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PVAL
PVAL
12:10 AM
@BalarkaSen This is certainly proved early in Milnor's h-cobordism notes (which are perhaps in my humble opinion the best example of mathematical writing).
(This is Thm 3.4 in Milnor's notes.)
 
Mike Miller
Mike Miller
@Ted: If I were working out transversality myself what your example would tell me is "Oh, for stability I need to work with closed submanifolds" as opposed to saying that stability fails in the noncompact world.
I like Milnor's books and papers but I am not sure if I would call the h-cob notes the best mathematical writing ever.
 
PVAL
PVAL
I like them. A lot.
 
Mike Miller
Mike Miller
Also @PVAL He probably needs a lot more training in the basic facts of smooth manifolds before reading a book that works with gradient flows on Riemannian manifolds.
 
PVAL
PVAL
12:27 AM
I don't know if thm 3.4 requires so much background (besides the fundamental existence and uniqueness for ODE's which will exist anything where flows are involved).
 
Mike Miller
Mike Miller
You in particular need to know what a gradient is and what a flow is.
 
PVAL
PVAL
The notes are actually somewhat eccentric in that they avoid using a metric until quite later.
Instead defining "gradient-like" vector fields.
 
Mike Miller
Mike Miller
I forgot about that. Sure.
In any case I still think one should learn about manifolds before the h-cobordism theorem, but perhaps that's just taste.
 
PVAL
PVAL
They also essentially define a flow in the proof of that theorem (though I think some basic DE's is probably assumed as they do not define integral curves).
@MikeMiller That seems reasonable...
@MikeMiller I have meant some people who were trying to learn derived algebraic geometry without studying varieties or Fukaya categories without studying symplectic manifolds. So it really might perhaps just be taste.
 
Mike Miller
Mike Miller
How well does it go for them?
 
PVAL
PVAL
12:39 AM
Dunno, at some point I stop understanding them.
 
Mike Miller
Mike Miller
Some talks I've seen by unnamed people who work in stuff like that makes it seem like proving theorems is also a matter of taste.
 
PVAL
PVAL
I read Milnor's notes immediately after a first course on diff. topology. You really do not need very much.
 
L33ter
L33ter
I keep hearing grassmians from my topology prof
I was wondering what is that ?
 
PVAL
PVAL
Do you know what projective space is?
 
L33ter
L33ter
no
 
Mike Miller
Mike Miller
12:44 AM
I think there's a Wikipedia article on Grassmannians.
 
PVAL
PVAL
$\Bbb P^n$ is the space of all lines through the origin. Grassmannians are the higher dimensional analogue of that (the space of all k-dimensional subspaces of $\Bbb R^n$ or $\Bbb C^n$)
 
Mike Miller
Mike Miller
@PVAL: I think the sexier Milnor book is the Morse theory one, but that I suppose really is taste.
 
PVAL
PVAL
@MikeMiller I think many of the theorems in the notes are identical to theorems in the Morse Theory book with homotopy equivalence replaced by homeomorphism or diffeomorphism.
 
Mike Miller
Mike Miller
The second half is the sexy part. I agree that I prefer the handlebody approach to Morse theory as opposed to the cellular approach.
I am perpetually annoyed that people pass to cells.
 
L33ter
L33ter
I see
where do normally Grassmannians appear ?
i.e what part of math ?
 
PVAL
PVAL
12:53 AM
Classification of vector bundles is the obvious one.
 
Mike Miller
Mike Miller
Geometry (algebraic, differential). Topology.
 
L33ter
L33ter
cool
 
PVAL
PVAL
I'd suggest a source though I think I'd take more flak for suggesting it then suggesting Milnor's notes to Balarka.
 
Mike Miller
Mike Miller
lol.
 
PVAL
PVAL
@MikeMiller Milnor proves strong Whitney embedding, Poincare duality, h-cobordism, that every contractible manifold bounding a smooth S^4 is diffeomorphic to $D^5$ and every contractible manifold bounding a topological $S^4$ is homeomorphic to $D^5$. Those notes are just full of "wow" results.
 
L33ter
L33ter
1:06 AM
I can't wait to understand all of that stuff
hopefully by before 2017 I will be able to go into such stuff and understand them
 
Mike Miller
Mike Miller
@PVAL: I read them before, so I remember those. I just didn't remember the setup.
 
PVAL
PVAL
I don't even know of another account of Whitney's theorem outside of the original papers and those notes.
 
Ted Shifrin
Ted Shifrin
2:04 AM
@MikeM @Balarka: You're right. I was giving (or helping Balarka find) a counterexample when the submanifold $Z$ is not closed. ($f\colon X\to Y$, transversality to $Z$, submanifold of $Y$) We can easily keep the domain compact by making it an ellipse tangent to the $x$-axis. We can modify the example, pretty easily, to mess up compactness of $X$ and making $Z$ closed.
 
mixedmath
mixedmath
@quid do you have strong feelings for math.stackexchange.com/questions/1554756/… to be on matheducators?
 
