Mathematics - 2020-11-23 (page 1 of 2)

12:01 AM

@Astyx So, by looking to the chain of $B_i$, each element in that chain must be finitely generated. Further, the union over the chain must also be finitely generated. So, the chain must stabilize. I think I might need to show that the $B$ for $\bigcup_iI_i$ is $\bigcup_iB_i$ itself, though, in order to make this argument

Astyx

12:11 AM

Ah yes you're right

Rithaniel

Excellent. I have too little experience with non-Noetherian stuff to be entirely 100% on my approaches with this stuff, so it's good to get confirmation

robjohn

12:27 AM

@TedShifrin I think the Hölder exponent of a space-filling curve needs to be at least $1/n$ (not Lipshitz exponent). But by the way a Hilbert curve fills a cube with subcubes, I think it needs to be $2^n:1$ at some points.

But I'm not as sure about the last part.

Rithaniel

1:40 AM

Question. Given a field $K$, then consider the ring $R=K[x,\{y_i\}_{i\in\mathbb{N}}]$ and the ideal $I$ generated by all polynomials of the form $x-\prod_{i=1}^ny_i$. Is $R/I$ Noetherian?

Writing it down, I'm pretty sure the answer is no, because the ideal $(\{y_i\}_{i\in\mathbb{N}})$ is not finitely generated. However, it doesn't satisfy the other property I'm looking for (All radical ideals are the same ideal).

I think what I want is something like $R=K[x,\{y_i\}_{i\in\mathbb{N}}]$ and $I$ is the ideal generated by all polynomials of the form $y_i-x^{p_i}$ where $p_i$ is the $i-$th prime in the integers.

(Disclaimer: Goal is to find a ring that isn't SFT but which still satisfies the ascending chain condition on radical ideals. My thought is to make it easy by having all radical ideals be the same ideal)

Yeah, my last idea doesn't work either. It's just $K[x]$, in the end

Maybe $y_i^{p_i}-x^{p_i}$? Is $(x,\{y_i\}_{i\in\mathbb{N}})$ still a prime ideal in that version? I think so. Then the radical is the entire ring, so any chain of radical ideals automatically stabilizes

(Thank you all for allowing me to rubber-duck)

Chris Steinbeck Bell

2:05 AM

@TedShifrin Sorry for the late reply. Um.. well I wish to have another choice, but I must solve these problems if I want to learn. It's not that I haven't tried. Its just that I came to a point where I cannot go further without help. Its kind of depressing for someone when your own ability doesn't let you solve more problems though. I think in this case, in the sense of euclidean geometry the thing is that is more than drawing because some problems require construction.

@TedShifrin But I'm having the impression that some problems requiring the use of other tools other than just typing mathjax gather some indiference or nuisance to some people because they think the OP doesn't have the desire to do a drawing of its own, which I do difer.

@TedShifrin Had just read your comment about drawing a perpendicular. Yes I tried doing this but I'm stuck. The official solution is that the angle is $53/2^{\circ}$. The question is how to get there. As it stands now the question remains unanswered and nobody can post answers. Which I do hope other people could come and vote to reopen.

Just in case I'm referring to the question math.stackexchange.com/questions/3916522/…

user2103480

Is there an adequate mathematical definition for a phase space in thermodynamics, especially for the sigma algebra that we integrate against? How does the discretization work here?

Chris Steinbeck Bell

mentioned earlier. Sorry buddy but I had to go, its late in my timezone. But I'll check it later. It seems that the indicated approach is trying to spot the right congruence set of triangles in the figure to get it done correctly.

user2103480

Our lecturer defined it rather mathematically but still vaguely, such that it seems like nonsense if I read carefully

"The event sigma-algebra is generated by the set of all micro states"

Microstates are identified with compact regions in space of some specified volume

Is that just a finite partition? Or are there infinite "possibilities" that are included as well, i.e. all such regions in space that have midpoints lying on some possible trajectory

user193319

2:39 AM

I'm trying to show that $M_n(R^{opp}) \cong M_n(R)^{opp}$ as rings. My thought was that $\varphi : M_n(R)^{opp} \to M_n(R^{opp})$ defined by $\varphi (A) = A^{T}$ is the desired isomorphism. But I'm a little confused as to what I have to verify...Clearly $\varphi^{-1} = \varphi$, so it is bijective, $\varphi (A+B) = (A+B)^{T} = A^{T} + B^{T}$, and $\varphi (A \star B) = \varphi (BA) = (BA)^{T} = A^{T} B^{T} = \varphi (A) \varphi (B)$...

