long exact sequences are one of the most useful things ever

Drugs are ok

Some people don't like algebra too much and I'm ironically echoing them

They seem to prefer pornographic photos of tori instead

Torusporn is the best porn

@MatheinBoulomenos hello

@Jacksoja hello

@MatheinBoulomenos I wanted to ask you if it is possible that someone does topology before real analysis

or at the same time

I did topology before real analysis but I don't recommend it

some motivations and examples of topological spaces come from analysis

I did topology after I knew real analysis and I do recommend it

it's not optimal, but possible

I only wanted to know if it is possible :)

the reason is

I did them at the same time and it works out pretty well

the profs of those 2 courses are good

@Daminark I assumed he was talking about point set

But idk

and not sure when next time they will retake the same course, usually after 2 years

at least you will have a more conceptual understanding of the intermediate value theorem

I meant metric spaces basically

@Eric m8

I did algebraic topology before point-set topology

@MatheinBoulomenos is that theorem that deep ?

metric spaces are what we did in real analysis

@Jacksoja no, but the topology proof is much better than the real analysis proof

Like, I was sitting in on 203 while doing the 160s so I basically learned metric topology and a bit of general topology while I was learning single-variable calculus

am not sure we do the same topics, but in real analysis we gonna use Rudin, and do some lebesque integral ect

@BalarkaSen tbh point set doesn't even matter

Beyond like basics

If any of you know discrete optimization...

That's true. But when I started learning altop I knew like 0.00001% of point set

I went through a completely fucked learning trajectory

Lol

@MatheinBoulomenos how would you describe going from group actions as matrices ?

@MatheinBoulomenos I met my prof and he said he gonna do that

I don't understand the question

As it stands I'm mostly black boxing a lot of the results in point-set. Like, the stuff that came up in analysis came up, which isn't perhaps completely non-substantial? But like, the business about compactly generated and function spaces, I just black box it to say that it allows you to make currying a homeomorphism

@MatheinBoulomenos linear representation theory is done basiclly with matrices right?

And have mostly ignored stuff like axioms of separation

@MatheinBoulomenos I could be totally wrong ofc, I did not do these topics yet

@Jacksoja uhm, you sometimes write things out in matrices, but for the theory it's sometimes enough to work with linear maps

@Eric Did you see my end of the year find on 4chan/mu/ that I posted here yesterday?

No

link

but group actions are more general than representations. Representations are linear group actions

@MatheinBoulomenos okay , I guess I have to stop worring and take the course in january :)

@MatheinBoulomenos thanks :)

Oh my god

knew you'd like it

@Jacksoja you should do it in February instead

:P

@Daminark not possible

Nah I was just messing with you

hi

how did this go? @BalarkaSen

@BalarkaSen do you know about Universal bundle ?

@Adeek you'll be on my side here against these pictorial plebs when I say that long exact sequences are the heart of math, right?

just want to discuss something about the map

yeah @Daminark

hi yall

HAH

@Kasmir yo

\o

@Daminark I say homological algebra provides a tool to capture a lot of the geometry

can someone explain to me automorphism and innerauto morphism in a good wa ?

@Daminark sup dami :D

the reason being is it codifies geometerical data through algebra

@Adeek Yes, what do you want to know?

@skull didn't try it today

Dammit Adeek you weren't supposed to engage with this sensibly

@Kasmir I can try

So an automorphism of $G$ is an isomorphism $G\to G$, yeah?

yes

its an isomorphism from G to G

1-1 onto homomorphism from the G to itself

Well, one example of an automorphism is this

another episode of Life is Cringe is out

I can't believe I want to watch the playthrough

Given $g\in G$, you define $L_g(h) = g^{-1}hg$, yeah?

\o @Semiclassical

conjugating by g

So $U_{r,n}(K)$ is defined as the disjoint union of r-dimensional subspaces of K. Then, we define so given $v \in U_{r,n}$ it is an element of some of the disjoint union. So $\pi(v)$ we send it to the r-dimensional subspace which contains v ?

@Daminark yes so far so good

$L_g$ is definitely a map $G\to G$, it's a homomorphism since $L_g(hk) = g^{-1}hkg = g^{-1}hgg^{-1}kg = L_g(h)L_g(k)$

@Daminark but that is also an inner automorphism

Inner automorphisms are defined as the automorphisms of this form

are they automorphims that are not inner?

Yes

Okay so

Let $A$ be an abelian group and consider the automorphism $A\times A \to A\times A$ $(a,b) \mapsto (b,a)$

Let's look at $\mathbb{Z}/p\mathbb{Z}$

Or you can do that

@BalarkaSen is that how it works ?

@MatheinBoulomenos mathein :D

@Adeek Give some context, dude. I can't possibly know your notation beforehand.

