random cohomology for quantum nerds

MatheinBoulomenos

00:12

@MikeMiller I've stumbled naturally and unexpectingly upon some PDE stuff recently which changed my opinion a bit

Mike Miller

:)

I think ODE makes people too grumpy

for good reason

MatheinBoulomenos

the thing was this: in our modular forms course, we proved the Eichler-Selberg trace formula which is a really cool theorem that related the traces of Hecke operators on cusp forms for $\mathrm{SL}_2(\Bbb Z)$ to (basically) class numbers of orders in imaginary quadratic fields (you can also think about it in terms of binary quadratic forms). I like the theorem a lot, but the proof we did (due to Zagier I think) was just lots of calculations without motivation

So I looked for a more conceptual proof

I don't understand how it works, but it turns out that you can also get this trace formula by studying the spectral stuff associated with the Laplace–Beltrami operator on the upper half plane which gives a more conceptual proof apparently

Mike Miller

Lol I assume you've seen Zagier's sum of two squares proof?

MatheinBoulomenos

I do

Balarka Sen

Ugh that proof

Mike Miller

00:17

If you explain to me all of the stuff above, I bet I can come up with an idea of how these things link together

What is Eichler-Selberg trace? What is a Hecke operator? What is a cusp form?

Lol

MatheinBoulomenos

It's kinda late here

and that would take a bit

Mike Miller

Also I just asked you to give me a full quarter's course

MatheinBoulomenos

but I can explain it some other time

Mike Miller

Only if you feel like it

Get some sleep :)

MatheinBoulomenos

okay

it would be fun to explain that stuff when I have more time

but I'm not really sure if you look at the Laplace-Beltrami operator on the upper half-plane or on the modular curve

probably the first one, since the modular curve is just a sphere and you want to have the hyperbolic metric in there

but it should be $\operatorname{SL}_2(\Bbb Z)$-invariant somehow, I feel

anyway, I'll follow your advice and get some sleep

Good night @MikeMiller @BalarkaSen and everyone else who is still here!

Balarka Sen

00:22

Guten nicht, if my German is correct

loch

goodnight

MatheinBoulomenos

"Gute Nacht" ;)

Nicht = not Nacht = night

Balarka Sen

Well rip

MatheinBoulomenos

@loch goodnicht!

Balarka Sen

Freudfreudfreudfreud

Mike Miller

00:32

@BalarkaSen That is indeed a vaid German word

Alex

@BalarkaSen Keennnnnn

Balarka Sen

@Alex Noice!

Alex

I've been wanting to learn perverse sheaves for a year, but never had the time/prerequisite knowledge

I'd say I am ready now to start

Balarka Sen

But I am not! I am going to sleep now

It's 6 AM

Mike Miller

lunatic

Alex

00:34

I think you'll find that it's 10:30am

Just look at my clock

Balarka Sen

The plan is we'll read lecture 1+2 and discuss on saturday/sunday

Mike Miller

my experience is that it's 9:30 am

where tf is alex

Alex

Sounds like a plan Balarka

Balarka Sen

Alex's timezone sucks

Alex

I am in the best place

Mike Miller

00:35

are you watching the world cup

Balarka Sen

I am

Alex

My country is already dead, so not really :P

Mike Miller

I was wondering if you were literally in Russia

I have no idea where you could eb

Balarka Sen

stralia m8

look down from where you are

Alex

-23.843138, 151.268356

Gladstone, Queensland, Australia atm

Where are you that's 9:30am though :O

Mike Miller

00:38

oh

fucking australians

I'm on an island as well

Alex

Japan?

Tokyo

My man

Mike Miller

Japan, not Tokyo

I am in Gunma prefecture, in the city of Numata, at the Tambara institute

Alex

Temperature looks good :P

Mike Miller

It was a fucking oven recently but a typhoon hits this area in this day and the next

loch

01:48

the classroom looks very nice

Mike Miller

It's what the whole building looks like (it's entirely wood). It's not much more than what you see; there's a lounge directly above that to drink in, and the bedrooms at the other end of a hall

Daminark

02:19

Ah so you live at that school, nice

Mike Miller

02:38

Yeah for another few days

2 more nights here

Daminark

Are you returning to the states after that?

