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00:00
idk cause they're all so different
a typical brazilian would put white but im a minority in brazil too
they have a "mixed" box there
Oh, well, it's not worth losing sleep and stressing over. Pick something that minimizes error :P
Similarly one defines the exponential series to be $exp(a)=\sum_{n\geq 0}\frac{a^n}{n!}$ and the claim is similar that this is convergence because for each $k \geq 0$, the term of degree k of $\sum_{n=0}^N \frac{a^n}{n!}$ has a limit in $E^{\otimes k}$ as N tends to $\infty$.
@TedShifrin I thought the secret might lies in that the exponential series and log series are maps to the space $T(E)$ which can be thought in basis form $\sum_{n=0}^\infty \sum a_{i_1,...,in}e_{i_1}\otimes...e_{i_n}$
Thus, what tey mean is that each coefficent to the base $e_{i_1}\otimes...\otimes e_{i_n}$ should be finite.
To be honest, I don't have the desire or the brain power to try to sort this out. Sorry.
Ok, I think I figured it more or less out. Thanks
00:07
Ted are you lectures in the natural order listed on Youtube ?@TedShifrin
Or can I chose the order?
@Jacksoja: Yes, with the exception of lecture 24 in the second set. It got messed up and there's a replacement one at the very bottom.
@TedShifrin Okay thanks !
ah ha @Ted i have found a way to input text in a drop down menu so my problem is solved
You can choose whatever order you want ... but there is a logical flow. Obviously, you can guess from titles and do what's important/interesting.
I can tell you enjoy teaching ! not like most proffesors I had so far ...
00:08
Oh cool, @EricSilva.
Am gonna skip the physics applications
like kepler 's law
these americans and their nonsense classifications
@EricSilva How dare you ...
There will be more physics-y things in the second semester.
@Jacksoja wut
00:10
Are they worth to watch if am a math student ? with very little background in physics?
I'm on Eric's side.
Who knows, @Jacksoja — you might learn something :D
I made it mostly self-contained. Most people know what Newton's laws are.
If I had time pressure can I skip ? Id love to see them ofc
Yes basic physics I can handle
You're in charge of your life. I'm not making you do anything :D
haha
I got it , thanks
Trump
00:18
I bid 6 NO trump.
By the way thank Ted. I don't know how I missed that
It was quite unexpected that I knew the answer so easily. Measure theory is NOT my forte.
@TedShifrin Well played! :]
well somehow the point is that if something holds as h->0 for all x, then it'd better hold for a fixed sequence h_n -> 0, like n^{-1}.
and that gives you exactly what you need.
I guess normal people write n^-1 as 1/n.
measure theory would be easy if it werent also extremely boring
00:25
my other solution used signed measures and Radon Nikodym derivative lol
Well, sometimes it can be quite nonobvious too.
though of course, you can say the same about lots of areas of math
:)
I love the starred question in capitals on the side! xD
idk, when it's nonobvious it's usually cause there's a non-measure theoretic interaction happening
Here's a fun one: Show that B \subset I (I a bounded subinterval of R) is Lebesgue measurable iff m*(I) = m*(B) +m*(I \ B), with m* Leb outer measure
* with m Lebesgue outer measure
Depends on your definition of "fun."
It's pretty tricky not gonna lie
00:35
When I was your age, I knew how to do such things. Now I'm old.
By the way, Ted. Do you know much Geometric Measure Theory?
I have dabbled, but, once again, now I'm old.
Eric does, though :P
throwing me under the bus i see
mostly was gonna ask about a good book to learn from
seen some results about Hausdorff measure looks fun
read frank morgan if you dont want to hate your life
00:37
Oh, to get an idea of what's going on, look at Frank Morgan's book.
