How was dinner?

fine

@BalarkaSen

So here is what I am thinking so far

I think we can construct a lift from $(X,x)$ to $(\bar{Y},\bar{y})$

I am pretty sure I can do that but let us forget about that for now

I want to talk about uniqueness okay ?

OK

Uniqueness is just uniqueness of lifts

how does uniqueness of lift imply uniqueness of a map which makes the outer square commute?

there cud be two maps which make the outer diagram commute but just one of them would commute with the lift

the question asks for a weaker condition right?

$\tilde{f}$ is just given by $g \circ p$ where $g$ is the lift.

If $g$ is unique that's unique

I don't see what's your problem

Unique? As in one and only one?

yes, unique for a given $f$

I mean if we are given let us say $g : (\bar{X},\bar{x}) \rightarrow (\bar{Y},\bar{y})$

which makes the diagram commutes

then why is it unique ?

I defined the $\bar{f} := h \circ p$ where h is the lift of f.

@BalarkaSen ?

@Adeek If $g : \widetilde{X} \to \widetilde{Y}$ is given then I claim $g$ factors through $p$, because $g = q \circ f \circ p$, so $g(x) = g(y)$ if $x, y$ lie above the same fiber of $p$. By universal theorem, claim is proved. Then $g = h \circ p$ where $h : X \to \widetilde{Y}$ is a map. By commutativity, $h$ has to be a lift of $f$ - invoke uniqueness.

There. I just literally told you the solution.

how did you get $g = q \circ f \circ p$ ?

Meant $g = q^{-1} \circ f \circ p$. Typo.

Wait a sec, that's nonsense.

$q^{-1}$ need not be well-defined!

That's why I said it's nonsense.

@Adeek OK. Write $h : X \to \tilde{Y}$ to be the lift of $f$. Then $g = h \circ p$, I believe. Do you agree?

yeah

@Hippalectryon hi, i have a book

@Alucard :-) nice username

Thank you very much

I'm sorry but I'm not sure where/when we've met before >.> haven't been on SE a lot lately @Alucard

@Adeek Well, $g = h \circ p$ is what you have to prove, because you defined $\tilde{f}$ by that, and you want to prove $\tilde{f} = g$.

yeah

@Hippalectryon that's ok, i don't even know what time it is, even tho i could look, but i don't feel like it. walked a lot today.

We know $q \circ g = f \circ p$. As $h$ is a lift of $f$, $q \circ h = f$. So $q \circ g = q \circ (h \circ p)$.

So $g$ and $h \circ p$ are both lifts of the same map $\tilde{X} \to Y$. Uniqueness of lift guarantees $g = h \circ p$ :)

oh I see

yeha right

It's a diagram chase argument. Nothing topological

A little confusing, but that's it

yeah I see thanks @BalarkaSen

@Alucard So, what's the book ?

@Hippalectryon It is about C++. Hello World and all that good Jazz...

@Hippalectryon since you hang around here in the math chat, maybe a book about math would be more appropriate?

@Alucard I haven't done C++ in a long time, Python's giving me all the tools I need already :-)

@Alucard Any book is fine

@Hippalectryon oh, ok I look what i have in my room, maybe something relaxing, like fantasy

@Alucard Found anything ?

Nice problem there. First I thought "Yet another small numbers coincidence" and looked for a counterexample. After searching for a while with Pari/GP without counterexamples, I looked closer and found a proof for a special case (where $H$ contains $p-1$), but not for the general statement. Sangchul Lee has found the general proof. Congratulations!

If I may a silly question. Why does this equal to $pi$ pbs.twimg.com/media/C8X8GJOUQAEGsQ0.jpg . I can't relate it to the Leibniz formula.

@Hippalectryon I got the "QUR'AN" in german, with a white cover

@Alucard Damn I can't read german

@Hippalectryon neither can i, it's too sick for me

but i like the cover, it's glitters

@CaptainGiraffe: Partition the interval into $(0,1)$ and $(1,\infty)$. Substitute $x\to\frac{1}{x}$ in the latter to turn the second interval into $(0,1)$ again, then combine the two parts again. You will find that the $dx/(1+x^2)$ remains, and the other parts combine to 1.

