Mathematics - 2018-08-22 (page 1 of 3)

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12:40 AM

@Isa: Thinking about argument per se isn't going to help you here. You have two problem points, both $+i$ and $-i$.

1:10 AM

how do you get to the list of all questions that you have favourited thus far? Like I just prefer to go thru all the ones that appear on the recommendations list for my current problem and favourite the ones I believe I need to return to. I would of course facilitate this in the browser's functionality, but it would be heaps better if I can go through them via my SE acct

@Adam Go to your profile, then the favorites tab

@TedShifrin yes, I know. But when defining the functions $f_1$ and $f_2$ I need to specify the interval where $\theta_i$ is. I think it is $f_1(z)=\sqrt{r_1r_2}e^{(\theta_1+\theta_2)/2}=f_2(z)$ and $\theta_1\in(-\pi,\pi)$ and $\theta_2\in(0,2\pi)$. Is it correct ?

1:28 AM

@B.Mehta ok thankyou

1:39 AM

@Isa: I don't understand this. You need to think about two angles: one the angle around the point $+i$ and the other the angle around the point $-i$. If you make a circle around one of those points but not about the other, then the function changes value by a factor of $-1$. So you need a domain that will force you to go around both of those points the same number of times. There are two possible domains that will achieve that.

2 hours later…

3:10 AM

[Random]

Pseudo, Quasi, Semi, Near,

Almost

Anti, A, Non, Retro, Inverse, Pre

A Quasilogic = my thinking style

The deduction rules follow a partial order. However unlike classical logic, there are seemly random, sometimes unpredictable, often emotional and memory driven jumps between steps

i am actually more like an artist in thinking than a scientist

The illusion of order and organisation when people saw my notes is likely from them perceiving this well order, but not the jumps

4:04 AM

@Semiclassical, I am doing this Rudin exercise: In metric space $X$, limit point compactness implies compactness. Hint goes like this: since $X$ limit point compact, it has countable base. And hence any cover of $X$ has countable subcover of $X$. I can't understand how this last line is derived.

user image

4:21 AM

@geocalc33 There are strictly more than 1

2 hours later…

6:15 AM

@LeakyNun, will you please look at my question above?

6:45 AM

is anyone know html coding?

Tobias Kildetoft

7:40 AM

I will be starting the algebra course on Monday. My plan is to go over some motivating examples in the first lecture, to give some idea of the sort of problems we will be able to solve at the end of the course. These will be ones that can be formulated without any of the theory from the course, and I am looking for more good examples.

hi @TobiasKildetoft

what do you mean by "algebra"?

Tobias Kildetoft

At the moment, each "major" topic has one such example planned. For basic number theory the goal is to explain how RSA works. For group theory we will explain how to count things up to symmetry using Burnside.

@LeakyNun Hi

Good noon

@LeakyNun the art of solving for unknowns as proposed by al jibr.

Tobias Kildetoft

For rings, we will count integer solutions to $x^2 + y^2 = n$. For polynomials we will classify finite fields (= simple commutative rings). And for representation theory, we will study the stochastic variable giving the number of fixed points when we shuffle cards.

@LeakyNun The topics just mentioned :)

@TobiasKildetoft so basically what our university offers as 20 courses is once course in your university?

one*

Tobias Kildetoft

7:43 AM

I am looking for some more good examples. Especially ones coming from other areas of math, so the students can get an idea that algebra is also good to know even if you don't want to specialize in it

@LeakyNun The contents are really not that broad

I think a lot of courses these days smash together a wide variety of concepts. Academia.SE talks about this from time to time.

@TobiasKildetoft I think an example is how the Weil group of a local non-archimedean field is canonically isomorphic to the units of the field

Tobias Kildetoft

@LeakyNun Right. Now explain that without using the terminology introduced in an introductory algebra course

you mean using?

