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00:00
@DanielCortild: Like consider $f(a,b,c,u,v,w)=\dfrac{ab+1}u+\dfrac{ac+1}v+\dfrac{bc+1}w$ subject to the constraint $u+v+w=4$.
Yes, but you can't rotate the function...
What do you mean?
@MatheinBoulomenos so we fix a projective variety X. Then $Z_d(X)$ is just the free group whose elements are generated by irreducible subvarieties of codimension d ?
Lagrange is when you make $u,v,w$ vary isn't it?
where the dimension is defined as the maximum length of ascending chains of irreducible subsets ?
00:01
And actually the constraint is $u+v+w\le 4$
Yes, but we should be able to argue that you minimize when you take the extreme case.
@Adeek yeah
cool
We actually then have further constraints that $u=(a+b)^2$, $v=(a+c)^2$, $w=(b+c)^2$.
Yea seems logical...
00:02
I dunno, @DanielCortild. I am no good at this stuff.
it's similar to the chains in cohomology theories in topology, if you've seen that
Yes, but those constraints just get us the the beginning case... :P
Ohh well... Thanks for trying :)
Mostly I don't like this flavor of math at all.
Unless I can see some geometric meaning to the question.
Wait did I hear cohomology theory?
But inequalities rarly have geometric interpretations...?
00:03
@Daminark :P
Lol jk I'll head back to the void
@MatheinBoulomenos I literally love your approach to every question I asked here, the explanation with the kernels was wonderful thank you. You always think about mappings!
@MatheinBoulomenos yeah I have seen those
@MatheinBoulomenos Yeah I am very familiar with algebraic topology
@Eran you're welcome
so essentially algebraic cycles your transfering the ideas of algebraic topology i.e cohomology theory to algebraic geometry ?
00:05
@Eran: I'm glad @Mathei was so helpful. And I'm sorry I wasn't helpful. Best fortune to you!! :)
@TedShifrin Thanks alot!
@MatheinBoulomenos do you know few things about Motivic cohomology ?
@Daminark Yup, algebraic cycles give you a cohomology theory
@Adeek yes, I'd say algebraic cycles are inspired by stuff like singular chains in algebraic topology
Those are called Chow groups
00:07
cycles, in fact.
yeah exactly @BalarkaSen
sorry @MatheinBoulomenos had a typo
It's very much motivated by intersection homology and the likes
I see
I've heard about motivic cohomology, but I don't know any details
yeah makes sense
00:08
It's a good exercise to figure out what homologous means for algebraic cycles, btw
I'm first learning étale cohomology, that's already difficult enough
@MatheinBoulomenos I have heard it is like a universal cohomology theory
I think intersection theory is easier than that stuff @Mathei
It's much more geometric in nature
that captures all cohomology theory
Oh, you mean motivic
hell if i know
00:09
yeah I wrote that @BalarkaSen
@Balarka aight that's motivation for me to actually try out algebraic geometry, let's go
@BalarkaSen i'm just happy to know what homologous means for complex analysis
Intersection homology is the bomb, @Balarka.
00:15
I know just enough about intersections to know that it's important
but the only actual result I know, I think, is the Picard-Lefschetz formula. And that's pretty esoteric
@Semiclassical You're give a picture frame and two pegs on the wall. How would you hang the picture frame on the wall in a way so that taking out any one of the pegs makes the picture fall? (The rope is frictionless)
something something trefoil knot?
Nobody except SemiC tells me the answer
we did intersection forms in our alg top course, but not more
oh. pochammer contour
00:17
Bam
aka why homotopy is usually overkill in complex analysis
@Mathei: Intersection homology was invented to make Poincaré duality work in singular spaces.
@TedShifrin do you know about this divisor map ?
@BalarkaSen Glue the pegs
What divisor map?
00:20
I am not sure I understand the notation in book I am reading
@MikeMiller lol
$div : \bigoplus (f,Z) \rightarrow Z^r(W)$.
they never defined this
I am not sure what this means
I bet they defined more than you're telling me.
they defined map as $div(f) = (f)_0 - (f)_{\infty}$ and $f \in k(Z)^{\times}$
$(f)_0$ seems like zeroes of f ?
Right, and $Z$ is an $(r+1)$-dimensional subvariety of $W$?
00:22
yeah
Yes, $(f)_0$ is the zero-divisor of $f$ and $(f)_\infty$ is the infinity-divisor of $f$.
$(f)_0$ is the zeroes of f and $(f)_{\infty}$ are the poles ?
So they're just considering these divisors as $r$-dimensional cycles on $Z$ and then setting them into $W$.
oh
oh okayyy
and what does it mean $f \in k(Z)^{\times}$ ?
rational function other than 0
00:25
ohhh
okayy cool
what does this give us though ?
why is this map useful ?
it's one way of generating $k$-dimensional cycles on $W$
divisors are always in abundance ... higher-codimension things are far more difficult.
I see
you can always intersect different divisors — that's called a complete intersection ... But lots of things are not complete intersections.
Is intersectional homology like a radical feminist version of homology?
I see
00:27
Good exercise for you: Take the twisted cubic $[1,t,t^2,t^3]$ in $\Bbb P^3$. Show that it is the intersection of $3$ quadric surfaces, but cannot be a complete intersection.
Obviously, I mean to close up that curve with the appropriate point(s) at infinity.
I haven't learned about intersection theory @TedShifrin
You don't need anything fancy. Just play.
what is complete intersection ?
oh okay I see
I was sloppy. A subvariety of codimension $k$ is called a complete intersection if it's given as the intersection of precisely $k$ divisors.
Back Grins
00:30
I see
So my curve in $\Bbb P^3$ is codimension 2, but you need 3 quadrics to define it.
0
Q: If $X_{n} \to X$ and $Y_{n} \to Y$ in probability, then $f(X_{n},Y_{n}) \to f(X,Y)$ in probability for $f$ continuous

