@JasperLoy You're Chris'sis now ? :P she used to have that profile pic

@JasperLoy Your videos aren't from Romania :-)

I'm a wizard

@JasperLoy or listen to you state your location

that's the opposite side of the world from the fortress of solitude

I had to decide whether to go out and drink or stay in and eat chocolate. Chose the latter.

I usually eat approximately 15 OZ milk chocolate in one serving and it gets me super high.

ok

Hi @arctic ... Long time no see.

hi

Hi @Jasper, @Hippa!

Things are very, very, very quiet.

\RidiculouslyHuge Oh?

LOL ... I've missed you, anon.

You ready for the year? Teaching again?

@TedShifrin o/

Tu vis toujours, @Hippa!! :)

On dirait :-)

Comment ça va ?

Mais c'est bien tôt le matin ...

7h30

Pour un jeune, c'est plutôt la nuit ... :)

En fait je n'ai pas dormi >.>

@TedShifrin nope, not ready. yes teaching again.

Oh oh ... quelque chose de grave?

Non juste les vacances

Ah, ok, @Hippa.

Ah d'ailleurs j'ai réussi mes concours :-D

大家好

Félicitations! Qu'est-ce qui se passe après? :)

LOL @idont

@TedShifrin Après j'entre à l'école polytechnique le 28 août

：）

There aren't any more math courses for you to take, @arctic ...

Merveilleux, @Hippa. Formidable!

\o/ thank you

Très fier :)

Tu vas passer ta vie à Palaiseau ... :)

Quatre ans

I remember the steep hike up the hill from the metro, @Hippa ... I did that for two weeks.

You taught here for two weeks ? Or were you just visiting ?

I was visiting ... but I gave four or five lectures (en français) ... back in 1980.

@TedShifrin I had to climb those too to go to the written exams :-)

Well, apparently the hike got the blood circulating to your brain, Hippa :)

I guess :D

Odd that I'd remember that 36 years later ...

Trying to decide whether to go visit Europe or Australia in a few months ...

I have some 1997 transcript of a series of conferences at the ecole polytechnique on divergent series and ways to make them converge and the utility of doing so

My former math teacher sent me those last year

@TedShifrin Where in Europe ?

Actually, one of my lectures appeared in print, @Hippa. Not exactly down your alley (yet, anyhow).

Don't know yet ... maybe Germany, Sweden/Norway, Spain ... But I find it impossible to go to Europe and not spend a few days in Paris.

@TedShifrin I was thinking. Every rotation in n-dim is a product of orthogonal 2D rotations. How well known do you think this fact is? I only learned of it in the past year or so, and it doesn't really seem to be in any discussions on SO(n) and such.

anon, it's well-known from the normal form (follows from the spectral theorem).

In fact, I used to assign that as an exercise when I taught differential topology in order to prove an isotopy fact.

canonical form also follows from fund. thm. fin. gen. mod. over PIDs

(most abbreviations in one phrase I think I've ever used)

but yeah, okay

Hmm ... how do you get it that way? I haven't thought about it.

I usually use the realification of the complex spectral theorem.

I know the usual Jordan and ratl can forms that way, of course.

if A is a linear map on a vector space V, one may treat V as a k[T]-module, then iso V with sum sum of k[T]/(f(T)^e)s, then choose an appropriate basis in this direct sum to get a canonical form

it's an application of the fund. thm. mentioned in most grad algebra books I'd think

Sure, I know all that. But how do you take particular advantage of orthogonality?

what do you mean?

That's just the usual way of getting rational canonical form. I don't see how you use orthogonality of T to get the complex diagonalization.

I was talking about the usual way of getting canonical form. never said I used orthogonality to get complex diagonalization.

Ohhh, I'm confuzled.

I said "canonical form also follows from fund. thm."

I said I used the spectral theorem and you made it sound like you had an alternative way to see the result.

