Mathematics - 2019-12-07 (page 1 of 2)

Shine On You Crazy Diamond

12:58 AM

@Ultradark math.stackexchange.com/questions/3466294/…

I contradicted twin primes, bro

They can't find anything wrong with it. It's a miracle!

Semiclassical

Yes, the 5 people who have viewed it since you posted it 10 minutes ago haven't found anything wrong with it.

Shine On You Crazy Diamond

lol

It's in the style of Furstenberg

as a topology is sort of formed, but we haven't used that fact

@Ultradark you there, mon?

Ultradark

1:16 AM

I’m here

Shine On You Crazy Diamond

Check out my post

0

Q: The set $X$ of all $A \subset \Bbb{Z}$ such that $A \cap (A + 2)$ is finite contains the primes?

Let $X = \{ A \subset \Bbb{Z} : A \cap (A + 2) = $ a finite set $\}$ where $A + 2 = \{ a + 2 : a \in A\}$. $X$ is closed under taking arbitrary intersections. Proof. Let $A_i \in X, i \in I$, where $I$ is an arbitrary index set. Then $(\bigcap_i A_i ) \cap (\bigcap_i A_i + 2) = (\bigcap_i A_i)...

it uses ideals

as we were talking about

$(n\Bbb{Z} + 2 ) \cap n \Bbb{Z} = \varnothing$ when $n \gt 2$.

Dannyu NDos

Hi everyone.

I have a soft question.

The direct product operation of finite groups up to isomorphism forms a commutative monoid.

Do you think it's possible to take this as multiplicative and form a ring?

In other words, is there an additive operation on finite groups?

Balarka Sen

1:33 AM

@Semiclassical Read the Call of Cthulhu last night, finally

Semiclassical

neat

Balarka Sen

It's very hard to get spooked by it because the plot idea is literally everywhere nowadays

Alessandro Codenotti

I suggest the temple and the rats in the walls by Lovecraft

Balarka Sen

But I can imagine how original it was back then

@Alessandro I'll try it out

Alessandro Codenotti

They're not part of the old gods cycle

I used to really like Lovecraft, I have a nice hardcover ~~brick~~ volume with his complete works, but now I haven't read anything by him in years

ÍgjøgnumMeg

1:37 AM

bit.ly/33UwCAk a very potent, meaningful, obviously not racist poem by Lovecraft

Balarka Sen

yeah we're aware

ÍgjøgnumMeg

lol nah I just wanted to post that because it's so silly

Balarka Sen

I was going to say that his distinct revulsion to non-European cultures in Call of Cthulhu itself doesn't help that much in reading it

Alessandro Codenotti

It's quite known that he had some racist opinions, but that doesn't mean you can't appreciate his works without xenophobic passages

ÍgjøgnumMeg

(I know, I was just being a dick)

Alessandro Codenotti

1:42 AM

Similarly studying Teichmüller theory doesn't make you a nazi supporter

ÍgjøgnumMeg

or basically any number theory lol

Alessandro Codenotti

I disagree with people who try to justify it as "it was early 1900 america, everyone was racist" though

ÍgjøgnumMeg

Idk, why do you disagree with that?

Alessandro Codenotti

I mean that doesn't justify being racist

Balarka Sen

@AlessandroCodenotti yeah, totally. but now that i read call of cthulhu, i think the "horror of xenophobia" plays a key part, much like Joseph Conrad's "Heart of Darkness"

ÍgjøgnumMeg

1:46 AM

ofc not, but at the same time, it's hard to imagine how one would act in such a situation; can you imagine being the ONLY person you know who doesn't believe in something that 99% of your country believes?

(maybe 99% is an exaggeration, but you know what I mean)

Balarka Sen

that the mysterious cult of ancient ages is lurking in voodoo cults of Africa or India, of which Europeans know nothing about, etc

its plays a part in building the horror

ÍgjøgnumMeg

@Balarka I like that, it shows how big a part ignorance plays in the racist psyche

skullpatrol

how are we defining "racism" here?