ALannisterFor a bivariate, real continuous function $f:\mathbb{R}^{2} \mapsto \mathbb{R}$, I need to show that $f(X_{n},Y_{n}) \to f(X,Y)$ in probability whenever $X_{n} \to X$ and $Y_{n}\to Y$ in probability. This is essentially a bivariate version of the Continuous Mapping Theorem for convergence in pro...

I see
Suddenly I see! This is what I wanna be. Suddenly I see! Why the heck this means so much to me.
Suddenly I see! Suddenly I see! Suddenly I see! Suddenly I see! Suddenly I see!
00:34
OK, guys, this is getting truly annoying. As bad as Demonark and Balarka and their thonky memes.
Maybe I should retire.
Or that one guy who calls everybody handsome
I would agree, although I happen to think I'm handsome.
yeah I agree @TedShifrin
@TedShifrin In my lab reviews I got a comment about how some student wants to marry me
haha
00:39
Somebody left me a $10 gift card to Dunkin Donuts in my office anonymously
Some of my classmates wrote "great hairstyle" in the evaluation of a professor who's bald
That's harsh.
That's bold
No, that's bald.
I feel like I'm in a Monty Python sketch
2
00:41
Yes, this is a cheese shop.
I guess I've gotten some gifts from students at the ends of class, but never a marriage proposal. A date or two, though ... :P
But writing stupid comments in evaluations is really common here. For example, several people wrote "More experiments!" as a suggestion for improvement in the advanced analysis class and also in the funtional analysis class
haha
@MatheinBoulomenos germans are mean
lol
History would agree with you, but I won't go there
<---- mean, but not German
00:43
@TedShifrin yeah I always gave gifts to prof's who were very nice and explained things nicely in a class
This is a Monty Python sketch, not Fawlty Towers
I wrote "More topological vector spaces, more banach algebras, more non-archimedean functional analysis" as suggestions in the functional analysis class, obviously
one of the best profs I had was someone in classical mechanics who always used to say define your coordinate system they are not god given @TedShifrin
haha
Another long shot... how many of you are familiar with Fourier analysis AND time series?
more algebraic esoterica
00:44
Then you add in after all ofthat stuff @MatheinBoulomenos "More cowbell"
in other news, I got a ribbon on my head
See how many people get the reference.
@TedShifrin none of these are particularly algebraic or esoteric
non-archimedean functional analysis?
I'm starting to wonder if I shouldn't have posted that question on Stats Exchange instead.
00:45
oh, does that mean functional analysis over $\Bbb C$?
No
Over $\Bbb Q_p$ and stuff like that
That's what I figured.
One of my profs uses that in his research
oh
I love functional analysis
it would be cool to do research in the boundary between algebra, analysis, and geometry.
Some stuff (e.g. Hahn-Banach) even works over any completely valued field, so you can give one proof for $\Bbb R$, $\Bbb C$, $\Bbb Q_p$, $\Bbb F_p((t))$
00:47
not the boundary
intersection
@MatheinBoulomenos !!?
Honestly, topological vector spaces or banach algebras would have had to do more with functional analysis than proving the convergence of numerical methods for PDEs ...
can you approximate vectors using functionals in same way as you can in regular functional analysis ?
i.e we know with regular functional analysis we have $\|x\| = sup_{f \in X^* : \|f\| \leq 1} |f(x)|$
Functional analysis is super important for all sorts of PDE stuff.
yeah
Hodge theory as well
I learned all about unbounded operators studying PDE.
00:50
nice
algebraic geometry is super important for number theory, thus you have to do number theory in an algebraic geometry course?
Same line of reasoning
I don't see why you expect everyone to share your particular slants on things, @Mathei. It's quite annoying.
On that note, I'm out of here.
@TedShifrin Can we define spectrum over p-adic fields as well ?
@TedShifrin If I sign up for a functional analysis course, I don't expect to be doing numerical analysis