Is that it?

Seems like a cheesy proof.

Thorgott

I don't think $\varphi^{-1}=\varphi$, given that codomain and domain are different

user193319

Ah, you're right. I thought something was off.

Does the definition of $\varphi$ even make sense?

Ted Shifrin

3:03 AM

@ChrisSteinbeckBell I believe that answer is impossible. It is in fact $30^\circ$, which is what I figured from the beginning. Did you read my conversation with @Astyx? You have to use the geometry I talked about and a bunch of algebra. Try to follow what I said about similar triangles and Pythagorean Theorem. And note that I gave intermediate results. Try to get them.

If you make sincere efforts and ask me a few questions, you should get it.

user193319

@Thorgott Actually, I think it's all right because the underlying sets are the same.

Rithaniel

3:20 AM

Question, if you have a prime element $x$ that divides a high enough power of all generators of a ring, then, for any particular element of that ring, can you find a high enough power of that element such that $x$ divides it?

I'm fooling around with the multinomial theorem, but I keep ending up doubting myself

Rithaniel

3:35 AM

The answer is "nope"

Leaky Nun

in the ring Z[x], x divides x, and {x} generates Z[x], but x does not divide 1

Rithaniel

A good point

user193319

3:51 AM

@LeakyNun Does what I wrote above about $M_n(R^{opp}) \cong M_n(R)^{opp}$ seem right?

love_sodam

6:19 AM

I asked this before and I thought I prove but turned out that my proof have some error.
$\Bbb Q(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3}) = \Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3})$

skullpatrol

8:58 AM

Games you can play with area:

robjohn

9:45 AM

@skullpatrol extreme gerrymandering

Sha Vuklia

10:07 AM

@MikeMiller Oooooh! That was my error indeed. Thanks a lot!

Rithaniel

I could use some help with a problem, if anyone has any knowledge on said problem

The problem is "Let $R$ be $1-$dimensional and Noetherian. Show that, if $\mathfrak{P}$ is a prime ideal of $R$ and $T$ is an overring of $R$ then there are only (at most) finitely many prime ideals in $T$ that lie over $\mathfrak{P}$."

I'm having trouble getting a good grip on it. I made a concerted effort to figure it out based on an incorrect assumption that $R$ was a domain, but that's not being assumed here, so that effort went out the window

Federico

sorry to bother you all, I need a brief help trying to recognize an equation.
I have a covariance matrix whose elements are computed in a given frame.
Now, a former colleague extracted the projection on the y and z axis of a different frame by appling the following:
[CoV22, CoV12*tan(phi);
CoV12*tan(phi), Cov33 - 2CoV13*tan(phi) + CoV11*tan(phi)^2]
Now, I can't find in the documentation how this transformation is derived, but I know from context that phi is the angle between the two frames, and the rotation is around the y axis (that should be in common between the two frames, and this seem…

skullpatrol

10:49 AM

Bangladesh-mandering? @robjohn :P

Thorgott

12:07 PM

@Rithaniel what's an overring in the context of not necessarily integral domains

epic_math

Do you people know about pari/gp

Daniel Sehn Colao

hi

epic_math

Hello @DanielSehnColao

Myprofileissaved

12:22 PM

Given a smooth oriented atlas A on a topological manifold M, can I construct a smooth oriented structure on M?

oriented atlas - the determinant of the differential of every transition map is positive

Thorgott

what's an oriented structure to you

Myprofileissaved

oriented structure - a smooth structure s.t the determinant of all the transition maps is positive

maximal oriented smooth atlas *

Mike Miller

Sure

What's wrong with "the collection of all charts so that the transition functions to any chart in A are smooth and have positive Jacobian determinant"

Thorgott

yeah, that

Myprofileissaved

But is it a structure? I mean, can't I add another chart that is compatible with all the charts in this structure but does not preserve orientation?

Mike Miller

12:34 PM

That doesn't sound like an oriented smooth atlas anymore

"Maximal oriented smooth atlas" means you can't add another chart and have it remain an oriented smooth atlas

Myprofileissaved

My definition of oriented smooth manifold is a a smooth manifold that have a structure s.t every transition map has a differential with positive determinant. If I understand this correctly, I must find a structure with orientation

smooth manifold that can admit a structure s.t every transition map ...

