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

Because I can do the same thing "multiple times"?

yes

the intuition here is that rings are like operators on something "like a vector space", we can add those operators pointwise and we can compose them

Oh yes I understand!

For example, the set of bounded operators on a Hilbert space forms a ring

I have once tried to "extend that to a field"

You know there is the "particle generation operator".

It is something from "second quantization".

9:05 PM

Sup @Daminark

You act it on the vacuum and get a particle: a |> = |1>

then there is the destruction operator a*|1> = |>

and so on.

Then my idea was to define an operator that you have to square to get 1 the "a" operator.

For that I tried to use a series expansion of the square root.

but around "close to zero"

and then you get some superposition of infinitely many generators with prefactors. But when you then try to calculate <0.5|0.5> you get some logarithmically divergent stuff.

Just reminded me in that ...

(<\psi|\psi>=\int_{\Omega} \psi* \psi d \tau, should always be 1.0 for a proper wave function \psi) but please continue I have interrupted you!

@Rudi_Birnbaum one more thing we can say is that like a group is a one-object category where each morphism has an inverse (i.e. a one-object "groupoid"), a ring is a one-object category where we can add morphisms, i.e. a one-object "preadditive" category

@MatheinBoulomenos: But not a field?

not sure about fields

they don't have simple categorical description I think

@MatheinBoulomenos: And do I find every group there?

9:18 PM

every group is the automorphisms of some object in a category, yes

Is it completeness?

nice

@Rudi_Birnbaum I don't understand that question

Well what makes the field special in comparison with the ring that its not (or not simply) to be described as a cathegory

?

Then my take is that is is connected with the "Vollständigkeit"

@Rudi_Birnbaum what do you mean by that?

Or how to name the "Dedekindscher Schnitt" in a phrase?

9:21 PM

you're thinking just about $\Bbb R$ not every field is complete

every convergent sequence has its limit in

?

(and it might not even make sense to ask if a field is complete)

Oh, how that?

@Rudi_Birnbaum what's a convergent sequence in an abstract field?

and I think you mean Cauchy sequence

you need a measure

9:22 PM

you need a topology, but not every field comes with that and even if it does, it might not be complete, like $\Bbb Q$

Oh sorry Q is field!!

So the field is a ring with mult. inverse, right?

and commutative

OK! Now I remember ...

you can phrase that in categorese, yeah but it's not really insightful or useful afaik

that at least covers some part of rings. On a theoretical level, all rings arise like that (i.e. as endomorphisms (= morphisms from an object to itself) in an category with addition on morphisms, but in practice, there seems another phenomenon which is more common for commutative rings. Suppose you have a set $X$ and a ring $R$ (say for example a field), then the set of all maps (of sets) from $X$ to $R$ is a ring with pointwise addition and multiplication.
It might be that you have some class of functions to a ring that is closed under addition and multiplication like continuous, different…

There are other examples: for every field $k$ (suppose it's infinite), we can think of the polynomials in one variable over $k$ as functions $k \to k$ and polynomials in $n$ variables can be thought of functions $k^n \to k$

in modern algebraic geometry, there is a method to view every commutative ring as the "ring of functions" on some "space"

but I'll stop here, as the details for that are quite intricate

> categorese

9:27 PM

thats quite interesting!!

@LeakyNun my favourite language

Since we have such a cosy chat, what I always wanted to know about ...

why is k[X] the functions from k to k?

@LeakyNun I didn't say all functions

just some functions

which functions?

9:29 PM

(and I was careful and assumed that $k$ is infinite)

the functions which are given by evaluating a polynomial at a point

oh lmao

i awlays think of lmaa ..when i read lmao

lets just say Hom(k[X],A) = A for any k-algebra A

i.e. the A-points of $\Bbb A^1_k$ is just A

Hey @Mathein! How's it going? Also @Leaky and everyone else!

hey @Daminark

9:34 PM

Hej!

@LeakyNun I was talking about more elementary things

So what is Grothendiecks starting point? Cathegories? And what does he want to do? Is it about solvabilities of equations in $\Bbb N$?

hi @Mathein, @ÍgjøgnumMeg, Demonark, @Leaky

@Rudi_Birnbaum the starting point is this "viewing commutative rings as functions on spaces" I think

hi @Ted :)

9:38 PM

of course historically, it was more about proving the Weil conjectures I guess

So then really actually what you described above how to get to rings with c.t.?

c.t.?

Hi @Ted

And the "Weil conjectures" are grossly about? (category theory)

There are about solutions of equations over finite fields

I see.

