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8:00 PM
Doesn't matter, first you calculate the first limit(the sum) for $k$ a number, then you take the limit
 
agreed
 
So for the sum $k$ is finite
 
It can be interpreted as a sequence in $k$.
 
btw this discussion started because of this
@Holo
2
Q: A rough proof for infinitesimals?

More AnonymousI discovered the following relation for arbitrary $d_r$: $$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \int_0^\infty...

 
8:02 PM
Yes, you can look at this as the limit of the sequence $$a_k=\lim_{n\to\infty} \sum_{m=0}^n f(k \frac{m}{n})k/n$$
 
Im using the intuitive idea of infnitesimals
 
and $d_r$ is the number of times it repeats
why not?
 
How do you define infinitesimal?
 
@Shobhit Not a fan of 0-dimensional topological vector spaces?
 
8:05 PM
Yes this was what I wanted to know. If this is a correct instance of the Riemann sum for the integral. Now I really believe.
 
because there is no point. There is a reason people stopped using infinitesimals for a long time until they were made rigorous
 
0-dimensional? i don't understand
 
but I have a counting formula which can be connected to geometry if infinitesimals exist
@TobiasKildetoft
 
jk @Shobhit. For dimension >0, you use the continuity of scalar multiplication on a well-chosen subspace.
 
So $d_r$ could be identified as a $g(r k/n)$?
 
8:07 PM
What if $$\sum_{r=1}^\infty b_r \left(\int_0^\infty f(rx) \, dx\right)^s = \infty?$$
 
@MoreAnonymous If infinitesimals exist where?
 
@Holo then RHS = infinty
 
I mean, we know there are nice fields containing the reals as well as infinitesimals, and with transfer allowing us to do nonstandard analysis
 
to achieve? @KarlKronenfeld
 
But you do not seem to mean anything as precise as that
 
8:09 PM
k ... would u agree sum f(1/n)1/n
with appropriate limits
 
You asked about showing that the topology is not discrete earlier @Shobhit and as far as I can tell nobody addressed it.
 
Can one interpret f(1/n) 1/n as an infinitesimal?
@TobiasKildetoft
 
yes, nobody did :(
 
@MoreAnonymous: What happens when you define a $g(r k/n):=d_r$ ?
 
@MoreAnonymous I don't see how or why
It is just a real number, which definitely makes it not an infinitesimal
 
8:10 PM
so choose a subspace and using continuity i should prove that given its cts. the topology could never be discrete? @KarlKronenfeld
 
so basically its only folklore u add infinitesimals to get area?
@Rudi_Birnbaum not sure ...
u have to be carefull of the limt
*limits
@TobiasKildetoft so basically its only folklore u add infinitesimals to get area?
 
@MoreAnonymous you would get the $\int g(x) f(x) dx$, I'd say. Then its trivial.
 
What does cts stand for? @Shobhit
 
You can use infinitesimals iirc, but it require a lot of definitions and proofs to show it is the same, and it create a lot of more work that I can't remember
So unless you love hyperreals no one do it
 
@Holo ... I don't think my formula can have an interpertation in geometry without it
2
Q: A rough proof for infinitesimals?

More AnonymousI discovered the following relation for arbitrary $d_r$: $$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \int_0^\infty...

 
8:14 PM
What I mean is define a $g(x)$ s.t. $g(r k/n) = d_r$.
 
@Rudi_Birnbaum Id advise u go thru the proof and see if ur allowed to do that
by brain feels a lil fried
 
continous @KarlKronenfeld
 
@Shobhit Aw, not carpel tunnel syndrome. :( Yeah, making a good choice of subspace makes the problem easier on yourself and scalar multiplication is where you bring in the topology of R or C.
 
bine as well
 
XD. Let me think and i'll get back to you. thanks @KarlKronenfeld
 
8:17 PM
@Holo I hope My ideas make sense to someone ...
 
@MoreAnonymous maybe you can formulate it without mentioning "infinitesimals"? :P
 
@Rudi_Birnbaum ... yes but then it would loose all geometric intution and would be a mere algebric formula
 
I can't quite follow why $$\lim_{s\to 1}\sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s \sim \sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)$$(Btw, I am also watching the international rn I am not super focused)
 
@KarlKronenfeld i can't think of a good subspace. help.
 
It's my boy Karl
 
8:31 PM
@Shobhit How about the subspace consisting of all scalar multiples of something nonzero?
 
thinking
 
@MikeMiller Ruling out possibilities like somebody else taking over the mse account.
 