@MatheinBoulomenos do we consider that an automorphism ?

well the smallest example is Z/3Z @KasmirKhaan

@MatheinBoulomenos because of AXA -->A , are not the same Group no ?

Are you writing down the tautological bundle over the Grassmannian, or what?

@KasmirKhaan it is an automorphism

hmm

Nope, $(a,b)\in A\times A$ and $(b,a)\in A\times A$

it's a map AxA -> AxA, not A

Maybe you're thinking of $(a,b)\mapsto ab$?

So we would like to define some kinda bundle over $G_{r,n}$ i.e over r dimensional Grassmannian manifold.

aha oups

its from AxA --> AxA ok am with you now

@BalarkaSen yeah if by tautological bundle you mean the Grassmannian bundle.

so the inner part

what does it inficises?

@Adeek No, I do not mean that.

That's why I asked you to provide more context.

@KasmirKhaan emphasize?

I guess it's called "inner", because the automorphism in some loose sense "comes from the group itself"

@BalarkaSen Book defines this as some kinda of Universal bundle

since you're just multiplying with elements from the group on the left and on the right

it's useful to look at examples. For symmetric groups $S_n$ an inner automorphism is just relabelling the elements you're permuting

@Adeek What book are you reading? Can you snapshot me the page?

sure

for $GL_n(K)$ an inner automorphism is a base change

@MatheinBoulomenos ah that one. if you embed your group via Cayley, an outer automorphism is conjugation by an external element

oh yeah, right

It is by Ribet

Complex geometry by Ribet

@Leaky shit that makes a lot of sense actually

@MatheinBoulomenos nice :D can you give me example of each ?

@Adeek did a frog write that?

@Adeek Yes, so this is the tautological bundle.

@BalarkaSen so we consider $v \in U_{r,n}$ we consider it as a normal vector not some equivalence class or anything and we just send it to r-dimensional subspace which contains it ?

@BalarkaSen I see

hi

hi

hi

@KasmirKhaan have you proved that Aut is a group?

@KasmirKhaan if $\sigma \in S_n$, then using cycle notation, $\sigma (a_1 \dots a_n) \sigma^{-1} = (\sigma(a_1) \dots \sigma(a_n))$

@Adeek That is correct. Every point $P$ in $G_{r, n}$ corresponds to an $r$-dimensional subspace of $K^n$ (tautologically; by definition). That is the fiber $\pi^{-1}(P)$ of this bundle projection $\pi : U_{r, n} \to G_{r, n}$.

the same works for products of cycles

@LeakyNun well not yet, but from what I understood it is a subset of S_n, and also a subgroup so it is a group

@MatheinBoulomenos working on thatnow :)

I'd say it's better to prove it directly tbh

^

I see cool

@BalarkaSen whenever I see Life is Strange in my Youtube feed I pause, consider watching it, and then decide to see if there’s anything else worth wasting time on

Thinking about it as a bunch of elements you can conjugate by when you embed it into $S_n$ is sorta unnatural here

If you have $r = 1$, this is the tautological line bundle on $\Bbb P^{n-1}_K$

@BalarkaSen one of the reason we use Grassmannians in order to capture more than just linear data of the manifolds right ?

@Semiclassical Hahaha. This is accurate on so many levels.

@KasmirKhaan go verify the group axioms

well aut(G) is not empty , the identity function is in there

I’ve seen one or two videos but I just can’t bring myself to give a f***

The RE7 dlc, that I’ll watch

@BalarkaSen Can we use Grassmannians to create some kinda of invariants on manifolds ?

@Semiclassical I believe that I just like the visual aesthetics and the musical aspect. But at the same time I can't stand watching these edgy 15 year old girls

associativity is just associativty of composition of functions

inverses in the same way'

hmm I need to do this properly

Also, what the fuck happened to when you could time-travel in the original game?

@KasmirKhaan yes, exactly

but proving it as a subset of S_n

@Adeek Well, you could say that.

why is that not good?

@KasmirKhaan because you haven't proved it

I mean

I see

alright it's a subset

you haven't proved that it's a subgroup

and that requires the same effort as if you don't embed it to start with

I mean Grassmannian of a manifold is a good parameter space/moduli space of the distributions on your manifold

That's reason enough why one should care

well if i start from there it is easiar

And even when you do that, I think you more often want to think about the elements of Aut(G) as maps

@KasmirKhaan it isn't

@Adeek Also, yes, there are ways to make invariants from the Grassmannians. Let me link you one of my answers.

