Mathematics - 2019-07-13 (page 1 of 2)

sani

05:56

Hey can some one help me in puzzling.stackexchange.com/questions/86116/… Not related to high level math but interesting

J. Doe

09:48

If $m$ is square free odd positive integer, then $x^2 \equiv a \pmod{m}$ has $\prod_{p|m}(1+\left(\frac{a}{p}\right) )$ solutions.

Can I say since $m$ is square free we can write $m = p_1p_2 \cdots p_n$; so we have $x^2 \equiv a \pmod{p_i}$ for $i =1,2, \cdots n$.

Since $p_i$ are odd, by Euler's criterion whenever $(a, p_i) = 1$ each of these has either $2 = 1+1$ solutions or $0= 1-1$ solutions;that's, $1+\left(\frac{a}{p_i}\right)$ solutions. So in total the equation has $\prod_{i=1}^{n}\left (1+\left(\frac{a}{p_i}\right) \right)=\prod_{p|m}\left (1+\left(\frac{a}{p}\right) \right)$ solutions.

Leaky Nun

@J.Doe "and since $p_i$ are distinct, solutions to $x^2 \equiv a \pmod m$ correspond bijectively to solutions to $x^2 \equiv a \pmod {p_i}$"

J. Doe

Right, I'm missing the $p_i$ that divide $a$; so I divide the final product by $\prod_{p_i|a}(1+\left(\frac{a}{p_i}\right) ) = 1$ as the then the symbol is $0$.

@LeakyNun Does it all sound right? With your addition.

Leaky Nun

yes

J. Doe

Good man, thanks.

Leaky Nun

10:22

@MatheinBoulomenos $V$ is smooth at $p$ iff $\mathcal O_{V,p}$ is integrally closed?

MatheinBoulomenos

@LeakyNun if $V$ is a curve, at least

Leaky Nun

why on earth?

actually could you remind me of the definition of the former

Poline Sandra

hello how to prove the continuity of exp with the $\varepsilon-\delta$ definition

Leaky Nun

@PolineSandra prove that in general power series is smooth in their radius of convergence, lol

MatheinBoulomenos

@LeakyNun it's a bit technical. Is $V$ a variety? For a one-dimensional Noetherian ring, being regular is equivalent to being a DVR

Leaky Nun

10:27

@PolineSandra but if you just want to do it for exp, prove that it is continuous at zero first, and then use the fact that $\exp(x+y) = \exp(x) \exp(y)$

Poline Sandra

how it is zero?

Leaky Nun

@MatheinBoulomenos let's say $V$ is a closed subvariety of $\Bbb A^2$

@PolineSandra I fixed the typo to "it is continuous at zero"

Poline Sandra

$exp(x)<\varepsilone $ how to find delta ?

ln(\varepsilone)

but for $|\exp(x)-\exp(x_0)|$??

MatheinBoulomenos

@LeakyNun If $V=V(f_1,f_2)$, then you can look at the Jacobian matrix of $(f_1,f_2)$

Leaky Nun

@MatheinBoulomenos surely $V=V(f)$

$V(f_1, f_2)$ would be $0$-dimensional right

MatheinBoulomenos

10:34

oh right

then you can just look at where the partial derivatives simultanously vanish

Leaky Nun

and that's the definition of singular?

sounds right

MatheinBoulomenos

right

Leaky Nun

how on earth is it connected to integral closed

MatheinBoulomenos

For a commutative ring $R$, a standard smooth $R$-algebra is an $R$-algebra of the form $S=R[x_1, \dots,x_n]/(f_1, \dots, f_n)$ such that the determinant $\det (\partial f_i /\partial x_j)$ is an invertible element in $S$. A morphism of schemes $f:X \to Y$ is smooth at $x \in X$ if there are open affine neighborhoods $U \supset x, V \supset f(x)$ such that $f(U) \supset V$ and the induced map $U \to V$ comes from a standard smooth algebra

@LeakyNun it's related to tangent spaces, first of all

Poline Sandra

@LeakyNun? $|\exp(x)-\exp(x_0)|<\exp(x)+\exp(x_0)$ how to continue?

