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14:03
to show it's surjective, suffices to show that $\frac1{(1,v)}$ is in the image
well $((1,v) - (1,1))(1,0) = 0$ so again $(1,v)$ is just the unity
ok so it's surjective
did you prove it
let me know when you do
@orientablesurface, btw, $f \ge 1$ a.e. only on $A$, it's $0$ everywhere else. You gotta work that into the proof
now suppose $x$ is mapped to $0$, so $(x,0) (y,v) = 0$ for some $y \in A - \mathfrak m$; since $y$ is invertible, this means $x = 0$
@BalarkaSen therefore my lemma is proved
that the localization of $A \oplus k^n$ at $\mathfrak m \oplus k^n$ is just $A$
@LeakyNun That's cool! That's all I need.
yeah
so what great theorem do you have in return?
14:13
If $f, g : (\Bbb C^n, 0) \to (\Bbb C, 0)$ are two germs of smooth functions then if you can change coordinates to write $f(x_1, \cdots, x_n) + x_{n+1}^2 + \cdots + x_k^2$, germ of a smooth function at $0$ in $\Bbb C^{n+k}$, as $g(y_1, \cdots, y_n) + y_{n+1}^2 + \cdots + y_k^2$, then you can change coordinates to write $f$ as $g$ as well.
When I say change coordinates to write A as B I mean the germs A and B are in the same $\text{DIFF}(\Bbb C^n, 0)$-orbit, where this is the group of germs of diffeomorphisms $(\Bbb C^n, 0) \to (\Bbb C^n, 0)$, acting on all germs in the obvious way
I'll look at your proof in greater detail later, @Leaky. Thanks again
and what application does this theorem have?
This is used in classification of all germs of smooth functions upto $\text{DIFF}$-orbit; these are exactly what are known as singularities
like groups, you cant classify all possible singularities. you can classify the "simple" singularities
and these simple singularities are in correspondence with simple Lie algebras
this is Arnold's theorem
how many types of singularities do you know?
I only know about node and cusp basically
Take a look at Arnold-Goryunov-Lyashko-Vasilev, "Singularity Theory I". They have tables of lots of singularities
You're talking about singularities of varieties, which are slightly different from those of maps.
Simplest kinds of singularities of maps are Morse ones, $f : \Bbb R^n \to \Bbb R$, $f(x_1, \cdots, x_n) = x_1^2 + \cdots + x_k^2 - x_{k+1}^2 - \cdots - x_n^2$
 
2 hours later…
16:42
howdy
vzn
vzn
any Tao and/ or Collatz fans? nice/ notable recent work/ news quantamagazine.org/… terrytao.files.wordpress.com/2020/02/collatz.pdf
I remember that Tao recently proved that the Collatz map $\rightarrow 1$ almost surely
17:04
I like that he includes "bragging rights" as reason for trying to solve the problem
17:31
@Thorgott Why else would you solve it
I wouldn't :P
@BalarkaSen So you're trying to prove that finite dimensional k-algebras are stably isomorphic iff they're isomorphic right?
Say a k-algebra A is n-stable if there's a retract k^n -> A -> k^n and unstable if it's not 1-stable
I would expect an n-stable algebra to not have too many such retracts
whos an algebraist in the crowd?
Thorgott
17:48
what's a good primer on algebraic geometry?
how do i characterise geometrically this eq x^3+y^3-1+3xy=0
i can use an identity on a^3+b^3+c^3-3abc
i can make it a product and say either one is zero
so i get points on either of the 2 curves
@JoeShmo Are you looking for a resource on scheme theory or on the classical stuff?
classical stuff for idiots
18:28
V a k i l
 
