« first day (2425 days earlier)      last day (2587 days later) » 

12:09 AM
Hi @skillpatrol
 
12:55 AM
Anyone here?
 
1:27 AM
Nope. No one's here.
 
@Ted @Danimark To me Stein manifolds are intersting because they are complex manifolds with a certain kind of function to $\Bbb R$ modeled after the square of the modulus function.
 
I just stopped by to see if anything had happened. But it hasn't.
 
Oh hey @Ted
 
Sure, of course, @PVAL. I'm acquainted with exhaustion functions — even though they exhaust me.
Actually, you should have pinged Danu.
 
Generically these functions determine interesting morse structures, which grant deep connections between contact,symplectic and complex geometry.
@Danu see above.
@Ted Similar names.
 
1:29 AM
LOL ... I should understand this more deeply than I do, but I get it ... :)
Yeah, @PVAL, but Demonark is but a sophomore :P
 
He's a rather good sophmore.
 
Oh, indeed. But not ready for complex manifolds and symplectic geometry just yet.
 
I think he could read MMoCM (if he desired to).
 
He's about to take his first course on manifolds (G&P).
 
is that a reasonable acronym?
to you?
 
1:32 AM
I'm not sure I know to what you refer.
 
Arnold
 
Oh ... duh.
But too obtuse.
 
There's certainly things to say in symplectic geometry that really don't have much to do with manifolds.
 
The Chicago curriculum is off in the never-neverland. He is learning functional analysis at the expense of a solid background in multivariable calculus, differential forms, etc.
 
Though the differential form calculus is pretty necessary.
 
1:33 AM
I have long been unhappy with their curriculum, but somehow they don't ask me for advice.
Yup. So he's seen 1- and 2-forms in the plane and no more. I recommended my book :P
 
Even Moser's proof of Darboux is still interesting in that case, and really doesn't require any manifolds.
 
Sure, Darboux is a good exercise in $\Bbb R^n$, nothing to do with manifolds. For the first grad course I taught (manifolds at MIT) I mistakenly trivialized the proof as an exercise and had to hand out a corrected exercise :P
Scary that the top 4 things on the star board are me or reference me. I should be more scarce.
 
Functional analysis is nice
at least in some categories
 
Not at the expense of a mastery of multivariable calculus (both theoretical and computational).
 
I'm sure I agree with you.
Certainly in my area I use multivariable calculus arguments way more than any functional analysis.
 
1:40 AM
The Chicago and Harvard MATH 55 obsessions truly upset me. But that's life.
 
In fact most of the functional analysis I use
I black box pretty heavily.
I'm sure @Mike knows it way better.
 
The physics majors who took my course felt well-served. They'd be screwed taking Chicago's course. And I have no patience for math majors who want to do theory, theory, theory and can't compute a damn thing.
@MikeM knows (and actually understands) an amazing amount of stuff. I've been impressed for quite a while.
But he doesn't like my saying that. :P
 
I sort of feel like I could have used functional analysis at an earlier age.
I learned (what I know of it) in grad school.
 
I only began to appreciate it in seminars (e.g., the $\bar\partial$-Neumann problem) in grad school. I learned some earlier.
 
And after some time barely understand the sort of things you can prove about spectrums.
 
1:44 AM
I actually never took any analysis in grad school, because I'd had so much as an undergrad. I sort of regret it now. But who cares :P
 
If I feel caught up on my writing, I might try and take PDE's next year.
 
That might actually do you well in future research.
 
Functional analysis is something which I only -sorta- know.
 
It's pretty intimidating with all the spawn of Cafferelli about.
 
Well, damn, @Semiclassic, you're a physicist. You feel like you need to know as much as a Ph.D. mathematician :P
 
1:45 AM
lol
 
@PVAL: You're learning for yourself. Not for anyone else.
 
It's not so much competing with them.
It's keeping up with them.
 
Seriously, @Semiclassic, I'm pretty impressed with how much hard math you've picked up. And you ask plenty of questions I fumble.
My point is that you don't need to keep up with anyone, @PVAL. You're learning for your own edification and for things you might want to do down the line.
If you're not the best in the class, so be it.
 
I probably won't have any of them for the moment; I'm working on a different problem atm, and for the time being it doesn't connect to anything geometric.
 
