So sometime we would like to see if a space has hodge structure as that provides certain kind of decomposition on cohomological level. For example with Khaler manifold we know it has Hodge decomposition.

memes are not whiskeys

they die with age

like mortal men

but it turns out (I am not sure of the details) that non-khaler manifolds don't have such hodge decomposition.

i.e nice decomposition on the level of cohomology

@Adeek Hm, I only vaguely know about Hodge structures. Can you give a intro?

sure

@BalarkaSen nah many memes age quite well

Though it's one of those things where in the first month or so you should be using them every 5 minutes

Depends on the target @Daminark

Then after that for a few months it's very rare

@BalarkaSen you can do cool things with Hodge structures

And after that you reintroduce it in moderation

@MatheinBoulomenos jesus

so essentially a mixed hodge structure is a linearization of a space. More precisely,

you're starting my PTSD again

The data given by hodge structure is given by an $A-module$

which has finite descending filteration for the complexified space which goes up and goes down

@Daminark Oh yeah?

@skullpatrol if targets don't agree then they're just wrong

and they interact through regular hodge structure

Hm... why doesn't the latex embed properly here?

~$realmusic

Finest piece of classical music of all time

Try an immersion instead

a hodge structure is as follows

@Pseudohuman Thank you!

@BalarkaSen You're welcome!

Let $A$ be a subring of $\mathbb{R}$

@Pseudohuman I have been slain and I accept it

a hodge structure of weight N is given by a finitely generated $A$ module

@Daminark you should use them more the closer you are to their inception. At the moment of meme-genesis you should be using it infinitely often.

It's a bit hard to read the latex properly when not embedded

@Adeek Okay

such that $V_{\mathbb{C}} = \bigoplus_{p + q = N} V^{p,q}$

$V_{\mathbb{C}} = V \otimes \Bbb C$ being the complexification?

yeah

so essentially hodge structure tries to capture information that is like hodge decomposition

So is there a grading in $V^{p, q}$?

yeah

I mean, maps $V^{p, q} \to V^{p+1, q}$ and $V^{p, q} \to V^{p, q +1}$ so that they form a double complex or something?

yes

Hodge structure don't really work for varieties, so in order to have some kinda of decomposition on cohomology level we introduce Mixed hodge structure

which is essentially breaking the space apart

For example if Y is complex variety, then $H^{i}(Y,\mathbb{Z})$ has canoncial mixed hodge structure

Let's take it slow there, I do not know anything about this. So what is the significance of a Hodge structure?

Ah

The significance is that we try get some nice decomposition for the cohomology

So if $M$ is a complex manifold, when does $H^\bullet(M; \Bbb R)$ admit a Hodge structure?

Is the Hodge structure coming from the doubly graded complex de Rham theory?

For non-khaler it doesn't yeah to second answer

Oh

I should paraphrase that it is spectulacted that non-khaler don't admit hodge structure with answer being 90 % true

I see

But with schemes they have more local algebraic data

Why do we get a Hodge structure on $H^\bullet(M; \Bbb R)$ if $M$ is Kahler? Is there a short answer?

yeah ofcourse

It is divided into p,q

this is by hodge decomposition theorem

But where do we need the Kahler structure specifically to --- Oh ok

I am sorry ? I don't understand.

I was going to ask why the Kahler structure is essential for this to work but you replied with "Hodge decomposition theorem" before finishing my question :)

@BalarkaSen This needs subtitles badly

:)

I think if you don't have a Kähler structure, then the spectral sequence associated to the complex de-Rham double complex doesn't necessarily converge (though I can't explain why exactly, I think there's quite a bit of elliptic PDE involved iirc)

I think your right @MatheinBoulomenos

Wow.

Can anyone tell me what a "Euclidean Vector Space" is? Is this just a vector space with an inner product defined?

It turns out those MHS and HS is linked to something called motivs

motifs

@AkivaWeinberger True

@Adeek What're those?

Hey folks - I don't use math.stackexchange much. I have a question about finding an algorithm that will efficiently identify a particular type of subgraph in a directed graph. Is that a legit type of question to ask on math.stackexchange?

which is something very deep, but I don't know anything yet of that thing. Essentially motifs is something associated with a variety

essentially it is like to break the variety into something like CW complex

@Brionius Sounds like a computer science question

@Brionius It's possible that the question will fit better on cs.stackexchange or maybe even stackoverflow depending on what exactly you want to ask

@William

I thought those are called stratifications

@William ok

thx

for example you know atiyah-hirzebruch spectral sequence for universal cohomology theory ?

I'm afraid not

What is it?