MatheinBoulomenos

10:38

at the singular points, the tangent space is two-dimensional, at the smooth points, the tangent space is just one-dimensional

Leaky Nun

@PolineSandra $|\exp(x) - \exp(x_0)| = \exp(x_0)|\exp(x-x_0)-\exp(0)|$

which reduces the problem to the case at $0$

MatheinBoulomenos

recall that the tangent space is the dual vector space of $\mathfrak{m}_x/\mathfrak{m}_x^2$

Leaky Nun

@MatheinBoulomenos I'm 50% convinced

if the partial derivatives vanish then why isn't the tangent space zero lol

isn't the tangent space spanned by the partial derivatives

Poline Sandra

oooo right I will try it

Leaky Nun

@MatheinBoulomenos what is $\mathfrak m_x$?

MatheinBoulomenos

10:41

the maximal ideal of the stalk at $x$

@LeakyNun no, it's basically dual to the partial derivatives I think

Leaky Nun

why?

Balarka Sen

@LeakyNun The tangent space is kernel of the Jacobian man

It's a level curve

This is calculus lol

MatheinBoulomenos

we can also say that a point is singular if $\mathrm{dim} _{k}\mathfrak{m}_x/\mathfrak{m}_x^2 > \mathrm{dim}(V)$

Balarka Sen

If $f : \Bbb R^2 \to \Bbb R$ with $0$ as regular value, then tangent space to $f^{-1}(0)$ at $x$ is $\ker df_x$. Jesus Christ.

So you define the same thing with algebraic functions except non-smoothness will increase the dimension of tangent space now

Leaky Nun

let's say $\gamma \subseteq f^{-1}(0)$

so $f \circ \gamma = 0$

MatheinBoulomenos

10:48

so suppose that $(R,\mathfrak{m})$ is a Noetherian one-dimensional local ring such that $\mathfrak{m}/\mathfrak{m}^2$ is one-dimensional over $R/\mathfrak{m}$, then you can show by some commutative algebra that $(R,\mathfrak{m})$ is a DVR

Leaky Nun

so $\mathrm df_x(\gamma(0)) = 0$ or something like that

MatheinBoulomenos

2

Q: equivalent characterizations of discrete valuation rings

Let $R$ be a commutative ring with identiy, then the following are equivalent: $R$ is a DVR $R$ is a local Euclidean domain that is not a field. $R$ is a local PID that is not a field. $R$ is a local Dedekind domain that is not a field. $R$ is a UFD with a unique irreducible element up to a uni...

Leaky Nun

ok it makes sense why it is $\ker \mathrm df_x$ lol

MatheinBoulomenos

(9) is the condition that $\mathfrak{m}/\mathfrak{m}^2$ is one-dimensional over $R/\mathfrak{m}$

Balarka Sen

A regular local ring is surely integrally closed, yes/

MatheinBoulomenos

10:50

yes, but the other direction may fail

Balarka Sen

Of course.

MatheinBoulomenos

but in dimension 1 we get equivalence

Balarka Sen

Aha.

Right, Weierstrass preparation theorem

They are all DVRs

MatheinBoulomenos

@Leaky does this answer your question?

Nemo

I have a question regarding the first comment under math.stackexchange.com/questions/1149288/…. It states that the boundary of the closed cylinder $C$ is $S^1 \times \{0,1\}$. But with what topology on $C$? Even if I consider $C$ as embedded (thus a subset) in $\mathbb{R}^3$, then it is closed and has no inner points; thus the boundary is the whole thing.

Balarka Sen

10:53

Because ring of germs of an algebraic curve at a point has a uniformizer $t$, in the sense that every element is like $t^k$ times a unit ($t$ is the parametrizing variable for the curve).

Leaky Nun

@MatheinBoulomenos so basically... $V$ is smooth at $x$ iff the tangent space at $x$ is 1-dimensional iff the stalk is DVR iff the stalk is integrally closed?