Tobias Kildetoft

no, without

The point is that an example only motivates the introduction of a bunch of theory and concepts if we can give the example without mentioning those concepts

7:49 AM

oh

how about p-adic numbers

Tobias Kildetoft

As a solution to what problem?

as a motivation of some algebra

I like to think of them intuitively as integers that go infinitely to the left

just like real numbers go infinitely to the right

Tobias Kildetoft

@LeakyNun I would prefer to start with some question that seems natural, and which then gets an answer (possibly just in an easier way) by applying the new concepts

Like "number of integer solutions" is often a good one (unfortunately, I will only be able to do the very simple one with sum of two squares).

Alessandro Codenotti

Speaking of p-adics... So we have $\Bbb Z_p\cap\Bbb Q=\Bbb Z_{(p)}$ and the latter is hence dense in the former

yes

7:59 AM

If I have $2^{-\frac{1}{3}}(-\frac{4}{3}x^{-\frac{7}{3}})$ Do I multiple the x expression by the 4/3 in the following manor: $-\frac{4}{3}\frac{x^{-\frac{7}{3}}}{1}$

I think a lot of the usual examples come from number theory - but for something thats not nt maybe (?) it would be fun to wave your hands a little bit (or maybe a lot) about how groups show up in topplogy and deduce certain different spaces are not homeomorphic

Tobias Kildetoft

@loch Unfortunately, they have not seen topology at this point

@AlessandroCodenotti maybe you should find the 4 idempotents in the ten-adic land

@TobiasKildetoft bunch of topology-illiterates

Of course i wouldnt expect them to :p i was thinking that its something that id be like woah thats pretty cool even without knowing how things work

Alessandro Codenotti

@LeakyNun But 10 is not prime, I don't like it

8:01 AM

i visit here simply for the search terms.

ἀγεωμέτρητος μηδεὶς εἰσίτω

@AlessandroCodenotti but ten-adic is in your blood

two-adic is more abstract

Alessandro Codenotti

Hmmm why is the quotient topology on $\Bbb Z_p/p^n\Bbb Z_p$ discrete?

because p^n Z_p is open

recall that { p^n Z_p | n } is a basis at 0

Actually for polynomials/ fields if linear algebra is assumed one it probbaly doesnt take too much time to do things like impossibility to trisect an angle etc

Tobias Kildetoft

@loch Hmm, that might be a neat thing to do, yeah

2 hours later…

10:27 AM

Terrence Tao needs help programming

10

A: What programming language should a professional mathematician know?

My answer is: TikZ This is a programming language, often used in combination with LaTeX, for producing high-quality graphics. I view this language as important for mathematicians, not because mathematicians will use it to solve their mathematical problems, but rather, because mathematicians wil...

10:51 AM

hi @AlexClark

Hey @LeakyNun

What're you working on?

nothing much

@AlexClark are you an Agalloch fan?

I am indeed!

cool

11:04 AM

Do you listen to them @user1732?

11:15 AM

This is most probably a silly query but is $\tan(x)$ a function ? Because if $f: K \rightarrow R$ with $K$ comapct then the image of $f$ is compact too, but then if you let $f = \tan x$ and $K = [0, 10]$ image of $f$ is not bounded and hence clearly not compact, right ?

@alxchen no it isn't a function

because $\tan(\pi/2)$ is not defined

@LeakyNun Why can't we let $\tan (\frac{\pi}{2}) = \infty $ ? Sorry this is probably dumb

@alxchen Because $\tan(\frac{\pi}{2} - \epsilon)$ is very large and positive, but $\tan(\frac{\pi}{2} + \epsilon)$ is very large and negative.

OK, but if you define $g(x) = |\tan x|$ and "define" $g(\pi/2) = \infty$ then why is $g$ not a function ?

Well it would be a function, but it would in particular be a surjective function onto the projective line, the totality of which is compact.

So the contradiction you raised doesn't exist: for $K = [0,10]$, the image of $K$ is just the whole projective line, aka the one-point compactification of $\Bbb R$.

11:24 AM

ok thanks for replying but I don't understand compactification lol

The very rough way to think about it is as taking "negative infinity" and "positive infinity" to be the same, and adding that point to the space. So the real line closes up into a circle, with this "infinity" added.

This circle is (homeomorphic to) a closed and bounded set in $\Bbb R^2$, and therefore is compact.