Mike Miller

Indeed

I fly out Saturday at 1PM and land Saturday at noon

Daminark

03:56

@MikeMiller nice, lmao

Alessandro Codenotti

08:15

@MatheinBoulomenos Hmmm is that a product of chains?

Balarka Sen

09:24

@Alessandro Tensor product of chain complexes. If $(K_\bullet, d)$ and $(L_\bullet, \partial)$ are two chain complexes you can construct the double complex $(K_\bullet \otimes L_\bullet, d, \partial)$, which is a big ass 2D commutative diagram with arrows going in and out of the $(i, j)$-lattice points which are $K_i \otimes L_j$

You can make a chain complex out of this double complex using the "total complex" construction

Namely, if $(A_{ij}, d, \partial)$ is a double complex, consider the $\text{Tot}(A)_k = \bigoplus_{i + j = k} A_{ij}$, the sum over the antidiagonal entries.

The total differential is $D = d + (-1)^i \partial$, and $D^2 = 0$ reduces to the fact that $d$ and $\partial$ commute.

That's the monoidal enrichment of the category of chain complexes of $R$-modules.

Alessandro Codenotti

09:43

I think I should really learn about tensor products of vector spaces/modules before

Balarka Sen

Didn't you learn that in Reid's commutative algebra?

That's chapter 2 material iirc

Alessandro Codenotti

Nope, they're not mentioned in his book

Balarka Sen

Really

Huh

Alessandro Codenotti

I'm pretty sure because I remember having a conversation in chat with someone (maybe Mathei) who was surprised a book on commutative algebra doesn't talk about tensor products

Balarka Sen

I studied Reid and Atiyah-MacDonald togather, so maybe I am confused.

A-M has tensor product in chapter 2

Wow it (Reid) really doesn't

That's outrageous.

Alessandro Codenotti

09:53

Can't fit everything into such a thin book! It still contains plenty of stuff for its size in my opinion

But I guess I'll get a pdf of A-M then

Balarka Sen

A-M is way thinner than Reid

That's the point of Reid: Elaborating A-M a bit more

Alessandro Codenotti

Really? I don't know why but I was picturing a big tome

Balarka Sen

It's unbelievably terse, to the point of being unreadable to some

Alessandro Codenotti

@BalarkaSen Maybe Reid just doesn't like tensor products :P

Balarka Sen

Possible!

MatheinBoulomenos

10:13

I think the terse-ness of A-M is managable, they're just very concise. you learn the stuff by doing exercises. But it can't hurt to have a second text that elaborates more, of course

but not doing tensor products is very strange

they're absolutely fundamental

MatheinBoulomenos

10:25

what I mean is that you can't just "read" A-M, you have to work though it: you skim a chapter (that won't take long, since they're very short), you try the exercises, you will probably get some and get stuck on other ones, when you get stuck, you look through the chapter and the previous chapters for results or proof ideas that can help you

I think that's the only way work with A-M. That's not exactly easy, but you can learn a lot

Mike Miller

10:39

Agree

Balarka Sen

+1

Alex

11:55

Should we do this in here @BalarkaSen

Balarka Sen

Yeah better

Alex

We don't want them finding our secrets

Balarka Sen

Correct, not because of ourselves, but because of their sake

We're too good

they wouldn't be able to comprehend how good we are

Alex

Hahaha

So have you read any of this yet?

Balarka Sen

Nope.

Alex

11:58

Presheaves on Top(X)

Let $\mathsf{C=Top}(X)$ and consider $\mathsf{Sets^{C^{op}}}$

Balarka Sen

What is Top(X)

Alex

Top(X) is the category whose objects are open subsets of a top space X, and whose morphisms are inclusions

Balarka Sen

Your notation sucks

Call it Open(X) like normal people

Alex

Who does that?

Filthy analysts?