Bye @Ted!
Hope I didn't scare you away
Slutsky's Theorem
00:57
@ALannister What about it?
@ALannister sounds promiscuous
I just like saying it
@MatheinBoulomenos
it isn't as much fun as the Cox-Zucker machine
2
LOL!
That's right, this is why none of y'all can have nice things.
@MatheinBoulomenos do you know what faces meeting properly means in the context of algebraic geometry ?
01:03
@Adeek no idea
$\gamma \in z^r(w \times \Delta^m)$ where faces meet properly
@MatheinBoulomenos
01:36
Hi!
I don't recall that, in general, the smallest eigenvalue of a matrix is useful for something or particular in some sense.
Do you know of any particularity or application of this smallest eigenvalue?
I could see spectral graph theory having use for it
@nbro well it's the minimum of ||T(v)||/||v|| where T is symmetric
Actually Leaky does mention something I should clarify with, if we're dealing with a matrix that might have real or complex eigenvalues, and we say alright, let's take its smallest real eigenvalue and then see what it does, well, that might not have much useful info. It might, but I dunno
@Daminark Yes, exactly, but I didn't want to mention it, so that to not influence your answers. Anyway, it is part of the theory behind the inertial partitioning algorithm. My question is: why the smallest eigenvalue?
But if you have a matrix with all real eigenvalues, like symmetric matrices (e.g. adjacency), that I could buy
01:40
If we two discrete random variables they show the length of a product which are produced by two machines. The normal length is 450 mm but there are some deviations. I have calculated the expected value for each machine, and both are equal to 450 mm. And it is asked which machine is more reliable. Since the expected value is at each machine the normal length 450 mm, do we have to check the probability $P(X=450)$ at each machine?
We have these information: http://www.directupload.net/file/d/4941/7kpx6kn4_jpg.htm
I'm not terribly familiar with spectral stuff, I do know that the largest eigenvalue is involved in some bounds involving degree
@LeakyNun What do you mean by that index notation? What's that operation?
So I was just kinda like, aight probably smallest will come up somehow
@nbro T is a linear transformation
|| . || is norm
@LeakyNun Taking the norm of a linear transformation?
Do you mean $T v$, where $T$ is the matrix representing that linear transformation?
01:42
@nbro ...
So, he did write $\|T(v)\|$
I use A for matrix and T for linear transformation
Av and T(v)
Because $T(v)$ is a vector, so that makes sense
@Daminark Why those parentheses?
@nbro because it's a godforsaken function
01:43
You can write it as $Av$, where $A$ is a matrix in the applicable basis and $v$ is some vector written out in coordinates
Ok, I got it.
You're applying the norm to the result of the linear transformation.
yes, I am
It is simply a notation I have not seen often.
But you can also say $T$ is a linear transformation, so it takes in $v$ and spits out $T(v)$. That's a vector, norm it
@LeakyNun Stay calm dude, not everyone looks at linear algebra notation all day!
01:43
But it turns out you do have a notion of operator norm
@nbro I just confirmed what you said
btw ||T(v)||/||v|| has a limit at 0 iff T in Z
where Z is the center of the GL group
Which is $\sup_{\|v\| = 1} \|T(v)\|$
You can prove that it's a norm and everything
maximum abstraction
Lol, not my finest moment :P
Thanks for the catch
@LeakyNun Why?
01:48
@nbro all symmetric matrices are orthogonally diagonalizable
This is called the spectral theorem
Wow, you know a lot about linear algebra
:D
I don't recall having learned orthogonally diagonalizable.
So, roughly, how does that imply your initial statement?
ugh
learn more about linear algebra
I'm out
daminark may explain to you
express each vector as a linear combination of the eigenvectors
I asked "roughly" because I don't have time to learn more about linear algebra right now.
