Mathematics - 2021-01-16 (page 2 of 4)

so similar setup as a little while ago, $k$ algebraically closed of characteristic $2$.

Is $f'' = 0$ for any $f \in k[x]$? I think so.

Also, $f(x) = \Sigma_{i = 0} a_i x^i$, $f'(x) := \Pi_{i = 1}^t ((x-b_i)^{r_i})^2$ is true.

Given this, why would $\alpha_i = \frac{\sqrt{f(b_i)} - \sqrt{f(x)}}{(x - b_i)^{r_i}}$ be integral over $k[x]$?

Astyx

From my (vague) understanding, questions as defining equality between types, and equality between equalities of types etc

Which has concrete applications to type checking programs and stuff

@BigSocks Deriving $x^{n+2}$ twice gives $(n+1)(n+2)x^n$ and one of $n+1$ and $n+2$ is even

BigSocks

2:21 PM

right, that's why I thought so, that's good.

user2103480

Simplicial sets and other topological structures give semantics for martin-löf type theory + univalence so if one reasons within the latter, one can not infer false statements (in the language of type theory) about these structures

Astyx

You can also use your previous question twice

BigSocks

yeah I just thought they were assuming it, but it's actually a consequence of char $2$, which is neat I guess

user2103480

"Encapsulate topology" is a strong statement though

Leonhard Euler

Why don't Mobius strip lovers need arguments?

Astyx

2:23 PM

granted

user2103480

ncatlab.org/nlab/show/Awodey%27s+proposal this is the more or less precise form of what HoTT encapsulates

Leonhard Euler

Because they are all on the same side

:]

Semiclassical

"So I suppose the real answer depends on what your definition of "is" is."

user2103480

@Semiclassical deep

Astyx

@user2103480 cool!

Alessandro Codenotti

2:28 PM

@user2103480 Damn Ayoade got a long way from the IT crowd

BigSocks

@Astyx use my previous question twice? for the integrality of
$\alpha_i = \frac{\sqrt{f(b_i)} - \sqrt{f(x)}}{(x - b_i)^{r_i}}$ over $k[x]$?

user2103480

Should've stopped before it was too late

Thorgott

something something homotopy type is determined by the fundamental infty-groupoid

Astyx

@BigSocks no, for showing f''=0

BigSocks

oh lol yeah

user2103480

2:31 PM

@Thorgott please no

big urs schreiber vibes emanating here

polite proofs

"Let $a,b,c \in \mathbb{Z}. Then at least one of the numbers a + b, a + c,$ and $b + c$ is even."

If we wanted to prove this by contradiction, we would prove the statement:

$\forall a,b,c \in \mathbb{Z}, a + b$ odd $\land b + c$ odd $\land a + c$ odd.

But what is the contrapositive of this statement?

user2103480

52

Q: Intuition behind Snake Lemma

I've been struggling with this for some time. I can prove the Snake Lemma, but I don't really “understand” it. By that I mean if no one told me Snake Lemma existed, I would not even attempt to prove it. I believe I am missing some important intuition that tells me that the lemma “could” be true b...

All these diagram chasing lemmas (snake lemma, 3x3 lemma, four lemma, five lemma, etc.) follow directly from the "salamander lemma" due to George Bergman, see salamander lemma.

And that is pretty transparent. It is so transparent that for instance it is immediate to see (which no textbook ever mentions) that there are just as easily 4x4 lemmas, 5x5 lemmas, 6x6 lemmas. etc. In other words: once you see the simple idea of the salamander lemma, you can come up yourself with more such diagram chasing lemmas at will.

The corresponding nlab article:

ncatlab.org/nlab/show/salamander+lemma

me, having seen the snake lemma shortly before and knowing singular homology for like two days:

Astyx

The ice wyrm lemma

user2103480

Semiclassical

@politeproofs Maybe: If $a+b,a+c,b+c$ are all odd integers, then at least one of $a,b,c$ is not an integer.

user2103480

2:35 PM

I'm sure this is all just diagram chasing but nonetheless this is completely opaque

polite proofs

@Semiclassical But that doesn't even make sense to me.

Semiclassical

what about it doesn't make sense?

polite proofs

Why would that be it? I don't see why that would mean at least one of $a,b,c$ is not an integer/

BigSocks

write it all out, add 2 equations, find the third one on one side, LHS is odd, RHS is even

Semiclassical

Because the negation of "a+b,b+c,a+c are all even" is "at least one of a+b,b+c,a+c is not even"

user2103480

2:38 PM

If $a,b,c$ are all integers, can it be that none of $a+b,a+c,b+c$ is even?

