Oh, I see, Novikov says it's not possible to decide if a given $k$-dimensional simplicial complex is homeomorphic to $S^k$ or not, for $k \geq 5$.

Looking back at the algebra, though, what I said before isn't quite right. The max value of $|x|\delta_\tau(x)$ is 1/2, not $\tau/2$.

Interesting.

to show $\delta(ax) = \frac{\delta(x)}{|a|}$, are we using the definition of the delta function $\delta_{\tau}(x)$ as $\tau \to 0$ again @semiC

Eh, you really don't need to. Just put it inside an integral and do a $u$-substitution.

@BalarkaSen matroids are usually defined as follows: you take a finite "ground set" $E$ and a set $X$ of subsets of $E$ such that all maximal subsets are the same size and any subset of an element in $X$ is also an element of $X$.

For $k = 4$, you can actually do this. Heh. This should be a consequence of Freedman's theorem.

i.e. consider $\int_{-\infty}^\infty f(x)\delta(a x)\,dx$

there are about ten million different definitions though

Well, for manifolds. If I have a 4-manifold, I look at it's $\pi_1$. If it's nonzero, rubbish, forget it. If it's zero, invoke Freedman's theorem - you just need to classify the intersection form.

good examples of matroids are: a) take a graph $G$, and set $E = E(G)$, $X =$ anything without cycles

@BalarkaSen where do I learn about this

For that, I am willing to believe there is an algorithm. But, it's not clear to me how to do this for general 4-complexes.

@0celo7 Hell if I know.

Where did you learn about it

Overhearing discussions.

you have a good memory

"intellectual osmosis"

Well, it's a surprising theorem.

b) take a matrix $M$ over an arbitrary field. Set $E$ to be the columns of $M$, and define $X$ to be the linearly independent subsets of $E$

Can't forget so easily.

in the answer to that question, what does C() mean?

sorry for just jumping in with that all of a sudden...

@SamuelYusim Ah, OK.

@Brett38 Ring of continuous functions to $\Bbb R$.

thank you!

c) take $E$ to be the 0-skeleton of an abstract simplicial complex (up to unions) with the property that all maximal simplices have the same size, and $X$ to be the set of simplices

so somehow these things generalize both graph connectedness and linear independence

there are analogues of the rank-nullity theorem for matroids in general

I see.

it's also common to see people doing projective geometry in terms of matroids

Are those used for situations like counting problems?

yeah, and other stuff too

Nice.

I only really have a surface-level understanding of matroids but I know I don't want to work with them because they're too annoying

(BTW, so Novikov's theorem says e.g. that I can't determine whether or not a 5-manifold is homeomorphic to S^5. That's an exciting result! I didn't know that)

yeah that's actually really cool

Wait a second, not quite. It says there's no algorithm when the class is all of 5-complexes. 5-manifolds is a subclass, it might be possible there.

But is it? I wonder.

Note that by Poincare conjecture one just has to check whether $M$ has the same homotopy groups as $S^5$. I.e., all zero except $\pi_5 \cong \Bbb Z$.

Homotopy groups are rather hard. I am not sure whether or not there is an algorithm for that.

@BalarkaSen what do you mean, "can't determine"

Undecidable.

possibly it's equivalent to the halting problem

that's not any clearer...it is or it isn't homeo

oh, there's no algorithm for determining it?

That's what undecidable means...

I don't speak English

apparently the problem is even undecidable for diffeomorphisms of $5$-manifolds

Oh, sure, it's not possible to decide whether two given manifolds of dim > 3 are diffeom or not.

That's weaker than Novikov's thing though.

It's essentially a consequence of the isomorphism problem for groups.

Idea is, every finitely presented group appears as fundamental group of a manifold of sufficiently high dimension. When I say sufficiently high, I mean >= 4.

whoa

You got me, thanks.

@BalarkaSen still whoa

Yes, surprising at a glance.

@Clement In your second bullet on the answer for "Conditional expectation of a product XY given Z with Y independent of Z" you don't need that X,Y, Z are pairwise independent to show the result. Neither the Rademacher densities. Just X, Y nonzero and independent Z=XY and E(X) or E(Y)=0. Yours was not exactly the answer to the question either.

@semiC For $\delta(x)$ that is, we can choose $f(x) = |x|$

@0celo7 This fact is the reason why one restricts to simply connected manifolds when studying homeomorphism problem for 4-manifolds. In which case, one gets the Freedman's theorem :)

@BalarkaSen Simply connected compact?

