Mathematics - 2024-12-17

leslie townes

00:00

i also have one of istvan fary's books, although he is less famous because he didn't write a famous textbook

ben: you should appropriate it and sell it online for big bux

Balarka Sen

@Thorgott I think so. Say $M$ is $n$-dimensional. Consider $(M \times I) \#_p X$ for some $(n+1)$-manifold $X$, where $p \in M \times I$ is an interior point. Consider the map $(M \times I) \#_p X \to M \times I \to M$, where the first map is $(M \times I) \#_p X \to (M \times I) \vee_p X \to M \times I$, where the first map is crushing the connect sum sphere to a point and the second map is projection to first factor, and the second map is projection to first factor

I hope you don't mind my recursive writing

Ben Steffan

@leslietownes I can just get things signed by Peter Scholze for the same effect

although I'm not sure anybody outside of bonn really cares

I could sell Scholze fan merch to the new undergrads, hmm

leslie townes

maybe more money in just stealing his stuff and collecting a "reward" for locating the lost property

Ben Steffan

I smell a capitalism

imbAF

Hello people, I come from the physics site but I have a math related question

leslie townes

00:04

yes, we got to "selling someone's stuff back to them" very quickly because markets are efficient

Jakobian

I think a lot of people are into algebraic geometry and what not, the novel things, and in particular they are Scholze fans (be it if they understand his work or not)

Ben Steffan

it's at least true here locally

Thorgott

@BalarkaSen ah you're right, I was somehow looking at a picture of taking connected sum of a cylinder and a torus and it (surprisingly) isn't immediately clear, but it does work out

Ben Steffan

I recently got stood up on a study date and instead attended a talk where I learnt what a solid abelian group is

Thorgott

the opposite of a liquid abelian group, duh

Jakobian

00:06

I think its true everywhere, to a lesser extent. A weird phenomena

Ben Steffan

@Thorgott ...I'm not entirely sure whether that's actually a thing or not :)

liquid vector spaces are a thing

Balarka Sen

what is a liquid vector space?

Ben Steffan

right now I think things are either liquid, solid, or condensed

Thorgott

if liquid vector spaces are a thing, surely liquid abelian groups are too

Balarka Sen

is there also a gaseous vector space

Thorgott

00:07

has Scholze invented gas yet?

Ben Steffan

@BalarkaSen haven't the faintest. you can read about them in Scholze's papers :)))

@Thorgott people have been joking about this here for a while

and what about plasmic groups, while we're at it

imbAF

when we consider a function f and a small perturbation, which would be a small added term in mathematical language, one write:
$f(x+\epsilon)=f(x)+\epsilon f'(x)$.

Now for a functional the expression is:
$J[f(x)+\epsilon\delta(x-y)]=J[f(x)]+ \epsilon \delta(x-y)$. Can someone help me understand this expression and the logic behind it?

Ben Steffan

what's a bose-einstein condensate ring

solid, liquid (as in states of matter)
condensed (as in milk)

Thorgott

just Scholze things

Jakobian

@imbAF calculus of variations

imbAF

00:08

but why delta?

Jakobian

I imagine they are trying to get you through Euler-Lagrange formula

imbAF

Of course, I am through, but I was looking at some functionals

I like and want to have a clean mathematical writing of stuff

I don't like half-assed writings, which plagues physics

But I cannot find the correct one

Ben Steffan

@Thorgott you shouldn't feel too safe: a lot of this is happening in $\infty$-category land and may crash somewhere near you not too far in the future

although I guess I'm more prone to it

Jakobian

well, I read about this from Gelfand and I don't think he provides the appropriate mathematical setting in which to make sense of such expressions

Ben Steffan

what with people talking about "animation" of rings and stuff

Balarka Sen

00:10

@BenSteffan i pass

Thorgott

I'm a fan of animation as an art form, can't wait to learn how to do it myself :)

imbAF

@BenSteffan near 0K boson "gas" shaped as a ring because of topological monopole defects

Balarka Sen

@imbAF words conjured up by the utterly deranged

imbAF

why?

