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00:13
Wow, this is a hell of an answer.
 
2 hours later…
02:16
@MikeMiller "Abuse of notation" is an ugly phrase for something that often is quite handy
heh, yeah
though i'm not sure i know a better name for it
it's sort of like the 'rigorous' versus 'formal' distinction people sometimes use in proofs
perhaps one could instead that to rigorously true vs. formally true t vs. morally true
i.e. even though it's not actually rigorous to think of something a certain way, one nevertheless 'should' think about it that way
anyone here familiar with quantum probability theory?
What kind've question do you have? That's a bit broadly stated.
I mean I just kind of am trying to find some good introductory to midlevel materials on it
02:32
@MikeMiller hmm, speaking of which, i just came across this online: cheng.staff.shef.ac.uk/morality/morality.pdf
@YashFarooqui can't say i know a good answer to that
on an unrelated note, has anyone come across any sort of paper exploring pseudorandom properties of permutations?
in a similar way to pseudorandom graphs and hypergraphs
The reason I ask is: consider the permutation on p-1 elements by sending them each to their inverse mod p
in many ways this family of permutations is far from random, but it apparently has the property that the longest increasing subsequence of this family tends to 2sqrt{p-1}
same as a random permutation
main thing i know about random permutations and the like is the link to the Tracy-Widom distribution
what's the connection? (I'm just a student and I've not heard of the latter)
tbh, i don't know how it works. but, take a look at the definition offered in the link and compare it with what you just said re longest increasing subsequence
this has some more precise remarks, while nto appearing too complicated: statslab.cam.ac.uk/~beresty/Articles/tw.pdf
the first section, anyways
the research i'm doing is evidently related to Tracy-Widom as well, somehow, but it's entirely beyond me how
I guess something that might be interesting to look into and possibly fruitful for my problem is to study the general theory of pattern avoidance for random permutations
02:40
sure.
they mention a few references on the first page which might be worth looking at
I think I recognize a lot of the mathematics from Dan Romik's book
I guess I'm fundamentally trying to look for something to cling to that implies the 2sqrt{n} upper bound (or lower bound) while being somewhat nonrandom to use as the basis for proving said pattern
ah, interesting
what i'm pointing toward, i suppose, is the larger question of how those lengths are distributed
mm
it's a good idea I think, it makes sense to understand things that behave almost randomly by studying actual random objects
right. that way it's more clear what counts as a deviation from randomness
on an semi-related note, here's a cute trick you might not have noticed - let X be a random variable corresponding to the number of fixed points of a random permutation
on say N elements
calculate the first N-1 moments of X
or if you don't want to
they turn out to be spoilers: the Bell numbers
02:47
nice. i've probably seen that at some point
the number of fixed points of the inversion permutation is fixed at 2 since x = x^(-1) implies x^2 = 1 mod p and F_p is a field, meaning x = 1, -1
it sounds like stuff i heard about in the realm of free probability theory, though i've never managed to crack into that
so that is a completely useless idea of randomness
what's free probability thoery? :0
confusing :p
side note: being detached from university, it's nice to have anyone to talk to who can understand math
02:49
yeah, this is a good place for that
and wow this just brings the discussion full circle to noncommutative probability XD
yep.
and given the link between Tracy-Widom and random matrix theory...yep, sounds about right
the line from that Wikipedia page which the above sounds reminiscent of
and the terrence tao article I was reading to better understand quantum probability was entitle Free Probability
I guess I'm just oblivious
"The free cumulant functional...plays a major role in the theory. It is related to the lattice of noncrossing partitions of the set { 1, ..., n } in the same way in which the classic cumulant functional is related to the lattice of all partitions of that set."
ahah. i was about to suggest that blog post as well :P
a lot of people I know do free probability!
02:51
that's cool Mike! :0
do any of them study permutations?
and increasing subsequences?
free probability theory seems quite interesting. but also quite beyond me
It will almost certainly be beyond me too haha
a solid one!
