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02:00
$C^N$ is just $R^{2N}$
I don't know anything about this stuff but I've heard that every smooth manifold can be made into a real analytic manifold
together with some extra structure
So I'm gonna guess that this means not all real analytic manifolds are necessarily complex manifolds
yeah but we can't embed compact complex manifold into $R^N$ for any N.
I am little bit confused
Well, they said for some N
02:01
@SimplyBeautifulArt your example doesn't work
@Daminark for any N we can't embedd compact complex manifold into it.
Wait I don't have any idea what you're talking about
What you said is that given an analytic manifold, you can embed it into some R^N
@Daminark so we can embedd any real analytic manifold into $R^N$ for some N. given any real analytic manifold.
What's wrong with that?
02:04
In particular given real analytic manifold of even dimension we can embedd it into $R^N$
i.e we can embedd complex manifold into $R^N$ I am little bit confused
I mean it would be a "real embedding" maybe?
ohhh
Like okay you can embed into C^n
okay yeah that would make sense yeah
sure
And then just be like, alright now we shed the complex structure and just do R^2n
02:06
sure yeah sorry for this silly thing
It's aight
oh wait
@Daminark it is real analytic embedding
the theorem is real analytic embedding
oh yeah
but it is not necessarily true that it admits it into $R2N$ for some N
@Daminark precisely we showed that any real analytic embedding must embedd into odd if the manifold is compact real analytic manifold.
odd M for $R^M$
Wait once you embed it into R^n, you can get every dimension larger than n
But just including R^n into R^m
So there's no requirement of oddness
No there is requirement
because
@Daminark we know that compact complex manifold can't be embedded into $C^N$ for any N. So in particular we don't have analytic embedding of complex manifold into $R^{2N}$.
Because compact complex manifold is in particular real analytic, so the theorem still holds, but we don't have embedding into $R^{2m}$ for some M, otherwise we would be have complex embedding. It must be odd.
02:19
1) I'm not sure you can extend real analytic embeddings into complex embeddings
analytic embedding is the same as complex embedding
2) The Riemann sphere can be embedded into C^2. Maybe it's not true that all complex manifolds embed
I mean they are the same notion
But the statement that none can is highly suspect
I mean analytic functions are functions which converges locally to f through the taylor expansion
02:21
@Daminark how are you embedding the Riemann sphere in $\Bbb C^2$?
you can't embedd Riemannian sphere into $C^2$
Real analytic and complex analytic are not the same though and Dami's (1) holds
Yea I agree (Riemann sphere; Riemannian might be interpreted as with metric as opposed to complex structure)
I am little bit confused isn't real analytic functions same as complex analytic ?
what is the difference?
02:23
think about that in the case of $\Bbb C \to \Bbb C$
yes
I mean does thinking about it just sitting in C^2 canonically not work?
$z \mapsto \text{Re}(z)$ is linear, hence very much real analytic
@Daminark canonically how
ohh
real analytic functions need not at all be complex differentiable, you still need to satisfy the CR equations
02:25
Oh wait hold on I'm dumb Liouville prohibits it. And I was thinking just the fact that it is a subset of C^2 as points of norm 1
That's S^3 :)
ohh
Oh... Okay this is embarrassing
Is there someone here who understands (polar) planimeters
oh complex analytic is that given any point we have an expression as power series. But, real analytic means that the function converges to f in a nbhd of all points to f..
02:27
No need to be embarrassed, I've made worse mistakes
Because I'm just getting more and more confused
Also Liouville won't kill it, max modulus might?
@Adeek power series in z vs power series in x and y
Are they allowed to read areas that surround the fixed point?
@MikeMiller ohhhhhhhhhhhh
ohhhh okayyyyy I get it noww
thanks a lot
this is a key point
yeah @Daminark
On compact complex manifolds functions out the manifold are locally constant you can prove that using max modulus.
02:29
@BalarkaSen Planimeters make no sense
and I need you to help me
Wouldn't it give a reading of 0 to a circle of radius $a^2+b^2$ about the origin?
'Cause the two sticks in the linkage would be perpendicular to each other, and you're just rotating it like that around the origin
and then the axis of the wheel is always parallel to the motion, so it just skids and doesn't roll
and so you're left with 0
Jan
Jan
Is anyone here familiar with matrix factorization?
What are some easy ways to constrain matrix factorization where the matrix has phase-specific information?
$\sqrt{a^2+b^2}$ I meant
DogAteMy: Re planimeter, see the exercise (with picture) at the end of section 8.3 ... :P
Oh, no, the wheel rolls fine in that configuration.
@Adeek: $x-iy$ is real analytic, but certainly not complex analytic :P
@TedShifrin I don't understand
Well, that's good. Neither do I :P
Oh, I see, you're pivoting on the origin, and not on the pivot of the apparatus, DogAteMy?
02:42
What?
Oh, if I'm doing a circle around the origin, don't both nonfixed things have to move?
The origin is the fixed point, as you've drawn it
So you're thinking about locking it with $\theta=\tau$. Then you'll get 0, yeah.
But you can't physically do that because there are actual metal bars (as pictured)>
No, $\theta$ moves
$\tau$ is locked at 90 degrees
Oh. Not what I was visualizing.
Or hold on, that's not what $\tau$ is
Sorry
You mean $\tau = \theta+\pi/2$.
02:45
I thought $\tau$ was the angle between the two sticks
@TedShifrin Yes
