Mathematics - 2024-09-20 (page 1 of 2)

Silly Goose

12:02 AM

I am confused by the definition of a section of a tangent bundle

consider a tangent bundle $(TM, M, \pi, T_pM)$

a definition of a section over $TM$ is simply a map $\sigma: M \to TM$ such that $\pi \circ \sigma = \text{id}_U$ over an open neighborhood $U$.

Ben Steffan

no

it's a map such that $\pi \circ \sigma = \mathrm{id}_M$

there should be no $U$'s in this definition

Silly Goose

hm wait does "section" mean "global section"

Ben Steffan

yes

Silly Goose

ah

Tu's diffeG book defines a "section over an open neighborhood"

okay so i will use the language section and local section to distinguish

so is a local section not just over any open neighborhood, it should be over a chart in the atlas that comes with defining the manifold at hand?

i am confused because if we define local sections over open neighborhoods in general, $M$ itself is always such an open neighborhood and so there is nothing distinguishing a local from global section. so i must be confusing something.

Ben Steffan

a local section around a point $p \in M$ is a map $\sigma\colon U \to TM$ defined on an open neighborhood $U$ of $p$ such that $\sigma(p) \in (TM)_p$

alternatively, if you are happy with identifying $TM|_U$ with $TU$, then it's a global section of $TU$

Silly Goose

12:12 AM

okay so i am seeing the difference is that a local section is vector valued and a global section is not (strictly speaking) vector valued

Ben Steffan

The distinction is precisely that a local section need not be defined on all of $M$, although it could be

@SillyGoose I really wouldn't put it this way

You could rewrite the condition for global section in the same way: It's a map $\sigma\colon M \to TM$ such that $\sigma(p) \in (TM)_p$

(check that this is equivalent to $\pi \circ \sigma = \mathrm{id}_M$!)

Silly Goose

hm so what is so special about $M$ as opposed to any other open neighborhood or union of open neighborhoods?

i think i am still caught up on the fact that $M$ itself is just an open neighborhood

what makes $U$ "local" and $M$ "not local"

Ben Steffan

well $M$ is the whole space, so "global" seems quite appropriate :^)
on a more serious note, giving global sections with certain properties (for instance being nowhere 0) is generally difficult, and whether such sections exist for any given manifold is a hard problem (which is often topological in nature)

for instance, $S^2$ does not admit a nowhere vanishing global section on its tangent bundle (this is the famous "hairy ball" theorem)

also, admitting such a global sections tells you something about the structure of $TM$ as a vector bundle

if you have a nowhere vanishing section, it splits as a direct sum, for instance

don't worry if these words don't mean anything to you yet :)

but the point is, we care about global sections for various reasons

on the other hand, local sections are much easier to understand: if $p \in M$ is any point, and $U$ is a chart domain around $p$, then $TU \cong U \times \mathbb{R}^n$ where $n = \dim M$

so a section $U \to TU$ is a simply a map $x \mapsto (x, \text{something})$

Silly Goose

but i mean are there not obstructions to defining a section over two (non-emptliy) intersecting open neighborhoods

Ben Steffan

and $U$ is diffeomorphic to $\mathbb{R}^n$ itself, so "something" is any smooth map $\mathbb{R}^n \to \mathbb{R}^n$

@SillyGoose no, not over 2

Silly Goose

12:24 AM

@BenSteffan ...3? :)

Ben Steffan

little trickier

Silly Goose

i guess i have heard of this character classes business, yes. but a priori i don't see why there should exist obstructions only to sections over $M$.

Ben Steffan

@SillyGoose that's not the case (also it's characteristic classes)

Silly Goose

oh oops

Ben Steffan

there are no obstructions to defining sections on charts

there are obstructions to gluing these together into larger sections

defined on larger open sets, eventually $M$ itself

Silly Goose

12:26 AM

okay okay i think that does resolve my conclusion

confusion

Ben Steffan

see the example above: you can cover $S^2$ with two chart domains, but you cannot glue two arbitrary sections defined on each together willy-nilly

since you cannot get something out that vanishes nowhere

but gluing local sections together is an important technique for constructing global sections

anyways it'll all probably become clearer as you work with these things

Silly Goose

so what precisely constrains whether one can define a section or not. is it a complicated answer :P

Ben Steffan

yeah

that's where stuff like characteristic classes come in

they at least let you decide questions in the negative: if certain classes are non-zero, then this implies there are no (nonvanishing) sections

or they bound how many linearly independent sections there can be, etc.

of course you can always define a section: the map $M \to TM$ sending $p$ to the 0 vector in the fibre over $p$ is a section, but not a very interesting one

Silly Goose

is a class in this context like a $\ker A / \text{im}{B}$ type object (as in a cohomology theory)

i might have flipped the quotient

Ben Steffan

it's not a cohomology theory, it's an element of... usually $H^*(M)$

so ordinary cohomology of $M$

Silly Goose

12:31 AM

oh

Ben Steffan

but the cohomology theories vary

Silly Goose

so this can be computed via like doing de Rham cohomology of $M$ or something?

