In the question there is a comment by Daniel Fischer, which states in the weak topology on Banach spaces you have compact iff sequentially compact. The space of measures is a Banach space, and if he is referring to the weak (and not weak*) topology you do have this property

but the weak topology of a Banach space is never metrizable

@neraj: If you want to think about a statement about A(S) for sets S that is true you could try to prove that if S is a finite set with n elements, then A(S) has exactly n! = n * (n-1) * ... * 2 * 1 elements.

@lush Do you think Herstein will give such simple question .Anyway please go to PDF i mentioned and read following - There is a bijection from S1 to a subset of T of S2, which enables you to construct a bijective isomorphism from A(S1) to A(T). Then you can embed A(T) into A(S2)

Goodnight

@neraj the proof in your link believes that an isomorphism is the same as an injective homomorphism

@neraj: that's true, but you still don't have an isomorphism A(S1) -> A(S2) what you claimed.

be careful with older non-standard terminology. some authors use "isomorphism into" for injective homomorphism

Yes I want the prove only the way PDF HAS MENTIONED

As used as I am to having a not-great sleep schedule

howdy @Mathein @Semiclassic

@neraj then read the pdf, what more do you want?

@Semiclassical soja bhai

I still feel it when that sleep schedule gets disrupted

Yes i will

hey @Ted

@s.harp read the pdf

And understood the solution

(And a night where I’m waking up every hour or so definitely counts as disrupted)

the solution is clear, but once again the author believes an isomorphism to be the same as an injective homomorphism, this is not the way this word is used today

@Mike yes but i have a meeting in a minute

The solution is clear if read in proper context

so rip

@s.harp write a book which can compete with herstein

hi/bye @Eric

what?

hibye

@s.harp “Forget it, Jake, it’s Chinatown”

well ive never muted somebody before...

@Ted can you... Stop this?

Did you ever ask robjohn or anon to stop idiotic behavior? Just curious.

@neraj: You will be silenced for a period of time if you cannot behave professionally.

@TedShifrin They weren't here.

Just use the damn ignore buttons, as I had to with people who annoyed me.

I do. It's still disruptive for everyone else.

A mod kicked him out for a minute. Like that'll do a lot.

I really have enough stress in my life. I don't come here for more stress.

It’s the line between “annoys me” vs “disrupts the room”

Well, there's all sorts of music discussions, politics discussions, and other non-mathematical topics that occur here.

There is a difference there but I’m not sure how to qualify it

Plus some of us speak in German, in French, etc. That's annoying to plenty of people.

I find it hard to have a uniform policy here.

Yeah

k

I personally disliked the pop music and movie discussions that Balarka and others engaged in.

Are we going to keep it 100% serious mathematics?

Harassment or abusive behavior is something else. That I won't tolerate, and a mod should kick someone off the site for that. (I've been the victim of that on main.)

I can apparently put the entire room in timeout for a designated amount of time.

I wonder how heavy-handed I need to be to get you guys complaining and wanting me bounced.

@TedShifrin Any good book recommendations for easy to understand intro to Galois theory? So far I have recommended some students to start reading Dummit & Foote, but if there is some beter resource it would be good to know of

Oh sure, Tobias. Two suggestions. One is Ian Stewart's book on Galois Theory. The other is a beautiful little MAA book by Hadlock, Field Theory and its Classical Problems.

These are more accessible than something like Dummit & Foote.

Cool, I will take a look. I recall reading a bit of Stewart and liking it, but it also felt like it might take a bit long to read if one has a more specific purpose in mind

Hi chat!

@TedShifrin If we just started talking about literature recommendations: Have you got a favourite book on etale cohomology?

@Tobias: Stewart has lots of short chapters. But I think it's well done.

In this case, I have two students that want to write a project on Galois theory, and I have set the assignment of fleshing out a proof sketch I gave them of how to prove that finite fields of the same order are isomorphic using Galois theory

I know nothing, @lush.

@TobiasKildetoft if prerequisities are an issue, Artin's lecture notes are really self-contained, he even covers the necessary linear algebra

Right, Artin's book is nice, too, but it's somehow much harder reading.