Thorgott

people here also care about Scholze's work, but moreso the actual algebra

Ben Steffan

00:12

@BalarkaSen wise choice (but for how long)

Jakobian

@BalarkaSen that's just physics

imbAF

You perform an experiment at near 0K temperature, you consider a bosonic gas, because paulis principle doesn't apply, which means all the bosons are in the same state=same energy

Thorgott

he apparently has this idea of turning a number field into a condensed stack over Berkovich space with an $\mathbb{R}_+$-action whose length $p$ orbits correspond to primes over $p$ or something

imbAF

and then because of defects in the structure, the physical configuration is similar to a ring

Balarka Sen

what is a topological monopole

Ben Steffan

00:13

@Thorgott I dare you to look me straight in the eye while saying this and not burst out laughing

Balarka Sen

@Thorgott still saner than Alain Connes

Thorgott

@Balarka but the result is nonsense, no? assume $N$ has stably trivial normal bundle, then you're asking for a normal map $f$ to simply mean $TM$ is stably trivial, right?

imbAF

@BalarkaSen simply said, its a framework where one associates a field (magnetic) configuration with a particle-like system

Thorgott

so take a manifold with stably trivial normal bundle, the trivial cobordism and take connected sum with something really ugly, the resulting space surely does not have stably trivial tangent bundle

imbAF

topology in itself applied in condensed matter physics is crazy enough

Thorgott

00:16

@BalarkaSen lol

@BenSteffan tough challenge, but I think I can do it

I've spent an entire day at an AG in Heidelberg once, you learn to persevere

imbAF

Can someone help me with my question?

Or even better , can someone write it in a more rigorous way and explain to me

how can one, make the proper jump from how we express f(x+\delta\epsilon)

to J[f(x)+\delta(x-y)]

Thorgott

I'm not equipped to explain this right now, but the keyword you should look up is "calculus of variations"

Balarka Sen

@Thorgott Hm, let's see. Normality of $f$ says $f^* \nu(M) \oplus T((M \times I) \#_p X)$ must be stably trivial. Already true along $M \times I$

Also true along $X$, because the retract map $f$ collapses $X$ to a point

imbAF

@Thorgott ok

Balarka Sen

Oh, guess not. I need $X$ to be stably trivial?

Jakobian

00:19

@imbAF if you want something rigorous then try Calculus of variations I by Giaquinta and Hildrebrandt

Thorgott

yeah, I think so, you should be left with $TX$

Jakobian

the textbook seems to derive everything formally, in a precise way

imbAF

@Jakobian wtf, I thought that was just a concept, that's a whole branch of math?

Well thanks anyway

Thorgott

if every topic that had a textbook written on it was a branch, there would be a lot of branches

Jakobian

trees in math should really be called bushes

Balarka Sen

00:23

@Thorgott Thanks! You're right, this is nonsense

imbAF

it's so painful to do physics, and in every step of the way

2 or 3 whole new maths are added

to the list of books that one needs to read to have an understanding

Jakobian

@imbAF I don't know, it might be a sub-branch of PDE/ODE theory/analysis, but I don't know much about it to tell you

imbAF

ok

Balarka Sen

And in virtue of this I also understood what the sensible thing should be. That helped a lot!

Jakobian

@imbAF minus the mathematics, that goes for every piece of science/knowledge

Thorgott

00:26

no problem, glad to have helped in spite of not really having geometric insight :)

imbAF

yeah but like, I mean in physics we take things at face value sometimes, since stuff are proven by mathematicians, but still

the more I read the less I know and the more I need to read

which is like filling a glass of water with a hole in the bottom

Jakobian

you can't know everything

thats why people specialize

imbAF

Well you need to know A LOT for QFT and idk how to do it

Jakobian

as for Euler-Lagrange, you need to ask yourself if you really need to have a proper derivation of the formula

or can you settle for hand-waving

imbAF

of course you can settle for hand-waving

But I find it quite a good thing to have a proper, rigorous, only once, derivation of the definition

and that is it

Thorgott

00:29

I remember the derivation not actually being that difficult, but, at the same time, I've also forgotten it, so who knows