Ben Steffan

some can, to some extent, yes

de Rham is relatively crude however; the topological side of things gives more information

Silly Goose

hm i see

mick

12:56 AM

2

Q: Asymptotic for counting irreducible $0-1$ polynomials?

Consider integer polynomials of degree $n$ such that all coefficients are $0$ or $1$. The so-called $0-1$ polynomials. For instance $1 + x^5 + x^{83} + x^{90}.$ We can list them and order them by setting $x=2$ because that is the binary expansion with $n$ bits. $$ 0 \implies 0 $$ $$1 \implies 1 $...

Is it about O ( n / ln(n)^2 ) or O( n - n/ln(n)^2 ) ?

Obliv

1:17 AM

I'm having trouble with this problem:

Fix $n\in\mathbb{N}$, and let $A_n$ be the set of algebraic numbers obtained as roots of polynomials with integer coefficients that have degree $n$. Using the facts that every polynomial has a finite number of roots and that $\mathbb{Z}^{n+1}$ is countable, show that $A_n$ is countable. Hint: Express $A_n$ as a countable union of finite sets.

Not sure how to partition $A_n$

Idk, this feels like it could involve some combinatorics or something

some finite number of possible forms for algebraic numbers for every $n$, then you have each coefficient being any integer

still ends up being countable

Xander Henderson

So .. in the original Star Wars movies, the Empire is the good guys, right?

Obliv

I think so.

Xander Henderson

1:35 AM

@Obliv Yeah, democratically elected government vs a that's group of guerillas defending a hereditary monarchy.

Obliv

Makes me wanna watch them. I feel like I'd appreciate them more having matured a bit.

Thorgott

1:53 AM

@XanderHenderson as self-declared Star Wars expert, I must raise objection. while the Empire may have been the good guys, there were no democratic elections. perhaps we ought to conclude from this that democracy is not good.

Xander Henderson

2:15 AM

@Thorgott Sure there were---how do you think the Imperial Senate was selected?

Xander Henderson

2:43 AM

(I will say, however, that Wars is low on my list of Stars. Trek is better. Gate, too. Even Search.)

zeta space

4:04 AM

Assume $X$ is a smooth convex open manifold and $\partial X$ is a convex regular polytope.

Def: An $\mathcal F$-completion, or foliational completion of a block, $\mathcal B:=X\cup\partial X$ over its vertex set $V$, is $\mathscr CX_V:=\cup_{v_{ij}\in V} \mathcal F_{v_{ij}}$ where $i,j$ index the vertices.

Is this comprehensible to anyone here?

Jakobian

8:33 AM

@XanderHenderson I didn't know those were the same universe

@Obliv $A_n = \bigcup_{a_0, ..., a_n\in\mathbb{Z}, a_0\neq 0} \{x: a_0x^n+...+a_n=0\}$

Countable union of finite sets

cOnnectOrTR 12

8:52 AM

Hi 👋. Can somebody tell me the points , lines , surfaces defined in Euclid geometry are real or hypothetical ? Do points, lines and surfaces exist in nature ?

Thanks

Jakobian

9:15 AM

@cOnnectOrTR12 hypothetical

as for your second question - I don't know

Thorgott

9:44 AM

@XanderHenderson choosing to answer this question as literally as possible: on a planet-by-planet basis and, empirically speaking, most planets shown in Star Wars are not democratic

Jakobian

I personally don't think democracy is good, because we can see what's happening in current politics, it's definitely not ideal and has room for crooks and politicians to gather money

That said, empire is not a country so it'd be normal for them to have a different political system, something like a "superdemocracy" or "interplanetary democracy"

cOnnectOrTR 12

10:28 AM

@Jakobian how can it be hypothetical when we know we can only think about those things which we have experienced. A surface I think do exist because when we touch something we are touching that 2D surface. Similarly points and lines.

Ben Steffan

@cOnnectOrTR12 "we know we can only think about those things which we have experienced" that's... a statement I think pretty much nobody would agree with

cOnnectOrTR 12

Can you think of something that does not exist ?

Ben Steffan

that's not what you said at all

but "does not exist" depends on your definition of exists

I can think of a lot of things that don't exist as physical objects of the real world

cOnnectOrTR 12

I meant that only. Points , lines and surfaces are objects

@BenSteffan like ?