Semiclassical

that said, i'm not sure i'm dealing with the universal quantifier properly

Thorgott

yeah man, the snake lemma is pretty clear once you understand spectral sequences

polite proofs

@Semiclassical That's not even what I wrote

Semiclassical

erm. you're...right

wait

user2103480

@Thorgott it's the most straightforward explanation

Semiclassical

2:40 PM

the negation of "at least one of a+b,b+c,a+c is even" is "none of a+b,b+c,a+c are even"

polite proofs

Indeed. That's what I wrote.

I claim that is the contradiction that we would need to prove, if we wanted to prove the original statement.

Semiclassical

So the contrapositive should be "If none of a+b,b+c,a+c are even, then at least one of a,b,c isn't an integer."

polite proofs

@Semiclassical I don't follow. What are $a+b, a+c, b+c$ elements of then?

Semiclassical

yeah, that's the chink in my statement

I figured I wasn't dealing with the quantifier correctly

you can take the case of them being real numbers but I'm not certain that's legit

polite proofs

Yeah, you're right about the fact that it doesn't seem legit.

Semiclassical

2:44 PM

(as in, "If a,b,c are real numbers such that none of a+b,a+c,b+c are even integers, then a,b,c can't all be integers." that's a fine statement but i'm not sure it's the contrapositive)

polite proofs

Since saying they're elements of R seems rather arbitrary, as "even" and "odd" are not really defined in R, but in Z

Semiclassical

I'm a bit puzzled how to think of this in $P\implies Q$ form, come to think of it. Verbally I see it but it seems like the correct statement is just $\forall a,b,c\in\mathbb{Z}(a+b\in 2\mathbb{Z}\vee b+c\in 2\mathbb{Z}\vee a+c\in 2\mathbb{Z})$

polite proofs

Sounds like you need to read an intro proofs book, just like me. ;)

BigSocks

contrapositive should be if all these sums are even, it is not the case $a, b,c \in \Bbb Z$

Semiclassical

that's what I said to begin with

BigSocks

2:48 PM

why was it wrong

polite proofs

@BigSocks Because what are they then?

Semiclassical

How do you describe a+b,b+c,a+c? What are a,b,c elements of?

BigSocks

could be rationals idk

polite proofs

No, again

Rationals is very arbitrary

BigSocks

I don't make the rules, that's the contrapositive

I didn't say they were but they could be

you have a diophantine predicate involving $a,b,c$

Semiclassical

2:49 PM

could take them to be reals as well, for that matter, which makes it seems as if the contrapositive is not unique

polite proofs

Clearly it's not the contrapositive since the contrapositive has to still make sense, but we have unquantified things in your statement

BigSocks

it's $D(a,b,c)$

the antecedent is $a,b,c \in \Bbb Z$

so you just do $\neg D(a,b,c) \Rightarrow \neg [ a,b,c \in \Bbb Z]$

polite proofs

But the antecedent isn't $a,b,c \in \mathbb{Z}$?

Astyx

A contrapositive is "$\forall a, b, c, \in \Bbb Z$, (a+b odd and a+c odd and b+c odd )$\implies$ False"

polite proofs

The $a,b,c \in \mathbb{Z}$ is a quantifier for a,b,c

I don't see how it's an antecedent.

It's like saying $\forall x \in S, P(x) \implies Q(x)$.

Semiclassical

2:53 PM

boddanda

polite proofs

$\forall x \in S$ is not an antecedence here.

Semiclassical

i don't know why that amuses me

polite proofs

@Semiclassical What?

Astyx

lol

Semiclassical

momentary typo in what Astyx had

polite proofs

2:53 PM

Oh okay, sorry didn't see

Semiclassical

yeah, it went quick

Astyx

The fastest latex correction in the west

polite proofs

I am going to start my undergrad soon, so I have to learn this basic logic stuff before I do.

Help me out, please. :D

Semiclassical

The good the bad and the ugly - Theme

►

Astyx

@politeproofs do you not agree with what I wrote ?

Semiclassical

2:55 PM

is $\forall x\in S (P(x))$ equivalent to $(x\in S)\implies P(x)$?

polite proofs

@Semiclassical From my intro to basic math proofs book, I would say so

@Astyx Well, I don't see why it would imply false.