I mean thats really the old Greek stuff isn't it? (no disrespect intended - quite the contrary)

9:42 PM

Let $T:P_2(\Bbb R)\to P_2(\Bbb R)$ be defined as $T(f)=f+f'+f''$, then the matrix of $T$ with respect to basis $\{1,x,x^2\}$ is $\begin{pmatrix}1&1&2\\ 0&1&2\\ 0&0&1\end{pmatrix}$, right?

Yup.

Diophantine equations (so these are not finite fields .. ??)

Oh wait, then its Restklassenkörper, isnt it?

yes, $\Bbb Z/p\Bbb Z$ is an example of a finite field

and all other finite fields are extensions of that, like $\Bbb C$ is an extension of $\Bbb R$

@TedShifrin But my solution says that $T$ is not invertible!

Gauss started that, didn't he?

9:45 PM

@Silent: What do you mean by "my solution says"?

So and Serre or who was that proved the Weil conjectures, right?

@Rudi_Birnbaum many people worked on that, Serre, Grothendieck, Deligne

Deligne proved the last remaining one

Hey Ted and @ÍgjøgnumMeg!

@MatheinBoulomenos: Oh yes he was the one "betraying" him (=A.G.), right?

@Mathein REU started today and it seems like the NT talks are gonna focus on modular forms

Maximus

9:47 PM

In definitions, if-then are considered iff statements. Axioms are considered definitions correct?

@TedShifrin i mean, the solution manual i have, says that $T$ not invertible.

@Daminark nice!

Alessandro Codenotti

Hi @Ted

modular forms are quite magical

Oh, well, solutions manuals are infamous for wrong things, @Silent.

Hi @Alessandro

9:48 PM

ok :)

It's invertible on that subspace

@Rudi_Birnbaum yeah, in his later years, Grothendieck made some statements like that

I read one or two biographies about him, really cool guy (and bit crazy, but who isn't?)

Note, though, that $f+f'+f''=0$ does have nontrivial solutions

It does?

9:50 PM

So the idea that lim_{p\to\infty} $\Bbb Z/p\Bbb Z = \Bbb Z$ doesn't quite work I guess :-)

On this vector space?

I was being sloppy

Hey everyone!

@Silent: Note that the third derivative vanishes on this vector space. So if $f+f'+f''=g$, then, differentiating, $f'+f''=g'$, so we can invert explicitly by taking $f=g-g'$.

Hey @TedShifrin :)

9:51 PM

What I meant was that there are functions which satisfy $Tf=0$.

You should check that with your matrix, @Silent.

@Daminark what will you do with modular forms in your REU?

hi @Perturbative.

But said functions aren't elements of the vector space

Q: Hidden technicalities in the definition of a derivative

1

I know that this is a fairly provocative title so before judging please read what I have to say. Below is a definition of a derivative taken from Analysis on Manifolds by James Munkres: Definition: Let $A \subseteq \mathbb{R}$ and let $f : A \to \mathbb{R}$ be any function. Suppose that $A$...

9:52 PM

They would be, though, if you were asked about the set $\{1,x,x^2,\ldots\}$

@loch will you be there on thursday and friday?

Yes, @Perturbative, derivatives can only be defined at interior points.

@TedShifrin If you don't mind could you take a look at my question above?

In my book, I typically took open sets when I defined differentiability.

Ahh okay

9:52 PM

which I guess would be $T:P(\mathbb{R})\to P(\mathbb{R})$

Note that in G&P they talk about differentiability on an arbitrary subset of $\Bbb R^n$ by saying there's a local differentiable extension to an open set in $\Bbb R^n$.

@Semiclassic: I bet you need exponential/trig functions, not polynomials.

@Ted Yeah I remember

Eh, Taylor series...

@TedShifrin that's how we did it in complex analysis and there was some exercise where being sloppy about that lead to some very subtle errors

@TedShifrin Would you say my definition of the derivative in the question was OK?

9:54 PM

Huh? @Semiclassic. The indicial equation is $t^2+t+1=0$.

I mean that the Taylor series of the solutions would still be power series solutions

@Perturbative: I don't see why you're unhappy with Munkres's definition. Why not allow a more general domain?

@Semiclassic, but $P(\Bbb R)$ is not Taylor series.

Ah.

It's the vector space of all polynomials.

Nuts.

loch

9:56 PM

@LeakyNun unfortunately not - good luck with your presentation though !