I don't know much of it but the gist that I get out of it. Is that it is universal cohomology theory on topological space, where you have some spectral sequence converging to all cohomology theories

don't quote me on this though.

but anyway in algebraic geometry we would like to have a universal cohomology theory

but that isn't known to exist yet and a lot of it is conjectured

I don't follow. What is a universal cohomology theory?

It is something that somehow captures all known cohomologies

I don't understand what that means

Adeek, IIRC I remember reading on the connection between (∞,1) -topos and cohomology

*universal cohomology

For example there is singular cohomology, de Rham cohomology, l-acidic, crystalline cohomology, etc

so one natural question to ask if there exist a universal cohomology that captures all of those things

It turns out from my understanding atleast now that in topology such thing exist through Atiyah-Hirzebruch

I don't see why that's a natural question. What does "capturing" mean?

See, this is the kind of thing I was objecting to

@BalarkaSen it means it has the properties shared by all of those cohomologies

isn't every cohomology theory just counting connected components in the hom set of some $(\infty,1)$-topos?

Singular cohomology and de Rham cohomology are already equivalent theories by the de Rham theorem

@MatheinBoulomenos exactly what I was referring to

What are the properties of the other theories that are not shared by these two?

Their is certain axioms for example that they must satisfy which is defined by Weil

I am not sure of the details yet as I am still learning this stuff

Those are all satisfied by singular cohomology and deRham cohomology in the smooth manifold category

The extra axioms are just Lefschetz hyperplane theorem and the fixed point theorem

yes, but for example l-acidic cohomology is hard to prove those axioms

But this does not explain your claim that you need a cohomology theory which shares properties of all the cohomology theories you listed

Because all of these theories satisfy the same properties as far as I can see, in the appropriate categories they are defined on

@MatheinBoulomenos a relation on a set and a partion of a set, i know they are the same thing, but can you explain to me more that idea?

In particular you haven't given me a concrete example of such a property

@WilliamOliver I think it just means $\Bbb R^n$

@Adeek Do you see my point? It's this kind of flabby motivation that I was objecting to. It's not a concrete intuition or point of view in any way

@BalarkaSen I am not sure yet of what exactly is this this universal cohomology theory

@KasmirKhaan you mean an equivalence relation

@MatheinBoulomenos Yes sir!

@Adeek Fair. You should learn it then, and then we can continue this discussion.

I find it hard to see that if we take a set , and partition it in any way we want

then there is an equiv relation that describe that

@BalarkaSen Alright, I will cover those things in my thesis

just define two elements as equivalent if they're in the same set from the partition

I could send it to you once I am done

say if we the integers from 1 to 10

can we define a partition this way

1-4 in one class, 5 in a class, 6-10 in a class

how would we describe that relation ?

do what I said

you have a partition

and then you say two elements are equivalent iff they're in the same subset from the partition

so it is something we claim

not somethign we "find"

like a function

well, it's something you can define

not every function is given by a nice formula

same thing with relations in general

that is what i was driving at :D

because it was very clean criteria

when we had agroup hom

a R b iff a'b in H

H being the kernel of the map

Hello!!

@MatheinBoulomenos thanks mathein as allways :D

@MatheinBoulomenos You're thinking of the spectral sequence of the del delbar double complex? The point being that taking homology with respect to the two derivatives gives the E^2 page of a spectral sequence, whereas the E-infty page is (the associated graded) of the homology of d = del + delbar.

yeah

Now you can identify Ker(d)/Im(d) with the kernel of $\Delta$, the Hodge Laplacian, on a compact smooth manifold (this proof is the elliptic PDE you mentioned), and similarly the Dolbeault cohomology is the kernel of $\Delta_{\partial}$, the complex Laplacian. If I'm not being foolish nothing so far has needed a Kahler structure

I think Kähler gives you a degeneration after the E^2 page?

The Kahler structure furnishes you with $\Delta_{\partial} = \Delta$

or something like that

In particular, the rank of the Dolbeault cohomology (summed over p+q = n) is equal to the rank of regular cohomology

Oh, so ker(d)/im(d) \cong \ker Delta is like, picking a harmonic representative out of a cohomology class? (the Hodge theorem)

Yes

The spectral sequence therefore must degenerate on E^2 - the rank of E^2 is the same as the rank of E^infty

It's just a dimension count

The higher differentials in the non-Kahler case are... mysterious to me

Okay that clarifies things, thanks @MikeMiller

Sure sure

@MatheinBoulomenos savage af

tHe wOrlD iS mAdE oF aLgEbRa

everywhere i go there's algebra

lord help us

sounds like a 10/10 world to me

the preimages of elements of T

give the partition

@MatheinBoulomenos can you please tell me what I wrote make sense or not :D

the non-empty preimages, yes (you don't allow empty sets in a partition), so if f is surjective, you're good

@lush that word belongs to the mavericks

don't ever say that

i like the new guy

@MatheinBoulomenos thanks :D

mavericks?

