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17:00
What to do next :/ ?
what is the constant?
(-a-b) ?
p
?
17:01
@Ste @Ale @Bal Ah, I see, I was confusing the ideal generated by $F\in R[x]$ with the subring generated by $F$. (The former includes $xF$; the latter doesn't in general. The latter is $\{1,F,F^2,\dots\}$.)
1?
yes, but I'm asking about the constant in x^2+(-a-b)x+ab
I don't know.
So, yeah, to deal with the ones of sufficiently high degree, find polynomials in the ideal whose leading coefficients generate the ideal of all leading coefficients. And a similar muckery deals with the ones of smaller degree.
(to i-deal with the ones of...)
17:03
@Abcd the constant term is ab
okay
ab = 1?
Hm, random observation: SteamyRoot, Alessandro, Ted, Balarka $\mapsto$ soprano, alto, tenor, bass
(Well, Steamy and Semi and Secret and…)
(…and Astyx and Arctic and…)
i'm confused
17:06
Hm. I guess Amin can be an alto.
@Astyx First letters of your names
@LeakyNun Is there any simpler way :( ?
Ah right
It's like the parts of a choir.
@Abcd yes. use vieta's formulas
Is there something inside undecidable languages which isn't RE and non-RE?
17:07
Makes sense for Ted since he's into differential geometry so he likes ten(s)or... snaps
@AkivaWeinberger I play the bass though :P (and I can't sing)
Is there any solution using my method? @LeakyNun
@Daminark it was nice knowing you
@Abcd I don't know
okay
17:07
tear but that was good @Astyx
@Alessandro It has been decided that you're alto so I guess you must change your musical ways now
Wait, no, that would make me an alto.
Eh, it's almost exactly an octave above my range. I can sing down the octave, no one will notice.
Perfect!
We should take lessons from Vitas (look him up on Youtube... how??)
First song is "Finite Simple Group of Order Two," obviously
That title is very redundant. Groups of order 2 are automatically finite and simple... :P
17:16
Though it seems the flow of the song necessitates that so we can roll with it
Does any one else know the solution?
@Abcd I believe you will need Vieta's formulas, which are not too hard to derive especially in the quadratic case. Leaky was walking you through it rather well. Think about what he was saying and review your formulas, this will make sense
@Daminark okay.
Here's a more general observation. Suppose $f(x)$ is a degree-n polynomial with roots $r_1,r_2,...,r_n$.
17:20
(Hey @Eric!)
Then $f(x)$ must be of the form $f(x)=a(x-r_1)(x-r_2)\cdots(x-r_n)$.
If you then imagine distributing that you'll find that
1) If you want to get x^n, you'll need to take x from each factor.
2) If you want to get x^(n-1), you'll need to take x from all but one of the factors, and take r_k from the one you don't take x from.
If you follow that logic, you'll find that $f(x)=a(x^n-(r_1+r_2+\cdots+r_n)x^{n-1}+...)$
So the coefficient of the first term after x^n tells you something about the sum of the roots.
@Semi I had this one fun problem in the REU last summer which related to this. So let's say you're a pirate, and the treasure map tells you the location based on the roots of a polynomial. Because it's ripped, you only see the first three terms, $x^{17} + 5x^{16} + 13x^{15}$. Show that not all the roots are real
Similarly: 3) If you want to get the constant term, you need to take r_k from all the linear factors; putting these together gives the constant term as $(-r_1)(-r_2)\cdots(-r_n)$.
So the constant term tells you about the product of the roots.
Hmm, neat.
How special are 5, 13 here?
Certain swaps can be made, but not all
17:28
Hey @Paul!
Yoyo @Daminark
This is the problem you liked so much @Eric! :P
@Daminark Hello
Decide if you are going to do a reading course?
In what sense $\frac{i}{v-i}=\frac iv-O(\frac{i^2}{v^2})$?
I'm still trying to see, next year there won't likely be much room, and also my experience level will be a bit low. 4th year, I could very much see this happening
17:30
That's the leading term in the Laurent expansion around $v=\infty$.
That'll be followed by (i/v)^2, (i/v)^3, etc.
Where there particular people you wanted to ask for a reading course? @Daminark
Oh shit ok that problem I hate that problem
Yes, $\frac{i}{v-i}=sum_k (i/v)^k$ exactly. And we view this as a function of $\frac iv$ to make the $O()$ meaningful, thanks
I've considered asking either one of the PDE people about doing calc of variations (probably Souganidis, already took a class with him and am about to do another). Another thing I have in mind is to see after algebra next year, I'll possibly ask maybe Emerton then if I find some topic in that which would be suitable to a good reading course @Paul
Calc of variations!