@MoreAnonymous with $g(rk/n)=d_r$ you would get that: $$ \lim_{s \to 1} \! \frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s} \int_0^\infty f(x) \, dx = \int_0^\infty g(x) f(x) \, dx $$
 
i think i have to use that {0} is open to come to a contradiction? @KarlKronenfeld
 
How's life?
 
8:38 PM
@Holo
By using zeroth order
s goes to 1
 
@MikeMiller Life's pleasant, moving forward and whatnot. How about you?
@Shobhit Give it a try.
 
ok
 
It's good, now is a stressful time since I need to write like crazy and apply for postdocs. But I've been in good if stressed spirits about it.
 
@MikeMiller Whats your field of research?
 
Sounds terribly stressful. Has grad school been good?
 
8:46 PM
Something between low-dimensional topology, algebraic topology, PDE, and differential geometry.
 
So what's the title of your thesis?
 
Grad school is and has been a good experience for me but it has its perils.
I think the grant application I submitted said it would be "equivariant instanton homology and applications"
 
@Shobhit What progress have you made?
 
@MikeMiller sounds quite physical
@MikeMiller So you apply for a post doc before PhD already?
 
I don't know any physics.
 
8:52 PM
So what dou you understand by "instanton" ?
 
In the US one always applies for postdocs the year they graduate, to start in the fall after graduation. Gotta feed yourself that next year.
A connection on a principal bundle over a Riemannian 4-manifold whose curvature form is anti-self-dual.
 
I see. Kind of makes sense. Over here (and in my field) graduation is a procedure taking some months (starting from first "submission" of the thesis). And people start to care about their postdocs usually after the thesis is written up, and eventually get a grant 3 months after PhD where they often still have contracts. At times you apply also for positions which are announced somewhere.
@MikeMiller sounds really mean
 
It proves theorems about topology. See "Donaldson's diagonalizability theorem".
 
Is it in anyway connected with the instantons in physics?
An instanton (or pseudoparticle) is a notion appearing in theoretical and mathematical physics. An instanton is a classical solution to equations of motion with a finite, non-zero action, either in quantum mechanics or in quantum field theory. More precisely, it is a solution to the equations of motion of the classical field theory on a Euclidean spacetime. In such quantum theories, solutions to the equations of motion may be thought of as critical points of the action. The critical points of the action may be local maxima of the action, local minima, or saddle points. Instantons are important...
Oh its the same. Oh boy, how different can you talk about one and the same thing ...
So the "applications" part in your project application is it in any way about connected with physics?
 
9:08 PM
It is not the same; I think my instantons are theirs for a specific functional.
Very specific.
The applications are to topology. :) I do not really care about physics.
 
hi demonic @Alessandro
 
@MikeMiller so what would be that functional?
 
Anti-self-dual connections minimize the Yang-Mills functional, but they are not the only stationary points. Despite this ASD connections are sometimes called Yang-Mills instantons.
Yang-Mills theory is a physical theory, but I do not think about the physics or know anything about how they use it.
(The Yang-Mills functional is the L^2 norm of the curvature F_A of a connection A.)
@Ted: Take an oriented hypersurface in an oriented Riemannian manifold. Choose an oriented unit normal field, and exponentiate. This gives you a diffeomorphism from (-s,s) x Y to an open subset of M. Pull back the metric. I think in coordinates it should be dt^2 + g_t, the latter being a time-dependent metric on Y. Do you know a reference?
 
hi @MikeM ... not to interrupt
So you're generalizing the Gauss Lemma, basically.
Um, do I know a reference.
 
9:16 PM
Basically.
It makes some formulas run easier for me.
 
I have so few books any more .... I wonder if this might be in Al Gray's Volume of Tubes book. This is the very simplest case (codim 1).
 
Yeah absolutely. It is probably obvious to someone who has thought a little about stuff like this.
I also wonder what one can say about geodesic coordinates on a tubular neighborhood in general but only need codim 1.
 
This effectively goes back to Hermann Weyl (thinking about tubes).
 
Wow, it does seem in his book.
I was just going to state it without proof. :O
 
I think I could write down a (moving frames) proof pretty easily.
Oh, you have looked already?
 
9:22 PM
I would like to see the moving frames proof.
 
You just want to make me work.
 
I found a generalized Gauss lemma in his book.
 