(being an undergrad)
that's not really the flavor of what they do, no
random matrix theory and free probability are related but to my impression really flavored very differently
02:53
ah okay
random matrix theory is related to integrable systems stuff, as i understand it
whereas free probability theory is more about combinatorics
03:09
hi
I need help proving: The equation $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$.
03:40
hi
 
1 hour later…
05:01
@MikeMiller Technically speaking, random matrix theory is a subset of free probability theory, correct?
I have no idea. Certainly not the "free probability" this guy claims to do.
@MikeMiller Wikipedia says free probability theory is the study of noncommutative random variables, which random matrices are
i'll take your word for it :) they would probably describe it to me as the study of a certain kind of von Neumann algebra
 
4 hours later…
08:49
Hello
 
2 hours later…
11:04
quick question: Just starting algebraic geometry... aside from its rigorous definition how and why is the word affine being used in affine space?
11:15
@PrinceM I think it's because you're solving a system of polynomial equations $P_i(x_1,x_2,...,x_n) = 0$ and the (vector) solutions $(x_1,x_2,...,x_n)$ need not admit $(0,0,...,0)$ as a solution, i.e. if you have a vector space of solutions it is most likely a vector space without origin, i.e. an affine space.
11:47
@PrinceM Affine spaces have a notion of parallelism in it. If you choose an origin, there is not such notion because say any two codimension 1 subspaces intersect at the origin. But if you don't choose an origin, you can "say" that codimension 1 affine subspaces are translates of the vector subspaces. So parallelism works.
This is precisely what happens in $\Bbb A^n$. There is no god given choice of the origin (hence coordinates) in there.
The point is that without a choice of an origin your making parallelism work. That's all.
This is one of the reasons that $\Bbb A^n$ and $k^n$ are set theoretically the same thing yet have different notation, ya know. The latter is a $k$-vector space - that structure is imposed. The former is a geometric object.
@Akiva Not really funny.
Hello guys very quick question regarding correct wording
$y=3x^{2}$ is this still considerered as a proportional relation?
No, $y$ and $x$ are not proportional in there.
(I suppose it would be correcter to say quadratically proportional?)
oh ok..
$y$ is said to be proportional to $x$ if $y/x$ is a constant. A fixed number; independent of $y$ or $x$. $y/x$ here is $3x$. Very not constant as $x$ varies.
so I guess $Q = \frac{1}{cos\theta}$ is not proportional neither?
*inversely proportional
11:56
Depends on which variables you are asking to be proportional.
Q and theta
No. They are not.
ok tvm :)
12:12
Hey everyone! Does anybody know if there is something like the first isomorphism theorem for groups that applies to differentiable manifolds? I am hoping for some result like : "let F: N-> M be a differentiable map then n1 ~ n2 iff F(n1) = F(n2) is an equivalence relation. If N/~ is a smooth manifold, then F': N/~ -> im(F) is a diffeomorphism"?
12:32
@WardBeullens There's an analogue: it's called the universal property of quotient maps in topological spaces. There's nothing special about manifolds in that context.
Thanks a lot, i will look it up!
12:44
If I understand i correctly, the universal property says that F':N/~ -> im(F) is continuous and bijective, but can we conclude that F' is a diffeomorphism given the assumtion that N/~ is a differentiable manifold?
Or maybe one needs some other assumptions as well
@WardBeullens It says that if $f : X \to Y$ is a surjective map and the topology on $Y$ agrees with the final topology with respect to $f$, then given an equivalence relation $\sim$ on $X$ defined by $x \sim x'$ iff $f(x) = f(x')$, the induced map $X/\sim \to Y$ is a homeomorphism.
Is that not what you want to prove?
$X, Y$ need not be manifolds here. Not at all.
13:06
I understand, but I don't want a homeomorphism, I want a diffeomorphism
Sure, then assume everything is smooth.
Same thing works.
If I assume everything to be smooth, is the induced map guaranteed to be smooth?
Yes, I think so. I suppose you also need the quotient map $X \to X/\sim$ to be smooth. You need to impose sufficient conditions so that everything works out.
@WardBeullens What does N/~ mean?
n1 ~ n2 iff F(n1) = F(n2) is an equivalence relation. N/~ is the quotient by that equivalence relation
13:15
I think Mike is asking what is the smooth structure you are imposing N/~ with.