MatheinBoulomenos

@LeakyNun right

Leaky Nun

@Nemo this is not boundary in the topological sense, but in the manifold sense

a manifold is a topological space locally homeomorphic to $\Bbb R^n$ or $\Bbb H^n$

the points where it is locally homeomorphic to $\Bbb H^n$ are the boundary points of the manifold

@MatheinBoulomenos the last two steps seem a bit magical but ok

Balarka Sen

A manifold is always locally homeomorphic to $\Bbb R^n$. A manifold-with-boundary is locally homeomorphic to either $\Bbb R^n$ or $\Bbb H^n$.

@LeakyNun See my uniformizer comment.

MatheinBoulomenos

@LeakyNun "magic" aka commutative algebra

Nemo

10:56

@LeakyNun Then, does this still answer the question if they are non-homeomorphic as topological spaces?

Leaky Nun

@BalarkaSen what's the uniformizer of $V(X^2-Y^3)$ at $(0,0)$?

MatheinBoulomenos

@Balarka I think I figured out why if we take the Zariski-Riemann space of the field of meromorphic functions on a compact Riemann surface $X$, then the initial topology that makes all meromorphic on it continuous wrt the analytical topology on $\widehat{\Bbb C}$, then it is homeomorphic to $X$

Balarka Sen

@LeakyNun Smooth curve. That curve isn't smooth at $(0, 0)$.

MatheinBoulomenos

$k[X,Y]/(X^2-Y^3)$ localized at $(X,Y)$ is not a DVR

Balarka Sen

$x^2 - y^3 = 0$ is parametrized rationally by $(t^3, t^2)$. This $t$ is the uniformizer at all smooth points, basically.

Leaky Nun

10:58

I see

Balarka Sen

@MatheinBoulomenos Nice!! I really liked that construction, but had hard time following it in Hartshorne

If you have a singular curve, Zariski-Riemann space defines the normalization, IIRC

MatheinBoulomenos

So let $Y$ be the Zariski-Riemann space of $\mathcal{M}(X)$ equipped with the initial topology described as above. Via the theory of extension of valuations and the fact that $Y$ is a ramified cover of $\Bbb{P}^1(\Bbb C)$,s o $\mathcal{M}(X)$ is a finite extension of $\Bbb{C}(X)$, I can show that the natural map $X \to Y$ is a bijection. It's also continuous by the universal property of the initial topology

Leaky Nun

can projective space be open subscheme of affine?

Balarka Sen

No. It has no regular functions on it.

Every regular function on CP^n is constant.

Leaky Nun

i'm talking about schemes

Balarka Sen

11:02

I think that shouldn't be a problem.

There are lots of global sections of an open subscheme of an affine space. But for CP^n it's only the constants

An isomorphism of schemes will give an isomorphism on the space of global sections, no

MatheinBoulomenos

so basically all that is left to show is that $Y$ is Hausdorff. Firstly, Riemann-Roch implies that $\mathcal{M}(X)$ separates points as follows: let $x \in X$ and consider the divisor $D=(g+1) \cdot x$, then by Riemann-Roch $L(D) \geq \mathrm{deg}(D)+1-g=2$, so there is a nonconstant function $f \in L(D)$. This function has a pole and the only poles which are allowed are at $x$, so $f$ has a pole at $x$ and hence at no other points, thus $\mathcal{M}(X)$ separates points

Balarka Sen

More precisely I think it's true that the category of quasiprojective varieties with regular maps between them and the category of quasiprojective varieties with morphism of schemes between them are equivalent categories (@Mathein?)

Leaky Nun

how should I think about birational equivalence?

MatheinBoulomenos

@BalarkaSen yes

the embedding is fully faithful

Balarka Sen

Nice.

Leaky Nun

11:05

@MatheinBoulomenos nice plug

MatheinBoulomenos

now to see that $Y$ is Hausdorff, let $(x_i)_{i \in I}$ be a net that converges to two points $x$ and $y$, then by continuity we get that for any meromorphic function $f$, $f(x_i)$ converges to $f(x)$ and $f(y)$, but as $\mathbb{P}^1(\mathbb{C})$ is Hausdorff, we get that $f(x)=f(y)$, hence $x=y$ as $\mathcal{M}(X)$ separates points by the Riemann-Roch argument above

Balarka Sen

@LeakyNun A rational parametrization, that's all.