Alessandro Codenotti

When you add $\infty$ to $\Bbb R$ you need to specify what's the topology on $\Bbb R\cup\{\infty\}$ to speak of continuity and compactness

lol i don't understand topology too

so many things to learn

so little time

3 hours later…

2:15 PM

@AlessandroCodenotti ok so the closure of $\{0\}$ is $\bigcap_n I^n$, and for a topological group, T1 iff T2

so T2 iff $\bigcap_n I^n = (0)$

2:43 PM

Hello!!

2 coins are thrown 20 times.
I want to calculate the probability
(a) To achieve exactly 5 times the Tail/Tail
(b) To achieve at least 2 times Tail/Tail

Is the probability at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?
Could you give me a hint for (b) ?

Alessandro Codenotti

2:59 PM

@ÍgjøgnumMeg It's probably in Milne's notes but I can't find it, what does $\Bbb N\mathfrak p$ mean where $\mathfrak p$ is a prime ideal of a number field $K$?

As used in theorem 7.14 for example

@LeakyNun I see, thanks

@LeakyNun if you get protonated will it make dative bond ?

@Fawad e.g.?

@LeakyNun h3o+

there is indeed what you would call dative bond in h3o+ right

Other websites were saying it has covalent bond

so I was little confused 🤷‍♂️

3:05 PM

a dative bond is a covalent bond

dative bond = coordinate covalent bond

But a covalent bond need not to be a dative bond

tq :)

@Fawad those are all names

they have no meaning in real life

@AlessandroCodenotti I think it's the norm?

@loch that makes sense, apart from the notation lol

i.e. $|\mathcal{O}_K/\mathfrak{p}|$

Alessandro Codenotti

3:14 PM

@loch It'd make sense, but he uses a different notation for the norm

oh welp

Alessandro Codenotti

Namely $N_{L/K}$ as every sane person

@AlessandroCodenotti no that's a different norm

$N_{L/K}$ is the norm of a non-zero element in $L$ and it outputs a non-zero element in $K$

on the other hand we are talking about the norm of an ideal

Alessandro Codenotti

Oh of course

Could be a weird notation for the ideal norm then

Ah, found it, $\Bbb N\mathfrak a=(\mathcal{O}_K:\mathfrak a)$

There is no "dative" bond in H$_2$O (nor in H$_3$O^+)

Its simply covalent bonds

"polar" ones though

Alessandro Codenotti

3:22 PM

So it's the usual ideal norm, good

@LeakyNun also that one is not the orthodox way to define "dative" bond

@Rudi_Birnbaum I would also say that there are no covalent bonds in H2O or H3O+

they are arbitrary names

@LeakyNun "coordinate bond" not neccesarily is "covalent".

in real life all 3 bonds in H3O+ are the same

Hmm ...

1. Types of bonds are not found in nature

I agree

Though they can be consistently defined

3:25 PM

and also I think the fact that H3O+ is an ion messes up the idea of dative bonds

dative bonds are supposed to be the middle-ground of covalent and ionic bonds right

Bonds exist in the world of the "chemical models" as I call it.

i.e. they are covalent bonds but with a considerable amount of dipole

not actually in the classic understanding.

well the classical understanding says that they are covalent bonds where the "two electrons come from the same atom"

every word in my quote is problematic

In the classic understanding a bond is dative if it (a) connects fragments that are "happy molecules" themselves

and

3:27 PM

3 mins ago, by Leaky Nun

in real life all 3 bonds in H3O+ are the same

this is wrong

(b) where one partner can be identified as

in real life you can't count "3 bonds"

electron pair donor

and the other one as

electron pair acceptor

I see

the "spectrum" ionic-covalent exists independent of that

thats all in a kind of classic (Lewis) based understanind

and different flavours exist.

@LeakyNun yes and no

3:30 PM

yes as in you can count that there are 3 bonds in total (eh, sort of) but no as in you can't identify the individual bonds

Its a difficult question ...

most of all the 6 electron pairs are "bound" to the molecule with different energy!

sorry 3 I mean

3 Electron pairs,

Thats one actually puzzling thing

in my humble opinion

user image

the only way to avoid huge confusion about the

YES!