Jakobian

@Thorgott did you read a formal derivation? I only read the one in Gelfand and Fomin and I remember the proof was a little bit of hand-waving

they only give you idea for why it holds iirc

Thorgott

perhaps the thing I remember doing is actually not whatever general statement that is meant here

imbAF

Actually apart from $\delta(....)$

my suspicion was correct, as in there is a correspondence

Thorgott

I just remember doing some basic calculus of variations in Riemannian geometry to characterize geodesics

imbAF

yeah I need Riemannian geometry in physics

differential geometry, Riemann geometry, group theory, complex analysis

linear algebra of course too

ModularMindset

01:11

Way to transport multigraph $\mathcal I_{\Gamma} \to \Bbb R^3$ enriched with line bundle to $\Bbb C$? Can derive meromorphic function (rational in fact) by forming $\zeta_{\mathcal I_{\Gamma}}(u)$ (Ihara zeta function for regular multigraph). Know how to associate multigraph (w/o embedding) to meromorphic function on $\Bbb C$, but can do complexification of line bundle in $\Bbb R^3$ to $\Bbb C$?

which will sit in $\Bbb C^3$ ideally

It will probably take a few months to read Hatcher and figure out how to complexify the line bundle. The more time consuming part is to verify whether $\mathrm{Hol}(\mathcal V_{\mathcal I_{\Gamma}})$ is dual to the poles/zeros of $\zeta_{\mathcal I_{\Gamma}}$

I know probably not a popular opinion

Whether the fixed points of the holonomy group of the line bundle are dual to the... etc.

and this is very clearly not about the Riemann zeta function

in fact the analogous riemann hypothesis for Ihara zeta function for k-regular graphs is already proven

Jakobian

07:21

Apparently as of 2023 its still unknown if the box product of countably many real numbers can be proven to be $T_4$ in ZFC (it's $T_4$ by a theorem of Rudin if CH holds)

psie

08:13

Consider the definition of the integral of a real-valued measurable function $f:E\to\mathbb R$, namely if $\int |f|\,\mathrm{d}\mu<\infty$, then $$\int f\,\mathrm{d}\mu=\int f^+\,\mathrm{d}\mu-\int f^-\,\mathrm{d}\mu.$$The author of my book claims this definition is consistent with the one for nonnegative functions. How? For nonnegative functions, we take the supremum of integrals of simple functions $\phi$ that are $\leq f$. How can these definitions coincide?

Jakobian

perhaps it would help if you thought about this

leslie townes

08:33

psie please note that this isn't "the" definition of the integral, it's "a" definition of the integral, under a presumption that one has already defined int [blah] dmu for nonnegative [blah], and this "definition" is only a definition of one in terms of the other, i.e. it assumes that int [blah] dmu has an understood meaning for nonnegative [blah], as a kind of black box

if f happens to be nonnegative, see jakobian's remark

Ryder Rude

09:46

Malament–Hogarth spacetime

A Malament–Hogarth (M-H) spacetime, named after David B. Malament and Mark Hogarth, is a relativistic spacetime that possesses the following property: there exists a worldline λ {\displaystyle \lambda } and an event p such that all events along λ {\displaystyle \lambda } are a finite interval in the past of p, but the proper time along λ {\displaystyle \lambda } is infinite. The event p is known as an M-H event. The boundary between events with the M-H property and...

i think, if spacetime were like this, one could explicitly verify things like the twin prime conjecture without having to prove it from axioms

Xander Henderson

11:44

@RyderRude and all that is required is that you cross the event horizon of a black hole and then survive beyond the heat death of the universe!

Thorgott

@XanderHenderson is this one of those "back in my days, to get to school I had to..." bits

3

nickbros123

12:17

lol

@RyderRude this seems interesting

time to reinvigorate my interest in physics

before that let me see the definition of reinvigorate

Ryder Rude

12:59

@XanderHenderson lolol

@Thorgott xD

@nickbros123 one must be careful with rigor

i also found this paper sciencedirect.com/science/article/abs/pii/S0096300305008398

i think Turing's initial idea assumed infinite time computations couldn't be done and that there was no true randomness

but the former seems to be ruled out by Relativity and the latter by Quantum Mechanics

so it seems the physical world is more than a computer. it is a hypercomputer

Xander Henderson

14:03

@RyderRude I don't think that it is unreasonable to assume that infinite computations can't be done. There is no practical or reasonable way of doing them.