Mike Miller

You or your source are being sloppy with definitions. Your confusion stems from the ambiguity in the word "structure".

[One definition of] an oriented smooth manifold is a smooth manifold equipped with a maximal smooth oriented atlas. A smooth oriented atlas is an atlas so that all transitions functions are smooth with det(Df) > 0. A maximal smooth oriented atlas is a smooth oriented atlas which is not contained in a smooth oriented atlas with more charts.

The definition is not a smooth manifold equipped with a maximal atlas (maximal among all smooth atlases), so that in this maximal atlas, we happen to also have det(Df) > 0 for every map. That's not the definition because no such thing exists.

As you say you could just include a chart which is oriented incorrectly. Easy to produce; take an oriented chart and precompose with a reflection on R^n.

Myprofileissaved

I really though I am losing my mind over it

thank you very much. This really helps

user2103480

1:11 PM

>when you look at a sheet, think that this is pretty chill, and then notice there is a page 2

Edward Evans

literally had exactly the same thought just now

first page was some nice calculations about units in local fields, second page "HENSELIANNESS COMMUTES WITH FILTERED COLIMITS"

Thorgott

that sounds nice

Edward Evans

lol

Astyx

hello chat

what's up

Edward Evans

yo @Astyx

geocalc33

1:33 PM

are mixture distributions well-understood?

geocalc33

2:23 PM

what kinds of norms can you put on a function space?

Thorgott

how do I show that $z-(x-a_1z)(x-a_2z)(x-a_3z)$, $-(x-a_2z)(x-a_3z)-(x-a_1z)(x-a_3z)-(x-a_1z)(x-a_2z)$ and $1+a_1(x-a_2z)(x-a_3z)+a_2(x-a_1z)(x-a_3z)+a_3(x-a_1z)(x-a_2z)$ can not be simultaneously zero, where $a_1,a_2,a_3$ are distinct complex numbers

there should be some exploitable symmetry, but all the arithmetic I do ends up very cranky

epic_math

So many questions but so few answers

Mike Miller

Where is this from?

There must be a better expression of this junk

Thorgott

trying to show the vanishing locus of $y^2z-(x-a_1z)(x-a_2z)(x-a_3z)$ is a smooth projective plane curve

above is just how it looks like in affine coordinates with $y=1$ and the derivatives

geocalc33

is this a norm? $f:\frac{s}{x}\mapsto s$ because it associates every function to a real number $s$?

Mike Miller

2:35 PM

Yeah there must be something better to say let me think

@Thorgott Seems like it might be better to think about the homogeneous polynomial in 3D instead

$y = 1$ seems to make this uglier

But Iunno

Thorgott

the only definition of smooth projective plane curve I have is that it's a smooth affine plane curve in every affine chart

it's horrible

Astyx

If you call $u_i = x-za_i$, equation 1 tells you $z = u_1u_2u_3$ and eq 3 tells you this isn't 0

Then equation 2 tells you $u_1u_2 + u_2u_3 + u_1u_3 = 0$

Finally replacing $a_i = (u_1-x)/z$ in equation 3, you find 4=0

Thorgott

how does eq3 say that isn't 0?

Astyx

If z is zero, x is zero by eq 1, and 1=0 in eq 3

I meant $a_i = (u_i-x)/z$ of course

Thorgott

2:51 PM

right, you get $-2=0$ in the end, but I got it now

thanks man

Astyx

adding integers is hard :(

glad to help

epic_math

Hey, I have a question

How can we prove that of p is an odd prime and gcd(a,p)=1, then the congruence $$x^2\equiv a(\mathrm{mod }p^n)\quad n\ge 1$$ has a solution if and only if (a/p)=1.

Here (a/p) is legendre symbol

I proved that the congruence has a solution implies (a/p)=1, but couldn't prove the converse

Can someone help me?

Any idea?

Lukas Heger

3:26 PM

@epic_math if $\left(\frac{a}{p}\right)=1$, then $x^2=a\pmod{p}$ has a solution. By Hensel's lemma this solution lifts to a solution $x^2=a\pmod{p^n}$

epic_math

Thanks

But I have to use induction.

Astyx

Lifting can be done inductively

Mike Miller

OK, so you want to prove Hensel's lemma by induction. You know there is a solution x^2 = a mod p. Prove that if there is a solution to x^2 = a mod p^n then there is also a solution of x^2 = a mod p^{n+1}.