@ccorn How is the 2017 related?

Just throwing a random idea out there for someone to latch onto. Matrices are grids of numbers. What if we instead thought out matrices as continuous grids.... i.e. Ranges of functions.

a 3x3 matrix is a function of two variables spanning 0-3 and 0-3

rwgulsr matrices are just step functions

now then... how does one interpret the "determinant"

?

@CaptainGiraffe It's not, nor is the pi on the LHS. Put any positive power (might have to be >1) and it works

Thanks.

@CaptainGiraffe: Not at all. That's the point: The second factor could be any $(1 + x^r)$, as long as the integral converges.

hi @BalarkaSen still around ?

Ever asked urself ;: " Am I a genius ? " . I can answer it for you ; " No , you're not ".

@TheGreatDuck I'm not sure what you mean. Please elaborate?

Just like there is no insane person that can ask himself ; " Am I crazy ? "

@JamesGroon The answer doesn't make anyone happy. That's why.

@Adeek Yes.

Not for long though.

@MeowMix I believe he means, instead of a standard 3 x 3 matrix, imagine a scalar field defined over all of $[1,3]\times[1,3]$. This "extension" is not meaningful I think.

I don't see a meaningful way to generalize the determinant in that scenario.

Oh...

Yeah, me neither

Perhaps that could represent a linear map between uncountably infinite dimensional vector spaces?

For instance, how would you define the antisymmetry?

Well, I don't even know if that's possible.

@Brody given integrable functions $A:[a,b]\times[c,d]\to \Bbb R$ and $B:[c,d]\times[e,f]\to\Bbb R$ we have a "product" $AB:[a,b]\times[e,f]\to\Bbb R$ given by $AB(x,z)=\int_c^d A(x,y)B(y,z)\,{\rm d}y$.

also matrix multiplication of vectors may be generalized using kernels

if you get a Hilbert basis for some nice space of function on which they act, you can define the spectrum (eigenvalues), and then get a zeta-regularized determinant

@arctictern Hmm. This is mostly above me. So does one index the rows/columns of a matrix by, say, the reals?

sure

Oh, neat

so, like scalar fields over closed "rectangles"?

And we could generalize tensors to products of $k$ closed intervals of $\Bbb R$?

Anybody familiar with interpolation spaces in functional analysis? I'm interested in this question math.stackexchange.com/questions/2213592/…

@TheGreatDuck I believe it would basically be the same as the determinant of infinite matrices, with a change of axis, $\mathbb{Z}\to\mathbb{Q}\cap[0,1]$

@MeowMix leave tensor products out of it

i was only wonderingggggg

@Hippalectryon ?

@arctictern I'm saying that having a "continuous" matrix on, say, $[0,3]\times[0,3]$ is the same as taking an infinite matrix (i.e. a matrix with infinitely many rows and culumns) and making its indexes fit in $[0,3]\times[0,3]$ instead of $\mathbb{Z}\times\mathbb{Z}$

no

@Hippalectryon i am impressed, hats off Sir

$\Bbb Z\times\Bbb Z$ and $[0,3]\times[0,3]$ are different measure spaces. aren't even the same cardinality.

@arctictern Yeah that's why I said $\cap\mathbb{Q}$ before, sorry I forgot that the second time >.>

then you're no longer talking about real intervals, which is presumably what TGD was talking about, and using the word "continuous" can only be misleading

While it's true it won't be real intervals, it's perfectly fine to define continuous functions on $\mathbb{Q}$

and you think TGD (and subsequently Meow and Brody) was talking about continuous functions on $\Bbb Q$ rather than $\Bbb R$ why?