Ryder Rude

@XanderHenderson it is not unreasonable, yes

but with our understanding of time modified, we should look out for potential changes in our understanding of computability

time was a big part of the notion of computability

Xander Henderson

@RyderRude From a mathematical point of view, this doesn't matter. Turing already could have assumed infinite time---the mathematical tools for that have existed for centuries.

He didn't, because that doesn't match any practical part of the real world.

Again, what you are proposing send to require crossing the event horizon of a black hole and surviving the heat death of the universe. And Hawking decay would have to stop being a thing.

Ryder Rude

14:30

@XanderHenderson i just think that hypotheses like Church Turing theses are about physics as much as they r about math. e.g. this paper sciencedirect.com/science/article/abs/pii/S0096300305008398

like, these hypotheses deal with physical limitations

@XanderHenderson it is fantasy for now. but I think the relevant part is that our understanding of time has been modified. this is one of the things that could lead to hypercomputability. physics does not have to stay computable. i think future quantum gravity theories could offer something beyond

there also seems to be true randomness in the universe which is non computable

@XanderHenderson i agree. it remains a reasonable assumption

Michael

15:15

What are some math jobs that make heavy use of calculus?

Besides any programming profession

(I quit programming after 10 years)

leslie townes

michael: other than teaching calculus? :)

think_meaning_buildß

ask at NASA

Jakobian

15:46

What are some math jobs

teaching math. And?

leslie townes

if you do it right, even teaching calculus doesn't make 'heavy use' of calculus

ModularMindset

algorithmic trader

quantitative analyst

but on second thought those probably use more of markov chains and statistical analysis

think_meaning_buildß

@leslietownes wdym

ModularMindset

Calculus Artist - someone who uses heavy calculus to make artwork?

mathematical artwork is a big thing

leslie townes

think: i am mostly being facetious but in my experience calculus is one of the easiest classes to prepare for because anyone qualified to teach it has seen literally everything that can come up in the class

and the only people filling page after page with calculus are students who are about to maybe get a D in it

think_meaning_buildß

15:55

i see

leslie townes

the non facetious version of my comment is that people who are good at calculus are often able to use less paper than people who aren't, hence, the more you know, the less 'heavy use' you make of the techniques

experience makes you more efficient at seeing what is necessary and what isn't

think_meaning_buildß

sure, it builds upon itself

ModularMindset

can you transport a holonomy group of a line bundle to complex space?

Its possible to complexify a line bundle (say its embedded in R^3) to a $\Bbb C$ one, which should be sitting in C^3

Xander Henderson

I think that it is kind of weird to expect there to be jobs which "make heavy use of calculus". Everything that is taught in an undergraduate calculus class is done pretty well by computers. Professionals rarely get their hands dirty with those tools directly. The goal is to be "good enough" to check the inputs and outputs.

No one "does calculus" in the professional world.

I mean, I'm an analyst, and I try to avoid "doing calculus" whenever I can.

ModularMindset

But how does the jibe with Hol(line bundle) extension?

If we assume it is so - the question becomes how do we interpret the complexified holonomy group of the bundle and what conditions ensure an embedding of the bundle into complex space?

nickbros123

16:12

there's some calculus in ML like maybe Lagrange multipliers etc, basic stuff (ML can, however, go higher than calculus , like topology, geometry etc), but the money is in the tensorflow, scikitlearn, keras, etc, so while many algorithms use calculus for proofs / derivations, no one is paid to do that.

Either you're paid to be good at programming- aforementioned scikitlearn, pytorch stuff etc, or you're paid (as an academic) to research new algorithms, and the math used here can go way higher than calculus (usually does). But also, idk what heavy use means

ModularMindset

Any physicists in the house?