Balarka Sen

Normal peoples

Alex

12:01

We are creeps atm though

Balarka Sen

Also true

Alex

And $A^B$ means the functor category whose objects are functors $B\to A$ and morphisms are natural transformations

Balarka Sen

That's indeed the category of presheaves on X

Alex

Or on Top(X) if you wish

Since one can say 'the category of presheaves on $\mathsf{C}$'

Balarka Sen

No

No Grothendieck fuckery here

Alex

12:03

:(

But we need those spicy stacks and topoi

Balarka Sen

is regretting inviting Alex

Alex

Hahaha

Balarka Sen

:P

Alex

Note: I still don't know what a stack or a topos is, beyond the definitions

I'd like to though

Balarka Sen

But yeah you can be a soyboy and say the category of contravariant functors from a category to Sets is the category of presheaves on that category

Alex

12:05

:')

Mike Miller

Is a soy boy someone who is salty

Balarka Sen

a soy boy is someone who likes soybeans

Alex

It's someone with 'low testosterone, who is hence not a man'

Mike Miller

I like my interpretation better

Balarka Sen

truly the pinnacle of 4chan insults

Mike Miller

12:06

It's actually a good joke

Balarka Sen

i dunno what soy has to do with low testosterone though

but ill take it

its surreal

Alex

Apparently drinking soy milk reduces test counts

Balarka Sen

lmao

Mike Miller

Meh

Alex

Here, have a totally rigorous study

mensjournal.com/health-fitness/…

Mike Miller

12:07

Maybe I'll read some K theory out of Balarka spite

Alex

Anyway I'll go read the first three pages :P

Good idea @MikeMiller, how about Rosenberg?

Mike Miller

@Alex The joke in main was off of perverse

That's the one I have on hand

Alex

Is that language normal @BalarkaSen 'presheaves on $X$ with coefficients in $\mathsf{C}$'

Balarka Sen

No

Presheaves are dumb

Alex

Is it to motivate sheaf cohomology you think?

Balarka Sen

12:15

I'm not sure what you're referring to.

Alex

Page one of the lecture notes

Balarka Sen

What is to motivate sheaf cohomology?

Alex

Starting to speak in terms of 'coefficients in $\mathsf{C}$'

Balarka Sen

Oh I see I didn't even read or notice that

Sure I mean usually you'll take values in an abelian category

I thought you're making it up just to annoy me

Alex

I wouldn't do that :P

Mike Miller

12:16

@BalarkaSen Probably not usually, you might want rings.

Balarka Sen

If you want to do cohomology theory you better take values in abelian categories, @MikeM

But it depends on the context of course

Alex

Aren't we going to start deriving cats for the purpose of regaining cohomology theory?

Balarka Sen

Maybe, I have no idea

Mike Miller

@BalarkaSen If you want to get a cohomology ring you'd better take values in rings :o

Balarka Sen

@MikeMiller ? That's not the value-category of the sheaf.

Mike Miller

12:20

Are you sure?

I guess your point is you just forget the ring structure when defining cohomology

And remember it when you want to multiply

Usually you have a product map from F and G-valued coho to F otimes G-valued

And eg if you have a sheaf of rings that comes equipped with a map from R otimes R to R

So ya good

Balarka Sen

Hmm, let's see. The singular cohomology is $S^\bullet$-valued Cech cohomology. $S^\bullet$ is the sheaf $U \mapsto S^\bullet(U)$. The reason singular cohomology has a product structure does boil down tot the fact that $S^\bullet(U)$ has a product structure... but the sheaf $S^\bullet$ takes values in not Rings but the category of differential graded algebras, right? Which happens to be abelian

Mike Miller

That is definitely not an abelian category.

You won't have kernels if you assume unital, and you'll never have cokernelw

Balarka Sen

How come? Kernels/cokernels make sense, not?

Mike Miller

The dg doesn't solve anything. How do you multiply in R/S?

Categories of rings/algebras are never abelian

But I also see the moral you make above: "Who cares?" Define cohomology for abelian categories, and show that you have extra structure for ring objects in those categories

I am a cancer object in the category of garbologists

Alex

@MikeMiller Hahaha

@BalarkaSen Why would kernels make sense?

Mike Miller

12:35

That's fine as long as you don't assume unital, yes?

