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12:27 AM
@rschwieb Details: 1. Let $R$ be a non-discrete valuation domain. 2. Let $\mathfrak p$ be a prime ideal of $R$. 3. $R_{\mathfrak p}\supset R$ in the field of fractions of $R$. 4. $R_{\mathfrak p}$ is a valuation ring. 5. $R$ is Prufer. 6. $R$ is local; let $\mathfrak m$ be the maximal ideal. So $R_{\mathfrak m}\cong R$ is not a discrete valuation ring. 7. $R$ is not almost-dedekind.
 
 
5 hours later…
5:20 AM
Why evolute of an involute is the original curve?
17:42
I don't understand his explanation fully.
Evolute means locus of all centre of curvature of points on the curve.
For Involute, tieing a string along the curve. Let P,Q be the points. When we move Q, we get involute right?
Attach a string to a point on a curve. Extend the string so that it is tangent to the curve at the point of attachment. Then wind the string up, keeping it always taut. The locus of points traced out by the end of the string is called the involute of the original curve
 
5:40 AM
Is the following equation valid ?

$ \int_1^{x^2+2} \sin(t^2+2)dt = \int_1^{x^2} \sin(t^2+2)dt + \int_{x^2}^2 \sin(t^2+2)dt $
$= \int_1^{x^2}\sin(t^2+2)dt - \int_2^{x^2}\sin(t^2+2)dt$
?
 
no you cannot split $x^2+2$ the way you wrote
 
hmm, thank you
 
 
3 hours later…
cdt
8:59 AM
@SubhasisBiswas Um I read the news, but I don't see why someone in class 10 would suicide for ISI admission which happens after class 12 ?
 
@flowian you can go from $[1; x^2+2]$ to $[-1; x^2]$ if you account for that in $t^2$, but I don't see how that would help you in any way
 
9:51 AM
@cdt obsession.
 
10:23 AM
23 hours ago, by skullpatrol
getting obsessed with one question is dangerous
 
When can we say that a local Hamiltonian vector field is global. Can I say that its global when $H$ exists and is defined properly on the entire manifold?
So this book says that a locally Hamiltonian vector field with no zero points on a compact symplectic manifold cannot be globally Hamiltonian. How does one show this?
 
10:59 AM
So, I have an associative binary operation $g:X\times X\to X$ and I want to characterize all unitary functions $f:X\to X$ such that $f\circ g$ is still associative. How would you approach that proof?
 
11:14 AM
Can every symplectic manifold be given a Kahler structure?
 
11:26 AM
@Albas Surely not.
Hard for me to cook up an example but Kahler is a very strong restriction!
I can give you a couple basic things off the top of my head. (1) Spheres never admit symplectic structures, because you can take the kernel of the quadratic form of the symplectic 2-form and that'd give you a nonvanishing vector field which contradicts hairy ball. So look for better examples
(2) Symplectic manifold always has an almost complex structure (in fact the space of almost symplectic structures ~ space of almost complex structures), so if you strategy is to look for "symplectic but not complex" examples you have to be careful. I was thinking of CP2 # CP2 but I learnt from here that this is not even almost complex, so admits no symplectic structure actually
I can't help you more unfortunately, my knowledge of examples end here
 
Man symplectic geometry is weird.
It's hard to cook up examples
 
Nah I'm sure this should be easy, I just don't know anything about the subject
 
11:41 AM
yet :-)
 
I should read more on this. Currently I am doing Hamiltonian vector fields.
1 hour ago, by Albas
So this book says that a locally Hamiltonian vector field with no zero points on a compact symplectic manifold cannot be globally Hamiltonian. How does one show this?
Anything on that @BalarkaSen
 
Can you remind me what Hamiltonian vector field means?
 
$X_H$ is called a Hamiltonian vector field on the manifold $(M , \omega)$ if for $X_H$ you define $i_{X) (\omega) = dH$ where $H$ is a smooth function on $M$
Umm I can't fix the latex, can you understand what I mean @BalarkaSen
Here $i$ is the interior product
 
$\iota_{X_H} \omega = dH$, yeah?
So like gradient, but in the symplectic context
 
Yeah
 
11:50 AM
\o @Secret
 
@Albas What's locally Hamiltonian?
 