Balarka Sen

yeah, but also without the political connotations, it does build up a genuine sense of horror :)

ÍgjøgnumMeg

Xenophobia I guess

Alessandro Codenotti

1:49 AM

@BalarkaSen I see your point, for sure he knew his audience, whether or not he actually believed in those opinions

ÍgjøgnumMeg

idk I guess so, just "hatred of the other"

Balarka Sen

@Alessandro Yeah

skullpatrol

ok

Balarka Sen

It's as John Carpenter would say, "right-wing horror" (horror of the "others") in its most honest form

ÍgjøgnumMeg

in any case, Call of Cthlulhu is cool af

so much shorter than I expected it to be somehow lol

skullpatrol

1:50 AM

yup

Balarka Sen

yeah I like it

but found hard to REALLY like it because I know the plot from so many horror literature, movies and videogames

skullpatrol

Uncle Tom's Cabin

Uncle Tom's Cabin; or, Life Among the Lowly, is an anti-slavery novel by American author Harriet Beecher Stowe. Published in 1852, the novel had a profound effect on attitudes toward African Americans and slavery in the U.S. and is said to have "helped lay the groundwork for the Civil War".Stowe, a Connecticut-born teacher at the Hartford Female Seminary and an active abolitionist, featured the character of Uncle Tom, a long-suffering black slave around whom the stories of other characters revolve. The sentimental novel depicts the reality of slavery while also asserting that Christian love can...

a must read^

Balarka Sen

the whole idea of cosmic horror, of revelation of the sort that would make reality fall apart at the seams, feels very akin to Borges somehow

to be specific I am thinking of Tlon, Uqbar, Orbis Tertius

apparently Borges thought Lovecraft was a parody of Poe lol

to which, well, which horror fiction author isn't?

skullpatrol

yeah, Poe was a giant

Balarka Sen

I remember The Pit and The Pendulum being the first serious fiction in English I ever read

blew my mind lmao

skullpatrol

2:03 AM

classic

@BalarkaSen hi

Hey

how's galois theory

Good, haven't thought about anything new yet though

Will tell you if something occurs

Leaky Nun

cool

Thorgott

2:11 AM

Shadow over Innsmouth and The Music of Erich Zann are also great Lovecraft classics, the latter at under 20 pages

2

skullpatrol

short is good, but not too short

imho

Balarka Sen

average Borgest fictions are like 5 pages so i disagree!

@Thorgott I'll check these out

skullpatrol

For sale: Baby shoes, never used.

shortest story^

:(

Balarka Sen

on a tangent, I read some Sam Beckett short stories a while back

they're pretty damn good

I think Imagination Dead Imagine has to be the shortest novel ever written or something

not counting Hemingway's baby shoes, that's a prank

skullpatrol

::nods::

Balarka Sen

2:26 AM

i love how all these three pranksters, Hemingway, Joyce and Beckett, knew each other very well

best crossover i know of

skullpatrol

Have you read Uncle Tom's Cabin?

Gaurang Tandon

@ÍgjøgnumMeg Thanks! I understood.

Balarka Sen

there's a poetry collection by James Joyce called "chamber music"

he claimed the name is derived from the sound of urinating on a chamber pot

@skullpatrol I haven't actually

skullpatrol

ok

Tesseract

2:43 AM

On the subject of Lovecraft, try "At the Mountains of Madness".

Balarka Sen

Yeah I have that in my list as well

I'll try it out

Balarka Sen

4:19 AM

@LeakyNun I was about to ask if I have two field extensions $K, L$ of $F$, and an $F$-automorphism $L \to L$, does this naturally extend to an $F$-automorphism $KL \to KL$? But then I figured $K = \Bbb Q(2^{1/4})$ and $L = \Bbb Q(2^{1/2})$ is a counterexample.

Non-normality is annoying

Semiclassical

@Thorgott I've read surprisingly little actual Lovecraft but Music of Erich Zann is great

Balarka Sen

The automorphism $\sigma : \Bbb Q(\sqrt{2}) \to \Bbb Q(\sqrt{2})$ given by $\sigma(\sqrt{2}) = -\sqrt{2}$ does not extend, because the only nontrivial $\Bbb Q$-automorphism of $\Bbb Q(2^{1/4})$ sends $2^{1/4}$ to $-2^{1/4}$, which sends $\sqrt{2}$ to itself.

I wonder what composite of fields correspond to in the covering space analogy. If $Y \stackrel{p}{\to} X \stackrel{q}{\leftarrow} Z$ are two covering spaces, the composite would perhaps be the smallest covering space which covers both $Y$ and $Z$?

So exactly the pullback diagram $$\require{AMScd}\begin{CD}E @>>> Y \\ @VVV @V{p}VV \\ Z @>{q}>> X\end{CD}$$

It's $Y \times_X Z$.

Nah, that actually corresponds to $K \otimes_F L$ (because $\text{Spec} K \times_{\text{Spec} F} \text{Spec} L \cong \text{Spec}(K \otimes_F L)$). I know that's a field iff $[KL, F] = [K, F][L, F]$, though, when $K/F$ and $L/F$ are both finite

In that case $K \otimes_F L \cong KL$, I think.