sorry it's a logan paul meme

i forget that not everyone is into youtube drama

You can think of $O(2)$ as the group of rotation matrices. Why is it that the set of clockwise rotations is not connected to the set of anticlockwise rotations? Cannot you always arrive at a clockwise rotation through an anticlockwise rotation?

ahh, I think I know where you are referring to :D

I mean rotating clockwise by $x$ is the same thing as rotating counterclockwise by $2\pi - x$

Can anyone help me with my latex?

yes exactly

\o

I am unable to fix it in an answer.

o/

@Pseudohuman Hey hey hey

@BalarkaSen composer | Mick Ralphs

So I don't see why you say the clockwise and counterclockwise rotations aren't connected

@Skortya The determinant of both is +1 and -1

You can't go continuously from det 1 to det -1

@Pseudohuman Interesting piece of useless information

@BalarkaSen I do not understand.

Since the determinant is a continuous function

~$learn cancer buy dat merch

@Slereah, ok I get that. I am still puzzled why my mental model of connectedness is failing

@BalarkaSen Learned command: cancer

Good

@Skortya O(2) is not a group of rotation matrices; that's SO(2). O(2)'s identity component is SO(2). The other component consists of reflection matrices

just "draw" a path in $O(2)$

And take the determinant of each point in that path

That will give you a path in $\Bbb R$ by continuity

Every rotation matrix has determinant +1, and the fact that they can be connected to the identity matrix through a path of rotations is essentially proof

@Mike Miller, right right right, my mistake

yeah I get it now, thanks

@AkivaWeinberger Thanks. I am asking in the context of natural transformations between vector space category and the dualization functor. So saying that there only exists natural transformations when the vector space is a euclidean space is basically saying that they exist only when the vector space is finite dimensional?

I don't know enough about category theory to answer that

@AkivaWeinberger Okay, thanks!

Hi @Ted

Hi Mathein

Hi @Ted

Hi Balarka, DogAteMy

Hey Ted!

Hi Demonark

~$learn ted Geometric Geometry: The Geometric Approach

@MatheinBoulomenos I am assuming that when people say this they mean to include dinatural transformations

@BalarkaSen Learned command: ted

~$ted

Geometric Geometry: The Geometric Approach

Beautiful.

Have we become an artificial intelligence room?

I recently found out that we can teach this SE-wide bot custom commands

I have been at it with the memes from then onwards

Lovely.

Will this bot be all over Facebook for the 2018 US elections?

Good call

@MatheinBoulomenos Also, I am kind of confused about why the dualization functor has to be contravariant. Like its basically defined to be the contravariant hom-functor right? But why is it defined that way and not some other way?

What is dual?

if it was defined in any other way, I wouldn't call it dualization functor

~$learn MAGA I have to have my China

@BalarkaSen Learned command: MAGA

smacks Balarka hard

Good thing I'm leaving to go to a concert

Hahah. Enjoy

Hi @Ted

@MatheinBoulomenos Well yeah, but the concept of a contravariant vectors existed before functors I am sure. How is that shown to be described exactly by the dualization functor?

I don't really understand the question, sorry

I'm off, cya guys

Tschau, @lush

Byebye @lush

I guess I am asking, before category theory, there was the concept of components of linear map representations changing in the "opposite direction" of things like position vectors when you changed the basis of the position vectors. These linear maps were called contravariant because they changed in the opposite way. Is this the same concept that contravariance in category theory describes?

specifically, contravariance of the dualization functor

I need a plan. How do you plan the planning? I guess I need a plan for planning the planning.

Planning is hard.

You need an Eilenberg swindle

A plan and a plan to destroy that plan and swindle it to infinity

Shift it up a little bit

and you have your plan

@WilliamOliver the contravariant vectors are ordinary vectors. dual vectors are covariant. so it's the opposite

the terminology is totally different

@EricSilva So is this to say that it is totally unrelated?

yes

@WilliamOliver contravariance of the dualization functor is just a statement how taking a map to its dual behaves under compositions

Ugh, so I wasted all this time trying to understand this when there was never a connection in the first place? There is no connection at all??

we use a few words to mean a lot of different things

Hey @BalarkaSen do you want me to tell you few things about K-theory ?

Go ahead.

@EricSilva Okay, so for example, if we restrict ourselves to finite dimensional spaces. Is it completely just convention to represent a bilinear map by $V^* \otimes V^*$ as opposed to $V \otimes V$

One thing I like to use is the following analogy

@WilliamOliver no, that's not just a convention

@MatheinBoulomenos Why??

so if I tell you the concept of the cup then in your head you will have the image of the cup. However, it is more accurate the describe the cup by its interaction with the enviroment.

So, for example we could use the same logic for categories.