Well the domain was never specified, and it doesn't make much sense imo to have a determinant for a matrix defined over a real grid, so I just discarded that possibility. I could totally be entirely wrong though :-)

when talking about intervals and continuity, that is by default reals unless otherwise specified. if you think the determinant didn't make sense should have said that to begin with. (and you need a "cofinite support" hypothesis on your functions to make sense of the determinant anyway)

I'm assuming you're using the Leibniz formula for determinant, or expansion by minors, etc. also, on $\Bbb Z\times\Bbb Z$, you can define a sign function, how would you do that on $(\Bbb Q\cap I)^2$ where $I$ is an interval?

Hm my initial argument was much more simple, but I probably wasn't clear. Basically the way I'm approaching this is by approaching the idea of a "continuous" matrix by recursively splitting a finite matrix until its values converge to the function's, assuming arbitrarily that the "edges" of the matrix correspond to the extremas of the finite interval where the function is studied.

That way, the determinant of the "continuous" matrix would be akin to that of a normal infinite matrix.

no idea what that's supposed to mean

Hm sorry I do realize I'm probably not very clear, but the idea behind it simple really. Let me do a drawing

Hi chat

Hi @Astyx

:]

@arctictern ^ sorry for the quality

What are we doing ? :) @Hippalectryon

Hi @Astyx

Hi @Brody, long time no see !

that seems really silly, and in no way implied at all by TGD

@Astyx I suppose it's been. Good to see ya :)

@arctictern That's how I interpreted it though. Maybe I'm just weird :P

Good to see you too @Brody

@Hipplectryon @arctictern Is the determinant of a endomorphism on an infinite dimensionnal space defined ?

no, but they may have zeta-regularized determinants depending on their spectrum

(examples are given on wikipedia)

@Astyx I don't know, but I was thinking of "infinite matrices" as limits of finite ones (like described in the beautiful image above) :P. Which of course would only exist provided the matrice's properties do have a limit

@TedShifrin Okay, so the values of any old, everywhere continuous function on a dense subset tell us the values of the function everywhere. To qualify, continuity over a dense subset alone guarantees nothing, as with Thomae's function. Right?

Oh, that's a coincidence.

Right, @Brody. I stipulated to start with that the function was (everywhere) continuous.

What is $\Bbb Ca P^2$ supposed to denote?!

(some kind of projective space)

It appears on page 180 of Besse's book on Einstein manifolds

being equal to $F_4/\text{Spin}(9)$

Cayley projective plane @Danu

Aaah, Cayley!

Well, Roger pulled it off. Now he can rest for a few months.

So $\Bbb O$ in modern notation?

Right.

@TedShifrin What a god damn legend, huh

I don't think we'll see someone as great in the coming decades

This match wasn't even that close either

Well, until he started to crash and burn it looked like Novak might catch him.

Oh it was pretty close.

Yeah, but Novak got friggin rekt over the past 9 months

@TedShifrin It was, but not Kyrgios-close!

Yeah, because Nadal doesn't serve like Kyrgios.

Yeah, that's right. That was the main differnce

I also feel like Nadal was a little loose on the forehand today

Bunch of unnecessary misses

Well, just like Roger used to miss when he shouldn't have ... those guys pressure each other unbelievably.

Yeah, maybe it's just the pressure. Roger had some really weird shots too.

I got lucky and found some super useful references for my thesis

Now I just need to decode their terminology

Lie algebra stuff involved...

But they give a flag description of those $G_2$-homogeneous spaces I'm studying

Well, that's definitely an important book for you to read. Plus probably some of Bryant's stuff.

Ah, that makes sense.

I'm tired of being sick :(

But I think in the end it's very geometrical: You can really "see" the subspaces :)

@TedShifrin you mean Besse?

Yeah, I meant Besse.

Mhm... I struggle a little. It's quite high-level for me.

Welcome to life :)

Also, just the fact that about 80% of the claims are not proven...

Often even without reference

Oh really?

Yeah, of course

I thought there were tons of references. I haven't looked at it in years.

Yes, tons!

But tons$^2$ things they claim!

Well, I won't challenge you to establish that claim.

My cardinal number skills aren't good enough.

I can give you plenty of examples

I still don't really know what cardinal numbers are... Not that I really care though...

