Mathematics - 2018-07-06 (page 1 of 3)

geocalc33

00:03

Would a question about tensor networks be appropriate for math S.E.

Failed to be a Mathematician

01:47

@Secret I got it, keep on studying, I forget about group action 😊 . Sorry for the rubbish Question.

user76284

02:20

Are generating functions best viewed as real or complex functions? Why?

Xander Henderson

02:57

Yes.

The Great Duck

03:27

0

Q: Proving that for a $g(x,y)$ such that $\partial_x g$ exists on a set that there is a $g$ such that $\partial_y g$ exists on the same set.

We define a step function in this context to be a function $C(x)$ such that there exists a uniformly discrete set $S$ such that for any bounded interval $[a,b]$ if $[a,b] \cap S = \emptyset$ then $C(a) = C(b)$. Now let $g(x,y)$ be a function such that there exists a step function $C(x)$ such tha...

real-analysis

Meow Mix

05:29

Problem: You start with 1, and each time subtract a random real number from 0 to 1 inclusive, all equally likely, until the number you end up with is less than 1. What is the average amount of times you will subtract?

Abcd

08:01

Hi @LeakyNun

Alessandro Codenotti

09:10

@MeowMix 1?

Failed to be a Mathematician

09:39

What do you meant by homomorphism of graded algebra?

I understood that it is an algebra Homomorphism preserves the multiplicative structure.

What do you mean by degree here?

I don't understand the term degree here.

Alessandro Codenotti

The elements of $A^k$ are called homogenous of degree $k$

Abdullah UYU

There was a feature that takes me to my last message. How was it?

Alessandro Codenotti

preserving the degree means that if the homomorphism is $\bigoplus_kA^k\to\bigoplus_j B^j$ it should send $A^n$ to $B^n$ for all $n$

@AbdullahUYU Scroll to the top of the chat, it's up there together with a button to open the full transcript

Abdullah UYU

Ah, thanks @AlessandroCodenotti

Leaky Nun

I finished chapter 1 of atiyah without skipping any questions :D

2

Failed to be a Mathematician

09:59

8

Q: Understanding of graded algebra

I am recently learning from Loring W. Tu's An Introduction to Manifolds the concept graded algebra, which is used for introducing exterior algebra. I don't understand the following definition: An algebra $A$ over a field $K$ is said to be graded if it can be written as a direct sum $A=\bigopl...

abstract-algebra exterior-algebra

but Yuan's answer said that $A^k$ , k is just a superscript.

Leaky Nun

did anyone say it isn't?

Alessandro Codenotti

Sure, I usually write $A_k$ to avoid confusion, but I was following the notation from the image you posted

Failed to be a Mathematician

@AlessandroCodenotti degree means what? can you explain?

Entrepreneur

Help please math.stackexchange.com/questions/2839745/…

Alessandro Codenotti

In general "degree $k$" means "belongs to $A_k$". If you think about polynomial rings as graded algebras for example the degree will agree with the usual notion of degree

Failed to be a Mathematician

10:05

okay. Thank you @AlessandroCodenotti

Alex

Another example: A superalgebra is said to be $\Bbb Z/2\Bbb Z$-graded, and has degree $0$ and degree $1$ elements, i.e. even and odd elements.

Leaky Nun

ring isn't abelian

but their modules is abelian

but their algebras isn't abelian

what a strange world

@AlessandroCodenotti what do we know about rings whose algebra is abelian?

Alessandro Codenotti

@LeakyNun Personally nothing :P

Alex

@LeakyNun You mean seeing commutative unitial $R$ as an $R$-algebra?

Balarka Sen

@Alex u wanna read Goresky

Leaky Nun

10:09

@Alex if a (commutative unital) ring $R$ satisfies the property that R-Alg is an abelian category, what do we know about R?

Alex

@LeakyNun let me think

@LeakyNun I should clarify, how are you defining algebras? Always associative unital, or?

@BalarkaSen Soon for sure

Leaky Nun

commutative associative unital

Alex

@BalarkaSen Was reading some Aguilar

Leaky Nun

in the sense of Atiyah-Macdonald

Balarka Sen

ah okok

Alex

10:11

@LeakyNun Sure

Balarka Sen

aguilar is good

Failed to be a Mathematician

Do I need more algebra for study geometrical stuff?