Alex

Sure

Mike Miller

An ideal is in particular a non-unital subring

Alex

Yep

I think I remember Balarka saying that there is no such thing as a non-unital or non-commutative ring

That was his philosophy :P

Mike Miller

But there are non unital dg algebras, maybe

Balarka Sen

My internet bunkled

@MikeMiller Yeah good point...

@Alex Well, these are algebras, right? Kernel of morphisms between algebras is an algebra, I think (if I am wrong I have forgotten all algebra)

Ideals vs rings distinction isn't there

Alex

12:41

Depends how one defines an algebra

If unital associative, which is what many just call an algebra, it's just $A\to B$

Balarka Sen

You have a nice product structure over a vector space. That's an algebra to me.

So yeah, non-unital.

Alex

Yep

Then Lie algebras are algebras again

Balarka Sen

@MikeMiller Right, fair point. Slightly weird, so the product structure isn't built into the homological algebra

The point is that differential graded algebras are a hybrid, so admit a forgetful map to an abelian category (chain complexes)

Use that forgetful map to do the homological algebra

Then use the structure you forgot to define a product on cohomology

Strange. Never crossed my mind I had to do that.

@Alex Right

Mike Miller

@BalarkaSen I guess it's not dissimilar from a more down to earth phenomenon: defining singular cohomology with coefficients.

The definition of cohomology groups only uses the abelian group structure, and chain complexes thereof (and to take homology, you need both kernels and cokernels)

Then you have functoriality for maps of coefficient groupsn

Alex

in Geometry & Topology: Dank Ass Ganja, 38 secs ago, by Martin Sleziak

???

Mike Miller

12:54

And rings have functoriality maps R otimes R -> R

Balarka Sen

@MikeMiller Right.

It makes sense. Usually the product is where all the geometry lies in most cohomology theories

intersection product: "miss me with that algebraic shit"

@Alex LOL

Alex

13:08

@BalarkaSen You might like Agrypnie's album exit, in particular the 4th-7th songs

Balarka Sen

Link?

Alex

Agrypnie - Exit [2008] (full album)

►

Balarka Sen

Thanks!

Alex

16:23 onwards are the songs in particular

But I do like the whole album

Balarka Sen

I'm listening in full

But I'll let you know what I think of 4th -7th in particular

Alex

13:11

:)

Luke

15:12

Alright gang, I have a question

So I am perfectly happy with why we can realise a morphism from a coproduct to a product in an additive category via matrices of morphisms

However in this situation

Balarka Sen

(Oh no another category theorist)

Leaky Nun

(Yay another category theorist)

Luke

Leaky Nun

stares at Balarka

Luke

we have a morphism from a coproduct to a coproduct

and the claim is that there is a matrix of maps

Alex

15:13

@BalarkaSen ;)

Balarka Sen

Why did we give access to Luke

Who did this

jk welcome @Luke

3

MatheinBoulomenos

@BalarkaSen I used to be an adventurer like you, then I took an universal arrow in the knee, now I'm a category theorist

welcome @Luke

Balarka Sen

LOL

Luke

thanks guys

Leaky Nun

@Luke how about this: use that matrix to define a map from the coproduct to the product

and realize that the image is contained inside the coproduct

(the coproduct is the elements of the product with finite support)

(and the matrix is locally finite)

MatheinBoulomenos

15:19

@Luke the matrix is just a way of representing what the morphisms do on each component. you always have morphism from the coproduct to the product and then you can use the projections from the product and compose them with inclusions from to the coproduct to get the components of your matrix

Leaky Nun

hey that's what i said xd

MatheinBoulomenos

are you sure that "the coproduct is the elements of the product with finite support" holds in any triangulated category?

it might, but I'm not sure

Luke

My problem is whether "finite support" means anything in the situation where it's not a concrete category

for example a homotopy category, which is my situation of interest in triangulated categories

But the other answer makes sense, so thanks!