Ah.
Goody. Now I get a credit in your thesis :P
 
Aw, man.
(You got it.)
But now you definitely owe me a moving frames proof.
 
Whats the books title?
 
9:24 PM
LOL
Such bribery.
 
Tubes
 
Hermann Weyl: "Tubes" / "Rohre"?
 
Sorry, Alfred Gray's book is just titled Tubes.
The only thing I actually needed is as early as Lemma 2.4.
oh, no, nevermind, that's not what I want
 
A Math Overflow answer suggests it might be in O'Neill's Semi-Riemannian Geometry.
But I'll try to write it out.
@Rudi: Hermann Weyl wrote a paper on the Volume of Tubes (introducing various functions of curvature now called Lipschitz-Killing curvatures). I don't think he had a book ...
 
yeah this is just a very special case of his generalized Gauss lemma.
 
9:31 PM
@Rudi: If you're interested in Weyl's paper, I can track down the exact info pretty easily.
@MikeM: Right ... I just wasn't sure if he stated stuff for an arbitrary ambient Riemannian manifold or just for space forms.
 
@Ted: Oh yes please, since I see a minimal chance to understand maybe that one :-)
 
Semi-Riemannian geometry... What does that even mean?
(Also hey everyone!)
 
Hi Demonark
 
It means you allow metrics of arbitrary signature (not just positive definite) and still do geometry.
 
9:34 PM
Interesting
 
@MikeMiller how comes you are not interested in physics? Is it a matter of opportunities or of principle?
 
@Rudi: "On the volume of tubes," American Journal of Math, vol. 61 (1939), p. 461-472.
 
Just taste.
 
muchas gracias
So I guess curiosity is no major driving force in maths?
 
It's very hard to understand everything and still make progress to get/keep a job, @Rudi. It's not lack of curiosity.
BTW, @MikeM, did you see the post about orientability for Gauss-Bonnet? This guy (who's apparently a UCSB physics grad student) was pestering me about it and listed almost a dozen web sources that state Gauss Bonnet for surfaces and don't require orientability. I told him they were all just wrong.
 
9:38 PM
I commented on a more recent question and then decided to stop engaging, the one asking "are isometries maps between smooth manifolds or Riemannian manifolds"?
 
oh yeah, that was silly
 
I wonder if there's any way to recover GB by twisting with a line bundle or something.
 
@MikeMiller What is it in physics that doesn't satisfy your taste?
 
Anyhow, I did work out details for the Möbius strip (which was an exercise I posed to him) just to see what happens. If you integrate $K$ over everything but a line segment you get all sorts of garbage for Gauss Bonnet because boundary orientations on the rectangle totally mess you up.
 
Oh, interesting. 2*C instead of C - C.
 
9:40 PM
Right.
 
hi @TedShifrin
 
hi @Shobhit
@MikeM: So are you serious about wanting me to try to work out a proof for you? Or should I not bother?
 
@TedShifrin It would just be for the benefit of my understanding how moving frames work better.
I've added the tubes citation and the thanks.
 
oh ... have you seen the moving frames proof of the usual Gauss Lemma?
 
nope
but surely this would do that for me :D
 
9:41 PM
That's in places like Spivak's book. But I'll include that.
Oh hell, I might have to LaTeX everything or else just scan handwriting.
 
@TedShifrin is it that one math.jhu.edu/~ajm ?
 
Yup, @Rudi, but who knows if electronic stuff goes back to the 30s.
 
I like sharelatex.com.
 
@TedShifrin lol
 
@KarlKronenfeld none, help please.
 
9:44 PM
oh, hi, Shobhit.
 
@Shobhit Well, where are you stuck?
 
I got it, its polynomials !
 
There's some invariant theory in there, I'm sure, @Rudi.
 
the interesting thing is "tubes" around complicated shape; if you pick a knot in 3-space, and set K(r) = {x, d(x,K) <=r}, can you find a formula for V(K(r)) when r is small?
 
Sure.
 
9:51 PM
Yes, but before I go to the interesting cases I like to understand the trivial ones :-)
 
It actually is very simple when you have a $1$-dimensional core, as there's only its length for geometry.
 
Just a strange habbit
 
No, that's a good habit, Rudi.
@MikeM: For small $r$ it's just $\pi r^2 L$. Knottedness is irrelephant.
But it gets more interesting when you start with a higher-dimensional core that has geometry to it.
 
@TedShifrin That was what I was hoping for !
 