Alex

@BalarkaSen I've read about 35 pages so far I think. It's really cool

Balarka Sen

nice

i havent went through it thoroughly, i just got exposed to bits of it when daminark was reading it

i liked it.

Leaky Nun

you havent gone through it thoroughly

Alex

10:13

Is this true? I haven't thought much about it yet:

>Let $f:X\longrightarrow Y$ be continuous. $f$ is nullhomotopic if and only if $f(X)$ is contractible.

Leaky Nun

I think it is true

Balarka Sen

garbage

Leaky Nun

what?

Alex

Rip

Balarka Sen

yes. think hard, there are easy counterexamples

Leaky Nun

10:14

oh right

f(X) contractible doesn't give you a series of maps

Alex

How do you visualise a nullhomotopic map?

Alessandro Codenotti

@BalarkaSen There are counterexamples with maps $[0,1]\to S^1$ right?

Alex

You warp f(X) continuously onto a point

Failed to be a Mathematician

youtube.com/watch?v=OwcgIHTcuFw <=====Young mathematician of indian origin :)

Leaky Nun

aha

map S1 -> S2 to the equator

Mike Miller

10:15

Hello

Leaky Nun

the forward direction fails too

Alex

I just wanted to visualise a nullhomotopic map in general

Balarka Sen

@AlessandroCodenotti Yes, surjective map S^1 --> S^1 can be nullhomotopic

Leaky Nun

aha

Mike Miller

You definitely do not warp f(X) continuously to a point. You need to picture the map with its domain and continuously modify the map

Leaky Nun

10:16

in fact the inclusion S^1 -> R^2 is nullhomotopic

Balarka Sen

f(X) is a subspace of Y. The warping happens to f, not it's image.

Alex

It is indeed

Mike Miller

There are surjective maps from [0,1] to any compact locally connected Hausdorff space

Leaky Nun

but S^1 is surely not contractible

Alex

It's not

You can map S^1\to R^2-\{0\}

Balarka Sen

10:18

Evening @Mike

Leaky Nun

we don't need -{0}

we shouldn't have -{0}

if you have -{0} then it fails to be a counter-example

because both the condition and the result fail

Balarka Sen

Depends on the map.

Alex

@BalarkaSen For pointed maps I guess that is false?

Balarka Sen

No.

Alex

Only the constant pointed maps = degree 0 winding - are nullhomotopic

Balarka Sen

10:21

You can be surjective and degree 0

Alex

?

Balarka Sen

Consider $S^1 \to [0, 1]$ by collapsing the upper and lower hemisphere togather. Then map $[0, 1] \to S^1$ by the quotient map.

Alex

Isn't it the case that $f:\Bbb S^1\to \Bbb S^1$ is degree $0$ if it is represented by $\hat{\varphi}(e^{2\pi it})= e^0$ so that $f(e^{2\pi it}) = f(1)e^0$

Balarka Sen

Namely, if you think of the circle as a rubber band, you're folding the rubber band, and then snaking it around the circle.

@Alex No, degree zero is a homotopical property, I don't know where you are getting these false statements from.

It's degree zero if it's homotopic to something that can be represented as that, whatever formula you wrote.

See my example.

Leaky Nun

ker(f oplus g) = (ker f) oplus (ker g)?

Alex

10:31

Thanks @BalarkaSen

That makes sense

Mike Miller

@Alex Without prodding do a degree that makes you uncomfortable, I am curious what your previous thought process was

Alessandro Codenotti

What's the intuition behind the excision theorem?

Mike Miller

Relative homology is supposed to ignore the behavior of chains in A. So if we get rid of some of A, why should that change anything?

Leaky Nun

@AlessandroCodenotti things behave nice locally

Balarka Sen

That's a trash description.

Leaky Nun

10:35

alright

Alex

@MikeMiller In aguilar he shows that all the pointed maps $\Bbb S^1\to \Bbb S^1$ are homotopic to some representative $g_i$ for $i\in \Bbb Z$, and then he does all computations w.r.t those - I just forgot that it was up to homotopy temporarily :P

Balarka Sen

@AlessandroCodenotti Are you reading the theorem statement, or are you reading the proof?