But how are the usual canonical forms remotely helpful in this case? Don't we want to know that normal operators are diagonalizable?

rotations are not diagonalizable over R, and you seem to know canonical forms are helpful (I assume by helpful you mean helpful to proving every rotation is a product of orthogonal 2D rotations)

Well, I never mentioned canonical forms, myself.

I said normal form ...

I guess I don't know how to do this without a form of the spectral theorem.

the spectral theorem $\ne$ the fund. thm. fin. gen. mod. over PIDs right?

Nope.

Or yup. English is ridiculous.

heh

was about to accuse you of thinking (1+1=2) = (fermat's last theorem)

You're the one who brought fin gen modules (i.e., canonical forms) into it, not I.

sure

But you won't answer my question that I've asked twice, so I'm giving up.

eh?

anyway, the way I'd show every rotation is a product of orthogonal 2D rotations is to first show a stronger form of semisimplicity (complement of invariant subspace also invariant, which can be refined into some kind of 'maximal' decomp into orthogonal minimal invariant subspaces), then use fact every rotation has an invariant 1D or 2D subspace (follows from fund. thm. fin. gen. mod. over PIDs), then analyze how rotation acts on 1D and 2D invariant subspaces

Well, the complement claim is immediate from orthogonality. How are you seeing the invariant 1- or 2-dim subspace from modules?

All in all, I think the spectral theorem proof is more elementary (if one knows the linear algebra result).

V is isomorphic to a direct sum of R[T]/(f(T)^e)s. from semisimplicity, we see each e<2. the only irreducibles in R[T] are linear or quadratic.

OK, I see. This makes sense coming from all the rep thy you've been doing. But my exercise was perfectly reasonable for undergraduates who knew some linear algebra ...

Thanks for explaining.

if there were a way to prove minimal invariant subspaces have dim<3 elementarily, I would very much like to know it

It follows from the fact that the operator diagonalizes over $\Bbb C$.

To me that's much more elementary.

true

Hey there, Jasper. I haven't been around MSE much recently.

implying disjointness

Well, I've had lots of visitors the last month or so, but I've also gotten a bit tired of being here.

Nope, not tired of seeing people. Tired of some of the stuff around here.

I did give a little lecture to 8 undergraduates and 3 faculty up around 110 miles away.

@TedShifrin

is this you?

does isometrically isomorphic mean existence of a bijective isomorphism?

an isomorphism which is also an isometry

isomorphism confuses me a lot, what does it exactly mean in this context?

in what context?

you didn't give any

I mean as in two spaces $X$ and $Y$ are isometrically isomorphic

spaces? well, in the category of topological spaces, isomorphism means homeomorphism.

isometry would seem to imply homeomorphism though, so that would be redundant

perhaps you could be more specific about what X and Y are

well the topic here is functional analysis, and to be exact, I'm talking about two subspaces of $L^2(\mathbb{R})$

ah, vector spaces

see, it's important to give context!

sorry

isomorphism means bijective linear map in the context of vector spaces

wait isn't $L^2(\mathbb{R})$ the space of all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\int_{\mathbb{R}} ||f||^2dx < \infty $?

sorry I can't get how that's a vector space

@arctictern

yes

show it satisfies the axioms of being a vector space

@SoumyoB Triangle inequality.

You can moreover show that it's a normed vector space, under the norm $\|f\|_{L^2} = \int\|f\|^2 $.

(indeed, that's why the "isometry" part makes sense)

@BalarkaSen hi

Hi @Danu

Wanna help out with an annoying detail in a proof? :D

Let's see.

Let $M$ be compact, $R$-oriented with boundary

What's the issue?

I want to show $H^k(M,\partial M)\cong H^k_c(M-\partial M)$

Using a collar I already have $H^k(M-\partial M,U-\partial M)$

where $U$ is the collar

I'm having difficulty translating this to $H^k_c$, mostly (I think) because I have no good hands-on way of dealing with the direct limit definition

Yikes. I am rusty on compactly supported cohomology.