They're what you use to count cardinal sins.

bwhaha

Or stupidities of politicians. ... Oh, never mind.

I'm gonna need me plenty of those

Oh lord, have I sinned...

If DogAteMy had returned my eyes, I'd roll a bunch.

@TedShifrin Does this seem right?

Wouldn't want it any other way

lol, extraordinary cohomology

what's funny

Mike, you think I know extraordinary anything? ... I did take a course 40 years ago on generalized cohomology theories, but ...

That's a cute name

Balarka, it's past your un-sleep time.

@TedShifrin Right. Thanks for the tidbit.

The word "extraordinary" is one you should never "claim" for a theory: What if something more extraordinary comes along next?!

@Danu: I think you can substitute "generalized."

Mhm...

What do they do again? No trivial homology for a point?

Yes

Like cobordism

@Danu then it's 2-extraordinary

These days I get lost in ordinary.

There is something poetic here

Hi, tern

I just can't quite find the wording

hi

My grandfather made some beautiful poems about losing touch with reality in his last years.

Regrettably, they're in Dutch so there's not much point in posting them here.

Luckily for you all, I've never been worth a damn as a poet.

words are too hard

hence: zang tumb tumb

I can write words. It's poems I can't write.

Maybe I should reread the Love Song of J. Alfred Prufrock.

Eliot is great.

My dad actually set a few of his poems to music, if I berember correctly.

on the note of losing touch with reality i like agape agape

I sometimes wish I'd make more time to read novels.

But since I don't, I guess I don't really mean it :(

@TedShifrin That's cool.

I should try to read Tlön, Uqbar, Orbis Tertius at some point soon

i just downloaded agape agape and in this horrible formatting it's only 20 pages

@Balarka: I think I was wrong. I just did research. There were poems by Yeats and by Thomas Hardy.

Ah

Also Rilke and Gerard Manley Hopkins and others.

Hey everyone!

Heya Demonark.

Balarka, I still say it's past your non-sleep time.

How's it going?

@Ted I agree. I'm going to bed.

Night

Night :

See you @Balarka!

@TedShifrin Didn't know there was a Cheshire cat smiley

Just for you. G'night!

So, pulled any fun April fool's jokes?

Nope, but I hardly even did that when I was teaching.

Did you?

I guess the humor escapes me.

Lol, fair

@Daminark Nice

Need quick verification for a simple limit. Clearly, $\lim_{x\to 2} (x^2+5x-2)=12$. Would $0<|x-2|<\min(\frac{\varepsilon}{2},1)$ guarantee $|(x^2+5x-2)-12|<\varepsilon$?

Whoa. Where did you get $\varepsilon/2$?

Typo. $\varepsilon /10$

Oh. That's a lot more correct. :)

Phew, ty

:)

Spivak did something kinda funny in solutions

Really?

As in the text, we have $|x^2-4|<\varepsilon\text{ for }|x-2|<\delta=\min(\varepsilon /2\cdot 2+1,1)=(\varepsilon / 5,1)$ so that $|x^2-4|<\frac{\varepsilon}{2}\text{ for }|x-2|<\bar{\delta}=\min(\varepsilon /10,1)$, and we also have $|5x-10|=5|x-2|<\frac{\varepsilon}{2}\text{ for }|x-2|<\frac{\varepsilon}{10}$.

Then Spivak concludes... So for $|x-2| < \bar{\delta}$ we have $|x^2+5x-12|=|(x^2-4)+(5x-10)|\le \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$.

Oh, he's being slick. Sure. Just use the triangle inequality.

Except the final $\le$ should be $<$.

And you or he left out a "min" on the first line.

You just did it directly by factoring $x^2+5x=14=(x-2)(x+7)$. No biggie.

But this proof models the way you prove the limit of a sum theorem.

@TedShifrin Whoops. Right.

Ah, I see the deal now.

Are we copacetic?

Not quite. I'm looking to see what he's referencing. Otherwise, all is fine. I'll have to study this approach more.

