Mathematics - 2018-06-18 (page 2 of 4)

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5:00 PM

I'm trying to include the data of "where $M$ hits $W$", so that we can run that rank stratification

If $f(q) = p \notin W$, can we not by definition assume that the rank is $\dim N$, because it's transverse to $W$ vacuously if it doesn't hit $W$?

wow interesting @anakhronizein

Very!

@BalarkaSen Hmm. I suppose so.

MatheinBoulomenos

@BalarkaSen I'm not sure I understand the topology on $C^\infty(M,N)$ you write that you take the union of the topologies under inclusion maps $C^\infty(M,N) \hookrightarrow C^k(M,N)$. But the inclusion $C^{k+1}(M,N) \hookrightarrow C^k(M,N)$ is continuous, so if you take the union of these topologies, you just get the subspace topology coming from $C^1(M,N)$. Do you mean the intersection of these topologies?

5:02 PM

@MatheinBoulomenos yes, intersection

MatheinBoulomenos

ah, so it is actually the limit topology

@anakhronizein That is dope. When can we integrate a cohomology of two symplectic forms to an isotopy? inb4 h-principles :3

@BalarkaSen So you want to first associate a map $C^\infty(M, N) \to \text{Map}(M, \Bbb N)$, associating a "dimension of obstruction space" to each point, which is zero at $q$ if $f(q) \not \in W$.

You want to say something about semicontinuity of this map, then take preimage of the least element, which should be open, or something like this

Here $\text{Map}(M, \Bbb N)$ should be maps of the appropriate type of semicontinuity

@MatheinBoulomenos I meant the basis of open sets are $\mathcal{V}(K, U, r) = \{f \in C^\infty(M, N) : J^rf(K) \subset U\}$ for a compact $K \subset M$ and $U \subset J^r(M, N)$. That doesn't agree with the intersection nor the subspace topology from $C^1(M, N)$, I seriously doubt that.

There are much smaller open sets than the ones coming from $r = 1$

Sorry about interrupting for you there

I think this is the same as what I was saying above, since the 'bad set' of a map $f: M \to N$ is determined by knowing $\text{rk}^{-1}(1, \infty)$ - you can only have 'bad rank' if you have image inside $W$

5:07 PM

I was confused about this point myself, no worries

I think the following is obvious (at least for finite dimension) but I wanted to be sure.

MatheinBoulomenos

but is that actually the limit?

Who even cares

It's the weak smooth Whitney topology

That's all I ever need

:P

MatheinBoulomenos

I just thought there's a neat version of an abstract "Mittag-Leffler theorem" which you can use to prove Baire and it seems like it might apply here

Oh, I retract my statement then

5:10 PM

Context: A Gram matrix of a set of vectors $\{v_i\}$ in an inner product space is, by definition, the Hermitian matrix of inner products i.e. entries $G_{ij}=\langle v_i,v_j\rangle$

Tell me about it

Problem: Let be a metric space and A any set for which there is a one-to-one mapping f of A onto the set X. Show that there is a unique metric on A for which f is an isometry of metric spaces. Question: Is $f : A \to X$ just injective or is it also surjective?

@MikeMiller Ah I see what you mean. Good formalism with $C^\infty(M, N) \to \text{Map}(M, \Bbb N)$, by the way!

Suppose I've got $G\in\mathbb{R}^{n\times n}$ as the Gram matrix of a set of $n$ real vectors in $\mathbb{R}^d$ with the usual Euclidean dot product.

MatheinBoulomenos

If $(X_{\alpha})_{\alpha \in I}$ is an inverse system of complete uniform spaces over a countable index set with transition maps $f_{\alpha,\beta}$ and suppose that for all $\alpha \in I$, there exists a $\beta \geq \alpha$ such that for all $\gamma \geq \beta$ $f_{\alpha \gamma}(X_\gamma)$ is dense in $f_{\alpha \beta}(X_\beta)$.
Then if we set $X = \varprojlim X_{\alpha}$ with canonical maps $f_{\alpha}:X \to X_{\alpha}$, then for any $\alpha \in I$ und $\beta \geq \alpha$ as above, we have that $f_{\alpha}(X)$ is dense in $f_{\alpha \beta}(X_\beta)$

5:14 PM

Wow

My question (which seems sorta obvious as I say it): This presentation of $G$ shouldn't be unique, right?