MatheinBoulomenos

thus $Y$ is Hausdorff as every convergent net has a unique limit

so the map $X \to Y$ is a continuous bijection from a compact space to a Hausdorff space

Balarka Sen

Lol this argument sounds awesome

Using Riemann-Roch and nets at the same time lmao

Leaky Nun

@MatheinBoulomenos now put GAGA into the mix

Balarka Sen

11:08

LOL

@MatheinBoulomenos Cool argument!

Leaky Nun

11:26

@MatheinBoulomenos we ignore the irrelevant ideal because it corresponds to the non-point $[0:0:\cdots:0]$ in the projective space case?

Balarka Sen

Ye

Leaky Nun

I've never thought about it this way

this is cool

Balarka Sen

The correspondence between projective varieties and homogeneous radical ideals of the coordinate ring is nothing special actually; you just think of a projective variety as an affine variety, namely, the affine cone on it

The cone on zero is zero which never appears in CP^n so you just chug it out

J. Doe

11:57

@LeakyNun Do we have a bijection in the more general case where $m$ is odd and $(a, m) = 1$?

I want to say the number of solutions to $x^2 = a\pmod{m}$ correspond bijectively to the number of solutions to $x^2 = a \pmod {p_i^{\alpha_i}}$ which correspond bijectively to the number of solutions to $x^2 = a \pmod {p_i}$?

Because I don't have the Chinese Remainder Theorem, I'm not sure.

sani

Hi chat

??

J. Doe

Hello

sani

Can some one help me to solve this puzzle - puzzling.stackexchange.com/questions/86116/…

christmas_light

13:25

hey chat, I have some notation (and underatanfing) algebraic problems

Let $R$ be a ring, $M$ a left $R$-module: what are the differences between End($M$) and $\text{End}_R(M)$?

loch

I would usually read them as the same thing

christmas_light

I thought it was same restriction on the set of the endomorphism; I mean, with $\text{End}_R(M)$ is set the of endomorphism od $M$, regarded as a left $R$-module, so it is the set of all maps $M\rightarrow$ made by the action of $R\times M\rightarrow M$, but I'm a bit confused

@loch you serious? ahahhaah

loch

yes

christmas_light

ok, thanks :)

loch

anyway usually it's clear from the context

Leaky Nun

13:36

@loch what is a finitely generated $\mathcal O_X$-module?

loch

@LeakyNun it's the sheafy version of finitely generated modules

Balarka Sen

a sheaf F over the structure sheaf such that F(U) is fg over O_X(U)

Leaky Nun

what's the definition?

loch

so it's an O_X module, where on each open U, it is a f.g. O_X(U)-module

Balarka Sen

"over" means... well you can guess

Leaky Nun

13:37

ok so you say "each"

since stacks says that at any point there is a nbhd on which it is fg

Balarka Sen

the module structures should be coherent as well: for every pair (U, V) of open sets, the colimit diagram for F on U, V, U \cap V and O_X for U, V, U \cap V are coherent

Leaky Nun

so stacks says "exists"

coherent... is another concept :P

Balarka Sen

I speak English not SGA

loch

hmm it's not clear ot me that theyre the same
so maybe im wrong?

Leaky Nun

@BalarkaSen I've never seen this on stacks

ok lemme dig up the stacks definition

Balarka Sen

13:40

Sheaf of modules

In mathematics, a sheaf of O-modules or simply an O-module over a ringed space (X, O) is a sheaf F such that, for any open subset U of X, F(U) is an O(U)-module and the restriction maps F(U) →F(V) are compatible with the restriction maps O(U) →O(V): the restriction of fs is the restriction of f times that of s for any f in O(U) and s in F(U). The standard case is when X is a scheme and O its structure sheaf. If O is the constant sheaf Z _ {\displaystyle {\underline {\mathbf...

Leaky Nun

@BalarkaSen I also don't see the colimit thing there

Balarka Sen

This implies the colimit thing

I mean obviously right

And if you take U to be empty set in my colimit thing it should give this :p

Leaky Nun

do we have an intrinsic definition like we have for vector bundle?

i.e. an OX-module M is a sheaf with scalar multiplication OX x M -> M etc

Balarka Sen

Well. What does OX x M mean? What is the value category of this sheaf

Leaky Nun

is an abelian-group-valued sheaf the same thing as an abelian group object in the category of sheaves?