L33ter
L33ter
@TedShifrin
 
Ted Shifrin
Ted Shifrin
now what, Karim? :)
 
L33ter
L33ter
can you explain to me laurent series ?
 
Ted Shifrin
Ted Shifrin
it's like Taylor series but what happens when you have a pole (or essential singularity) or want to work with a function that's holomorphic outside a certain disk ...
I don't know what you mean by "explain" :)
 
L33ter
L33ter
2:12 AM
I see
I just want some quick intro
 
Ted Shifrin
Ted Shifrin
Well, that was a quick intro. I can't type you a lecture here.
 
MickLH
MickLH
Oh ted, don't be shy
 
Ted Shifrin
Ted Shifrin
glares at Mick
 
L33ter
L33ter
I want to understand laurent series and poles
 
MickLH
MickLH
I think everyone wants to understand poles.. I mean seriously why do people put diacritics in their language.
 
Ted Shifrin
Ted Shifrin
2:16 AM
Karim, you understand $1/z$ and $1/z^5$ ... and $e^{1/z}$ has an essential singularity.
I love accents in lots of languages, @MickLH, so stop your complaining :D
The trickier thing with Laurent series is to give the expansion for something like $\dfrac1{(z-1)(z-2)}$ in the annulus $1<|z|<2$.
 
L33ter
L33ter
what is the defn of essential singularity @TedShifrin ?
 
Ted Shifrin
Ted Shifrin
It's like a pole of infinite order ... infinitely many nonzero negative terms in the Laurent series.
 
L33ter
L33ter
I see
I will see this video youtube.com/…
 
Ted Shifrin
Ted Shifrin
OK, so my work is done. Time for me to cook dinner. Bubye.
 
L33ter
L33ter
cya !
 
Mike Miller
Mike Miller
2:33 AM
@PVAL: Me neither, but if someone tells you the idea, it's not hard to carry it out. It's just that the idea is very smart.
 
L33ter
L33ter
lets say I want to compute the laurent series of sinz
is it just the taylor series?
 
anon
anon
what do you think and why?
 
L33ter
L33ter
1 sec I am just reviewing the definition now need to understand some stuff
we haven't introduced it in class yet
but I would like to understand it
before class
 
L33ter
L33ter
2:58 AM
oh I see I understand now
@anon the reason both laurent series and taylor series agree is that we don't encircle a singularity
 
anon
anon
because 0 is not a singularity specifically
 
L33ter
L33ter
yeah
what about sin(1/z) it will just be the laurent series on domain $\theta = \{z \in C : z \neq 0\}$
in this case it will also agree
as 0 isn't in our domain
 
L33ter
L33ter
3:20 AM
very nice videos
 
 
1 hour later…
L33ter
L33ter
4:45 AM
:D
finally understood poles and all of that stuff
 
Mike Miller
Mike Miller
5:17 AM
@anon suppose I have two unitary representations, conjugate via a matrix in $SL_2$. are they unitarily conjugate?
to be precise they're reps to SU(2)
on the level of matrices specht's theorem says yes
 
L33ter
L33ter
if I have $f(z) = \frac{1}{z^3(z + 4)}$ the "singular points" are the singularities yes?
 
Rajesh Dachiraju
Rajesh Dachiraju
6:12 AM
Hi @all
 
Enjoys Math
Enjoys Math
6:50 AM
There's a result saying that for some large number $K$ there is an infinite number of twin primes separated by $2K$. Does this imply that larger separations are in infinitude as well?
 
Alan
Alan
peeks around Anyone around for a topology question? Been too long
 
Mike Miller
Mike Miller
i can try
 
Alan
Alan
Is it sufficient to show something is a homeomorphism that it be bijective and that it send basic open sets to basic open sets? That seems to be what this book is implying...
But I'm not seeing how that gets us that the inverse image is continuous.
(tries to find where he put his Munkres Topology book.....)
 
Mike Miller
Mike Miller
any open set is a union of basic open sets, so if f(B) is open for all basis elements B, then f(U) = f(union B_i) = union f(B_i) is a union of open sets, so is open
so f is an open map; and then what you're saying is that open bijections are homeomorphisms, which is true
 
Alan
Alan
ahh, right. Doh. Thanks :)
So don't even need that the targets are basic open sets, just that they are open when the input are basic open sets.
 
Mike Miller
Mike Miller
7:07 AM
yup
 
L33ter
L33ter
nice
 
Alan
Alan
Now to show that these sets actually form a basis for the same topology. X compact Hausdorff, the claim is that for each continuous real valued function f in C(X), U_f being the set of all points where f does not vanish forms a basis for X.
 
L33ter
L33ter
@MikeMiller which year are you in ?
 