17:41
Also I've considered doing some amount of harmonic/Fourier analysis, since if Soug teaches fall quarter I'm not sure how much of that we'll be getting
I think there are a lot of mistakes with this answer:
2
A: How to solve $ty' + y = t^2 \ln t$

TheDeadLegendIn big words. BERNOULI'S $$y'+\frac{y}{t}=t \ln t$$ Integrating Factor: $e^{\int\frac{dt}{t}}=t$ $$\int d(yt)=\int t^2\ln t.dt+C $$ To solve $\int t^2 \ln t$, take $t=e^x$ and proceed. This shall give your answer $$\int e^{2x}.x=\frac{1}{2}(x.e^{2x}-e^{2x})$$

This will come as a mix of reading courses and DRPs since there very much won't be time to do it all
Hey @Mike and @Adeek!
@arctictern check this proof I made it is very nice
0
Q: Let A be a ring and A[x] be polynomial ring. f is a zero divisor iff there exists $a \neq 0$ such that af = 0. (Proof verification)

AdeekSuppose f is a zero divisor. Choose polynomial $g = b_0 + b_1x + \ldots + b_mx^m$ of least degree m such that $fg = 0 \implies \Sigma_{k \geq 0} \Sigma_{i + j = k} a_i b_j x^k = 0$. From the formula of multiplication of two polynomials we must have $a_nb_m = 0 \implies a_n g = 0$ (Because $a_n g$...

hey @Daminark @MikeMiller
@MeowMix @AkivaWeinberger
@Daminark Do they not have any interesting algebra classes you can take (maybe a grad class or topics class?). I mentored someone for a little bit ind a DRP, stopped after a couple weeks though
17:45
Oh there are a lot. There's grad algebra, which I hope to do, and then there are undergrad classes in rep theory, algebraic geometry, commutative algebra, algebraic curves, algebraic number theory, etc
Commutative algebra and Rep theory sounds interesting.
also Algebraic geometry but you need commutative algebra for that
Normally that'd be true
But for some reason it's done in a weird order here
I really like my proof
Algebraic geometry is a fall class, and it requires a year of analysis and 2 quarters of algebra only. They do recommend you do third quarter algebra and topology/complex analysis as well
Commutative algebra is a winter class which requires the full year of algebra
That's cause the algebraic geometry class is usually classical over $\mathbb{C}$ so all the stuff you need is taught in the undergrad algebra
17:47
you should read it. It is very nice.
no required high level knowledge to understand the proof.
Hi guys. Is there any mable expert here tonight?
I'm trying to subsitute some values into an equation, but for some odd reason it doesn't seem to wiork ?!
I dunno, they should probably just swap the two quarter and make commutative algebra a prereq or something, but I guess. Whatever the case, the first priority is grad algebra, it'll subsume the others anyway
@Daminark there's a reason it doesn't use commutative algebra as a prereq
Hi Demonark, @Eric, Karim, et al.
Heya @Ted
17:51
Perhaps, I don't know enough math to really say anything, everything I've gotten has been limited to vague impressions thus far
@DanielGuldbergAaes 1) Ew, Maple; 2) Try to understand what the error says
hi @TedShifrin
Hey @Ted!
I mean, you clearly forgot a comma
@TedShifrin check out my proof it is nice.
0
Q: Let A be a ring and A[x] be polynomial ring. f is a zero divisor iff there exists $a \neq 0$ such that af = 0. (Proof verification)

AdeekSuppose f is a zero divisor. Choose polynomial $g = b_0 + b_1x + \ldots + b_mx^m$ of least degree m such that $fg = 0 \implies \Sigma_{k \geq 0} \Sigma_{i + j = k} a_i b_j x^k = 0$. From the formula of multiplication of two polynomials we must have $a_nb_m = 0 \implies a_n g = 0$ (Because $a_n g$...

17:51
@Danu: I saw your question earlier. I've never worked this out, but I can tell you it is true for the quadric surface in $\Bbb P^3$. This is the Veronese embedding of $\Bbb P^1\times\Bbb P^1$ and the embedding is an isometry. So you can check it easily in that case. I'll ponder more later.
@Adeek Nice
@SteamyRoot I have tried for almos an hour know -_- and no matter of what i Can't get it to work
There's no comma after [2] ...
