12:03 AM
heya @Mathein ... how's the magnum opus?

I'm working on it :)

Back to Balarka-ish hours.

1:00 AM
@RyanUnger im gonna sign up for the optional elections yeah

4 hours later…
5:04 AM
hey chat. it’s been a time

8 hours later…
1:05 PM
Good morning

@ÉricoMeloSilva dude I am already getting spam emails on my princeton account
who tf has that email

1:58 PM
Alright, I have this group $\langle x_i, i\in\mathbb{Z}\mid x_i^2=x_{i-1}x_{i+1}\rangle$ and I'm trying to determine whether $x_ix_j=x_jx_i$ or not. I'm unsure there is enough information to decide this, to be honest.

that does seem rather thin

2:14 PM
@RyanUnger lol im not

Is there such a thing as uniform convergence almost everywhere?
I guess $f_n \to f$ uniformly almost everywhere means there exists a set $E$ of measure $0$ such that $f_n \to f$ uniformly on $E^c$...right?
Is there an example of a sequence of functions $f_n$ converging pointwise a.e. to $f$ but NOT uniformly a.e. to $f$?

2 hours later…
4:05 PM
What about $x^n$ on $[0,1]$?

Hey @loch
@RyanUnger You around

@Thorgott Doesn't $x^n$ converge uniformly on $[0,1] \setminus \{1\}$?

I need either an algebraic geometer or a differential geometer
@user193319 No. The rate of convergence of $x^n$ to $0$ depends pretty explicitly on $x$.

oh god

There you are
You good with jets, right?

4:15 PM
I know what they are...
this is going to be some h-principle thing

Nah, I have a pretty garbage question. Let me spell it out.
I have a fiber bundle $p : E \to M$ where $\dim M = m$ and $\dim E = m+k$. Usually a normal person defines $J^r E$ as follows: for any point $x \in M$ look at local sections of $p$ over $x$.
For two local sections $s_1, s_2$ defined on some nbhd of $x$ with $s_1(x) = s_2(x) = y$, say $J^r_p s_1 = J^r_p s_2$ if with respect to some choice of coordinates $(x_1, \cdots, x_m)$ near $x$ and $(x_1, \cdots, x_{m+k})$ near $y$ such that $p$ is projection to first $m$ variables in these coordinates, $D^I s_1(0) = D^I s_2(0)$ for all $|I| \leq r$.
This is a coordinate-independent (chain rule) equivalence relation on local sections of $p$ defined near $x$. So let the set of equivalence classes be $J^r_x E$ which inherits a natural topology after identifying it with $J^r_0(\Bbb R^m, \Bbb R^k)$ which is space of $r$-order Taylor expansions at $0$ of functions $\Bbb R^m \to \Bbb R^k$ preserving origin.
Then declare $J^r p : J^r E \to M$ is the bundle whose fiber over $x$ is $J^r_x E$, and you can set up the transition functions etc no problem so all topology is set. This becomes an affine bundle.
@RyanUnger All good?

yeah

I propose an alternative definition which avoids coordinate bullshittery.

what on earth is J

4:31 PM
A jet bundle

Define the $r$-jet sheaf $\mathscr{J}^r_E$ to be the sheaf which assigns to every open set $U \subset M$ an $(r+1)$-tuple $(s = s_0, s_1, s_2, \cdots, s_r)$ where $s$ is a section of $p : E \to M$ over $U$, $s_1$ is a section of $dp : TE \to TU$ over $U$, $\cdots$, $s_r$ is a section of $d^r p : T^r E \to T^r U$ where $T^k X$ is the iterated $k$-fold tangent bundle of $X$, and the tuple satisfies the following commutation relation for all $0 \leq k < r$
$$\require{AMScd}\begin{CD} T^{k+1} E @>>> T^k E\\ @AAA @AAA \\ T^{k+1} U @>>> T^k U \end{CD}$$
How do I CD

@user193319 It converges uniformly on $[0,r]$ for any $r\in(0,1)$, but not on $[0,1)$, cause deleting a measure zero set won't prevent you from getting arbitrarily close to $1$ (for a non-degenerate interval has positive measure).

are you going to be taking higher order germs

The top and bottom maps are tangent bundle projections, and the left and right maps are $s_{k+1}$ and $s_k$.
@RyanUnger Well I am going to dispense with the bundle altogether and work with the sheaf, is the idea.
The presheaf is $U \mapsto \mathscr{J}^r_E(U)$ where $\mathscr{J}^r_E(U) \subset \prod_{k = 0}^r \Gamma_{T^k E}(T^k U)$ consists of all the $(r+1)$-tuples of the sort I described
It's easy to check that this is a sheaf, because basically sections of a bundle form a sheaf, and when you glue two of those $(r+1)$-tuples of the sort I describe, you still get an $(r+1)$-tuple that preserves the commutation relation
The stalk of $\mathscr{J}^r_E$ over a point $x \in M$ is clearly the same as $J^r_x E$, consisting of all possible $r$-order Taylor series expansions of sections of $E$ defined near $x$ possible.
There is no issue with this formalism, yes?