Alan
Alan
(That these U_f form a basis for the X, that is).
I know the general rule to show it forms A basis is to show that every element of X is in at least one, and that if x is in the intersection of two basic open sets, there's a third baisic open set that fits inside both that contains x....how do I show that this is actually the same topology as X. Show that each are finer than the other?
oh, doh
nm, just show that any open set is the union of basic open sets. Right
 
MathNovice
MathNovice
I am wondering if the following integral is conv or div $\int_a^{\infty} e^{-\imath x} dx$, (a is constant)? please comment/help.
 
Mike Miller
Mike Miller
7:22 AM
@L33ter: I'm a second year grad student.
 
Alan
Alan
Hrm, tried clicking the latex in mathjax button on side, and it's not working.
 
Mike Miller
Mike Miller
It should open a new tab, with instructions on how to activate TeX.
It only works on things encased in dollar signs like $\int_a^b$.
 
Alan
Alan
there, got it
 
Mike Miller
Mike Miller
@Alan: Right. I think this is the better way of thinking of a basis (as opposed to the roundabout one you alluded to): it generates the topology in the very obvious sense of generate, just like you can pick a generating set for a group, say.
 
Alan
Alan
Umm...that looks divergent to me by instinct, since $e^{-ix}$ is just tracing around the unit circle
 
Mike Miller
Mike Miller
7:24 AM
@MathNovice: Well, what does that mean? It means, precisely, $\lim_{t \to \infty} \int_a^t e^{ix}dx$. Can you explicitly compute this integral?
 
MathNovice
MathNovice
I did, and I find it divergent
 
Alan
Alan
There's pretty much no way for that to converge, since your "limit value" isn't going to 0, it's circling around the origin
 
MathNovice
MathNovice
for the reason $lim_{t \to \infty} e^{\imath t} $ does not exist
 
Mike Miller
Mike Miller
So you have your answer :)
 
MathNovice
MathNovice
I meant $lim_{t \to \infty} e^{-\imath t} $, with the negative
 
Alan
Alan
7:27 AM
Same with positive or negative
still just tracing around unit circle, just in opposite directions :)
 
MathNovice
MathNovice
a set of notes I am consulting doesn't seem to bel that
 
Alan
Alan
Throw it into Euler's formula $e^{-it}=\cos t -i\sin t$
 
MathNovice
MathNovice
here is what they write
$1/ 2 pi \int_{-\infty}^{\infty} \Phi(u) [\int_a^{\infty} e^(-i u x) dx] du$ = 1/2 + 1/pi \int_0^{\infty} Re[\frac{e^(-i u a) \Phi(u)}{i u}] du$
$\Phi$ is pdf of a r.v.
 
Mike Miller
Mike Miller
$$1/ 2 \pi \int_{-\infty}^{\infty} \Phi(u) \left[\int_a^{\infty} e^{-i u x}dx\right] du = 1/2 + 1/\pi \int_0^{\infty} \Re\left[\frac{e^{-i u a} \Phi(u)}{i u}\right] du$$
I have nothing intelligent to say about this.
 
MathNovice
MathNovice
Why I don't see mathematics displayed, I did run ChatJax, and MathJax
 
Mike Miller
Mike Miller
7:34 AM
Did you run them on this page?
 
MathNovice
MathNovice
Yes did it again now. it refereshed but still the same
got the maths display
 
Alan
Alan
What kind of value is u?
oh, real
 
MathNovice
MathNovice
$-\infty < u < \infty$
 
Alan
Alan
Must have something to do with the properties of that r.v., no clue
 
MathNovice
MathNovice
sorry $\Phi$ is the Characteristic Functional actually
 
Alan
Alan
7:38 AM
Hmm....So I have an open set in a compact hausdorff space. I'm trying to make a continuous real valued function be 0 outside the set and not zero inside....
No clue what the characteristic functional is
hits head, trying to think if charctersit function would work
nope, not that simple, cause just the stuff outside might not be open.
 
Mike Miller
Mike Miller
7:52 AM
@Alan: This is the sort of thing I might expect you need more regularity than Hausdorffness for.
But I'm no expert, just guessing.
 
Alan
Alan
Yeah....all I have is compact hausdorff, alas. Actually what I have is the intersection of the sets of where two continuous functions don't vanish.
and I need a new continuous function that wherever it doesn't vanish, neither of the two starting functions vanish.
Tried doing algebra tricks like $f^2+g^2$, etc.
Putting that on pause, and trying the next part, which I think requires me to show that for an open set $U\subset X$, if $x\in U$, then there exists a continuous function f such that $x\in U_f \subset U$, where $U_f=\{x \in X:f(x)\ne )\}$
err, not equal to 0, that is
 
Mike Miller
Mike Miller
I'm told partitions of unity do exist for compact Hausdorff spaces. I guess this is maybe what you're proving.
 
Alan
Alan
Hmm....maybe, I was very lousy in multivariable analysis when we went over that. goes to peek
doh, someone suggested $h=fg$ for the first queston.
 