@Semiclassical Nah, let's work with $G = \text{Id}$ as one example (the only one I'm competent to come up with). Then this arises for any orthonormal basis

MatheinBoulomenos

This is neat, because you can apply it to topological vector spaces more general than just Banach etc. a friend of mine did a technical proof in a seminar on multiple complex variables where it was crucial that this also applies to Frechet spaces

i.e. if I consider a different Hilbert space of appropriate dimension then I should be able to obtain $G$ as the Gram matrix of a set of $n$ vectors in that space instead

@MatheinBoulomenos have you read this?

MatheinBoulomenos

5:15 PM

Not sure if it's actually useful here

I hope it is and I hope Balarka works it out for me

MatheinBoulomenos

@LeakyNun yes

@MatheinBoulomenos Oh very nice

@Semiclassical Wouldn't rotating the vectors induce the same gram matrix?

Is that what you are asking about?

Sure, but I don't want to be in the same Hilbert space

5:16 PM

I am going to save the theorem statement in my files

in particular, I don't want to assume I've even got the same inner product anymore

I'm guessing it doesn't matter, though, because inner product spaces of finite dimension are isometric(?)

As such, the isomorphism between two Hilbert spaces of the same dimension should automatically give me an example in the second Hilbert space

MatheinBoulomenos

I hope I didn't mess up the theorem, I translated it from French, but it should be okay

@MatheinBoulomenos how would you notate initial/terminal/zero object?

INITIALOBJECT and TERMINALOBJECT. ;)

(the example I've got in mind is something where you go from the inner product space being Euclidean vectors to, say, the Hilbert-Schmidt inner product on matrices)

5:20 PM

Another interesting math fact I learned today: if $X$ is a topological space such that $X\times\mathbb R\cong \mathbb R^4$, then $X$ is not necessarily homeomorphic to $\mathbb R^3$. Most notably, $X$ can be a contractible 3-manifold and this still doesn't hold.

MatheinBoulomenos

@LeakyNun depends on the category I'm working in. If it's an abstract category, maybe 1 for initial and 0 for terminal

(the other example in my head being if you take the vectors to be random variables $\{X_k\}$ of zero mean and the inner product to be the pairing $\langle X_i,X_j\rangle =\mathbb{E}[X_i X_j]$)

@MatheinBoulomenos I thought 1 is for terminal

@anakhronizein what about INITIALOBJECT and TCEJBOLAITINI

@mercio this is good.

I wrote a paper once on F-algebras and F-coalgebras. The category of F-algebras I called F-alg, then for F-coalgebras it was gla-F.

5:29 PM

wouldn't that imply the map is idempotent

MatheinBoulomenos

@LeakyNun is it? Maybe just go for I and T or something like that

no, not idempotent

what's the other one

@MatheinBoulomenos godforsaken notations

where applying it twice gives the original object

involution

5:30 PM

there we go

Hello all! Here is my "final" question about these representation things. After a while of consideration I decided to first post it on MO. But I would very much like to discuss it here a bit as well. With @MatheinBoulomenos @mercio, @LeakyNun @anon and anyone else who has ideas? mathoverflow.net/questions/303066/…

take R to be the set of all sets x such that $x \notin x$

:)

who volunteers to police Rudi that he doesn't ask any more questions about this?

The problem with Russel's paradox is when you let x = R, correct?

5:36 PM

the problem with Russell'

's aradox is assuming $R \ne \varnothing$

MatheinBoulomenos

@Rudi_Birnbaum if you're thinking about real representations you should mention that. People will assume complex unless stated otherwise, probably

It just needs a hint, I will never ask again (though the why would be interesting)

@Rudi_Birnbaum I think it's true if you detail how $[(V_1 \oplus V_2) \otimes (V_1 \oplus V_2)] = [V_1 \otimes V_1] \oplus [V_2 \otimes V_2] \oplus W$ where $W$ is isomorphic to $V_1 \otimes V_2$

rather, uh

yes $R$ i was being silly Max

@MatheinBoulomenos: In this formulation that unseeing problem is eliminated (\rho branches into two, thats it)

5:39 PM

so it's how I said it

correct?

yes

Thank you!

with $(v_1 \otimes v_2) \in V_1 \otimes V_2 \mapsto v_1 \otimes v_2 - v_2 \otimes v_1 \in W$

so why can't we just restrict this issue so we dont have russel's paradox

that's what we do

we impose rules for what kind of sets you can assume exists

the most common rule set are the zermelo-frankel axioms

5:40 PM

like axiom of comprehension

proofwiki.org/wiki/Definition:ZFC yup

@GFauxPas ??

???

@mercio: Uuh This is a bit fast ..

though Halmos summarizes them , let me find

5:41 PM

@GFauxPas I don't think what you said is right

find his summary

what , that the most common is ZFC?