Just do it on $C^2(\Bbb R)$ :P

yeah, yeah

I was thinking about projective space when I saw $P(\Bbb R)$

@loch ok thanks

Mostly I just wanted "oh, $P_n$ with $n\to\infty$"

9:56 PM

Whatever that means.

yeah, well

Colimit in the category of vector spaces

Hi @MikeM

One notion of limit is polynomials, the other (with appropriate topology) is convergent power series, another is formal power series, another ...

Yeah, fair enough

if you take the colimit along inclusions, you get polynomials, if you take the limit along projections, you get formal power series and not sure about convergent power series

9:58 PM

Mostly I was just trying to find a setting in which $T$ would fail to be invertible

@TedShifrin Well the more general domain $A$ needs to contain an open set $U$ in $\mathbb{R}$ anyway, so I was thinking for any function $f : A \to \mathbb{R}$ then we just restrict $f$ to $\operatorname{Int}(A)$ and talk about differentiability for points in $\operatorname{Int}(A)$ (because $f$ won't be differentiable at non-interior points anyway)

@Mathein not sure, our professor just asked for the audience background and was like alright, modular forms would be a good topic.

But you can't even discuss a function defined on a closed interval.

@Secret didn't respond to my explicit inverse.

what is a projective limit?

Probably need a completion of a local ring, @Mathein :P

10:00 PM

@TedShifrin But a function defined on a closed interval is only going to be differentiable on it's interior

So I don't at the moment see what we're missing out on if we're only talking about differentiability

@Perturbative we did define differentiable on subsets of $\Bbb R$ such that for every point, the subset contains a non-degenerate closed interval containing that point. This allows one to talk about differentiable functions on e.g. a closed interval and it's possible for functions to be differntiable or not differentiable at the boundaries of a closed interval

Yes, but you're not allowing me to discuss differentiability of $f\colon [0,1]\to\Bbb R$ because its domain isn't open, @Perturbative. Get over this.

@TedShifrin that just gives you formal power series

Oh, rats, right, @Mathein.

I guess algebra doesn't know from convergence, anyhow.

yeah

you need some analytic input for that

10:04 PM

@TedShifrin What if I define differentiability in the following way then? Let $A \subseteq \mathbb{R}$ and let $f : A \to \mathbb{R}$ be any function. Pick $a \in \operatorname{Int}(A)$ and choose $r > 0$ such that $B(a, r) \subseteq \operatorname{Int}(A)$. Define $\phi : B(0, r) \setminus \{0\} \to \mathbb{R}$ by $$\phi(t) = \frac{f(a+t) - f(a)}{t}.$$ Then we define the derivative of $f$ at $a$ as $$f'(a) = \lim_{t \to 0} \phi(t)$$ provided that the limit exists.

Mike Miller

Hi @Ted

@Perturbative the function $f:[0,1] \to \Bbb R$, $f(x)=1$ is differentiable at $0$ and $1$ and the function $g:[0,1] \to \Bbb R, g(x)= 1$ for $x \neq 0$ and $g(0)=42$ is not differentiable at $0$, but is differentiable at $1$. Your definition doesn't cover that

guate nocht mitanond

There's absolutely no difference between that and his, @Perturbative.

Q: Prove that $\frac{x^2y}{1+x^4+y^2}$ has no global minimum

3

I am not sure how to approach this problem. The usual methods do not work to find a minimum, I can see that, but how to show that there must not exist a minimum?

real-analysis multivariable-calculus maxima-minima

why do people prefer calculus over elementary arithmetic? @Ted

10:08 PM

@ÍgjøgnumMeg guadinocht

:D

:D

@MatheinBoulomenos I'm confused, Ted agreed with me when I said that a function defined on a closed interval is only going to be differentiable on it's interior (well based on Munkres' definition anyway)

@ÍgjøgnumMeg: Tirol/Pinzgau?

@Perturbative ah sorry, I was thinking about the definition I learned

10:09 PM

@Rudi Vorarlberg!

@MatheinBoulomenos No worries

Gut's Nächtle @ÍgjøgnumMeg

baschd aa

A schwob!

hahaha hoi

Naja, net wirklich

10:10 PM

alemannisch halt

:D I hau mi aa hi - God natt (och händerna på teken).

I bin scho da ufg'wachse, aber gscheid Schwäbisch schwätze tu I net

:D I am anyway impressed&convinced

@TedShifrin I guess I wanted Munkres to explain a bit more, but thanks for all the help! Final question when you said in your book that you took open sets when you defined differentiability I take it that you didn't require the domains of the functions to be open sets?

@Rudi_Birnbaum Gut's Nächtle dir au!

10:20 PM

@Perturbative: I pretty much did, I think. But I didn't intend my book to be 100% pedantic.