Mike Miller

@Alex Ah, I see. That makes sense

While it is nice to have canonical representatives of homotopy classes, this is extremely lucky to this special case. I would endorse spending time trying to think about tbe kind of homotopic maps you can cook up on S^1 as preparation for the more wild things ahead!

Alessandro Codenotti

@MikeMiller Ok, that does make sense

Mike Miller

I think that is supposed to be the idea. The proof if I recall is a bit more involved than the clean idea

Balarka Sen

10:38

The usual theorem statement is a bit ucky. The proof boils down to the following sequence of ideas. Suppose $X$ is a topological space with a cover $\mathcal{U}$. Denote by $S_\bullet^\mathcal{U}(X) \subset S_\bullet(X)$ to be the subcomplex of $\bullet$-simplices that fit inside an open set of the cover $\mathcal{U}$.

Alessandro Codenotti

@BalarkaSen The statement, but I already read the barycentric subdivision business which seems to be the main ingredient of the proof

Balarka Sen

Then the chain-level excision theorem is that the inclusion $S_\bullet^\mathcal{U}(X) \to S_\bullet(X)$ is a chain-homotopy equivalence.

Or more specifically, the homology of those complexes are isomorphic.

I find that much more intuitive

Alessandro Codenotti

@BalarkaSen Right, I read the proof of that fact (which I also find rather intuitive, I'm just cutting the chains into smaller pieces, what could go wrong?)

Balarka Sen

Right, the barycentric subdivision is to obtain a chain-homotopy inverse $S_\bullet(X) \to S_\bullet^\mathcal{U}(X)$, which is given by taking a singular simplex in $X$ and then subdividing it (with Lebesgue number lemma) so that it fits inside the cover $\mathcal{U}$.

If $\mathcal{U} = \{U, V\}$ is your cover, then some manipulation recovers that the statement that those two complexes have isomorphic homology is the same as $H_n(X, U) \cong H_n(V, U \cap V)$ if I remember right.

The former is the homology of $S_\bullet(X)/S_\bullet(U)$, the latter is the homology of $S_\bullet^\mathcal{U}(X)/S_\bullet(U)$.

And that's the statement of excision, pretty much.

@Alessandro Here's in my opinion a better reformulation of the cycle of ideas surrounding excision.

Alessandro Codenotti

Skimming the proof in tom Dieck's book it seems like a lot of diagram chasing (the Five lemma? What's that) and a couple of applications of the isomorphism induced by $S_\bullet^{\mathcal{U}}(X)\hookrightarrow S_\bullet(X)$, I'll read the details later

Balarka Sen

10:53

fam dieck works too hard

Alessandro Codenotti

@BalarkaSen I'm leaving for lunch in two minutes, but I'm interested in hearing this

Balarka Sen

I'll write it down so you can come back and read it

Alessandro Codenotti

Ok, I'll be back soon! Thanks

Balarka Sen

Bon apetite

or however the spelling goes

Kari

*bone apple tea

Alessandro Codenotti

11:22

I'm back

Balarka Sen

I ended up procrastinating instead of writing lmao

Alessandro Codenotti

lol that's very understandable

Alex

@BalarkaSen That's been me the last 3 days

Balarka Sen

Suppose $\mathcal{U} = \{U, V\}$ is a cover of $X$. Then there is a chain map $S_\bullet(U) \oplus S_\bullet(V) \to S^\mathcal{U}_\bullet(X)$ which sends $(\zeta, \xi) \mapsto \zeta + \xi$ where $\zeta$ and $\xi$ are singular chains living entirely in $U$ or $V$ respectively

Alessandro Codenotti

@BalarkaSen That's actually an isomorphism, right?

Balarka Sen

11:26

Mmm no, there is kernel!

What if $\zeta + \xi = 0$?

Alessandro Codenotti

Oh, right

Balarka Sen

The kernel of this chain map is $S_\bullet(U \cap V)$, included in $S_\bullet(U) \oplus S_\bullet(V)$ as the antidiagonal copy $(\zeta, -\zeta)$ for $\zeta \in S_\bullet(U \cap V)$.