Ben Steffan

unicorns

or vector spaces

or the real numbers

cOnnectOrTR 12

10:33 AM

A unicorn is a concept that has been derived from reality only. A horse with horns and wings. Horse, horn and wings do exist.

Ben Steffan

irrelevant. unicorns do not physically exist, and this is what I claimed I can do

but that's also why I pointed out that it depends on what "exists" means

are mathematical concepts ultimately derived from physical reality? probably yes, in some sense

cOnnectOrTR 12

I am asking about simple objects. Unicorn is a creation from all those elements which we have seen in past. Similarly I think lines and surfaces are real and with which we can make such figures which I don’t think can exist like 12D , 18D figure. Like unicorns

Ben Steffan

even basic mathematical objects are abstractions, not things in the physical world

for instance, a line consists of infinitely many points

but as far as we know there aren't infinitely many things in the world at all

n.b. that the ontological status of mathematical objects is a controversial topic in philosophy of mathematics. most mathematicians I know take a stance like this, however

cOnnectOrTR 12

@BenSteffan like what ?

Ben Steffan

?

what do you mean?

oh, the stance?

like what I'm trying to argue, is what I'm trying to say :)

cOnnectOrTR 12

10:40 AM

Yeah

Ben Steffan

and what Jakobian said above

i.e. that math objects don't have a "real" or "physical" existence of their own

they have a theoretical existence as ideas or concepts

cOnnectOrTR 12

Ok. But don’t you think as Euclid said that a surface is a 2d which has length and breadth. So a surface of a material has length and breath. Similarly he says a surface is made of infinite lines. And so we can see that if we stack infinite lines of that surface we will get that surface. And a point has no part. So the line on which it exist has no part

Ben Steffan

that seems fine, if rather far removed from the modern way we do mathematics

I see nothing there that would be at odds with my position :)

Jakobian

@cOnnectOrTR12 well so did lines and so on

but mathematical lines are infinitely thin and so they're not like real life lines

same for surfaces, points

Thorgott

@Jakobian it used to be in the name. the 'Galactic Republic' is a republic constituted by the galaxies' planets in the same way that a federal republic constituted by a nation's federal states, just one level higher. the Empire, however, is simply fascist.

cOnnectOrTR 12

10:47 AM

@BenSteffan you mean hilberts way ?

Ben Steffan

I guess

I'm talking about the way we talk about these things nowadays, mostly

if you open a modern textbook on linear algebra, you will find definitions that look quite different

in particular Euclid's statement only passes as semi-mathematical at best under modern scrutiny

cOnnectOrTR 12

@BenSteffan Definition of lines and points ??

Ben Steffan

yes

Jakobian

How did the Greeks know what a line is? They were drawing those lines. But those lines always had some width. This is in contrast with how they thought about geometry though (or at least I think they did) in that those lines were supposed to be infinitely thin

cOnnectOrTR 12

I will search it.

Ben Steffan

10:50 AM

in the context of linear algebra, a point is simply an element of a vector space, and a line is a one-dimensional subspace

this of course presumes you know what a vector space is

Jakobian

@cOnnectOrTR12 Geometry got formalised by Hilbert

Hilbert's axioms

Hilbert's axioms are a set of 20 assumptions proposed by David Hilbert in 1899 in his book Grundlagen der Geometrie (tr. The Foundations of Geometry) as the foundation for a modern treatment of Euclidean geometry. Other well-known modern axiomatizations of Euclidean geometry are those of Alfred Tarski and of George Birkhoff. == The axioms == Hilbert's axiom system is constructed with six primitive notions: three primitive terms: point; line; plane; and three primitive relations: Betweenness, a ternary relation linking points; Lies on (Containment), three binary relations, one linking points...

cOnnectOrTR 12

@Jakobian ok

Jakobian

we didn't have a proper set up for those geometric concepts until then

Euclid's way wasn't the proper axiomatic way, yes, it was somewhat rigorous, but not as rigorous. It was Hilbert who presented a satisfying theory of geometry

cOnnectOrTR 12

@BenSteffan I don’t know but you are talking about same space that we live in. Right ? That is vector space ?

CroCo

good morning everyone, I had to prove $\dot{u}(t) \geq -\eta u(t)$ where $u(t)\in\mathbb{R}^+_*$ to my supervisor. He said the proof correct but he said you need to understand that this is a special case of Grönwall's inequality lemma which has numerous variants. When I look it at Wikipedia, I see one case $\dot{u}(t) \leq \beta(t)u(t)$. Where is the rest of variants for this lemma?

Ben Steffan

10:53 AM

define "space we live in." that's not a mathematical object a priori
if you're talking about 3-dimensional euclidean space then the answer is yes, that's a vector space

but there are many, many other vector spaces for which the definition also makes sense

Jakobian

Note this was in 1899 which was in the 18th (almost 19th) century, while Euclid lived in 300 BC

So it took us 2200 years to even have a proper, axiomatic, understanding of geometry since Euclid's times

cOnnectOrTR 12

@Jakobian hmm. Did we solve problems that Euclid geometry couldn’t answer ?