1 hour later…
19:37
@JoeShmo: Vakil is still extremely commutative algebra heavy. Try Shafarevich for a little lighter, or, for plenty of examples and classical approach, Joe Harris's introductory book (not Griffiths/Harris, although I love that).
Thank, Ted. I'll check Harris and Shafarevich out.
Wikipedia attributes anti Semitism to Shafarevich.. I'll stick to Harris for now
20:06
Shafarevich book is very good
Iirc it was translated into English by Miles Reid
20:24
if $f$ is measurable is it true that for each $0<p<\infty$, $f^p$ is measurable?
surely its true for integer p
20:40
note $f^p$ does not denote composition
20:54
the composition of measurable functions is measurable
@orientablesurface yes, because $\{ x \mid f(x)^p < r \} = \{ x \mid f(x) < r^{1/p} \}$
21:38
In a PDF, the probability is the area under the curve. This means that the probability of any particular value occurring is either zero or a vanishingly small limit.
However the probability of some event occurring is 1.
How can these two things be compatible? It looks almost as if the first is universally quantified over while the second is existentially quantifying over the negation of what the first is.
This isn't really a probability issue, you're asking how the area under a curve can be positive if the area under every single point is zero
It's easier to consider for probabilities case, because we can say things like "the probability that some event will occur is 1", which we can't do the rephrasing.
I understand the rephrasing in the sense of summing lots of infinitesimal quantities, but I can't seem to extend this definition to probability.
the fallacy is in your interpretation. probability just gives you proportions of areas (out of $1$ unit of area). You chose to interpret it so that it means that a certain event should occur -- mind you, I'm not saying that that's unreasonable, but the buy-in is that then it follows that the probability of any one (discrete) event occurring is $0$
@user400188 Sure you can, if you know that the area under the curve is 1, as in the case of a pdf
I would say that it is unreasonable, as you could express that some event occurs as $\exists x P(x)=1$, and that any particular even occurs as $\forall x P(x)=lim_{k\rightarrow 0} k$.
Sorry I keep mis-clicking enter.
21:54
no, $\exists xP(x)=1$ says that there exists some event occurring almost surely, not that surely some event occurs
Since $lim_{k\rightarrow 0}$ does not equal one, we can also write $\forall x P(x)\neq 1$, so we arrive at a contradiction by making this claim.
@Thorgott So how would I express the fact that some event will occur?
$P\{X = \omega\} (= 0)$
@JoeShmo I'm not sure I understand the notation here.
$\mathbb{P}(\Omega)=1$, where $\Omega$ is your probability space and $\mathbb{P}$ the probability measure
@Thorgott I am unfamiliar with probability measures as well, but the Wikipedia page on them is surprisingly brief and clear. How would I represent the probability of a particular event occurring in the same notation?
22:02
$\mathbb{P}(A)$, where $A$ is your event
It might be an interesting project to go through and write proofs for the equivalence of all the different definitions of Prüfer domains
I wonder how difficult it would end up being
@Thorgott Would $\Omega=\bigcup^{}_{i\in I} A_i$?
$\Omega$ is the event space
@Rithaniel I have never come across the term "Prüfer domain" before.
the set of all possible events
22:05
what's $A_i$
That comment wasn't intended for you
A particular event.
I was trying to write that $\Omega$ was the union of all possible events.
well, that would be true, because $\Omega$ itself is an event
namely, the event that something occurs, which surely happens
In that case I must have been wrong, as the union would have to contain itself.
every set contains itself as a subset
(no set contains itself as an element, but that's a different matter)
22:08
Oh, silly me I was thinking of the second point.
Prüfer domains are mainly in ring theory. I was just making an idle comment, apologies.
Still, this new explanation just seems to be a rewording of the original question. The probability of some event occurring from the event space is still 1, and the probability of a particular event occurring still isn't 1.
Unless $\mathbb{P}(A)=1$ for the some particular $A$.
yes, cause there can be multiple outcomes
if you throw a coin, you will always get heads or tails (let's assume we live in an ideal world where coins don't fall on the side), but it's not like you will always get head or always get tail
But aren't the two statements negations of another? $\exists x\in \Omega P(x)=1$, and $\forall x\in \Omega P(x)\neq 1$.
How would I express that the probability of some event occurring from the event space is 1 in symbolic logic? I think this is the heart of my problem.
what's insufficient about saying $\mathbb{P}(\Omega)=1$
22:16
@Thorgott I can see that my first statement is still falling for the same error mentioned before, but I can't think of another way to write it in symbolic logic.
@Thorgott It is a goal of mine to be able to write it in symbolic logic, because then I can precisely translate it to English.
So my vague statement "The probability of some event occurring from the event space is still 1" could be rephrased into something precise and correct.
I believe that's what the statement $\mathbb{P}(\Omega)=1$ does
It says that the probability of any possible outcome occurring is $1$
Then what would $\mathbb{P}(A)\neq 1$, where $A$ is a particular event, translate to? That the probability of a particular event occurring is not 1?
If so, doesn't that seem like a problem?
$\mathbb{P}(A)\neq1$ means that the probability of $A$ occurring is not $1$
This is not a problem
Think of a coin toss, the probability of flipping heads is not $1$
But if heads occurs it was 1.
@Balarka @Alessandro This is the music of 2020

https://www.youtube.com/watch?v=xeRFfyD72dE
22:28
I think you're misunderstanding how probability works
I think so too.
Probability is about how probable a certain event is, not whether it occurs or not in a specific iteration
The probability of flipping heads in a coin toss is 1/2, because it is just as probable to flip heads as it is to flip tails. This is universal. It does not care about whether you flip heads or tails in one specific coin toss.
@Thorgott Even the second part of this sentence isn't making sense to me. "not whether it occurs or not in a specific iteration" seems to be claiming that it is not the case, that an event will occur, or it will not occur. But this is a contradiction.
No, when you carry out an experiment, such as a coin toss, an event either does occur or does not occur. But we're not concerned with the deterministic post hoc side of things, but the probabilistic pre hoc side of things.
We don't even flip the coin, we just wonder what would happen if we flipped it and how likely each of the possible outcomes is.
My problem is, the event has to either occur, or not occur.
If it occurs, then the probability that it occurs is 1. If it does not occur, then the probability that it occurs is 0. But this completely excludes every other probability.
22:38
That's not what probability is, though
In the literal sense of the word
Maybe a stupid question, but given a problem statement which says:
> For which positive integers n is it possible to find n integers whose arithmetic mean equals their geometric mean?
Do the phrase "positive integers n" and "n integers" refer to the same thing or different things?
no
If there exist 3 integers whose AM is their GM, then 3 is one of the n integers (first occurence in the statement)
I understand them as different, but one guy keeps telling me they are the same and therefore the n integers must be understood as positive
That's not what's written
Although I think the Geometric mean is only defined for positive integers (?)
@Thorgott Thank you for your patience explaining everything to me. I haven't reached an understanding yet, but I sincerely appreciate all the help you have given me.
This goes for @JoeShmo and @AlessandroCodenotti too.
I plan to ask a question on the main site to help clarify my understanding regarding this.
22:47
So n=3, n integers = (8,-1,-1) is a valid answer, right? (if we allow GM to be defined for negative numbers too)
Yes
Anyone else? Do all of you agree?
23:13
I agree with me that we take the geometric mean only of positive numbers. (Do you know why it's called that?)
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