I wouldn't come to these chats if I had time to learn the material...
It'll be tricky to do this quickly but I will try
So two matrices $A$ and $B$ are similar if there's an invertible matrix $P$ such that $B = P^{-1}AP$
You can see right away that this is an equivalence relation
01:55
Oh, yes.
You say $A$ is diagonalizable if it's similar to a diagonal matrix
Now, one thing to note about similarity is this. If you think about an invertible matrix, all it really does is change one basis to another, right?
@Daminark Of course, I heard and used SVD, and it seems to be related. But by the name "spectral theorem" I was not connecting...
Spectrum of an operator is the set of $\lambda \in\mathbb{C}$ such that $\lambda I - A$ is invertible. Meaning, the set of eigenvalues
So that's where the name comes from
But moving on
So, because invertible matrices are change of basis, the idea is that similar matrices are really just representations of a linear transformation in two different bases
You take one set of coordinates, change them, do the transformation in those coordinates, and change back
Now, you say a matrix is orthogonally diagonalizable if that change of basis matrix is an orthogonal matrix
But back to the general notion of similarity, a lot of properties are invariant under similarity. In particular, the characteristic polynomial
Oh, ok
Now, a matrix is diagonalizable iff its linear transformation can be represented in some basis by a diagonal matrix
But that just means you should have a basis of the space consisting of eigenvectors of the transformation
(eigenbasis, for short)
Now, that transformation is $T(v) = Av$ for our matrix $A$ and vector $v$. Thing is, this is the representation of $T$ in the standard basis for $\mathbb{R}^n$. In particular, this is an orthonormal basis
So if our change of basis matrix is orthogonal, this means it takes the standard basis $e_1,\ldots,e_n$ to another orthonormal basis
02:04
Last night dream there was a strange real analysis problem in a maths fair:
Scenario: There is a maths fair where one of the stalls contains acrylic models of curve, and one girl then use a bunch of triangles to illustrate numerical differentiation in the form of newton's formula. Later on in a completely white cutscene, there's a smooth hill shaped curve which is then slowly filled with more finely divided trapezoids which is said to illustrate numerical integration in the form of trapezoidal rule. I then asked one thing that always confuses me a lot in real analysis is given some integral
Reality check: Obviously, me real life counterpart have no idea what my dream self is talking about given I have not even started real analysis, but if I read the dream correctly, it seemed to suggest there are smooth functions that cannot be integrated by the trapezoidal rule since the area given by the trapezoids will always underestimate the area under the curve
Whether it is the case remains to be checked by some googling...
Hey sorry I trailed, I'm somewhat busy right now @nbro
So the rest of the spiel is this, a matrix $A$ is orthogonally diagonalizable iff it has an orthonormal eigenbasis (there's an orthonormal basis of the space consisting of eigenvectors of $A$)
@Daminark Hey, thanks a lot dude!
In the spectral theorem, you use Lagrange multipliers to prove that this is true for a symmetric matrix
It always nice to recap these concepts.
So that's it. In particular, adjacency matrices are symmetric, so you can spectral theorem them
02:16
@Daminark You meant not invertible, right?
@Daminark So, in the case of $A = PBP^{-1}$ the orthogonal matrix should be $P$.
Yeah
@Daminark I am not completely sure how this statement follows from your previous ones...
I need to go to sleep
I can't think now :D
Yeah fair. But when you can think, think about what a diagonal matrix involves
Meanwhile, thanks for your explanations!
:)
02:45
Almost forgot, in real life, newton's method is used to find roots, not to differentiate things numerically
But see Newton Raphson:
 