Mike Miller
Mike Miller
They should be separate constructions. The one you know from analysis is valid for subsets of R^n and uses, say, an integral to construct them. A construction for a compact Hausdorff space certainly can't use derivatives or integrals.
 
Alan
Alan
Yeah.
 
 
1 hour later…
Alan
Alan
9:00 AM
Sheesh, finally finished this problem. You'd think I'd have done it at some point before 7 hours before I'm supposed to present it, for 30% of my grade :)
 
quid
quid
@mixedmath In principle the question could be alright. But it has several answers here, some very highly voted here, I rather not want all this moved to MESE.
 
Balarka Sen
Balarka Sen
@PVAL Thanks, but I think I kind of agree with @MikeMiller that I should learn more G&P before that. Prof actually told me to read first few chapters of G&P and jump to Milnor's Morse theory book, but I don't think I'd follow that advise either. I like to be thorough at what I learn :) Thank you for the link, though!
 
Frank Science
Frank Science
@TedShifrin I was stupid. In fact, for example, consider $f_n$ is a sequence of constant functions (trivial case), or multiplying a sequence of constant functions with a fixed holomorphic function. But I still doubt whether things like Abel-Dirichlet will be effectively used in holomorphic settings. Oscillatory stuff, like oscillatory integrals, to me, it's more a real-variable technique.
 
Balarka Sen
Balarka Sen
I was asking that question merely out of curiosity - if it requires facts which I don't know yet (Riemannian manifolds, etc) then I probably wouldn't learn the proof right now.
 
Frank Science
Frank Science
@BalarkaSen It seems to me that Milnor's Morse Theory is quite self-contained, but you need to be familiar with some properties about homotopy groups.
 
Balarka Sen
Balarka Sen
9:07 AM
Oh, that I know to some extent :)
But do I not need to know something about smooth manifolds?
I hardly know anything.
 
Frank Science
Frank Science
You needn't.
And a bit Riemannian Geometry is taught in that book.
It's a wonderful book, so thin but still informative.
 
Balarka Sen
Balarka Sen
I'll keep that in mind, thanks. But maybe I'll stick to non-advanced stuffs for now :)
 
Frank Science
Frank Science
Most part of Milnor's book isn't advanced.
Except that of Bott periodity.
I was suggested to read this book in the junior year. It's okay for (strong) sophomores to read it.
Personally, I think it's better to read Milnor's Morse Theory and Topology from a Differential Viewpoint, in a very early year. Even if we cannot understand much, we will appreciate how thing could be treated very beautifully.
It's very difficult to get things understood, but for appreciating a masterpiece, it's not that difficult.
There is a list of great writings.
You can see that three of Milnor's books are suggested with ($\times2$).
@MikeMiller But locally compact Hausdorff spaces are regular, and compact Hausdorff spaces are normal, where there's Urysohn lemma. A direct proof is explained in Rudin's Real and Complex Analysis. @Alan
 
Balarka Sen
Balarka Sen
9:29 AM
@FrankScience Ah, I see. But Milnor's differential topology book does not have any exercises?
Thank you for that list, btw.
 
Frank Science
Frank Science
@Alan And it might be better to prove Urysohn's lemma on your own. Here is a material online, if you can read French.
@BalarkaSen Milnor has no book on differential topology, but a lecture note. That Topology from a Differentiable Viewpoint contains exercises at the end of the book.
 
Balarka Sen
Balarka Sen
Ok, so topology from a differential viewpoint is a lecture note? I wasn't aware.
 
Frank Science
Frank Science
No, it's a book, but not definitively a book on differential topology.
 
Balarka Sen
Balarka Sen
Yeah, but I have heard the exercises are less harder than G&P's, thus I was recommended to read G&P instead of that.
@FrankScience Oh?
 
Frank Science
Frank Science
No idea on G&P. I was recommended to read masterpieces, or books from masters, so Spivak's book on differential geometry was deprecated.
 
Balarka Sen
Balarka Sen
9:37 AM
Unfortunately, just reading theory and not doing hard exercises is not my thing :) I can't understand the theory if I don't use it in proving something.
So I usually look for books with lots of exercises.
 
Frank Science
Frank Science
You can try to prove theorems on your own.
Anyway, well-organized exercises are an important part of a book.
 
Balarka Sen
Balarka Sen
Eh, there's a difference between proving theorems by my own and computations, I think. But yeah, proving theorems by oneself helps - I have tried it, and when the idea clicks, it feels great.
 
Frank Science
Frank Science
And I should admit that I cannot prove every theorem on my own for any book I've read.
 
Balarka Sen
Balarka Sen
Me neither :) I have tried it for a very small proportions of the theorems I have studied.
And I don't think I did any of them by myself. Maybe someone told the idea before, and I recalled it.
 