I am really sorry but I am suddenly caught up with urgent work. As soon as I finish this, I will get back to you @BalarkaSen
 
12:21 PM
Can any one please tell me, What can we say about the dynamics of a system for which the baby mandelbrot set is deformed ?.
 
12:43 PM
Urgent work
Sounds like you're working in a hospital :P
 
Three week summer schools should be banned
This is too much math
 
What math are you learning in those three weeks?
 
if he's working in a hospital he wouldn't be able to even type that sentence
 
Why?
 
@Albas I suppose locally Hamiltonian would just mean $\iota_X \omega$ is, around any neighborhood of a point, $df$ for some locally defined function $f$ on that neighborhood
That's the same as saying $i_X\omega$ is closed
If $X_H$ is a global Hamiltonian vector field corresponding to some $H : M \to \Bbb R$ and $M$ is compact, then $H$ has a critical point on $M$, which means $X_H$ vanishes somewhere. Actually, it has to vanish at two points.
So that proves your claim
 
 
1 hour later…
2:03 PM
@user681391 not learning too much
would have been better a year ago
but it's just a lot of shit
 
@RyanUnger Oh I meant, what sort of mathematics are they doing in those three weeks? I saw a summer school that was three (may have been six) weeks long that was meant to go through derived algebraic geometry and algebraic K-theory for undergraduates, completely nuts
 
lol
 
The students who go to those sort of things seem to learn lots of fancy words
(but do they learn much more lol)
Balarka would
 
Surely depends on the student, but I personally find the simplest slice of modern algebraic geometry (eg, scheme theory) hard to understand - I think these sort of things require a lot of background on much more classical mathematics
 
Seems fair
I thought I understood scheme theory, until I had to find examples of things
 
2:15 PM
Is there a way of deriving the formula $\dim (U+W) = \dim U + \dim W - \dim (U \cap W)$ from the rank nullity theorem or codimension equation? It feels like there should be one.
 
Historically things like scheme theory or stable homotopy theory or ... were developed as a context in which one can rephrase classical questions in algebraic geometry/algebraic topology/... in a way that decidedly gives better results and resolves classical questions
So I don't think it's possible to fully understand them without knowing concrete examples in which one can ask concrete things
 
I definitely agree. Unfortunately it's hard to find good concrete motivation without searching endlessly through ancient papers
And then you have to self-translate the old-wording carefully into the modern wording
 
There are definitely books on these subjects which build up to the abstraction from actual concrete math
 
@user193319 $U,W$ are submanifolds of a specific manifold that intersect?
@BalarkaSen Like Shaf?
 
No, just linear subspaces.
 
2:18 PM
@user193319 Oh, right
 
@user681391 no this is brainlet tier stuff mostly
Sobolev inequalities on surfaces right now
Moser-Trudinger inequality
 
@user681391 I can't recommend a text on algebraic geometry because I don't know much about it. Shafarevich is a good book but it's not scheme theory.
 
That doesn't sound that brainlet-tier :P
 
analysis is mostly trivial
2
more and more involved applications of the fundamental theorem
 
@BalarkaSen It's the only that comes to mind that builds up the abstraction from actual concrete mathematics
 
2:20 PM
quite tiresome really
Sobolev inequalities are literally proved by using TFC
 
Shafarevich part 1 is just varieties, no abstraction there
 
Moser-Trudinger is FCT plus a neat power series argument
 
part 2 is complex geometry but I haven't read it
 
@RyanUnger Analysis is a way of thinking :P. Modern algebra is about endlessly building a tower of abstraction. Analysis is about mastering latent inhibition
Shaf has scheme theory in it
 
Algebra is much nicer
 
2:22 PM
Shaf has two books that break up into 5 parts, and the third part has scheme theory, whichever books those are
I borrowed such a book out a year or two ago
 
maybe falls in part 2 i never read it
 
@BalarkaSen do you want to learn about nodal sets of eigenfunctions
it's very Russian math
 