Balarka Sen

4:48 AM

Yeah, that should be right. $[KL, F] = [K, F][L, F]$ implies $K \cap L = \{0\}$, because otherwise choose an $F$-basis $\{\alpha_1, \cdots, \alpha_k\}$ for $K \cap L$, extend that to a basis $\{\alpha_1, \cdots, \alpha_k, \gamma_1, \cdots, \gamma_{n-k}\}$ of $K$ and a basis $\{\alpha_1, \cdots, \alpha_k, \beta_1, \cdots, \beta_{m-k}\}$ of $L$.

Then you can make an $F$-generating set of $KL$ with less than $nm = [K, F][L, F]$ elements by taking mutual products of $\alpha_i, \beta_j, \gamma_k$.

With $mn - k^2$ elements.

Then choose some primitive element $\alpha \in K$, so $K \cong F(\alpha) \cong F[x]/(f(x))$. Then $K \otimes_F L \cong L[x]/(f(x)) \cong L(\alpha)$.

But if $\{\beta_1, \cdots, \beta_m\}$ is an $F$-basis for $L$, then as $\{1, \alpha, \cdots, \alpha^n\}$ is an $F$-basis for $K$, jointly they form an $F$-basis for $KL$, which shows $KL \cong F \langle \beta_1, \cdots, \beta_m, 1, \alpha, \cdots, \alpha^n \rangle = L(\alpha)$.

I guess I used the primitive element theorem, so separability, somewhere. Unsure how to argue cleanly without that.

OK, same stuff. If $K = F(\alpha_1, \cdots, \alpha_n)$, then $K \cong F[x_1, \cdots, x_n]/(f_1(x_1), \cdots, f_n(x_n))$ where $f_i$ is the minimal polynomial of $\alpha_i$.

Then $K \otimes_F L = L[x_1, \cdots, x_n]/(f_1(x_1), \cdots, f_n(x_n))$, which is still a field because of given hypothesis, $K \otimes_F L \cong L(\alpha_1, \cdots, \alpha_n)$ which is in turn $KL$.

For completion: $[KL, F] = [K, L][L, F]$ implies $K \otimes_F L$ is a field because an irreducible polynomial over $F$ which has a root in $K$ cannot be reducible in $L$, otherwise $[KL, L] \leq [K, F]$. Then you run the same argument as before.

I think disconnected covers correspond to product of fields. For example, if you take the double cover $S^2 \to \Bbb{RP}^2$ and do fibered product of it with itself, $S^2 \to \Bbb{RP}^2 \leftarrow S^2$, then you get the disconnected $2$-sheeted cover $S^2 \times \{0, 1\} \to S^2$.

Similarly, $\Bbb C \otimes_{\Bbb R} \Bbb C \cong \Bbb C \times \Bbb C$.

Can we classify finite dimensional $F$-algebras? They are not always product of field extensions of $F$ because of weird things like $F[x]/(x^n)$.

If $A$ is a finite dimensional $F$-algebra then $A$ has finitely many prime ideals, because $A \otimes_F F$ is a finite $F$-vector space, and since Spec of that is fiber of the map $\text{Spec} A \to \text{Spec} F$ over $(0)$, this implies this is a finite-to-one map. So $A$ has finitely many prime ideals, and $A/\text{nil}(A)$ is a product of finite extensions of $F$ by CRT.

That's all I can say I suppose.

Leaky Nun

5:31 AM

interesting

I know less than I should about compositum of fields

Balarka Sen

You always have the consolation of knowing more than me in anything about algebra :)

If I take $K = \Bbb F_p(t^{1/p})$ over $\Bbb F_p(t)$, and do $K \otimes_{\Bbb F_p(t)} K$, what do I get? $\Bbb F_p(t^{1/p})[x]/(x^p - t) = \Bbb F_p(t^{1/p})[x]/((x - t^{1/p})^p)$.

A local algebra like $F[x]/(x^n)$

This is the sort of stuff that never occurs in covering spaces; it's like covering $S^2$ by $S^2 \times (-\varepsilon, \varepsilon)$.

A cover with a multiplicity on the sheet

If we stick to characteristic 0, every finite extension $K/F$ is separable, so $K \otimes_F L = L[x]/(f(x))$ where $f$ is the minimal defining polynomial for $F$, would never have multiplicities in the factor, so we can safely say $K \otimes_F L$ is a product of field extensions of $L$.

Separable field extensions are like connected covering spaces. Field extensions are like connected covering spaces with "multiplicities on sheets"

Product of field extensions are like disconnected covering spaces

So in general a finite dimensional $F$-algebra is like a non-necessesarily-connected covering space with multiplicites on the sheets.