Well right here you have 2 and 1

electron pairs.

But not 3 same level ones

I see

the only way to avoid huge confusion about the nature of the chemical bond

is not retstrict it to a mdel existence. and not to try to search for "real bonds" in nature.

3:34 PM

what is the purple vs green?

amplitude of the orbials

positive or negative

since time independent wave functions are purely real

(can be made purely real)

but they still have amplitude

you have to take $|\psi|^2$

to get the observable density $\rho$

does the sign mean anything?

Well its the amplitude of the wave function ...

like a light wave it has an amplitude

what is a negative amplitude?

Can't observe it, you only observe $|\psi|^2$ and not $\psi$

$\psi$ is real valued but also can be written complex valued.

3:39 PM

6 mins ago, by Leaky Nun

user image

how should I interpret the orbitals in terms of the "original" electrons?

For time dependent problems the wavefunctions are inherently complex

Its mostly a basis.

What you mean by "interpret"?

how do I recover the individual bonds?

Its NOT possible

There are no individual bonds ...

You need a model for that

there are of course models

froem which you can "choose" but its nothing generic

You use the model to explain chemistry

But you do not search for the model in chemistry. Its probably very uncommon a view for a mathematician.

In the MO model you could search for overlap

WHat you also do is to "rotate" the orbitals.

why are there not three orbitals of the same potential?

Thats what I say!

There are no three equivalent two-electron bonds in the Oribtal picture

Just 1 and 2 (the three highest orbitals)

Now you can form arbitrary linear copmbinations of the orbials

for example.

That wont change any QM observable

3:46 PM

so I'm wondering why there aren't 3 equivalent 2c2e bonds

Thus its already within the Orbital piture ambiguous

I understand that

You can form 3 spatially equivalent bonds

but they will be no longer eigenfunctions

with an energy eigenvalue

rather with an expectation value

and when you kick out electrons and measure the energy you need you get the same result as the diagramm suggests

(roughly speaking)

so you're saying that experimentally the first two electrons are easier to kick out

Yes

I see

Though you cannot distinguish the three bonds

by any means

For that I say one has different models in mind

they even contradict themselves

3:50 PM

47 mins ago, by Fawad

@LeakyNun if you get protonated will it make dative bond ?

but each can be used to understand certain aspects of nature

let's go back to here

should we even talk about dative bond in an ion?

@Rudi_Birnbaum that sounds an awful lot like complementarity, lol

@LeakyNun In principle its a correct example

because H$_2$O can live on its own

and H$^+$ VERY roughly speaking can as well

but not every protonation is converting H2O to oxonium

3:51 PM

For that It can indeed be considered as a kind of simple dative bond.

But in my lectures I avaoid that since it can lead to confusion about covalence

Its better to regard it as covalent bond imho

@LeakyNun What do you mean?

NH3 + HCl -> NH4Cl

But here you don't have H2O

?

do we?

@Semiclassical When models contradict I dont think its complementary

Tell that to Bohr.

Complementarity (physics)

In physics, complementarity is both a theoretical and an experimental result of quantum mechanics, also referred to as principle of complementarity. It holds that objects have certain pairs of complementary properties which cannot all be observed or measured simultaneously. The complementarity principle was formulated by Niels Bohr, a leading founder of quantum mechanics. Examples of complementary properties that Bohr considered: Position and momentum Energy and duration Spin on different axes Wave and particle-related properties Value of a field and its change (at a certain position) Entanglement...

3:55 PM

@Semiclassical No thats unrelated

@Semiclassical Thats a deep princple of quantm mechanics

@LeakyNun you have written : "but not every protonation is converting H2O to oxonium"

But in "NH3 + HCl -> NH4Cl" is no H2O.

right

Oh sorry I misunderstood

I thought you mean not every protonation of H2O ...

leads to H3O+

You man you can protonate other stuff but H2O

right

thats right!

H2 can be protonated

to give the most aboundant ion in the universe!!