There's no canonical one given an arbitrary equivalence relation ~, I think.
I mean, it doesn't actually make sense. The quotient of a manifold by a random equivalence relation is usually not a topological manifold.
He's assumed that N/~ is a manifold above.
But I'll step away from the conversation in case I am being a noise.
Fine. If $N/\sim$ has a manifold structure such that the map $N \to N/\sim$ is a submersion (this is, of course, quite rare; more or less this happens when $\sim$ is the quotient by a group acting freely), then you have your theorem, fairly straightforwardly.
@BalarkaSen You're fine.
Been busy with non-math lately, I assume?
Yeah. Results of my finals weren't exceptionally good - 80% is considered more or less average. A bit worried about admission, etc., but it seems I will survive.
Yeah the example i was thinking of is where N/~ is the quotient by a group acting freely
13:22
Actually, sorry, can you restate your precise goal again?
@Balarka Sorry to hear that. If admissions is anything like the US, you'll be fine.
Ok, i'll be a bit less general: I have a transitive lie group action from G on M, the stabilizers happen to be the orbits of an action of H on G. I would like to conclude that G/H is diffeomorphic to M. Im quite new to these things, so im sorry if this is really easy
@MikeMiller The reason people get more scores in here is that after we take the 10th grade exams the sheets go to different schools throughout the states for evaluations. But it seems our school got low scores in general, thus there is hope.
Back here: chat.stackexchange.com/transcript/message/16267539#16267539 I asked whether anyone had searched for matching strings of digits in e and pi. Well based upon: superuser.com/q/3810/39835 I generated the first million digits of both and there's 13 matching digits starting at 799096 in e and 295318 in pi: "6425963474335" :-)
Alright, I gotta go.
13:32
bye! and thanks
@Ward: Ok, I was just asking for your original statement and I couldn't see it. If you have a transitive Lie group action of $G$ on a space $M$, then $M$ is diffeomorphic to $G$ mod the stabilizer. As you say, the proof is quite clean: $G \to M$ factors through $G/H$. Working locally we check 1) the map $G/H$ is smooth 2) It's a submersion. That proves it.
@Balarka See ya. I posted an answer the other day about when a manifold, after you quotient out a submanifold, is still a topological manifold. You might like it.
All of this is written down carefully in the appropriate chapter of Lee, I think.
@MikeMiller Thanks i will give it a look
13:48
@MikeMiller I'm having a look.
Of course there are a lot of examples. One can just blowup something at a point and then quotient that exceptional submanifold. But admittedly I am curious about a classification; let me read on.
@MikeMiller @MikeMiller Im sorry to bother you again, but could you explain a bit how to prove 1) ?
@Ward I would prefer if you did it. Have you tried much yet?
@ Mike: Neat, I read a bit and the idea was quite interesting.
@Balarka You were much quicker in finding that example than I was.
I started with the intent of proving such objects did not exist. Oops!
14:04
@MikeMiller I'm afraid I don't see how to start
@WardBeullens How do you show anything, ever, is smooth?
Hehe, that's what I thought at first read too.
by choosing charts and showing that the composition is differentiable as maps from R^n to R^m
@WardBeullens So what are the charts on $G/H$?
Ah, we use that G -> G/H is a submersion? to get a local normal form.
?
14:07
It doesn't even make sense to say that your map is a submersion without having charts on $G/H$
Wait are we showing that G -> G/H is smooth or that G/H -> M is?
The latter.
But you haven't told me what charts on G/H are yet, so we're at an impasse.
Can we somehow use charts on G to get charts on G/H?
Yes, but if you don't know how to do this, that's a place to start. Find your favorite reference on how to get smooth structures on the quotients of manifolds by freely acting compact Lie groups, and read that.
@BalarkaSen In any case, there is a full classification of fake $\Bbb{RP}^n$s and $\Bbb{CP}^n$s easily findable. I couldn't find a classification for $\Bbb{HP}^n$ and $\Bbb{OP}^2$. I posted a question about this on MathOverflow; the cases $\Bbb{HP}^2$ and $\Bbb{OP}^2$ have been resolved.
