$\mathrel{\lessgtr\mkern-8mu\mid\mkern8mu}$

$\mathrel{=\mkern-8mu\mid\mkern8mu}$

I am now thinking it is not possible.

@anakhro What does this actually mean geometrically

Why is that expression appearing

Kind of long-winded. It's something that I need to replicate a proof from a GL(n) case to an analogous Sp(2n) case. So if I can get this Y, then I can complete the proof analogously. The geometry behind it is to express decomposition of matrices through optimal mass transport.

Howdy, DogAteMy @Akiva!

Hello

So, here's something gross from the site

@MikeMiller Well, but we can read off the canonical basis for the column space from the pivot columns of the REF, anyhow, but yes. I actually taught my students (influenced by Strang) about the left nullspace, as well, as its basis gives you the constraint equations a vector must satisfy to be in the column space.

Haven't seen you in ages, DogAteMy!

oh oh @Semiclassic

i came across a nice little geometry problem on site a few days ago, and it got me to put up a question related to it

Right, that makes sense.

Again, frustrating that I have so little time. I could spend a month on dimension, kernel, and image, and how these play off each other.

come today, i find out that said question (and several others on the main site) appeared on yesterday's AMC 12B competition.

Maybe only 3 weeks.

@MikeMiller If you wanted a more geometric statement, I have a bundle $\pi\colon Sp(2n)\to P(2n,A)\colon X\mapsto XAX^\top$ where P(2n,A) is the image of this map (it's actually the collection of p.def. symmetric matrices with the same symplectic eigenvalues of $A$), and I want to show that the subset of the symmetric p.def. symplectic matrices is a section of this bundle.

note the timing there...

Yeah, I ended up with pretty efficient treatments of this stuff in my books, @MikeM. But applications will use it, anyhow, so it's not like they get to forget it.

"left nullspace"? is that a weird way of saying cokernel?

I doubt that's a fiber bundle, but sure.

so someone was putting up exam questions on MSE for a math contest -before- the competition went up......

@Thorgott It's the set of row vectors $f \in (\Bbb R^n)^*$ so that $fA = 0$.

@Semiclassic So how did kids get the question ahead of time? But yeah, tons of competition questions and AoPS competition questions too

AKA the set of constraints on the image of $A$.

@TedShifrin that indeed is the question

@Thor: Or kernel of the transpose, if you prefer. But it has geometry just as the usual kernel does.

This is definitely the hardest I've had to work to write a halfway decent class.

the user who noticed it thinks they were a proctor for the exam

@MikeM: Well, first time thinking about it and teaching it, whereas topology you think about all the time, etc.

if so, i feel gross to get caught up in it unintentionally

WTF is a proctor doing publicizing? That's grounds for ...

quite

@MikeM: The first few times I taught linear algebra, I followed the textbook I was assigned (although I complained about the first one a lot and the second time was a different book I liked more), but I certainly didn't get around to writing my own course until the sixth time or so. :)

for context, like everything else they're having to do the AMC exams online and remotely this year because of COVID

I like the book I use pretty well. I don't like to follow a book closely. If I am repeating, with the same set of ideas, and the same language, I see little value added.

so more vulnerable to cheating because of taht

Of course I'll try to have corresponding sections so they can learn something from both of us.

@MikeM: Well, giving insight and emphasis and different examples one can still add lots to the book. Plus, most students are of one type — either learn from lecture or learn from book — rarely both.

Better learn to be both or you're burning $$$. Just my view.

ah ok, so it's the kernel of the dual

No, kernel of the dual.

yeah, my order was mixed up

@MikeM: I always learned more from lectures, but of course I am capable of reading a book.

But I've had students who, despite my flawless and entertaining lectures, said they preferred to stay home and read the book. When I wrote the book, I suppose that's a valid approach, but otherwise I rarely followed the book identically (even with books like G&P and Munkres, which I like a lot).

@MikeMiller Is there an easy way to tell if it is not?

Oh, I'm sorry, I misread.

It's very much a fiber bundle, because that's the orbit of a group action. The map $G \to G \cdot p$ is always a fiber bundle.