@mercio: I start seeing what you mean ...

> the problem with Russell's aradox is assuming R≠∅

I corrected myself

@Maximus sorry not Halmos, it's another book I have

summarizes the important consequences of the rules

1) a set will never contain itself

2) for every set of sets there is the union of all the sets

3) for every set there is the power set

4) any condition on the elements of a set can be used to define a new set, a subset

5) choice

the relevant sentence from Wikipedia: "ZFC does not assume that, for every property, there is a set of all things satisfying that property. Rather, it asserts that given any set X, any subset of X definable using first-order logic exists. The object R discussed above cannot be constructed in this fashion, and is therefore not a ZFC set."

5:44 PM

@GFauxPas Thanks!

this is an appendex of a book I have on functional analysis

@GFauxPas I am reading through this book by Hrbacek

it's not all the axioms but those 5 are pretty much all you need outside of studying set theory itself

So Russell's paradox ceases to be a possibility in ZFC.

this is "Functional Analysis, Spectral Theory, and Applications" by Einsiedler and Ward

I personally would add to that "$\mathbb N$ exists"

which is more or less the axiom of infinity

$\mathbb N$ exists and you can do induction on it

which book by Hrbacek @Maximus ?

does he talk about Peano axioms?

5:48 PM

a little bit

@mercio: Where comes the restriction into play in this argument?

Introduction to Set Theory by hrbacek @GFauxPas

@mercio: Is it the V1 x V2 versus V1 or V2 alone?

ah okay he talks about the natural numbers

that's how I personally think about the axiom of infinity. there exists $\mathbb N$, you can do induction on it, if $n$ is a number then $n+1$ is a number, and so on

Here's a claim which I don't know how to prove and I find that embaressing

Assume $m\leq n$. For a real matrix $m\times n$ matrix with entries $c_{ij}$, the following are equivalent:
(A) There exist two sets of unit vectors $\{u_i\}_{i=1}^m, \{v_j\}_{j=1}^n\in\mathbb{R}^d$, $d\leq m+n$ for which $c_{ij}=u_i\cdot v_j.$
(B) There exist two sets of vectors $\{u_i\}_{i=1}^m, \{v_j\}_{j=1}^n\in\mathbb{R}^m$ for which $\|u_i\|\leq 1, \|v_j\|\leq 1$ and $c_{ij}=u_i\cdot v_j$ for all $i,j$.

A more compact formulation of condition (A) is that $C=U^T V$ where $U$ is a real $d\times m$ matrix and $V$ is a real $d\times n$ matrix, each with unit column vectors.

5:59 PM

hmm

part of what's confusing in the source for this is that it's not clear to me whether $d\geq m$ or not

I sorta feel like it should be

Oh. I guess I can always add zeros to the ends of vectors to increase their dimension without changing their norm

So $m\leq d$ should be trivial, and $d>m$ the nontrivial case

@Rudi_Birnbaum when you look at $V_\sigma \subset [ V_{\rho} \otimes V_{\rho}]^-$ and you restrict to $G'$ you get $V_{\sigma'} \subset [(V_{\rho'_1} \oplus V_{\rho'_2}) \otimes (V_{\rho'_1} \oplus V_{\rho'_2})]^-$

and then you uuuh "develop" the thing on the right

and you get $V_{\sigma'} \subset [V_{\rho'_1} \otimes V_{\rho'_1}]^- \oplus [V_{\rho'_2} \otimes V_{\rho'_2}]^- \oplus V_{\rho'_1} \otimes V_{\rho'_2}$ after a suitable isomorphism

Hi guys

I'm not sure why it should be in the third piece, though

How do we know that $x+\frac{2^2x^2}{2!}+\frac{3^3x^3}{3!}+\dots$ is convergent if $x\in(0,1/e)$?

6:07 PM

root test, maybe?

stirling approximation perhaps ?

@mercio: because the other ones are only the total symmetric ones since they are "1-dim"?

If you can take $(1+1/n)^n\to e$ as $n\to \infty$ for granted, then the ratio test is enough

@mercio: sorry: "trivial ones"

@mercio: In deed that easy!! Thank you!!!

well you would need more context in order to know their dimensions

6:10 PM

@mercio: Yes its not "in-there" but it comes out - lol

@Semiclassical is this for me?

Yeah.

ok

@mercio: Whats that "suitable isomorphism" thing?