I don't like that one restricts to compacts all the time

Sure, sure, whenever I say manifold, I mean compact without boundary.

I guess I mean orientable here too.

@semiC actually it's pretty trivial if $\delta(x)$ follows that triangular path constructed from before. $\delta(.01*x) \geq \delta(x)$ for positive $x$, so it makes sense that $\delta(ax) = \frac{\delta(x)}{|a|}$

Huh, it seems there is a full classification of simply connected 5-manifolds upto diffeomorphism. I looked in the chat, Mike mentioned this before.

It was done in the 60's.

So maybe there is a way to decide if a 5-manifold is homeom to S^5 or not. Who knows.

@Obliv Hint: $u$ substitution

impossible

why

tell me what to substitute for in $\int_{-\infty}^{\infty}\delta(ax)f(x)dx$ for some smooth function $f(x)$

Do you know what $u$ substitution is

...

OK, I gotta go.

don't patronize me

cya @balarka o/

@BalarkaSen Cya.

@Obliv Do you want the proof

$\delta(u)$ where $u = ax$ now what

I mean if it's long I don't need it

I understand conceptually why this is true.

I might as well do it.

First note that $\delta(ax)=\delta(|a|x)$

Do you see this

no

It's an even function

yeah i see it now

Exercise: show that $f(ax)=f(|a|x)$ for any even function

c'moonn i was just joking

I see it :)

Ok, so $\int \delta(ax)f(x)\,dx=\int \delta(|a|x)f(x)\,dx$

mhm

Let $u=|a|x$, then $dx=du/|a|$

So we have $\int\frac{\delta (u)}{|a|}f(u/|a|)\,du$

$\int \frac{1}{|a|}\cdot\delta(u)f(\frac{u}{|a|})du$

yeah

SEE

The set of all sets that don't contain themselves what an idea

Well...I might have lied or you need physics logic

The problem with physics is that you never know how much bullshit is too much

well okay $\int_{-\infty}^{\infty}\delta(x)f(x)dx = f(0)$ regularly, right? So then

Give me a second

$\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(u)f(\frac{u}{|a|})du =$

Ah

I messed up

What you want is $\int \delta(x)\,dx=\int\delta(au)\,d(au)=\int \delta(|a|u)a\,du$

crap, where does the $|a|$ come in

I don't follow.

even if you prove that the integrals are the same, is that sufficient to show they are the same function

In physics, yeah.

Oh

Ah, I got it.

If $a$ is negative then $d(ax)=-a\,dx$

So you really get $d(au)=|a|\,du$.

Alright, let's take this from the top

ok

@0celo7 What? $d(ax) = adx$ regardless of whether $a$ is negative or positive.

$$f(0)=\int \delta(x)f(x)\,dx=\int \delta(au)f(au)\,d(au)$$

@BalarkaSen you have to account for the integral bounds flipping

we're always integrating from $-\infty$ to $\infty$

OK, fair.

Now...

But $d(ax) = -adx$ is craptalk, still.

With that, I'm off.

@BalarkaSen and the infinite spike isn't :P

another way to do is it always to do the substitution with $|a|$, and then just observe that $\delta(-x)=\delta(x)$

$$\int \delta(au)f(au)\,d(au)=\int \delta(au)|a|f(au)\,du$$

I'm not sure if you can do much more than handwave here, sadly.

I'm not seeing the trick

$\int\delta(au)|a|f(au)du = \int\delta(x)f(x)dx$

so now

take away the integral

divide the $a$ and f(x)

You can't take away the integral

Maybe you have to for the physicist proof

you probably can

No, you can't.

wait dude

$\int_\Bbb R$ is not invertible

Certainly not injective.

$x = au$?

$$\int f(x)\delta(ax)\,dx=\int f(u/|a|)\delta(u)\frac{du}{|a|}=\frac{f(0)}{|a|}=\int f(x)\cdot \frac{\delta(x)}{|a|}\,dx$$

what was the substitution again

@Semiclassical Yes, that's what I needed.