Alan
Alan
10:21 AM
I will go back through Urysohn's Lemma when I have more time....between semesters I'm scheduling myself for a crash review in topology, since it's been a while since I did General Topology and I'm doing Algabraic topology next semester
 
SadStudent
SadStudent
Hi
 
Alan
Alan
Greetings
 
SadStudent
SadStudent
Is there any source that I can learn maths from the beginning? I am a beginner but I really love maths and want to learn more about it.
 
Alan
Alan
That depends on what you mean by "From the beginning."
Some might argue you should start with Euclid's the Elements....I'm not one of those :)
Also, what do you mean by 'maths'? Beyond the very basics, the word "mathematics" is as precise as the word "Science"....there's tons of separate fields
 
SadStudent
SadStudent
Well, what you mean by very basics might be complex for me
generally, subjects like equations, elementary numbers, functions...
 
Alan
Alan
10:28 AM
A good starting point for how to think mathematically would be Polya's "How to solve it"
It looks like you're talking about things at a basic algebra level, for which pretty much any algebra text should do....online, you can google paul's online math notes, he has an algebra section
 
Overflowh
Overflowh
Good morning
 
I'm mostly just an idiot
I'm mostly just an idiot
Hello @Overflowh.
 
Overflowh
Overflowh
Hey @I'mmostlyjustanidiot
LOL, this turned out to be funnier than I though
 
I'm mostly just an idiot
I'm mostly just an idiot
@Overflowh What turned out to be funny? Are you Jasper Loy?
 
Alan
Alan
peers over this proof he finished The proof was that for a compact hausdorff space, the space is homeomorphic to the subspace of the zariski topology on the ring of real valued functions on that space consisting of the maximal ideals. All I used in the proof was that it was completely regular and normal though....curious if compact hausdorff was actually needed, since that is stronger than CR+N, right?
 
Overflowh
Overflowh
10:31 AM
Put a comma after the "hey" and read it out loud xD
 
I'm mostly just an idiot
I'm mostly just an idiot
@Overflowh Haha, yes, that wasn't the intention when I set the name however.
 
Overflowh
Overflowh
@I'mmostlyjustanidiot Sure don't worry, I didn't mean that :P
 
I'm mostly just an idiot
I'm mostly just an idiot
@Overflowh The name was to indicate that I'm mostly just an idiot.
 
Overflowh
Overflowh
@I'mmostlyjustanidiot Aren't we all? If you consider the amount of things (in general) that a single human knows compared to the amount of things to know :P
 
I'm mostly just an idiot
I'm mostly just an idiot
@Overflowh I guess so, that's my hope anyway.
 
Overflowh
Overflowh
10:36 AM
@I'mmostlyjustanidiot Hope that you are not alone, or that we are all idiots?
 
Balarka Sen
Balarka Sen
@Alan I am pretty sure compact Hausdorff is needed up there somewhere for that particular proof. Anyway, here's a generalization: it is true that if $V$ is an affine algebraic variety, $V$ is homeomorphic to $\text{mSpec} k[V]$.
 
I'm mostly just an idiot
I'm mostly just an idiot
@Overflowh Both(I think they may be equivalent:D)
 
Overflowh
Overflowh
Yeah I think you're right :p
 
SadStudent
SadStudent
Thank you, Alan.
Is this chat place somewhere I can ask math questions that I am struggling with?
 
Antonio Vargas
Antonio Vargas
@SadStudent, from the room description: Associated with Math.SE; for both general discussion & math questions alike. Just ask; don't ask to ask.
 
SadStudent
SadStudent
10:44 AM
Oh thank you :)
 
Antonio Vargas
Antonio Vargas
👍
 
I'm mostly just an idiot
I'm mostly just an idiot
Does an infinite dimensional vector space still have a basis, or some other concept comes in to play in this case?
 
Alan
Alan
If you accept the axiom of choice, yes
 
Antonio Vargas
Antonio Vargas
That doesn't really look like a thumbs up, but it is.
 
Alan
Alan
Every infinite dimensional vector space having a basis is equivalent to AC
 
I'm mostly just an idiot
I'm mostly just an idiot
10:46 AM
So I could say that my infinite dimensional vectorspace has basis $\{v_1,v_2,\cdots\}$
 
Alan
Alan
not one that you can index by the natural numbers
that implies the basis is countable
 
I'm mostly just an idiot
I'm mostly just an idiot
Okay, thanks, that makes sense.
 
Alan
Alan
You can say it has a basis $\{ v_\alpha \}_{\alpha\in I}$ for some index set I
Calling it a basis means that there is a FINITE Linear combination for every vector, fyi.
 
I'm mostly just an idiot
I'm mostly just an idiot
I wondered why I had seen those indexing families, now I understand the motivation.
 
Alan
Alan
Now, if your vector space happens to be a Hilbert space (A complete inner product space), and separable, then there's also a notation of a Hilbert basis, which allows for infinite sums
Yep :).
 