@BalarkaSen idk, set?

Balarka Sen

13:44

Lol

Leaky Nun

U-small sets

Balarka Sen

Meh you have to state all those module conditions locally in any case

I wouldn't do this

Leaky Nun

is the condition on wiki something like "restriction is linear"?

Balarka Sen

It is precisely that. The restriction maps preserve the module structures

loch

intrinsic defn for what?

Leaky Nun

13:46

@BalarkaSen hey stacks is saying the intrinsic one! :P

stacks.math.columbia.edu/tag/006Q

so we start with abelian groups

@BalarkaSen I'm slightly confused about the "colimit" that you mentioned

Balarka Sen

I'm just saying there's a diagram F(U) -> F(U cap V) <- F(V) which glues to F(U cup V). This should be compatible with the diagram O(U) -> O(U cap V) <- O(V) which glues to O(U cup V)

Leaky Nun

3 mins ago, by Leaky Nun

is an abelian-group-valued sheaf the same thing as an abelian group object in the category of sheaves?

also this

@BalarkaSen oh so the coequalizer exact sequence?

Balarka Sen

In the sense that there's an action of the latter diagram on the former diagram (there's an action termwise which commutes with the maps)

Leaky Nun

when do we use F? M?

Balarka Sen

@LeakyNun I don't know fuqualizer shit

Leaky Nun

13:50

whenever $\{ U_i \to U \} \in \mathrm{Cov}(\mathcal C)$ we have an exact sequence $0 \to F(U) \to \prod F(U_i) \rightrightarrows \prod \prod F(U_i \times U_j)$

@loch is an abelian-group-valued sheaf the same thing as an abelian group object in the category of sheaves?

loch

stacksproject says yes

should be easy to see once you write out the defn anyway

(but it's easier to stacks project)

Leaky Nun

tag?

loch

006J

Leaky Nun

thanks

@loch I like how they included - and 0

unlike the usual introductory texts to group who hide them under an existential quantifier

@loch I couldn't find the corresponding statement for sheaves...

but I suppose an abelian presehaf is a sheaf iff the presheaf of sets is a sheaf

loch

@LeakyNun looks true to me

Leaky Nun

13:59

yay intrinsic definitions

ringed-valued sheaf also?

Ultradark

Yo

To

Yo

Leaky Nun

$\uparrow$o

Balarka Sen

underlying set of pullback of commutative rings is indeed the pullback of the underlying sets

so your equalizer or whatever is ok

loch

@LeakyNun yeah should be

Leaky Nun

@BalarkaSen yay

@loch is the definition you initially mentioned the one taught in 18.726?

loch

14:04

for f.g. O_X modules?

i dont rmb lol

Leaky Nun

@loch I don't think Vakil ever uses finitely generated to describe O_X modules right

we talk about finite type right

but the GAGA notes for 18.726 used f.g.

yeah stack says coherent module is one of finite type and etc

and 18.726 says f.g. and etc

so I guess finite type = f.g.

stacks.math.columbia.edu/tag/01B4

in which case stack says "there exists"

loch

i guess there exists is more natural anyway

Leaky Nun

@loch in that case can't some V have M(V) be not f.g.

but M(U) and M(W) are f.g. for some W subset V subset U

loch

probably
but i dont know an example

MatheinBoulomenos

14:37

@LeakyNun I think you misread stacks. $\mathcal{F}|_U$ is a sheaf on $U$, that's not the same as $\mathcal{F}(U)$

user131753

15:14

Hi @MatheinBoulomenos.

user131753

Hello @AlessandroCodenotti.

