the construction I mentioned is general construction

it works with any category doesn't have to be topological one

any exact category

Why is it interesting? <-- this is the question you should answer then

Oh because one can do intersection theory on the level of K-groups

Then, you don't need moving lemma's and such

Tell me how you'd do it.

@MatheinBoulomenos Well, I cant give you another way, but I can't prove that. maybe I just need more practice to see the full picture

I do not know how to do it, for one. Can you do it for topological K-groups as an example?

Just plain old $X \wedge (BO \times \Bbb Z)$ if you want.

@BalarkaSen taking the algebraic K-theory of rings of continuous/smooth functions gives you topological K-theory

That's the spectrum

@MatheinBoulomenos I know you know good answers to the question. I'm trying to get Adeek to tell me his answer.

(That's part of the Serre-Swan correspondence, of course)

wait

isn't algebraic K-theory the same construction for any exact category @MatheinBoulomenos ?

it is

@WilliamOliver but do you see that the usual thing of how the dualization functor acts on morphisms is kind of a "natural" thing to do? You have a linear map $f:V \to W$ and you have these dual spaces $V^*$ and $W^*$ and you want to define somehow a map between them. Well, if you have an element in $W^*$, i.e. linear map $W \to K$, then composing that with $f$ gives you a linear map on $V \to W \to K$, so an element of $V^*$

It is given by $\lambda$ operation I don't know details ring operation on K-groups yet

This construction is how many functors works

rings of continuous/smooth/holomorphic/regular functions, differential forms etc.

I just wanted to discuss why I find K-groups interesting for now @BalarkaSen because they give dictionary between cycles and K-groups

in all these situations, you make these into a functor by the kind of construction I described above

if you do that, then contravariance is a consequence

@Adeek What is this dictionary? Explain.

@MatheinBoulomenos I don't have much experience with functors so I think I am missing the intuition that you have

So there is an isomorphism between $K_0(X)$ tensored with Q and cycles

i.e $A(X) \otimes Q$

@WilliamOliver but can you agree that the construction I described is kind of "the obvious thing to do"?

this is given by GRR

What is the category you're taking homotopy groups of the nerve of when you say $K_0(X)$?

oh no $K_0$ is seperate construction

than higher K-groups

@MatheinBoulomenos I don't know...

So you still haven't told me why your construction is useful

Well, we can define higher algebraic cycles

and then $A^{i}(X) \otimes Q \cong K_i(X) \otimes Q$

@MatheinBoulomenos Really? This feels off to me

Algebraic cycles are divisors (linear combination of codimension 1 subvarieties). What are higher algebraic cycles?

@MikeMiller isn't this Serre-Swan-stuff?

so we can model things like homology

@BalarkaSen we define the following

@MikeMiller I feel like any K-theory of C(X) should agree with the theory of projective modules on C(X)

Freudian slippery

It holds for $K_0$ at least

Yeah, I'm worried about any statement other than K_0

but maybe I've been to quick with higher $K$-groups

Ah

hmm, you may be right

we define $\Delta^m$ in similiar way as homology

I have no intuition for higher K groups other than formalism

I think I've heard that it's actually hard to recover topological K-theory from algebraic K-theory

@MatheinBoulomenos Sorry.. I really think its an experience thing.

i.e $\Delta^{m} = spec(k[t_0,\ldots,t_m] / (1 - \Sigma t_j)$

Then we define higher chow group to be homology group of the $\partial$ map

@MikeMiller oh, you're probably right. I think I had the $K_0$ thing in mind and incorrectly extrapolated it

it turns out that such thing when tensored with Q is isomorphic to higher Chow group @BalarkaSen

again I don't know yet the details

@Adeek I do not understand. Your Delta^m is a hyperplane.

It's not a simplex, how do you define any simplicial thing out of it?

yeah so we define the following thing mimicing the regular classical cycles

(In particular, what does "boundary of a simplex" mean?)