What I need is:

$$\varinjlim_{K\subset M-\partial M\text{ cpt}} H^k(M-\partial M,(M-\partial M) - K)\cong H^k(M-\partial M, U - \partial M) $$

Right.

One thing I was thinking was

So one has to take bigger and bigger compact sets in the interior which eventually covers all of M - U.

$H^k(M-\partial M,U-\partial M)=H^k(M-\partial M, (M-\partial M)-(M-U))$

That uses compactness.

Anyways that doesn't seem to help

@Danu So you have maps $H^k(M - \partial M, (M - \partial M) - K_i) \to H^k(M - \partial M, U - \partial M)$ where $K_i$ is an increasing sequence of compact sets covering $M - U$.

Universal property of direct limits gives you a map from the direct limit.

All you need is to verify that's an isomorphism, then, yeah?

$K_i$ should cover $M-\partial M$

But that doesn't matter due to defo retr

I'm not so sure yet

Well, if $K_i$ cover $M - \partial M$ it automatically covers $M - U$.

Of course

I'm not worried about that

You still have the map from $\varinjlim H^k(M - \partial M, (M - \partial M) - K_i) \to H^k(M - \partial M, U - \partial M)$.

But I am worried whether those extra $K_i$'s covering more than what is necessary doesn't prevent the map being an isom.

So elements of $\newcommand{\interior}{\operatorname{int}} H^k(\interior M,\interior M-K_i)$ vanish on all chains outside $K$ right

Surjectivity should be straightforward. Take a relative cochain in $H^k(M - \partial M, U - \partial M)$, restrict to $(M - \partial M) - K_i$ for each $i$ and send this sequence of cochains to the direct limit.

@Danu Yes.

Now, using that $U-\partial M= \interior M - (M-U)$ that gives inclusion-induced maps into $H^k(\interior M,\interior M-(M-U))$ whenever $K\subset M-U$

So if the $K_i$'s just cover $M-U$ we indeed get maps into the thing we have

OK, fine

Yes.

and those maps give a directed system, because they're inclusion-induced so we indeed get to apply the universal property

Yes.

Coolbeans

Be sure to check if my surjectivity argument is right: it should be verified that the map I gave is indeed a section.

Now, I am seriously worried about injectivity. Hmm.

Whoops, I gotta go. Sorry.

Bye

If you haven't solved the problem by when I am back, ping me and I'll see if I can help.

I think you can figure the rest out by yourself better than I can, though.

This meta.SE post might be of interest for math.SE community: What is the appropriate site for questions about stochastic processes?

@BalarkaSen: what you up to?

@Huy stuff. mostly analysis, need to catch up on topology

@BalarkaSen Hi

So I did it by a different approach

Hello again.

Ah?

Mike suggested it to me

OK, then I needn't ask. Neat.

@BalarkaSen 1 sec lemme finish writing it up, then I'll show it

No need; if Mike suggested it, it should be good.

Well, he only told me the first step and then I worked it out myself

plenty of room for errors

The final little step I am a little iffy about is the third vertical map in the last diagram

$\interior M - K_n$ is like a thickening of $\partial M$ but with $\partial M$ also taken out

I was thinking: The full thickening (i.e. including $\partial M$) deformation retracts onto its middle, lets say, and so does $\interior M- K_n$, so it should still work, no?

Whaddaya think

A collar neighborhood of $\partial M$ is of the form $\partial M \times [0, 1)$. If you take out $\partial M \times \{0\}$, it's still homotopy equivalent to $\partial M$ (def ret to $\partial M \times \{1/2\}$ say)

Yeah, exactly

That's in symbols what I said above

AWESOME

just one more proof and I'll have finished these notes

@BalarkaSen: I've been trying to make the whole "the chain of curves is bounded by a circle, and cutting along a circle results in at the most two disconnected components, so cutting along the chain itself cannot disconnect the surface" argument rigorous but haven't quite managed yet. do you have a suggestion?

I've produced about 210 pages of notes (just on topology) over the past year!

@Huy I have forgotten what that was. What did you try to prove?