Jakobian

unfortunately

nickbros123

damn jakobian didnt know u were a physicist

Jakobian

because its not true

ModularMindset

say you have a piecewise defined connection, that fails to be gauge equivalent because the connections fail to coincide, and are loosely speaking 'undefined'

but otoh the line bundle with that connection vanishes the points where the connection(s) are undefined

are these compatible?

and is it physical?

mo-_-

16:31

hi

think_meaning_buildß

hi

psie

16:42

If $f:E\to\mathbb R$, then we have $\left|\int f\,\mathrm{d}\mu\right|\leq\int |f|\,\mathrm{d}\mu$. If $f$ is complex-valued, my book argues that $$\left|\int f\,\mathrm{d}\mu\right|=\sup_{a\in\mathbb C,|a|=1}a\cdot \int f\,\mathrm{d}\mu=\sup_{a\in\mathbb C,|a|=1}\int a\cdot f\,\mathrm{d}\mu,$$where $a\cdot z$ denotes the Euclidean scalar product on $\mathbb C$ identified with $\mathbb R^2$.

I wonder
1) What is the author using in the first equality? Why does the second equality hold?
2) Is the final expression on the right simply $\int |f|\,\mathrm{d}\mu$?

leslie townes

what book is this? what does "sup" mean for complex numbers?

if mu is a probability measure and f is identically 1 this is an assertion that 1 = sup_{a in C, |a| = 1} a, but what is "sup" independently of this assertion

psie

@leslietownes it's this book. Previously it was established that $f\mapsto\int f\,\mathrm{d}\mu$ is a linear functional on $\mathcal L^1(E,\mathcal A,\mu)$, the space of all real-valued integrable functions. I don't know what $\sup$ means here, but we are taking the $\sup$ of a set of real numbers. It reminds me of the operator norm, but I'm not sure.

leslie townes

you can't just say "we are taking the sup of a set of real numbers" if int f dmu is 1

i'd junk the book

i suggest reading that for whatever it is worth and not reading this

i'm assuming that whatever this person means by their stuff they are just using the fact that given any complex number w there is a unimodular c so that cw = |w|

you should be focusing on concepts (which are eternal and unchanging) and not the particulars of textbooks (which are always subject to errata and revision)

Sine of the Time

16:58

nothing is eternal :(

mo-_-

17:31

Consider the vector space $\mathbb{R}^3$ and the following subspaces:
$V = \{(x, y, z) \in \mathbb{R}^3 : x - 2y = 0 \text{ and } y + 2z = 0\}, \quad
W = \{(x, y, z) \in \mathbb{R}^3 : x + y - z = 0\}$
Determine the dimension and a basis for $V + W$.

leslie townes

mo: have you done anything? e.g. computed dimensions of V and W? do you know the possible dimensions of V+W?

mo-_-

yes, im having problem typing

wait

leslie townes

the trickiest part of this type of exercise is maybe that there isn't just one way to do it, but a lot of ways to do it

mo-_-

$x - 2y = 0 \implies x = 2y, \quad y + 2z = 0 \implies y = -2z, \quad x = -4z$
$(x, y, z) = (-4z, -2z, z) = z(-4, -2, 1)$
Basis of V : $\{(-4, -2, 1)\}, \quad (\dim V = 1)$

leslie townes

if ted were here he'd say to think geometrically. V is a line through the origin. W is a plane through the origin. V+W contains both V and W and for geometric understanding reasons will therefore be either W (if V is contained in W) or all of R^3 (if V is not contained in W)

mo-_-

17:37

$x + y - z = 0 \implies x = z - y$
$(x, y, z) = t(-1, 1, 0) + s(1, 0, 1), \quad t, s \in \mathbb{R}$
Basis of $W$: $\{(-1, 1, 0), (1, 0, 1)\}$, $\quad (\dim W = 2).$

I've reached this point (if I've done it right)

leslie townes

the dimensions sound right to me (i haven't done any calculations) and the (assumed but checkable) fact that (-4, -2, 1) is in V and not in W would then imply that V+W = R^3

if you were to feed an exercise of this type into a machine, one way of framing it would be to taking as inputs two matrices A and B with the same number of columns (whose nullspaces define V and W respectively) and returning a basis for V + W by e.g. using some recipe for computing a basis for the nullspace of a matrix as a black box, putting those basis elements as rows [or columns], and then using some recipe for computing a basis for the row (or column) space of a matrix as a black box

mo-_-

@leslietownes so I just need to join the bases of V and W

leslie townes

sure, in this case you can get a basis for V+W by separately computing bases for V and W and just putting those two lists of vectors together

in general, "a basis for V together with a basis for W" will only be a spanning set for V + W, i.e. it might be linearly dependent, and you may need to do calculations to identify how you can remove vectors from that list (or otherwise modify it) without changing the span

mo-_-

so I have to check if the union is linearly independent?