H$_3^+$ a cool guy!

its symmetry is D_{3h}!

Insert CH$_5^+$

3:59 PM

gleichseitiges Dreieck

@Secret Yes another fascinating guy!

gleichseitiges Dreieck = equilateral triangle

thx .. keep mixong those up learned the terms much too late

like edge and vertex ...

so are you saying that H_3^+ has no dative bonds?

That again depends on the model in which you decribe it. But e.g. in QM in the orbital picture there anyway only one doubly occupied orbital

But that transforms like the trivial representation in $D_{3h}$

Each orbital can be assigned to exactly one specific irrep

So QM suggests its only "one bond"

13 mins ago, by Leaky Nun

should we even talk about dative bond in an ion?

4:04 PM

The Lewis model suggests its three 1/3 bonds

@LeakyNun Yes definitely

32 mins ago, by Leaky Nun

user image

also LUMO -2.9????

That ion or not (almost ) never makes any difference

Yes,

the orbital energies do NOT correspond to any real energy

in the first instance

You only have ONE energy for the whole system of 18 electrons

the orbitals are merely a "vector basis" (Hilbert space)

to decribe the n-electron wavefunction

to SOME approximation

well MOs don't exist because we stole the idea from the hydrogen ion

some energies however can be used as a rough approximetion to some physics

All the eigenfunctions of the hydrogen atom form a (almost) complete basis

neglect the continuum

Here is a point-set topology fact I just learned that I have found useful; @AlessandroCodenotti might care.

4:08 PM

Thus we can use these to compose some "trial" wavefunctions

and then we minimise the total energy expectation

by variation of coefficients

what we get is the orbitals of a molcule

@MikeMiller I think I would also care

It is well-known that a proper map to a locally compact space is closed. But actually, this is true in much more generality: a proper map to a compactly generated space is closed. This means that a set $Z$ is closed if and only if $Z \cap K$ is compact for any compact set $K$. Every metric space is compactly generated.

what is a proper map?

Inverse image of compact is compact

It's like "compact fibers" but better

Assume $Y$ is compactly generated, so a compactly closed subset is closed. We want to show that if $f: X \to Y$ is proper, and $Z \subset X$ is closed, then $f(Z)$ is compactly closed, which would imply closed.

So the actual lemma should be "A proper map sends closed sets to compactly closed sets."

Then the point is that $f(Z) \cap K = f(f^{-1}(K) \cap Z)$. Of course the first term $f^{-1} K$ is assumed compact, so $f^{-1}(K) \cap Z$ is compact, so $f(Z) \cap K$ is compact, as desired.

in fact "Every proper map to $Y$ is closed" is equivalent to $Y$ being compactly generated. I learned this (and this proof) from a 1-page article by Dick Palais titled "when are proper maps closed?"

@Rudi_Birnbaum talk to me about 3c2e in H3^+

Alessandro Codenotti

4:15 PM

Unluckily "compactly generated" is not among the properties known to the pi-base so I can't easily look up a compactly generated not locally compact space

@LeakyNun Yes thats what the QM orbital picture predicts 3c2e.

@LeakyNun Both (the only) two electrons are in one orbital

@AlessandroCodenotti Any metric space that is not locally compact gives an example. So for me I care about Banach spaces and Banach manifolds

@Rudi_Birnbaum why do 2H tend to become H2?

and this orbial has the shape of the trivial irrep

son it encloses all three nuclei in the same way.

thus its one bond of two e^- for al three nuclei.

because it lowers to total energy of the system.

Now:

Alessandro Codenotti

@MikeMiller Oh, of course, that's a nice example

4:18 PM

In the orbital picture its because of "overlapp" thats a bit lengthy to explain.

Alessandro Codenotti

I had never met compactly generated spaces until 5 minutes ago but it seems that a space needs to be really ugly to not be compactly generated

Essentially for you as a non chemist and mathematician it might be a good way to think like:

I start with one orbital of the shape of a ball

But when I form the bond I need a linear combination of the two balls.