Alessandro Codenotti

So we have $S_n^{\mathcal{U}}(X)=S_n(U)+S_n(V)$ (sum as in sum of $\Bbb Z$-modules), but the sum is not direct

Balarka Sen

Right!

So there is a short exact sequence of chain complexes $0 \to S_\bullet(U \cap V) \to S_\bullet(U) \oplus S_\bullet(V) \to S_\bullet^\mathcal{U}(X) \to 0$ where the first map is $\zeta \mapsto (\zeta, -\zeta)$ and the second map is $(\zeta, \xi) \mapsto \zeta + \xi$.

This gives... a long exact sequence in homology

mick

0

Q: About $ f(x) + c \space f(g(x)) = h(x) $

Let $g(x),h(x), f’(x) $ be functions that can be expressed by radicals , log and exp , but $f(x) $ can not. Now consider functional equations like $$ f(x) + c \space f(g(x)) = h(x) $$ Where $c^2 = 1$ The only solutions with h(x) not identically 0 I know are based on $g(x) = a + b x $ with sol...

calculus functional-equations cyclic-groups

Balarka Sen

11:29

Remember that $H_*(S_\bullet^\mathcal{U}(X)) \cong H_*(X)$ by excision.

So we have a long exact sequence $\cdots \to H_{n+1}(X) \to H_n(U \cap V) \to H_n(U) \oplus H_n(V) \to H_n(X) \to \cdots$

This is known as the Mayer-Vietoris sequence, and is the most concise and axiomatic form of the excision theorem

It's an analog of Siefert-van Kampen theorem for homology groups, if you wish.

Danu

Hey @BalarkaSen here's a question

Alessandro Codenotti

@BalarkaSen Aha, so this is the famous Mayer-Vietoris sequence everyone talks about!

Balarka Sen

Yup.

Danu

I'm sure you know the picture of the dense (irrational slope) curve on a 2-torus.

Balarka Sen

Right.

Danu

11:33

Can we similarly have one-parameter subgroups of $T^n$ whose closure is all of $T^n$ for any $n$?

Answer is, I think, yes, because of some assertions I'm seeing in some paper

but it is not clear to me

Balarka Sen

For sure. $T^n$ can be realized as $\Bbb R^n/\Bbb Z^n$. Pick a line on $\Bbb R^n$ which has irrational slope projected to each coordinate 2-plane. I think that works.

Danu

Yeah, that works?

cool shit

Balarka Sen

Yup.

Danu

so in fact it's generic

Balarka Sen

Yeah.

Danu

11:35

(which also agrees with what I think must be true)

excellent

thanks

Balarka Sen

Do you want to hear an odd question which somebody asked me recently, relevant to this picture?

Danu

That establishes the correspondence between two definitions of flag manifolds :)

Balarka Sen

Oh, what are those definitions?

Danu

(i) Quotients $G/H$ where $G$ is compact, semisimple and cnonected, and $H=C(T)$ the centralizer of a torus

(ii) Orbits of the adjoint action $G\curvearrowright \mathfrak g$

Given a point $w\in \mathfrak g$, the closure of $\exp(\Bbb R w)$ is the torus

Alessandro Codenotti

I see that Hatcher spends a few words on examples and "why is excision useful" instead of just giving the statement, I'll read that as well! I have to leave for a while now though, bye everyone and thanks Balarka!

Balarka Sen

11:40

Hm, you know what, I might be wrong about the construction I gave. I think you need the direction vector of the line in $\Bbb R^n$ to have coordinates $\Bbb Q$-linearly independent.

Otherwise your closure will collapse into a subtorus of $T^n$

Danu

yeah

sure

still generic

Balarka Sen

Yep.

@Danu Ah I see.