Seems fine, yeah. I'm surprised people don't think of it this way

4:43 PM
Right? It's so much more easier
You can talk about two sections being close as well, because $\prod_{k = 0}^r \Gamma_{T^k E}(T^k U)$ obviously has a topology.
This is basically all one needs
There's a sheaf homomorphism $\Gamma_E \to \mathscr{J}^r_E$ sending $s$ to $(s, ds, \cdots, d^r s)$. The image is the presheaf of holonomic sections.
(The sheafification is nearly the same, except it consists of all $(s, ds, \cdots, d^r s)$ such that $s$ will be a $C^r$-section of $E$)

@BalarkaSen there is some discussion of the relationship between the r-th order tangent bundle and the jet bundles in Kollar-Michor-Slovak
but not using sheaves

Can you tell me page no.

@BalarkaSen uhhhh actually their notation is fucked
they have $T^rM$ and $T^{(r)}M$

oh shit
whats the difference

This book is impossible to read dude
Do you know of it?

4:52 PM
Nope
Looking through it

page 121 seems relevant
somehow the jets arise as a kernel in an exact sequence of iterated tangent bundles?
i dunno man
I've tried to read this book at various times and have failed

so complicated lol
This seems like Global Calculus but on LSD

@BalarkaSen the goal of the book is to somehow classify natural transformations of bundle functors

LOL

turns out they're all Lie derivatives

4:57 PM
Oh wow

I'm sure that's wrong lmao but it's the spirit of the book
it's a legitimately good reference for basic differential geometry haha

Oh I remember you reading this once and talking about it in hbar

those parts got turned into Michor's book

Category over manifolds lmao
page 169 is where funk starts

@BalarkaSen they tease a category of Riemannian manifolds, too
I don't know if this is explored

5:03 PM
What a cool book. Michor is a cool guy. I doubt most differential geometers care about this sort of things.
Damn @Ryan

yeah it's a very cool and niche book
@BalarkaSen it's funny...if $f_i$ is a sequence with bounded Lipschitz norm, then it doesn't subconverge in the Lipschitz norm
but it does subconverge uniformly to a Lipschitz function
the same thing is true for any Holder space, you get convergence for any smaller power
quite useful

subconverge = a subsequence converges?

yeah

That's very nice.

1 hour later…
6:15 PM
What

lol
I hit keys to wake the pc from sleep

Oh lol

how I hit enter is unclear

Looks like I'm officially doing C* stuff for my thesis after all

This is how you become a crack addict
Kids: Don't do drugs

6:23 PM
D:

6:36 PM
@BalarkaSen do you see any easy way to do this
it's intuitively clear buy I can't prove it without machinery
4

Let $M \subset \mathbb{R}^d$ be a compact smooth $k$-dimensional manifold embedded in $\mathbb{R}^d$. Let $\mathcal{N}(\varepsilon)$ denote the minimal cardinal of an $\varepsilon$-cover $P$ of $M$; that is for every point $x \in M$ there exists a $p \in P$ such that $\| x - p\|_{2}<\varepsilon$....

Oh interesting @Ryan
Is it true?

I was thinking optimization problem that relates to programming and mathematics. Has any mathematician ideas for this one? python-forum.io/…

pretty sure it's true Balarka
there's a comment in Lawson-Michelson to this effect
I never sat down and wrote a rigorous proof
I am wondering if it follows from the definition of Hausdorff dimension.

Oh maybe

but then the same thing would be true for fractals
I'm not sure about that
oh god
I'm led to chapter 3 of you know what

6:48 PM
F-bomb?

yes

or the F-daddy whatever you want to call it
RIP

"spherical Hausdorff measure" comes to mind

If you end up writing an answer let me know, I'd like to see one

@BalarkaSen One way that I'm sure works is to use elliptic theory to write $M$ as a Lipschitz-small graph over a finite family of uniformly sized balls in tangent spaces
collection of graphs I mean
then prove it explicitly for small graphs

6:50 PM
@BalarkaSen a brave choice

One might need better than Lipschitz, I'm not sure.

Just say that it follows from Theorem 4.5.9, what doesn't?

@RyanUnger $M$ inherits a natural Riemannian metric from the background, and locally the induced metric and the induced length metric aren't that different, so maybe you can dispense with all of that?

The same result should be true for abstract Riemannian manifolds. Do you know how to prove it in that case?
I think there you really do need some kind of PDEs to construct good charts.
I might be way overcomplicating this.
If we define $\tilde{\mathcal H}^k_\delta$ to be the $\delta$-Hausdorff "measure" but instead of $diam(U_i)\le\delta$ we set $diam(U_i)=\delta$, does this converge to the usual Hausdorff measure as $\delta\searrow 0$?
I think so by the squeeze theorem or something.
this is a larger "measure" than $\mathcal H^k_\delta$ and that increases to $\mathcal H^k$
but then we can replace all of those $U_i$'s with balls, incurring some fixed error
like a factor of $2^k$
then argue by contradiction
I'll write this up
unless you see some obvious mistake

@RyanUnger Join the points in the $\varepsilon$-net by geodesics. The Riemannian volume is sum of the volumes of these geodesic simplices, which are each like of order $\varepsilon^k$.
So you can guess that the number of those simplices is of order $1/\varepsilon^k$
Estimates are too hard for me though

7:00 PM
ok you need comparison geometry to relate the volume to the radius
that probably does work
can you write it up?