I'm mostly just an idiot
I'm mostly just an idiot
10:48 AM
I am trying to show that if $E$ is a finite dimensional subspace of a normed space $X$, then $E$ is closed.
From memory I have proven that $E$ being a finite dimensional subspace of a normed space, means that it is complete.
 
Balarka Sen
Balarka Sen
@Alan If $x \neq y$, $\mathfrak{m}_x$ and $\mathfrak{m}_y$ are distinct. So there is a function which vanishes on $x$ but not on $y$. Let $f$ be such a function. Pick disjoint nbhds of $f(x)$ and $f(y)$. Take preimage to get disjoint nbhds of $x$ and $y$. There you have Hausdorffness. And my proof of surjectivity definitely uses compactness.
 
Alan
Alan
This is definitely true, and in this case, you can take your arbitrary basis, and then since $E$ is finite dimensional, there are a finite number of bais elements in them
 
Balarka Sen
Balarka Sen
If you haven't used Hausdorffness somewhere, your proof is thus definitely wrong.
 
Alan
Alan
Oh right, the surjective needs compact, forgot that part
Check, thanks ;)
 
Balarka Sen
Balarka Sen
Sorry for the late reply, it took me some time to reproduce the proof and I don't have Atiyah-MacDonald in front of me.
 
Alan
Alan
10:51 AM
no problem :).
 
Balarka Sen
Balarka Sen
Sure thing. Are you working through A-M?
 
Alan
Alan
And yep
 
Balarka Sen
Balarka Sen
Cool. You saw the generalization (well, analogue) for varieties I mentioned up there? That was the next exercise, I believe.
 
Alan
Alan
I know the next exercise introduced the varieties, I'll go look at it in a sec
Pondering the vector space problem of I'm mostly's, and wondering why my brain is fried :)
 
Balarka Sen
Balarka Sen
Oh, right, you won't know about varieties yet.
 
Alan
Alan
10:53 AM
Actually been through artin rings, this is the end of the semester 'project' where I'm presenting on a longer problem
though we skipped a bunch of things, including varieties
and primary decomposition
 
Balarka Sen
Balarka Sen
Infinite dimensional vector spaces are screwed.
 
Alan
Alan
laughs THat's going to be my area of research. Yeah functional analysis
I start research next semester, finishing up Kryzeig's Functional Analysis this semester. Gulp
 
Balarka Sen
Balarka Sen
Boo, varieties are cool stuff. Commutative algebra should be introduced using varieties, it makes the geometry apparent.
@Alan oh, lol, no offense then. I don't know any functional analysis :)
 
Alan
Alan
grin Oh, they are messed up spaces
 
I'm mostly just an idiot
I'm mostly just an idiot
So $E$ is complete with basis $\{v_1,v_2,\cdots,v_n\}$. I guess I just need to get a cauchy sequence to converge on every point in $E$.
 
Alan
Alan
10:55 AM
I actually took commutative algebra this semester as a somewhat sidetrek
One look at the Tor functor confimed that I don't want to continue :)
 
Balarka Sen
Balarka Sen
I kind of hate that no books starts by talking about varieties and sheaves on varieties. That's a major motivation for a lot of commutative algebra for me.
 
Alan
Alan
that was for the vector space one
 
Balarka Sen
Balarka Sen
Varieties are an algebraic analogue of complex manifolds. A variety over field $k$ is intersection of zero locus of a bunch of polynomials on $k$.
 
I'm mostly just an idiot
I'm mostly just an idiot
@Alan People giving me hints here is infinitely better than seeing the full proof :D>
 
Alan
Alan
nod I also don't understand manifolds at all. NExt semester I'm sitting in on the class in multivariable analysis, because when I took it, I didn't get it at all
True :)
 
Balarka Sen
Balarka Sen
10:58 AM
Zero locus of 1 single polynomial $f$ in $k^n$ is called a hyperplane. Every variety is intersection of (possibly infinitely many) hyperplanes.
 
Alan
Alan
Use the limit point definition of closed (That it contains all of its limit points)
 
Balarka Sen
Balarka Sen
Now, a lot of theorems can be interpreted geometrically using varities. E.g. Hilbert's basis theorem is equivalent to saying any variety is intersection of finitely many hyperplanes.
Cool, eh?
@Alan Ah, manifolds are nice spaces. Yes, I guess motivation for them comes from the implicit function theorem.
 
I'm mostly just an idiot
I'm mostly just an idiot
@BalarkaSen Where should I learn algebraic geometry?
 
Balarka Sen
Balarka Sen
@I'mmostlyjustanidiot Lurie's Higher Algebra.
Kidding, I don't know algebraic geometry.
 
I'm mostly just an idiot
I'm mostly just an idiot
Aren't varieties algebraic geometry?
 
Balarka Sen
Balarka Sen
11:00 AM
Yes, but of the elementary sort.
 