Ben Steffan

yes

the greeks had a few problems they couldn't solve, and that were open for a long time

squaring the circle, angle trisection for instance

some of these only got resolved in the 1900s

cOnnectOrTR 12

Don’t you think the the hilbert way of defining lines and points are just another way and that Euclid was not wrong ?

Jakobian

it lead to better understanding of the parallel lines axiom of geometry, so it opened geometry to non-Euclidean settings e.g. on the sphere or the various hyperbolic spaces

Ben Steffan

10:58 AM

a definition doesn't have truth content

Jakobian

I don't exactly know if Hilbert's axioms were before or after the whole dispute with parallel lines, but it was around that time I think

Ben Steffan

I don't think that Euclid was somehow wrong and Hilbert were right, but rather that Hilbert's definitions are a recasting into a better, more rigorous, modern setting

picking up something classical, dusting it off and giving it new life, so to speak :)

Jakobian

A mathematician who only uses their intuition in doing mathematics is not wrong in doing mathematics

but formalising those mathematical settings is just as important, because it allows us to see what we actually mean by those things

Ben Steffan

nevertheless we would not accept Euclid's definitions you gave above as mathematical definitions nowadays, because they rely on non-rigorously defined concepts

cOnnectOrTR 12

@BenSteffan If definitions don’t have a truth then the theorems we develop how do they work and that we apply all those in understanding physics. And that the physics is real.

@BenSteffan yeah 👍🏻

Ben Steffan

11:02 AM

@cOnnectOrTR12 A definition is just a convenient name for a bunch of properties

Truth stems from applying logical arguments to mathematical objects using the axioms of your theory

...or objects

For instance we usually define $\mathbb{N}$ to be the set of all natural numbers. This is free of any truth content, and we do it because you don't want to write out, say $\{1, 2, 3, \ldots\}$, everytime you use it

Jakobian

Definition to me is a way to talk about information in a concise way

we could, in theory, write any set theory statement only in terms of language of set theory

cOnnectOrTR 12

@BenSteffan let me get this straight

Jakobian

but it's better to write it in terms of definitions which we then can substitute into

Ben Steffan

as for physics, that's out of my scope, but let me note that "why is mathematics so effective in physics?" is a controversial problem in (philosophy of) physics

cOnnectOrTR 12

So we have defined something like a curve line say a parabola study all its properties and then we use it to understand projectile motion.

Jakobian

11:06 AM

We apply mathematics to physics in a way that is this or that, because that leads to good predictions about our universe

I don't think any good physicist really thinks that those objects reflect real world one-to-one

cOnnectOrTR 12

@BenSteffan don’t you think it’s effective because all that we talk about has some meaning or existence not physical always ?

Ben Steffan

I have no opinion.

Ryder Rude

@BenSteffan the modern formulation of Euclid's axioms is by Hilbert, i think

Ben Steffan

I don't care much about physics

Ryder Rude

that formulation does not talk about real numbers or vector spaces

Jakobian

11:09 AM

@RyderRude yeah, scroll up

Ryder Rude

but vector spaces are more like a model of those axioms

@Jakobian oh

but again, i also think it is fine to directly formalise it using vector spaces

Jakobian

Doing geometry from axioms alone is called synthetic geometry

and it has its merits, similarly to how synthetic differential geometry (I think its called) has its merits in modern day mathematics

Joe

I would say: many mathematical concepts are inspired by physical concepts. For example, a "line" in mathematics is an idealisation of the concept of physical line. But that doesn't mean that every property of mathematical lines is going to have an analogue in the physical world.

For example, the real number line may have a subset which is neither in bijection with $\mathbb N$ nor the continuum, depending on whether the continuum hypothesis holds. I cannot imagine for one moment that the continuum hypothesis is important in physics.

Ryder Rude

@Jakobian many physicists have this viewpoint...

especially the Many Worldians

but quantum mechanics cannot be thought of in that way in an undebatable sense

cOnnectOrTR 12

@BenSteffan 😅 it came to my mind the existence of lines and points because the geometry that there is we use in physics at least as far as I know in classical Newtonian physics. And so it feels like if we are able to deduce laws that are real then we could only do that because we have deduced it from some reality. We cannot deduce a truth from a lie .