Well, he did $\lim\limits_{x\to 2} x^2=4$ in the text and got $\delta = \min(\varepsilon/5,1)$.

Oh, another error. It should conclude with $|x^2+5x-14|$, not $-12$ oc

@SimplyBeautifulArt Just saw your answer here. Looks nice!

@AkivaWeinberger Thanks

But the idea of the proof is that if $|f(x)-\ell|<\epsilon/2$ when $0<|x-a|<\delta_1$ and $|g(x)-m|<\epsilon/2$ when $0<|x-a|<\delta_2$, then $|(f(x)+g(x))-(\ell+m)|<\epsilon$ when $0<|x-a|<\min(\delta_1,\delta_2)$.

Oh, DogAteMy has returned. I needed some of my eyes back.

and I just had the moment where I realized that using negative numbers could be very useful for creating large numbers through array stuff

סּסּסּסּסּסּסּ

@TedShifrin

They appear to have been damaged :(

@SimplyBeautifulArt ?

@TedShifrin ?

@AkivaWeinberger ?

@TedShifrin I somehow understand this general procedure better.

@SimplyBeautifulArt Oh, I see (I was confused what you were talking about here but now I get it)

:P Hope I got this

Well, he's using $f(x)=x^2$, $\ell = 4$, $g(x)=5x-2$, $m=8$. :) @Brody

The idea is to break up something complicated into manageable bites.

The sneaky part was substituting $\varepsilon/2$ for $\varepsilon$ in the known expression for $\delta$ to get $\bar\delta$.

Mmm. This is actually clever (to me at least)

The idea is useful if you wanted to do something like $\lim\limits_{x\to 0} x^2+5x + 2 + x\sin(1/x)$, @Brody.

@Daminark My history teacher said that April Fool's day is the only day he will be completely serious

(Unfortunately, this year it was a Saturday)

Serious as in not sarcastic.

Lmao @Akiva

I see you come by your snide wit with good cause, DogAteMy.

I don't think I'm as mean as he is. Or as funny

At one point, he claimed that his class was a social experiment

Well, there's always room for improvement, I suppose.

I guess you should ask him at the end whether — so far as you're concerned — the experiment was a success or a failure :P

He seems pretty entertained, which I guess is what counts as a success

I think I'm sick.

I continue to be sick, Zach, despite antibiotics :(

Not good for having surgery in a few days.

That's unfortunate... What surgery?

I don't know anyone in this part of GA who hasn't been sick in the past few weeks.

A lot of it seems to be just seasonal allergies, though, with the pollen explosion.

This respiratory/sinus thing's been going around, for sure.

Dental implant, Zach. Exciting, I know.

I had that about a week ago. Was not fun

The sickness, not the surgery

:(

hi chat

@AkivaWeinberger I was sick just last week

'Tis the season

Every Hilbert space has an orthonormal basis. Therefore, every Hilbert space has a unitary on it taking values on some $\ell^2$ space. Since $\ell^2$ is complete, every Hilbert space is complete.

Is there a shorter way to prove Hilbert spaces are complete?

It's the definition, @Aetos.

BTW, did you ever work out that locally integrable question in $\Bbb C^2$?

@TedShifrin I tried, but I am hopelessly stuck. :(

Hmm ...

"Hopelessly" has such a dire sound to it.

Did you try following up on my suggestion?

What does $dV$ look like (roughly) in spherical coordinates in 4 dimensions?

@TedShifrin LOL. I think I need a vacation, hahahaha. Indeed, it is complete by definition.

Where's vacation? :)

Hi

@mick It is an honor to meet you

Lol thanks I Guess , but Why

Because i need more training, i do now my stomach excercises, thank you bro

@TedShifrin Consider the map $T^3 \to SO(3)$ given by sending $(r, s, t)$ to $\text{rot}_1^r \text{rot}_2^s \text{rot}_3^t$, where these are rotation by the appropriate angle about the appropriate axis. Is this map degree 1?

@mick your questions are always absurdly difficult. Could you give me a break on them?