@BalarkaSen: that any chain of scc of even length is nonseparating

chain = sequence of scc such that consecutive elements have geometric intersection number 1

Ah, right, that.

you suggested considering a small tubular nbhd of it and prove that its boundary is a circle

which can be done via induction

Yes, so the whole deal is to rigorously prove a chain of even length has tubular nbhd with boundary a circle.

That's what you want to do, is it?

no, I already know the boundary is a circle

OK, great. So what's next?

but how do you rigorously argue that then cutting along the chain of scc itself cannot disconnect?

maybe I'm missing something elementary topological here

The tubular nbhd of the chain deformation retracts to the chain, yup?

yes

And you know complement of the tubular nbhd of the chain is connected.

(because that is a connected component of complement of boundary circle of the tubular nbhd)

urm, wait

ah, yes I understand now what you mean

yes, I agree with all that

I see what's happening

basically, when retracting, the connected component of the nbhd becomes smaller and smaller until it vanishes (when the boundary "coincides" with the chain)

Yes, the picture is clear, but there's some argument needed here. Deformation retract doesn't induce a deformation retract of the complement always (you can probably come up with a ctrexample)

but I'm not comfortable with this argument because it feels too hand-wavy

right

I agree :) That's why you try to make it rigorous. Let me ponder a bit.

no rush, I'm just trying to as rigorously as possible prove missing parts of F&M because I sometimes feel like too much hand-waving makes me come to premature and wrong conclusions sometimes (or simply misunderstand proofs)

^

This every time :P

so dangerous in topology :P

yes, occupational hazard

quoting a topologist here :P

@BalarkaSen thanks for the explanation

by the way, surely, you're doing a PhD in topology aren't you?

\

@Huy One thing is clear: one needs the fact that we're looking at a tubular nbhd here, not just some arbitrary nbhd.

Tubular nbhds have a very rigid structure.

@SoumyoB Nope.

man you're such a mystery

@Huy For example. $M$ be your surface, take a circle inside $M$. $T$ be a tubular nbhd of that circle. $M - T$ is homotopy eq (indeed, def. ret.) to $M$ - circle. This is because $T$ minus circle is $S^1 \times (0, 1)$ minus it's center circle, so we can straightline homotope $X$ minus circle along those $x \times (0, 1)$'s to $X - T$.

(well, smooth that up appropriately by a bump function, but that's easy to do and a technical detail)

So you know this for circles. Trouble is, tubular nbhds of a chain is slightly harder because a chain is not technically speaking a submanifold.

Idea: do what I said for each circle on the chain individually and compose all the deformation retracts. If necessary, do them simultaneously (Hatcher's trick in chapter 0, homotopy extension property, IIRC)

I'll have a look at that

This trick is in page 15, prop 0.16.

do you understand my argument for a (smoothly) embedded circle btw?

I can write it up better if you want to.

I'm thinking about it, if you want to give more detail, feel free to

(just reading the Hatcher part atm)

It is this. $F : (S^1 \times (0, 1) - S^1 \times \{1/2\}) \times [0, 1] \to S^1 \times (0, 1)$ be the deformation retract of $S^1 \times (0, 1) - S^1 \times \{1/2\}$ to $S^1 \times \{0, 1\}$, $F(x, 0)$ the inclusion and $F(x, 1)$ the retraction (the def. ret. is finished).

Now, consider an embedded circle (or as you call it, simple closed curve) $C$ in $M$. $T$ it's tubular nbhd: as $M$ is orientable, $T$ is topologically the same as $C \times (0, 1)$.

Then $H : (M - C) \times [0, 1] \to M - C$, given by $H(x, t) = x$ for all $x$ away from $T$, and $H(x, t) = F(x, t)$ for all $x \in T$ is the desired deformation retract.

Such an $H$ can be made smooth by gluing the identity map sufficiently far away from $T$ with $F$ on $T$ by a bump function.