Physicist proof

doing the substitution as $x=u/|a|$

Oh i get it

$f(\frac{u}{|a|})$ doesn't matter

just treat it as $f(x)$ so that the integral is $f(0)$

the point is more that $x=0\implies u=0$

yeah

but yeah. regardless of how one writes it, the evaluation must give $f(0)$

in retrospect, the proof using $\delta_\tau(x)$ is pretty nice as well

@Semiclassical what

yes, at the level of proof we're talking about anyways

Taking $a>0$, note that $\delta_\tau(ax)$ would by construction 1) have max value of $1/\tau$, 2) vanish when $|ax|>\tau$ aka $|x|> \tau/a$.

@Obliv Or, $\int \delta(x)\,dx=\int\delta(ua)|a|\,du$, which shows that $\delta(ax)$ has area $1/|a|$

So it's got an area of $1/a$ instead of $1$. So $\delta_\tau(ax)/a$ has area of 1.

And it's a spike, too

so it should be proportional to $\delta(x)$

@Obliv But we've beaten this to death

Read on

on it

Well, it has to converge to the same functional as $\delta_\tau(x)$ does since they've equal area.

But yeah.

I'm actually having to deal with some amount of delta function stuff right now, namely in the realm of Green's functions. haven't done that in a while, so i'm having to refresh my memory re: kernels and various boundary conditions.

a change of variable for the orbit of $r(\theta)$ by $u = \frac{1}{r}$ produces this orbit modeled by $\frac{1}{2\dot{u^2}} + \frac{l^2u^2}{2} - GMu = \epsilon$

where can I get the solution manual for zee @0celo7

@0celo7 I don't even know what you're trying to prove, but just wave your hands with enough convinction

good evening everybody by the way

hello @alessandro

@Obliv: so how come you've studied algebra as a physics major

@Obliv I don't have it.

@huy well I disliked the idea of using mathematical objects without looking further into the motivation of the use of these objects as well as the construction of the objects themselves. Specifically, when our class developed notation for vectors and 'cross/dot' products, I decided to learn a bit more about this. I tried to learn some linear algebra to get familiar with the math in vector spaces etc

@Obliv: did you not learn about cross/dot products in high school?

but then I learned about abstract algebra and that there are more than just vector spaces as abstract algebraic objects. So i dug myself into an algebra book for a while. But, it's so vast and dense that I wouldn't be able to get through it all haha

no

@0celo7 Do you have your notes at least?

may I ask what country you went to high school in?

U.S.

interesting

what algebra book did you dig yourself into?

@huy I'm aware that the schools you teach at are far more advanced

For high school, anyway.

Abstract algebra by dummit & foote

ah, ok. I'm sure you could get through it if you had a lot of free time and tried really hard :)

That's interesting, in Italy people studying physics have to take quite a few math courses

same here, like real analysis, complex analysis, linear algebra

Yep, those 3 are mandatory, I think there also some statistics and maybe some differential geometry, but I would have to check

are we still talking about high school?

no, that's university level

Hi @Alessandro

:)

I was referring to people studying physics in a university

@Obliv no

Good evening @isaac9A

@theoGR You don't need it, but I gave it as a concrete example. The point of my answer is to show that there is no general simplification, by giving 3 examples leading to different (and non-equal) simplifications.

@0celo7 Alright, he said we'd use this result later anyway so I'll figure out eventually if it's correct (I'm pretty sure it's a simple substitution of variables so I can't imagine how I'd get it wrong)

@Obliv: so how come you study physics if you dislike the idea of using mathematical objects without having a bit of a background on them?

@Obliv There might be a Fourier analysis proof that's more convincing

@0celo7 I'm talking about #2 about the orbit not the one we just did with semic

oh

@huy I would hope most physicists try to understand the background of the math they use in modelling nature. I'm not that concerned with math anyway. I see it as a language of abstract logic. I'd prefer to use it for something.

@Clement OK but I think you were also expanding beyond the original question

@Obliv The wiki article has a full derivation.

I'll look at it later. I have to go to class :[

anyone here can help me clear a couple of basic doubts about ìè

o90

that was my cat jumping on the keyboard by the way

anyone here can help me clear a couple of basic doubts about an exercise in mathematical logic? About first-order theories to be more precise

Now there is a different life after reinstalling some stuff here. It was a ridiculous situation, I couldn't even edit (the messages) or see my icon on the right panel.

hehe, cool, all is back to normal!

@robjohn sorry for the delay, I had some issues with my connectivity. It's about some hybrid stuff consisted of integrals and series, very complex and very hard to calculate (and the series cannot be calculated separately, because I suppose no one knows how to do that at the moment - that's based on my knowledge).