Alan
Alan
Yeah. The problem I had was the teacher used a completely different set of notation than the book we used for manifolds, then used erase and replace a lot to make it impossible to follow, and went very, very fast.
so hopefully a second sit through the class will help.
 
Balarka Sen
Balarka Sen
And Tor functors aren't so bad :)
But again, motivation comes from algebraic topology.
 
Alan
Alan
This semester was the first time I started to get into particularly funky vector spaces. Like Sobolev spaces, or the topology on compactly supported smooth real valued functions on an open subset of $\mathbb R ^n$
The actual definition of the open sets on $C^{\infty}_0(\Omega)$ is weird
 
Balarka Sen
Balarka Sen
Oh, I don't know anything about Sobolev spaces. I have heard they are useful.
 
Alan
Alan
They are....You know $L^p$ spaces?
(THe basic lebesgue integrable spaces...)
Sobolev spaces show up in PDEs a lot. You get functions which aren't actually differentiable, but have a "weak" derivative that allows them to satisfy PDEs.
 
Balarka Sen
Balarka Sen
11:09 AM
@Alan Nope.
My real analysis is sort of weak.
@Alan ok, I see.
 
Alan
Alan
nod $L^p(\Omega)$ is the set of all functions $f$ such that $\int _\Omega f^p=L<\infty$
ie, the p'th power of the function is integrable in the Lebesgue sense
 
Balarka Sen
Balarka Sen
mhm, ok.
 
Alan
Alan
Sobolev spaces are subspaces where the functions also have weak derivatives up to the k'th order....weak derivatives are unique up to sets of measure 0, and are defined as making the integration by parts formula work against all "test functions", which are smooth compactly supported functions
 
Balarka Sen
Balarka Sen
Not sure I understand that. Can you elaborate on the definition of weak derivatives?
 
Paradox 101
Paradox 101
If we have an euler-cauchy differential equation which is : $x^2 y"+xy'-ny=0$ how do we get an answer in terms of $cosh$ and $isinh$? The values of $r$ I get are $\sqrt n $ and $-\sqrt n $ which should give a simple answer in terms of $x $ and its coefficients. What am I doing wrong?
 
Alan
Alan
11:18 AM
to keep it simple, go with the one variable case.
we say $g$ is a weak derivative of $f$ if for all smooth, compactly suported test functions $h$, $gh'=fh$
or more accurately, that they integrate to the same funciton
$\int gh'=\int fh$ over the region
 
Balarka Sen
Balarka Sen
mhm, ok
 
SadStudent
SadStudent
11:30 AM
A = {x: 20 ≤ x < 253, x = 4k, k is a positive integer}
B = {x: 25 ≤ x < 300, x = 6k, k is a positive integer}

How many components are there in (A U B)?
I keep finding 95 but the answer is 85
^^
 
Overflowh
Overflowh
Guys I have a question. The subject is Linear Algebra.
Take this thing:
$$U = \{(x,y,z,t) \in \mathbb{R^4}: x + y + z + t = 0, −x − y + 2t = 0, x + y − z = 0\}$$
$U$ is a linear subspace of $\mathbb{R^4}$.
What are exactly the equations after the colon? And what's their purpose?
And why there are three of them and not, say, 2 or 4?
 
I'm mostly just an idiot
I'm mostly just an idiot
@Overflowh These just describe the nature of the points in your set.
 
Overflowh
Overflowh
@I'mmostlyjustanidiot Which means? They are generators?
 
I'm mostly just an idiot
I'm mostly just an idiot
Most of the time your geometric understanding will fail with even things as simple as this.
@Overflowh Generators? No. They just say for some $(x,y,z,t)$ to be in your set, the positions in the 4-tuple must follow those rules.
As in just looking at the first rule, $x+y+z+t=0$, we could have $(1,1,-1,-1)$
But the other rules may 'forbid' this point.
(Which they do)
 
Overflowh
Overflowh
Oh, so I can use them to create vectors $\in U$?
 
I'm mostly just an idiot
I'm mostly just an idiot
11:39 AM
Yep.
 
Overflowh
Overflowh
How? I need to put them in a system to verify all three at the same time?
 
I'm mostly just an idiot
I'm mostly just an idiot
Specifically $U$ has all $4$-tuples that follow those rules, and take values in the reals.
Well you have $x+y+z+t=-x-y+2t=x+y-z=0$
 
Overflowh
Overflowh
Is that mathematically legal?
 
I'm mostly just an idiot
I'm mostly just an idiot
Why not? They all equal zero, so they all equal eachother. As long as you apply operations on all of them at once always.
 
Overflowh
Overflowh
Well, it is logically legal, but you can't solve it this way in order to get a 4-tuple.
 
