffiths/Harris, etc.

@shred: Right. So if we had $(x,y_1)$ and $(x,y_2)\in R_1$, would that contradict one of your compositions?

Mike Miller

@PVAL I know this. I said examples where the codomain isn't a sphere.

Any simply connected rational homology sphere is an example.

Anubhav

@TedShifrin I've read 1st 12 chapters of Milnor's book, now I'm thinking the I should first go through the appendix before starting Charn classes.

Ted Shifrin

I don't like that book, honestly, but I'm a differential geometer, not a topologist. I would learn from Chern or Griffiths/Harris or other places ...

PVAL-inactive

@MikeMiller For some reason I thought we were still assuming the manifold has universal cover a sphere.

Mike Miller

Milnor's book is awful but it's the best place for a topologist to learn.

Ted Shifrin

17:54

@shred: Are you drawing any pictures to help?

@MikeM: I put my non-topologist disavowal.

Anubhav

@TedShifrin personally I prefer the language of topology over geometry.

PVAL-inactive

I do not think that book is awful.

Ted Shifrin

Then you shouldn't talk to me, @Anubhav :P

Anubhav

bcz most of the time, while doing geometry, I lost inside heavy notations.

Don't take it as offence @TedShifrin

Ted Shifrin

I think differential forms and moving frames make geometry much more geometric than the traditional approach, but most mathematicians avoid them.

Alessandro

17:56

good evening

Ted Shifrin

I'm not offended, @Anubhav; I'm just saying my viewpoint disagrees with yours and my taste disagrees with yours, so you might not want advice from me.

Good morning, @Alessandro :P

Alessandro

Hi @Ted

shredalert

@ted Does it contradict the definition? I'm not really sure. I've not been taught how to draw graphs for diagonals.

Mike Miller

@PVAL I think a good book would have an actually readable section on obstruction theory, a discussion of G-bundles as well as vector bundles, a good discussion of classifying spaces and some discussion of curvature and geometry.

I also don't much like their writing.

Anubhav

@TedShifrin I dont have as much knowladge as you. You know the global picture of math way more than me, so taking advice from you is always he;pful.

Mike Miller

17:58

The obstruction theory is probably the thing that upsets me most.

Anubhav

@MikeMiller I agree with you

PVAL-inactive

@MikeMiller You're asking for the scope of the book to increase.

Ted Shifrin

Just think of $X=Y=\Bbb R$ and what's the easiest relation that's not a function (that you know from high school)? @shred

PVAL-inactive

I like the fact that it isn't Steenrod at least.

Mike Miller

That's true.

Ted Shifrin

17:59

I liked Steenrod, @PVAL, so we disagree.

PVAL-inactive

I think I might have had I been taught by people who use that language.

Alessandro

I have yet to give the paper a careful look but I've been thinking about yesterday's problem @Ted, a poset could have some infinite descending chains and also some minimal elements, so the fact that we can't extract a minimal basis from the rational balls one does not necessarily imply that there are no minimal basis at all

PVAL-inactive

It just felt completely at odds with much of the stuff I've read.

Ted Shifrin

@Alessandro: I thought we got to start with a basis for the topology. We're only starting with the topology?

PVAL-inactive

Like I think it was probably a much better book nearer to when it was published.

Ted Shifrin

18:01

@PVAL: Steenrod (and Chern) preceded the Bourbaki-ing of mathematics. For that I am grateful.

Alessandro

And if I'm not mistaken the fact that countably generated $\sigma$-algebras have minimal generating set is not enough to conclude the existence of a minimal basis for the usual topology on $\mathbb{R}$

shredalert

@ted $xRx^2$?

Mike Miller

Mathematics has not successfully been Bourbaki'd.

Ted Shifrin

Ah, right, @Alessandro, I forgot we had switched to $\sigma$-algebras.

@shred, right. Now use that and its inverse relation and write down the compositions.

Alessandro

my first comment was about topologies though, what I meant is that it's possible that there is no minimal basis which is a subset of the rational balls basis, but there could be other basis from which a minimal one can be extracted

Ted Shifrin

18:03

@Alessandro: I would still bet you $10 that that can't be true. :P

Alessandro

I still think you would win that bet, but I'm still looking for a proof :P @Ted

Ted Shifrin

Good. I'm glad you are :)

PVAL-inactive

@TedShifrin :/

shredalert

@ted so we have $xR_1x^2$ and $x^2R_2x$?