Yeah, that also doesn't work. The integral turns into something with $d\theta\wedge d\tau$.
So there are some physical limitations, yes.
You're right — that just drags the wheel along.
What do you mean it doesn't work?
OK
It works in somewhat generic conditions, but not always.
If you play with one physically, you see that.
Is it OK to just assume it doesn't wind around the origin?
I think it can wind, generically.
BTW, I also think #25 is pretty cool.
02:50
Why is this not generic?
Because you're fixing $\tau-\theta$ constant.
If you work it out, you actually need $\sin(\tau-\theta)\ne 0$ and $d\theta\wedge d\tau\ne 0$ to get a reasonable solution.
DogAteMy: As often happens with "applied" problems, you impose certain genericity conditions along the way ... :P
Out of here for now, DogAteMy. Take care.
03:40
hi
@AkivaWeinberger The .pdf I gave you does it :)
Well, uh, for CW manifolds. I don't know how Mike plans on generalizing it further to all manifolds.
Bottom line, this shit is not easy. TOP is much much harder than PL or DIFF
@MikeMiller Yup, I got that.
04:03
Hi, I have a quick question. I just finished papa and baby Rudin and I am looking for a book recommendation. Should I just go ahead and get myself grandpa Rudin or explore something else like topology instead of more analysis (I am a high school student and I will be going to university next year)?
Papa would be the next logical choice @RajivKaipa
He said he finished Papa Rudin
If you have studied baby Rudin thoroughly you don't need to know any more point-set topology
So there isnt any point in getting myself something like Munkres?
Can you construct a perfect subset of $\Bbb R \setminus \Bbb Q$?
04:07
A single point
Very not perfect, Dami
Have you serously mastered papa rudin? @RajivKaipa
I know you're making a dumb pun
Will there be any which is not a single point?
Are you sure? I'm not doing a pun here, it's a closed set such that every point in it is the limit of a sequence in it
04:09
Yes, there will be many. A single point is also not an example.
Or wait does it have to be a limit point in the sense of, it has to intersect a punctured ball?
@Daminark Yes, that.
In that case yeah it's no good
Empty set
Lol jk really though
I'm p sure you can construct a Cantor set that does it
Empty set is a valid example. Give another one.
@Rajiv If you really read Rudin you should be able to do this one.
This is the hardest exercise in chapter 2.
@Daminark in fact it's really easy to do this
04:13
There are many ways to do it actually
Construct the Cantor set the usual way by cutting out intervals from each component, but do it in such a way that the $n$th stage removes all rationals with denominator $n$
i can think of 3 off the top of my head, two involve cantor boiz
I guess you want to start with an interval whose endpoints are irrational as well
Alternatively, mess with the decimal expansion definition of the middle-thirds Cantor set
@Eric There's a secret proof given by taking limit set on $\partial \Bbb H^2$ of an appropriate group action on $\Bbb H^2$
one of them is super nonconstructive
@Balarka oh id like to see this
04:15
I'm using the word "decimal" in the generic sense of, like, writing real numbers in a base system. Not specifically the base-10 thing
My favorite way to do it is to take cantor cross cantor
@Eric Ok, let me recall this
that's the super nonconstructive one i was thinking of
@Akiva Yeah that works
Hello all!
Wait what do you do with Cantor cross Cantor
04:17
@Akiva That's homeo to Cantor
So send your rational bois to C x C
chuck out countably many {x_0} x C vertical fibers
you get uncountably many cantor sets in R - Q
all perfect
@BalarkaSen ?
in fact uncountably many which lies inside a given dude
Oh wait I remember Soug assigned a problem to us about this
My solution was a bit jank
Something like
04:18
@Akiva you flibbackadoob the rationals to C x C
is that clearer?
Do 0.0_0__0___...
Hi @CookieToast welcome :-)
Yeah I think that's the way Akiva was talking about too.
Right that was the decimal thingy I mentioned I think @Daminark
Where you fill the blanks with 1 or 2
I see
But yeah you can prove that's a perfect set and it's all irrational numbers
04:21
Yup, I clearly have work to do. See ya in a couple of months :)
Munkres should be a good book
Do read it.
Yeah I'll get that as well
@EricSilva Okay, so $SL_2(\Bbb Z)$-orbit of infinity on the boundary is all of the rationals, right?
So I guess you need a subgroup of SL_2(Z) without unipotents
@robjohn thanks :-)
Maybe I'm wrong
04:30
$T_{\sigma\delta}$ and $T_{\delta}$ subsets of the reals are hard to visualise
04:41
if an irreducible element $a$ of an integral domain is the product of primes, then $a=rp$ where $r$ is a unit and $p$ a prime?
04:54
yesterday, by Daminark
So sometimes I say "So this guy dies"
Assasinator
Assimilator
Sans
No ketchup, just sans, raw sans @Daminark
But my favourite: Nuking a theorem
Given a formal system S. A theorem T in S is nuked if a suitable supermodel M that contains S such that T is unprovable in the super theorem S'
In other words, if a weakened version S' of S exists such that T become unprovable, then T is nuked in S'
Nuking is necessary (to be proved) to divide by zero
> It is almost surely absolutely definitely ... infinity ... always usually never that A exists and does not exists
When a topologist proves something: the ting goes skrrrra
@Balarka
@Daminark Big Shaq is the world's leading topologist
Man knows math
05:16
If an integral domain has the property that every element can be factored into a product of primes, then every irreducible element is prime?
is that true or false?
Seems true
ok
I'm trying to prove the second problem
if that is true I think I finished...
06:13
0
Q: how to deal with "the unknown unknown"