Artin père, of course.

ah ok, thx though

hi @AbdullahUYU

What is the title of the Artin notes?

@TobiasKildetoft you might give one student to read abelian galois extension or cyclotomic extension

Galois Theory: Lectures Delivered at the University of Notre Dame by Emil Artin

if possible

thanks

I have a third student who is also doing a project, but he has read a lot more stuff than the others, so I set him to read the chapter on symmetric algebras in Geck-Pfeiffer with the aim to show the orthogonality relations for characters of finite groups by first showing the analogous ones for symmetric algebras (with suitable additional assumptions)

@TobiasKildetoft wow

hi @Abdullah

I sounds like he is finding it challenging, but doable so far

symmetric algebras are something I want to study when I find the time

Suppose $a_n \to a$. Define $s_n = \frac{1}{n} \sum_{k=1}^{n} a_k$. Prove that $s_n \to a$. I started with writing by definition: $\forall\epsilon>0,\exists N,\forall n\in N, |a_n -a|<\epsilon$.

seems to put a lot of stuff in rep theory in a more general context, from what I've heard so far

Oh hey, Cesaro sums

@MatheinBoulomenos Yeah. I am not sure how wide-spread they really are. I mainly know about them from reading that Geck-Pfeiffer book

is it works for him @TobiasKildetoft? That third student otherwise give him some choice of other topic if not coverd so far.

But the nice thing is that once you have the setup, it immediately applies also to Iwahori-Hecke algebras and for example gives a nice proof that these are actually isomorphic to the group algebra for a generic parameter

ok @TobiasKildetoft bye

On a slightly related note, I have completed my overhaul of the rep theory notes, so now those proofs that used the group algebra in disguise have been replaced with proper character theoretic ones, which come about much more naturally in the setup I have

My initial goal was to make it similar to the sum but I think I stuck at here.

Well, you're trying to bound $|s_n-a|=\left|\left(\sum_k \frac{a_k}{n}\right)-a\right|$

Yeah

Hello there

Nerds really do come in flocks. Heya @Dami and chat at large.

Hey @Daminark!

Can you see any resemblance between the terms of $|s_n-a|$ and the $|a_k-a|/n$ you were working with?

Yes indeed. I actually tried to use triangular inequality, but it applies in the wrong direction, there.

I'm not sure I know what you mean.

What did you actually write out?

We have $\frac{|a_1-a|}{n}+\frac{|a_2-a|}{n}+\dots+\frac{|a_n-a|}{n}<\frac{\epsilon}{n}n‌​=\epsilon$, right?

Sure, but that's not the triangle inequality. That's a bunch of individual inequalities

The left part can be transformed to the very sum.

Ok. Write that out, please

Ok, let me ask you about a little point. We have $|a+b+c|\leq|a|+|b|+|c|$, don't we?

Yep.

And we can increase the number of terms.

and more generally $\left|\sum_k c_k\right|\leq \sum_k |c_k|$

Indeed.

So, about the left part, say $x$, we have $x\geq |s_n|$.

sure. and then your proceeding discussion showed that you had $x<\epsilon$

If we had to have $x\leq |s_n|$, then we were done.

so $|s_n|\leq x<\epsilon\implies |s_n|<\epsilon$

Ah, wait a sec.

Oh, yeah, we're done.

:)

Actually, I've wrote this proof and my elder brother confused me. Said that I applied the triangular inequality in the wrong direction.

In fact, it's true.

:)

lol

No lol in this chat.

Ricci curvature is the trace of the curvature tensor

Here does trace mean the usual linear algebra trace on the curvature tensor as a linear Tran

Or is it the trace of the curvature operator

As a bilinear form

@Semiclassical I think there is a problem in our proof. We can't say $\frac{|a_1-a|}{n}<\frac{\epsilon}{n}$ in the first place.

the zeta zeros are the zeros of the graph

@AbdullahUYU can't we?

Publish and perish.

Because, we're saying that it's true for the $n$'s that are bigger that $N$.

hi everyone

Maybe $N$ isn't 0. If it's 5 for example, we don't have the inequalities till the 5th one.

@MatsGranvik Well, most people go with "or", but if that is your preference.