MatheinBoulomenos

Hi @user170039

user131753

@MatheinBoulomenos When are you planning to post your blog post on adjoint functors online?

user131753

I was wondering whether the natural transformation as homotopies between morphisms of $\mathbf{Cat}$ can be generalized to arbitrary categories and this has led me to ask the following question: Is it possible to define a notion analogous to the notion of homotopy of continuous functions for any two morphisms of a category $\mathbf{C}$? I had some discussion regarding that here. Probably you may find it interesting.

loch

15:33

1

Q: Finite generation of sections of a coherent sheaf

Let $X$ be a quasi-projective scheme over a noetherian ring, $\mathcal F$ a coherent sheaf of $\mathcal O_X$-modules, and $U$ an arbitrary open subset of $X$. Is $\Gamma(U,\mathcal F)$ a $\Gamma(U,\mathcal O_X)$-module of finite type?

algebraic-geometry

Leaky Nun

@MatheinBoulomenos what does it mean then

MatheinBoulomenos

@user170039 I'm a bit busy writing my bachelor thesis atm

Alessandro Codenotti

For $V\subseteq U$, $F|_U(V)=F(V)$

Leaky Nun

but what does it mean to be finitely generated

loch

@LeakyNun if you read the proof of the lemma after the defn you'll see that they meant there exists a surjection of OXmodules O_X^n -> F

Leaky Nun

15:41

i.e. surjective at all stalks?

loch

yes

Leaky Nun

great

loch

but of course in the defn of finitely generated -- you want this surjection to exist locally

if it exists globally then your sheaf is 'globally generated'

note that a map O_X -> F is determined by the image of 1 under O_X(X) -> F(X)

Leaky Nun

cool

Semiclassical

trying to work on my CV/resume right now

it's like pulling teeth lol

anakhro

16:29

Hi friends.

Balarka Sen

Hey @anakhro

Ryan Unger

years later and I still can't answer this

5

Q: $L^2\cap C^\infty$ Hodge/Helmholtz decomposition on $\Bbb R^n$

In Majda and Bertozzi, Vorticity and Incompressible Flow on page 32, the orthogonal decomposition of a vector field $v\in L^2(\Bbb R^n)\cap C^\infty(\Bbb R^n)$ is proved, namely $$v=w+\nabla q, \quad w\bot_{L^2} \nabla q,$$ with both $w$ and $\nabla q$ in $L^2\cap C^\infty$. The proof is stated ...

real-analysis pde fluid-dynamics

feels bad

anakhro

How are you doing, @BalarkaSen?

Ryan Unger

I swear they need to add something like $\nabla v\in L^2$ as a hypothesis or something

Balarka Sen

@anakhro Not bad. Thinking about things

Ryan Unger

16:33

or at least, you need more sophisticated ideas to prove than they indicate

anakhro

@BalarkaSen I am not sure if you still wanted to/had time, but I can try taking a look at that Gromov thing or whatever with you

Balarka Sen

@anakhro I do still want to, but pressed with other things at the moment, unfortunately. I'll be completely free after 24th when semester starts, but we can start before that a little bit. I'll let you know when I have time.

You done with thesis?

anakhro

Sure! If I don't appear on here for a bit, ping me on discord. And yeah I am done my thesis.

I just need to defend sometime in August.

Balarka Sen

Excellent, send it to me

On discord if you want. I currently don't have it open but I'll check soon.

Ryan Unger

pdfs.semanticscholar.org/ecfa/…

much harder ideas

anakhro

16:38

I am a little embarrassed about it. I want to see how the defense goes and then I will send you the finished copy in September after I make edits after the defense.

Balarka Sen

For sure. I think it'd be good for me to read it as we dive into symplectic/contact h-principles because I don't know anything about symplectic/contact topology.

Ryan Unger

@BalarkaSen are you ready for the rush on area 51

Balarka Sen

wut

Ryan Unger

all the kids are getting ready to storm area 51

to find the alien gfs

Balarka Sen

What

I'm out of the meme loop it seems

Ryan Unger

17:09

@BalarkaSen ok

so let $B^2_\rho$ denote the disk of radius $\rho$ in $\Bbb R^2$

let $0<r<R$

then $\partial(B_R^2\times B_r^2)=(S^1_R\times B_r^2)\cup (B_R^2\times S_r^1)$ and they intersect along $S^1_R\times S_r^1$, right?

Balarka Sen

Yes. What an unnecessarily clunky notation to say something simple.