$a(-4, -2, 1) + b(-1, 1, 0) + c(1, 0, 1) = (0, 0, 0)$

$\begin{cases}
-4a - b + c = 0 \\
-2a + b = 0 \\
a + c = 0
\end{cases} $

yes i get a = 0, b = 0, c = 0

The union $\{(-4, -2, 1), (-1, 1, 0), (1, 0, 1)\}$ is linearly independent. So it is a basis for $V + W$

@leslietownes right?

leslie townes

17:54

i haven't double checked the calculations but if they are correct that is a right answer

note that (1,0,0), (0,1,0), (0,0,1) or literally any other basis of R^3 would also be a right answer

i'm not sure if you 'have to' check that the union is linearly independent, there are surrounding facts that make this redundant (e.g. the fact that V intersect W is {0}, which is geometrically clear from the fact that V is a line through the origin that does not lie in W, guarantees this without calculation)

but that "without calculation" is leaning on a geometric understanding of subspaces of R^3

your calculations (if they are correct) are a more concrete verification of the algebraic fact

mo-_-

oh okay, thanks!

psie

18:49

Is it true that for $v\in\mathbb R^2$, $$|v| = \sup_{\substack{u \in \mathbb{R}^2 \\ |u| = 1}} \left\langle u, v \right\rangle$$where $\langle\cdot,\cdot\rangle$ is the Euclidean dot product? By Cauchy-Schwarz, we have $\left< u, v \right> \leq |v|$ for any $u\in\mathbb R^2$ with $|u|=1$, and equality holds for $u=v/|v|$. I just don't see why one would take the $\sup$ in the expression above for $|v|$?

Jakobian

its true\

leslie townes

psie: are you distressed about the sup not being a max? or what is the issue exactly

the sup literally is in this instance a max, but there's no reason why i would indicate one instead of the other

psie

ok, that's what I was suspecting, that the sup is a max.

Jakobian

its a statement about the linear functional $f_x:X\to \mathbb{R}$ given by $y\mapsto \langle x, y\rangle $ having norm $1$, which holds for any inner product space

leslie townes

the sup formula has meaning for any hilbert space, including hilbert spaces in which the closed unit ball is not necessarily compact

i would not expect formulas in analysis textbooks to be 'maximally specialized' to tell you as much as possible, usually they are written only to tell you whatever needs to be true to allow for what the author is using

psie

18:57

alright, the sup's more general then (as always)

Jakobian

@Jakobian norm $|x|$

its a statement about the map $x\mapsto f_x$ being an isometry

thus establishing an embedding of an inner product space into its dual

leslie townes

in particular i would not interpret a "sup" in an analysis textbook to be "a sup that is not a max," this is the exact opposite of the way you should read an analysis textbook

Jakobian

if your inner product space is a Hilbert space, then so called Riesz theorem gives you a converse, that this isometry is surjective

Zekai Chen

19:32

is Tom Dieck's proof of Proposition 5.1.10 incomplete? I could not work out why $k$ is a h-equivalece. But I found another proof which I think make more sense though.

Thorgott

20:21

it's part of the statement of Theorem 5.1.9

if you cobase change two homotopic maps along a cofibration, you get two homotopy-equivalent spaces under the new base

Huibong

20:53

Hi

Has anyone heard of the Workshops at Oberwolfach?

Ben Steffan

...yes?

Huibong

Are those hard to get into?

Ben Steffan

Well, you can't "get into" them, really

you have to be invited

Thorgott

almost every mathematician knows of these

Xander Henderson

@Thorgott What's the sigma algebra?

And measure?