$z^{r}(W,m) = \{ \tau \in z^{r}(W\times \Delta^m) : \tau \ meets \ all \ faces \ properly\}$

then define

What is $z^r(W \times \Delta^m)$

the way I visualize it is something like a triangulated variety

I am not sure if that is the correct way to visualize it

I don't want a visualization. I need a definition.

I just defined it

oh

Where?

$z^r(W \times \Delta^m)$ is the regular cycles

@MatheinBoulomenos Oh wait, actually I think I totally get it now, I think it all just clicked!!

i.e irreducible subvarieties of $W \times \Delta^m$

dimension $r$ or codimension $r$?

codimension

Ok, so I know $z^r(W, m)$. Now what?

there is a boundary map

Wait, what does saying $\tau$ meets the faces even mean?

What faces?

$\tau$ meets faces $\{t_{i1} = \ldots = t_{il} = 0\}$ $l \geq 0$ properly

You should have explained all this in one sentence rather than ten million sentences but no matter. So you have a boundary map, which I can guess.

I am not entirely sure why $\partial^2 = 0$ for that boundary map, but I'll take it

This homology group is $A^r(X)$?

yeah

OK great

Now the Grothendieck-Riemann-Roch theorem is more interesting.

If that's what it is

Maybe that's for $r = 0$, whatever

for r = 0

yeah

for lower Chow group this is achieved through GRR. For Higher Chow groups this is achieved by Bloch conjecture.

I would like to understand both

This is nice, this is nice.

I like this.

I don't understand them yet. characteristic classes for varieties are mysterious for me as of yet.

anyhow, for my thesis I am looking at the complexity of both lower Chow groups and higher Chow groups.

@WilliamOliver glad to help

@BalarkaSen From the point of view of transcendental methods and arithmetic methods.

@MatheinBoulomenos Omg okay, so if you throw out the idea of these things being maps, you could think of it as just taking a matrix to the same matrix basically, because thats what composition means in this context right?

I am not interested in those words but I like how this gives a way to define intersection of cycles inside varieties.

@MatheinBoulomenos media.giphy.com/media/75ZaxapnyMp2w/giphy.gif For some reason I just never thought about it like that.

yeah It was mentioned to me that in Arakelov geometry people use K-theory to define intersection of cycles

@BalarkaSen Relevant [14:27]

@MatheinBoulomenos This is what I was thinking of

anyway I am gonna go back to study for my exams and do work

cya all

@BalarkaSen @MatheinBoulomenos and everyone

@WilliamOliver to think of everything in terms of matrices, you need to choose bases for $V$ and $V^*$ if you do the sensible thing and choose a basis for $V$ and the corresponding dual basis for $V^*$, then you can check that this construction I described just gives you the transpose

cya @Adeek

bye

@MikeMiller I see, thanks for the correction

Are there any advantages of 1k reputation?

Problem: Give an example of a collection of sets $\mathcal{A}$ that is not locally finite, such that the collection $\mathcal{B} = \{\overline{A} \mid \mathcal{A} \}$ is locally finite. Attempt: Consider $\mathcal{A} = \{\Bbb{Q} + r \mid r \notin \Bbb{Q}\}$. Clearly this is not locally finite: since translation is a homeomorphism, $\Bbb{Q} + r$ is dense for each $r$; hence, given any open nbhd of some point $x$, it will intersect $\Bbb{Q}+r$ for every $r \notin \Bbb{Q}$, so $\mathcal{A}$ is...

...is not locally finite. But $\mathcal{B} = \{ \overline{\Bbb{Q} + r} \mid r \notin \Bbb{Q} \} = \{\Bbb{R} \mid r \notin \Bbb{Q} \} = \{\Bbb{R} \}$ is locally finite, since it only contains one set, so any open set always intersects finitely many sets in $\mathcal{B}$....How does this sound?

@MatheinBoulomenos Thanks

$\overline{\Bbb Q+r}$ can't be $\Bbb R$ @user193319

@AkivaWeinberger Why? Don't homeomorphisms map dense sets to dense sets?

Oh sorry I thought the line meant complement, not closure

You're right