Like, here's one example of a group automorphism which isn't inner

automorphism are 1-1 maps

@KasmirKhaan you still need to prove that the embedding preserves the structure

i mean bijective maps

i.e. composition of automorphisms give you composition of the corresponding permutations

@BalarkaSen thanks

@Daminark I was going to tell him to find Aut(Z/nZ) next lol

@Adeek Read question + answer and then read this answer. Ignore the $BO(k)$ bullshit; I can explain that later.

cool

thanks a lot

okay kasmir will return in short :)

No problem

thanks yall :)

See you! And yeah work out what Leaky's saying, you'll see why I sorta think it's probably better conceptually and for the sake of examples to look at these things as maps in their own right rather than to find some element in $S_n$ whose action by conjugation achieves the same effect

btw guys

is there a reason why they use into and onto

how are they different?

G onto G'

and G into G'

Onto is how evil people say surjective

aha that makes sense

and into ?

a map of G into G'

Into is just a generic term for telling you the target space

so onto has a mathematical sense

and into is just a way to express from what to what

okay thanks :D

Yeah exactly. Though please don't become one of the people who says "Oh this map is onto"

Like, I'm alright with people saying "So let's say $f$ maps this onto this", but at some point ideally the word surjective, or some other variant, should appear

that really makes more sense

this map is onto.

or just saying let phi be an epimorphism

or a surjective hom

@BalarkaSen yeah

Yeah that works too

epimorphism can have different meanings though, so one has to be careful

It's so easy to miss "onto" or read it as "into" (that I probably have mild dyslexia doesn't help)

@MatheinBoulomenos and Mathein is back with his evil category theory

@MatheinBoulomenos okay ><

ill use surjective hom to be safe

but this into was bad wording

though i can't agree with that without being a bit hypocritical: i've been watching a lot of videos on the progression of various speedrunning world records, and not a small amount of that is angsty teens wanting to be the best at something

it's not really relevant at your level, but it matters if you go deeper into the rabbit hole of abstract algebra

und der Sylowmeister ist zurueck mit seiner boese Kategorietheorie @MatheinBoulomenos

Well I want to do that once I get these stuff in my head :D

@Leaky boesen

@MatheinBoulomenos headache

hm? bsen?

adjective declinsion in German is evil, I'll give you that

most Germans don't even know how it works theoretically

@AlessandroCodenotti

he will not divide us! he will not divide us! he will not divide us!

MAGA

get MAGA'd

praise kek

So the 4chan people DID IT, if you know what I mean

they JUST DID IT

lol

So @BalarkaSen about this...

How would you approach this? My guess is to look for obstructions from (co)homology

The actions on homology are not so hard to determine

So the Z/2 action on the double torus is rotating the whole picture along an axis going perpendicularly through in between the holes, right?

I guess so

So $\{a_1,b_1,a_2,b_2\}\mapsto \{a_2,b_2,a_1,b_1\}$

So there's just two fixed points

(where these are the standard generators of $H_1$

@Danu Maybe some minus signs?

I don't think so, since it's rotating the orientation is preserved no?

I think if you rotate some orientations of the loops change?

no, they're preserved

Nah, some signs definitely change

I just drew the picture

Orientation-preservation of doesn't have much to do with orientation preservation on a single loop.

it preserves the orientation of the loops is what I meant

how could it not

Think of this arrow >

Rotate it 180 degrees along the z-axis

You get <

yes, but a circle

stays the same oriented circle

obviously

just 180 degree rotated

just the top becomes the bottom, if you will. But the orientation is of course the same

Ah, I see, I was drawing the wrong pic

you're literally rotating the surface

OK, you're right

how can you flip the orientation of the circles

I agree with you

ok

and on $RP^3$ it i the isomorphism which sends the generator of $\Bbb Z_2=H_1$ to itself

Equivariance gives us $f_*a_1=f_*a_2$

and similarly for $b_j$

the annoying thing is that the irng structure on cohomology is no good since there's torsion

Uhh. What's the Z/2-action on RP^3? RP^3 itself is a quotient of a Z/2 action on S^3...

It's just a 90 degree rotation of the sphere

so it's iso to the identity

(the isotopy is given by $e^{i\varphi} z$ for $\varphi\in [\pi/2,0]$)

Hm. I guess I don't see what that does on RP^3 level

it's homotopic to the identity map

It fixes a point and "rotates", OK

so it's just the identity map for all purposes

@Danu That's not very descriptive

But sure.

...but now I'm not sure how to proceed

Actually it's a bit tricky---the map on $RP^3$ has no fixed points so forget about me calling it a rotation (that's only true for $S^1$)

Oh, I see. You're fixing an RP^1

Because the rotation of S^3 fixes a circle, right?

I'm not exactly sure whatcha mean

Rotations of n-sphere fixes a codim 2 subsphere

If you restrict to the $z_2=0$ coordinate plane in $\Bbb C^2$, you'll see a rotating circle. Is that what you mean?

Yes, any rotation of S^3 fixes a circle.