Yoneda lemma, sure.

no? im just saying the way the coordinates transform and the terminology used for that has nothing to do with the category theory term

Categories are really described not by the objects of the category, but by the morphisms.

Using the data of morphisms we can build simplicial set out of a category.

That's a bad example but sure go on.

But first in order not to deal with huge category we would like to linearize it more

@EricSilva Could we not switch $V^* \otimes V^*$ and $V \otimes V$ and be totally fine? I know by definition this is wrong, but is there any other reason why this wouldn't produce a consistent theory that matches intuition?

First we apply some construction which breaks down the category apart

by linearizing it

then after that from that category we build a simplicial set

ofcourse, from simplicial set we can topologize it

then the K-groups are defined to be the homotopy groups of that final construction

I'd say that's an awful definition but whatever floats your boat.

this has nothing to do with coordinates @William i dont understand where the confusion is coming from

Chalkboard, whiteboard, paper, or plain old latex editor to do your math guys?

what does this have to do with the fact that contravariant means a different thing when youre talking about components vs functors

@WilliamOliver Honestly I'm not very familiar with how geometers use "covariant" and "contravariant", so I'll talk about vectors and functionals instead. Let $V$ be a finite-dimensional vector space over $K$. The dualization functor induces a map $f:\operatorname{End}_K(V) \to \operatorname{End}_K(V^*)$. Assume you know two things: 1) the dualization functor is contravariant 2) the double-dual functor is naturally isomorphic to the identity functor.

I find myself switching back and forth between those media

@MatheinBoulomenos Wait, thats where I am getting confused, assumption 1) why are we making that assumption?

I can't seem to fixate myself on just one particular setting. Sometimes I do my work on an editor, sometimes through a whiteboard (thinking of switching to chalkboard though), and sometimes on paper/pen/pencil

@MatheinBoulomenos basically contravariant means that when you change basis, you have to multiply by the inverse map to get the new components for your vector in your new basis. So if you take an ordinary vector in $\mathbb{R}^{3}$ and multiply the standard basis by $2$, to represent your vector in the new basis you have to divide the components by 2

i.e. contravariance

opposite happens in the dual

so functionals have covariant components

because 1) it is a fact and 2) I want to show how covariance in the category-theoric sense gives you the behaviour under change of basis

And I am sorry @EricSilva haha I realize that I haven't explained my motivations very well I've just been stuck on this for a while and I am frustrated :(

@Adeek Was that all?

I was just typing

@MatheinBoulomenos but its a fact by definition, could we not have defined it any other way?

these two properties imply that if you choose an isomorphism between $\operatorname{End}_K(V)$ and $\operatorname{End}_K(V^*)$, then the map $f$ satisfies $f^2= \operatorname{id}$, $f$ is $K$-linear and $f(AB)=f(B)f(A)$

@BalarkaSen Then, ofcourse as we know K-groups are hard to compute because homotopy groups are hard to compute.

No, this is garbage

But then came Grothiendieck-Riemann-Roch which tells us that $K_0(X) \otimes Q \cong A(X) \otimes X$

Such maps are completely classified: math.stackexchange.com/questions/1618223/…

Stop. You're not listening. What you just said is a garbage statement.

also one can achieve higher K-groups as well from higher chow groups

Balarka, what in particular is garbage?

Singular homology is homotopy group of a certain spectrum. Any generalized cohomology theory is.

@WilliamOliver it's fine i was just confused

That doesn't make them hard to compute

Adeek are you a grad student in AG/AT

@MatheinBoulomenos Maybe this is going a bit over my head.

so, up to a suitable choice of isomorphism between $V$ and $V^*$ which gives you the isomorphism between $\operatorname{End}_K(V)$ and $\operatorname{End}_K(V*)$, $f$ must be transposition

It's the sphere spectrum that's hard to compute the homotopy groups of/

yes @ZalTukhara

Ah I see.

@WilliamOliver I'm just saying that you can prove that these two facts give you how the matrix representation of an endomorphism changes

if you take the dual

I'm sure you can generalize this to stuff which is not just an endomorphism

@MatheinBoulomenos Ah okay

@Adeek Whatever you said so far sheds no light on what K-groups are

Try to answer that.

man that's a lot of words to do linear algebra

I gave you the definition didn't I

Yes, and not all definitions are good.

@WilliamOliver you claim that it could be any other way, but how would you do that? remember that the dual is $\operatorname{Hom}_K(V,K)$ by definition. How you define a functor (i.e. an action on morphisms satisfying the necessary axioms) that does exactly this on objects and is not contravariant? I don't see any way to do this, other than brute-force-axiom-of-choice-shenanigans

@Adeek Besides, you haven't even described to me what's the spectrum you're taking homotopy groups of