Danu

so that's saying that the generic orbit yields a "full flag manifold" $G/T$ where $T=C(T)$ is a maximal torus

all this stuff is really well-understood in terms of Lie-algebraic data; it corresponds to these Weyl chambers; interior means full flag manifold, inside a wall means not full

Balarka Sen

Ah gotcha

Danu

but I don't understand that POV haha

I gotta fucking learn representation theory

but it's just so tedious to me

Balarka Sen

11:43

I know absolutely nothing about that stuff

@AlessandroCodenotti Yup it's a good idea to read Hatcher more on that note

Danu

There's so much cool stuff you can do with it

all these homogeneous space things

their characteristic classes and stuff are all computable in terms of Lie-algebraic data

Balarka Sen

Very interesting

Danu

Hirzebruch & Borel, Characteristic classes and Homogeneous Spaces; famous series of papers

Also all these flag manifolds are apparently importatn to a lot of different groups of people, many of which study them because you can use them to study representation theory

Mike Miller

@Danu Do you know if every homogeneous space G/H is diffeomorphic to the interior of a compact manifold

Alex

@Danu The beautiful geometric realisation of the classification of fin dim irreps of a reductive algebraic group

Alessandro Codenotti

12:00

@BalarkaSen I'm reading everything from Hatcher as well to get a second perspective

Balarka Sen

Relief!

You won't turn into a fucken algebraist after all

Leaky Nun

:( where have all the algebraists gone

Alessandro Codenotti

lol

But no, I don't think I will become an algebraist :P

Mike Miller

Alessandro, soon

Balarka Sen

jesus

the answer is of course (2)

"I am essentially looking for a formulation of knot theory that looks like it came from nLab." O_O

the category theory of that question is bullshit as well

Danu

12:19

@MikeMiller So you're asking if non-compact ones are interiors of compact with boundary? I don't know.

Balarka Sen

Just pass to $\Pi_1 C^\infty\text{Emb}(S^1, S^3)$ like Will wrote. These nlab people are so annoying.

Danu

@Alex I'd like to learn that

Alex

@Danu I think Luries argument is pretty cool math.harvard.edu/~lurie/papers/bwb.pdf

(Theorem 2)

Mike Miller

@Danu Yeah. It seems plausible to me, as the Lie groups themselves have this property.

curiosity

Hi guys quick question: im reading a physics paper and the author arrives after some calculation at this: $$ \sum _{l=0}^{\infty } \frac{(2 l+1) \Im\left(f_l\right)}{k}=\sum _{l=0}^{\infty } (2 l+1)
|f_l|^2 $$

and then concludes that $$ \Im(f_l)/k=|f_l|^2 $$

but im not sure that this follows just because the serieses are the same

Balarka Sen

12:26

@Alex "Blackwater Park" is actually a pretty good album

curiosity

but maybe because both series contain tis (2l+1) summand we can conclude it? or does the fact that both sides contain the $ f_l $ make sure that all the summands have to be the same?

Alex

@BalarkaSen I'll smash it now, and do some goresky

Balarka Sen

I'm like 1/3 through it

Alex

The album, or through lectures 1&2?

Balarka Sen

Album. I pretty much haven't looked at the lectures

I'm a procrastinating bum

Alex

12:33

Naughty rascal

!

Balarka Sen

did you suddenly turn into an Englishman from the nineteenth century

thats not how Australian insults go as far as I know

Alex

Depends on how many layers of irony one has ascended

Balarka Sen

also tru

Alex

Lol, goresky sends me back to learn AT some more :P

Balarka Sen

Where are you rn

loch

12:37

@LeakyNun this almost always doesn't happen right]

Alex

I haven't checked some statements, but at 1.7.d)

Balarka Sen

Ah

Well i don't know why the image presheaf is not always a sheaf

but im sure its just a matter of coming up with an example

Oh

consider the sheaf of 1-forms over the circle

d should be a sheaf morphism from the sheaf of smooth functions to the sheaf of 1-forms

Alex

Yeah I think thats the normal guy

Balarka Sen

the image is the presheaf of exact forms on the circle... but that's not a sheaf, right?

exact forms don't glue to exact forms

glue them over the circle

blat

12:52

Hi

We define the hodge operator $\ast: \Lambda^p \to \Lambda^{n-p}$ by $a \wedge \ast b = \langle a,b \rangle vol$. How do we know this is well defined, i.e. why does such $\ast b$ exist and is unique?