Ryan Unger

how is this clunky

it seems like a very complicated thing to me

this seems to describe a canonical way to do a Dehn surgery on a hypersurface in $\Bbb R^4$

user131753

@MatheinBoulomenos What is the topic of your bachelor's thesis?

Ultradark

@MatheinBoulomenos What is the topic of your bachelor's thesis? (I'd like to know as well)

Balarka Sen

@RyanUnger OK. One guy has a thin boundary torus, and other guy has a fat boundary torus. What's the big issue

Ryan Unger

17:17

whatever, sorry for bothering you

this observation makes what I want possible

Balarka Sen

It's ok, thought you had deep metric shit happening

Ryan Unger

@BalarkaSen no, my whole question was even if I get control over the singular region, I don't know how to actually do the surgery. this gives me a hint

actually picturing it is hard though

it would be interesting to construct an example of an S^3 where the flow develops a singularity like this and you get an S^1 x S^2

this seems doable

Balarka Sen

17:37

@RyanUnger So the singularity is achieved at an S^1 x S^1, and you have to cut along that? Didn't you tell me a week ago that Hamilton-Ivey proved that the singularity is achieved at a submanifold of positive sectional curvature?

I don't know anything, just trying to understand

Semiclassical

boring question: Suppose I start with a point [x,y,z] and want to replace it with [1,x,y,z]

the intent is that I'm going from affine to projective space. Is there a name for that operation? I thought there was but I'm blanking

Alessandro Codenotti

That's a standard embedding $\Bbb A^3\to\Bbb P^3$, I don't know if it has a special name though

Balarka Sen

It's one of the "affine charts of $\Bbb P^3$"

Ryan Unger

@BalarkaSen not talking about Ricci flow

the singularity is along an $S^1_1\times B^2_{1/\epsilon}$ for mean curvature flow

Balarka Sen

oh ok

spooky

Ryan Unger

17:48

well, it probably is

I don't quite know how to prove that yet

but I need to work on toy models first

$\epsilon<10^{-3}$

Balarka Sen

Lol

Semiclassical

hmm

Ryan Unger

@BalarkaSen well you want epsilon really small compared to 1

so you have a lot of room in the "flat direction"

Leaky Nun

18:31

@loch in your link, how is $i_\ast \mathcal O_L$ a $\Bbb P^2$-module?

I don't see how scalar multiplication is defined...

loch

If you have a morphism of ringed spaces f:(X, O_X) -> (Y,O_Y) and F an O_X module, then f_*F has a natural O_Y module structure

Ultradark

19:17

QUESTION (on notation): $\phi: X \times \Bbb R \to X$

this supposedly defines a "flow" on the set $X$

and the flow is a group action of the additive group of real numbers on $X$

Ryan Unger

yes

Ultradark

and the mapping I gave defines a flow (with some other conditions)

what does $X \times \Bbb R$ mean

oh

it's the cartesian product against the reals

How would I define this flow?

So I know the manifold is $S^2$

the integral curves are several longitudinal geodesics on $S^2$ originating from a point source, $p.$

evenly spaced

Albas

19:41

@Ultradark So like are these integral curves looping back to $p$. If yes then I guess you can think of it as the action of S^1 maybe in a given direction?

Ultradark

19:51

okay

@Albas what do you mean by the "action of $S^1$"

do you mean the paths of least action

Albas

20:06

As in group action of S^1 generating those flows.

Ultradark

20:40

oh

you said $S^1$??!

Ted Shifrin

@Albas: But there's a sink at $-p=q$.

Ultradark

Ted, I think he was considering the flow looped around

Ted Shifrin

That can't happen.

Euler characteristic forbids it.

Leaky Nun

@TedShifrin hi

Ted Shifrin

Unless you have an index 2 singularity at $p$, which you don't. You have a source.

hi @Leaky

Ultradark

20:46

leaky!

Leaky Nun

is the source Hatcher?

Ultradark

source is $p$

Ted Shifrin

@Ultra: Figure out how to do a single circle (parametrized by $\phi$). Then it will be the same in spherical coordinates for all values of $\theta$.