I already can guess the shape of the linear combination, by using group theory

It can be either trivially symmetric

or it can be anti-symmetric

It turns out that the trivially (we say totally) symmetric has much less curvature

than the "anti-symmetric" one.

and that's the BO

@AlessandroCodenotti Indeed so

Yes

If you out in two more e^-

the go into the ant-symmetric one, which has a nodal plane

and the total system has more energy than two individual H^-

so thats not happening

The point group of H$_2$ we call $D_{\infty h}$.

Its a continous group, but for us only the trivial and the anti-symmetric (sign) irrep matter

Alessandro Codenotti

4:23 PM

@Mike I'm thinking about some exercises from Hatcher. $H_n(X,A)\simeq H_n(X)$ for all $n>0$ when $A$ is a finite subset of $X$, right? But we need $X$ nice enough to say what happens with $n=0$, for example it should be $T_1$ or at least the induced topology on $A$ should be discrete I think (Hatcher asks about $X=S^1\times S^1$ so it's as nice as it gets)

One way to see covalence is to see it as a minimzation of curvature

curvature (the Laplacian) is one part of the Hamilton operator

that "measures"/determines the energy

@AlessandroCodenotti Yes, I suppose so

Alessandro Codenotti

Hmm, actually I'm now thinking we need $A$ discrete even for $n>0$

Can a finite space have nontrivial $H_n$ for $n>0$? I suppose so

Another complementary picture -- which is kind of missleading -- is to say that the electrons between the nuclei make the glue - but its better to forget that model.

Alessandro Codenotti

The pseudocircle works

4:26 PM

Its old and grossly abgelehnt

(word) but some still "believe" in it ...

why is it wrong?

It contradicts QM, you can partition the total energy into kinetic energy ($\nabla$) and the e^- potential energy and nuclear repulsion term

and then you see that the nuclear repulsion is not overcome by the e^- ic potential

then why does the BMO have the lowest energy among the linear combinations of the AOs?

Of course you can do that stuff only for 2-Atom molecule because you cannot vary only 1 internuclear distance in a big molecule ...

e.g. Because its total curvature is smaller than in the anti-symmetric

But also because of the potential (in part)

of courcse

you cannot separate these either ...

@Alessandro was your question answered? lol

4:32 PM

if you change the shape of the wave function you always change kinetic and potential Energy ...

Alessandro Codenotti

@ÍgjøgnumMeg Yep, turns out it's a (debatable) notation for the norm

@Alessandro I see, it was about $N\mathfrak{p}$?

Alessandro Codenotti

$\Bbb N\mathfrak p$ actually using Milne's notation

Ah right I see

The anti symm looks like the infty symbol while the symmetric looks like an O. ;-)

Alessandro Codenotti

4:33 PM

Which is why I call it debatable :P

Hi @ÍgjøgnumMeg!

Sure, not a fan either hahaha

Hi @Rudi :)

@Rudi_Birnbaum so why does the BMO have the lowest potential?

@Rudi my bike is fixed and ready to goooo

because both kinetic and electrostatic rep are lower, though the e-stat repulsion is differs less

"O" has less curvature than "8".

@ÍgjøgnumMeg great man!

4:36 PM

why is the curvature connected to the energy?

$\hat{H}=\hat{T}+\hat{V}$

$\hat{H} \Psi = E \Psi$

$\hat{T}=c \nabla$

@Rudi yis!

Happy

@ÍgjøgnumMeg :-)

@LeakyNun rough physical idea might be that to flying a curve you need acceleration ...

boi

$E$ ist the energy

4:39 PM

user image

This looks NOTHING like a circle

(Not to mention on a circle you won't eventually get stuck at two points)

Btw the Hamiltionan $\hat{H}$ that determines the energy by the Schrödinger equation commutes with the symmetry elements of the molecule.

Eigenfunctions $\Psi$ have minimal energy "expectaion" value $\overline{E}= \frac{<\Psi|\hat{H}\Psi>}{<\Psi|\Psi>}$

Random PhD comment: The PES makes no sense

Realising that x and y axes get swapped

Ah... NOW it makes a lot more sense!

@LeakyNun did that make sense to you?

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