Leaky Nun

we aren't representing the matrix inside the category anyway

Luke

Although that seems to depend on the existence of arbitrary (or countable at least in this case) products

Leaky Nun

15:23

alright

MatheinBoulomenos

@Luke you're right that it depends on the existence of products, I have no idea how you get the matrix components otherwise

Luke

My initial thought was to just do this in the finite case

then use the fact that the product can be realized as a directed limit of the finite case

but then the same problem arises - why does that directed limit exist

Leaky Nun

again, the matrix isn't in the category, so let's just do everything outside

Luke

I guess I'll just put it down to it being an imprecise statement, and that the matrix representation is just an intuition

sure, the matrix isn't in the category

but to generate the elements of the matrix we need to look at inclusions and projections from things inside the category

MatheinBoulomenos

@Luke yeah, I was about to say, is the matrix thing really essential? You can construct the morphism just by universal properties and the matrix seems to be more to illutstrate what is going on

Luke

15:30

nah it's really not essential at all

I just thought I was missing something since the book tends to be fairly precise with language

MatheinBoulomenos

@Luke of course, you could just embed the category into its free completion (that's just the dual of the usual Yoneda embedding, see ncatlab.org/nlab/show/free+completion), noting that the functor into the free completion preserves colimits and the product exists in the free completion, so we can use the projections in the larger category

but that seems overkill to justify an intuition

Leaky Nun

there is no kill like overkill

MatheinBoulomenos

@Luke incidentally, which book are you reading?

Luke

Amnon Neeman's Triangulated Categories

MatheinBoulomenos

Is it good?

Luke

15:41

Yeah I quite like it

MatheinBoulomenos

I see, thanks!

Luke

I mean I'm really using it as a reference to read a paper of Krause

but it's a good introduction to the basic properties of triangulated categories

MatheinBoulomenos

maybe that'll help me with working with derived categories

Luke

Mmnon Yekutieli online notes on derived categories quite good

most references I've found seem to kind of just sweet the set theoretic issues under the carpet for the localisation process though

sweep*

MatheinBoulomenos

I'm fine with ignoring some set theoretic issues in category theory

you need AoC for proper classes (that has another name, I forgot it), to even prove that essentially surjective and fully faithful functors are equivalences

Luke

15:46

For the most part I am

although it kind of matters when you have situations where your hom sets aren't sets and you want to apply Yoneda

MatheinBoulomenos

ah true

I remember a talk on algebraic K-theory where the only reason that hom sets in a localization were sets was because there was a generator in the category

Luke

Interesting

Are you familiar with Grothendieck's argument that the existence of a generator gives you a small collection of subobjects?

I assume it's related to taht

MatheinBoulomenos

yeah, it was basically a variation of that

I did that once as an exercise

I think the proper setting for categories and homological algebra is Grothendieck universes, like it's done in Kashiwara-Schapira

that allows you to basically overcome all set-theoretic issues without ignoring them

Luke

I mean most of what I know about grothendieck categories comes directly from Tohoku

I might have a look at Kashiwara-Schapira

MatheinBoulomenos

it doesn't ignore any set-theoretic issues, but of course Grothendieck universes are less standard than ZFC (although people happily cite the SGA volumes where Grothendieck worked with that without worrying about it)

@Luke I think Kashiwara-Schapira is a great book which lays nice foundations for homological algebra and they do some stuff that is rarely covered in other homological algebra books: versions of Brown representability for triangulated and derived categores, Ind-categories, abelian sheaves on a site etc.

Luke

15:57

this sounds like something I should read

What I'm doing now is really leading up to Brown representability

MatheinBoulomenos

I can't say if Kashiwara-Schapira is easier than what you're currently reading or not

Luke

how would it compare difficulty-wise to Hartshorne?

MatheinBoulomenos

I haven't read Hartshorne, so I can't say

it's technically even self-contained starting with categories and limits etc. I found it to be more detailed on some stuff like localization than Weibel (but it has less concrete applications than Weibel, so it's not really comparable), so it should be doable if you have some experience with categories and homological algebra and general mathematical maturity (which you seem to do)

it always bothered me that if you take e.g. a functor category of a locally-small, but not small category, it's no longer locally-small

the system with Grothendieck universes eliminates those annyoing parts

I have only worked through about 50%, so maybe it gets significantly harder later on, but probably not

Alessandro Codenotti

17:28

When you think about a tensor product of modules do you think about the universal property or a particular construction?