Semiclassical

13:03

If two such $n-p$ forms $c_1,c_2$ satisfied $a\wedge c=\langle a,b\rangle vol$, then $a\wedge (c_1-c_2)=0$

And since this would need to be true regardless of $a$...

blat

I see thanks

How about the existence?

loch

@BalarkaSen i think in general whenever you have a surjective morphism of sheaves $F\rightarrow G$ (i.e. image = $G$) which is not surjective on every open, then the image presheaf is not going to be a sheaf (if it was then sheafifying it does nothing, but it's not G now, which contradicts image = $G$)

Balarka Sen

@loch I don't understand the sheafification argument.

What is not $G$ now?

loch

i guess i was being sloppy both notatino and argument

Balarka Sen

Are you sheafifying the Hom presheaf, $U \mapsto \text{Hom}(F(U), G(U))$?

loch

13:17

I'm saying - let $f:F\rightarrow G$ be a morphism of sheaves. We define $im(f)$ to be the sheafification of the image presheaf $im(f)^{pre}$, and say $f$ is surjective if $im(f) = G$.

If the image presheaf $im(f)^{pre}$ is a sheaf, then sheafifying it doesn't do anything. The image presheaf is a sub-presheaf of $G$, and if $f$ is not surjective on every open it is not going to be equal to $G$.

Balarka Sen

Ah.

Gotchu

My example is a special case of this, of course, because $d : \Omega^0 \to \Omega^1$ is a surjective morphism of sheaves - it's surjective germ-level because every closed form is locally exact. Germinally closed forms are exact.

But it's clearly not surjective on every open set (eg the full $S^1$)

loch

yep

Mike Miller

I imagine that instead of cooking something up via general principles, we should be able to think vaguely about how being a sheaf should fail and cook up an example by hand that does it

I do like the d

Example though

Balarka Sen

His general principle generalized my concrete example so it passes the "algebra? bullshit or not" test

loch

yeah i dont think the remark i made was important - i was just happening to think about it

Leaky Nun

13:22

@loch so when does it happen?

Balarka Sen

@MikeMiller r/nocontext

loch

@LeakyNun you want your category to have kernels etc., but so in particular you want your ideals to be R-algebras, which by your definition also contains $1$..

Leaky Nun

aha

Balarka Sen

Algebras should not be unital

Leaky Nun

fight with me

Balarka Sen

13:23

But even then cokernel fails in general

Leaky Nun

1

A: Test for equality of quadratic extensions

The notation $\operatorname{Gal}(F(\sqrt a)/F)$ assumes that $F(\sqrt a)/F$ is separable, but $X^2 - a$ has derivative $0$ in characteristic $2$. For a counter-example, set $F = \Bbb F_2(t)$, then $F(\sqrt t) = F(\sqrt{t+1})$ but $\sqrt{t(t+1)} \notin F$

How to show that $\sqrt{t(t+1)} \notin \Bbb F_2(t)$?

loch

i havent read the post carefully, but arent they assuming char $\ne$ 2

Leaky Nun

you haven't read the post carefully

loch

aha

i didnt read the first line!

Leaky Nun

but neither did I, because I missed "please check for correctness"

loch

13:28

@LeakyNun anyway i think you cna just check by hand? i.e. take any element in $\mathbb{F}_2(t)$, and square it, and say it's not $t^2+t$

Leaky Nun

eh... there's infinitely many elements

Mike Miller

@Balarka ;)

If A is a finite C-algebra presenting an affine complex variety, what is the algebra corresponding to blowup at a point?