Leaky Nun

5:49 AM

@BalarkaSen the term is "etale algebra" or something like that

Balarka Sen

Yeah, it makes a lot of sense. I vaguely know the term from Szamuely, which I skimmed through once

Leaky Nun

@BalarkaSen isn't it a big example in Galois category

it's like etale morphism <-> covering space

or something like that

Balarka Sen

Something like that, but I think you know this better than I do

I am a beginner struggling to understand field extensions hah

Let me try to get this off my chest anyway, at the cost of sounding pretentious:

If $K/L$ is a field extension, we think of the map $f : \text{Spec} L \to \text{Spec} K$. Choose a geometric basepoint $x : \text{Spec} \overline{K} \to \text{Spec} K$ and a neighborhood of $x$, which would be an etale neighborhood, a map of schemes $U : X \to \text{Spec} K$ along with a geometric point $x_U : \text{Spec} \overline{K} \to X$, such that the triangle involving $x$ and $x_U$ commutes. Say $X = \text{Spec} A$, so $A$ is a $K$-algebra.

The lift of this neighborhood is the fibered product $U \times_{\text{Spec} K} \text{Spec} L \cong \text{Spec} A \otimes_K L$.

So it would be really nice if $A \otimes_K L$ broke off as a product of $L$-algebras. Then we can say $f$ satisfies the evenly covered property, just like classical covering space theory

But if $A$ is an etale algebra, like you say, it's a product of finite separable extensions of $K$, $A \cong K_1 \times \cdots \times K_n$. Then $A \otimes_K L$ is also a product of finite separable extensions of $L$, by doing the tensor product termwise, and using the fact about separability and tensor product I said earlier.

It's like if you diagrammatize everything happening in classical covering space theory it translates word by word to Galois theory if you know what are the right words

Leaky Nun

cool

Balarka Sen

Very fascinating, but about time I get back to concrete math instead of philosophy :P

Leaky Nun

5:58 AM

lol

@BalarkaSen so you get things like

Jun 17 '18 at 19:45, by MatheinBoulomenos

$K/F$ is separable iff $K \otimes_F \overline{F}$ is reduced

Balarka Sen

Yeah.

Leaky Nun

this is easy to see if $K/F$ is simple

i.e. $K = F(\alpha)$

Balarka Sen

Right

It's easy to see that if $K/F$ is not separable, then $K \otimes_F \overline{F}$ is not reduced, right? We can take an element $\alpha \in K$ whose minimal polynomial does not split into distinct linear factors over $\overline{F}$. Then $F(\alpha)$ is a simple subextension which is not separable, so $F(\alpha) \otimes_F \overline{F}$ gives a subalgebra which is not reduced.

Leaky Nun

oh nice!

K/F is a direct limit of simple extensions... lol

I didn't see this

Balarka Sen

@LeakyNun Is it? $K/F$ count be uncountable dimensional, not?

It's not like it's algebraic

Leaky Nun

6:12 AM

$K = \bigcup_{a \in K} F(a)$

Balarka Sen

Ah OK

Leaky Nun

but I just realized that this isn't directed

Balarka Sen

Yes, I got confused. Thanks.

Hm.

Leaky Nun

(there's a generalization of primitive element theorem that says if $b$ is separable then $F(a,b)$ is simple; unfortunately no such assumption exists in our situation)

Balarka Sen

Oh cool. I'll think about that when I get to primitive element theorem.

Leaky Nun

6:14 AM

so it's a union of simple extensions nevertheless, if that has any categorical meaning

Balarka Sen

If $K/F$ is finite, separable does imply $K \otimes_F \overline{F}$ is reduced. I'm thinking why it's true for infinite.

Leaky Nun

because... :P

Balarka Sen

Your union idea works?

Leaky Nun

let's say K/F is algebraic

(is $K(x)/K$ separable?)

Balarka Sen

Algebraic is fine

It's a direct limit of finite algebraics, as you pointed out

Leaky Nun

6:16 AM

indeed it is

Balarka Sen

Oh, algebraic is in-built in separable, right?

My bad

This is what I get when I talk before reading. Back to Dummit-Foote.

Leaky Nun

lol

you can generalize separability using that very equivalence

$K(x) \otimes_K \overline{K} = \overline{K}(x)$ is reduced so $K(x)/K$ is "separable"

Balarka Sen

@LeakyNun I can't help but point out that if $K/F$ is finite (always a safe assumption to have when comparing with covering spaces, I think), $K \otimes_F \overline{F} \cong \prod \overline{F}$ where there are $[K : F]$ many factors. So a geometric point $\text{Spec} \overline{F} \to \text{Spec} F$ lifts to $[K : F]$ many geometric points $\text{Spec} \overline{F} \to \text{Spec} K$

The "constant fiber property" of covering maps, along with establishing the degree of the field extension as degree of the cover.