PVAL-inactive

I think bringing up the Bourbaki in a pedagogical discussion is basically equivalent to bringing up Hitler in a political discussion.

5

Mike Miller

18:09

lol

Ted Shifrin

LOL

You mean equivalent to bringing up Trump?

@shred: Yes. I guess the domain of $R_2$ is just non-negative reals, but proceed.

PVAL-inactive

@TedShifrin Is there a difference?

shredalert

@ted so that would mean the composition $R_2 \circ R_1 = X \rightarrow X$?

Ted Shifrin

Yes, @shred, and $R_1\circ R_2\colon Y\to Y$, where $Y$ is nonnegatives.

Danu

Again, let's not draw parallels between WW2 Germany and the current state in the US! It takes away from the gravity of that war.

Mike Miller

18:15

You take away from the gravity of the current situation in US politics. It's hardly a triviality.

Ted Shifrin

American society (but, frankly, most of European, as well) is now at violently toxic levels.

Danu, did you see my postscript on that symplectic form thing?

@PVAL: In all honesty, I don't think Bourbaki is anywhere in that league. It's very much against my style, but I am not revolted by Bourbaki and have even read the occasional passage.

Mike Miller

I have to cite Bourbaki.

shredalert

@ted so two elements from the domain of $R_1$ map to one element in the domain of $R_2$ but those elements get mapped to only one element for the inverse relation?

Ted Shifrin

No, the inverse relation has both as values, @shred.

$1R_21$ and $1R_2(-1)$.

shredalert

@ted Ah, I see. So $x^2R_2x={(x,-x)\inX}$?

Ted Shifrin

18:23

@shred: What you wrote doesn't make sense, but I think you're on the right track.

Danu

@MikeMiller compared to WW2? How US centric and absurd

Ted Shifrin

Which composition are you looking at, @shred?

Danu

@TedShifrin no, sorry

shredalert

@ted not really up to grips with all this notation. I was meaning that the inverse relation is the set of ordered pairs of negative and positive $x$ in the set X which is the domain of $R_1$.

Danu

I'm on my phone

Ted Shifrin

18:25

I pinged you up there. I believe the right statement is that $\mu$ has to be orthogonal to the symplectic form.

Danu

Yeah,at least if you have a Hodge star I know it for sure

Ted Shifrin

@Shred: So tell me this. $1 R_1\circ R_2 ?$ and $1 R_2\circ R_1 ?$?

Alessandro

Does $2^\omega$ usually denote the set of binary sequences?

Danu

Time to board! Bye guyZzz

Ted Shifrin

That's fine, @Alessandro. I prefer to write $\{0,1\}^\omega$ for clarity. What you're writing is an ordinal, I guess, giving the cardinality.

Bye @Danu.

Alessandro

18:30

I'm asking because I found it in the $\sigma$-algebras paper and the next sentence is "If $X=2^\omega$ and $B$ is it's natural Borel structure" but I'm at a loss about what that natural Borel $\sigma$-algebra is supposed to be

shredalert

@ted Is it $1 R_1\circ R_2 1$ and $1 R_2\circ R_1 1$?

arctic tern

I would write $\{0,1\}^{\Bbb N}$ for binary sequences. The symbol $\omega$ connotes ordinals, and in ordinal arithmetic $2^\omega=\omega$ holds, which is not the set of binary sequences.

Ted Shifrin

For which one of them do you also get $1 R\circ R (-1)$?

Hi, tern. So you're totally agreeing with me. Yippee :)

arctic tern

hello

Mike Miller

But if you want to stick your head in the sand, keep it there where I can't hear you.

shredalert

18:32

@ted $1 R_2\circ R_1 (-1)$ I believe.

Ted Shifrin

@shred: $1 R_1 1$ and $1 R_1 (-1)$, but $(\pm 1)R_2 1$ only. Right?

So $1 R_2\circ R_1 1$ only.

(It helps me to draw $y=x^2$ and $x=y^2$ :))

Alessandro

Ah, I should learn how to read, what exactly is meant with $2^\omega$ is defined a few pages earlier

shredalert

@ted but according to the definition of a functional relation in my book, it says that the elements of the range must be unique

arctic tern

@Alessandro if it's power set of w, then that's already a sigma-algebra. that it?

Ted Shifrin

But remember that we're trying to understand what goes wrong if we do not start with a functional relation, @shred.

Alessandro

18:38

The fact that I've found a definition does not mean I understand it yet :P, give me a moment @arctic they're defining it as a particular case of a product between Borel spaces

shredalert

@ted looking at our relations we map from two elements to one, but on the composite we are mapping back to only one element.