@EricSilva: On balance, you owe me a few :P
you mean Federer is not the best place to start ;)
It's the place you'll end.
i tried to read federer once and now im legally deceased
@TedShifrin fair
I actually needed a result from Federer in my second paper.
why are we talking about tennis
00:38
I have never seen it anywhere else.
francesco maggi's book also is not bad if you like analysis a lot more than i do
apparently there are conjectures in Federer that still don't have proof
Hi, JoeShmo. It's the US OPEN. We should be.
fair point
@Drew: You have a world expert on GMT right there to bug ... talk to Brian White.
00:39
yeah I know. trying to preview some stuff before talking to him
I was supposed to have been visiting at Stanford a week ago, but I canceled my trip because of all the fires/smoke. Ugh.
oh damn ur at stanford?
Brian White gud ive read some of his papers
Brian doesn't really talk to undergrads..ever lol
00:40
*tried to read
Really? I don't believe that.
I'm sure he would if I bugged him. I talked to him about a reading course once and he said "maybe" lol
I don't know him well. So I can't really pass judgment. I knew Rick Schoen pretty well, but he moved on ... and I've known Rafe Mazzeo since he was a sophomore in college. :P
i think schoen is still active but at UCI
Yup, he's at UCI.
00:43
only reason im applying there :p
Meh. I wouldn't.
André told me i should so i figured id at least send one in
Leon Simon used to teach a GMT course, but has since vanished
isnt he old
Leon? Yeah.
Rick is older than I am, so he's no spring chicken.
00:44
Yeah Leon said he might be back in January lol
Leon's great.
When do you graduate, Drew?
He taught the first math class I took at stanford
Next year
Ah, he originally was using my book in the honors course, but then wrote his own and dumped mine :P
You wrote an analysis book?
Leon Simon has good GMT notes
00:45
Yeah he said "I'll send them to you once I edit them more"
I guess he's working on a new version
I wrote an integrated linear algebra/multivariable calculus/analysis book. Mine was a bit more grounded than Leon's. We can't all be blown away like Stanford students.
oh @TedShifrin
good that your here
heya Karim
Blown away?
My book is considered very hard by most students/schools, but it's too easy for Stanford, Drew :P
00:46
cuz y'alls is 2 gud
I was wondering do you know if classifying holomorphic vector bundles is same as reals ? That is holomorphic one are one-to-one correspondence $[X,G_n]$ ?
They used it many years at Yale, and I think they're still using it at Vandy (and UGA of course).
NOOOOO, Karim. You should know that's wrong.
ohh
because of partition of unity
I see
Well, that's the starting point. But don't you know about the exponential sheaf sequence for line bundles?
00:47
yeahhh that as well
$0\to\Bbb Z\to\mathscr O\to\mathscr O^*\to 1$?
yeahh
So run the exact cohomology sequence and you'll answer your own question :P
crap I have to remove certain part of my thesis
ROFL
That's a serious error, my man.
What gives you bundles with the same Chern class (hence topologically isomorphic) and therefore measures holomorphic structures?
00:49
@EricSilva I'm inviting you (based on that comment) to a race you probably haven't considered.
I was wondering, Eric.
yeah haha I mean in later part of my thesis I work with smooth projective varieties over C. I classified real vector bundles but it is not related to other parts of my thesis because that statement isn't true for holomorphic vector bundles @TedShifrin
Karim: Topological classification of complex vector bundles is fine. But holomorphic structure is much harder. Same thing with different complex structures on diffeomorphic complex manifolds.
yeah
their is a family
00:50
Right.
Haha!
And the Picard variety $H^1(X,\mathscr O)/H^1(X,\mathbb Z)$ gives you the torus that parametrizes different complex structures on the same complex line bundle.
Hello. How to derive $sin(x)=4sin(x/4) cos(x/4) cos(x/2)$ from sin2x=2sinx cosx ?
@Symposium idek what this comment means
Do it twice, @Alt..
00:52
I see @TedShifrin
is there a general statement for holomorphic vector bundles ?
Write down the exact sequence and you'll see that, Karim. Look at $\ker c_1$.
That's way harder.
@TedShifrin how? I could derive sinx=2sin(x/2)cos(x/2)
OK, @Alt., now use it on the sin again.