Hmm

@TobiasKildetoft lol ha, ha

Hmm

in $K[A^n]$define $f_i = x^i-x_1$. i want to show that if $f(t,t^2,\dots,t^n) =0 $ for all $t\in K$ then $f\in <f_1,\dots,f_n>$ someone can help ?

yeah, i see what you mean. I don't think that should be impassible but it does render the proof ineffective as it stands @AbdullahUYU

Actually, publish and perish makes sense to me as well.

@MatheinBoulomenos maybe you got an idea?

Once you have published enough, you don't need to publish anymore.

I feel like the reasoning should go something like this. Suppose for convenience that $a_0$ is farther from $a$ than everything else in the sequence

Hi @MatsGranvik how is your work on RH?

Then you're right that we have no obvious bound on $|a_0-a|$.

its suppose to be $f_i = x_i^i -x_1$

However, what we're really interested in is $\frac1n |a_0-a|$

@Liad is $k$ an infinite field? I don't think it works if $k$ is finite

so $\cap Z(f_i) = \{(t,t^2,\dots,t^n) : t\in K\}$

yes k is algebraically closed @MatheinBoulomenos

okay great, so we can apply the Nullstellensatz

@Semiclassical Hmm. I think I see what you mean.

So if we make n large enough, then we can make $1/n|a_0 -a|$ arbitrarily small

@MatheinBoulomenos ah. we didn't learn it and i think it helps with another exercise of mine

@MatheinBoulomenos can you tell me what this statement says?

I’m not sure how to make that into a proof tho

@Semiclassical Allright, that means we have to find another way to show it.

Thanks

@Liad the fact that I want to use is that if $J \subset K[x_1, \dots, x_n]$ is an ideal, then $I(V(J))=\mathrm{rad}(J)$

I think one can formalize that with enough care, to be clear

here be have $V(J) = \cap Z(f_i)$ right?

Oh, I guess I have a way.

yeah

We know that if $c_n$ is null, then $kc_n$ is null, $c\in \mathbb{R}$. We know that $(\frac{1}{n})$ is null and $|a_n-a| \in\mathbb{R}$. So $\frac{1}{n}|a_l-a|$ is null for all $l$. @Semiclassical

ok so $I(\cap Z(f_i) ) \sqrt{\cap Z(f_i))}$

@MatheinBoulomenos I think that is a nice little Dover book now.

@Liad $\sqrt{\cap Z(f_i)}$ doesn't make sense, you can only take the radical of an ideal

isn't it $I(Z(a)) = \sqrt{a}$ ?

yea but that's what you wrote didnt it?

ah.. ok

apparently you use $Z$ instead of $V$

we learned it with V , dont know why

In the first sentence, there is a typo; it should be $k\in\mathbb{R}$ not $c\in\mathbb{R}$.

@Liad: I think this works without the Nullstellensatz. In case you want to bypass it because you didn't learn it yet

I’m a bit distracted right now, but you should run that by someone else—it seems too easy

i want to show that if $f\in I(\cap Z(f_i) ) $then $f\in <f_1,\dots,f_n>$ didn't you showed the other way around ? @lush

Ok, no problem, you were very helpful.

Good luck

@lush maybe i will see how to use the Null.. can you tell me again please what this statement says?

@lush I think you're implictly using an isomorphism $k[x_1,...,x_n]/I \cong k[x_1]$

which is correct of course

@Liad: Mathein already wrote you

@lush where?

@Liad: For any Ideal J one has that I(Z(J)) = rad J

ok. so we have $I(\cap Z(f_i)) $

Do you know radical ideals and the I-operator?

yes

@Liad $\{(t,t^2,...,t_n) \mid t \in K\}$ is the vanishing set $V(f_1,\dots, f_n)$, you want to compute the ring of all functions that vanish on the set, i.e. $I(V(f_1,\dots,f_n))$, which the Nullstellensatz tells you is the radical of $(f_1,\dots,f_n)$, now you're not done yet, but you are if you can show that $(f_1,\dots,f_n)$ is prime

you can do that by showing that $k[x_1,\dots,x_n]/(f_1,\dots,f_n) \cong k[x_1]$

but im not sure why you say that $I(\cap Z(f_i)) = \sqrt{<f_1,\dots,f_n>}$

we know $I(Z(J)) = \sqrt{J}$

why do you keep writing $Z(J)$ as an intersection?