1:47 AM
whew :P
 
(Hopefully it will eventually, but for now no.)
 
Hello. I have what i think is a really simple question... if a "matrix" is 2 dimensional, has "rows" and "columns", and is "rectangular" (because all rows are same length etc), what are the equivalent mathematical terms for higher dimensions?
 
Tensors.
 
@Ted I totally agree with you, but somehow that doesn't fix the issue for me.
 
I tried to do my best :)
 
1:51 AM
A matrix $A$ has matrix elements $A_{ij}$ for $i,j=1,\cdots n$.
 
Yup, @Alistair, tensors of higher rank.
 
That makes it a rank-2 tensor.
 
hmm okay, and what do you call the "rows" in a tensor?
 
You don't :)
 
A rank-3 tensor, by contrast, would have elements of the form $A_{ijk}$.
Yeah. I guess the most you might do is, say, refer to the vector $A_{12k}$.
 
1:53 AM
But you also have more confusing things with some lower indices and some upper ones. A matrix actually represents a tensor of type $(1,1)$ if you think of it as a linear map; it represents a tensor of type $(2,0)$ if you think of it as a bilinear form. It gets more complicated in higher ranks.
 
in which case you're looking at the vector consisting of elements $(A_{121},A_{122},\cdots)$.
Yeah.
 
Hey everyone! So there's this really funny line from Rotman I think you guys would appreciate.
 
So are tensors "rectangular"?
 
Yeah.
 
It better be good, Demonark.
Did you crack up in public? :D
 
1:55 AM
"Given a homomorphism $f$, one must always salivate, like Pavlov's dog, by asking for its kernel and image"
 
@Ted $(V^\perp)^\perp$ is the whole space iff $V^\perp = \{\vec{0}\}$?
 
Yes, Zach.
 
(There is such a thing as a 'spherical tensor' but those are weird.)
 
Say what, @Semiclassic?
 
@Daminark what is this book you are reading
 
1:55 AM
You mean periodic?
 
Well, spherical tensor operator.
 
I'm at home so I'm more willing to lol, though the inclination was, admittedly, not as much as with the weekly convergent stuff
 
I wasn't remembering it well enough.
 
@Eric Rotman's Intro to the Theory of Groups
 
If rank-2 tensors are linear maps, what are rank-3 tensors?
 
1:56 AM
ah for Babai's class I presume
 
You run into them in advanced quantum mechanics / group theory stuff. They're annoying.
 
Yup, it's got a more combinatorial mindset
 
hi @Dami
 
For example, Zach, bilinear forms with values in a vector space, instead of $\Bbb R$.
 
Plus I never really managed to get far into DF, I just get bored so fast
It sucks the enjoyment out of what is currently one of my favorite subjects... :(
Hey @Meow!
 
1:57 AM
D&F is really dry and big
 
@Semiclassic: As one who's worked with tensors his whole life, I'm curious to know a definition :P
 
Are there definitions in advanced quantum mechanics?
 
Demonark, @Eric: You'll be amused that one of the reviewers of my algebra book (my first published book) complained that I had humor — and said that humor has no place in a mathematics text.
 
LOL
 
?!
 
1:58 AM
wow humor is one of the things I wish more books had
 
I strive for that
 
Zach, when you learn some differential geometry, a good example is the second fundamental form (which measures twisting of a surface, say) in higher codimension than $1$ (so you get twisting in each normal direction).
 
While I was a hopeful algebraist, I enjoyed the sort of individual subject books like A&M and Pierce a lot more then the big texts.
 
@Eric: I'm on your side :P
Books and courses.
 
@Ted Aren't differential forms covered in your book?
 
1:59 AM
Of course.
 
Like having a book which is 50% memes might be excessive (though that wouldn't be beyond me tbh... lol jk) but I find humor to be great
 
I don't want memes in serious books, thanks ... unless they're 6th grade.
 
I have a joke environment set up in TeX for my class notes
 
LOL, @Eric. You're great.
 
I try
 
2:00 AM
I wrote a LaTeX template for books for fun a while ago.
If anyone wants it, I can e-mail it to you
 
I typically inject more humor (no surjection) in class than in books. I wish I'd put "Let bigons be bigons" in my diff geo notes. Maybe I should work that in somewhere.
 