Astyx

Because P is equivalent to $True \implies P$

Semiclassical

then one would think that the correct contrapositive is $\neg P(x)\implies \neg (x\in S)$

hrm

what's the latex notation for negation

BigSocks

neg

polite proofs

\neg

Semiclassical

2:57 PM

But how do you negate $P(x)$ without knowing the domain of $x$

Astyx

I'm not sure you can instantiate elements out of thin air though

polite proofs

Sounds like we all need to read an intro to proofs book. :(

Semiclassical

for reference, i'm a physics guy

so i'm never so happy as when i can shut up and calculate

Astyx

@politeproofs But why do you not agree with what I said ?

polite proofs

@Astyx You haven't justified why it implies false

Astyx

2:58 PM

I have

Because P is equivalent to $True \implies P$

Thorgott

too bad our logician just left, I'm sure this can be made more complicated

Semiclassical

I think the issue is that I'm not remembering how to deal with quantifiers when doing contrapositives

polite proofs

@Semiclassical But I think that it's actually $\forall x \in S (P(x)) \equiv x \in S \implies P(x)$

And then the contrapositive would be, $x \in S \implies \neg P(x)$

Astyx

To define the contrapositive you need to know what implication you're dealing with

The contrapositive is simply the statement that $P\implies Q \equiv \neg Q \implies \neg P$

polite proofs

Since the contrapositive of, for example, let a, b be integers. if a + b > 0, then a > -b would be Let a,b be integers. If a ≤ -b, then a + b ≤ 0.

We don't magically turn a,b into real numbers, or rational numbers, or some other magical thing

Astyx

3:00 PM

Here the only implication I can see without expanding the ambiant set of a,b,c is $True \implies P$

Semiclassical

one assertion I'm seeing online is that the contrapositive of $\forall x\in S(P(x)\implies Q(x))$ is $\forall x\in S (\neg Q(x)\implies \neg P(x))$

Astyx

its contrapositive is $\neg P \implies False$

Which is what proof by contradiction is

polite proofs

@Semiclassical That's exactly my point

Semiclassical

in which case the form of your statement, per Astyx's point, is that the statement is $\forall a,b,c \in S(\top\implies P(x))$

polite proofs

Grr. Why T, though?

Semiclassical

3:02 PM

if you mean notation, it's because I dunno how to do "true" in latex

other than \text but that's ugly

Thorgott

I'm gonna channel my inner Ted for a second and remark that you are very likely making this needlessly complicated by writing it in terms of opaque symbols

Astyx

How are you meant to look clever if what you say is not opaque ?

BigSocks

yeah, $\forall x \in S [P(x)] \leftrightarrow \forall x [x \in S \Rightarrow P(x)]$

Semiclassical

as a matter of logic, though: you've specified the domain of a,b,c as even integers. once you've done that, it's a tautology that a+b,b+c,a+c are even

so you don't need any premise other than "true"

in which case, per Astyx, the correct contrapositive is $\forall a,b,c\in S(\neg P(x)\implies F)$

polite proofs

I see. I guess it makes sense. But that basically means contrapositive = contradiction in this case

Semiclassical

3:05 PM

yeah

BigSocks

I don't think that's right

Astyx

But the problem is badly phrased, because you never take the contrapositive of a statement, but of an implication

polite proofs

Okay, that's fine. Thanks for the help, everyone.

@Astyx It was just my idea to take the contrapositive of this.

Semiclassical

i guess the upshot is that, if your statement is a tautology when expressed as an implication

polite proofs

Because ultimately, it could be done, right? But I couldn't tell for sure how to do it correctly

Semiclassical

3:05 PM

then the contrapositive is necessarily that the contradiction is false

BigSocks

yeah you somehow have to phrase it as an implication and the only way to do that is to interpret universal quantification as containment of elements in a set

Astyx

As I already wrote above, the contrapositive is the logic equivalence between $P\implies Q$ and $\neg Q\implies \neg P$

Semiclassical

@Astyx right, and we're specifying to the case when $P$ is true and therefore $P\implies Q$ is a tautology

so the question is basically "how do you take the contrapositive of a tautology"

Astyx

It depends what you mean by tautology

BigSocks

and I don't think there's anything wrong with the initial thing. saying something is not in a set does not say where they are. I don't think there are any ontological debts to be paid

Semiclassical

3:07 PM

one way to get around it is probably to modify P(x) so that it already has a domain

which i take our rational/reals thing to be

polite proofs

@Semiclassical Have you taken measure theory, as a physicist?