Overall, research is absolutely amazing these days!

(and it has to be like that!!!)

You're crying wolf

And that's not helpful

@Krijn I'd appreciate when you address to me to use @1618033. So did you address to me?

Yes. @1618033

@Krijn Well, try to keep your remarks for you, I'm not interested. Besides, my saying above is specifically addressed to robjohn.

It's okay if you're not interested @1618033. However, that's not a reason to keep my remarks to myself.

@Krijn Then, I remember that you did in the past some horrible mistakes while evaluating an elementary integral. Did you improve in the meantime?

Folks, quick question

Is this meant as an ad hominem or as an interest in my mathematics?

I have two linear equations in three variables

Hit us!

$ax+by+cz=A$

$dx+ey+fz=B$

and I need a solution that's as symmetric as possible in both equations and all three variables

I'm sure there's a neat formula using cross products or whatever, but I can't find it

You can assume the planes are nonparallel

@Krijn It means that it would be better for you, since you are on a math channel, and I didn't address to you (and as this is already known), you might do something to improve your math in integration area better than commenting messages that are not addressed to you.

@user1618033 I will take that advice with me and I see that you imply that I need to mute you. That is okay, although I am disappointed it has to be that way.

@EmilioPisanty Does this math.stackexchange.com/questions/232715/… help?

lol cmon guys this is a math stack exchange, not a dick measuring stack exchange

@Krijn let me tell you one thing. You're not very different from others I had such discussions here with. You see, I'm highly optimistic, enthusiastic about my math and about everything I do around it, but if you're bored and you cannot imagine I simply stated real fact, trying to spread the same joy, I cannot convince you and I do not want to convince you of anything. Just admit that some people had an amazing amount of joy and satisfaction daily while doing math.

@Krijn Sort of. I already have the kernel of the homogeneous equations as the cross product of the normal

I'm looking for a single symmetric particular solution

And it's also very true some do an insane amount of research and have amazing results daily (well, at least amazing for them - not claiming to be amazing for the whole world)

@EmilioPisanty You will get a solution of the form $v + wt$ where $v,w \in \mathbb{R}^3$. What kind of symmetry are you looking for?

@Krijn I suppose you're just a kid, and then I let things that way. Over years you'll see who is actually right. ;)

@Krijn Something that doesn't single out any of $x,y,z$ in the expression for $v$ as a function of $a,...,f$.

Whoa what drama is this

Back to my research (these discussions are tiresome these days for the simple fact that I work very hard and a lot lately)

@0celo7 No drama, some are simply bored, that's all.

Hi @Krijn

Math chat produces an unusual amount of drama.

@user1618033: aren't you a girl? :>

My mood is too good for drama tonight.

@Huy :-)

@Krijn How about a tragedy? Black humor?

@BalarkaSen Oh, I can always enjoy a good tragedy; it's why I follow the election of the next president

@Krijn try Macbeth

I read that

I read Othello. Eh.

When shall we three meet again?

'Twas alright.

@Huy Tomorrow and tomorrow and tomorrow.

@BalarkaSen: I've actually never read Othello. One Shakespeare was enough for me.

BBL

Everyday's the same day.

@Huy Understandably.

I need to read Hamlet soon, I think

I also know the Shall I compare thee to a summer's day poem

I'm going to watch a movie today. Hmm. What to watch.

well I and Mike would recommend Lost Highway

Seen that.

Did I never mention that to you?

have you? didn't you read the summary and end up "not liking" it?

Yes, but I saw it on a whim.

ok

no, last time we spoke about it, what I said happened

did you share Mike's and my opinion of it?

here

welll

I agree Lynch's a genius

why did you feel somewhat sick?

but the plot is sorta crazy

peeking into a gutter of a mind is not an exciting experience

all of his movies are sorta crazy

@BalarkaSen probably because the people in it are all awkward

Lost Highway is awesome, I saw it in a local cinema, they also called the writer of a book about Lynch's cinema for some discussion, I didn't share her opinion though

@BalarkaSen You should read Notes from Underground

@Alessandro: glad you liked it. I was forced to watch it as a homework for my music course and for the first time in my life enjoyed "doing homework" for two hours. :P

I agree it's an excellent movie. But I probably won't watch it twice.

I'd say cinema is my second interest, right after math

@BalarkaSen: so what kind of movie are you in the mood for? maybe it's time for one of my amazing recommendations again. :D

@Krijn I will!