I remember when I first started learning about classical diff geo of surfaces, and I learned about the second fundamental form, but the person teaching me referred to the first fundamental form as the metric, and I was very very confused
 
@Ted Bigons don't sound like geometry to me :)
 
@Eric: You might get this, although no one else will. One of the horrendous lectures in undergrad diff geo is the day I prove the Codazzi and Gauss equations for a surface. One of my students, aghast at all the indices, said "I'm throwing in the tau."
@PVAL: A polygon with two sides?
 
ahaha fantastic
 
2:02 AM
It shows up in Gauss-Bonnet applications.
@Eric: Every class after that, I quoted him during that lecture :P
 
Nevermind
 
Uh huh, curvature.
LOL.
The hint was differential geometry :P
 
I gave the lecture introducing them at my summer program last year (out of your book actually) and I could just see the life leave everyone's eyes
 
Bigons don't exist in $\Bbb R^n$, right? Well, they're degenerate, I mean.
 
That's where differential forms and moving frames win, hands and feet down.
Right, Zach. But think on a sphere.
 
2:04 AM
Yeah
 
yeah, having the classical formulation side by side with the moving frames formulation is magical
 
Or on some other curved surface
 
Good luck on negatively curved :P
 
@Ted I'm so ruined that I think a bigon is a region between two curves (not necc. geodesics).
 
You're not ruined, PVAL. You're right. :)
Mine was a geodesic bigon :)
 
2:05 AM
Wait, so how do we define a "line segment" on a surface? Do we take some mapping from $\Bbb R^2$ to it and map the line segment? Or do we take the shortest path?
 
Yeah, Eric, that's why colored chalk and lots of energy are good for lectures. But it's also good to have notes and say — here's the idea, but it's just bookkeeping and you can check it out in the notes.
 
@MeowMix the locally shortest path.
 
Typically, we're talking about shortest paths, but it needn't be.
G'morning, @MikeM.
 
You flatter me. I wish I could take that and turn it into a paper, though.
 
Hi @AkivaWeinberger
 
2:06 AM
Oh geez, and now DogAteMy redescends.
 
Well, if you take antipodal points on a sphere now we have infinitely many :(
 
I was sincere, @MikeM. You know I don't lie.
 
@PVAL-inactive I liked the PDE I've learned but there's just a ton of stuff. Does UT have anybody that mostly thinks about elliptic stuff?
I wish I had a Gilbarg and Trudinger class.
 
Correct, Zach.
 
Most people here think mainly about Elliptic stuff,
 
2:07 AM
There was a class on harmonic analysis this quarter that taught Calderon-Zygmund that I found out about in its last week
Nice
 
Wait @PVAL you're at UT? How long have you been there?
 
Eventually, Zach, you'll get to differential geometry (or to the last section in my chapter on geometries in the algebra book) ... :P
 
It's Cafferelli's research group, and my understanding is hes firmly on the elliptic side.
 
Just wondering if you know anyone by name of Souganidis
I know he used to be there
 
@Daminark this is my 4th year
 
2:08 AM
PVAL is not infinitely old :P
 
in grad school
 
@Daminark Souganidis has been here in Chicago for like
9 years
 
I know, I guess I didn't know how old PVAL is
I actually didn't know you were a grad student still
I mean I had no expectations but if you told me that you were deeper in, I'd have totally believed you
 
A week argument, Demonark.
 
@Ted @Mike AR is leaving it seems like DF will too.
Big problems here.
 
2:13 AM
Fair, fair
 
I'm not quick enough with the initials.
 
Also I've gotta peace out real quick, see you in $\mathbb{\ell}$ units of time guys!
 
That is one thing I'd rather not write out.
 
No problem, @PVAL.
Bye, Demonark.
 
You certainly know DF.
 
2:14 AM
You can delete that.
 
Yes, @PVAL, I know one, but not the other.
 
I'm surprised about those.
 
UGA lost one of our superstars 15 years ago. These things happen. But academia is a bit more unstable now, I think.
 
@Ted Sorry, what was the thing you wanted me to ponder about $(V^\perp)^\perp$?
 