There's a source at $p$ and a sink at $-p$.

This is the usual Morse theory on the sphere.

Leaky Nun

yay Morse theory

Alessandro Codenotti

Hi @Ted

Ted Shifrin

20:48

Hi, demonic @Alessandro. I'm taking a break from packing ... which I so love to do.

Albas

Hmm didn't know this. Interesting@TedShifrin

Ultradark

I thought it was $\vec F(\theta,\phi)={1,\phi}$ but I'm stuck now

Ted Shifrin

I have no idea what those coordinates mean on the right.

Leaky Nun

@TedShifrin I'm self-learning classical algebraic geometry

Ted Shifrin

You want something that goes from $0$ length at $\phi = 0$ to length $1$, say, at $\phi=\pi/2$, and then back to $0$ length at $\phi=\pi$.

That could mean so many things, Leaky. Are you reading the Italians from the 19th century? :)

Ultradark

20:51

oh I'm

thinking sine?!

Ted Shifrin

OK, @Ultra. And it's in the $\phi$ direction, i.e., tangent to the circle $\theta=\text{constant}$.

Leaky Nun

@TedShifrin a quasi-projective variety is the intersection of a Zariski-open and a Zariski-closed of $\Bbb P^n := (k^{n+1}-\{0\})/k^\ast$ where $k$ is an algebraically closed field

speaking about theta and phi parametrization:

Ted Shifrin

LOL, so this is (late) 20th century "classical".

Leaky Nun

Borel–Kolmogorov paradox

In probability theory, the Borel–Kolmogorov paradox (sometimes known as Borel's paradox) is a paradox relating to conditional probability with respect to an event of probability zero (also known as a null set). It is named after Émile Borel and Andrey Kolmogorov. == A great circle puzzle == Suppose that a random variable has a uniform distribution on a unit sphere. What is its conditional distribution on a great circle? Because of the symmetry of the sphere, one might expect that the distribution is uniform and independent of the choice of coordinates. However, two analyses give contradictory...

Ted Shifrin

American mathematicians and (Europeans and American physicists) reverse $\phi$ and $\theta$. We don't need to argue about this.

Oh, there are always paradoxes on the circle.

Leaky Nun

20:53

what did the Italians do?

Ted Shifrin

You know about the famous Bertrand paradox(es)?

Érico Melo Silva

@TedShifrin classical is everything before me personally

Leaky Nun

I've heard about one where some probability is 1/2 and 1/3 and 1/4

Ted Shifrin

Lots of classic algebraic geometry. It's not an accident that most of the names on theorems are Italian.

Leaky Nun

I only know about Nullstellensatz which is German lol

Ted Shifrin

20:54

LOL, @Eric. You're almost as egocentric as our deluded dictator.

Leaky Nun

I do know that $\mathcal O$ is Italian

Ted Shifrin

Severi, Pieri, Castelnuovo, ....

Leaky Nun

oh, the names on theorems, not the names of theorems, lol

Ted Shifrin

Bertini

Érico Melo Silva

im pretty sure he’s actually a solopsist

Albas

20:55

This vector field that @Ultradark mentions with this emanating from a point and sinking at another seems a lot like magnetic fields caused by a dipole. Wonder if they could be interesting as a mathematical aspect of electrodynamics.

Ted Shifrin

So I just wished Danu a happy birthday, and he said he's at PCMI too, @Eric.

Érico Melo Silva

oh damn

Ted Shifrin

@Albas: I'm sure this is nothing new.

You can find him and talk complex geometry @Eric.

Leaky Nun

there's an italian geometer in imperial @TedShifrin

Ted Shifrin

I didn't list modern names like Arbarello, Cornalba, ....

Ultradark

20:58

Yo yo yo, A ringed space locally isomorphic to a ringed space of the form Spec(A)- the set of all prime ideals of a commutative ring A- should be considered as a generalization of an algebraic variety

Érico Melo Silva

@TedShifrin word

Ultradark

and now it's known as a scheme

Ted Shifrin

@Ultra: There's no need to duplicate Wiki in here, please.

@Eric: What's the best lecture/course you've been to?

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$Mathematics$ Mathematics