Semiclassical

taken? no

been trying to learn a bit of it myself tho

polite proofs

Read a book, whatever

Semiclassical

mostly so that i have a grasp on measure-theoretic probability theory

Kolmogorov foundations etc

I do QM stuff, especially stuff in the realm of history

so having a grasp on foundations isn't a small thing

(I'm not very good at it.)

Astyx

At this point in physics, not understanding the maths is a plus amirite

user2103480

3:10 PM

Hm, I need to deal with $(-\Delta)^\gamma$, where $\Delta$ is the laplacian and $\gamma > \frac14$. Is this just the function to the power of $\gamma$ pointwise? Or are there other sensible definition for this

Semiclassical

@user2103480 context?

that could be some kind of fractional derivative

e.g. en.wikipedia.org/wiki/Fractional_Laplacian

the upshot is that you define $(-\Delta)^\gamma f$ in terms of what it's Fourier transform looks like

user2103480

Not much more context. It's a simple setting so we look at functions in "$L^2(0,1)$" (I'm not sure whether this being an open interval is intention) for which the laplacian on $[0,1]$ with dirichlet boundary conditions is defined

Semiclassical

hmm. my impulse would be that it's a fractional derivative

user2103480

And we should prove that this fractional laplacian is a hilbert-schmidt operator

ah this looks sensible

damn prof, define the stuff

Semiclassical

^

polite proofs

3:14 PM

@Semiclassical Is there anywhere where I can ask you that's slightly more ontopic about physics textbooks?

Semiclassical

the obvious hot-spot is the physics chat room: chat.stackexchange.com/rooms/71/the-h-bar

but here is probably fine

user2103480

@Semiclassical thanks btw

Semiclassical

np

@user2103480 waving my hands a bit, is it obvious how "finite H-S norm" translates from $f$ to its Fourier transform

if so, then it'll presumably just amount to "prove it on the Fourier transform side"

user2103480

I have not tried yet. I'm watching a recording right now and the prof states that minus the laplacian is positive definite so one can take arbitrary powers and I'm not sure how that works

user586228

@LHC2012 Please describe in complete detail

user2103480

3:22 PM

"Taking the root" of a pos def operator I know, but I haven't seen any generalizations of that

user586228

Why should the first derivative be 0.If it is zero then why check the second derivative?

Thorgott

can't you do the same thing for arbitrary powers as for roots

user2103480

Apparently?

Semiclassical

not having a definition doesn't help here

polite proofs

Who was the logician that just left, by the way?

The one you referred to earlier who was here

user586228

3:29 PM

@politeproofs Did u mention me?

user2103480

6

Q: Fractional powers of positive self-adjoint operators

Consider two positive unbounded operators $A$ and $B$ densely defined on a Hilbert space $H$ self-adjoint on a domain $\mathcal{D}(A) = \mathcal{D}(B) = H_1$. By the spectral theorem, we can define the fractional powers of $A$ and $B$ as self-adjoint linear operators on $H$. My question is, is $\...

functional-analysis reference-request operator-theory

polite proofs

I don't think so?

user2103480

By the spectral theorem, apparently. I'm getting closer

polite proofs

Or was that supposed to be a joke? In which case, it wasn't too funny.

Astyx

I think $(-\Delta)^\gamma$ refers to the operator $\tilde f\to\tilde k^{2\gamma}f$ in Fourier space

The point being that in general, the action of an auto adjoint operator is isomorphic to the multiplication by a function in a certain function space

user2103480

3:33 PM

@politeproofs impolite proofs

polite proofs

@user2103480 I'm only polite during proofs. ;)

Astyx

For $-\Delta$, this function is $k^2$ and the space is the fourier space

As a matter of fact, for any function on $f:\Bbb R_+\to \Bbb R$ you can define $f(-\Delta)$ (because $-\Delta$ is positive definite)

(if my memory doesn't fail me completely)

user2103480

You need to phrase this a tad bit more down to earth since my functional analysis knowledge mostly ends after the spectral theorem for compact, self-adjoint operators

Astyx

What's your statement for the spectral theorem ?

