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00:04
Hello, everybody.
I just have a small question.
Hi @Ted.
rehi @MikeM
Is it "Noetherian ring" or "noetherian ring"?
Generally capitalized, Krijn. She was a person, after all.
00:07
Abelian group and abelian group :/
yeah, good point.
Don't ask for logic.
Either is fine.
Done tutoring for the day. Guess I should get some writing done tonight.
Now that Balarka has gone to bed and you don't have to harass him, that would be a good idea.
Ciprian told me to start writing at our meeting. It's a bit daunting.
Start writing what?
00:12
My project. I'm nowhere near done but a) Better to not write all of it at the end b) that way I don't forget how to prove things I know how to prove c) catch errors in my logic
he's right ... at least write stuff in detailed outline form.
I don't disagree. Just sort of "eek!"
Should start setting deadlines for me to write little microscopic bits of it, so that it feels less daunting.
don't you start acting like Semiclassic now.
get used to it ... detailed outlines are less daunting.
then you don't agonize over wording.
I wrote a fair bit in undergrad... I know the agonizing over wording is for when you edit. :)
Writing philosophy and literature papers is somehow very different from writing mathematics.
00:16
True enough
I'm going to start auditing some art history classes in the Fall, I think... never had time to do much of that in undergrad.
go for it
you do much of that at Berkeley?
nope, none.
but I was a French major as an undergrad :P
I guess I'll find out if that extra couple hours/week is a significant productivity drain
I took/sat in on a lot of math courses, though.
You could cut out MSE time to compensate, you know.
00:24
I could, for sure.
BTW, @MikeM, my sister and her husband are both art historians ...
I took a course on the history of postmodern art (mostly US-centric) which was fascinating in my last qtr of undergrad. Wished I'd engaged with the dept more then.
Didn't know that @Ted.
nope, I don't think I've mentioned it before — no context.
True enough.
 
3 hours later…
03:46
@TedShifrin i'd be upset by my name becoming a byword for writing procrastination were it not entirely appropriate @MikeMiller
03:57
hi @SemiC
maybe mikes just don't like to write
Hi, what is the probability of 1? i.e. $P(1)$ = ? (where 1 is the integer 1)
 
2 hours later…
05:30
Is there a nice way to integrate some $2$-form over the ellipsoid $x^2+y^2/4+z^2/9=1$? I parameterized it using spherical coordinates, but I got a long and ugly expression.
What 2-form
$\omega=zdx\wedge dy-ydz\wedge dx$
Try using Stokes.
I'm new to this, I tried using Stoke's theorem and $d\omega$, but I cannot seem to figure out how to integrate over the interior of that ellipsoid.
Yeah, that's what I thought. Hmm...
What's $d\omega$?
05:35
Oh
I got $0$.
@MikeMiller arxiv.org/abs/1604.06597 Pretty "miraculous" construction to produce slice 2-knots.
$d\omega=dx\wedge dy\wedge dz-dx\wedge dy\wedge dz$
(All 2-knots are slice.)
@Josue easy integral :)
05:57
@MikeMiller Yeah, it's just zero. For some reason, I thought I was computing for the volume of the ellipsoid, and that threw me off a bit.
06:47
If $f(x)$ is a real valued function such that $2f(x) + 3f( x) = 15 - 4x$ then $f(2)$ is ?
2f + 3f =5f = 15 - 4x
f = 3 - (4/5)x
f(2) = 3 - 8/5
=7/5
 
1 hour later…
08:05
Nice avatar @anon
Hi @Krijn
Hey @BalarkaSen, you didn't get much sleep, did you
7 hours. Does that classify as not much?
No, that's quite enough actually
I miscalculated
So, what're you studying?
08:12
I have Alg Geom class in about 15 minutes
But I have to finish my coffee first so I'll be there a little later
@Krijn You mean you can't bring the coffee to class?
@TobiasKildetoft The bike ruins that
08:15
You have a bike? Your nerdiness gauge went down a bit.
In reality I'm much less nerdy than you'd expect, I guess
(and that's not a compliment from me)
@BalarkaSen How do nerds commute, then?
Unicycle?
They don't. It's a noncommutative ring.
@BalarkaSen Commuting nerds sounds like something the football team does for fun in an American high school movie.
08:28
That's actually a good observation. You're really onto american high school movies aren't you?
@BalarkaSen Onto in what sense? You mean surjective?
OK, I should stop chatting garbage for now and get to work. :)
And I'm going to commute
@Krijn With what :)
His bike, obviously.
08:37
Hmm, so does that mean it does not matter whether he sits on the bike or it sits on him?
(I'll stop now)
 