Thorgott

20:56

counting measure :)

Xander Henderson

Then I don't think that your statement is correct.

Huibong

@BenSteffan and how does that work? What should one do to get invited?

Xander Henderson

I mean, there has to be at least one mathematician who doesn't...

Ben Steffan

@Huibong be an expert in your field, or a promising student, or..

things like that

Xander Henderson

@Huibong Are you a superstar?

Ben Steffan

20:57

basically be very good at what you do

Huibong

So it's not for upcoming researchers, and I'm no superstar, I'm a pseudomathematician

I claim to understand math

Ben Steffan

yeah fat chance lol

Xander Henderson

@Huibong The workshops are for acknowledged experts in a field. People with many highly cited publications.

Ben Steffan

they have no business in inviting people who claim to understand math, and certainly not pseudomathematicians

Huibong

understood

Thorgott

21:04

yeah, invitations are handled by the director personally

Huibong

is there any website that has links to math workshops for upcoming researchers? Where do first-timers publish their work typically?

Xander Henderson

@Huibong These are questions which would be appropriate for an advisor.

Generally, if you want to publish, you should be reading a lot of papers, too, and getting a sense of which journals publish papers in the area you are working in.

For example, there are about half a dozen good journals which publish the kind of fractal analysis, fractal geometry, and analysis on fractals results that I am interested in, as well as a few very good generalist journals that have published this kind of work (the only paper that I have published in mathematics, for example, is in Advances, which is a very good generalist journal).

Huibong

@XanderHenderson I did find some journals, but I thought workshops would be first place to start? I am from a Computer Science background and rely on workshops, conferences usually

Xander Henderson

@Huibong There are workshops in mathematics, but conference proceedings and workshops don't have the same status in mathematics as they do in CS.

But you can also look for conferences and/or workshops in your specific field.

For example, when I have time, I attend the One World Fractals meetings (this is a virtual lecture series).

Huibong

@XanderHenderson I see. That makes sense. I was looking to target a few workshops in topology specifically, without the interlacings of CS/Engineering

ModularMindset

21:46

$$ f(x)=\frac{-1}{\log\big({1-e^{-\frac{1}{x}}}\big)}$$

and take $\int_2^t f(x)dx$

Is this easier to integrate from the standpoint of computational time than $\int_2^t 1/{\ln(t)} dt$

same computational time - nvm

basically the same integral

except f(x) is an involution

and 1/ln(t) aint

Thorgott

22:12

@Ben I just proved something I don't understand

suppose we have two towers $\dotsc\rightarrow\mathcal{C}_{n+1}\rightarrow\mathcal{C}_n\rightarrow\dotsc\rightarrow\mathcal{C}_0$ and $\dotsc\rightarrow\mathcal{D}_{n+1}\rightarrow\mathcal{D}_n\rightarrow\dotsc\rightarrow\mathcal{D}_0$ of $\infty$-categories whose transfer maps are limit-preserving and with sequential limits $\mathcal{C}$ and $\mathcal{D}$ respectively.
if we have a natural transformation $L_{\bullet}\colon\mathcal{C}_{\bullet}\rightarrow\mathcal{D}_{\bullet}$ that consists of object-wise left adjoints, then the induced functor $L_{\bullet}\colon\mathcal{C}\rightarrow\mathca…

(no assumption on compatibility between the right adjoints and transfer maps!)

Thorgott

22:37

and I have no clue whatsoever whether this generalizes to diagrams of more complicated shape

Ben Steffan

wdym by transfer map?

Thorgott

the maps in the diagrams

perhaps that's a nonstandard choice of words

Ben Steffan

hm, that's really interesting

no further assumptions on the categories involved, e.g. presentability?

Thorgott

none, I construct the right adjoint by hand

the idea is actually not that complicated

notation: $p_n\colon\mathcal{C}_{n+1}\rightarrow\mathcal{C}_n$ and $q_n\colon\mathcal{D}_{n+1}\rightarrow\mathcal{D}_n$ are the transfer maps, $R_n\colon\mathcal{D}_n\rightarrow\mathcal{C}_n$ the right adjoints
you get a natural transformation $p_nR_{n+1}\rightarrow R_nL_np_nR_{n+1}\simeq R_nq_nL_{n+1}R_{n+1}\rightarrow R_nq_n$, where the first trafo comes from the unit transformation for $L_n\dashv R_n$ and the last trafo comes from the counit transformation for $L_{n+1}\dashv R_{n+1}$.
now, if you have an object $d=(d_n)_n\in\mathcal{D}$, you do this for all $m\ge n$ and project down via …

Xander Henderson

22:56

Yeah, but where is the epsilon?