Ted Shifrin

Not so, @shred. Let's look at $R_1\circ R_2$.

Start with $1$. Where does it go?

Oh, hell. I have my R's labelled wrong for our discussion.

Sorry.

shredalert

@ted so $x^2 R_2 x$ goes from 1 to 1 and then $x R_1 x^2$ maps from 1 to 1?

@ted it's okay. I've spent long enough on this problem, don't mind spending more haha.

Ted Shifrin

I'm sorry. I messed up the indices. Let's do $R_2\circ R_1$. $1R_1 1$ and $1 R_2 (\pm 1)$, so ....

(What I said far above was using the switched indices. Sorry.)

shredalert

That's fine @ted I appreciate all the help I can get

Ted Shifrin

18:44

So you were absolutely correct. You have $1 R_2\circ R_1 1$ and $1 R_2\circ R_1 (-1)$. So you do not get just the diagonal for the graph of $R_2\circ R_1$.

shredalert

@ted because the diagonal would only map 1 to itself, yes?

Ted Shifrin

Yes. The diagonal relation is just $x R x$.

shredalert

So this composite relation does not give us the diagonal. So this implies that if we have a composite relation which gives us the diagonal, the first relation and second relations would have unique elements in the range for every element in the domain?

Ted Shifrin

Right.

Easiest way to prove this is to do the contradiction argument we just did the example of. You think you can do it?

shredalert

So the first part of my proof to show that $R_1$ and $R_2$ are functional relations is complete?

Ted Shifrin

18:51

Well, you have to write a general argument. Suppose $x R_i y$ and $x R_i y'$. Show that $y=y'$ must follow.

(Do $R_1$ for concreteness. Choose the right composition. Say an analogous argument works for $R_2$.)

shredalert

Just trying to arrange all of this in my head, sorry if I am a bit slow to respond

Ted Shifrin

I recommend doing concrete examples often in learning mathematics.

shredalert

This is a new topic for me and I should have started with it when I started the section tbh

started with concrete examples I mean

Ted Shifrin

Yes, even if your teacher doesn't, you should!

shredalert

My teacher is the book here. Self-studying this :P

And you're teaching me too ofc :)

Ted Shifrin

18:58

Ah. I don't know what book you have, but I like to recommend Kevin Houston's How to Think Like a Mathematician.

shredalert

I'm using Zorich's Mathematical Analysis I. I had no trouble in the last section's exercises. I managed to push through, but this section is giving me trouble.

Ted Shifrin

The proofs in analysis get logically quite complicated — layers of quantifiers. Have you done some proofs in linear algebra and/or abstract algebra?

shredalert

I've done a few proofs with groups. Very elementary though.

Ted Shifrin

My personal recommendation is to do some linear algebra proofs (and abstract algebra proofs, if possible) before trying to learn analysis.

shredalert

I've noticed that. As soon as I started I was bombarded with logical quantifiers.

Ted Shifrin

19:03

Yuppers.

shredalert

The ones in linear algebra are a lot more straightforward, the ones I've encountered so far at least.

Ted Shifrin

Right, usually just one quantifier, not two or three.

shredalert

I've been using Vinberg's A Course in Algebra. I don't know if you've heard of it?

Ted Shifrin

Nope.

Alessandro

I remember the look of resignation on my professor's face as he was about to negate the triple quantifier in the definition of uniform continuity for a proof...

Ted Shifrin

19:04

Yup. That's one of the standard ones. Or equivalence of continuity definitions.

Alessandro

he actually said that writing the negation correctly was the hardest part of the proof and he was right

Ted Shifrin

What I observed is that most of the students in my Honors multivariable course were smart enough that they caught on quickly. But I remember spending an hour on this when I taught advanced calculus back 30 years ago.

Absitively, @Alessandro.

Alessandro

I just learned a new word!

shredalert

Haha

Ted Shifrin

You should never learn words from me, @Alessandro :)

mercio

19:07

negating quantifiers is easy

negating implications is less easy

Ted Shifrin

"easy" for people who go to grad school in math, not "easy" for average or below-average students of proof mathematics

shredalert

Newbie here

Ted Shifrin

don't make the mistake of assuming everyone is at or above your level of competence or intelligence

@shred: Take a look at that Houston book if you can.

And try to look at a proof-oriented linear algebra course.

shredalert

It's on the online library I have access to. I can check it out now.