@mercio: According to my experiments, I guess it could be as well in the other two, but then its again explicitely in the third. Would that make sense?

well that kinda depends how you define tensor products

6:14 PM

I feel like I'm learning so much math on here

if you fix a basis for each vector space

@mercio: Oh I see. The "multiplicities" don't occure normally because its about "the space".

and you define $V \otimes W$ to be a vector space with basis the set of $e_i \otimes f_j$

where $\{ e_i\}$ is a basis of $V$ and $\{f_i\}$ a basis of $W$

I see

then it's just a bit of juggling

i don't know what you mean with "multiplicities" or "the space"

6:18 PM

Why does it only work with the antisymmetric part?

@Semiclassical I can't go further of $\frac{nx}{\sqrt[n]{n!}}$

also you have to do the same with $\oplus$, that you kinda concatenate two basis together in order to get the basis of the sum

modulo annoying details when you're making a direct sum of something with itself

etc

Denominator is discrete.

well the symmetric part also has an expression like that

So, how to take limit?

6:19 PM

just replace $-$ with $+$ everywhere and change the isomorphism with the third piece

the third piece would still be isomorphic to $V_{\rho'_1} \oplus V_{\rho'_2}$

@Silent that's the root test. what do you get with the ratio test?

(Also, presumably one has $n/\sqrt[n]{n!}\to e$ as $n\to\infty$)

Can someone please help me with a question

@mercio: So now you're the only one in the world who understands why symmetry breaking in some molecules comes with paramagnetic response.

I'm pretty sure I don't

lol

6:28 PM

@mercio But more than anyone!

x)

MatheinBoulomenos

@mercio when did you become such a chemistry expert?

I read the cotton book

(not)

actually earlier today I read the wikipedia page about the hydrogen atom

because I wanted to know what are the representations they are talking about

and I left confused

@Semiclassical Thsnk you very much

@mercio: Some molecular (symmetry) groups contain in all squares of the non-1-dim. irreps the rotation around the main axis.

maybe I can enlighten you about the H-atom

??

6:30 PM

Ratio test gives a wiered looking expression

Hmm. Following up on what I was saying earlier, I guess I can project the column vectors of U onto the column space of V, and vice versa

Its sperical harmonics

and I don’t think that changes the inner products? Hmm

if I rotate a harmonic, is it still a harmonic ?

that is moded out

6:32 PM

:o

@MatheinBoulomenos check out my latest question specifically the edit part

makes no sense?

it makes a bit of sense

but hmm

$\displaystyle \dfrac{n}{\sqrt[n]{n!}} = \exp\left(\ln n - \frac1n\sum_{k=1}^n \ln r\right) = \exp\left(\frac1n\sum_{k=1}^n \left(\ln n - \ln r\right)\right) \\ = \displaystyle \exp\left(-\frac1n\sum_{k=1}^n \ln \frac r n \right) \color{red}\to \displaystyle \exp\left(-\int_0^1 \ln x \ \mathrm dx \right) = \exp(-1)$

@Silent @Semiclassical ^

how do you decide which is the nicest reference frame

like, are you aligning an axis with some magnetic axis

6:34 PM

Kewl

In the H atom I'd say its choice.

@Semiclassical I must have made a sign error somewhere

But its a bit more complicated

surely it is larger than 1

or is it that when I rotate a harmonic, it decomposes as a sum of harmonics

6:35 PM

The orbitals as they come out as a solution of the Schrödinger equation are complex.

I have neevr solved the Schrodinger equation

@LeakyNun check out my latest question specifically the edit part

And then you can transform them so that you get reals

n!<n^n, so

right, the only error is the last step

everything is fine

just replace $\exp(-1)$ with $\exp(1)$

6:36 PM

Nice

done

Ratio test is probably still easier here tho

The eigenfunction with the lowest energy (eigenvalue) is spherically symmetric its e^(-r)

okay, and the next ones ?

The next one as well but it has a radial node (c-r)e^(-r) but, then ....

6:39 PM

@LeakyNun, how do we know that $f(x,y)=\frac{xy}{x^2+y}$ when $x^2\ne y$ and $0$ otherwise, is not continuous? i tried $y=x$.

and $y=-x$

@GFauxPas check out my latest question

@Silent shouldn't you get 1 and -1

@Silent “is not continuous at (0,0)”, presumably

There seems to be a discrepancy over the derivative of arcoth(x) between wikipedia (en.wikipedia.org/wiki/Hyperbolic_function) and this calculator (derivative-calculator.net). Which one is correct?

yes

6:43 PM

@user10478 I don't see any discrepancy

@mercio: and then you get some exp(r k i) prefactor to the exp(-r) where k=-1,0,1

@LeakyNun but limit $\lim x\to 0\frac{x^2}{x^2+x}=0$, right?

ok

so all straight lines to the origin go to 0

is i the complex unit ?

the problem would have to be the "speed" at which they go to 0

which can be detected by the r-theta approach, I hope

6:45 PM

oh!

yes

should the function be zero when $y=-x^2$

... so they are still radial ?

aha

so you can transform the three to get all reals pointing in three different directions with a central node

6:46 PM

As written, the exceptional case doesn’t address what happens when the denominator is zero

@Semiclassical so sorry! you are right

Kk

you get the sine and cosine expressions.