I'm mostly just an idiot
I'm mostly just an idiot
11:43 AM
I.e. $a=b=c=d\implies a+q=b+q=c+q=d+q$
 
Overflowh
Overflowh
Oh cool, didn't know equations could have more than two members at the same time
Mh... It doesn't create vectors in that way anyway :/
 
I'm mostly just an idiot
I'm mostly just an idiot
$x=-y$
Or maybe I am just being dumb, I don't want to do any more hand computations while I should be doing a proof :D.
But if that is the case, then $z=0$
and $t = 0$
 
Overflowh
Overflowh
I don't get it :/ so yeah, I am definitely dumb
 
Paradox 101
Paradox 101
12:11 PM
@I'mmostlyjustanidiot have you done a course in differential equations?
 
SadStudent
SadStudent
is there a simple way to solve a problem with Euclid algorithm?
 
Paradox 101
Paradox 101
12:40 PM
Given the euler-cauchy equation: $x^2y''+xy'-ny=0$ how is its solution in terms of $cosh$ and $sinh$? The $r$ I get is $\sqrt n$ and $-\sqrt n$ so the solution should be simply $y=a x^{\sqrt n} + b x^{-\sqrt n}$ where $a$ and $b$ are constants but I don't get the actual answer. Can anyone let me know what am I doing wrong?
 
Paradox 101
Paradox 101
1:07 PM
@Huy can you check the above question?
 
Rajesh Dachiraju
Rajesh Dachiraju
1:52 PM
0
Q: Placing delta's at maxima, Is there any smart equation based expression?

Rajesh DachirajuLet $M$ be the set of maxima of the function $k:\mathbb{R}\to \mathbb{R}$. We define the function $$L(t) = \sum_{y\in M} k(y)\delta(t-y)$$ and there by the step function $$\Gamma(t) = \int_0^t L(\tau)d\tau$$. I'd like to know, if there is any single equation based (kind of closed form in terms o...

calculus real-analysis analysis distribution-theory
 
iwriteonbananas
iwriteonbananas
Hey @BalarkaSen @MikeMiller
 
r9m
r9m
@RonGordon: to say that your eccentricity is greater than $1$ would be hyperbolic. — robjohn ♦ Mar 6 '14 at 16:33
@robjohn LOL :P
 
robjohn
robjohn
@r9m just sayin'
@r9m Thanks for the feedback (and upvote?)
 
Balarka Sen
Balarka Sen
2:10 PM
hi @iwriteonbananas
 
r9m
r9m
@robjohn (+1)'ed :-)
 
robjohn
robjohn
@r9m the proximity in time sort of hinted at that :-)
 
r9m
r9m
@robjohn hehe! :-) cool answer!
 
iwriteonbananas
iwriteonbananas
2:25 PM
@BalarkaSen, you computed $H^*(\Bbb RP^\infty; \Bbb Z_m)$ before, right?
for $m>2$
 
Balarka Sen
Balarka Sen
I think so.
(I pinged you in the alg top chat)
 
Mike Miller
Mike Miller
@Alan You really, really want to do your best to work with sequences as opposed to open sets here, for sanity purposes. (Luckily the moment you pass to a different Sobolev norm you don't need to think about opens again, since now you have an actual metric...)
Morning @iwriteonbananas.
@FrankScience: I tend not to remember most of these point-set things.
 
Huy
Huy
@MikeMiller: I seem to get completions confused. Let's look at $C_c^\infty(M)$, $H_0^1(M)$ and $L^2(M)$. The completion of $C_c^\infty$ wrt. the $L^2$ norm yields $L^2(M)$, and the completion of $C_c^\infty$ wrt. the $H^1$ norm yields $H_0^1$, right? Is there any way to get from $C_c^\infty$ to $H_0^1$?
 
Mike Miller
Mike Miller
Remind me what your $H^1_0$ norm is?
I forget the different names here.
 
Huy
Huy
2:40 PM
I want to define it as adding the norm of the function and the norm of the differential.
 
Mike Miller
Mike Miller
ok
 
Huy
Huy
(or total derivative if you will)
 
Mike Miller
Mike Miller
Is $M$ a manifold with boundary?
 
Huy
Huy
no
we can also do it in $R^n$, I just want to do something analogous on a Riemannian manifold
 
Mike Miller
Mike Miller
Sure, you may as well work locally with open subsets of $\Bbb R^n$
 
Huy
Huy
2:42 PM
yeah
I think I need to even do a step in between iirc
iirc $H_0^1$ is the completion of $H^1$ wrt to the $H^1$ norm, where $H^1$ is the space with function and differential in $L^2$
is that right? or can I get from $C_c^\infty$ directly to $H_0^1$ with some completion?
 
Mike Miller
Mike Miller
@Huy: The way I remember it $H^1_0$ is supposed to mean functions that vanish on the boundary, where your domain is a bounded open subset of $\Bbb R^n$ with smooth boundary. I'm trying to prove that something in the completion w/r/t to the $H^1$ norm automatically has this property.
 
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