I know, but that's a special case of Weyl's formula, I think, right? The topology of the embedding doesn't matter.
 
9:54 PM
The geometry matters. You get various symmetric functions of the second fundamental form.
 
But the given radius (of the covering spheres) shall be some given $a$, how does that make sense?
 
OK, I'm tuning out to work on Mike's homework assignment to me.
 
Oh, I see. I misread the intro of Tubes.
 
Oops.
 
I thought he was saying the integrals ended up being characteristic classes.
 
9:55 PM
In the complex case, you get Chern forms. In the real case, it's not topological.
This is all related to my thesis many decades ago ...
 
OK now I see its somehow mostly about the dimensions
 
you said to think of some good choice of a subspace to arrive at a contradiction, but i cant see how to use the subspace you gave me as hint. @KarlKronenfeld
 
@Shobhit Ah but then you chose the singleton set {0}, which is open.
It's a good strategy in my opinion, though the details might not work out.
 
can you tell me what i should be trying to prove? @KarlKronenfeld
 
"Arrive at a contradiction" is rather vague, isn't it. You will find a subset of R or C that is not open, but would be open if your topological vector space was discrete.
 
10:06 PM
thinkinh
g*
 
10:24 PM
@KarlKronenfeld i dont know what to do, should i do it like for $\alpha = 0$ and $x \ne 0$ we have $<\alpha , x> = 0$ and this function is continous so its inverse image on $\{0\}$ should be open?
i don't know what to do here :(
 
Find the inverse image!
 
it would be $<0, x>$ for all $x \in X$, no?
@KarlKronenfeld
 
Yeah. You should try making some adjustments. For instance, why is $\alpha$ fixed?
 
BTW, hi @Karl
 
Hey @Ted
 
10:35 PM
@KarlKronenfeld i dont know, i just took it to be 0 :(
 
Yeah, it's a simplifying choice. It just is too restrictive in this case because $\{0\}\times V$ is discrete.
 
{0} x V is discrete, so? what are you getting at? @KarlKronenfeld
 
Well, if $\alpha$ is free, you are dealing with $X=F\times (V-\{0\})$, where $F=\mathbb R$ or $F=\mathbb C$ with the usual topology. Note that $X$ is not discrete.
 
why is it not discrete? discrete means every subset is open, which of $X$ subset is not open? the set with $\alpha =0 and x \ne 0 $? @KarlKronenfeld
 
10:51 PM
That one is open, recall the product topology @Shobhit. How about a singleton in $X$?
 
how can i get a singleton in $X$? it has open sets of the form $P$x $Q$ where $P$ is open in $F$ and $Q$ is open in $(V - {0})$
 
That's a base for the topology, rather. A singleton is $\{\alpha_0,v_0\}$, where $\alpha_0\in F$ and $v_0\in V-\{0\}$ are arbitrary.
 
oh
then i can take any singleton $\{0,x\}$
i need to show that this is not open
 
Yeah, this is an important step.
 
but scalar multiplication function is continous and inverse image of open sets is open for continous function and the set above is inverse image of $\{0\}$ so why will it not be open?
 
11:03 PM
Because, the topology of $X$ dictates it is not open - you have to fill in the details here. Then you will have a contradiction - you have to fill in the details of what you just said though.
 
ok, i'll think about all of this and get back to you. Actually i have to go to sleep now, i have a class at 10:00 am and its 4:40 am now so i have to sleep. I have what you told me i'll solve this. Thanks for the help, you were very patient with me :) @KarlKronenfeld
 
Ah, yes, better get some sleep. Gladly.
 
:)
 
Night, Shobhit.
 
Jesus, classes don't start for me for yet another week and a day. This is the latest my classes have ever started.
 
11:09 PM
Heya @Fargle. I start AoPS on Sunday morning.
 
Fun!
 
Mike assigned me homework to type up a proof of something for him, so I'm working on his assignment.
 
Even more fun!
 
Hope you're doing OK. I got your message a few days ago.
 
Hey Fargle!
 
11:10 PM
Heya @Daminark
 
Have fun with classes, and hope you have fun with AoPS as well Ted!
 
@Ted Wait, which one?
 
Your reply to my hello. :) You said you'd had issues. Hence my good wishes now.
 
I also start later than usual this year, though not by much (October 1st)
 
Ah yes. It's just been a very busy time, mostly because everyone I know has been moving.
But the well-wishes are appreciated, as always. :)
 
11:12 PM
I've never started that late, Demonark. Even when UGA was on quarters, we started mid-September.
Well, I'm always ready to talk math, Fargle, when you feel like it.
 