I thought your map was $(X, A) \to \cdots$

@TedShifrin there's also the learn from neither type

Okay, so I am not crazy then. :P

Yeah, @Alessandro, I'm afraid that's true, although often that type doesn't try very hard.

the just don't learn type

One of the reasons I retired earlier than expected is that the number of students I could not motivate to work and pass my courses had increased, and that was too frustrating to me. Not the honors multivariable course, but the other courses I typically taught.

@Semiclassic You certainly can't blame yourself. That said, I'm posting more and more that people need to make effort and/or that they certainly shouldn't be posting verbatim homework or EXAM questions.

What is $\Gamma_A$, anakhro?

Stabilizer of $A$

under your action $X \cdot A = X A X^\top$

what is the best way to learn assuming you have 1 Billion dollars

As we've just been saying, every student learns somewhat differently from every other student (and some don't learn well at all).

@MikeMiller It's non-specific. But that stabilizer gets used to make it a principal bundle.

If it's non-specific then I am confident the answer to your question will depend on $A$.

Probably on the symplectic eigenvalues of A.

Maybe I'm wrong. Is the answer to the analagous question for GL(n) always yes?

Independent of A?

I could answer that if I wanted to but I'm looking at this offhand

For GL(n) it was a little different because my bundle was $GL(n)\to P(n)$, so there was no need for A.

Since GL(n) acts transitively on P(n), A didn't matter.

@TedShifrin yeah.

So it was specifically because of symplectic eigenvalues that this mess was brought up anyway, but I thought I could maybe work past it (seemingly everything does, until this section part).

Right, OK, that's what I am forgetting.

it's one reason why find it hard to agree with this meta post:

because a lot of it is not "this person doesn't know how to ask a good question" but "this person is trying to get something for nothing"

I was trying to break it by finding some sort of relation between symplectic and ordinary eigenvalues but there are very few theorems I could find about these two.

@anakhro Am I supposed to know what a symplectic eigenvalue is?

@TedShifrin it's not a popular term, but if you have a p.def. symmetric P then you can find a symplectic X so that $XPX^\top$ is diag(D,D) for a diagonal nxn matrix D.

The diagonal values of D are the symplectic eigenvalues.

But what does it mean :(

$P=A$?

@TedShifrin :) Good catch!

@MikeMiller Not sure what they actually represent. But they are unique up to reordering and not a lot of things to say about them according to Google.

So if $P$ is diagonal to start with, what can happen other the entries of $D$ being the square roots?

Hmm, no, maybe not.

I've never ever never ever encountered this.

@anakhro is this for even-dimensional P specifically? otherwise X doesn't seem possible

It reminds me (but inappropriately) of Pfaffians.

@Semiclassical Yes. 2n dimensional.

Yeah, @Semiclassic, $P$ has to have even "dimension."

wacky

The theorem is called Williamson's theorem, if you liked it.

the paper here links it up to a classical-mechanics parallel for how the quantum harmonic oscillator is solved with ladder operators

which is something i'm well-versed in

so that's neat

I am thinking this might only end up working in analogous cases where I get some sort of transitive action.

Rather than this one where it heavily depends on $A$.

Surely there is some geometric meaning to these.

To the symplectic eigenvalues?

Or to the setup?

Symplectic eigenvalues.

@Semiclassical could you discern any meaningful geometric meaning of them from your understanding of that paper?

not without more thinking

Has anyone worked out the $2\times 2$ case?

the paper does include one 2-by-2 case for reference:

actually, not sure i understand how it's an example yet

I am working on 2x2 right now.

Oh they are the eigenvalues of JA.

I am trying to use The Method of Frobenius to solve $xy'' + y = 0$. I plugged in the guess $y = x^s \sum_{n = 0}^\infty a_n x^n$ and got the indicial equation $s^2 - s = 0$ with solutions $s = 0, 1$, and the recurrence relation $a_n = -\frac{1}{(n + s)(n + s - 1)}a_{n - 1}$.