Zach: How might that be related to the smallest closed set containing $V$?
 
2:16 AM
OH
 
In particular, ponder this: If $V$ isn't necessarily closed, must $V^\perp$ be closed?
Maybe you want to wait until Chapter 2 to think about this.
 
Uhh, maybe.
 
Make a note of it :)
 
*takes off sticky note from computer and appends "Why does $V^\perp$ have to be closed?"
 
Good boy. :)
 
2:20 AM
I do have one, and it lists various exercises I need to do and what not.
 
Are you enjoying learning stuff more thoroughly?
 
Of course.
I actually never gave a second thought about infinite dimensional vector spaces
 
Well, I'm just asking. Usually people have to get older to enjoy that.
 
I have a question however. Is the metric always dependent on our inner product?
 
Trust me. I never gave them a thought until college. I didn't know what a vector space was until college.
 
2:21 AM
Because if we don't have some external metric (and it isn't defined on our inner product), how can we define closedness / open-ness?
 
The metric is the inner product.
 
Well, the standard construction is that the norm (hence metric) is defined in terms of the inner product. You could define a norm a different way. (Try this on $\Bbb R^2$. Can you think of different norms that don't come from a metric?)
@PVAL: He means distance (coming from norm).
 
So, let me look up what a bilinear form is. Because I know an inner product is a positive definite, symmetric bilinear form (Courtesy of Balarka)
 
Zach: For future reference — you can define open sets with absolutely no notion of metric. That's what you'll eventually learn in a topology class. But for now, we can stick with $\Bbb R^n$ and the notion of open set coming from the usual metric.
Night for today, Zach.
Bye, @PVAL, @MikeM, et al.
 
@Ted Yeah, don't we have a "topology" of open sets?
 
2:24 AM
@MeowMix Note how @Ted uses the word metric for inner product even when correcting me for doing so.
:P
 
I did not (in this conversation) !!
 
Anyways, have a nice night @Ted :]
 
@PVAL: Zach is not doing Riemannian manifolds, damn it.
 
"Can you think of different norms that don't come from a metric?"
 
Oh crap.
Mea maxima culpa.
hides in shame
That was a typo, dammit.
Explain it to him, PVAL. :P
 
2:26 AM
You might be able to get me to do that. I dunno if it's so easy with PVAL.
 
I feel like a child in the middle of two adults arguing
 
@Meow There is a natural way to go from the data of an inner product to the data of a norm, and to go from the data of a norm to a distance function (i.e. the structure of a metric space on your vector space).
 
Actually, that's true. Nevermind
 
The last has an obvious notion of open and closed.
 
What does "norm" mean again? Pardon me.
 
2:28 AM
In linear algebra, functional analysis, and related areas of mathematics, a norm is a function that assigns a strictly positive length or size to each vector in a vector space—save for the zero vector, which is assigned a length of zero. A seminorm, on the other hand, is allowed to assign zero length to some non-zero vectors (in addition to the zero vector). A norm must also satisfy certain properties pertaining to scalability and additivity which are given in the formal definition below. A simple example is the 2-dimensional Euclidean space R2 equipped with the Euclidean norm. Elements in this...
 
Oh, I know
 
You should figure out how to go through all those steps.
 
In any vector space, is the distance given as $||\vec{x}-\vec{y}||$?
 
is that a distance function?
 
Well, it sure looks like it.
Assuming $||\vec{x}||$ is our norm
 
2:30 AM
"it sure looks like it" is not the correct response.
It is.
 
It takes two vectors, is always greater than 0, satisfies symmetry, and subadditivity
 
but you should check (mentally or on paper)
that it satisfies all the axioms
ok
So now how do you go from an inner product to a norm?
 
Take the inner product of a vector with itself. Well, that's what $\Bbb R^n$ does
 
No
It is not what happens in $\Bbb R^n$
(you are close)
 
Well they also take the square root of it
 
2:33 AM
good.
Now come up with another norm on R^n
 
R^2 even
 
All norms are 0. :}
I'll find one a little less joke-y
 
that's not a norm
 
Oh, it isn't, I just realized
 
and show it doesn't come from an inner product.
(I don't know how to do this last bit without a little calculus)
 
2:36 AM
Oh I have one
$||\vec{x}||$ is the sum of its components
 
||(-1,1)||=0
 
I mean
sum of the absolute values of its components
 
ok
now show that doesn't come from an inner product.
in the same way you constructed your first norm.
By taking the inner product with itself and then the square root.
 