Oh, and $C$ here is naturally identified with the center circle in $T$ (that is, under the idntification $T \cong S^1 \times (0, 1)$, $C$ goes to $S^1 \times 1/2$).

@Huy The part I was emphasizing is the infinite concatenation of homotopies bit. The $[1/2^n, 1/2^{n+1}]$ trick.

thanks for writing it out, I'll have a look at it and see if I can digest it

Sure, sure. The gist is, deformation retract cylinder minus center circle to the boundary of the cylinder.

(which is usable since a small tubular nbhd of the circle looks like a cylinder)

sorry for changing the topic, but just curious... is anyone here preparing for GRE?

General as well as Subject

What's GRE ?

Not a French thing, I guess?

Graduate Record Examination, if I remember correctly

it's an exam that students from non-English speaking countries, like mine, have to give in order to qualify for admission into most US and Canadian universities

why would you go there

of course, getting good marks in that exam is necessary, but not sufficient

UK is cooler

well... just... personal reasons

I don't like what the refugee immigration has turned the EU into

lol

UK is leaving EU

that's a good thing, I know Brexit happened

I support that decision wholeheartedly

ok

I just kinda hate the idea that while I'm having to work hard to qualify myself to go live in a first world country, people out there get a free refugee pass

lol

yeah, I too envy a refugee's life. /s

doors should never be opened fully like this, there should always be a bar set up that people need to pass

thats an odd point of view, imo

but eh, whatever

yeah I lean to the far Right

so if a man is an engineer and is fleeing from war with his family, his wife and child need to stay home if they're uneducated?

well... about that... let's just say the world is not fair

@SoumyoB three cheers for Trump and Putin

nothing's fair

so what

get over it

I do indeed support Trump

:P

shrug

the case of women and children is different though, and I wouldn't have much of a problem with the refugee immigration if most of the people coming in were actually helpless women and children

problem is, most people coming in are adult males

so you're in India but you know that exactly because you work as a frontier-guard located in the EU I suppose?

not that they shouldn't be coming in just on the basis of their age and gender, but it just never seems like they're the truly helpless ones

it pains me to see history that happened with India a few centuries ago (about 10th-14th centuries) now repeating itself in the EU

aka the foreign invasion

people getting run over by trucks and getting shot at en masse inside clubs and people of certain orientations fearing their lives wasn't something that was known inside the EU prior to the refugee immigration

ok

happens in America even without terror

not to mention what happened in Cologne this new year's eve

lol

@SoumyoB i'm sorry, what are you referring to? i have no idea how or why british invasion of india is comparable to refuge immigration

not the British invasion, I'm talking about the Islamic invasion

that's how Buddhism disappeared off the face of this country

are you a Buddhist?

nope, but I believe our country would have been much more peaceful if only Buddhism was one of the predominant religions like it was before the Islamic invasion

why aren't you a Buddhist

@SoumyoB this i doubt very much

we had a glorious history, everything turned inside out upside down after that invasion, that's what I believe in

"glorious history" lol? are we talking about india?

yeah

I feel sad when I look at the Nalanda university ruins

GloriousIndiaMasterRace ?

you sound like a supporter of hinduism through and through

lol

nope, trust me I'm agnostic

why should he have trust if you don't

sorry but because I'm against one religion shouldn't mean that I support the others?

I mean I do support Buddhism and Christianity

yes, Christianity, the true religion of peace

and I'm not even that big of a fan of Hinduism

I've come in at the wrong moment I see

it's been reformed a lot lately, it doesn't commit any acts of terror en masse now

^^

at least as far as I know, I could be wrong

I'll go discuss mathematics over here then chat.stackexchange.com/rooms/15189/hinduism

hello

sorry I'm not gonna talk about it any more

@Krijn

I'm just listening to some dope Daredevil OST

how are you ?

@SoumyoB Thanks! :)

@Huy i would have linked a tom lehrer song at this point but it's better to leave the topic be.

Although I can imagine that during the holiday season, not that many people are doing mathematics

So that other discussions arise