My understanding is that I am supposed to solve the recurrence relation without plugging in values of $s$, multiply the result by $s$, and take the partial derivative of that result with respect to $s$ and evaluate it at $0$. However, I'm not sure how to approach the recurrence, since knowing that $s = 0$ "behind the scenes," I can see that starting at $n = 1$ would blow up the fraction.

I didn't notice that before.....

Should I ignore this problem since technically the solution process doesn't plug in a value of $s$ yet, or do I need to do something else here?

But the square of the product should be the product of the original eigenvalues. That's why it felt Pfaffian to me.

I taught the method of Frobenius in 1974. I haven't thought about it since.

So in the 2x2 case, the idea, positive trace is enough for the positive definiteness in the presence of symmetry.

Oh, but then we are done.

Because it's impossible since Sp(2) is SL(2,R).

Welp

Back to the drawing board. Thanks @MikeMiller @TedShifrin and @Semiclassical!

Oh wait, I think I was too early to that.

It's not impossible. I am just impossibly dumb.

Sp(2) is $SL(2,\Bbb R)$? I doubt that.

Enough rambling, I will return to this. BYE FOR NOW.

Yeah, lunchtime for this bonzo, too.

@TedShifrin never doubt. n=1 case is special for Sp(2n) where Sp(2) = SL(2,R).

@TedShifrin You don't doubt that. Sp(2n) are the matrices that preserve a certain 2-form. In 2D, that just means preserving area.

I'm trying to minimally understand something about vector bundles. So as far as I understand it, $n$-vector bundles over $S^4$ are classified by homotopy classes of maps $S^3\rightarrow\operatorname{GL}(4)$ since they trivialize over each hemisphere and this map describes how the fibers on their intersections are glued together. Now $\pi_3(\operatorname{GL}(4))$ has a group structure, what does it correspond to on the level of vector bundles?

You can take connected sum of two (manifold, vector bundle) pair, just throw disks out and tube by identity on the boundary

Because on the disks the bundles are trivializable

So that's the operation that corresponds to the group structure

@Thor: How did you turn $n$ into $4$?

(Same dimension, same rank, but that's clear from my description)

Hi Ted

He started with rank $n$?

Hi, a @Balarka.

Yes, that was a typo

I suppose I should quit reading :)

so you remove discs in the base and then glue together the bundles over the boundary? I'm not quite sure what "tube by identity" means

and yes, n=4

it's frustrating to work with students when you ask questions which are seemingly simple and...silence

(in this case, it's the equivalent of "if y-x=5x, what is y". i'm not exaggerating by much)

Am I missing simple examples or is actually hard to come up with examples of Polish spaces of positive finite dimension for which all compact subsets have empty interior? I know that the is there Baire space in dimension zero (and in fact is the only example up to homeo) and the complete Erdös space in dimension one (all Banach spaces work in infinite dimension)

@Semiclassic: If it was $y-x=11x$, it would be a much harder problem.

I have no idea what you're talking about, @Alessandro.

@TedShifrin it did take me a bit to pick up on one misconception they had

I think you know what a Polish space is :P

No, I never use the term.

And the Baire space is just a fancy name for $\Bbb N^{\Bbb N}$

You've told me before, but I'm not going to remember.

Oh ok, they are completely metrizable separable spaces

there were a lot of people who responded to "the upward acceleration is 5g" with "well, free-fall acceleration is -g (downwards). so clearly the total acceleration is 5g-g=4g!"

....no. the acceleration is what i said it was: 5g

In most cases you can just think about $\Bbb R^n$, not here unfortunately

So you want interior coming only from noncompact subsets? Weird.

So only compact subsets are totally disconnected?

Not necessarily, in a Banach space every compact set has empty interior for example

You said finite-dimensional.

And what you just said is a lie.

You meant infinite-dimensional Banach space.

Right

Anyway I think the whole space will have to be totally disconnected

So this sounds pretty Cantorish.

Yeah but Cantor is zero dim unfortunately (in fact the only perfect metrizable compact zero dim space up to homeo)

@Thorgott yes

do the obvious thing

Is Cantor of Lebesgue measure $1-\epsilon$ also $0$-dimensional?