that's not helpful

@Liad: \cap Z(f_i) = Z(f1,...,fn)

that's the set im working with @MatheinBoulomenos

@Liad then use what lush said

ok but now its not making any sense becasue $\{f_1,\dots,f_n\}$ is a set of n elements and not an ideal ^^

ok nvm

@Liad: V( M ) = V( <M> ) for any set M

$Z(T) = Z(<T>$

@lush :)

ok so $I(Z(f_1,\dots,f_n)) = \sqrt{<f_1,\dots,f_n>}$

so if $<f_1,\dots,f_n>$ is prime we are done

Some quick geometry refresh for myself:

im asked first to find generagtors to $I(Y) $ (Y is the set from ealier) and then show $K[Y]$ is isomorphic to $k[x]$

Suppose I've got two non-collinear vectors $a,b$. for simplicity, I'll assume $a$ is a unit vector. I can decompose $b$ as $b=(aa^\top)b+(I-aa^\top)b$

one manifestly has that $(aa^\top)b=(a^\top b)a$ and $a^\top(I-aa^\top)b=(a^\top -a^\top)b=0$, so the first term is parallel to $a$ and the second is perpendicular

Now suppose I reflect $b\to b'$ across $a$. Then $b'=(aa^\top)b-(I-aa^\top)b=(2aa^\top -I)b=2(a^\top b)a-b$

Okay, and that's what it should be.

@lush okay yeah I see, you lift g to an element of k[x_1] and then show it's the zero polynomial

yup

@MatheinBoulomenos although one should show that the quotient is isomorphic to k[t] anyway :D

maybe it's simpler to say: one must have $b=b_1 a+b_2 u$ where $a^\top u=0$. then $a^\top b=b_1$ and under reflection we'll get $b'=b_1 a-b_2 u=2b_1 a-b=2(a^\top b)a-b$

hi everyone

so I read that the spectrum of an element of a Banach algebra is never empty

hey Leaky Nun

and I wondered if this is a new proof of the fundamental theorem of algebra

Hi @LeakyNun

@MatheinBoulomenos macht es sinn?

(continuing my talking to myself) The real point: Suppose $f(a,b)$ is a symmetric bilinear function of vectors $a,b$, and I demand that $f(a,b)$ be invariant under simultaneous rotations of both vectors i.e. $f(a,b)=f(a',b')$

@LeakyNun it's correct, but it's not a new observation

@MatheinBoulomenos ok

hmm

I'll suppose without loss of generality that $a$ is a unit vector. If I take this simultaneous rotation to be a 180 degree rotation around the $a$ axis, then I get $f(a,b)=f(a,b')=f(a,2(a^\top b)a-b)=2(a^\top b)f(a,a)-f(a,b)$

And therefore $f(a,b)=(a^\top b)f(a,a)\propto a^\top b$

@MatheinBoulomenos I don't understand a step in the proof that the spectrum is nonempty

So $f(a,b)=a^\top b$ is the only such function up to an overall constant

kann ich dich fragen?

sure, not sure if I can help ofc

so it says that the function $R : \lambda \mapsto (\lambda e - x)^{-1} : \Bbb C \to A$ is differentiable, which is ok

and bounded, which is also ok

and then it says that by composing with an arbitrary $\varphi \in A^\ast$ we see that $R$ is zero everywhere, which is not so ok

which $\varphi$ should I compose it with?

can I just distinguish every point?

Okay, question for the chat: Is there a more familiar name for what I just showed? i.e. if $f(a,b)$ is a bilinear function of vectors $a,b$ and $f(a,b)$ is invariant under simultaneous rotations, then $f(a,b)$ is the dot product up to an overall constant.

@LeakyNun that's a Hahn-Banach argument actually

:o

My brain is saying Schur's lemma but I'm bad with this kind of argument

sup @ÉricoMeloSilva

i ain’t here

I see Eric changed his name.

so you can prove a Banach space version of Liouville's theorem, that's what happening here. If $f:\Bbb{C} \to A$ is bounded and complex differentiable, where $A$ is a complex differentiable, then $f$ is constant

@Semiclassical what is "rotation"?