Ted Shifrin

cool ... I wonder if my books are available in that library :P

shredalert

19:09

Let me have a look

bolbteppa

Why is it that when you quotient out the center of a compact Lie algebra you get a semi-simple remainder, but for a non-compact Lie algebra you have to use solvable ideals and not the center again?

Ted Shifrin

Agh, I don't remember enough to answer that, @bolbteppa. Maybe ask @arctictern?

shredalert

@ted There are a few articles with your name on it. But I don't see if there are any books. Might have to look deeper in the system. The library has a terrible search engine.

bolbteppa

Really really interesting to see how using solvable ideals in lie algebras seems to be motivated by the failure of the center

arctic tern

iunno. a little jarred by the word "remainder" used to mean "quotient" though :P

Ted Shifrin

19:13

OK, @shred. There is a proof-oriented linear algebra book (coauthor Adams) if you can find it without having to spend horrid numbers of dollars.

bolbteppa

haha

I nearly referred to quotienting as 'throwing away'

Ted Shifrin

@bolbteppa: If @TobiasKildetoft shows up, you might ask him.

bolbteppa

A comment of his on a deleted question motivated this!

Ted Shifrin

All the more reason, then :)

I'm going to have lunch. See you all later on !

Good luck, @shred!

Alessandro

buon appetito @Ted!

shredalert

19:15

@ted thanks. Have a good lunch :)

Ted Shifrin

Grazie, @Alessandro.

Alessandro

I'm having trouble seeing the intuition behind the definition of $B$ in the product below. If $B$ is a $\sigma$-algebra and $B\subseteq X$ we call call the pair $(X,B)$ a Borel space.

If $\{X_\alpha,B_\alpha\}_{\alpha\in\Delta}$ is a sequence of Borel spaces we define their product as the pair $(X,B)$ where $X=\bigotimes\limits_{\alpha\in\Delta} X_\alpha$ and $B=\sigma\{A_\alpha\times\bigotimes\limits_{\beta\neq\alpha}X_\beta:A_\alpha\in B_\alpha,\alpha\in\Delta\}$

It seems to me that we're picking only an element from every $B_\alpha$ but that can't be correct

Secret

19:31

Hey guys, are there classes or examples of nonassociative algebra that has this rock paper scissors like property. More generally, if an algebraic structure display such property, is there any special names for the elements that participate in said rock paper scissors property?

shredalert

Going to head off for a bit. Good luck with sorting out your problems everyone. :)

TheGreatDuck

19:50

0

Q: In differential equations, do we ever use the assumption that the integration variance C is a constant?

Indefinite Integrals vary by constant values. This is a fact. In differential equations, these C values end up being mixed around all over the place. My issue (or rather my question) is based on the fact that some integral/derivative pairs change what C is and even make them into functions of th...

differential-equations

can anyone answer this?

even a naive answer would be good to some degree

It's still got a while but it would suck to waste a bounty when the question clearly seems... useful to others.

arctic tern

@TheGreatDuck I have no idea what you're asking.

TheGreatDuck

@arctictern the various "cn's" we write in the general solution to a differential equation are obviously constants, but do we ever take advantage of that fact during the solution process

arctic tern

@TheGreatDuck if you integrate more than once, you may end up using that fact.

seems super weird to put a large bounty on this

TheGreatDuck

in general, would our solutions to differential equations change in form completely if the nature of c was a different function, or would it just be a substitution

@arctictern it's actually a pretty important question, believe it or not.

arctic tern

@TheGreatDuck if you replace a constant C with something that's not constant you almost surely are no longer talking about a solution to the question

TheGreatDuck

19:55

@arctictern there are integrals that treat certain types of functions as constants

arctic tern

@TheGreatDuck what makes you say so?

TheGreatDuck

@arctictern what makes me say what?

arctic tern

@TheGreatDuck sure, if you're integrating a function of one variable with respect to a second independent variable, then it is constant (with respect to the second variable)

@TheGreatDuck "it's actually a pertty important question" (the thing I specifically replied to - see the gray arrows in front of our comments)

TheGreatDuck

@arctictern no... I mean there are alternate definitions of the integral that treat certain functions as constants

arctic tern

@TheGreatDuck how so?