I wish my homework were like that @Huy

When will you talk about interest in dates?

I borrowed the copy from father yesterday. Will delve into it soon enough, @Krijn.

my window blinds just smashed against the wall (storm's starting) and my windows are open. I almost peed myself.

(I'm sitting directly next to the window)

BBL

Yay. :D

@Alessandro well, I did have to prepare a talk of 2 hours about its music/film score afterwards, so it wasn't just watching the movie

@Huy I don't have a particularly well-defined category - I like stuff which seriously gets one to thinking, or stuff which emotionally affects the person watching it. Maybe just tell me about some arbitrary category, or some movie.

@Huy I like the song at the beginning and the end. Very appropriate.

ah, I see @Huy I understand next to nothing concerning music, but I'm trying to learn, I picked up my old bass again recently so I've been looking into music theory in the last days (without too much success nor too much effort if I have to be completely sincere)

@BalarkaSen: one of my favourite movies is Memento. have you watched it? a newer one I really love is The Grand Budapest Hotel, which is a visual masterpiece.

yep, I have watched Memento.

@Alessandro: haha, don't worry, I don't understand anything either, but apparently, I have a good sense of judgement and that always helped me

Nolan's stuff are often quite good.

@Alessandro: the best way to get a bit of work done to me is to pick a song you really want to be able to play and then practice

also the other movies by Wes Anderson are very enjoyable in my opinion, he has a distinctive style that I really like

@BalarkaSen: yeah, I think I recommended the Batman series to you a while ago, from Nolan?

mhm

I'm sure you've seen Inception too?

those were certainly the masterpieces, but Inception and Interstellar are good too

that's what I'm doing, but most of the songs I like are waaaay above my level at the moment (that's a good incentive to practice more though), what do you play @Huy?

@Alessandro: mostly the guitar, sometimes the piano. you can always check out covers on YouTube, you'll find many alternative arrangements which might be simpler to start with

@BalarkaSen: just out of curiosity, what did you think about Slumdog Millionaire in case you've seen it?

I haven't seen it, honestly.

Only heard of it.

ok. I found it terribly overrated and couldn't understand the general liking of it

@Alessandro: what songs are we talking about, btw? also, have you seen Whiplash?

@Huy i don't expect anything from it

I did and found it very good, actually my cousin has a projector in his flat so we set up some kind of home cinema and that was the film we inaugurated it with! A drummer friend of mine criticized the realism though, but I can't judge that

@BalarkaSen: what do you think of the MCE movies if you've watched any?

MCU, you mean? Enjoyable.

@Alessandro: what part of the realism? I have quite a few close friends who play in an orchestra, and they would always tell me stories like that

Not stuff I'd put too ahead in my list of favorites.

@BalarkaSen: yes, sorry

@BalarkaSen: have you checked out the TV shows?

nope

@BalarkaSen: personally, I find Agents of SHIELD to be extremely well done, funny and emotional, but I don't know if you'll share that opinion

Mostly funky stuff with a lot of slap, but I'd like to learn how to play normally first ^^ Apart from this songs by Jamiroquai and progressive rock bands, pink floyd, rush, and the usual

@Huy I don't remember exactly, that was a while ago and I don't know much about drums (or orchestras)

re:slumdog millionare. I don't think anything remotely related to bollywood can be good. however, surprisingly there are some very amazing tollywood films which are not at all well known

mostly by Satyajit Roy and Ritwik Ghatak.

@Alessandro: I felt like while the movie was pretty good, the ending kind of opposes the message the movie/director's trying to deliver

@Huy hmm, I'll have a look

@BalarkaSen: the problem is that the first few episodes aren't that exciting, but as soon as the plot starts to build, it gets really good.

Oh, I did watch Whiplash lately

I thought it was good although I drum myself a bit and my own experience was quite different

I'll keep that in mind while watching it, @Huy. Thanks.

@Huy I really liked the ending, it would have ruined the movie imho had they shown the public applauding at the end, it was all about the student and the teacher (sorry, I can't remember names...)

@Alessandro: I mean the fact that the student manages to pull it off in the end. Chazelle made the movie to criticise the abusive way of the teacher, but the ending basically confirms that the abusive way works.

Wel, yes. But that doesn't take away the criticism

@Krijn: does for me. if it works, it works. nobody's forced to go down that path.