Lol no but you should

use compactness to bound Ricci from below to use comparison

I don't know this stuff, but I can learn from what you have written

did you read my word vomit

7:02 PM
do you know what the delta-Hausdorff "measure" is

vaguely. you take infimum of sum of diameter^d over open covers with open sets each of diameter < delta?

yeah

@RyanUnger ah yeah volume of geodesic ball, that has a Taylor series expansion in terms of radius
the Ricci curvature crops up
it's all compact so you can bound w/e

oh what I'm saying isn't quite right
hmm
I conjecture that $\tilde{\mathcal H}^k_\delta\to \mathcal H^k$

convergence in what sense. as premeasures?

7:07 PM
yeah, tested on any compact set
if the Hausdorff measure is finite, that is

what's the definition of Hausdorff again? isn't it sup over delta

yeah, or lim as $\delta\to 0$
but note the difference between $\tilde{\mathcal H}^k_\delta$ and $\mathcal H^k_\delta$
we always have $\mathcal H^k_\delta\le \tilde{\mathcal H}^k_\delta$

ah your diam = delta thing

right but since $\delta\to 0$ anyway, that shouldn't matter in the limit

makes sense

7:11 PM
@BalarkaSen actually they're outer measures, not premeasures
stupid terminology

Ah OK, premeasures extend to outermeasures by taking covers by elements from the semialgebra
Read this shit a semester ago, forgot it all promptly lol

Federer doesn't use this terminology
his measures are outer measures
but then there's a sigma-algebra on which they're additive, hence actual measures
In fractal geometry, the Minkowski–Bouligand dimension, also known as Minkowski dimension or box-counting dimension, is a way of determining the fractal dimension of a set S in a Euclidean space Rn, or more generally in a metric space (X, d). It is named after the German mathematician Hermann Minkowski and the French mathematician Georges Bouligand. To calculate this dimension for a fractal S, imagine this fractal lying on an evenly spaced grid, and count how many boxes are required to cover the set. The box-counting dimension is calculated by seeing how this number changes as we make the grid...
@BalarkaSen what is this
ok but this does confirm that what I'm trying to do is wrong haha
one needs smoothness

Aha

In mathematics, Hausdorff dimension (a.k.a. fractal dimension) is a measure of roughness and/or chaos that was first introduced in 1918 by mathematician Felix Hausdorff. Applying the mathematical formula, the Hausdorff dimension of a single point is zero, of a line segment is 1, of a square is 2, and of a cube is 3. That is, for sets of points that define a smooth shape or a shape that has a small number of corners—the shapes of traditional geometry and science—the Hausdorff dimension is an integer agreeing with the usual sense of dimension, also known as the topological dimension. However, formulas...
I've never heard of this condition
@BalarkaSen ok should I nuke it

7:30 PM
Ya do it

@BalarkaSen do you believe the result if the $M$ is a $C^{2,\alpha}$ graph over a ball?

I need $C^2$ regularity to pull off Riemannian arguments, so I guess
The Ricci curvature stuff pops up like you said

this is a hard problem lol
must be way overthinking it
turns out my $\tilde{\mathcal H}^k_\delta$ is a thing already
they state this result as an easy proposition
fml
ah, it follows from the fact that it's Lipschitz invariant
nice

7:54 PM
@BalarkaSen did you know that singular homology for locally contractible spaces coincide with the sheaf cohomology of the constant sheaf $\Bbb Z$?

I'm sure he does lol
You don't even need locally contractible, locally cohomologically contractible works
each point has a nbhd with the cohomology of a point

how do you prove it?

Locally cohomologically contractible means the cochain complex gives an acyclic resolution of the constant $\Bbb Z$ sheaf.
So you can use the abstract de Rham theorem.

that means if your space is locally cohomologically contractible then your space is connected iff path connected?

if $H^1=0$ locally then this is obvious right
you don't need sheaves to prove that

8:18 PM
@BalarkaSen here
a little sloppy but the idea is right
I posted it
let me know if I need to fix it haha

3 hours later…
11:45 PM
Let $a,b \in \Bbb{R}$ be fixed, and let $n \in \Bbb{Z}$. If $[\cdot]$ denotes the greatest integer function, is it possible to bound $|[abn] - [a[bn]|$ by a constant that is independent of $n$? Are there any nice inequalities with the greatest integer function?
I am trying to show that $n \mapsto [abn]$ and $n \mapsto [a[bn]]$ are equivalent quasi-isometries of $\Bbb{Z}$...that's the motivation.