Alex

Yesterdays mistakes :P

loch

the blow up isn't affine in general though

actually

Mike Miller

@loch oh rip

That surprises me I guess

Balarka Sen

13:33

It's the P R O J C O N S T R U C T I O N

loch

^

haha

i guess it's not too surprising, if you blow up $\mathbb{A}^2$ you get a copy of $\mathbb{P}^1$ sitting inside the blow up as the exceptional divisor

which can't happen if your thing is affine

Balarka Sen

It's taking connected sum with something projective my dude

it's not an algebraic construction

of course algebra fails

loch

it's an algebraic construction! well maybe there are notions of blow ups away from alg geom

Balarka Sen

It's a geometric construction, the algebraists had to save grace so they discovered the obscene proj

Mike Miller

Somehow the picture to me seemed like I should be able to make it affine one dimension up but I am v clueless

Balarka Sen

13:37

The extra dimension is projective

Mike Miller

Anyway who even cares about geometry

2

Leaky Nun

I don't think you can embed P^1 into any affine space

Balarka Sen

Blowup of A^2 at a point sits in A^2 x P^1

loch

@BalarkaSen forgetting the algebra means that you can't blow up a fat point!

Mike Miller

If you can't make it an infinity functor I'm outta here

@BalarkaSen aha of course

Balarka Sen

13:43

@loch Those need fitness training, not blowing up

Mike Miller

Lmao

loch

anyway if you're blowing up an ideal $I$ in an affine thing (say $\operatorname{Spec}(A)$) then the blow up is $Proj(\bigoplus_i I^i)$ , i.e. taking proj of the Rees algebra

which looks kind of scary, but if $I=(f_0,\ldots,f_n)$, then you have a surjection of graded rings $A[x_0,\ldots,x_n] \rightarrow \bigoplus_i I^i$ mapping $x_i \mapsto f_i$ in degree $1$

which gives you a closed embedding of the blow up to $\mathbb{P}^n \times A$, in particular the blow up is cut out by the equations defining the kernel of the surjection of graded rings

Mike Miller

[eyes bulge, bowtie swivels 720 degrees]

3

loch

@BalarkaSen lmao

i think some of my recent SE answers are on blow ups

Mike Miller

I don't know how to fit this with the geometry yet, but one day

Leaky Nun

13:50

@loch eh... were you typing something about my question?

loch

i think what i said is just a helpful way of computing examples

i was then i forgot

Leaky Nun

but you deleted your message

loch

oh i made a mistake

Leaky Nun

@loch do you by any chance league?

loch

what about this :$\frac{f^2}{g^2} = t^2+t$. If $t^i$ be the max $i$ such that $t^i$ divides $g$. Then we have that $t^{2i+1}$ is the max $i$ dividing $f^2$, but that can't be true?

long time ago - nowadays i only follow the competitive scene

Leaky Nun

13:54

my university is so inefficient

I still don't have my results

loch

they should be coming out soon

Leaky Nun

very inefficient

loch

if i remember correctly in my third year it came out on something like july 11

Leaky Nun

very inefficient

Balarka Sen

@Alex u don't like 1.7.(d)?

thats stuff is jam my dude

Alex

14:01

I didn't say I didn't like it, it's just that I've been meaning to properly read Hatcher 1.3, so I went to do that before reading on

Balarka Sen

Ahh

Alex

Typically I'd just pick up what's needed and then move on, but since I have some spare time, I might as well try to cross this off my list

Balarka Sen

It's not too terrifying, if you have a representation $\rho : \pi_1(X, x_0) \to \text{GL}_n(\Bbb C)$, you can make a complex vector bundle rank $n$ over $X$ by taking a good cover $\{U_\alpha\}_{\alpha \in I}$ of $X$, taking patches $\coprod_{\alpha \in I} U_\alpha \times \Bbb C^n$ and gluing according to the "monodromy information of $\rho$"

i.e., if $\gamma$ is a loop at $x_0$ and $U_1, \cdots, U_n$ is a cover of $\gamma$ by the elements of the cover, patch $U_i \times \Bbb C^n$'s togather so that the composition of the (linear) transition functions $\varphi_i : U_i \cap U_{i+1} \to \text{GL}_n(\Bbb C)$ is $\rho([\gamma]) \in \text{GL}_n(\Bbb C)$

Alex

Sure

Seems reasonable

Alex

14:29

In 1.7.e I guess he just gives up on the twiddlification :P

Balarka Sen

I am not even going to read that example :P

Alex

It just ends with [to be completed]

Balarka Sen

Yeah saw

who even cares about Spec R

Alex

You can just whip open your desk copy of Hartshorne to II.5

Or even your pocket copy if you're wearing it

Balarka Sen

Instead I am whipping open Hatcher to revisit on homology with local coefficients

Alex

14:31

What a chad

Balarka Sen

:GWChadThonk:

Alex

@BalarkaSen What's the easiest interesting choice of $X$, $x_0$ and $n$ here?