1 hour later…
03:47
Newton's method is nice in higher dimensions where there isn't a good alternative, but I prefer more stable algorithms for finding roots
like false position method
04:39
NO!
Mathematicians are NOT allowed to sleep.
2
???
Not when there's so much fun to be had...
@Secret!
05:30
Szechuan sauce
06:17
@Daminark
Can you get me some of that Szechuan sauce
I see Rick and Morty references
Lmao
07:03
I have a doubt:
Can $x^4+y^4=1$ have rational solutions ?
No
This is a special case of Fermat's last theorem
(The proof is elementary, though)
But that article there says something else
It cant so what do they mean by rational solutions
To be fair, there are some trivial solutions where $x=0$ and $y=\pm 1$ or $x=\pm 1$ and $y=0$
I don't see anything in the article that suggests that $x^4+y^4=1$ has nontrivial rational solutions
"if you plot (complex) solutions to the Diophantine equation x4 + y4 = 1, you get the three-holed torus. The rational points on this torus lack geometric structure"
It says that, could you tell me what that means
so you have a three-holed torus which consists of all the complex solutions which means that they have a nice geometric structure. Inside those complex solutions, the rational solutions are just 4 random points, so they don't have an obvious geometric structure which makes them easy to find among the complex solutions
07:12
@Albas From that article: "The set of rational solutions to an equation doesn’t have any symmetry and doesn’t form a group". So the writer has no idea what they are talking about
Okay.
And this is even after having mentioned Falting
That sounded funny , true
So what are they exactly trying to aim at in that article?
I was at a talk by Minhyong Kim on this topic
I wrote up some of my impressions in this reddit post: reddit.com/r/math/comments/7gxkk4/…
@MatheinBoulomenos Neat. So what does he actually do?
Ahh, that sounds cool
07:18
@TobiasKildetoft TLDR: he associates to each variety over $\Bbb Q$ a larger algebro-geometric object (an étale sheaf over $\operatorname{Spec}(\Bbb Q)$ to be precise) which is inspired by physics and he calls "space of gauge fields over that variety", he does the same thing with the base changes to $\Bbb Q_p$ of the variety and gets a nice commutative square which can actually compute all solutions to some diophantine equations
at least that's what he was talking about in the talk I attended. He gave other talks in which he talked about other aspects of the theory which I didn't attend
really cool stuff
While correcting the last one of the final homework for my course (the final one was graded by myself rather than the TAs since it was technically after the course had ended), I discovered an error in the assignment that technically made the statement false. Fortunately none of the students who had done that assignment had noticed this, so they had correctly solved the problem I had intended to pose.
projective plane is awesome
point-line duality :o
(it is descended from orthogonal complement in R^3, right?)
 