I think he continues to call things varieties, but still properly defines them
 
I need someone to read this book with me
 
@RyanUnger oof
why tho
dude i have like 50 things on my reading list already
i wont read any of them in reality but
 
2:25 PM
Well what are you reading then
other than this lovely chat dialogue
 
nothing lol
 
ded
 
I should read a bit about H-spaces so I can fix this answer and I should read a bit of hyperbolic geometry so I can fix this one
Trying to trick myself by writing garbage that I have to fix at some point
and cannot fix without learning something
 
@BalarkaSen but you like Russian math
I think Sasha is great but his lectures are incomprehensible
 
@BalarkaSen How are you saying that it has a critical point?
 
2:33 PM
@BalarkaSen I don't know anything really about this, but I thought spheres are suspensions and suspensions are H-spaces?
 
@Albas Continuous function on a compact manifold has a global maximum. That's a critical point if your function was smooth
@user681391 Loopspaces are H-spaces, which are adjoint to suspensions :)
 
@BalarkaSen I have a vector field $X$ transverse to $S^2\subset\mathbb R^3$, and then I have an embedding $e$ of a sphere into a 3-manifold $M$. I compose this embedding with an orientation preserving diffeomorphism $f\colon e(S^2)\to S^2\subset\mathbb R^3$. The embedding $f\circ e$ is transverse to $X$, too, right?
 
Suspensions are not H-spaces. You can call them co-H-spaces
 
Oh right, so they're co-H-spaces?
 
Yes :)
 
2:34 PM
:D
 
co-H-spaces, pronounced "coh-spaces"
 
The power of vaguely remembering things
 
@BalarkaSen Ahh okay, functions from a compact set
Thanks
 
@anakhro Transverse means it's a section of the normal bundle of $S^2$?
 
@anakhro Ya
 
2:35 PM
yo
 
Hey @SemiC
 
I don't know what this book is saying then. So back to the drawing board.
@user681391 I would suppose $S\pitchfork X$ means (at the most basic level) that $T_pS + \text{span}\{X_p\} = T_p\mathbb R^3$, where $S$ is the embedded surface.
 
@anakhro But you have a vector field transverse to $S^2$?
 
Yes.
 
That's equivalent to user68's interpretation. Sure, @user681391, you can talk about the vector X_p being transverse to T_p S in T_p R^3
 
2:38 PM
But that mean $X|_{\iota(S^2)}$ is a smooth section of the normal bundle?
Oh right
I didn't read that correctly because $X$ was the vector field lol
My bad
Doesn't your orientation preserving diffeomorphism do nothing to $X$?
 
how to prove that the cube of any integer can be written as the difference between two squares?
 
A point $a \in S $ is called an accumulation point or limit point of $S$ if every deleted $\delta$ neighborhood of $a$ contains points of $S$.
Can we regarard these accumulation points $a$ as complements to $S$?
 
@Misha.P What does complements to $S$ mean?
 
If $B \subset A$ the $A-B$ is called complement of $B$ relative to $A$
 
@Mathphile Have you seen this: $n^3 = \left(\frac{n(n+1)}{2}\right)^2 - \left(\frac{n(n-1)}{2}\right)^2$
 
2:47 PM
@user681391 That's cute.
 
@user681391 i haven't seen this formula before
how did you derive it?
wait i understand
$n^3=(1^3+2^3....n^3)-(1^3+2^3....(n-1)^3)$
 
I've seen something like that in Spivak. It's probably derrived in some way relying on the fact that $\sum_{i - odd}^{n}i=n^2$
 
$1^3+2^3....n^3= \left(\frac{n(n+1)}{2}\right)^2$
 
You can see it geometrically with some trouble by thinking about squaring triangular numbers
 
$(1^3+2^3....(n-1)^3)=\left(\frac{n(n-1)}{2}\right)^2$
 
2:52 PM
@Misha.P How do you want to regard them as such? What are the limit points of $\{(x,y)\in\Bbb R^2\mid x^2+y^2< 1\}$
Or in $\Bbb R^1$ what are the limit points of $(-1,1)\subset \Bbb R$
 
@user681391 There's a cute generalization, though. For any $n$, $\sum_{d|n} \tau_0(d)^3 = \left ( \sum_{d|n} \tau_0(d) \right )^2$.
$\tau_0(m)$ being the number of divisors of $m$
For $n = 2^k$, you get back the sum of cubes of the first k guys = square of their sum thing
 
@BalarkaSen That is nice
 
@user681391 neither of your examples has any discontinuities and therefore have no limit points.
 
