Mathematics

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

skull petrol

6:10 PM

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.

ramanujan_dirac

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

6:21 PM

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

6:31 PM

@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

6:35 PM

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

6:53 PM

@Alucard Found anything ?

ccorn

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!

Captain Giraffe

7:09 PM

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

ccorn

@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.

Captain Giraffe

@ccorn How is the 2017 related?

TheGreatDuck

7:23 PM

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

Captain Giraffe

Thanks.

ccorn

@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 ?

James Groon

7:42 PM

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?

James Groon

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.

Semiclassical

7:53 PM

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?

Semiclassical

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?

8:03 PM

sure

Oh, neat

so, like scalar fields over closed "rectangles"?

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

s.harp

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

8:09 PM

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.

8:13 PM

@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?

8:22 PM

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

8:32 PM

What are we doing ? :) @Hippalectryon

1 hour ago, by TheGreatDuck

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.

1 hour ago, by TheGreatDuck

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

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 :)

8:35 PM

@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