He means topological dimension I'm assuming

Yes, compact Hausdorff spaces are zero dimensional iff totally disconnected iff they have a basis of clopen sets

I never studied dimension theory, either. Only know Hausdorff dimension.

I don't know what the obvious thing is. There isn't a canonical way of identifying these boundaries so as to glue together along them.

Well, @Alessandro, I stand by what I said originally. I know nothing.

@BalarkaSen right (and the spaces I'm interested in are metrizable so Lebesgue covering dimension and the inductive dimensions all agree)

@Thorgott there is, the bundles are trivializable over the disk. Trivialize, then glue by identity.

@TedShifrin that makes two of us with regard to my question then I fear, because all I know was in its statement

trivialization isn't canonical, but I guess you're saying this choice doesn't affect isomorphism type

That is correct

The obvious thing works. No need for a 2hr debate.

ok, this sounds nice enough to work with explicitly

Ah yes, the partial derivative w.r.t. the 3N dimensional vector

In the solution the prof switches to partial derivatives w.r.t. x_i and x_j and I still dont really get whether these are supposed to be 3-dimensional or actual coordinates

maybe $\frac{\partial}{\partial x}$ is supposed to be the vector consisting of the actual partials $\frac{\partial}{\partial x_i}$ and then they take a dot product?

could be the case, but could also not be the case

probability

It's also not clear whether the first derivative concerns everything afterwards or just the D(x)

@Thorgott Note that Balarka's construction literally gives the usual sum on homotopy groups when applied to the clutching construction

I guess it's everything since the prof gets the adjoint by some form of integration by parts

@Thorgott physics

The real $\frac{\partial}{\partial x}$ is the friends you made along the way

and your velocity

I'm in a symplectic mood...

@AlessandroCodenotti What if, like, you took the subspace R x Q^infty-1 of R^infty and gave it subspace topology?

@BalarkaSen how does that have finite dimension?

>0

Q has 0 dimension, R has 1 dimension

You need to argue because it's not a product

R^infty here is topologized as direct limit of R^n's by the way

@MikeMiller I don't quite follow

Does everybody here like "dark metal" music?

@BalarkaSen thought the $\infty$ was on the other guy

dyslexia + no dollar signs :/

That was actually a typo originally lol

So my fault

oh lol ok I didn't imagine it

It seems to me that compact subsets are empty interior because of the same reason that R x {0} has empty interior in R x Q under subspace topology from R^2

You have accumulating lines coming at you from infinitely many dimensions

But to be compact you can only afford to take finitely many into account

@geocalc33 Do you like "dark metal" music?

I respect it but I don't listen to it @user2103480

oh

Maybe you mean black metal

Then I guess not everybody here likes it

yeah

Black metal ist Krieg

the most logician answer smh

Hmm that doesn't look very Polish though

couldn't help

@AlessandroCodenotti yeah maybe it isn't. I said the idea anyway, hopefully you can modify

@BalarkaSen more like nerd music

If one more mathematician tells about superior metal music with 27/11 rhythms imma flip

I just finessed a sick track

7/11 polyrythms with a missing 5th

@BalarkaSen oh wait what does this mean again

open iff intersection with each R^n open

colimit

its the obvious thing

true

@BalarkaSen "musics gotta be complex how else will I enjoy it"

obvious topology for the obvious CW structure

@user2103480 edm chad

@BalarkaSen EDM? too complex man

progressive house

tropical house

meet me vibing to some 11 minute pure noise track

Noise is good

Merzbow intensifies

hiphop is sometimes good

Sunn O))) intensifies

every-other-beat-removed tracks

@AlessandroCodenotti I think I categorically defeated you

Sunn O))) is like Lana del Rey compared to Merzbow

I guess Sunn O))) are drone not noise anyway

Brrrrrrrrrrzzzzzoiiiiinggggbrrrrrrrrr

BrrrrrrrrrrDRRRRRRRRRbrrrrrrr

This is the opening track of monoliths and dimensions

actual lyrics

better delete that the title is too macabre lmao

My favorite Sunn O))) project though is Scott Walker + Sunn O))), "Soused"

Me and Ed make music on soundcloud for fun

we are so underrated

(some of the residents play drone fairly often according to their own statements)

not that I've ever gotten in lmao

You gotta listen to the REAL /MU/SIC man

@Thorgott If you choose the connected-sum disc along the equator (and your three trivializations to coincide on the connected-sum disc), then you can identify the clutching function (transition function along equator between north pole trivialization and south pole trivialization) of the connected sum as being the clutching function on the LHS and the clutching function on the RHS.