@MatheinBoulomenos how?

$a\mapsto RaR^{-1}$ where $R$ is some orthogonal matrix

@LeakyNun have you heard term called devissage filtration?

no

(I should probably be insisting that det(R)=1 to make it a proper rotation but I'm not convinced it matters)

Anyone heard it devissage filtration?

for context: en.wikipedia.org/wiki/D%C3%A9vissage. ("The word dévissage is French for unscrewing.")

Proof: suppose that $f$ is not constant, take $z_1,z_2 \in \Bbb C$ with $f(z_1)\neq f(z_2)$, then consider the subspace of $A$ spanned by $f(z_1),f(z_2)$, this is finite-dimensional, so we can just define a linear functional $\xi$ on the subspace such that $\xi(f(z_1)) \neq \xi(f(z_2))$, extend $\xi$ to an element of $\hat{\xi}$ $A^*$ by Hahn-Banach, then $\hat{\xi} \circ f:\Bbb C \to \Bbb C$ is holomorphic and non-constant, contradicting the classical version of Liouville

that probably makes sense, if you understand Grothendieck :/

I guess what I want is "G-invariant bilinear form"?

specifically G=O(n)...or should it be G=SO(n)

@Semiclassical thanks

@MatheinBoulomenos why is $\widehat{\xi} \circ f$ bounded?

@Semiclassical thanks

@WillHunting That square wave with Riemann zeta zeros as zeros is my latest plot/work. I am mostly watching Netflix these days. I like this series called Once upon a time. The time travel part was a bit annoying but I am past that now.

it's hard to avoid time travel being a really annoying plot device

though somehow I find 'multiple universes' more so

@LeakyNun it suffices to show that a bounded linear map $g$ between Banach spaces such as $\hat{xi}$ maps a bounded set such as the image of $f$ to a bounded set, note that this just a consequence of $\|g(x)\| \leq \|g\| \|x\|$ from the definition of the operator norm

@Semiclassical is this a new train of thought or related to your previous messages? XD

bounded is a really confusing term here

it was a reply to what Mats just said

since bounded linear maps are not bounded as a non-necessarily linear map (you know what I mean)

A question. The highlighted statement clearly shows that $n=(n-1)+1$ which B, as a whole, contains, but as argument unrawels contradiction is shown through a simple appearence of the $n-1$. Is this because we Have To account for Every object that pops up?

oh my bad, i didnt read that fully it seemed completely out of left field

@Semiclassical you can derive it from Schur's lemma yeah

@MatheinBoulomenos Mind refreshing me on that?

@MatheinBoulomenos ok thx

If $T$ is an operator on the FDVS $V$ which has a basis $B$ for which the matrix representation of $T$ with this basis is upper-triangular, does it follow that $T$ represented with respect to the basis obtained from $B$ by Gram-Schmidt is also upper-triangular?

FDVS = finite-dimensional vector space?

Yes

mmkay

@Semiclassical no wait, Schur's lemma doesn't directly imply it, since we're not working over $\Bbb C$

ah, drat

I mean, I like my argument: $f(a,b)=f(a,b')=f(a,(b+b')/2)=(a\cdot b)f(a,a)\propto a\cdot b$

But I feel like I'm reinventing something standard

if you want an abstract interpretation of your result: bilinear maps correspond to linear maps $V \to V^*$, and $G$-invariant bilinear forms to homomorphisms from $V$ to its dual representation, you've shown that the space $\mathrm{Hom}_G(V,V^*)$ is one-dimensional

hmm. That sounds awfully like Schur's lemma. (Though "sounds like" != "equivalent to")

yeah, I agree

but over $\Bbb R$ there are irreducible finite-dimensional representations with endomorphism ring $\Bbb C$ or $\Bbb H$

@MatheinBoulomenos so how can i show that $k[x_1,\dots,x_n] $ over $<f_1,\dots,f_n>$ is isomorphic to $k[x_1]$ ?

hmmm

just send $g + <f_1,..,f_n> $to $g(x_1,1,..,1)?$

without an example I'm pretty blind, unfortunately

@MatheinBoulomenos and why do you know about these stuffs lol

or 0 instead of 1 ?