14:41
@robjohn hey, did you manage to take a look over my article I sent to you some weeks ago?
@robjohn Based upon that integral I derived a far harder one that I didn't fully finished till now. I know what to do, but there is still a long way to go.
15:08
heya
@robjohn it was already accepted (like many others). I was just asking if it was anything interesting to you.
Anyway.
@N3buchadnezzar How is it going?
@Semiclassical btw, let me know if you're done with the simple version of my limit. There is a very easy way to finish that.
15:27
AB is tangent to the circle. Intersection point between AD and point in circle is C.
$\frac{AB}{AC} = m$
I need to find the ratio between area of BCD and ABC using just $m$.

I feel that there is not enough data to find it but I'm mistaken. Can someone give me a hint?
$AB = m \cdot AC$
15:59
@PichiWuana Intersecting sectant theorem: |AB|^2=|AC|*|AD|
vzn
vzn
16:25
↑ going to see in 1hr =D
hey @BalarkaSen miss you in the number theory room :(
16:39
@GridleyQuayle Where can I read about that theorem?
Have we this inequality : $\int_0^1 |t| |f(t)| dt\leq \int_0^1 |t| dt \times \int_0^1 |f(t)|dt$
can someone help me ?
Does anyone know of an easier (or less algebraic) way of showing that the matrix of the composition linear maps is the product of the matrices of the linear maps?
@GridleyQuayle thanks
16:55
@Vrouvrou Is that not just Chebyshev's inequality?
@GridleyQuayle i don't know it
@GridleyQuayle i know Cauchy-chwartz and Holder
that's all
frankly, just writing out the definition of linear maps tends to work well for the composition thing
L(e_i) = v_i = sum a_i e_i
M(e_i) = w_i = sum b_i e_i

M(L(e_i)) = M(v_i) = sum a_i w_i = sum a_i (sum b_j e_j) (meaning the coefficient of e_j in column i is v_i*w_j)
but also I think it's easier to just see what you wrote as the definition of matrix multiplication
and use the properties of linear maps to compute matrix multiples
Hi @iwriteonbananas
@Vrouvrou Chebyshev is just Holder with p=q=2 I think.

@YashFarooqui That's the way I know how to do it but that is quite wordy (even if short) and I don't think it helps with insight much.
17:10
Matrixes are just tools to represent linear maps?
because you can effectively just write down where bases go and that determines completely?
that's what I would say, yes
I think that's the only major insight
I suppose so yes.
@GridleyQuayle I think that the two triangles are not similar. So how can I find ratio?
use power of a point theorem
17:27
Hey @MikeMiller
how's math?
I wasted so much time today on two ridiculous exercises, and now I've fallen behind on my schedule
"Yikes!"
Tried proving directly that a CW complex $X$ is contractible if it's the union of an increasing sequence of subcomplexes $X_1\to X_2\to...$ with each inclusion nullhomotopic
I know it can be done with Whitehead
But it's an exercise from my course and we haven't done Whitehead yet. So I spent like 4 hours trying to find a direct proof
Sounds like the sort of thing Hatcher does a lot, something something telescope construction?
17:30
I tried that too
Can show that $X$ is htp equiv to some infinite union of mapping cylinders
But I couldn't conclude that $X$ is contractible
Maybe we're doing Whitehead in the next lecture and we're just behind schedule lol
o well
this is not so exciting that I am going to get real into it with you :)
are you still an undergrad? i forget.
17:32
Man, I don't want to think about that problem anymore...it got so annoying lol. I'm handing in my Whitehead solution no matter what
Yeah, still an undergrad
Final semester though
You're taking that algebraic topology class, anything else? You're graduating this year?
Where you headed, can you say?
I guess maybe not publicly, but if you want to you know my email.
Actually since yesterday I can say. I'll email you. I'm also taking an intro to algebra class. It's a freak coincidence that I haven't taken that course in a previous semester.
Need to write my thesis and do a couple of seminar talks too
intro to algebra lol
Sounds like you're doing pretty well.