Jakobian

11:12 AM

@RyderRude perhaps, but certainly not for the concept of lines

Ryder Rude

Sean Carroll straight up says things like "The universe is a quantum state"

it is a naive way of interpreting physics imo

Jakobian

"perhaps, but certainly not for the concept of lines" - me

Ryder Rude

@Jakobian what do u mean

Jakobian

I am citing myself

Ryder Rude

@Jakobian yes, but what did u mean by this

Jakobian

11:15 AM

That no physicist understands lines to be real world lines

they are approximation of the real world

Ryder Rude

yeah

but again, many physicists would say that spacetime is a manifold

Jakobian

well, it might be that an infinitely thin line exists in real world, I am not arguing that, but its also not the topic of discussion

cOnnectOrTR 12

@Jakobian what harm it is if we think of real world lines and mathematical lines as same ?

Jakobian

@cOnnectOrTR12 They're just different concepts. I can't point towards any harm out of context

Ryder Rude

@Jakobian i understand u now. No physicists identifies geometric lines with lines on a paper, in an exact way

because that's just wrong

but physicists may say that geometric lines exist in the real world cuz they may say that the real world is a manifold

Joe

11:18 AM

I think physicists who say "the real world is a manifold" are just using that language as a shorthand for "the real world can be closely modelled by manifolds".

Jakobian

I'd say its not important if mathematical lines exist in real world

Ryder Rude

@Joe no, really! Many physicists think that physics models ARE the world

especially Many worldians

Many wordlians won't say that universe is a manifold. They would say that universe is a wavefunction

and they think it describes all there is about the universe

Jakobian

and even if mathematical lines were to exist in the real world, how would you verify that the two are the same thing? Or that one is model for the other?

Its a meaningless question to me

Joe

@RyderRude What would that even mean? A manifold is a topological space $(X,\tau)$ satisfying various axioms. Does anybody think that the real world is an ordered pair?

Ryder Rude

it can never be verified, yes. Some physicists may take it on faith if a theory works well @Jakobian

cOnnectOrTR 12

11:22 AM

@Jakobian you mean that mathematicians saw an straight object with sharp edges and corners and defined a mathematical object line and point ?

Ryder Rude

@Joe I think they would just say that this ordered pair exhaustively represents all the properties of spacetime (after adding some other structure)

so it "represents", not that it is

because different notations and formalisations can ofc describe the same information @Joe

Joe

Fine, but in that case it is surely an abuse of language to say that the real world "is" a manifold.

Jakobian

@cOnnectOrTR12 Let me ask you this, if we know Newtonian mechanics isn't perfectly reflective of our reality, why do we still use it?

Ryder Rude

@Joe maybe. but i find this other viewpoint pretty weird too

Jakobian

to me, mathematics is like Newtonian mechanics, but it can be actually adapted to whatever setting that physicists would like

cOnnectOrTR 12

11:24 AM

@Jakobian because it works

Ryder Rude

@Joe it's basically saying that universe is the completely described by the model

Jakobian

@cOnnectOrTR12 and mathematics works. End of story

cOnnectOrTR 12

Ok 👍🏻

zeta space

I subscribe to the viewpoint that fields are transversal sections of bundles over spacetime

Ryder Rude

@zetaspace it is just a formalisation of fields

cOnnectOrTR 12

11:26 AM

Thanks man and @BenSteffan

Joe

@RyderRude Do you think then that it is significant to physics whether there exists a set of real numbers which has cardinality strictly in between $\aleph_0$ and $\mathfrak c$? After all, whether or not this is true affects the mathematical properties of manifolds (though not in a particularly interesting way to most mathematicians, I admit).

Ryder Rude

@Joe I'm not familiar with the effects of continuum hypothesis in making physics predictions. Maybe some physicists have worked on it... but it must be an obscure topic

Jakobian

@Joe But you are talking about this since this is the context of how to construct those things. But now let us work in synthetic differential geometry. You can't use that argument anymore

Ryder Rude

physicists don't use rigorous set theory. they don't really care about technical axioms

Joe

I think the fact that physicists don't use rigorous set theory demonstrates that there are aspects of mathematics which are irrelevant to their work.

Ryder Rude

11:30 AM

@Joe yes

Jakobian

I mean, we use set theory to model our mathematics

If we chop of the parts of how we got to it, then it can still be valid

just take this as axiomatic

Ryder Rude

or more like, physics models, if one wants to axiomatise them rigorously, can be formulated in ZF, or maybe a much weaker theory. so additional axioms are irrelevant to making predictions @Joe

so i guess u only need those properties of manifolds that are derivable in ZF

nickbros123

There was this theorem which stated that: $f:X \to Y$ where $X,Y$ are metric spaces, with $X= \cup_{i=1}^N F_i$ (where $F_i$ are closed sets), and $f$ is continuous on each of these closed sets, then $f$ is continouos on the whole set $X$. This I was able to show. Now the statement becomes false if $N=\infty$, (a countable union). To show this, I had the following counter example:

Let my collection of closed sets be [0], and all sets of the kind [1/(n+1),1/n]. On each [1/n+1,1/n] closed interval, let me define the function to be 0 at the endpoints, 1 at the midpoint, and the function is a …

Jakobian

@RyderRude you don't need set theory for everything. It just so happens that its convenient to discuss mathematics with.