@TedShifrin btw I decided to write everything in latex. My solutions for books and also my notes and stuff. I solved first 3 chapters half of the questions for Michael atiyah, but I lost the notebook.
at least, as I can see in the screenshot
17:52
and having everything in computer is much more organized than having million of notebooks
@DanielGuldbergAaes Even though the error doesn't say "you forgot a comma", it helps you narrow down the problem. In thise case, it tells you that there's something wrong with the input {false,x = 4-t}. x = 4-t is indeed what you want there, but "false" isn't - you want y = ... and z = ... . Hence something is wrong with those input thingies
@Danu: I misspoke. I meant Segre embedding. Sorry.
Hi @Ted
Hi @Balarka
18:04
hi @BalarkaSen
I guess I want to understand the geodesics in SO(n) now
hey
Hey @Balarka!
Good exercise. When you have the bi-invariant metric, geodesics through $e$ are all the $1$-parameter subgroups.
I think I proved this last time I taught Riemannian geometry.
Hi @Daminark.
@Ted: "..."
18:06
@MikeM: "...?"
Oh, that's interesting, @Ted.
He coulda proved that!
@Balarka: True in general, what I said.
Oh. Well, he still has to prove it.
I don't know how I would discover it easily.
It's the statement that's hard to think of, IMO, not the proof. :)
The proof, as I do it, still requires some nontrivial computation using structure constants.
18:07
Definitely not necessary.
Hi
Well, OK, Balarka can find a better proof.
Hi @Astyx
Was your doctor nice ?
Hey @Astyx!
@MikeM: I guess my natural approach is to write down what the connection is in terms of the structure constants.
18:09
@EricS Done already? :P
@Ted: Different tastes, I suspect. I facebooked you my approach.
I'm terribad with matrix groups. But I guess I can work with general Lie group.
(compact, connected)
@Daminark i thought I had put you on ignore after that terrible far-fetched pun :p
I like having connection 1-forms explicit.
Obviously, we have orthogonal tastes.
You need biinvariant metric, @Balarka. False otherwise.
We both like orthogonal groups, you mean?
18:10
The main thing I know about Lie groups is that the Baker-Campbell-Hausdorff formula manages to simultaneously be useful and a huge pain.
@TedShifrin I saw from your comment :) That's why I added compact, connected.
I've never in my life used that, @Semiclassic.
You run into it a ton in quantum mechanics.
Still, @Balarka, you have to specify the metric when you mumble geodesics.
18:11
Just sayin, Semiclassic. We all have different interests/uses for things.
@Astyx That ought to have expired by now
I only like connected, simply connected, nilpotent Lie groups :o
Oh, too bad :(
Since the time-evolution of an operator in the Heisenberg picture is given by $\hat{A}(t)=e^{i t\hat{H}/\hbar}\hat{A}(0)e^{-i t\hat{H}/\hbar}.$
Would someone mind helping with this problem: math.stackexchange.com/questions/2262743/…
18:12
@MikeM: Do you have a nice proof that the smooth quadric hypersurface in $\Bbb CP^n$ with the induced metric is Kähler-Einstein? (Danu was asking.)
Comes in real handy for such matrix exponential stuff...when it's simple enough to be useful, at any rate.
Also @Ted turns out we will do some differential forms, since we'll be finding how to integrate in order to find the degree of a map
Well, Demonark, to be honest, I prove that theorem, but rarely do I use that to compute degree.
Or at least that's what I recall was said
@Ted Almost certainly not.
18:14
@MikeM: Do you know if it's true? I have checked it for the surface case (isometric to $\Bbb P^1\times\Bbb P^1$).
Now that's a question I'm more likely to find the answer to. Let me look at Besse
@Astyx it truly is unfortunate, now you'll have to be subject to even more wonderful punning
New series from 3B1B \o/
Aaargh ...
"The Essence of Calculus"
18:16
Yeah, I've been keeping up-to-date
(unless he's released one today that I haven't seen yet)
But have no fear, I'm generous today and may partially spare you
wonderful?!
Demonark: Would that it were even meagerly wonderful.
Every (Hausdorff) topological space generates a notion of a limit, right?
they are like bad
18:17
Can two nonhomeomorphic spaces generate the same limit function?
What are the requirements a function like that must have to be the limit function of some space?
I have no idea what you're talking about, DogAteMy.
@Ted Is the quadric naturally homogenous?
You can define what $\lim_{n\to\infty}a_n=L$ means if the $a_n$ are in some Hausdorff space.
(Maybe by some subgroup of PU(n)?)
With non-Hausdorff spaces, the limit need not be unique.
18:19
Sorry, forgot what to answer where, @MikeM.