Nicholas Roberts

Hey everyone, if i wanted the community to verify a proof of mine, would it be better to post it here or post it as a question on stackexchange?

@Silent then set $y=a-x^2$

6:47 PM

@Rudi_Birnbaum ugh, associated Legendre polynomials

But as long as there is "perturbation" it doesn't matter where the reference frame is.

Nicholas Roberts

Im worried that proof verification posts get downvoted

Or is it Laguerre

@Silent if it is continuous at 0, then I believe you can commute the limits of $a$ and $x$

you can use either one if I am not mistaken ...

6:49 PM

$f(x,y)=\dfrac{xy}{x^2+y}$, letting $y=a-x^2$ gives $g_a(x) = f(x,y) = \dfrac{ax-x^3}a = x - \dfrac{x^3}a$

Another tack would be to pick a path which gets closer and closer to the $y=-x^2$ path as you approach the origin

@Semiclassical that's what I said ^^

but usually Laguerre

maybe I should choose $y=-ax^2$ instead and have $a \to 1$

Not really. Your case has $a$ as an external parameter

6:50 PM

oh, sorry

why don't we try $y=ax-x^2$

I was thinking something like $y=-x^2+x^3$

what's a perturbation ?

Not sure about the coefficients or the particular power tho

hmm

The Hydrogen atom is SO(3).

6:52 PM

if $y=-x^2+x^3$ then $f(x,y) = \dfrac{-x^3+x^4}{x^3} \to 0$

If you switch on a homogenous magnetic field it has a direction, the field is a perturbation.

Hrm

If its weak

ah

3 mins ago, by Leaky Nun

$f(x,y)=\dfrac{xy}{x^2+y}$, letting $y=a-x^2$ gives $g_a(x) = f(x,y) = \dfrac{ax-x^3}a = x - \dfrac{x^3}a$

right, this should be the answer

6:53 PM

Then the otherwise "degenerate" orbitals "split" (branching)

$\displaystyle \lim_{a \to 0} \lim_{x \to 0} g_a(x) = 0$

$\displaystyle \lim_{x \to 0} \lim_{a \to 0} g_a(x) = \pm\infty$

I think that works, sure. I just prefer having some path along which the limit is explicitly nonzero

Then the "magnetic quantum numbers" (orientation of the orbitals) suddenly play a role for the energy.

If $y=-x^2+x^K$ then $f(x,y) = \dfrac{-x^3+x^{K+1}}{x^K} = -x^{3-K} + x$

@LeakyNun, @Semiclassical thank you so much for all this help!

6:55 PM

So why doesn’t K=3 work?

because i'm an idiot

3 mins ago, by Leaky Nun

if $y=-x^2+x^3$ then $f(x,y) = \dfrac{-x^3+x^4}{x^3} \to 0$

$\color{red}{\to -1}$

Ah, there we go

@Semiclassical how did you spot k=3?

@mercio: Here in table 2 you see the correct "orbitals" www1.udel.edu/pchem/C444/spLectures/04152008.pdf what I said was not correct, you get in addition to the complex phase also Polynomials as factors.

Tobias Kildetoft

For the physics minded: Would you consider it close enough if I described quantum groups as being a model for symmetry in quantum mechanics?

6:57 PM

:o

tbh I got lucky. I figured I should pick a power bigger than 2 so that the path would be asymptotic to $y=-x^2$

@TobiasKildetoft: What is a quantum group

?

I originally had x^4 in mind, but I figured I’d start small

Tobias Kildetoft

@Rudi_Birnbaum A certain type of Hopf algebra

In quantum mechanics you use point and space groups.

6:58 PM

Had I done that, the limit would’ve been infinity rather than zero

I think you talk about different stuff, thats more like qed.

quantum electro dynamics

@Semiclassical lmao

1 in infinity chance

So still an example. But the fact that the limit in the case of a cubic term was pure luck

you can win a lottery

Lol

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