Leaky Nun

$\Bbb F_p(t^{1/p}) \otimes_{\Bbb F_p(t)} \overline{\Bbb F_p(t)} = \overline{\Bbb F_p(t)}[x]/(x^p)$

Balarka Sen

$K$ is finite separable :P

Because we were thinking about separable stuff before

The geometric fiber is exactly the pullback $\text{Spec}(\overline{F}) \times_{\text{Spec} F } \text{Spec} K = \text{Spec}(K \otimes_F \overline{F})$.

@LeakyNun But yeah your example illustrates that the extension $\Bbb F_p(t^{1/p}) \to \Bbb F_p(t)$ is like an infinitisimal neighborhood of $0$ in $\Bbb C \to \Bbb C$, $z \mapsto z^p$

The geometric fiber contains only one geometric point, but with multiplicity $p$

Leaky Nun

6:26 AM

what does $\overline{K} \otimes_K \overline{K}$ look like

nothing?

Balarka Sen

If $K$ is a perfect field, that's going to be a product of $\overline{K}$'s as well, right?

Direct limit should behave well with tensor product.

But I don't think it's useful to compare infinite extensions with covering spaces; isn't taking stuff for finite etale coverings and then inverse limiting how Grothendieck constructed the etale fundamental group? It's not a great analogy to compare $\overline{K}$ with the universal cover of $K$, I think, albeit meaningful.

But I don't entirely understand and treading on muddy waters

OK I am actually leaving now because I can't resist myself to rant on Grothendieck's Galois theory without understanding classical Galois theory, and that's bad for me

Leaky Nun

lol ok

ÍgjøgnumMeg

7:10 AM

@Balarka this sounds a lot like you are reading Szamuely's book

Elior Dadon

8:12 AM

Hey, someone can help me with some discrete math? need to prove some function is onto

Mary Star

8:53 AM

Hello!!

Let $x \in \mathbb{R}^n$ be a fixed vector and $U$ a subspace. How can we show that
$$\left \{y\in \mathbb{R}^n\mid x-y\in U\right \}=x+U$$

Balarka Sen

@ÍgjøgnumMeg I skimmed it once, but I plan to read it seriously soon.

Leaky Nun

nice

ÍgjøgnumMeg

9:21 AM

@Balarka saaaame I'd very much like to

I have a copy, just haven't gotten round to it

Mats Granvik

Can the 256 cellular automata by Stephen Wolfram be expressed through addition, subtraction and multiplication?
It is possible for rule 30 and it seems that it is also possible for the rest 255 cellular automatas.

Shootforthemoon

10:22 AM

hi, anyone can help? math.stackexchange.com/questions/3464216/…

Thorgott

12:38 PM

@Shootforthemoon Well, you can derive the implicit equation in the neighborhood where the implicit function exists and solve for its derivative. However, the derivative of the implicit function will usually depend on the implicit function itself, which you usually don't have a handy expression for (otherwise, you wouldn't need to apply the IFT), so this information doesn't help you very much.

In one dimension, you could integrate this derivative, but note that you would then have expressed the function as an integral over an expression involving itself, which also isn't handy. Even finding primitives in higher dimensions, which is where you usually apply the IFT, is a much more difficult task, but calculating an implicit function like that would be akin to solving a differential equation, which, in general, can be very hard.

For example take $F(x,y)=x+y+y^5$. For fixed $x$, this increases monotonically in $y$ from $-\infty$ to $\infty$, so for each $x$, there is exactly one $y=g(x)$, such that $F(x,y)=0$. You can check this implicit function $g$ is differentiable using the IFT. Then you have $g'(x)=-\frac{1}{1+5g(x)^4}$. This tells you, for example, that $g$ is monotonically decreasing, which is good to know, but I think it will be difficult to reconstruct $g$ from this.

However, it should be noted that you usually know the value of the implicit function at one point, namely your initial point and if your original function is analytic, then the implicit function will be analytic as well (this is a more difficult theorem), so in that case you can theoretically expand the implicit function into a Taylor series around the initial point, which may or may not help you.

Shootforthemoon

1:36 PM

@Thorgott Thank you very much for the detail and the help given!

Do you want to post this as an answer?

Thorgott

2:20 PM

I've posted a more detailed version of the above as an answer, hope that helps!