@Ted but they are! You just need sufficient exposure to them and you'll see why they are great. In the first few years it might be hard to see, but give it time
Well, topology is only equivalent to sequence stuff in a first-countable space, DogAteMy, so one really needs to be careful.
Demonark: Great like the Orange Cheeto is making things great.
I promise, it isn't Stockholm syndrome
@Ted what happened this time??
18:20
@Ted Besse 8.95, a theorem of Matsushima, says every simply connected homogeneous compact Kahler manifold admits a unique invariant Kahler-Einstein metric.
One cataclysm after another. More religious bigotry this morning and he thinks he's going to be the one to broker peace in the Middle East — "since it isn't that complicated"
Indeed I suspect that the metric given as a sub space of the complex projective space is itself homogeneous. I don't know how to check KE from there, but it seems plausible.
@MikeM @Danu: But it isn't a priori obvious that the induced Kähler metric as it sits in $\Bbb P^N$ is that one.
We just said the same thing. I guess one needs to check equivariance. I'll think about it.
Annoying — I forgot to close some apps on my phone last evening and the battery ran down to 0. So I can't go run errands until it charges up enough, cuz I need it somewhere along the errands.
Third-world problem, I realize. Life was so much easier when I was 20 yrs behind the times.
Equivariance should be easy. The automorphisms making it homogeneous extend to CP^n, right?
I feel like this was an easy theorem when I learned a little algebraic geometry.
But I don't think they do in an obvious way, @MikeM, as it's an $SO(n)$ action and we're in $\Bbb CP^{n-1}$. Well, maybe.
18:25
@TedShifrin lol
SO(n) sits inside SU(n) projects to PU(n) injectively
But the quadric is given by complexifying $O(n)$, not thinking unitarily, @MikeM. I need to ponder.
Oh god... This isn't even a TV show at this point... I'll just revert to math, makes more sense than politics
It usually has, Demonark. That much is not new.
Oh, and the House did pass the regressive bill screwing everyone but the rich on healthcare.
Heya @Gridley
Hi,
I'm having some problems with the following question about sequential compactness
Why is the last space compact? I have a feeling it is something to do with the fact the the sum of $\frac{1}{n^2}$ is convergent
18:30
@Ted This is beyond my pay grade. I'm just pattern matching with things I understand.
The first one isnt if you take the sequence (1,0,0,0...) (0,1,0,0,0...), (0,0,1,0,0...) etc
That's fair, I guess this is the first time that I've actually been aware of politics so my entire view is Trumpian.
As for healthcare, I haven't thought much about it. I know in Texas everyone hates it because of the mandate and because most doctors there reject it anyway, but I've also heard that it very much was a step in the right direction
The second one is because you can just think about it as R^2015. It comes out through bolzano/heine borel. For the third take the sequence (1,0,0....) (1,1/2,0,0...)(1,1/2,1/3,0,0,0....) which converges to (1,1/2,1/3,.....1/n,1/(n+1),....) etc. So it is not compact.
So this is very unfortunate...
I've been told that the last one is but I am struggling for a proof.
18:35
Yeah, @Gridley, it is, because it's metrizable and is compact.
@TedShifrin Could you elaborate? This question theoretically should only need the sequential definition of compactness, Bolzano-Weierstrass, topological invariance of sequential compactness.
You should be able to write down a bare-hands argument that a sequence has a convergent subsequence. You probably will need to use the Cantor diagonal argument somewhere.
How does the l^2 norm and the restriction that $x_n \leq \frac1n$ come into play?
Well, note that the sequence given by right-hand endpoints converges in $\ell^2$ :)
What do you mean by right-hand endpoints?
18:43
$1$, $1/2$, $1/3$, $1/4$, \dots, $1/n$, ... :)
Oh right yeah.
So by comparison, you know that every element of this product space is actually in $\ell^2$.
So we can say that the sup-norm is bounded by 1/n, which converges in l2?
That doesn't make sense.
But you raise a good question. You need good notation to work with this problem. How're you going to write a sequence of sequences?
You need to know that if a sequence converges in $\ell^2$, that it converges "pointwise," i.e., coordinate by coordinate.
Usually I define it component wise. Something like this
18:48
OK, cool, just so long as you have your own notation you're comfortable with.
@Ted apparently we'll be finishing off the parts of do Carmo we'll be covering this quarter by next class, then doing a bunch of comparison theorems
So try to explicitly construct a convergent subsequence. How would you do it in $[0,1]^2$ directly?