@Daminark Compared to both schools I've been to, that's unprecedentedly late, lol
@TedShifrin I'll probably have good time to do that, considering I only have one course this year.
 
This last year we started on September 25th and were surprised at how early it was (this was also inconvenient because second quarter had to lose 2 days, Monday of that week was January 1st)
 
Only one course??!!
 
It's all I need to finish off my degree.
 
How indolent of you. What's the one course?
 
11:15 PM
The GRE has prepared me to read your message
 
Number theory.
 
:D
 
ROFL Demonark
 
lol
 
11:16 PM
Ah, well, then talking to me about geometry/topology/algebraic topology whatever will be a good complement, Fargle.
 
@Fargle learn Riemann surfaces with me
 
Has the math GRE prepared you to read my math, Demonark? :)
Are you learning it all algebraically or somewhat analytically, Demonark?
 
@Daminark I dunno, that sounds complex. :P
@TedShifrin I look forward to doing that. This is the first semester where I've actually had time to scratch the autodidact itch that I always scratch anyway.
 
I'd like a mix, I've found a book called Donaldson
Mike and PVAL mentioned it (it also hopefully will make me sit down and do covering spaces)
And regarding the math GRE, hopefully it will :P
 
Oh, Donaldson won't be super algebraic in spirit.
@Fargle: I won't even think about off-color responses to your statement.
 
11:22 PM
So it seems. I'm hoping to take AG this fall so hopefully that'll get the algebra side of the picture
 
I still have more to type for Mike.
 
sup nerdos
 
heya Eric
 
Howdy @Eric
 
just emailed my rec letter writers gulps nervously
 
11:22 PM
Oh, I should send Eric a copy of the homework Mike's making me do.
 
Oh yeah I should do that
 
You'll be fine, Eric. Re Stanford and Duke, if you want me to send an email putting in my two cents worth, let me know. I don't think the folks in question would discard you because of my input :P
 
i would appreciate it greatly Ted
@Daminark schlaggy's yale email kicked back on me lmao
 
Just send me statement of purpose stuff when you're working on things, Eric.
 
What's the term for people who do combinatorics? My guess would be "combinatorist" or something like that but I honestly have no earthly idea.
 
11:26 PM
Rip, guess it's not up yet
 
combinatorist is right, Fargle.
 
ya ofc, thank you kindly Ted!
 
One of my good friends at UCSD does that.
 
one of my best friends is a grad student doing combinatorics at ucsd
 
@TedShifrin Hey Ted.
 
11:27 PM
hey all!
 
@Fargle i call them county-bois
 
@TedShifrin Lucky me.
 
hi @quallenjäger.
 
I'd like to do more involving it--all I've ever done with it is a little course in discrete math.
 
Oh, Eric, I wonder if I've met him/her.
 
11:27 PM
Did someone say combinatorics?
 
Can anyone gimme their 2 cents (thoughts) on:
https://math.stackexchange.com/questions/2888976/a-rough-proof-for-infinitesimals
 
I did, but I retract it, Demonark.
 
this year was his first year
 
@TedShifrin Why is it natural to equip a tensor norm with $\parallel x \otimes y \parallel \leq \parallel x \parallel \parallel y \parallel$
 
I know one who just graduated, but I met some younger ones, Eric.
 
11:28 PM
@Daminark the proof uses combinatorics to some extent :P
 
Sam Spiro is his name
good dude
 
@quallenjäger: For example, for matrix products you want $\|AB\|\le \|A\|\|B\|$. This is of the same flavor.
Hmm, let me see if I can find a picture.
No, don't think I've met him, Eric.
 
@TedShifrin What is the general principle behind it to require such inequality?
 
@TedShifrin a shame
 
there must be some reason or some structure I need to have
 
11:32 PM
@quallenjäger: You want such things for control and estimates, as you always do when there are norms. For the matrix example, you want to argue, for example, that $e^A$ is given by a convergent power series. How do you do that? ... I haven't worked with the tensor norm enough to have an intelligent answer.
 
So it is natural to ask for such control for something similar to multiplication
 
i feel like the answer to why you want it isnt more complicated than "that's what numbers do"
 
Because i want to make this operation continuous?
 