@Misha.P Tell me a neighbourhood of $0$ in $(-1,1)$ that doesn't contain another point in $(-1,1)$?
 
3:08 PM
@user681391 there is none
Although, I would like to get back at my previous answer to your two examples. Perhaps there can be infinitely many neighbourhoods if we take arbitrary point of either set with arbitrary small $\delta$
 
cdt
Here's a fun little puzzle:
Prove in $\mathbb{R}^n$ there can be atmax $2^n$ points no three of which determines an obtuse angle
 
@Misha.P Do you still claim that $0$ isn't a limit point in $(-1,1)$?
 
@user681391 it is a limit point, for $\delta = 1$. However not the kind of point you asked for. Its neighbourhood will overlap with some other point in $(-1;1)$
 
It's either or a limit point or it's not a limit point right? It's not dependent on $\delta$
 
It has to be a limit point. But then again if it has to satisfy that"deleted $\delta$ neighborhood of $a$ contains points of $S$", then certainly "deleted neighborhood" implies that some point Is Not in $S$
 
3:23 PM
The deleted point thing is that:

$x\in S$ is a limit point of $S$ if for every $\delta>0$ the neighborhood has nontrivial intersection with $S$ after deleting $x$: $N_\delta(x)\cap (S-\{x\})\ne \emptyset$.
It's $x$ you're deleting, not the $\delta$-ball
 
Your definition is better
 
Then as you said, $0$ is a limit point
 
But just as any other number of that set, right?
 
Yes
But additionally
if $S\subset X$ is a subset (with $X$ a topological space, here $\Bbb R^1$), then $x\in X$ is a limit point of $S$ if for every $\delta>0$ the neighborhood has nontrivial intersection with $S$ after deleting $x$: $N_\delta(x)\cap (S-\{x\})\ne \emptyset$.
Can you find a neighborhood of $1$ that doesn't intersect $(-1,1)$?
 
That would be a number outside the set
 
3:31 PM
A limit point of a set doesn't have to be in the set (look up the definition of a closed set)
 
So I am correct?
 
About?
 
That a neighbourhood of $1$ that doesn't intersect $(-1,1)$ surrounds a point that is outside of $(-1,1)$
 
Are you claiming that there does exist a $\delta>0$ such that $N_\delta(1)\cap (-1,1)=\emptyset$?
Note that this delta ball is nothing other than $(1-\delta,1+\delta)$
 
Only if $\delta $ has an upper bond
 
3:36 PM
$\delta>0$ is a number, it doesn't have a notion of boundedness
 
Set of all possible $\delta$s
 
What is $(-1,1)\cap (1-\delta,1+\delta)$?
 
It's an intersection of a point $(-1, 1)$ and range of length $> 2$
 
But it equals $(1-\delta,1)$ right?
 
No. $(1-\delta,1) \subset (1-\delta,1+\delta)$
Unless $\delta = 0$ which it can't be
 
3:43 PM
?
$(-1,1)\cap (1-\delta,1+\delta)=(1-\delta,1)$ in which case for any $\delta>0$ this is a non-empty intersection, and thus $1$ is a limit point of $(-1,1)$
(well if $\delta\geq 2$ then $(-1,1)\cap (1-\delta,1+\delta)=(-1,1)$ but that stills gives nontrivial intersection)
 
Hey @AlessandroCodenotti A little question
Prove (from definition): $T= \{1,1/2,1/3, ...,1/n,...\} \cup \{0\}$ is compact. Here's my approach
 
@SubhasisBiswas In what space?
 
@user681391 real numbers
From the definition of limit: For any $\epsilon>0$ $\exists m \in N$ such that $|1/n| < \epsilon$ $\forall n \geq m$.