I mean, it isn't real math if there is no epsilon...

:P

Ben Steffan

there is an epsilon, it's just not spelled out

see the words "counit transformation" :)

Thorgott

indeed, the epsilons are in my handwritten notes

Ben Steffan

I guess ultimately this makes sense: having a right adjoint is a pointwise condition

Xander Henderson

@BenSteffan Lies.

Ben Steffan

@Thorgott there's no reason to expect the right adjoints to commute with the transfer maps automatically, is there?

Thorgott

23:04

they absolutely will not, in general, in fact I wanna apply this to such a scenario

where $\mathcal{C}_n=\mathcal{D}_n=\mathbf{Cat}_{(\infty,n)}$ and the left adjoints are product functors

so the point is the core functors don't preserve exponentials

you can already see that for $n=1$, the map $\mathrm{Fun}(\mathcal{C},\mathcal{D})^{\simeq}\rightarrow\mathrm{Fun}(\mathcal{C}^{\simeq},\mathcal{D}^{\simeq})$ is typically not an equivalence

Ben Steffan

yeah, alright

Thorgott

@BenSteffan so something weird happens, it suffices to check $\mathcal{C}_{/d}=\mathcal{C}\times_{\mathcal{D}}\mathcal{D}_{/d}$ has a terminal object and this is the limit of the $\mathcal{C_n}_{/d_n}$ via the induced transfer maps...

but these induced transfer maps need not preserve limits anymore and it's not true in general that a sequential limit of categories with terminal objects has a terminal object

(these transfer maps don't even preserve the terminal object in general; cause that would be equivalent to being compatible with the right adjoints)

anyway, I'm trying to look for an example right now, but I think the analogous result for pullbacks should not be true

perhaps it's a cofiltered limits type beat

2

Jakobian

coffeeltered

I have something interesting to think about

5

Q: Is $\ell^\infty$ with box topology connected?

Let $X = \ell^\infty$ be all bounded real sequences and equip $X$ with subspace topology $X\subseteq \square_{n=1}^\infty \mathbb{R}$ where $\square_{n=1}^\infty \mathbb{R}$ is box product of countable amount of copies of $\mathbb{R}$, that is the basis for its topology is $\prod_{n=1}^\infty U_n...

It seems that $A(x)$ are connected components of $X$

ModularMindset

23:30

Are there definite integrals that have visual proofs but not analytical proofs? I mean that their closed forms can be verified but not with analytical techniques, only visual and geometric reasoning.

and what would that mean for mathematics?

Ben Steffan

You cannot verify anything visually so no

Few people consider visual proofs valid

ModularMindset

23:44

@BenSteffan I guess that is reasonable. Given a surface L of positive gaussian curvature embedded in R^3 is it always possible to slice L with a plane to get a quartic variety?

sorry with cone points ,2

Ben Steffan

I haven't the faintest idea so I'll just say "yes" :^)

ModularMindset

sphere doesn't work that is why i changed it lol

do carmo says that we essentially have two cases here: no cone points, and cone points.

Jakobian

apparently $c_0$ is even written on wikipedia. I was basically outed for not reading my sources

(or at least wikipedia, I did read my sources)

ModularMindset

Given a surface of revolution, $L$ of positive constant gaussian curvature embedded in $\Bbb R^3$, with two cone points, is it always possible to slice $L$ with a plane to get a quartic variety?

(that is the correct statement)

leslie townes

if that were true, what would you do with it

ModularMindset

23:52

Should probably say "does there exist a slice" yielding a quartic variety

leslie townes

where is any of this stuff headed

Theorem. Let X be a [surface with whatever hypotheses]. Then __ <- what goes here

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