1010011010

Hi all

I'm not very confident with matrix stuff

I have a determinant that is of the form det(I+epsilon * A)

Shootforthemoon

@Thorgott Wow, I am really obliged to you

1010011010

I'm considering my system at some "meso"-scale meaning epsilon is not yet infinitesimal but rather it is small, so I'd like to use the secnd order expansion in epsilon

I found this: math.stackexchange.com/a/457315/155608

Which should be quite useful, my question is in in regard to the applicability based on the assumptions of the matrices and the scalar

In some answers it is referred that it is important what the field properties are of the variables considered

Field with characteristic 0 for example in another answer:

26

A: Series expansion of the determinant for a matrix near the identity

Note that the determinant of a matrix is just a polynomial in the components of the matrix. This means that the series in question is finite. This also means that the functions $f_1,f_2,\dots,f_N$ are polynomials. We start by computing the term which is first order in $\epsilon$. Let $A_1, A...

So my question is in regard to the applicability of the idea that the second order expansion in epsilon is (tr^2 A - tr (A^2) )/ 2

As in, for a general n by n matrix

No field charactersitic required in the consideration

Thorgott

2:36 PM

Well, in characteristic you would be dividing by $0$, so that certainly doesn't work

Secret

A class in programming is a blueprint for making objects, which are like custom types that are made in terms of other predefined types along with functions
A universe which obeys the simulation hypothesis will mean all types are composed of some fixed collection of predefined types
Such is sufficient to enable emergence as new behaviour results from combining arrangement of these types
However, there may be types that are inherently unspecifiable in terms of code. There existence of at least one of these will disprove the simulation hypothesis

Abhas Kumar Sinha

2:53 PM

can anyone prove this?

1

Q: $ \dfrac{d}{dz} \left \{ (1-z^2)P'(z) \right\} + \left\{ \beta - \dfrac{m^2}{1-z^2} \right\}p(z) = 0 $ Conversion of the DE into other forms.

I guess it's a bit related to physics, but the doubt is in the mathematical part. I was just watching some stuff on Quantum Mechanics for the solution of the Hydrogen atom. The lecture ends at.. $$ \dfrac{d}{dz} \left \{ (1-z^2)P'(z) \right\} + \left\{ \beta - \dfrac{m^2}{1-z^2} \right\}p(z) =...

calculus ordinary-differential-equations derivatives differential-geometry polar-coordinates

Astyx

In this answer : math.stackexchange.com/a/2212554/377528, it says to compute the cohomology of $0 \to \Bbb Z_m \to \Bbb Z_m \to 0$. What does this explicitly mean ?

Nevermind I got it

Shine On You Crazy Diamond

3:29 PM

Anyone know of a good book on type theory + implementation details (code)

Like making a basic proof assistant

Shine On You Crazy Diamond

3:43 PM

@Ultradark let's code some stuff, mon

:D

Ted E

4:07 PM

@Thorgott I definitely liked The Music of Erich Zann, although I felt similar to how Balarka described in regard to not finding it scary (and the reason why). I think the 'can't find the place ever again' theme has been copied one million times.

anakhro

4:36 PM

@TedE Hi!

Ted E

Hey there!

anakhro

How have you been?

Abhas Kumar Sinha

@ShineOnYouCrazyDiamond Is your name inspired by Rihanna's Diamonds?

Ted E

I'm well, just recapping some math I did awhile ago right now.

I wish I had typed up notes on some of this old stuff when I knew it well.

anakhro

What stuff in particular are you recapping?

Ted E

4:39 PM

Cech cohomology atm.

Balarka Sen

You're Ceching your Cech cohomology knowledge for potential flaws?

Ted E

What about you? What are you working on?

Bad Balarka :P.

There are more than potential flaws :).

I have to check a few things that might be obvious (or not, or they might be false even, but they were claimed in talks):
1) intersections of (probably open) affine subschemes of a separated scheme are also affine
2) Cech agrees with standard (zariski) sheaf cohomology when quasicompact and separated (which I think relied on the above being true)

anakhro

I am studying for exams.

I guess you are an algebraic geometer of a sort?

Balarka Sen

I never really ended up understanding what separated actually means, though I understand it's trying to enforce some sort of Hausdorffness

Ted E

Well, I wouldn't put myself into any camp yet really.

Balarka Sen

4:46 PM

(As in, the classical example of the line with two origins scheme not being separated)

Ted E

I haven't captured the true meaning of it either yet.

We could try to work that out now if you want.

Balarka Sen

I guess $\Bbb C^2$ with two origins (two copies of $X = \text{Spec} \Bbb C[x, y]$ glued to each other by the identity map along the subscheme $X \setminus \{0\}$) is an example where intersection of two affine subschemes is not affine

anakhro

Isn't Hausdorff = separated?

Or do you mean not in a strictly topological sense?

Balarka Sen

This is for schemes, not topological spaces.

anakhro

Wild.

Ted E

4:49 PM

@BalarkaSen Do you mean glued along identity map on $X\backslash\{(0)\}$ or on $X\backslash\{(x,y)\}$?

anakhro

(<-- knows nothing about schemes) maybe it is equivalent in the Zariski topology?