2 hours later…
09:01
Icarus
Should I stop?
Should I stop flying?
 
1 hour later…
10:10
Hello all! I have an assignment for school and wanted to pick peoples brains about possible areas of study to help me complete the project!
Anyone care to chat?
@DanB , from Hanuary or Hune ?
@mit
@MithleshUpadhyay Not sure I understand your question?
@DanB , starting month?
@MithleshUpadhyay For the project? It's a class assignment due by January
How long were you given to do the project?
10:15
@skullpatrol hix month
We got the assignment today
@DanB , xerox it and distribute to everyone
 
2 hours later…
12:04
We recently detected a massive anormally
AsafKaragila was known to not came to chat for almost an eternity, and then suddenly this happens
It is still unclear what result in this resurfacing and hence the anormally, but rest assured more data will arrive soon
So... with so many resurgence recently, it seems clear something big is going to happen
12:56
hi
$f \in L\ ^ 1(\lambda)$ and we define $g(x) = \int_x^1 f(t)/t dt$ , i need to show that $g \in L \ ^ 1 $. so i can show that $\int_0^1 f = \int_0^1 g$ but im not sure how to show that $g$ is measurable
i think it follows somehow from Fubini's theorem, but not sure how..
 
3 hours later…
15:34
1
Q: Spelling Out the Details that $[0,1]^\omega$ is not Locally Compact

user193319I am having trouble understanding gnometorule's answer to this question: Show that $[0,1]^{\omega}$ is not locally compact in the uniform topology, the uniform topology being induced by the metric $\displaystyle p(x,y) = \sup_{n \in \Bbb{N}} |x_n - y_n|$ Particularly, I am having trouble fo...

16:15
I'm a little bit confused. Let $A x = 0$ be a linear system with $A \in \mathbb{Q}^{n \times m}$. If this has a solution $x \in \mathbb{Q}_p^n$ for some prime $p$, does it follow that it has a rational solution $x \in \mathbb{Q}^n$?
both existence statements are equivalent to the determinant being zero
err, if it were square that is
my latex isn't on
I just found here "More generally, if a system of linear equations over some field $F$ has a solution in its extension $E$, it also has a solution in $F$"
7
Q: System of linear equations having a real solution has also a rational solution.

TrancotI saw this question Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $b ∈ \mathbb{Q}^m$. Suppose that the system of linear equations $Ax = b$ has a solution in $\mathbb{R}^n$. Does it necessarily have a solution in $\mathbb{Q}^n$? and I thought I'd give an interesting, possibly wrong, approach t...

But I don't know how to construct the solution in $F$ given the solution in $E$
okay, write a basis for E as an F-vector space
so E^n = (F^n)b1 + (F^n)b2 + ...
a matrix with entries in F acts on each (F^n) component independently
in order to be zero in E^n, all the F^n components need to be zero
Do you need that $E$ is a finite extension of $F$? Is this the case for $E=\mathbb{Q}_p$ and $F=\mathbb{Q}$?
no you don't need it to be a finite extension, no Q_p/Q is not a finite extension (it's an uncountable extension I think)
nix the "I think." internet connection interrupted, edit window elapsed.
16:32
So you are saying that if $x \in \mathbb{Q}_p^n$ is a solution of $Ax=0$, then any component of it (considered as vector in $\mathbb{Q}^n$) will be a solution of $Ax=0$?
yes
Thats supprisingly simple :)
And this argument works for $\mathbb{Q} \subseteq \mathbb{R}$ too? It requires axiom of choice, right?
I can't really write down a basis for $\mathbb{R}$ over $\mathbb{Q}$
the argument uses choice for a basis, yes
Interesting, I wonder if there is a pratical algorithm to get from a solution in $\mathbb{R}$ to one in $\mathbb{Q}$ then
hmm
well, there are algorithms for solving it in the first place, so that would technically count as such an algorithm even if it doesn't use the solution in R^n. you'd have to outline in what way you want it to depend on the one in R^n, and presumably there is no way for that to be done in a way that depends "continuously" on the solution
16:43
trying to remember how to do \sum with two arguments on the bottom. substack or something?
test: $\displaystyle \sum_{\substack{a\\b}}$
hrm
okay, that works
did someone just answer his own question
17:28
Hey all! I was wondering if my question lacked clarity? Would be totally up for some constructive feedback?
https://math.stackexchange.com/questions/2573290/variant-of-gilbreaths-conjecture-implications-for-a-series
00:00 - 18:0018:00 - 00:00

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