I see the result their is some twists
going on we have to detect that somehow
I will remove all of that together I don't want to bother
I was gonna suggest you might look at Griffiths & Adams at some point.
Yellow Princeton book.
But definitely hard.
00:56
sounds good @TedShifrin. I am just doing final revisions to my thesis. I will finish on Tuesday. I will go hang out with fellow graduate students on Thursday and drink wine and discuss philosophy. Starting from Friday I will start preparing for my PhD.
LOL ... better not to ask me questions, Karim :P
:D
I am excited though I mean my knowledge in transcendental aspects and arithmetic aspects of algebraic geometry is good, but I still have long way to go.
I always had a long way to go.
haha. I was thinking maybe I should once a month give lectures to other graduate students about my work. I think that would probably help clarify certain things.
@EricSilva It was joke regarding your "cuz y'alls is 2 gud" comment, because it sounded African-American English in my ear (mainly because of the usage of the auxiliary "is" instead of "are", otherwise it may have been just Southern).
00:59
ok.
You worked with the great Shiing-Shen Chern geometer right @TedShifrin ?
Sorry, just a bad joke! :(
Yup, Karim. I hope he wouldn't be too ashamed :)
@TedShifrin I don't know how. To derive sinx=2sin(x/2)cos(x/2) I wrote A=x/2 in the equality sin2A=2sinA cosA. What value of A should I take to get $sin(x)=4sin(x/4) cos(x/4) cos(x/2)$ ?
@Alt.: Write down $\sin(x/2) = ... ?$ So what should $A$ be now?
01:03
@TedShifrin You know it is so weird that I have very close to photographic memory, but I tend to lose my keys and wallet all the time. I just lost my credit card yesterday.
Not good, Karim. Make an effort to pay more attention.
@Symposium it is ok
yeah
Oh oh .. it's a Demonark.
gasp where?
01:06
@EricSilva: Out of curiosity, email me your list of schools. I may or may not have a comment.
no one move their vision is based on movement
@TedShifrin totally, was meaning to ask for your comments too
I'm happy for Demonark to ask me, too, but I'm less knowledgeable re your areas of interest.
Whatever input you think is worth giving, I would absolutely appreciate. I don't have a full list put together yet but I'll send you the places I've started looking at recently. Thank you!
u havent asked ur bois yet boi?
01:09
One of them never responds to email, so I'll have to wait until I get back to ask in person
i feel
Lol I love measure theory tricks
We need to geometrize Drew.
Ted do you know Donald Cohn's MT book
nope
Damn, Harvard's math dept website absolutely sucks rocks.
2
01:14
Right?
it truly truly does
It's more challenging than Folland I think. It has some tricky ones lol
@TedShifrin got it, so I just needed to change the value of x (x=x'/2,x'/4,etc) in sin(x)=2sin(x/2)cos(x/2)
It looks like it was done by a sixth grader.
Right, @Alt. Well done.
Hey I am trying to come up with a good name for this section
01:14
I took MT from Brian White and we used Cohn's book. I'm pretty sure he used it just as a f* you lol. Anyways it turns out its much easier to do the exercises once you actually know something about measure theory.
"Vector bundles and its relationship to cycles"
This doesn't seem to be a good name for some reason
any suggestions to edit it ?
Well, the English syntax sucks for starters.
When I told him it was tricky in OH he grinned and told me it would "build character"
lol
yeah
How are you envisioning relating vector bundles to cycles? It's not quite like line bundles and divisors.
@Drew: Well, I had colleagues who told students (both undergrad and grad) not to take classes from me because I expected too much work/commitment. I think you've done just fine.
Karim: Assuming the title is appropriate, I would word it "Vector bundles and algebraic cycles"
01:17
@TedShifrin K-theory of vector bundles is the same as K-theory of coherent sheaves over a smooth scheme X.
For fun I'm trying to solve all the exercises in Cohn. I think I'm learning something. On Chapter 2 now.