@Eric: I just cannot imagine a 10-week course. In Berkeley, this course ran 2 quarters (basic manifolds being the first quarter) when I took it there.
we've went so faaaasst
I teach fast (according to my students), but nothing like what you guys are doing, I think.
Comparison theorems are neat, but again I've never been a Riemannian geometer at heart.
it's only possible to go this fast cause neves is relegating entire sections of do carmo to read so that we can solve the problems
18:52
I'm not quite sure...
Well, for a grad course that's not sooo horrible, but it's lots of work for students.
he said he'd cap off the course with a sketch of the proof of the positive mass theorem which seemed interesting
@Gridley: Yeah, so how do we prove a sequence of points in $[0,1]^2$ always has a convergent subsequence? If you figure this out, it gives you the idea.
Oh, Schoen-Yau's theorem. I've never seen the proof, @Eric.
yeah I mean, I've found it to be an acceptable pace for me, but my classes other than the ones with neves have basically been prior knowledge for me, so haven't had to work much outside of geometry
meh why am so good at procrastination
18:54
and then I distracted you with alternative math, @Eric :P
@Balarka: At least you're good at something :D
on the brighter note yup
lol at least it gives me perspective on the math im doing otherwise @Ted
Ok let's see if I can prove the thing you said
I'm leaving shortly ... have a ton of errands to do.
I'll ping stuff with you if I figure it out.
18:56
@Gridley: Can you just take a convergent subsequence of $(a_n)_1$ and a convergent subsequence of $(a_n)_2$?
I'm going to be dumb and understand the (torsion-free) Riemannian connection of a bi-invariant metric first
That's a reasonable strategy.
Not dumb.
He's thinking more like you than me.
Thank goodness :D
Although he was thinking more like you a week ago when he asked me about the proof that the fixed-point set of an isometry is totally geodesic.
I wonder if Mike would understand it via exponential map.
But eh whatever. I'm going to wear myself out with formulas first, just because I can
19:03
OK, I need to get going. @Gridley, I hope my leading questions have gotten you started. If you get totally stuck, I should be back in a few hours.
Oh, oh, the formula looks promising. I actually know that $\langle X, [Y, Z] \rangle = \langle [X, Y], Z \rangle$ for biinvariant dudes.
Does anyone have an idea why Wolfram Alpha evaluates this limit at 0http://m.wolframalpha.com/input/?i=limit+as+n+goes+to+infinity+of+sum+i+from+1+‌​to+n+of+%28%285%5En-1%29%2F%28n*n%21%29%29&x=0&y=0
Maybe I can mess around with that.
The sequence is always positive, so it can't sum to 0
And evaluating it at various n's suggests that it converges to about 36.68
@Rance Wolframe did not understand what you asked
And you'd need $i$'s in that anyway, not $n$'s
19:24
@Astyx thanks!
Glad to help
19:38
Is the image of a connected component under a continuous map a connected component? Clearly the image will be connected, but I am having trouble showing that it will be a component.
@Eric So we've been lagging the syllabus big time. As it stands we've defined intersection number today, so next class we're gonna define degree in general, and then get into finding degree through differential forms. After we get through that stuff we'll be done with the technical theorems, and then move on to applications like Poincare-Hopf
i love poincare hopf
good stuff
wait @Daminark do you have class on tuesday
@user193319 That's false
@AlessandroCodenotti Hmm...well part (b) of exercise 2 in the following link seems to rely on this: isites.harvard.edu/fs/docs/icb.topic98903.files/Assignment_5/…
I must be misunderstanding the solution.
@Eric Yeah, our TA will be taking over that day
Also I think we're gonna do Hopf Degree Theorem? And if we have time maybe Gauss-Bonnet
19:46
GP's proof of GB relies crucially on PH
@AlessandroCodenotti Can you make sense of what they are saying?
I'm not sure where is that "fact" used in this solution
nice acronyms. it looks like a coded message now
ah @Daminark ok i dont think we're gonna have class
@Balarka What do you think of when you see "BCT"?
19:48
and he didnt want to commit to class on thursday, he said if he's too jetlagged we just wont
Baire Category
Baire Category
@AlessandroCodenotti The part I am referring to is this: "So the connected component of y must be a translate (by fy) of the connected component of the origin."
@Eric Our TA will be taking over then as well, Neves will just sit in the audience
yeah grad classes dont really have TAs just graders
19:51
@Balarka and @Alessandro Good
They have an homeomorphism there, not just a continuous map
@AlessandroCodenotti Ah, so in that case connected components get mapped to connected components?

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