(that you have in mind)

(to see this monodromy information)

Mike Miller

Ask me about monodromy

Balarka Sen

@Alex Here's an interesting example

Semiclassical

Like, the monodromy matrix in Floquet theory? :P

Mike Miller

14:43

No

Semiclassical

lol

Balarka Sen

(And Semiclassical you might like this example)

Semiclassical

huh. check out the first reference on the wiki page for the monodromy matrix (in the floquet sense): en.wikipedia.org/wiki/Monodromy_matrix

That title escalates quickly.

Balarka Sen

Say $X$ is a Riemann surface and $A$ is an $n \times n$ matrix of holomorphic $1$-forms on $X$. Consider the differential equation $dF= A\cdot F$ where $F$ is an $n$-vector of holomorphic functions on $X$.

This in general won't have a global solution: $F$ will be multivalued over $X$

So pass to what happens locally. Say $(z_1, \cdots, z_n)$ are local coordinates in a holomorphic chart on $X$. Then our differential equation looks like a system $\partial F_i/\partial z_j = A_{1j}F_1 + A_{2j}F_2 + \cdots + A_{nj} F_n$ of holomorphic ODEs in that chart.

This has a unique solution on that local chart (Think of it as holomorphic Picard-Lefschetz, but it's much easier to prove).

Semiclassical

I think I know the name of what this is getting at

Mike Miller

14:49

I don't really get the humor @SemiC

Semiclassical

It wasn't really intended humorously.

Balarka Sen

The problem, of course, is that if you have a loop $\gamma$ in $X$ based at some point, say, $z_0 \in X$, and try to analytically continue $F$ around that loop (using that it is well-defined locally), we might end up at two different $F$'s at a local holomorphic chart around $z$.

And this precisely happens when $\gamma$ is not null-homotopic.

Semiclassical

Except perhaps in the dark sense of "I guess that's one interpretation of 'optimal control'"

Balarka Sen

So what you have is a solution defined over the universal cover $\widetilde{X} \to X$ of the Riemann surface, i.e., the analytically continued $F$ is an element of $\mathscr{O}(\tilde{X})^n$.

Mike Miller

I can't see that phrase anywhere there

I'm really confused by what you mean

But it doesn't matter much

Semiclassical

14:52

"Optimal Control of Nonlinear Processes: With Applications in Drugs, Corruption, and Terror"

3

that's the first reference in the monodromy matrix wiki page

Mike Miller

Ah I am sorry, I was looking at See Also and the related pages instead!

Semiclassical

ah, lol

Mike Miller

"This page on Floquet theory seems completely tame..."

Semiclassical

ikr

It makes the subject go from seeming technical and innocuous to rather sinister

Balarka Sen

$\pi_1(X)$ acts on $\tilde{X}$ by deck transformations, therefore acts on the values of $F$ on $\tilde{X}$ as well. Precisely, if $\gamma$ is a non-nullhomotopic loop in $X$ based at $z_0$, $\gamma$ acts on $\Bbb C^n$ by linear transformation that takes the $n$-tuple of values of $F = (F_1, \cdots, F_n)$ defined at $z_0$ to the other solution $G = (G_1, \cdots, G_n)$ that one gets after monodrom-ing around.

That gives a representation $\pi_1(X) \to \text{GL}_n(\Bbb C)$

This is the "monodromy of the system"

Semiclassical

14:56

nice

what I was thinking about was Jacobi's inversion theorem

Balarka Sen

The question now is "which representation (read: flat complex bundles) of a surface group appear as monodromy representations of systems of holomorphic differential equations?"

The Riemann-Hilbert correspondence says all of them

The category of complex linear representations of a surface group is the same as the category of local systems on the surface.

Alex

@BalarkaSen Why are there n coordinates on a complex 1-manifold?

Balarka Sen

@Alex I have a system of n (first order) differential equations.

Transcript for

$Mathematics$ Mathematics