Joe

I would suspect that physics models can be axiomatised in a theory much much weaker than $\mathsf{ZF}$. After all, most mathematics can be formulated in a theory much weaker than $\mathsf{ZF}$ (though you might need something like dependent choice).

Ryder Rude

11:34 AM

@Jakobian yes

@Joe also, physicists won't care much about generating proofs about manifold theory. physics is more concerned with setting up equations to make predictions based on initial conditions

so they just need a manifold to set up a differential equation

proofs are not that relevant. Physics is focused on modeling experiments

Jakobian

@nickbros123 yes its valid

also note that this theorem does have a generalization

Ryder Rude

@Joe but it also depends on the physicist. mathematical physicists may care about proofs

Jakobian

just not to countable families, but locally finite families

Your family is not locally finite at $0$, since there is no neighbourhood of $0$ which intersects only finitely many elements of your cover

The theorem is: if $f:X\to Y$ is a map between topological spaces such that $f\restriction_{A_i}$ is continuous for each $i\in I$, where $(A_i)_{i\in I}$ is a cover of $X$ which is either 1) an open cover or 2) a locally finite closed cover, then $f$ is continuous

this is the general form of gluing lemma

nickbros123

could u define locally finite

ok so we have a neighbourhood that only intersects finite sets of the collection

Jakobian

A cover (or any collection) $\mathcal{U}$ of $X$ is locally finite at $x\in X$ when there is a neighbourhood of $x$ that intersects finitely many of elements of $\mathcal{U}$. The cover is locally finite if its locally finite at each of $x\in X$.

nickbros123

11:42 AM

I see. Thanks

Jakobian

Also maybe its good to mention a discrete family, which is a family $\mathcal{U}$ such that each $x\in X$ has a neighbourhood which intersects at most one element of $\mathcal{U}$.

Bing's theorem says that any open cover of a metric space admits a locally finite, $\sigma$-discrete open refinement. That is, refinement which is a countable union of discrete families

this is a statement similar to paracompactness, but slightly stronger

nickbros123

damn, you really do love topology :)

Jakobian

it can be used to prove various metrization theorems

nickbros123

one day i hope ill have as expansive a toolkit as u do in topology

Jakobian

I hope you'll have a more expansive toolkit than I do in topology

don't settle for mediocrity

nickbros123

11:54 AM

maybe youre undermining yourself here. but I agree with the sentiment that one musn't strive to be like someone else, but rather should just focus on trying to get as big a toolkit as one can. gotta keep grinding

Jakobian

no, although my toolkit probably is pretty average

it was a positive message, in any case. I don't have confidence problems

nickbros123

well then, thanks :)

Ryder Rude

@Joe i also found this thread philosophy.stackexchange.com/questions/96498/…

i would say no model of physics can be trusted to describe the universe without further modifications. manifolds may not even be relevant in some upcoming model

zeta space

12:13 PM

I like physics because nobody cares about proofs

Ryder Rude

Feynman calls this "peeling an onion with infinite layers"

there is always something new beneath

some people would invoke the "evolutionary universe hypotheis", in which the laws of physics are the way they are as a result of evolution of universes. but this is too speculative, on par with simulation hypothesis stuff

zeta space

Can fundamental forces be mixed?

Jakobian

12:28 PM

@zetaspace you like physics because of one of its negatives?

Physics cannot justify bad mathematics

Ryder Rude

@zetaspace i think electromagnetic and weak forces mix in the electroweak unification

@Jakobian it is not necessarily bad math

i read on overflow (i think) that physicists are like the consumers of math. And they care about different aspects of math

but what physicists actually do with math, they may do in a handwavy way

zeta space

yeah exactly Ryder

@Jakobian Interesting. I'm curious how you view "bad mathematics." Do you think it's bad mathematics if said math is incomplete for 10 years and then it becomes rigorous after 10 years of development?