Oh, I see
 
Hint: If it did some function would be differentiable.
and that should take you a little longer than the other parts so I am going to get some food.
 
2:43 AM
@PVAL-inactive You can see it from geometry too. Though I like your answer.
 
Sorry, I think I have to go.
 
@Mike I think we have the same thing in mind
The way we proved it in finite dimensions was to show that any inner produce could be given by <Ax,x> where A is symmetric
 
@MikeMiller It turns out this differentiability criterion is actually sufficient for any Banach space.
 
@TedShifrin @PVAL-inactive Had to step away. It's these bloody things: en.wikipedia.org/wiki/Tensor_operator
I haven't touched them in like...five years.
But my recollection is that the Wigner-Eckart theorem could go f* itself.
 
So you'd get that the unit ball for a finite dimensional inner product space is a smooth manifold, while for l1 and l^infinity it isn't
 
2:49 AM
@Daminark that sounds analogous to what I hinted at.
 
@TedShifrin You also do talk about 'spherical tensors' in the context of electrodynamics (for similar reasons, I suspect). Namely, you've got tensor spherical harmonics with appropriate multipole moments, and these form a spherical tensor (apparently).
 
@MeowMix hi pal, wazzup?
@Semiclassical have you read The Quark and The Jaguar by Gell-Mann?
 
Sorry about the double ping.
 
2:58 AM
:-)
 
@PVAL I imagine
In the second part of that problem, we proved things more generally, since Hilbert spaces have to satisfy the parallelogram law, and $\ell^p$ did iff $p=2$
 
3:41 AM
@Daminark The argument I hinted at works for more general Banach spaces.
And as a challenge, I invite you to try and prove the converse
That a norm comes from an inner product iff a certain function is Frechet differentiable.
 
 
1 hour later…
5:00 AM
preparing for my dissertation defense like the invasion of Normandy
 
all the best!
 
Dream logs analysis: When I had dreams involving maths, it often has the mathematical object floating and moving in an endless white void
 
Oh..nice..
 
What's it on @Forever?
 
I think it is related to topology @Daminark , but Forever should say it!
 
5:12 AM
One example in last night dream is the integers with cofinite topology floating through the void as if it is a moving platform, a theorem then pop up saying that we define an ordered relation given some number $r$. Then $r < s$ if it is copositive to one of the sets (which in the dream's language, it basically means r is in a closed set in this topology (specifically the finite sets))
 
I wish I could have a mathematical dream!
 
So basically, it means that the smaller a number is under this ordered relation, the smaller the closed set the number is found, so one can talk about closed sets nested by inclusion and a sequence of numbers $r_1 < r_2 < \cdots$ sort of index this orering
O and btw, copositive is actually a well defined term, it just does not mean what the dream thinks:
In mathematics, specifically linear algebra, a real matrix A is copositive if x T A x ≥ 0 {\displaystyle x^{T}Ax\geq 0} for every nonnegative vector x ≥ 0 {\displaystyle x\geq 0} . The collection of all copositive matrices is a proper cone; it includes as a subset the collection of real positive-definite matrices. Copositive matrices find applications in economics, operations research, and statistics. == ...
 
@Daminark strange connectedness properties (topology)
 
This is roughly what the scene looks like in the dream as the theorem is uttered
where the gray pairs that is opposite to each other are the open sets and the white part are the closed sets
 
 
1 hour later…
6:38 AM
@Forever That sounds p dank, good luck!
 
I was promised dankness.
 
Lmao
 
 
5 hours later…
12:05 PM
Freedom at last, comrades.
 
Hooray!
 
how did it go?
 
More time for Bowie?
 
stat went great. chemistry was not bad
@skill well, a lot of other stuff too.
 
Cool.
 
12:12 PM
kinda want to keep away from the 70/80's for a while
not just in terms of music, i mean
 
Yeah, there's a whole new millennium out there pal.
 
Great, when will you get the results of your exams back?
 