I figure there was some reason why Brian chose this book.
Karim: So K-theory should be in your title. But where are you linking to cycles?
Because of Grothiendieck-Riemann-Roch
That is way too vague an answer.
I reject it totally.
$K_0(X) \otimes Q \cong A(X) \otimes Q$
01:19
So you're doing only $0$-cycles?
Hmm.
sorry adjusted.
Grothiendieck-Riemann-Roch gives us $K_0(X) \otimes Q \cong A(X) \otimes Q$
So ... given an arbitrary element of the Chow group, how do I get a bundle?
I don't know the details of the explicit isomorphisms involved.
I know G-R-R and I don't see how this is helpful.
I know that G-R-R gives us somehow the isomorphism above. I don't know the details though.
$K_0(X)$ here being the K-theory of coherent sheaves on X.
and I know the result which state that topological K-theory over X is the same as K-theory of coherent sheaves on X.
01:23
If you can't answer my question better than that, this probably shouldn't be in the thesis.
yeah
But your title is totally misleading unless you actually explain stuff explicitly.
your right
yeah I will remove it
Better to have less stuff and do it well, than to try to be a dilettante.
Yeah I mean my thesis is already 350 pages.
yeah I agree.
01:25
Though probably # pages != measure of quality
That's 3 times too long.
Indeed, @Drew.
yeah I agree
For reference this book on measure theory is 350 pages long ;)
and probably your thesis has less content than this book does ;)
the savagery
I mean it is possible of course that Cohn is just crazy terse
01:27
How do you know @DrewBrady ?
I don't, it was just a suspicion.
Let's not make this too personal. But, Karim, 350 pages is ridiculously long.
unless you have my thesis you can't say that.
yeah I will remove certain aspects of it.
After all, the results in this book were developed over the course of, I dunno, at least a century. And you're right. It wasn't meant to be personal. Just a demonstration that you can fit a lot in 350 pages.
Karim, chill.
Drew, enough.
sigh
01:29
don't worry I mean I don't really take this personal it is okay
glad your enjoying your measure theory book.
and good luck with the thesis, of course!
thx
see, @Ted we can get along :)
I mean I am done already
the thing about a thesis is that it's not in the same category as a published article/book
01:30
just doing revisions atm
besides this is not PhD thesis it is just MSc thesis.
so having excess material isn't really a problem, as long as you can manage it
LOL ... Yeah, unless the adviser reads each word carefully and fills the thesis with red ink (as I have done numerous times), the quality is generally poor. General statement.
the reason I covered a lot is because I wanted to be prepared for my PhD that is all.
Karim: You know that you have a tendency to be sloppy and say things that are wrong, so I worry that you haven't had this sufficiently criticized. But that's all I'll say.
I have improved on that.
01:37
One of my professors for next semester has expertise: nilmanifolds, almost-complex manifolds, and manifolds with exceptional holonomy.
That makes sense, actually.
what I am missing now @TedShifrin is some organizational skills.
I always rely on my brain for everything
brb
@TedShifrin Everything else listed I've heard before, e.g. twistor spaces, harmonic maps, complex/conformal - quaternionic/octonionic geometry etc but not manifolds with exceptional holonomy etc. They sound amazing!
I'm already sold!
This is a Bryant-Salamon thing. Exceptional Lie groups as holonomy groups.
Holy molly!
01:47
LOL, moly or moley? Hmm ...
I don't know, haha!
thanks for the words of wisdom-y-ness @Ted, I am out so bye chat
Bubye, Eric. Keep me posted.
The professor I was talking about is Simon Salamon! @TedShifrin
I figured it might be, Symposium.
Yeah, I don't know about his teaching, but take his course.
01:49
@TedShifrin huh, that's a neat statement. Not that I know what it means
Do you know who the exceptional Lie groups are, Semiclassic?
Sorta. E6,E7,E8
Right, $G_2$, $F$ somebody.