@MatheinBoulomenos so the rep $O(n) \to GL(n)$ is irreducible?

@LeakyNun yes

cool

@MatheinBoulomenos maybe send g to $g(x_1,x_1^2,..,x_1^n)$

@Liad right!

that works

what I'm wondering at this point is something like: "What's a simple example of a group G such that $f(a,b)$ need not be the dot product?"

But that's not a good formulation, since there's no particular reason why you'd expect the dot product if you're not dealing with rotations

(That and I feel like something is wooshing over my head)

@MatheinBoulomenos the injective part is the only non-trivial part for me..

i want to show $g-h\ in <f_1,..,f_n>$ if they both agree on $(x_1,..,x_1^n)$

I want to say this is related to the following (stolen from WP): "If g is a simple Lie algebra then any invariant symmetric bilinear form on g is a scalar multiple of the Killing form."

i'm not convinced i'm right about that tho

@Liad construct an inverse map, you know that $k[x_1,\dots,x_n]/(f_1, \dots, f_n) \to k[x_1]$ is well-defined from the easy inclusion

the inverse is just given by sending $x_1$ to the class of $x_1$ in $k[x_1,\dots, x_n]/(f_1,\dots,f_n)$

@MatheinBoulomenos easy inclusion of what?

you have map $k[x_1, \dots, x_n] \to k[x_1]$ given by sending $x_i$ to $x_i^i$, $(f_1, \dots, f_n)$ is included in the kernel of that map

so by the homomorphism theorem, you get a well-defined map on the quotient

it's easier to construct an inverse here rather than checking that the kernel is exactly $(f_1, \dots, f_n)$

ok. just to make sure, $f_i = x_i^i -x_1$ right @MatheinBoulomenos

oops yeah, sorry

i couldn't see how its in the kernal ^^

maybe sending $x_i $ to $x_1^i$ ?

oh lol

yeah

I'm too tired

ok but then $f_i$ would be sent to $x_1^{2i}-x_1$ O_o

we need i-th root?

the $f_i$ should be $x_i-x_1^i$

but we want to get the set $\{t,t^2,..,t^n)\} $

ahh . ok

im tired too ! @MatheinBoulomenos :-)

yea now im looking at what i wrote yesterday

it is actualy $x_1^i - x_i$ ..

ok now its ok. thanks! @MatheinBoulomenos

np

I'm off. Cya all, gn8 :-)

now i want to show that $k[Y] $ isomorphic to $k[x]$ . does $k[Y]$ mean the polynomial ring with coefficients in $Y$ here?

@lush goodnight!

What is Y Liad?

And to make @LeakyNun jealous: see you tomorrow at 11am again, @MatheinBoulomenos :-D

lol

see you

Hey

@Liad I would assume that $k[Y]$ is the notation for the coordinate ring of $Y$

@Liad: If Y is sth. like Y = V( ... ), then k[Y] usually means: k[X_1,...,X_n] / I (Y)

$Y= \{(t,t^2,..,t^n : t\in k\}$

@Liad the good thing is that you already showed that

that's very good

^^

so why did they ask to find generators of $I(Z(..)) $ before O_o

I'm looking for very elementary source material on the modulus of continuity

I(Y) actually

@Liad it's harder to compute a quotient $k[X_1,...,X_n]/J$ for some ideal $J$ without knowing generators for $J$

but we showed $Y $ as $Z(f_1..) $

but we also showed that $I(Y)=(f_1,\dots, f_n)$

@lush >:(

so by showing rad f1..fn is prime i show what they asked latter, and that also found generators of $I(Z(f..) $

@MatheinBoulomenos yes, and in proving that, we finished the latter part of the exercise

lush gave you an alternative solution for finding generators for $I(Y)$

it's not that uncommon that there different orders in which you can do things

yea right. so we just had a shortcut

ok im tired too.

thanks and good night @MatheinBoulomenos @lush

good night @liad