12:39 PM
@AlessandroCodenotti dunno yet
probably not very soon
@skill by 70's i just meant the counterculture era. the new millennium looks complicated to me.
i don't think i understand it
 
 
2 hours later…
2:46 PM
"Let $(X,\mathcal{A},\mu)$ be a measure space, then there exist an outer measure $\varphi:2^X\to[0,\infty]$ such that every $A\in\mathcal{A}$ is measurable wrt $\varphi$ and $\varphi_{|_\mathcal{A}}=\mu$"
Does this theorem have a name? I have it in my notes without proof nor a googlable name
 
3:06 PM
Good morning
 
How are you?
 
Fine thanks
What about you?
 
OK I guess.
Still no reply.
:(
 
What reply are you waiting on
 
3:18 PM
From the math supervisor at my school district.
I asked to skip a few years of math, since they were superfluous, lol.
Substitution?
 
lol
 
?
 
I only took a maximum vote in math once, it was when I studied and I slept 8 hours lol
 
what's the maximum there
 
the scale changes from high school to universities..
 
3:27 PM
@AlessandroCodenotti you might call this theorem the Caratheodory construction of $\mu$ or something but I've never seen someone actually give this a name.
 
anyway, this maximum grade was during my last year of highschool... I had studied with my ex-gf: yes, we actually studied that time ahahah
 
maximum vote? huh?
 
I meant grade
and you know that
 
how many modules is that
 
no...
I don't even know what you are talking about.
 
3:31 PM
really? I'm sorry for you then
 
@Eric doesn't the Caratheodory construction extend measures on an algebra to measure on a $\sigma$-algebra?
 
that's the Caratheodory extension theorem
Caratheodory construction is the general procedure of getting a measure out of an outer measure
 
I used to go to school with 5 to 6 hours of sleep and that's why I was good at the arts course, apart from the fact that I'm also good at drawing, but that alone doesn't make you a good artist
ok
enough about me
ahahah
 
@Eric I think you mean the other way around? To get a measure from an outer measure it's enough to restrict it to the $\sigma$-algebra of measurable sets
 
Right that's what Caratheodory construction is.
 
3:37 PM
Right, but I was asking about the opposite, I already have a measure and I want to construct an outer measure extending it
 
right well, every measure (also every pre-measure) induces an outer measure in a natural way (the infimum of coverings by sets in $\mathcal{A}$)
This theorem is just a sanity check to make sure that you can't take the outer measure induced by a measure, and then go backwards by restricting to $\mathcal{A}$, and get something new
 
However if I start with a measure $\mu$ with $\sigma$-algebra $\mathcal{A}$, take the induced outer measure $\varphi$ and then take the measure obtained by restricting $\varphi$ to its measurable sets this might have a $\sigma$-algebra bigger than $\mathcal{A}$, right?
 
Yeah this happens.
An example is a incomplete measure, once you do this extension and restriction you get something bigger, i.e. the Borel measure becomes the Lebesgue measure.
 
anyone here has some familiarity with numerical methods, numerical computing or computational science in general?
 
Yeah, I found a simple example, take a measure with $\mathcal{A}=\{\varnothing,X\}$ which is 0 on the first and 1 on the second, this can be extended to the outer measure $\varphi(A)=0$ is $x_0\not\in A$ and $1$ otherwise (for some $x_0\in X$), this outer measure agrees with the first on $\mathcal{A}$ but everything is measurable wrt to it
 
3:53 PM
Yup, good example, Dirac measures are usually good to keep in mind when thinking about properties of measures.
 
Or I can just take any measure and restrict it to a sub $\sigma$-algebra of its $\sigma$-algebra, derp
 
yeah lol
 
@AlessandroCodenotti How do we construct a norm from an inner product space?
Is it always square root of inner product with itself?
 
By setting $||x||=\sqrt{\langle x,x\rangle}$
@MeowMix yes sorry, my internet is super slow today
 
Do all norms have a corresponding inner product?
 
3:58 PM
No
 
Hello everyone
I am trying to understand an easy formula:
$ \frac{d}{dt} ( \frac{\partial T}{\partial \dot r } ) $
is this a triple derivative?
the dot, the partial derivative and the 'dt'
 

« first day (2425 days earlier)      last day (2587 days later) »