2 hours later…
10:39
Hi @AkivaWeinberger.
Yo
I commute by train, by the way
Subway (probably called "metro" over there if I had to guess)
Yep, it's metro.
And if it's a particularly old system, it's a retro metro
@AkivaWeinberger Can we please move on from these commuting jokes? :)
Our subway map looks like a net. I wonder if it converges
(Are filters types of nets? I don't actually know what nets are)
(But I'm pretty sure they converge occasionally)
10:43
I don't know what nets are either.
nets are just sequences with the naturals replaced by a directed set
no idea about filters, though I know they were mentioned in a course I took a long time ago
"Directed"? Is that a totally ordered family with a bunch of inclusion functions… or something?
I have no idea
Filters are cool.
Ultrafilters more so
10:44
bleh
@AkivaWeinberger It is a partially ordered set such that for any two element $a$ and $b$ there is some $c$ which is larger than both
Oh
I was very confused. Sorry
That makes more sense
So, like, power sets ordered by inclusion
for example, yes
Hmm. I am pretty sure I have done this before but I am having trouble figuring out why $\int_{\gamma} \omega = 0$ for any loop $\gamma$ implies $\omega$ is exact ($\omega$ here is a $1$-form)
Ah, now I remember how I did it.
11:00
@TobiasKildetoft (Nonprincipal) ultrafilters are essentially when you cheat at the "Guess My Number" game.
(Principal ultrafilters are when you don't cheat)
Where, the game is, you think of a number, and I get to keep on asking you yes-or-no questions to figure out the number
Clearly, I can win, just by asking, "Is it 0?" "Is it 1?" "Is it 2?" etc. until I find your number
@AkivaWeinberger I think I recall the definition of an ultrafilter and what a principal one is. Just not what a filter actually is (nor why we should care about them)
Oh.
They're just ultrafilters without the "for every set, it's either in the ultrafilter or its complement is" condition
@AkivaWeinberger My number was $\sqrt{2}$. Sorry pal.
@AkivaWeinberger Ahh
So, like a Guess My Number game that's not done yet
Where, if I ask "Is it in the set $A$?", if you say yes I put $A$ in the filter and if you say no I put its complement in it
but I haven't asked about every set $A$, so there are some sets such that neither they nor their complement is in it
11:05
hello
Hello.
Your username is out of date, a bit
:)
I have here the definition of decreasing exponents: i.imgur.com/oGXR3We.png?1[/img]
and D is the operator of derivation on the polynomials of the MAX rate 3
What is the matrix D according to standard basis in the real polynomials of rate 3
@ akiva
can u assist here ?
thank you
I'm confused. How can we have $x\mapsto$ more than once?
@Algebra2015 There does not seem to be a question there
What is the matrix D according to standard basis in the real polynomials of rate 3
11:11
So what's the bottom image mean
in the second link
@Algebra2015 It is called degree, not rate
this reffes to 2nd question
1. question:
What is the matrix D according to standard basis in the real polynomials of rate 3
OK
Ah.
So, you know what the standard basis is here, right?
D is the operator of derivation on the polynomials of the MAX rate 3
Right
But the standard basis is…
11:13
yes: matrix for D in the standard basis
according to standard basis
Right. But I'm asking you what is the standard basis?
What are its elements?
Forgetting D for a moment
1, x, x2
?
Yup.
$\{1,x,x^2\}$
really?
Pretty sure
11:14
ok
then?
is you want max degree 3 then also $x^3$
Oh, right
Yeah, $\{1,x,x^2,x^3\}$
Sorry
and now ?
So what does $D$ do to each basis element?
@Algebra2015 why $a_3\neq 0$?
11:16
decrease the rate?
forget that link
@Algebra2015 stop posting images and just write stuff yourself
You want everything with max degree 3.
So it includes stuff with smaller degree
11:17
yes.
so, the matrix for D us then 2x2 ?
In any case, what's $D(1)$? $D(x)$? $D(x^2)$? $D(x^3)$?
No, 4x4. 'Cause it's a map from a four-dimensional space to a four-dimensional space
but we havee derivation ?
of 4x4 to 3x3 or?
$D:\Bbb R_3[x]\to\Bbb R_3[x]$, as said in one of the images
@Algebra2015 forget that the map has anything to do with derivatives
11:19
Answer my question above