I vaguely recollect that Lie algebras are classified by Dynkin diagrams
and that those ones are weird because they don't fit in the usual families
So the question is to find a manifold and a Riemannian metric on it with that holonomy group.
01:51
@TedShifrin I will. Thanks! I was amazed when I saw the name in your reply xD
My knowledge starts to run out there, unfortunately (manifold, yes. Riemannian metric, mostly. holonomy group, nope)
Well, I'm going to cook dinner, so you're spared a small lecture. :)
my knowledge of holonomy is limited to the vague notion of 'you go around something once and see what changes"
lol, okay
Well, for example, if you have a generic hermitian complex manifold, instead of the usual $SO(2n)$ you get $U(n)$.
It's not about once, actually, but "going around" means parallel translating around (not too large) closed loops and looking at the resulting linear transformation on the tangent vectors.
Right
Which I can sorta see connecting with curvature
01:57
Absolutely.
OK, I'm gone for now. Bye :)
 
3 hours later…
05:04
Am I missing something on this measure theory exercise?
Suppose (X, A, m) is a measure space, and f_n, f are measureable functions such that f_n -> f a.e. Show that there exists functions g_n such that g_n = f_n a.e. and g_n -> f everywhere.
Solution. Take g_n = 1_N * f + 1_N^c * f_n, where N = {f_n -/> f}
how do you define f_n -> f?
05:29
I am pretty sure strong inaccessibles are those $\alpha$ such that for all $\beta < \alpha$, $\mathscr{P}(\beta) < \alpha$
@Secret that's not enough, since $\alpha = P(P(\cdots(\aleph_0)\cdots)) = \beth_\omega$ already does the job
so we need a regular cardinal cofinality to prevents them from reached from below by power set?
@AlessandroCodenotti why do we need them to be regular cardinals?
f_n -> f means that f_n(x) -> f(x) point wise.
I usually write f_n => f for uniform convergence.
what does f_n(x) -> f(x) mean?
05:44
other things I have in mind is whether the powerset operation is a computable function, and whether there exists a hierarchy of definable functions that grow faster than the powerset operation
@Secret depends on your definition of computable
Leaky Nun, the usual thing
Oh have I not defined the codomain
I think that the procedure terminates in finite steps, that is obviously not possible if the input set is infinite
@DrewBrady that's not the problem
oops, I mean [-\infty, \infty] endowed with the usual topology.
05:46
the problem is that the domain has no topology
No, the problem is that the codomain has no topology.
But now it does.
the domain has no topology
that's irrelevant.
@Secret the problem is that when we say computable function, our domain (and codomain) is countable
f_n(x) -> f(x) is defined regardless of whether X has a topology or not.
05:47
ok
Ah, that will rule out sets of cardinality $\aleph_0$ being a valid input then, because then the output is necessary uncountable in ZF
then under this definition, $\mathscr{P}$ is computable since all its inputs are finite sets, and the members of their powerset can be enumerated in finite steps
hi @loch
06:08
The Burning Ship fractal, first described and created by Michael Michelitsch and Otto E. Rössler in 1992, is generated by iterating the function: z n + 1 = ( | Re ⁡ ( z n ) | + i | Im ⁡ ( z ...
pretty
06:56
Gotta be glad that any nontrivial division by zero pretty much requires you to break commutativity and associativity, thus this article actually makes sense
thus order matters in general
@LeakyNun because otherwise there are plenty of not really interesring examples I think
You cannot really have a singular cardinal to be not reachable from below by definition, or that even in ZFC, there are infinite cardinals that are neither singular nor regular?
singular iff not regular
@AlessandroCodenotti why |V[kappa]|=kappa for inaccessible kappa?
I know that kappa <= |V[kappa]|
07:38
Division by zero as a concept: The relation that takes a singleton to the universe
Or more accurately: $R : \{0\} \to U$
So technically speaking it is the inverse relation of the zero map
and that, is why it causes a lot of problems
2
Q: Prime twins and $ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $

mickLet $p$ be a prime such that $p+2$ is Also a prime. Define $$ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $$ For $z \neq 1 $ and $re(z) > 1$ this $f(z)$ always converges. If there are infinitely many prime twins , and prime twins grow like $O (n * ln(n)^2 ) $ as i...