user2103480

I honestly don't know for this case of densely defined unbounded positive operators. I only know there exists a spectral theorem which apparently quickly gives the construction of the power of the operators

I'm following the SE link above and what the statement is, is what I'm currently looking for

Astyx

3:41 PM

The most general statement I know says that an auto adjoint operator $(A, D(A))$ on a separable Hilbert space $\mathfrak h$ behaves isomorphically to the multiplication by some real valued function $a\in L^{\infty}_{loc}(B, d\mu)$ on $L^2(B, d\mu)$

user2103480

That may even be too general

Astyx

Where B is a subset of $\Bbb R^d$ for some $d$

Semiclassical

better than it is for physicists, including myself. a footnote on a book draft i'm involved in:

"Physicists tend to use these terms interchangeably but there is a subtle difference between self-adjoint and Hermitian operators. When in the 1960s, a mathematician (Kurt Otto Friedrichs of the Courant Institute in New York) mentioned to Heisenberg that another mathematician, von Neumann, had helpfully clarified this difference for physicists, Heisenberg purportedly asked him: ``What's the difference?''"

(my own answer to that question is roughly "boundary conditions")

Astyx

So in the specific case of $-\Delta$, the $a$ in question is $k\to k^2$ and B is the Fourier space

user2103480

Is there any way to get the power of the operator given an eigenvector basis? There is the obvious easy eigenvector basis of sines

Astyx

3:44 PM

That means that $\overline{-\Delta f} = k^2\bar f$

user2103480

Behaves isomorphically in the sense unitary equivalence?

Astyx

Now if you want to define $u(-\Delta)$ for some function $u$, this will be $\overline{u(-\Delta)f} = u(k^2)\overline{f}$

user2103480

and what is $k$ here?

Astyx

Oh sorry, yeah that's not clear

k is the variable

user2103480

frequency?

Astyx

3:46 PM

If you want, yes

Semiclassical

in physics you'd say that $x$ is position and $k$ is wavevector

or $t$ is time and $f$ is frequency

Astyx

What I shoudl write is that $\overline{u(-\Delta)f}$ is $k\mapsto u(k^2)\overline{f}(k)$

Semiclassical

specializing to the 1D case momentarily, i'd want it to be the case that $(-\Delta)^\gamma \sin(kx)=k^{2\gamma} \sin (kx)$

Astyx

In your specific example $u:k\mapsto k^{\gamma}$

Semiclassical

i'm not wholly sure my statement makes sense tho

Astyx

3:48 PM

(I too thought this was dark magic when I first learned it)

user2103480

and the overline is the fourier-transform?

Astyx

Yes

user2103480

so what, do I fourier transform and then back-transform to obtain the fractional thing

Astyx

I'm not sure what you want to do

Semiclassical

if you're trying to prove that $(-\Delta)^\gamma$ is a H-S operator, then i still think it makes sense to prove that its Fourier transform is a H-S operator

and then transfer that statement back to the original space

user2103480

3:52 PM

I just want to find a defintion of the fractional operator that I can work with. Since you defined $f(-\Delta)$ using the fourier-transform, I assumed I get the actual operator by transforming this back

Semiclassical

which I think is plausible, because going from position space to wavevector space usually plays nicely with the norm

Astyx

Yes you do, but it's way nicer to work in Fourier space

user2103480

@Semiclassical that sounds indeed like a plausible and probably pretty nice, high level approach

I'd need to know a few results that ensure that this works, though. I may try to just figure out what happens to the basis vectors to define this shite

thank you people in any case!

That helped a lot

Semiclassical

what's the latex symbol for logical equivalence. not the three bars but the left-right arrow

(and not \leftrightarrow)

user2103480

$\iff$?

Astyx

3:55 PM

\iff

Semiclassical

neat

user2103480

Working with a basis maybe makes hilbert-schmiddity easy

I reserve the rights for that adjective

Semiclassical

the whole momentum-space vs. position-space representations of the wavefunction in QM basically revolves around the fact that $\int |f(x)|^2\,dx<\infty \iff \int |\overline{f}(k)|^2\,dk<\infty$

Astyx

Those are equal right ?

Semiclassical

which i think is all you need here

depends on your convention

Astyx

3:57 PM

Yeah right

Semiclassical

which i hate but i have to leave open

i do prefer $\int |f(x)^2\,dx=\int |\overline{f}(k)|^2\,dk$ though

user2103480