Leaky Nun

@AlessandroCodenotti the universal property is not the only important property

the other important property is not categorical

it says that the tensor elements generate the tensor product

if you also think about that, then you are actually just thinking about the particular construction

so I would say that both the universal property and the construction are important

loch

17:48

yeah use whatever is convenient to your problem

Alessandro Codenotti

19:02

fair, thanks

loch

19:52

@AlessandroCodenotti For example, Let's say $A=k[x,y]$ and $I=(x,y)$. You might be interested in proving that $I$ is not flat as an $A$-module, and in proving this you might want to prove that $I\otimes I \rightarrow I$ given by $f\otimes g \mapsto fg$ and extending linearly is not injective.

You might see that oh - $x\otimes y - y\otimes x$ is mapped to $0$, but then you have the problem of proving that this element is non-zero in $I\otimes I$ - and I think it's easiest by using the universal property.

MatheinBoulomenos

21:03

@loch it's not really annoying using Yoneda and the universal property of tensor product and polynomial ring

loch

i figured so i said 'slightly' :p

i still think it's easier to convince someone that the statement is true w/ elements

well, to me anyway

MatheinBoulomenos

idk, using universal properties feels more conceptual

if you properly write out the proof with elements by defining a map in both directions and checking that they're inverses to each other it's not much shorter

Mike Miller

you feel more conceptual

loch

sure i guess
but if i had to convince someone in 5 seconds why that is true i think it's easier to say what i said
and then i'd probably say it's a good exercise to check that you can prove this also using universal properties

MatheinBoulomenos

everytime you define a map from a tensor product, e.g. $I \otimes I \to I$ you're using the universal property

unless you want to check by hand that's it's well-defined

loch

21:16

sure

Mike Miller

It is clear by hand that it is well defined

It is also clear by universal property

So whatever

loch

so maybe my first example is a better example of where having actual elements is useful (to produce a candidate which maps to 0, and then check using uni prop that it's non-zero)

but i still think my second example is fine as it is lol

Mike Miller

Agree and I see it as easier on the element level

Just easier for me to conceptualize

MatheinBoulomenos

elements and universal properties are both useful, often in conjunction

I think we can all agree on that

loch

yep

MatheinBoulomenos

21:23

in the second example I'd just say that the base change of $\Bbb A^1(A)$ to $B$ better be $\Bbb A^1(B)$ cause everything else would be weird (if we're just talking about convincing someone)

Mike Miller

We can all agree with that

Some people really hate elements I find

But those people suck

MatheinBoulomenos

I just wish my students for the course which I TA for would use universal properties a bit more

most of them check by hand that a homomorphism defined from a quotient is well-defined

and then they compute the kernel after that anyway

that's really unnecessary

they're basically reproving the homomorphism everytime

I mean the homomorphism theorem, of course

Mike Miller

For sure

MatheinBoulomenos

21:43

@MikeMiller are you up for some modular forms stuff? It would be cool if you can explain the connection with spectral theory of the Laplace-Beltram dude

Mike Miller

Hmm, can I take a rain check for one more day? I would like to do that on my laptop and it is dead right now

And I don't want to wake up my roommates to get it

MatheinBoulomenos

sure

Mike Miller

I am also enjoying reading Rosenberg this morning

MatheinBoulomenos

hmm, he wrote a book on K-theory, right?

I think that's the only book by someone named Rosenberg I know

Mike Miller

Yup

I figured you would approve

MatheinBoulomenos

21:49

I checked the table of contents, it looks really cool

he even has something on Milnor $K_2$ in number theory which is something I've been wanting to learn for some time

Mike Miller

I'll probably get through only a small fraction before I get distracted

MatheinBoulomenos

hmm, I know that feeling

but I'll put Rosenberg on my (everything but short) short list

Mike Miller

In a few days I will return to the states and have to get back to writing

Transcript for

♦ $Mathematics$ random cohomology for quantum nerds

Transcript for

♦ random cohomology for quantum nerds

♦ $Mathematics$ random cohomology for quantum nerds