J.D.

Hi, is there someone that can help me with a control theory/robotics question? I am really stuck. Thanks in advance.

Mike Miller

I would guess we have no such experts here but you can of course ask

J.D.

3:37 PM

I have posted on the forum ,I don't know if I am doing wrong asking it here, it is the first time I write on this chat, so in case please tell me. The probem is the following: math.stackexchange.com/questions/3919495/…

Mike Miller

Hopefully somebody will be able to help. You may want to look for tags that have more subscribers --- your top tag there still has very few questions

So few people are likely to see your post

epic_math

Does someone know any journals for analytic NT?

J.D.

I will try to put a different tag. Thank you for the suggestion

epic_math

I found many journals for NT, but they mostly contained algebraic NT.

Not satisfactory

@J.D. it is a long post, you surely put effort in it

J.D.

yes I have been working on this a lot, but I cannot find a way to solve this problem

Mike Miller

3:43 PM

the journal of number theory eg has many papers on both analytic and algberaic number theory. In any case most people do not find/read new results by browsing a journal; they follow tags on the arXiv.

epic_math

Yeah I am not searching only for new results

But I want journals that are completely on analytic NT

Mike Miller

ok.

epic_math

Fine

I would be glad if someone recommended a good journal

Mike Miller

that's a great journal. what do you actually want this for?

epic_math

Research (I want to collect info)

Mike Miller

3:47 PM

ok. well, I'll let you do that

epic_math

For information I will probably ask here if I don't get journals

Right now my research is on primes of various forms

And those forms are number theoretic functions.

I post my notes on my analytic number theory room

Analytic number theory

For people interested in analytic number theory, a field of ma...

CowperKettle

user193319

0

Q: Matrix Algebra Over the Opposite Ring

I'm trying to show that $M_n(R^{opp}) \cong M_n(R)^{opp}$ as rings. My thought was that $\varphi : M_n(R)^{opp} \to M_n(R^{opp})$ defined by $\varphi (A) = A^{T}$ is the desired isomorphism. But I'm a little confused as to what there is to verify... Clearly $\varphi^{-1} = \varphi$, so it is bije...

geocalc33

4:03 PM

p-addicts are crazy when you have to deal with them

Edward Evans

@epic_math why not listen to @Mike's advice?

why do you want a journal that is specifically on analytic number theory? lol

and why don't you just read an analytic NT book lol

and why don't you quit analytic NT because it's lame

/s

Astyx

not really /s tho

Edward Evans

(not really /s tho)

but I gotta be diplomatic

user2103480

4:34 PM

@EdwardEvans you can erase the adjective before NT

Astyx

Someone's gonna get hurt

Edward Evans

4:50 PM

@user2103480 you wanna fight mate

anakhro

algebrawl

user2103480

5:02 PM

@EdwardEvans no thanks I'm busy studying useful topics

Mike Miller

useful to who

user2103480

physicists, mathematical biologists and finance people I guess?

Astyx

yuck

Mike Miller

Hmm 1/3 useful ain't bad

user2103480

@MikeMiller that third refers to physics I suppose

tbh the neuroscience stuff I'm learning seems far from being useful to normal people

"i Do mAtHEmatIcAl nEuROsCiEnCE" aka electronic circuit analysis on steroids

geocalc33

5:09 PM

kirchoff's rule?

J.D.

Hi, do you know if there is a chat for control theory? Thanks in advance.

user2103480

@geocalc33 yeah literally

Astyx

@user2103480 The jewel of mathematics

user2103480

@Astyx we put stochastic noise terms on it so it gets to bad SDEs pretty quickly

Mike Miller

@user2103480 absolutely not

user2103480

5:12 PM

@MikeMiller wait, so mathematical biologists are useful? Or finance people?