stil 4x4 despite derivation of rate polynom ?
my view:
we have
$a*x3$
and with derivation we get $xˇ2$ ?
Remember that R_2[x] is a subset of R_3[x]
The derivative of everything with max degree 3 is something of max degree 2, but…
…everything with max degree 2 also has max degree 3
11:22
"Max degree 3" means "degree 3 or lower"
In any case, one of your links specifies what the codomain is
So, what does $D$ do to each basis element?
what is then matrix for Daccording to the standard basis in R_3[x] ?
5 mins ago, by Akiva Weinberger
In any case, what's $D(1)$? $D(x)$? $D(x^2)$? $D(x^3)$?
(I seem to use "in any case" a lot)
@akiva
D decreases the rate?
11:24
Yeah
But it's OK
See the top image there?
?
yes
it does nothing here?
D decreases the rate, but it's still in R_3[x] even if it has a smaller rate
because R_3[x] is the set of polynomials of rates 0, 1, 2, and 3
4x4 matrix...
11:26
@Algebra2015 You may as well get used to calling it the degree, because there is no such thing as the rate
with zeros in the last column ?
(We'll end up having a zero row.)
ok . degree
Pretty sure it's row, but I might be wrong
still, no final record of matrix..
11:27
So figure out what D does to each basis vector, and make those the columns of the matrix
3 mins ago, by Akiva Weinberger
5 mins ago, by Akiva Weinberger
In any case, what's $D(1)$? $D(x)$? $D(x^2)$? $D(x^3)$?
You need to do that at some point
$D(1)=x2$? $D(x)$=x2? $D(x^2)=x4$? $D(x^3)=x6$
?
…no. Isn't D differentiation?
What's the derivative of 1?
From calculus
but we hav at D of also x2
derivative of the 3 degree polinom is
with x_2
11:31
23 mins ago, by Algebra 2015
and D is the operator of derivation on the polynomials of the MAX rate 3
D(1) = 1_2
+1
=3
..?
I think you completely forgot what a derivative is
D(X)=X_2 + X +1
D(X_2)=(X_2)_2 + X_2+1
Where did it say that?
Maybe you're looking at a different question?
no
derivative of degree 3 polynom is
a1*x_2+a2*x+a3
??
and in this above
we put D(1), D(X), D(X_2) and D(X_3) ?
or....?
otherwise then above I do not understand
11:34
The derivative of
please help
The power rule says the derivative of $x^n$ is $n x^{n-1}$
Remember that?
And the derivative of a constant is zero
So the derivative of $1$ is $0$;
the derivative of $x$ is $1$;
the derivative of $x^2$ is $2x$;
derivative is :
3*a1*x_2+2*a2*x+ a3 ?
11:36
and the derivative of $x^3$ is $3x^2$
Yes
ufffff...
I gotta go for a bit, sorry
so.. D(1) =3*a1*1+2*a2*1+ a3 ?
??
x=1
if u put in 3*a1*x_2+2*a2*x+ a3
it is not 0
11:37
The derivative of a constant is zero
@Algebra2015 No, $x$ is not $1$
@Algebra2015 the derivative of $1$ is what you wrote if you put in the correct values for the coefficients
D=(1) = derivation in x=1 ?
I am getting a feeling you do not actually understand what those coefficients mean
11:39
no
@Algebra2015 No, these are not functions to be evaluated at some point
but... ?
what does it mean D(1) ?
@TobiasKildetoft Can you finish this? I got to go finish lunch
and how to put it in polynom and calculate then: 3*a1*x_2+2*a2*x+ a3
That's the derivative of a1*x^3+a2*x^2+a3*x+a4
11:41
ok
1 is 0*x^3+0*x^2+0*x+1
1 is a polynomial here
and D(1) i understant to put x=1 into this 2nd degree polynom
It's an element of R_3[x]
so D(1) actually means to do what?
what in what..
@Algebra2015 No, you are never ever in this exercise going to put in x = anything in these polynomials
11:42
thank you
58 secs ago, by Akiva Weinberger
1 is 0*x^3+0*x^2+0*x+1
It's the polynomial where all coefficients are 0 except for the last one
now u understand mw
these are not functions (or at least they should not be thought of as such for this purpose)
a1=0, a2=0, a3=0, a4=1
my wrong path at seeking for the solution
me
D(X) means what, and what is the "record" within 2nd degrree polynom
11:43
Plug those coefficients into 3*a1*x_2+2*a2*x+ a3
of D(X)'
?
D(p) is the derivative of p, where p is a polynomial
(In calculus, you take the derivative of a function, not of a number)
D(1) means then, put
a1=0, a2=0, a3=0, a4=1 into 3*a1*x_2+2*a2*x+ a3 ?
uffffff.
11:45
And, as I said before, D(1)=0
D(X) means then, put
a1=0, a2=0, a3=1, a4=1 into 3*a1*x_2+2*a2*x+ a3 ?
and D(X) = 1?
sure?
D(X_2) means then, put
a1=0, a2=1, a3=1, a4=1 into 3*a1*x_2+2*a2*x+ a3 ?