Ryder Rude

but sometimes they handwave because rigorous way of doing the computation doesn't even exist

e.g. rigorous path integrals in QFT don't exist yet

Jakobian

12:45 PM

@zetaspace I don't care about setting up a straw man with you

Its one thing to not be overly rigorous, and another to not care about it altogether, as physicists can, and do do that

physics cannot justify bad mathematics

Ryder Rude

@Jakobian yes. I'm just saying that not caring about proofs is minority of the part of the bad math in physics. The bad math is more in the computations

physicists care more about definitions and computations, than theorems

e.g. u define a manifold and u compute predictions using it. and u don't bother to prove more than a few key theorems

mathematicians and physicists explore different aspects of math

Jakobian

@RyderRude I don't think so, because its prevalent in how physicists teach mathematics

besides, computations and proofs go along together

Ryder Rude

i agree theorems r there, but much fewer compared to math

@Jakobian maybe...

e.g. a math book would have theorem after theorem. maybe even a mathematical physics book. but a usual physics book would have few key theorems

Jakobian

1:01 PM

Whatever. More generally, I'm just talking about any sort of mathematical negligence

one cannot expect a physicist to present good mathematics, but I would hope for them to at least keep the standard

but some people use it as an excuse for bad math - that's not okay

zeta spaces does post here about concepts which are often undefined
so thats the idea of what that message meant to me, i.e. they like it because they can be negligent about their mathematics. This is not fine with me, they should strive for rigor especially since they want to be a mathematician from what I see

thats why I objected

Ryder Rude

@Jakobian i also think that the bad math is in the definitions even more than in the computations. e.g. a physics book would write basis vectors $|x\rangle$ indexed by a continuously varying index $x\in R$ without properly defining what kind of mathematical object it is

Jakobian

1:16 PM

@RyderRude I don't think they really need to. Braket notation is a bit dangerous, but I actually sort of like it

Ryder Rude

@Jakobian yes, the notation is fine. but the notion of this basis vector is weird because it lies outside the Hilbert space

they don't define their mathematical object and proceed to do computations with them...

Jakobian

I don't exactly understand your issue with notation here

Ryder Rude

the notation is fine... but this basis vector has the property $\langle x|x'\rangle =\delta (x-x')$, so it is unclear in what sense it is a basis vector

cuz inner product must be a number

I'm good with the notation. u can write $|v\rangle$ for a vector without any issues

this basis vector has rigorous definitions, but most physics book wouldnt define it rigorously

Jakobian

@RyderRude That's another thing I feel like. And this actually does raise red flags

But you can understand it in certain way, that $\delta(x-x')$ here is supposed to be like a matrix where you have $1$ in a certain place and $0$ in all others. The problem is that they are mixing in the context of talking about spaces of functions so they use $\delta$ and now its confusing

Ryder Rude

yes. physics books do give the intuition that this is supposed to be a generalisation of the kronecker delta

Jakobian

1:25 PM

well, also because $x, x'$ are just indexes

I would definitely call this bad math, yeah

Ryder Rude

this is one approach to make these well defined en.m.wikipedia.org/wiki/Rigged_Hilbert_space

i haven't studied it...

zeta space

2:51 PM

@Jakobian except whenever I write something well-defined there's no response lol

I only hear about the incorrect things i do

which is great because I learn

Sourav Ghosh

Give an example of a non degenerate representation of a C* algebra that is not faithful.

Jakobian

3:21 PM

@zetaspace I don't sit in differential topology/geometry. But given that others aren't helping you too, maybe those topics you mention are too niche for them to answer/know about

ILikeMathematics

5:05 PM

@BenSteffan Do you happen to know if there are any stipends specifically aimed at math students in Bonn, btw? For example at Kaiserslautern, there is, but Googling doesn't give any meaningful results for Bonn

I'm searching for something where they look at mathematical commitment, certificates, mathematics grades, mathematical extracurricular things etc., not social or political stuff

Xander Henderson

5:27 PM

Oh, my jeebus... I am supposed take this training on AEDs... it is a bunch of vidoes. I cannot play them in the background, I cannot make them go faster, and I can't just read a transcript. ARG! What a waste of my time!

Pizza

👋

Sine of the Time

hi

Pizza

@SineoftheTime I was repeating the cases of differential equations, all the cases are quite clear to me except this one

The Second in the table

I don't understand what $i\lambda$ is

$\lambda$ would be the coefficient found in the sin/cos

And $i$?

Vladimir Lysikov

I don't know italian
But I believe that this is the case when the pure imaginary number $i\lambda$ is a root of the characteristic equation
$i$ is the imaginary unit

Pizza

For example I had $y'' + 4y' + 3y = \sin(3x)$ , so $\lambda^2 + 4\lambda + 13 = 0 \rightarrow \frac{-4±6i}{2}$

The solutions are $-2 + 3i$ and $-2 - 3i$

Vladimir Lysikov

5:41 PM

Then you use the first case

The second case is for example for $y'' + 9y = \sin (3x)$, so the roots are $\pm 3i$

Pizza

I'll try to see what changes with this exercise you suggested to me

@VladimirLysikov But then in the first case I shouldn't have had -2 and +2 in the solutions?