To be fair, I think you cannot get better intuitive explanations of SDEs than on quant stack exchange (sad state of affairs)

Edward Evans

@user2103480 you won't be laughing when number theorists accidentally stumble across the access codes to the Zion mainframe and destroy humanity

Astyx

You can't laugh if you don't exist any more

Edward Evans

i literally couldn't think of something real to spite you with

so I just

did the matrix

user2103480

@EdwardEvans you bet I'll laugh

it was long overdue

number theorist DESTROYED by nihilism and logic

Edward Evans

hahaha

Mike Miller

5:16 PM

@user2103480 Imagine how useless the other ones are if mathematical biologists count as useful

geocalc33

can anyone name me set of stochastic partial differential coupled elliptic-hyperbolic nonlinear equations?

user2103480

@MikeMiller Then how useless must I be

Astyx

Turns out the zeta function is encrypting the access to the CPU

Edward Evans

i have a wound that's looking pretty infected

getting kinda unsettling

user2103480

@geocalc33 no

Edward Evans

5:16 PM

there's a small civilisation growing around it

geocalc33

can any pde be recast as spde?

Mike Miller

@user2103480 Useful is a silly word

We are engaging in a pastime

to pass the time

Edward Evans

until the Earth reclaims us

Thorgott

5:28 PM

the only thing more silly than usefulness is finance

geocalc33

can any partial differential partial differential equation be coaxed into a stochastic partial differential equation though ?

Lelouch

Let $A$ be a ring, and $k$ be a field which is also a $A$-module. Is it possible for a $A$-module $B$, the $A$-module $k \otimes_A B$ is nonzero as a $A$-module, but zero as a $k$-vector space?

robjohn

5:44 PM

@Astyx Controlling Parallel Universe?

Tobias Kildetoft

@Lelouch No, being zero is determined purely by the abelian group

Astyx

@robjohn Shhh, don't say it out loud, you're going to anger them

robjohn

@Astyx Say what out loud? !-)

Khallil

The mean square is now a circle :O

user2103480

6:11 PM

@geocalc33 I'm pretty sure not every PDE can be easily cast into one one of the forms of SPDE

there are various issues of well-definedness

from arxiv.org/abs/2004.09336

Making sense of SPDEs is actually a huge research topic

There are, like, three approaches to defining several kinds of solutions to SPDEs

Thorgott

imagine studying a topic where the topic is finding out what the topic is

Mike Miller

6:27 PM

@user2103480 What text did you use in that weird corurse where you proved Brouwer by way of Sperner

robjohn

@Khallil pie are round; cornbread are square.

geocalc33

6:58 PM

@anakhro I recall you stating before that any manifold can be written as several transversal manifolds. Correct?

Astyx

What is the geometric intuition behind "geometrically integral/irreducible/reduced" schemes?

Rithaniel

7:40 PM

@Thorgott Well, in not necessarily integral domains, the equivalence relation on fraction is as follows: $\frac{a}{b}=\frac{c}{d}$ if and only if there exists $t\in R$ such that $t(ab-bc)=0$

So, in effect, if $c$ is a zero divisor, then there exists $t$ such that $tc=0$ and so $\frac{0}{1}=\frac{c}{d}$ where $d$ is arbitrary.

Thorgott

I don't follow, are you trying to localize?

Rithaniel

Well, an overring is a ring between a base ring and it's quotient field. So I always imagine an overring as adjoining fractions to the base ring.

Thorgott

Yeah, but quotient field only makes sense for integral domains. So an overring in general is a subring of some localization of the base ring?

Rithaniel

Ah, yeah, that's why I was bringing up the equivalence relation. In non-integral-domains, the quotient field simply can simply map all zero divisors to 0.

Lukas Heger

@Rithaniel it seems like you're localizing at the whole of $R$, this will just get you the zero ring

Rithaniel

7:49 PM

Yeah, that's why you generally exclude zero divisors from the denominators.

Like, even if you're in a non-integral-domain, you still include that restriction

Thorgott

you need to take $t$ only from some multiplicative subset

otherwise that might not actually be an equivalence relation

Rithaniel

I think that might be right. I should look up the definition again

Yeah, I actually made an incorrect statement just now, I think

Lukas Heger

yes, you need $t \in S$. If you allow $t=0$, then all fractions are equal

where $S \subset R$ is some multiplicatively closed subset

Rithaniel

Yeah, you don't necessarily exclude zero divisors, but you do exclude zero.

That's what I misremembered

anakhro

@geocalc33 no, I think the thing you asked amounted to wondering if any manifold can be written as the transverse intersection of two manifolds.

geocalc33

7:59 PM

@anakhro okay that works. So if that's true, for a simple example, consider $\Bbb R^2$. This is the intersection between two $\Bbb R^3$'s?

wait no

$\Bbb R^2$ can be written as two transversal submanifolds of $\Bbb R^2$?

anakhro

I don't know what that means.

geocalc33

in other words how does one write R^2 as two transversal manifolds based on the fact that any manifold can be written as the transverse intersection of two manifolds

Thorgott

hint: $\mathbb{R}^2=\mathbb{R}^2\cap\mathbb{R}^2$

nbro

2

Q: How should we interpret this figure that relates the perceptron criterion and the hinge loss?