Let $V_1, V_2, ...$ be an arbitrary open cover of the set $T$.

So, there is an open set $V_t$ that contains $0$. Now, we take an open interval $Q \subset V_t$ such that $0 \in Q$.
now
 
@user681391 why is $(-1,1)\cap (1-\delta,1+\delta)=(1-\delta,1)$ and not $=(-1,1)$
 
@KarlKronenfeld Thought so, thanks for the reassurance.
 
3:52 PM
For all $n \geq m$, $1/n \in Q$. We consider the worst possible scenario, i.e. none of $V_i$'s (other than $V_t$ ) contains more than one element of $T$. But, there can only be $m-1$ such open sets. So, we get a finite subcover.
 
@Misha.P Well if $\delta\geq 2$, then it is what you are saying (which I wrote above). But say $\delta =1/2$, then $(-1,1)\cap (1-\delta, 1+\delta) = (-1,1) \cap (1/2, 3/2) = (1/2,1)=(1-\delta,1)$ right?
 
is the proof good?
 
@N.Maneesh I can't speak to the usefulness of these videos, but the speaker is regarded as something of a crank in some circles, and that may explain some habits of unclear explanation he may exhibit. Perhaps just find a different speaker?
 
@user681391 I misread
 
@user681391 can you verify?
 
3:55 PM
@rschwieb okay. Thank you.
 
just found a copy of hubbard and hubbard on the free books shelf at our math library
yoooink
 
I want to learn differential geometry in a short period of time. Can you suggest some good lecture videos that bridge between D.G and G.T.R? @rschwieb
 
@SubhasisBiswas yes
 
Thank you. :)
 
waddup nörds
 
4:09 PM
Hi @ÍgjøgnumMeg
 
How's it going :)
 
They just published the courses for the next semester here, looks like there will be a cool number theory seminar
 
ooo
Nothing from Heidelberg yet
but according to Mathein ANT1 is on offer so that's nice
 
officially it's called "trascendental number theory", apparently the aim is to prove Thue-Siegel-Roth, but it will also go through class numbers and stuff from algebraic number theory
 
something about diophantine approximation?
 
4:13 PM
Yes, all algebraic irrationals have irrationality measure 2
 
Nice :)
I had a nice discussion with one of my old lecturers today who knew Atiyah quite weel
well*
said that the whole RH proof controversy had nothing to do with him being scenile or anything
and that in fact he was kind of used by the Heidelberg forum
and that he was very depressed because his wife and best friend had both died at similar times so he wasn't thinking properly and such
 
@Ryan @Érico and other analysists, what should I expect from a course called "semigroups and evolution equations"?
 
@ÍgjøgnumMeg there will be a lecture on affine algebraic groups as well
 
Prove that there exist real numbers which are not algebraic:

Proof: $\mathbb{R}$ is perfect, so it is uncountable. [Since it is closed and every point of it is a limit point]. The set of algebraic numbers, say $A$ is countable. Therefore, $\mathbb{R} \setminus A$ must be uncountable.
is it correct?
 
also a lecture on adic spaces: lsf.uni-heidelberg.de/qisserver/…
 
I'd say there are easier ways to show $\mathbb{R}$ is uncountable than relying on the fact that it's perfect, but yes, this is correct
 
4:55 PM
@SubhasisBiswas: There are far more elementary proofs that $\Bbb R$ is uncountable.
LOL, oops, Thorgott already said that.
hi @Mathein and devilish @Alessandro
 
hi @Ted
 
@Mathein nice, thanks :)
Hi @Ted
 
Too vague a title to answer, @Alessandro: Could be a serious PDE course.
heya @ÍgjøgnumMeg.
 
Hi @ÍgjøgnumMeg @Alessandro
 
4:57 PM
hi @ted
 
Googling brought up this notes that seem relevant
 
heya @Semiclassic
 
where does Hubbard+Hubbard line up in the realm of vector calc books
math library had a copy available on their free books shelf so I snagged it
 
@N.Maneesh Nope: i don't know much about differential geometry.
 