That sounds like an important thing to argue, @quallenjäger.
@MikeM, @EricSilva: Just sent my homework.
 
received
 
11:42 PM
I'm not sure if you've seen the moving frame argument for this.
 
oh this is the question Mike asked me
 
He asked me, too, and I was fortunate enough to give a reference. But then he wanted to see a moving frames argument.
 
i thought about it w jacobi
 
Blah. Overkill.
 
to write the $g_{t}$ term
 
11:43 PM
It's just a trivial case of the Gauss Lemma. For a higher codimensional submanifold, it would be just like the Gauss Lemma.
Ah, I see.
I love the Gauss Lemma proof by differentiating the tensor with respect to $t$.
 
i have never seen this w frames
 
@Daminark is a field extension separable over the purely inseparable subextension? I think F_2(t)[X]/(X^2+X+t) over F_2(t) is a counter example
 
BTW, @EricSilva: I dunno if you've emerged from your probabilistic horror, but you might be able to improve or correct this answer of mine.
Eric, I need to continue brainwashing you :P
 
my probabilistic horror continues until the 27th unfortunately
 
Oh :(
 
11:45 PM
im reading some crazy notes about infinite dim SPDE
it's actually killing me lol
 
Well, you get to see you stochastically struggled all summer ... :)
Yeah alliteration.
 
Stochastic DE's are terrifying, tbh
 
Well, the thing I linked you to is basics on currents.
 
Lol isn't is Hairer's work?
about regularity and this kind of stuff
 
My supervisor gave me the option to read more of Hairer's stuff or read his stuff and i chose to postpone reading any of Hairers stuff cause there's too much like
details to do in a short amount of time
 
11:48 PM
True
 
oh this Gauss lemma thing is a nice idea
 
@TedShifrin What could occur if such multiplication is not continuous?Does something wild happen with the algebraic structure of the "multiplication" $\otimes$
 
i guess it's kind of easy lol
 
Just another week Eric, you got this
 
yeah i wanna be done w this
 
11:51 PM
Yup, @EricSilva, and from the curvature structure equation you get Cartan-Hadamard quite beautifully.
 
good stuff
 
@quallenjäger: It's always important to have good intertwining of algebra and analysis or things are really messed up.
 
I see thanks.
I am currently doing Hopf-algebra and all this kind crazy stuff.
 
im prepping a lecture on that minimal surface in S^3 thing rn, forgot how nice it is to do some geometry
 
I am totally lost
 
11:52 PM
sad that it's all i have time to do between stochastic
 
@EricSilva Are you doing SPDE?
 
Ah is that what they voted on? Nice
 
that's what my summer project has been about yes
@Daminark no but i only wanted them to vote so id have an idea of what they were interested
 
Ah nice
 
i decided not to lecture on the most voted on thing cause it was too annoying
 
11:54 PM
@EricSilva: I was hoping to make you nostalgic when I sent you Mike's homework.
 
@Daminark was i successful at baiting u to read my post?
 
You succeeded at baiting me to open the link
 
What was the too annoying thing, Eric?
Demonark is easily baited.
 
Once I saw the length I was just like yeah maybe not
 
the too annoying thing was to talk about general theory of connections
 
11:55 PM
@TedShifrin Thank you , I will work on that tomorrow.
 
sigh ... :(
 
just some basic stuff
 
Sorry I can't be much help, @quallenjäger.
 
I was hoping to get the views of a pro
 
i didnt want a talk that was all definition no example, so i just decided to do the minimal surface boi
 
11:56 PM
Make them think it's democracy and then pull the plug
Uh, calling me a pro is perhaps a bit of a stretch
 
@Daminark Only 3 of them have filled out my doodle poll so it seems like they're kind of not interested anyway
 
Well, you could have done some Lie groups and done some examples, Eric, but there's a ton of content lurking in what you're doing.
 
@TedShifrin Sometimes you give me some new point start to thinking. That helps me alot.
 
@TedShifrin yeah i thought of this but i dont know what algebra they know
 
bah, matrices and take derivatives :P
 
11:57 PM
well in essence that's what im doing now lol
 
but, anyhow, I think this is deeper and you can do Weierstrass, too.
 
this is also just sexier
no mega fanciness to muddy the waters
 
Playing with connections in $S^3$ is already more general for them.
Just remember that they're not all little yous :P
 
i guess i do also need to talk about riemann surfaces for a sec
since the whole proof is about proving a quadratic differential on S^2 is holomorphic and therefore vanishes
 
well, you need complex structure earlier, too
 

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