How's math for you?
@GridleyQuayle $p=q=2$ is Cauchy Schwartz
17:38
It's ok, I'm writing an email to you right now actually. I have work to do today (see the imminent email) but right now I'm doing unrelated math because that's fun sometimes.
I just emailed you too
and i need to see if $\int_0^1 |t| |f(t)| dt \leq \int_0^1 |t| dt\times \int_0^1 |f(t)| dt$
@GridleyQuayle
without 2
I want to talk about this, but up to you where. How does the program work? Is it more classes? Do you work with someone?
We can talk about it here. Which program do you mean? The seminar?
17:42
@GridleyQuayle I Need $\leq$ not $\geq$
I meant the program you were accepted to. It'd be cool to hear about the seminar too though.
Ohh, the program is an ordinary European master's degree. I need to get a certain amount of ECTS credits, which I get by taking exams or doing seminars
or by writing a thesis
What do you want to do?
I want to learn as much algebraic topology as I can soak up in my time there
Hi bananas, @MikeM.
17:44
Hi @TedShifrin
Long time no see @GridleyQ
Hi @Ted. Are we friends again?
Did you decide we weren't?
@TedShifrin Hi
I thought I pissed you off enough one day that you left for a while.
17:45
How does it work in the US, @MikeMiller?
heya @Tobias
Graduate school
well, I wasn't overjoyed by your condescending tone, Mike, but I've been having some very stressful times, so decided it best just not to be here.
Fair enough. I hope you're feeling better.
Hi DogAteMy @Akiva ... I assume you're safely home and done celebrating Pesach :)
@MikeM: Dealing with a good friend who's badly messed up his life and trying to help is stressful at the very least.
But I figured I'd come say hi, since I'll be traveling soon for 2+ weeks and won't be around much at all.
17:48
Where you going to Ted?
Chicago and Atlanta
Maybe Europe in the autumn
@TedShifrin Hi. Exams are coming up so I've just been studying. Luckily the content isn't too complex so it's just down to practising questions/learning proofs.
@iwriteonbananas It is not uncommon (I would say ~70% of the time? Probably a bit more) that people go directly to a PhD instead of doing a Masters first, though two of my good friends have masters degrees. A masters program is usually mostly classwork and sometimes has a thesis component. The PhD program at UCLA has a year or so's worth of coursework required (you have to take n classes, more or less whatever you want), and then "go to town on research".
Working problems and understanding examples is always good, @Gridley, although I know some professors expect students basically just to memorize the textbook... :(
17:49
Holla at me when you have plans for Europe. We definitely need to meet up Ted.
For all you algebraic topology people, can you tell me if the problem I put on my topology final is wrong? I suspect it is.
@TedShifrin I know what you mean, I'm sorry to hear about your friend. Good luck to both you and him.
How has your back been lately?
If $p\colon E\to B$ is a covering space and $E$ is path connected, true/false that $\pi_1(B)$ acts on $p^{-1}(b_0)$. Of course we all know it's true when $E$ is simply connected.
Absolutely, bananas. I look forward to that!
@TedShifrin I've got one lecturer who actually sets questions (only a couple), based of jokes he has told in his lectures. One example is:
Let f(x) = c and g(x) = e^x. The product function fg is afraid of which differential operator?
I still don't get it.
LOL ... that's pretty pathetic material for a test, @GridleyQ. :)
17:51
(I don't go to many of his lectures)
@TedShifrin Well, the natural thing that acts is the deck transformation group; if $G$ is the subgroup corresponding to the cover, and $N$ its normalizer, this is $\pi_1(B)/N$. So $\pi_1(B)$ acts as $\pi_1(B) \to \pi_1(B)/N \to \text{Sym}(p^{-1}(b_0))$?
@MikeMiller Ah, okay...can't imagine going from bachelor straight to PhD lol. But I think your undergrad thing is a year longer than a typical bachelor. Masters here is pretty much coursework + thesis, too.
I assume he means $d/dx - I$, but where's the humor, GridleyQ? ... At least my students used to know to groan at my horrid puns, but they never showed up on exams.