Any ideas ?
Hi chat!
Suppose we have a map $f$ of two variables that is $x$ and $y$, then is there any relation between the determinant of the Jacobian of $f$ and $f$ beng locally invertible?
@BAYMAX a lot
Ohh @LeakyNun!! how ?? any reference for this?
07:54
@BAYMAX it's called "inverse function theorem" and I'm trying to find a reference
$f$ is locally invertible iff the determinant of the Jacobian is nonzero
08:35
Hi @LeakyNun
@loch how is your counting stuff
08:50
Lol havent been spending too much time on it recently
Hi all! For a set $M$ of points in $\Bbb R^3$ with point group symmetry $G$, is the concept of "symmetry adapted" local coordinates something which is familiar to the mathematician?
09:21
@LeakyNun Thank you for the reference!! I will go through it! I want to just state where I was thinking of using them -
Say $f$ is a piecewise function
that is $f(x,y;\chi) = f_{1}(x,y;\chi) $ if $y \leq g(x;\chi)$
and
$f(x,y;\chi) = f_{2}(x,y;\chi) $ if $y \geq g(x;\chi)$
Now I am trying to understand this statement -
I am asking because I'd like to understand how the distortion along such a coordinate effects symmetry. The gist of the SACs is that each one can be assigned to an irrep. Now the question is how distorting in that way "along" an irrep affects the symmetry.
If the product of $det(Df_{1})_{|(0,1;0)}$ and $det(Df_{2})_{|(0,1;0)}$ is positive, then $f$ is locally invertible
Well if the product of the determinant is positive then neither of the determinant is non-zero and hence locally invertible by Inverse function theorem
But I am thinking whether saying this would be wrong?
If the product of $det(Df1)_{|(0,1;0)}$ and $det(Df2)_{|(0,1;0)}$ is negative, then $f$ is locally invertible
?
10:09
Hello there! I wanted to find properties of parabola (vertex and focus) from its standard form of equation. How can I do it? Any answers from stack exchange? I can't find though.
Anyone?
I hope this will be of some help
10:35
@BAYMAX Thanks... But I wanted to know for an equation like $(x+y)^2+6x+4y+3=0$ ??
10:59
Hmm
I was thinking that you may want to take $x+y = t$
That is we try to bring the above equation to standard form
so $t^2 + 4t + 2(t-y)+3 = 0$
I think this will be a parabola in $t-y$ plane and is in standard form
and you can find it out
If any mistake from my side, please let me know!!
11:18
@Rudi_Birnbaum evolution isn't about the fittest organism surviving and reproducing
11:34
@BAYMAX thanks... I'll try it out...
 
1 hour later…
12:44
@LeakyNun Did I claim that? I think its about the best adapted which can produce the highest number of fertile offspring,
13:07
@AlessandroCodenotti why does Rudin use $\delta=d(q,p_{n_1})$? Would $\delta=1$ work?
13:35
@LeakyNun, will you please look at that?
@Silent on quick glance it looks like he does that to guarantee that $d(n_i,q) \leqq d(n_{i+1},q)$
13:58
@anakhro hmm, please explain a little more! how does $\delta=1$ does not give desired result?
Suppose $\delta < 1/2$, can you find an $x$ such that $d(x,q)<1/2$ and $d(x,p_{n_2})<1/2$ but yet $d(q,p_{n_1}) \leqq d(q,p_{n_2})$?
Drawing pictures help in this case.
14:34
@anakhro Oh! I see your point. Thanks a lot.
Not to say that you can't do it with 1 at first, it's just not immediately clear what the starting of your "algorithm" does until some n_i further down the line where it reduces to doing what Rudin's does, roughly.
Rudin just does it the crisp and clean way. :)

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