Let $f:[0,\pi/2]\to\Bbb R$ be continuous and satisfy $\int_0^{\sin x} f(t) dt=\frac{\sqrt3x}2$ for $0\le x\le\pi/2$, then how do we derive $f(1/2)=1$?

Differentiate at $x=\pi/6$.

@MatheinBoulomenos Was für ein Dialekt gibst Du uns? :)

@TedShifrin I'm going to prove weak Bezout this thursday in the seminar. The proof is not very geometric, just properties of differents for homogenous polynomials

@TedShifrin soll Schwäbisch sein, aber ich kann das nicht wirklich

@Mathein: Well, unless you're going to do homological stuff, ultimately it's all a fancy version of the fundamental theorem of algebra, which I don't find "geometric." :P

The Italian proof was to specialize to $d$ lines for one curve and $e$ lines for the other ... and put them in general position :P

Well, actually, I guess we only have to do it for one family.

hmm, I see that ill-defined notion of general position that the Italians had

10:28 PM

Well, this is why fancy modern stuff like MacPherson, et al., makes enumerative geometry much more powerful.

But homology says the Italians are right (working over $\Bbb C$, which they were).

you can do it for algebraically closd fields in characteristic p with étale cohomology, too

maybe I should instead give a talk on that

LOL

@Silent: You got it?

You have a bad habit of not following up when I tell you things.

@TedShifrin no, i am trying.

You know the Fundamental Theorem of Calculus about differentiating $\int_a^x f(t)\,dt$?

@TedShifrin i have heard of it, i never applied it

10:33 PM

Oh, well, that's what you need, but you need to use the chain rule, since the upper limit of the integral is $\sin x$.

I have no idea what you actually know and what you don't.

@TedShifrin I am so sorry. I will try to be better student.

LOL ?

?

I wasn't saying you were a bad student. I was saying I have no idea what you actually know.

@TedShifrin but some of this classical projective (even pre-classical AG) stuff is quite nice. I like the correspondence between algebra and geometry. You can define "projective plane" axiomatically and then you can show that for a projective plane $X$ you have:
$X=\Bbb P^2(K)$ for some field $K$ iff Pappus' theorem holds in $X$
$X=\Bbb P^2(D)$ for some associative division algebra $D$ iff Desargues's theorem holds in $X$
$X=\Bbb P^2(A)$ for some alternative, but not necessarily associative division algebra $A$ (octonions satisfy this for example) iff "little Desargues's theorem" holds in $X$

thus at least from the point of view of synthetic geometry, noncommutative and even nonassociative algebra is a natural thing to consider

10:51 PM

$(\ln f)_x = \frac1f f_x$?

sanity check

@TedShifrin, is this correct: After applying fundamental theorem of calculus, i get $\frac{\sqrt3}{2}\cos x=f(\sin x)$?

No, @Silent. Chain rule works how?

I am differentiating $\frac{\sqrt 3}{2}\sin x$ with respect to $x$, so $\frac{\sqrt 3}{2}\frac{d\sin x}{d\sin x}\frac{d\sin x}{dx}$

No.

Where did $\frac{\sqrt3}2\sin x$ come from?

What would the answer be if it were $\int_0^x f(t)\,dt = \frac{\sqrt3}2 x$?

@TedShifrin $f(x)=\frac{\sqrt 3}{2}$

Q: Prove that a subset of $\mathbb{R}$ with positive outer measure contains almost a whole interval

11:00 PM

OK, right.

Now, what if it's $\int_0^{x^2}f(t)\,dt = \frac{\sqrt3}2x$?

Nicholas Roberts

Hi, can anyone verify my proof? Would be much appreciated

1

This is exercise number 28 from chapter 1 of Stein & Shakarchi's Princeton Lectures in Analysis III: Real Analysis. Let $E$ be a subset of $\mathbb{R}$ with $m_*(E) > 0$, where $m_*(E)$ denotes the outer (or sometimes reffered to as the exterior) measure of $E$. Prove that for each $\alp...

real-analysis measure-theory proof-verification outer-measure

@TedShifrin $f(x^2)=\sqrt 3$, i think.

I will read this thing, the fundamental theorem of calculus and then come to you.

No, @Silent. Here's where you need chain rule. Let $F(x) = \int_0^x f(t)\,dt$ and $g(x)=x^2$. You're considering $F(g(x))$.

oh!

So, in my original problem, $F(g(x))=\frac{\sqrt3}2x$ where $g(x)=\sin x$?

11:16 PM

Yes.

So, is it $\frac{\sqrt3}2=f(\sin x)\cos x$?

Yes.

Thank you so so much!