@iwriteonbananas You are better prepared for a US PhD than I was, or for that matter anyone in my cohort.
We also had a question which was to find a parametrisation of a mobius strip (5 marks out of 100) which I thought was slightly unfair in exam conditions (exam only 2 hours long.
17:53
Keep in mind that in the US PhDs are usually five years, and for you that's usually separated as a 2year masters and a 3 year PhD.
OK, @MikeM, that's what I'd originally thought. But one of my two students convinced me he couldn't argue well-definedness without simple connectedness.
@GridleyQ: That's pretty sneaky unless you've seen the idea before.
@TedShifrin The thing you actually want to write down (lift paths) works fine. You don't need simple connectedness for it, unless I'm missing a subtle subtlety. What's his argument?
Yeah, bananas, you're way ahead of most US grad students, except other Europeans and the really gifted kids who went to MIT, Harvard, Chicago, Princeton as undergrads.
Come on, don't flatter me like that
Ted can confirm I don't make any efforts to flatter much anyone. I'm stating what I see as fact.
17:55
I came out of MIT (zillions of years ago) having taken something like 10 graduate courses, but most students at US grad schools haven't done that.
I sent that email.
Doesn't $\pi_1(B)$ always act on $p^{-1}(b_0)$ whenever $p:E\to B$ is a fibration?
No, you have to be much more subtle in what you're trying to say.
Ok, I guess what I mean is the following:
You can say that $\pi_1(B)$ acts on the homotopy groups of $p^{-1}(b_0)$; in Ted's discrete case this is what we're saying.
17:57
Yes, bananas, I thought so. Interestingly, though, I find statements in Munkres and Hatcher about the quotient $\pi_1(B)/p_*\pi_1(E)$ being in bijection with the fiber, and no one says $G$-set.
@MikeM: What he wrote me was: "I'm wanting to use that if alpha is a path from e_0 to e_1 and beta is one from e_1 to e_n, then alpha followed by beta is homotopic to any path from e_0 to e_n" ...
You could also say that there is an "action" of $\Omega B$ on $p^{-1}(b_0)$; this is actually true. You recover the previous statement by passing to homotopy groups.
But $\Omega B$ is a grouplike topological monoid, not a group, so this gives some people incl. me stomachaches.
But I guess the point is that we don't need a homotopy upstairs. We can push the closed loop downstairs.
hi all
welcome back, @ted
heya @Semiclassic
@TedShifrin I agree with your assessment, but I'm impressed with your student's attention to detail.
17:59
There's a group homomorphism $\pi_1(B)\to Aut(F_{b_0})$ which takes $\beta:I\to B$ and send it to a lift of $F_{b_0}\times I\to I\to B$
It's only temporary :)
What is $\text{Aut}$?
These two kids were amazing, @MikeM. Actually, the other one made a few mistakes but did superbly on the exam. Got some very hard problems. ...
I guess I mean self homotopy equivalences of $F_{b_0}$
18:00
speaking of exams, my students have one starting in half an hour
That's not a group. What you really want on the RHS is a flavor of mapping class group; you want $\pi_0 \text{Aut}(F_{b_0})$.
If you're going to mod out by homotopy in your source, you need to mod out by homotopy in the target.
They both were a bit stymied by my "Reeb foliation" leaf space question. I took the lines $x=\pm\pi/2$ and the level sets of $y-\sec x$ and asked what the leaf space was. They both disagree with me.
and because of that, i'm having to miss three talks in a row (there's a conference on campus today)
Oh yeah, makes sense Mike
so booooo
18:01
Reeb foliations are my friends.
I'm claiming that the two "unusual" points are dense. They're both disagreeing. They're only admitting non-Hausdorff.
Well, they're wrong. Are they not thinking about the leaf space picture?
I don't think they correctly understand the quotient topology. Admittedly, Munkres makes things unnecessarily opaque.
@iwriteonbananas If you want to work without modding out what I was saying before still applies: you have a monoid homomorphism $\Omega B \to \text{Aut}(p^{-1}(b_0))$.