@MikeMiller As opposed to 0?

Or 7.

o_0

More likely 2 or 8.

Ah, no, the degree is zero.

Is it not surjective?

Actually you can just write out the image in terms of a 3x3 matrix explicitly

Yeah, but that doesn't make the degree obvious.

Consider this as a map $S^1 \times T^2 \to SO(3)$, defined as the product of maps from each factor to $SO(3)$ using the group structure. Because $H_2(SO(3)) = 0$, the map from $T^2$ bounds a map from some 3-manifold. So I can extend this to a map $S^1 \times Y \to SO(3)$ and hence the degree is zero.

I thought the whole point of physicists' Euler angles is such a decomposition for every rotation.

@mick only for you this little sweety lovethispic.com/uploaded_images/160875-White-Cat.jpg 1+1=2 again.

That's the argument I needed.

So there must be an orientation-swapping almost double cover going on?

Nah, it's just the fact that $H_2 = 0$, I think. Or the fact that the rational homology is zero, if you like.

But it is surjective, so a regular value must get hit +- ...

Sure. My comment is very inexplicit

I'm just wondering if we can see my comment directly ... like $f(x,y,z)=f(-x,-y,-z)$, but orientation swaps.

Not that I necessarily believe myself yet.

I was mostly interested in the more general case and using this test case, but the explicit counts would be interesting. What's in the preimage of the identity?

I wonder if this is related to the belt trick for $\pi_1(SO(3)) = \Bbb Z_2$.

No, the question is IMO no more obvious inside of $S^3$.

Ah, nevermind, it is.

I'm not quite following you yet.

Alucard have we met Before ? @Alucard

@TedShifrin So... $f(x)=x^2+5x+2,\, \ell = 2,\, g(x)=x\sin(1/x),\, m=0$ and take $\delta_1=\min(\varepsilon /6,1)$ and $\delta_2=\varepsilon /2$.

The lift of this map is to send $(r,s,t)$ to $e^{ir} \cdot je^{is} \cdot ke^{it}$. You can make the above technique about bounding discs quite explicit inside $S^3$ - find a disc that $e^{ir}$ bounds.

@mick oh yes, that was some time ago where i was Null

repl.it/GrFZ/1

Pretty cool, eh, @Brody?

@SimplyBeautifulArt im still holding back actually :)

Any Ruby programmers know if theres anything wrong with the above code?

@TedShifrin For now, just want to verify that I get it. But it is neat, thanks

@TedShifrin Seems pretty easy to show that the only preimages of the identity are (0,0,0) and (1/2,1/2,1/2).

Yeah, sure, @MikeM.

(In the SO(3) case.)

I think I might assign this to students the next time that's a thing I do. It's a really nice problem.

@Alucard an honor to meet you again :)

I'm not sure why the second one gives the identity, though.

The matrices are diagonal and each is -1 on exactly two diagonal entries.

Ah, excellent.

Being sick doesn't help my thinking.

Now we'd have to check the orientation issue. :)

You need to count orientation, still, but that's not hard, since at (1/2, 1/2, 1/2) the differential is pretty explicitly calculable.

But that should be easy.

This is a great question.

Please vote to reopen this question, it was unwarrantedly closed: math.stackexchange.com/questions/2215087/…

I'm surprised I haven't seen it before.

@SimplyBeautifulArt Not that I know ruby, but: what's the code supposed to do?

Ohhh, @Alucard is the former Null. That explains a lot.

@SteamyRoot :-)

Make me a large number is all

@TedShifrin It came from a more general but less explicit question in a calculation I was doing, so you've helped me with my work :)

*coughs innocently*

:D

@CausingUnderflowsEverywhere Do you Ruby?

huh

I didn't do much other than mumble 2 or 8 :P

accdiently joined the chat but hello so nice to be welcomed

do you do Ruby?

Now for a couple more complutations. Getting pretty close to d^2 = 0.

no I java and have basic understanding of c++

Well, the code seems trivial enough that you don't need to know Ruby specifically...

Cool @MikeM.