Heya Ted
 
4:58 PM
It's the book that made me write mine, @Semiclassic. Very idiosyncratic, but some good stuff in it. Exercises are not so impressive.
hi @Rithaniel
 
mmkay
 
Wagwan
 
@TedShifrin Made you write yours as in "it inspired me" or as in "It was so terrible I had to replace it"? :)
 
@Alessandro Ho visto i mei compagni di tandem e ho visto dei video italiani per imparare la lingua. Aiuta molto. Sto migliorando il mio italianio :)
 
@BalarkaSen ayyy there's some kids doing algebraic geometry on the board in the lounge
this place is crazy
 
5:01 PM
I started teaching a brand new course out of it, @rschwieb, and after two months realized it was unusable in a serious way. Of course, lots of people use it and apparently love it, but the exercises are unpredictable, not enough middle-level problems, and the text has stuff that's interesting for pros but not — IMHO — appropriate for first-year college students. At any rate, I started writing the second half of my book during the fall, and then wrote the first half during the spring/summer
@Ryan: By kids you mean ... ?
Why is that crazy?
 
like really young people
algebraic geometry is too hard
 
LOL
is it like an REU?
 
idk they seem to be doing homework
maybe some summer reading course
 
You could be brave and ask them?
 
no, algebraists are scary
 
5:02 PM
Meh.
 
Of course, there are different approaches to algebraic geometry. My research in it was slightly algebraic, but all over $\Bbb C$.
 
We could make a horror movie about algebraists.
 
idk they're literally talking about zero sets of polynomials
 
Could make plenty of horror movies about analysts too.
 
5:04 PM
so not like etale cohomology
 
WTF is wrong with zero sets of polynomials?
 
lol
 
Logicians would be the real terrifying ones, I think.
 
omg someone gave a logic talk at the Notre Dame open house
I have never been so confused
 
I know someone who took grad alg geo as a sophomore and then etale cohomology as a senior
 
5:05 PM
next time I was so confused was during a "Hilbert schemes over points" talk at Yau's birthday conference
 
am i gonna die in Heidelberg
hahaha
 
Heidelberg sounds like the best university in the world for courses
 
@ÍgjøgnumMeg probably not
 
absolutely crazy
 
@MatheinBoulomenos But you never know
 
5:05 PM
Hey @Tobias
 
Hey @Tobias
 
heya @Tobias
 
Hi @ÍgjøgnumMeg @TedShifrin
 
@Mathein Hope not, thankfully I can do it with no interruptions so maybe I'll just have to get used to it for the first couple of weeks
 
5:06 PM
I came up with a neat proof that $R$ and $M_n(R)$ are Morita equivalent, using the algebra on a small category @Tobias
 
Paul Yang's minicourse on CR geometry is also very confusing
 
@MatheinBoulomenos nice
 
He didn't print out enough notes and I got completely lost
 
Go read the original Chern-Moser paper, @Ryan :P
 
I never imagined to hear "contact form, Reeb vector field, psedoconvex, and positive mass theorem" in the same talk
also isoperimetric inequality, Moser-Trudinger inequality
o.o
 
5:08 PM
Hmm, I wouldn't expect those all in a talk on CR.
 
Isoperineal
 
Ted I'd send you the notes but I don't have them in either physical or electronic form
he said he'd print me a copy haha
 
I have been writing up a test case today. The first steps can be summarized as 1. Create a new person. 2. Make that person dead (might require god mode)
 
OK, cool, thanks @Ryan.
 
@TedShifrin ok added to the infinite reading list
 
5:09 PM
@Rithaniel D:
 
LOL @ god mode.
 
Not you, Alessandro, you're cool.
 
oh and Sobolev inequalities adapted to the contact structure
using the subgradient
 
This sounds like GMT meets CR geometry.
 
no GMT, more like metric geometry somehow
 
5:10 PM
@TedShifrin But I think it will turn out that god mode is not necessary for this actually
 
he said a lot of the intuition comes from 4D spaces
and somehow contact 3D behaves like 4D
also Q curvature, Q' curvature, CR Yamabe problem
absolutely insane talk
 
sorta like classic contact geometry embedded in symplectic manifolds ...
 
he talked a lot about the Heisenberg group too
it's easier to list what he didn't talk about
 
I only know classic CR stuff (Chern-Moser was about invariants to detect when a CR manifold is a standard hyperquadric).
 