I am currently studying the textbook Neural Networks and Deep Learning by Charu C. Aggarwal. Chapter 1.2.1.2 Relationship with Support Vector Machines says the following: The perceptron criterion is a shifted version of the hinge-loss used in support vector machines (see Chapter 2). The hinge lo...

loss-functions support-vector-machine perceptron binary-classification hinge-loss

geocalc33

@Thorgott can you delve into further details?

anakhro

8:13 PM

@geocalc33 what does it mean for two submanifolds of M to be transverse?

Thorgott

sure, $\phi\vdash\phi\land\phi$

4

anakhro

Thorgott is rippin people to shredz

Alessandro Codenotti

Thorgott is finally doing some logic

geocalc33

@Thorgott I am trying to follow your logic

@anakhro it means that the tangent spaces added together generate the dimension of the ambient space

Thorgott

small brain topology vs big brain tautology

6

anakhro

8:21 PM

@geocalc33 So pick your ambient space. You seemed to be suggesting R^2 before, so maybe start there.

Alessandro Codenotti

Earlier I discovered yet another very awful space: The Cook continuum is a compact metric space with the property that all continuous functions to itself are either constant or the identity

Thorgott

that sounds kinda cool

tell me how to define that space and I will retract this statement

geocalc33

@anakhro if the ambient space is R^2 then we need the tangent spaces of the two submanifolds to add up to 1+1=2

so $\Bbb R^2 \cap \Bbb R^2$ is the answer?

I literally have no idea

anakhro

Why can't you just apply your definition to this to find out yourself?

geocalc33

I have no idea how to apply my definition. Maybe it's R^1 cap R^1

Alessandro Codenotti

8:30 PM

@Thorgott I don't know, I haven't checked the details. The paper is quite short though, we can look at it if you want

anakhro

@geocalc33 What is stopping you from applying you definition?

Thorgott

"Continua which admit only the identity mapping onto non-degenerate subcontinua" is such a fun name

wow, there's even commutative diagrams in there

Alessandro Codenotti

continuum (plural continua) is just a very fancy sounding name for compact connected metrizable space

Thorgott

not necessarily with cardinality continuum?

geocalc33

@anakhro I am trying to generate the tangent space of R^2 using two 1-manifolds, so I can take two 1-manifolds that intersect, and add the tangent spaces at that point. With this I can get two vectors that span R^2

Alessandro Codenotti

8:35 PM

@Thorgott That's a consequence of those properties, isn't it?

Thorgott

singleton?

Alessandro Codenotti

yeah ok that's the only exception

Often people assume that they are dealing with nondegenerate continua (continua with more than one point)

Thorgott

why can't you get larger than continuum?

anakhro

Transversality (mathematics)

In mathematics, transversality is a notion that describes how spaces can intersect; transversality can be seen as the "opposite" of tangency, and plays a role in general position. It formalizes the idea of a generic intersection in differential topology. It is defined by considering the linearizations of the intersecting spaces at the points of intersection. == Definition == Two submanifolds of a given finite-dimensional smooth manifold are said to intersect transversally if at every point of intersection, their separate tangent spaces at that point together generate the tangent space of...

@geocalc33 Note that you have put yourself in the second case.

Where the sum must be direct.

Alessandro Codenotti

@Thorgott There is a countable dense set $D$ and every element of the space is the limit of a sequence in $D$, of which there are continuum many ($|D|^{\aleph_0}$)

anakhro

8:38 PM

What does it say there about the dimension of their intersection?

Thorgott

why's there a countable dense set? are we assuming separable?