jasonhttps://www.youtube.com/watch?v=GiPe1OiKQuk alot of people make fun of Rumsfeld for the "unknown unknowns", but i'm wondering how a person like him would start to try to get his hands around the "unknown unknown". With known unknown, you can probably setup a strategy to explore the unknown. My...

My questions that asks are often specifically used to attack 3rd level unknowns as described there
This is why the stuff often reminds of category theory because the 3rd order unknowns live in a place much larger than anything we commonly work with in our scale
Things that interested me the most is not what is the proof, nor what is the question, but all the possible paths and constraints of a given question on a topic
> Unknowns of the kind "almost zero probability but almost infinite impact if occurs" are also best disregarded. Multiplying the near zero probability by the near infinite impact if occurs, to get an expected impact value, doesn't really work in this case. Very small differences in the numbers can produce very different expected values, so there's no real information about probability to be had.
> E.g. Blaise Pascal (contemporary of Descartes) used an expected impact value argument to apparently prove that one should better believe in a specific god. But it's in the nature of near-zero-probability things that there are zillions of in principle possible such things, so e.g. Pascal's argument failed when other possible gods were considered.
So it seems, an event of weighting unbounded above but with a probability tend to zero can have a indeterminate expectation value, hmm
So that means, it is possible to approximate something that is so unpredictable that it is not even probabilistic
Meanwhile, an unknowable is something that is of order $\omega$
It will be interesting to see if logic can prove the existence of an unknowable
Because that will be the upper limit of what us humans can possibly know and comprehend
 
3 hours later…
09:26
@Twink all primes are irreducible, so if every element factors into primes it also factors in a product of irreducibles. You're missing uniqueness though, think about the first one
10:01
So I actually read and understood the main theorem of a non-expository paper first time in my life.
#feelsgood
[Random]
Maths challenge:
Suppose x is some resource such that you want more and more
Normally, by the bottomlessness of human desire, you want a better x or more x each time
That is, $f(x_{n})<f(x_{n+1})$
however, such desire is not sustainable, thus your employer place the following restriction on f
1. If x is the amount of resource, then for all x f(x)<x
2. Forall x, the frequency of a positive value of f(x) cannot increase with n
Or in English:
1. You cannot get the same or better resource each subsequent step
2. You cannot increase the frequency of getting the resource each subsequent step
Find an f such that 1,2 is obeyed and still fulfill the requirement of bottomless human desire
You will be surprised such f exists and thus trying to stop human desire is nigh impossible
10:22
Oh, and if you think you can escape to the Continuum, then no. Not even The Continuum can save you now
hint hint
 
2 hours later…
12:28
0
Q: Iterations of $x^2 + y^2$

mickWe construct a sequence $S$ of distinct positive integers as follows 1) the sequence $S$ starts as $1,2,3$ 2) If and only If $x,y$ are in the sequence and $ x^2 + y^2 > 3 $ Then $x^2 + y^2 $ is also in the sequence. 3) the sequence is strictly increasing. 4) the sequence is completely determi...