Balarka Sen

By 0 I of course mean the ideal (x, y)

Ted E

Right

Balarka Sen

@anakhro Zariski is totally nonHausdorff it implies nothing about the underlying topology really

Ted E

@anakhro Schemes have more than just a topology, they have a sheaf of rings whose stalks are local rings (really more has to be said to make this precise, since really this is just a 'local' scheme)

Balarka Sen

4:50 PM

The forgetful functor Sch -> Top is very badly behaved

anakhro

That's right, the Zariski-open sets are all very big, right?

Balarka Sen

Yep

anakhro

I forgot about that.

Ted E

@BalarkaSen What are your two affines here?

Alessandro Codenotti

@TedE You don't need open in 1 iirc

@TedE The two copies of $\mathrm{Spec}\Bbb C[x,y]$

Balarka Sen

4:52 PM

@TedE So $Z = X \times_{X \setminus 0} X$. The two affines are $X \hookrightarrow Z$

They intersect at $X \setminus 0$, which is not isomorphic to an affine scheme

Ted E

Oh yes, of course, funnily enough I proved that ages ago using cech cohomology lol

If that was affine, then by leray vanishing theorem it would have trivial higher cech cohomology, but it doesn't

Alessandro Codenotti

That looks like massive overkill

Ted E

It's an exercise in hartshorne

You can also do it by considering $X\backslash\{(x,y)\}\to Spec(\Gamma(X\backslash\{(x,y)\}))$

Alessandro Codenotti

I guess

I shouldn't talk about algebraic geometry, I know nothing about it

Balarka Sen

I really doubt you need this. You can argue $\Bbb C^2 \setminus 0$ is not affine by noting that the inclusion map $\Bbb C^2 \setminus 0 \hookrightarrow \Bbb C^2$ is an isomorphism on the structure sheaf by the Hartogs extension theorem

Alessandro Codenotti

4:57 PM

@TedE What isn't though

@BalarkaSen That doesn't tell you that it's not isomorphic to an affine scheme though

Ted E

I meant serres vanishing theorem

Alessandro Codenotti

That can be argued in a completely elementary way by computing the regular functions on $\Bbb C^2\setminus 0$ by hand

Balarka Sen

@Alessandro Since it's an isomorphism on global sections, if it was an affine scheme, then the inclusion had to be an isomorphism

That's impossible

Ted E

@BalarkaSen That was my 'you can also do this by considering' argument

Balarka Sen

Similar argument should push through for $\text{Spec} \Bbb C[x, y]$ minus $(x, y)$

Alessandro Codenotti

4:59 PM

Hmm I think I don't understand your argument

Why doesn't it apply to $\Bbb C\setminus 0\hookrightarrow\Bbb C$?

Balarka Sen

Because Hartogs doesn't apply

There are regular functions on C \ 0 which doesn't extend to C

Alessandro Codenotti

Oh ok I see now

Balarka Sen

@TedE Ah yeah I missed it when I was frowning at your Cech cohomology overkill

Ted E

Haha, I just had that overkill in mind because I am recapping Cech cohomology

Hey @TedShifrin!

Ted Shifrin

Howdy, a @Balarka, demonic @Alessandro, other @TedE

Balarka Sen

5:02 PM

Hi @TedShifrin!

Alessandro Codenotti

Hi @Ted

Ted E

Hi @AlessandroCodenotti

Alessandro Codenotti

Ah, I was trying to ping @TedS

I suggest a western style duel because this chat is too small for two Teds

Ted Shifrin

Multiple Teds are a pain.

<--- stakes a claim to prior existence (5+ years)

Ted E

@AlessandroCodenotti I know, I was just joking about that

@TedShifrin It's okay, they can ping me with it

Ted Shifrin

5:04 PM

So I see Hartogs was mentioned. :)

Balarka Sen

In the algebraic category, far easier than the analytic Hartogs to your disappointment

Ted E

@BalarkaSen I think you can't think of separatedness as being 'topological' solely, since it is a relative notion

Alessandro Codenotti

Is this the same Hartogs of the Hartogs number in set theory?

Ted Shifrin

in the analytic setting, it holds for any compact subset. So I forget how that's handled in the algebraic category.

Balarka Sen

yeah thats so surprising

Ted E

5:06 PM

Like your example above isn't $\Bbb Z$-separated, but maybe it is separated over itself or something

Balarka Sen

@TedE Yeah I don't know this stuff

$S$-separated means the diagonal $X \to X \times_S X$ is a closed immersion?