TheGreatDuck

19:57

they hold certain function forms constant

which in general, means that the "c" is now some arbitrary function type dependent on x

arctic tern

if you're going to be mysterious and not tell me what you're referring to I'm not sure what the point of talking is

TheGreatDuck

im not being mysterious

it's a general case

there is no particular alteration worth mentioning

arctic tern

well good luck then

maybe someone out there will know what you're talking about

TheGreatDuck

but they all have in common that c changes to a type of function (linear, quadratic, piecewise constant, etc.)

so the question boils down to

"aside from the integration do we ever assume c is a constant in our algebraic steps"

we assume it's constant, obviously

but do we ever use that assumption when finding the general solution

@arctictern now does that make sense?

it's an integration theory thing

distort the definition of the integral and see what follows

(granted, integration theory is more heavy handed on how we integrate)

arctic tern

wondering if there's a nice way to see that $$\det\left(\sum_{g\in G}\rho_V(g)X_g\right) $$ is an irreducible polynomial in the formal variables $\{X_g:g\in G\}$, where $G$ is a finite group and $\rho_V$ an irrep

TheGreatDuck

20:04

whoa

i think that is something beyond my reasoning

:p

Secret

$$\int f(x,y,z,\cdots) dx = F(x) + C(y,z,\cdots)$$

no way can C be a function of x

Secret

@arctictern would you like to have a look at my rock paper scissors algebra question?

arctic tern

@Secret can you define "rock paper scissors property"? in the image you posted, it seems x is the usual group operation, but I don't understand + or its relationship to RPS

Secret

Ok, basically in the + cayley table (which is commutative) there's this interesting property where a+1=a, b+a=b, 1+b=1

arctic tern

as I understand it, in the rock paper scissors game, we have the set {r,p,s}, and the binary function outputs the winner of a round. but that implies every element is idempotent, unlike in your structure

Secret

20:18

So it's kinda like a literal rock paper scissors where one element absorbs another and together this absorbing property form a cycle

arctic tern

@Secret so it's the usual RPS structure but with squares of elements replaced by their superiors?

you'd have to decide how you want to generalize RPS to multiple hand gestures. also, is there any relationship between + and x in that structure?

Secret

I think so, the square of each element absorbs the corresponding nonsquared versions.

The only relationship between + and x is that distributive law holds as usual, only the + associativity is broken in all 27 except the listed 9 cases

Also, x forms our usual cyclic group of 3 elements

arctic tern

if you think RPS corresponds to quaternions, there could be an analogue for octonions. take the fano plane mnemonic - every two elements a,b is in some kind of cycle a,b,c with a third element c and we have "local" rock paper scissors. then you'd need to figure out what to do with squares though, as elements don't have unique superiors.

Secret

Sorry mistake, I mean, the double of an element absorbs the undoubled one, not squares. Let me flesh that out more clearly...

arctic tern

you seem to have a symmetric bilinear map $\phi:\Bbb Z_3\times\Bbb Z_3\to\Bbb Z_3$ satisfying the two properties $\phi(a,b)\in\{a,b\}$ and $\phi(\phi(a,a),a)=\phi(a,a)$.

Secret

20:28

yeah

So I guess it is kind of similar to quaternions, except that to get the next element, you only need one element instead of 2
The "generating equations"(a better term is needed) seemed to be the following:
1+1=a
a+a=b
b+b=1
The equations that give the RPS looking behaviour, and also the symmetric bilinear map you pointed above are
a+1=a
b+a=b
1+b=1

arctic tern

@Secret you're familiar with the cycle mnemonic for remembering i,j,k's multiplication table? I am using that as a model for which wins over which in rps.

Secret

Well, in physics we have the levi citiva symbol to do the bookkeeping for us because we use it daily in angular momentum calculations

The only thing I am known to get wrong is the sign

ij=k, jk=i, ki=j

arctic tern

suppose we take the oriented fano plane mnemonic for octonions and label them 1-7. there's now the Z/7Z operation, and then we can define a rock-paper-scissors operation for a,b by simply picking out the directed line that a,b are a part of and using the usual rock-paper-scissors operation.

not sure if distributivity will hold for any choice of 1-7 labelling though

I guess your $\phi$ has the additional property that if $\phi(a,b)=b$ and $\phi(b,c)=c$ then $\phi(c,a)=a$. this implies the "cycles have length 3" I guess.

Secret

yeah, that's what I noticed when I try to map where the elements gone in that triangle diagram in the slide, except the "cycle" is not the notion of cycle as that in cyclic groups, but a cycle formed by this interesting RPS like behaviour

meanwhile carrying out the fano plane suggestion to see what I get

syzygy

20:59

the basic equation is

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