Any ideas ?
your sequence isn’t uniquely defined
maybe you meant set
13:27
Hello!!
I want to show that $\nabla\times (f\nabla g)=\nabla f\times \nabla g$.

We have that $f\nabla g=f\sum\frac{\partial g}{\partial x_i}\hat{x}_i$, therefore we get $\nabla\times (f\nabla g)=\nabla\times \left (f\sum\frac{\partial g}{\partial x_i}\hat{x}_i \right )$.
Is it correct so far? How could we continue?
14:01
Please someone take a look here: (In my notations) Exercise 26's hint says to go this way: $A\implies C\implies D \implies B$. Does not that mean $C\implies B$? But $C\implies B$ is not true since $\Bbb R$ is separable but not compact.
hi @BalarkaSen
14:28
Hello guys. I'm trying to prove that a polynomial is injective. I take $f(x) = f(y)$ so $x^7 + x^5 + x^3 + x = y^7 + y^5 + y^3 + y$. But how can I arrive at the conclusion $x = y$? There is no easy root taking or squaring...
Hey @Bala
I wanna hand in my thesis today
Hey, long time no see
Wanna help me with the last issue before I hand it in?
Nice!
Well if it's number theory I am unlikely to be able to help but I'd like to hear what the issue is
It's more algebraic geometry
Let $C$ be a smooth algebraic curve
And look at $C^d$ for some positive number $d$
Divide it out by the action of $S_d$
14:31
So the symmetric product
Points are then just formal sums of points on $C$
Yeah
Now what I want to say is that if you take for example $d = 4$
And you look at the subspace $2C_1 + 2C_2$ with distinct points
Then the "closure" of such a space would include $4C$ as well, as a sort of limit when $C_1$ and $C_2$ get close
But this "convergence" I can't really write down properly
What's $C_1$ and $C_2$
Points on $C$
Oh, look at me making weird notation
$C$ is a point on $C$ as well
14:33
lol
Okay, let's say the curve is $\mathcal{C}$
For sure, I getchu
Wait, I don't get it. What does $2C_1 + 2C_2$ mean? A divisor?
@philmcole I upvoted your question. I'm not sure how to do it either
@BalarkaSen Yeah, you can think of it as an effective divisor
You can think of $\mathcal{C}^d$ as effective divisors of degree $d$ on $\mathcal{C}$
So you somehow want a variety of all divisors or something, and then get that the subvariety of divisors of the form $2C_1 + 2C_2$ is a Zariski open set inside that big moduli variety?
14:37
Hmm
Okay, let's get into more details
Let $\omega = (m_1, \ldots, m_k)$ be a partition of $d$,
I write $E_\omega := \{ m_1 P_1 + \ldots + m_K P_k \}$ as a substratum of $\mathcal{C}^d$ where the $P_i$ are distinct points
So in my example, $\omega = (2,2)$
Someone here was aware of this? In an infinite dimensional space, the complement of a compact set is path connected.
Of course it fails in a finite dimensional setting (unit sphere)
Now, I want the "closure" of $E_\omega$ to include $E_{(4)}$
I see, yes.
As we could intuitively have that two points get really really close, and thus that $2C_1 + 2C_2$ "converges" to a point $4C \in E_{(4)}$
14:40
Right, right.
I want to say something like, Zariski neighborhoods of $4C$ in $\mathcal{C}^4$ always intersect $E_{(2, 2)}$
But Zariski is a really shitty topology for this, I thought
Yeah
I was wondering if that's just dumb
Yeah, so, intuitively it's clear, but writing it down is hard :(
Yeah I don't know how fine a topology you want, or if you want to write this even in a topological sense
@philmcole Hint: Try to prove it's increasing.
Note that the function $x^7$ is increasing; that is, if $x>y$ then $x^7>y^7$.
(Note that the conclusion is not true if $x$ and $y$ are allowed to range over the complex numbers, so you need to use the fact that they're real numbers somehow. One thing that distinguishes $\Bbb R$ from $\Bbb C$ is the existence of an order $>$.)
14:54
@AkivaWeinberger wow, that slipped my mind, lol
I wonder if you can prove it without using ordering though
@Bala Does it make sense to make the points on $C$ converge in an affine sense?
@Krijn What's the topology on your affine space? You're in a general field, no?
The curve is over a finite field
But taking the algebraic closure would work I think
Terrible, terrible.
Yeah :)
Looking at the affine part of my curve as points in $k^2$ or something
14:59
I still don't understand what topology you have for $k$.

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