Hatcher's notes are good
Right, that makes sense. I think I saw that in May's book. Good point though, if we pass to the fundamental group we need to apply $\pi_0$ on the right
Which of Hatcher's notes do you mean? The bit on fibrations in his book, or are there some other notes he put out?
18:18
I was responding to Ted re: wuotient spaces
Yeah, I figured as much.
It's surprising that Munkres made that section singularly stiff and unintuitive.
I don't know Munkres very well, unfortunately.
I admire most of the book, but not that section.
@TedShifrin est ce que $f(t)=\sqrt{t}$ est un bon contre exemple pour montrer qur l'inégalité $\int_0^1 |t| |f(t)| dt \leq \int_0^1 |t| dt \times \int_0^1 |f(t)| dt$
est fausse
s'il vous plait
Moi je ne sais pas. Tu peux bien faire le calcul.
18:27
je trouve $\int_0^1 t \sqrt{t} dt=\frac25$ et $\int_0^1 tdt \times \int_0^1 \sqrt{t} dt=\frac13$
Donc, il ne faut pas me le demander.
ok merci
With both you and bananas here, maybe I will advertise a question I answered which I had a great time with. It's not completely answered; I asked an MO question about the last detail.
I would also like to advertise studiousus's incredible answer here.
I don't have time to read those carefully now. Studiosus is impressive. And I'm glad you're getting to know Andy Hwang. He's a good friend and enjoyable mathematician.
@iwriteonbananas Does your masters have a focus? Is there something you specifically want to do?
Yes, I enjoy most everything he writes. I desperately want to know why studiousus is.
18:35
why? who, I presume. :)
Yeah, I've been curious for a long time too.
Well, I think why someone is, is also an interesting question. But you are correct that I made a typo :)
I didn't realize you were an Existentialist.
I'm not much of anything.
Perhaps that makes me a half-assed nihilist.
Knowing you, I think it has to be at least 3/4-assed.
18:44
@MikeMiller Emailed you. I'm not sure how exactly it works there. Whether or not I need to choose a focus or whatever. But I don't think so. There's a ton of really cool sounding lectures that I want to do, let me see if I can find a list
@MikeMiller Asaf Schachar always asks good question, I'll have a look at your answer
I think it's more that people usually tend to start choosing things to pay more attention on, doing coursework that will prepare them for research in some area
I think the Guillemin & Pollack course my 3rd year of undergraduate is what aimed me toward geometry ... and then a complex manifolds course my 4th year sealed it.
i sometimes wonder why i went down the physics route vs. the math route for grad school
@MikeM: I just got my ballot. Agh ... 9 propositions ...
Ciprian taught a course my first year here that made me decide to do gauge theory over complex geometry. Feels very butterfly effect.
@Ted: Could be worse.
18:49
i preferred the problems in physics to the problems in math, i guess, despite being very mathematically-minded in how i solve problems
You sure seem drawn to math, @Semiclassic.
Applied math might have been a good compromise. :)
yeah
though physics isn't too far from that at times :)
heck, the stuff i'm doing lately vis-a-vis the KPZ equation has been pretty fun
But physicists seem to sacrifice any notion of precision/rigor most of the time.
@iwriteonbananas 1) Thanks for the email, I found it helpful. I am on my phone so will probably not respond now, but will later. We should correspond more.
2) I just looked at the faculty list in topology at your upcoming university. I think you'll be very happy there.
it's a different kind of standard
18:51
That said, I love physics and I have a lot of respect for my friends who are physicists, @Semiclassic.
OK, time for lunch and then going to 4 1/2 hours of Beethoven piano trios for the afternoon/evening.
i have a lot of respect for experimental physicists in particular, despite not being one
have fun
don't go off on a tangent :P
18:53
@MikeMiller Definitely. Funny, I just emailed you a link to the topology faculty thing where you can see graduate courses.
Have a great day and travel.
No hyperbole today, @Semiclassic.
@iwriteonbananas You've expressed a taste in both geometric and (sometimes computationally grounded) algebraic topology. Your new place has some of the world's best people at using algebraic topology for geometric purposes.
Nice. That's both intimidating and exciting.
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