Thorgott

@TedE Yeah, that's understandable. I think there is more to the Cosmic Horror aesthetic than just the literal horror though. Ideas present in Lovecrafts work that may have been novel at the time have been replicated time and time again since, but I find there to be little fiction that is actually Lovecraftian as a whole. To me, reading Lovecraft is not as much about the horror as it is about diving into the unique worlds and headspaces he creates, exploring the boundaries of our existence.

Ted E

Yes, where $X$ is an $S$-scheme and that fibre product is over the structure map both times, i.e. the structure map is separated

Balarka Sen

Gotcha

@TedS I asked Georges about an analytic vs algebraic question in this comment today

Hopefully there's a good answer and I am not making a fool of myself

Ted E

@Thorgott Is some sense, that's why I read math :-).

Alessandro Codenotti

5:09 PM

If you think about $X$ being separated as $\Delta:X\to X\times X$ being a closed immersion, where the product is categorical then being $S$-separated is being separated in $\mathsf{Sch}/S$

Ted Shifrin

He's a good one to ask, a @Balarka.

Ted E

@AlessandroCodenotti That's a restatement yes, since that's just saying we can denote the fiber product over the structure map to $S$ as a genuine product, as long as we intend for it to occur in the category of $S$-schemes

Alessandro Codenotti

(This is just a complicated way to say that $\mathsf{Sch}$ is the same as schemes over $\mathrm{Spec}(\Bbb Z)$ and that the product in $\mathsf{Sch}/S$ is a fibered product in $\mathsf{Sch}$)

Ted E

I know, I just meant that being separated can't really be totally topological, since it's a relative notion

It's like you can take a (absolutely) singular scheme over itself (with the identity map) and it's smooth over itself

Thorgott

Cthulhus lair was described as Non-Euclidean, after all

Balarka Sen

5:12 PM

@Thorgott That was a good touch

Alessandro Codenotti

@TedE That's a good point

Balarka Sen

I hope Cthulhus are hyperbolic

Very trippy

Ted E

@Thorgott I think my favorite part of Lovecraftian horror is that he describes it in such a way that you have your own mental image, and your own mental image can typically be much more horrifying than anything they describe.

Alessandro Codenotti

@BalarkaSen When chtulhus gather in a group they have solvable word problem

Ted E

Describing the horrible creatures as moving in ways that are self contradicting

Balarka Sen

5:15 PM

@Alessandro ayup

its very hard to imagine cthulhus differently after seeing them depicted as tentacle faced gentlemen through time immemorial tho

Thorgott

Yeah, that is definitely an important aspect. "Pickman's Model" is an entire story about an incredibly horrific painting, yet it's very minimal in actually describing what's being depicted.

Ted E

Yep, that's actually exactly the one I had in mind :D

I thought that one was actually genuinely scary, and I haven't really seen it copied

Mainly the twist at the end

Balarka Sen

Lovecraft seems very disapproving of futurism and cubism in Call of Cthulhu

Ted E

What do you mean by that?

I don't know what cubism is I guess

Balarka Sen

he ascribes the horrific pattern of the bas relief to cubism and futurism

or something like that

he passingly mentions it but maybe im biased because i know Lovecraft hated the vorticists

Ted E

5:21 PM

Oh, I don't remember that

Thorgott

Yeah, Pickman's Model is really cool. I guess it's lesser known than his most popular ones. Finding otherness in oneself, coupled with descent into insanity, is another great theme recurrent in lots of Lovecrafts writing

Balarka Sen

the major people involved with the BLAST magazine were Ezra Pound and T S Eliot, both of whom Lovecraft has been known to mock

Thorgott

Also, I don't think having a clear image of Cthulhu as it is always perpetuated is that detrimental. Lovecrafts descriptions are, in a way, designed to transcend your understanding, they let you know that you don't know. The contrast between the little concrete that you have and the inconcrete that is implied is, as Ted alluded to earlier, where a lot of the horror lies.

Ted E

@AlessandroCodenotti For what it's worth (and this doesn't imply that it really is an essential assumption), I found that Hartshorne says: Let $X$ be a separated scheme over an affine base. Then for $U,V$ affine open subsets of $X$, $U\cap V$ is also affine.

Alessandro Codenotti

Makes sense, I was thinking about separated as separated over $\Bbb Z$ earlier

Ted E

5:25 PM

So you think working specifically over $Spec(Z)$ removes the openness assumption on $U$ and $V$?

This might be nonsense, but I think if that were true, then since separated is preserved under base change, that might imply the openness of $U,V$ is irrelevant in general (for affine base as in the exercise I allude to above)

Alessandro Codenotti

No openness is actually needed

I was wrong

Ted E

Ahh ok, cool

I guess there's no reason why the intersection of two (arbitrary, i.e. not necessarily open or closed) affine subsets should even be a scheme?

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