Vladimir Lysikov

Okay, let's go from the start.
This table is for finding a special solution (soluzione particulare) to a non-homogeneous linear ODE

Pizza

Yes

Vladimir Lysikov

It says that if the right-hand side is a linear combination of $\sin (\lambda x)$ and $\cos (\lambda x)$, then we search for a special solution in the same form
UNLESS it so happens that $i\lambda$ is a root of the characteristic equation $ay^2 + by + c = 0$, in which case we search for a special solution in the form $x(A \sin(\lambda x) + B\cos (\lambda x))$

Pizza

Yes

Vladimir Lysikov

5:53 PM

So we find the roots of the characteristic equation (we probably find them on the previous step to get a general solution of the homogeneous equation) and check if $i\lambda$ is a root

Pizza

It would be that we find the solutions

Vladimir Lysikov

For the equation $y'' + 4y' + 3y = \sin 3x$ the roots are $-2\pm 3i$, the value $3i$ is not a root, so we search for a special solution $A \sin 3x + B \cos 3x$

For $y'' + 9y = \sin 3x$ the roots are $\pm 3i$, so $3i$ is a root and we search for a special solution $x(A \sin 3x + B \cos 3x)$

Pizza

Oh ok! I understand, thank you very much!

Vladimir Lysikov

Maybe you are confusing $\lambda$ in the right hand side with $\lambda$ in the characteristic equation.
It's better if the letters are different

Sine of the Time

6:13 PM

@pizza did you understand or is something still unclear?

@Pizza It may be useful looking at this example using the exponential representation of sine and cosine, are you familiar with that?

Ben Steffan

@ILikeMathematics don't really know, sorry

I'm sure there's something

Pizza

@SineoftheTime I think I understood $\lambda$ in this case it would be the coefficient found in $\sin(\lambda x)$

If I had for example $\sin(4x)$ then I had to check if $4i$ was a solution

Sine of the Time

yes, it's the coefficient of $x$ in the argument of $\sin$/$\cos $

@Pizza correct

Pizza

So there must not be for example $1+4i, 2+4i$ , Only $4i$ I need to have

Sine of the Time

are you familiar with these formulas: $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos x=\frac{e^{ix}+e^{-ix}}2$ ?

Ben Steffan

6:23 PM

@ILikeMathematics there's no information about mathematics in particular I can find, but this page may be helpful uni-bonn.de/de/studium/beratung-und-service/…

Pizza

@SineoftheTime Not very much

Sine of the Time

ok

Pizza

Ah anyway when he asks me to study the differential form, I have to check the domain, closure, say if it is exact, and find a primitive

Ben Steffan

if you don't have much to show in terms of social engagement then this will exclude you from many if not most of the "big" programs, but there's lots of smaller stuff for which it might not be relevant

Jakobian

@AlessandroCodenotti Do you know whats this property called? For every locally finite open cover there is a finite subcover

Sine of the Time

6:25 PM

the fact that you multiply by $x$ follows from the fact that $i\lambda$ has multiplicity $1$

if you remember when you have exponentials you multiply by $x^m$ where $m$ is the multiplicity

Pizza

Yes

Sine of the Time

@Pizza this depends from the professor actually

Pizza

But for example I could also have $\sin(-2x)$ so $\lambda = -2$ , I should have found the solution $-2i$ not $2i$ , right ?

Sine of the Time

actually, it does not matter

since $2i$ and $-2i$ are conjugate

moreover, since $\sin x$ is odd, $\sin(-2x)=-\sin (2x)$

Jakobian

This is equivalent to pseudocompactness for completely regular spaces, but I wondered if it has a name e.g. weak pseudocompactness or something. But weak pseudocompactness refers to something different

Pizza

6:30 PM

In the product between polynium and sine/cosine , it is required to verify whether α+iβ it's a solution, not α-iβ, its the same ?

Sine of the Time

@Pizza if $a+ib$ is a root of a polynomial with real coefficients, then also $a-ib$ is a root. Are you aware of this?

Pizza

Yes

Sine of the Time

@Pizza it's sufficient to check if $\alpha+i\beta$ is a root

Pizza

Ah ok

Thanks

Sine of the Time

other doubts?

CroCo

6:49 PM

sorry for asking same question again in this chat but it seems the question got buried. I would like to ask it again.
I had to prove $\dot{u}(t) \geq -\eta u(t)$ where $u(t)\in\mathbb{R}^+_*$ to my supervisor. He said the proof correct but he said you need to understand that this is a special case of Grönwall's inequality lemma which has numerous variants. When I look it at Wikipedia, I see one case $\dot{u}(t) \leq \beta(t)u(t)$. Where is the rest of variants for this lemma?

Ben Steffan

this should be in any text treating ODEs

Transcript for

$Mathematics$ Mathematics