@TobiasKildetoft the idea is this: let $C$ be a category with finitely many objects and $R$ be a ring, then $R[C]$ is defined as a free $R$-module on the set of all morphisms in $C$, with multiplication defined on the basis elements by $0$ if they are not composable and as composition if they are composable. Then one can show that $R[C]$-modules correspond to functors $C \to R\mathrm{-Mod}$, note that this implies that equivalent categories have Morita equivalent algebras.
Take $X$ to be the category with $n$ elements and for each pair of objects, exactly one morphism, i.e. the trivial rela
 
5:13 PM
ahh, neat
the first construction is very similar to something I have worked quite a bit with, except it started with a $2$-category instead
and you got a category rather than an algebra
 
One esoteric historical thing that's been bugging me: the phrase "Hilbert space of random variables"
on the one hand, that phrase can't go back farther than "Hilbert space" itself
which dates back to von Neumann 1929 afaik
on the other hand, the books on prob/stats where it shows up treat it as sufficiently 'classical' as to not need any discussion
 
Is there a higher order derivative test for local extrema of multivariable functions?
 
What do you have in mind?
 
5:29 PM
I'd guess if all partial derivatives vanish and if the Hessian is positive definite, you have a local minimum or something like that?
(not sure if that holds)
seems like the natural generalization, so it's probably true
 
@MatheinBoulomenos not sure if what holds?
 
@Thorgott: You have Taylor's Theorem, of course. So you look at the cubic form, but I don't know classifications for cubic and higher as I know for quadratic.
 
the thing about vanishing gradient + positie definite Hessian implies local minimum
 
Oh, I assumed @Thorgott was asking what to do if the second derivative test fails. It would be nice if one replied to questions like "What do you have in mind?" ...
@Ryan: I was out of the room and missed your pithy removed remark. Or is that a removed pithy remark?
 
5:48 PM
@TedShifrin someone from the summer school is stalking me on stack exchange now
@user189456
 
LOL ... makes you feel important.
 
no, he's sitting next to me
reading this as I type it
 
Oh, so unimportant.
 
it's a very meta conversation
he doesn't have enough rep to talk here
 
Does he have enough rep to talk to you in person?
 
5:51 PM
he says he doesn't
 
Sorry about that. I was afk for a bit. Yeah, I wanted to know about the scenario where the second derivative test fails, but it seems the terms do grow too complicated. I also considered reducing the multi-dimensional case to the one-dimensional case, but composing with arbitrary smooth curves doesn't seem too fruitful. Thanks for the reply.
 
You have to be careful with such approaches. There's a classic example of a smooth function on $\Bbb R^2$ that has a minimum at the origin along every line through the origin but fails to have a local minimum at the origin.
I used to run into a number of my students/former students in chat here.
 
he's only here because he saw me typing here
SPYING on me
 
As I said, it makes you feel important.
 
what?
 
5:55 PM
Being spied upon.
 
Something like $(y-x^2)(y-2x^2)$, yes. I think considering smooth curves rather than just straight lines should suffice, but that doesn't look like it would give rise to a practical test.
 
is there actually a good way to distinguish between a manifold and its immersed image
notationally
 
No, @Thorgott, it definitely doesn't, as we can soup up that example to make it fail along using all curves $|y|=|x|^n$ or vice-versa.
 
would something like x^2/(y+x^2) work?
 
@Semiclassic ??
@Ryan: Introduce a name for the immersion?
 
5:56 PM
as an example of "continuous along lines but not curves" ? I'm probably being silly
 
Yes but I'd get made fun of for writing that all the time
 
yeah, i am
bleh
 
But can it be improved so much as to fail for every smooth curve through the point?
 
No, not every smooth curve. But you're never going to get any sort of criterion out of this. Ultimately, it has to be an algebraic criterion in terms of the cubic, quartic, ... forms.
@Semiclassic: Try the curve $y=3x^2/2$.
 
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