Alessandro Codenotti

compact metric spaces are always separable

Thorgott

huh, I didn't know that

Alessandro Codenotti

Cover with balls of radius $1/n$, pick a finite subcover, pick a point in each ball. Repeat with $n+1$ and so on

geocalc33

@anakhro it says the dimension of the intersection is a point, or a zero-manifold

Thorgott

8:40 PM

right, I just thought of that as well

that's good to know

Alessandro Codenotti

This is really a proof that totally bounded implies separable, but compactness obviously implies totally bounded

In fact a metric space is separable iff it has an equivalent totally bounded metric

Lukas Heger

infact, for metric spaces separable is equivalent to Lindelöff (every open cover has a countable subcover) which implies compact metrizable => separable as well

Thorgott

actually, why can't we get anything less than continuum

anakhro

@geocalc33 What is the dimension of R^2?

geocalc33

2

I know that one!

Alessandro Codenotti

8:43 PM

fix $x_0\neq x_1$, and look at $f\colon X\to\Bbb R$, $f(y)=d(x_0,y)$. The image of $f$ contains the interval $[0,d(x_0,x_1)]$ since $X$ is connected

Thorgott

ah, neat

anakhro

@geocalc33 So apparently if you want to get R^2 as a transverse intersection of submanifolds of R^2, you need to change your approach.

Alessandro Codenotti

@AlessandroCodenotti here one direction is obvious, the other is that if $X$ is separable, then it embeds into $[0,1]^\Bbb N$, and the metric it gets as a subspace from there is totally bounded since the Hilbert cube is compact

geocalc33

@anakhro so how do you write R^2 as two transverse manifolds intersecting?

Lukas Heger

I wonder if every non-compact metric space admits an incomplete metric

Alessandro Codenotti

8:46 PM

yes

user2103480

@MikeMiller I think the main text for that lecture is armstrong

Lukas Heger

@Alessandro how do you show this?

anakhro

@geocalc33 Did you try applying your definition to the case which Thorgott suggested?

Alessandro Codenotti

It's nontrivial. There is a proof in Engelking but I remember reading a question on MSE on that same topic, let me see if I can find it

geocalc33

I don't understand how $\Bbb R^2=\Bbb R^2 \cap \Bbb R^2$ @anakhro

anakhro

8:48 PM

What does the right hand side say?

geocalc33

how can $\Bbb R^2$ intersect itself transversally?

anakhro

Well try out the definition.

What do you have to check?

user2103480

@Thorgott lmao finally

Alessandro Codenotti

20

Q: Is there always an equivalent metric which is not complete?

I have seen that completeness is not a topological property like compactness or connectedness. I have seen some examples also showing that there are two equivalent metrics one of which is complete and the other one is incomplete. I want to know some general result. Consider any metric space $(X,d...

real-analysis general-topology analysis metric-spaces

geocalc33

@anakhro gotta check that $T\Bbb R^2 + T\Bbb R^2=T\Bbb R^2$

Lukas Heger

8:50 PM

thanks

anakhro

@geocalc33 O.K. so have you checked that?

geocalc33

@anakhro yes and I found that $2+2\ne 2$

Alessandro Codenotti

@LukasHeger The answer by Eric Wofsey is kind of black magic but quite nice

user2103480

although I seriously don't want to see Logic-Thorgott

anakhro

@geocalc33 Look at this again. What does the + mean?

user2103480

8:52 PM

he'll do topos theory or some shite smh

Lukas Heger

@geocalc33 in the definition of transversal intersection, there is no requirement that the sum is a direct sum

Thorgott

@user2103480 this isnt even my final form

geocalc33

the plus sign means addition of the tangent spaces

anakhro

@geocalc33 So what does the left side equal?

geocalc33

4

user2103480

8:55 PM

"You can't defeat me" - "I know, but he can"

anakhro

@geocalc33 Why do you say 4? That isn't a tangent space.

That is a number.

Rithaniel

Thor's final form is "inventing new axiom sets"-Thorgott

geocalc33

@anakhro the tangent space of R^2 is straight lines from a point

so it looks like $\star$

anakhro

@geocalc33 I think you mean something a lot more specific than that. However, I don't know what this has to do with how that sum is equal to just "4". Do you know what a vector space is?

geocalc33

it's a space of elements closed under scalar mult and addition

anakhro

8:59 PM

That sounds more like a vector subspace, but close enough.

What does it mean to add two vector subspaces?

geocalc33

you just add the elements together

anakhro

Okay, let's do a little example.

user2103480

@Rithaniel that is easier than it sounds

geocalc33

$U+V=\{u+v|u\in U, v\in V\}$

user2103480

the hard part is showing that they add something new, or don't add something new

anakhro

9:01 PM

Consider R as your vector space. R is of course a subspace of itself.