and D(X) = 2?
Well, a4=0 in that case, but it's the same answer
@Algebra2015 No, a1=0, a2=1, a3=0, a4=0
so, D(X) = 1?
11:47
Indeed it is
And D(x^2) is 2x
D(X_2) = 3 ?
It's easier just using the power rule directly, by the way
so "numbers" at coefficient a-s?
11:48
@TobiasKildetoft In one of our first conversations you mentioned that projectivity over a group algebra gets more interesting in weird characteristic. Do you know of any references of results in that direction? It suddenly became relevant for my undergraduate project.
But, yeah, 3*0*x^2+2*1*x+0=2*x
D(X_3) = 3*X_2 ?
Yep
So, now
ufffff
thanks
answer than
The basis is {1, x, x^2, x^2}
(I've been writing x^2 for what you've been writing x_2 — usually, _ means subscript — but it doesn't really matter)
11:50
M=\pmatrix{0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\\ 0 & 2*X & 0 & 1 \\ 3*X_2 & 0 & 0 & 0}
?
But $'s at each end
$M=\pmatrix{0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\\ 0 & 2*X & 0 & 1 \\ 3*X_2 & 0 & 0 & 0}\\$
is what you wrote
@AndrewThompson Well, it actually becomes a question at all. In characteristic $0$, all modules over a group algebra are projective
this ok ?
11:52
Once this fails, there are a lot of other tools one can use to understand what happens
$$\pmatrix{0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0}$$
@TobiasKildetoft That look right to you?
@AkivaWeinberger yeah
Can you explain, I really got to go
For real this time
@tobias.. ?
thank you
(Essentially, you write $a+bx+cx^2+dx^3$ as $\pmatrix{a\\b\\c\\d}\\$, so that the basis of $\Bbb R_3[x ]$ matches that of $\Bbb R^4$)
(And make each column the derivative of each basis thing)
12:00
@TobiasKildetoft Yesh, that's the case I've been working with (characteristic not dividing the order of the group.) By 'question at all', you mean there are cases where there are no nontrivial projectives?
@AndrewThompson well, no projectives at all, but that is not for finite groups
I mean that studying projectives is not interesting when everything is projective
@TobiasKildetoft Indeed. So, to rephrase my question: for a finite group and a field of characteristic dividing the order of the group, under what conditions is a module over the group algebra projective? (Asking for references in that vague direction, not concrete answers.)
@AndrewThompson Ahh, I think that is usually covered is most texts on finite-dimensional algebras and Auslander-Reiten theory
Ah, I see there's no way around the Auslander-Reiten world.
Then I'll try Auslander-Reiten-Smalø again; thanks!
and of course also in any text that deals specifically with modular reps of finite groups
it might be possible to avoid, but I have no idea
12:05
2nd question
That already helps a bunch, thanks!
@AndrewThompson Not sure of a good reference for modular reps of finite groups in general
Find the matrix D with respect to the base of falling potentials
@TobiasKildetoft Wikipedia gives some pointers, I can also try Serre's linreps. Auslander-Reiten-Smalø didn't turn out to be overly helpful.
In general I am interested in, under what assumptions the induction functor induces a stable equivalence.
@AndrewThompson yeah, it is probably hard to get into if you are not already familiar with that way of looking at things (though as I recall, it does have some results on this precise question)
@AndrewThompson by which you mean an equivalence of the projectively stable categories of modules?
12:09
Yes.
Is your induction left- or right adjoint to restriction?
Left. (More than willing to change to coinduction if that helps.)
@AndrewThompson No idea, I was just curious which one it was
sounds like an interesting question
Thanks. The endgoal is to say something about Hochschild homology of group algebras.
I do find something in Auslander-Reiten-Smalø about relative projectivity for group algebras of finite reptype, however I am unable to see how that's immediately helpful.
Hmm, I think there is something about the dimension of projectives always being a power of the characteristic. But this may be just for simple projective and possibly just in the special case I am familiar with of finite group of Lie type in the defining characteristic
12:16
We do have an algebraic geometer working in finite characteristic I might be able to ask about this stuff.
oh, and I assume you are already aware that projectives and injectives coincide for these algebras
anyway, I need to go now.
Yup, and alright! Thanks a bunch as always!
 
2 hours later…
14:33
4
Q: Approximate spectral decomposition

Valery SaharovSee attempt below I am interested in effective computations in finding approximate spectral decompositions in some suitable format. Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can ...


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