Sure

ok

So anyways

Oh just this morning I learned that rotations of the sphere rotate the sphere. What a wondrous thing to learn

is there anything we can conclude from this?

I've never thought about existence of equivariant maps. I'm pondering it

Well

Except for maps between CP^n's and RP^n's of course

in our case it just tells us (together with the action being id on RP^3) that $f_*a_1=f_*a_2$ and similarly for $b_j$

and similar conclusions for higher homology groups

Hm. So there are not many maps $\Bbb Z^4 \to \Bbb Z/2$

well we're essentially reducing it to $\Bbb Z^2\to \Bbb Z_2$

There's all like, $\Bbb Z \to \Bbb Z/2$ on a factor or zero

one image for $a_j$ and one for $b_j$

Right

first observation: since the map on RP^3 has no fixed point, f may not be constant

@Danu Ah, a_j cup b_j is nonzero. So they have to both map to the generator of H_1(RP^3), right?

Z/2 coefficients

Hmm... good point

so you're sayin

We gotta be a bit careful cause you're using PD and stuff

Yeah but we're in field coefficients and everything is orientable so it shouldn't be an issue

ah you wanna work with Z_2 coeff

Right

mod 2 intersection number of a_j and b_j is zero; the only way for that to happen is for both of 'em to map to an [RP^1] in RP^3, because that has mod 2 self intersection number 1

No, garbage.

It's a 1-submanifold. It self intersects trivially in a 3-manifold

So we have $PD(f_*(a))\smile PD(f_*(b))=\alpha^2$ where $\alpha\in H^1(\Bbb RP^3;\Bbb Z_2)$ generates

Right. But $\alpha^2 = 0$

What? No

the generator generates in all degrees

I'm confused

it's generator, viewing $H^*$ as a ring

not just of $H^1$

You're right

I'm in a sleep deprived mode; it's 7 AM

But I don't think that the equation I wrote above is clear

why does that even hold

Naturality of cup product?

all I see is $PD(a)\smile PD(b)=PD(a\smile b)$

ok

but pullback goes the other way

and we don't know how $f^*$ acts on $H^*(\Bbb RP^3)$

but there should be some compatibility with $f_*$ so maybe it still works out

in fact, what we wrote above doesn't even make sense

because $\Bbb RP^3$ is 3-dim, we'd end up in degree 4

so we gotta be a bit more careful

@Danu I'm actually very confused. Say $f_*(a)$ and $f_*(b)$ are both homologous to an RP^1 in RP^3. But two RP^1's in an RP^3 never intersect transversely.

No, they don't. That's right.

That's just restating what I said above

the PD classes are deg-2

their cup is deg-4

Yeah yeah

which is of course trivial

I have forgotten the cup formalism

me too

but I'm getting there

Hey

we just gotta find out how to express the relatoins between $f_*, f^*$ intersection and cup in this specific case

So, is there a relation between intersection number of $a$ and $b$ and intersection number of $f_*a$ and $f_*b$?

well, no

Depends entirely on $f$ I guess

because the latter is always zero

since the target is 3-dim :P

Yes, that's what I said three lines above

But what I am saying is

yea

Can I conclude one of $f_* a$ and $f_* b$ is zero?

naw

I should compute the cohomology $H^i(\mathcal{P}^1_R,\mathcal{O}_{\mathcal{P}^1_R})$ of the projective space for all $i\in\mathbb{Z}$

you mean the cohomology of the trivial (holomorphic) line bundle @lattice? What tools do you have?

or wait, what's $R$? Real?

No, $R$ is some ring, and I mean the projective space as a scheme.

never mind me, then

lmao

cohomology of schemes nopenope

anyways @Balarka let's figure this out

That is, you somewhat glue the affine schemes $Spec R[X/Y]$ and $Spec R[Y/X]$

jesus I'm rusty on hatcher chapter 3. I reminded myself a chunk of the orientation double cover theory while writing an answer

a few days ago

right

Anyway I got $H^0(\mathcal{P}^1_R,\mathcal{O}_{\mathcal{P}^1_R})=R$ and $H^i(\mathcal{P}^1_R,\mathcal{O}_{\mathcal{P}^1_R})=0$ for all other $i$.

sounds good to me

Okay :D

it's the right answer for $\Bbb C$, anyways

Some affirmation was all I needed! :D Thanks and good luck with your stuff^^

So @BalarkaSen what we could do is

$(PD(f_*a))(f_*b)=?$

this should relate to a computation in the domain

Right, I'm pretty sure that $C(f_*A)=f^*C(A)$

So we can say $f^*(PD(f_*a))(b)=?$

hi @BalarkaSen

@Danu I think I agree. I don't know what that dude is though

Hi @Adeek

Yeah, might not be helpful