What is R+R, as the addition of vector subspaces?

geocalc33

R

Rithaniel

There's also a thing about showing that they can never contradict themselves, right?

anakhro

Great.

So try looking at your question again.

geocalc33

so R^2 + R^2 = R^2 when considered vector spaces

user2103480

@Rithaniel I was half joking. If they're contradictory, they can prove anything, and that's a known axiom system, and if they're (assumably) non-contradictory, they need to not be equivalent to known axiom systems

Rithaniel

9:06 PM

Interdasting

I always really enjoy particularly abstract topics in math

anakhro

@geocalc33 So what is your conclusion?

user2103480

This is overly simplifying though. Showing equivalence between different statements and axiom systems is a huge topic

geocalc33

@anakhro my conclusion is that R^2 can be written as, R^2 transversal to R^2

user2103480

@Rithaniel you may get over that someday. There's almost always a point where you ask whether certain abstractions are actually necessary, sensible or useful

and tbh at some point almost any topic is pretty abstract

Rithaniel

I've kind of seen that already. Like, talking about magmas. They're honestly way too nebulous

How do things like the group axioms fit into axiom sets? They're in addition to the base axioms, right?

user2103480

9:10 PM

@Rithaniel nope, they're just the group axioms

skullpatrol

@Rithaniel Your vid was nebulously well done :-)

user2103480

but you can't do all the considerations of subgroups etc without a set theory as a metatheory

Rithaniel

Ah, the talk I gave the math club about groups and the monster group? Danke schön

skullpatrol

Yup

user2103480

the group axioms are there to reason within the group

to reason about groups, you need more

and it's not like you just take the union of the axioms

Thorgott

9:13 PM

remark about gRoUP oBjeCts

Rithaniel

(Yeah, I still can't bring myself to watch the video, myself)
Alright, so you could add axioms to the group theory axioms that would facilitate those metatheory axioms?

user2103480

but rather that you start with the set theory axioms, and within that formal system, you define groups and study them

@Rithaniel not exactly. sets are our building blocks of the universe

if you add group axioms, you say that our universe is a group to some operation

which is an interesting concept but uhh not your goal I think

Rithaniel

Okay, so axioms only talk about the thing they are inside, and to analyze that thing, you need axioms from outside it?

user2103480

basically

Rithaniel

Again, interdasting

geocalc33

9:16 PM

@anakhro do you know how to combine metrics $ds_1^2 \otimes ds_2^2= \bigg (\frac{dxdy}{xy}\bigg)\bigg( \frac{dxdy}{y-xy} \bigg)$?

user2103480

Philosophically, logically? All very interesting stuff

Rithaniel

So, to analyze axiom sets, you kind of need axioms from outside the space of all universes? That's mind-bendy

user2103480

For doing serious mathematics? Mostly very useless, unless you're a model theory god

Serious = everyday mathematics. Not to offend anyone, I very much like logic

and respect the immense difficulty of the area

Rithaniel

Eh, it's all in pursuit of solving problems in abstract environments. Solve something in generality and you understand the world better

user2103480

Bold claim

maybe infinitesimally

Alessandro Codenotti

9:24 PM

@user2103480 triggered

user2103480

@AlessandroCodenotti you still have time to become that model theory god

Alessandro Codenotti

Right now I'm trying to understand the Hrushovski construction, so I feel very far from even a remotely decent model theorist :P

Ted Shifrin

Howdy, demonic.

Alessandro Codenotti

Hi Ted

user2103480

@AlessandroCodenotti try to find the computability aspect of it :P

Ha! As always, there's a computability aspect

conferences.leeds.ac.uk/lc2016/wp-content/uploads/sites/4/2017/…

Alessandro Codenotti

9:32 PM

There is also some work on the computability of Fraisse limits, which are like the easier, more classical case, of the Hrushovski construction (or so I'm told since I don't understand the latter)

user2103480

RIP

sounds harsh

Alessandro Codenotti

It's not something that I need to worry about so it's fine

Thorgott

9:58 PM

"In mathematical logic, specifically in the discipline of model theory, the Fraïssé limit (also called the Fraïssé construction or Fraïssé amalgamation) is a method used to construct (infinite) mathematical structures from their (finite) substructures. It is a special example of the more general concept of a direct limit in a category."

nice

Analytic number theory

Transcript for

$Mathematics$ Mathematics