Not EDM which is just 5 notes and two beat pattern over a Synth

@BalarkaSen the forbidden forum

Guess I will not need all the metallic music ...

/lit/ was, in fact, not that lit...

BRO IM about to sing on a EDM BEAT!?!?!!

Vivaldi is basically heavy metal

gotta respect edm

Summer is such a heavy track

@geocalc33 as a broad term, definitely

sure

@BalarkaSen and kraftwerk classical music

but I can see why people don't like the more poppy edm styles

@BalarkaSen n e r d

@MikeMiller ah, ok, yeah

If E has clutching function f_E in pi_{n-1} SO(k), then one can identify f_{E # E'} with f_E + f_F, the sum in the fundamental group. Or equivalently with the pullback of E vee E' by the pinch map S^2 -> S^2 v S^2.

four seasons landscaping?

My life is empty since trump's gone

News aren't morbid fun anymore

Just morbid again

I'm the president in 9 years and I will make the news great again

There will be no news

There are better irony rabbitholes than US politics

@BalarkaSen and better ways to lose brain cells

Yeah that's what I meant

@BalarkaSen Nargaroth Jahreszeiten is the best version of four seasons

Haha yeah

also I still listen to Mirror Reaper daily

Oh wait I have a video for you

@user2103480 if I make a YouTube video and post it will you click on it?

make or link?

but yes I will click on it, it's not a virus

Goniloc is a great channel by the way

well I have a soundcloud link but I don't have it on YouTube yet

@user2103480 you have to look at British news now (sorry @Ed)

@MikeMiller having the clutching function take values in GL^+ or SO doesn't make a difference since the latter is a deformation retract of the former, I wager?

That's okay, I don't identify as British

@AlessandroCodenotti Italian news are equally confusing and depressing, but less fun

also I only just meldete mich an in Mannheim two days ago

and managed not to have to pay a Strafe

ah edward you missed this one since I deleted it, but as a br*tish person you at least get the reference

woah

is this

noise

yes

@EdwardEvans this reminds me I have to go to Münster's city hall to argue with them but my will to do so is strictly negative

and the band is from the eighties so its not even like they took 20 years and then came up with the edgy name

@Alessandro I just charmed the young lady with my fine British wit

they jumped right into the burning mess

either that or she was just new and didn't know I had to pay a fine or smth

@EdwardEvans If they ever catch me I'm fooked

do you have to sing in German in Germany's music scene to be listened to?

the classic british wit

Mate I've been living in Mannheim for like 1,5 years and managed to not pay a fine

@geocalc33 no but you gotta use autotune and the syllables "lelele"

@EdwardEvans 5 years in cologne

bish

damn

okay I use some auto-enhancer

the trick is not to go to the Bürgeramt in the Innenstadt

I'm still hoping that I just never existed there when I leave the flat

tbh you could just wait until you move

lmao

and then dich anmelden

@EdwardEvans yeh thats the plan

is your accent a mix of british and austrian edward

nah my accent when speaking English is English and when speaking German it's Austrian

then it was your ... austrian charm?

Yeah it was my austrian farmer charm

Weird words, together in one sentence

I was like "und denn sind d'Kühe vum Berg abeku und mir hond zemmat milch trunka weil ma des tuat do wo i herkum"

I LOVE Franzl Lang

edward on his way to the ausländerbehörde

@Thorgott Yeah, same thing

Franzl Lang is the MAN

jolly bloke

Der Heini in that video is definitely 5 Krüge deep

and magnificent lad

haha

@user2103480 what a classic