point-set-topology

[orchid forge]

you should prove this statement carefully, and check to make sure that it doesn't matter if the sets are open or not-- just that they are both closed

[sullen lynx]

which statement ?

[orchid forge]

"a product of closed sets is closed"

or, if you want to be more concrete, you can draw a picture of this set, and remember a subset X of R^2 is open iff every point x is an interior point, i.e., you can fit an open ball around x and contained in X

so for example, a line is not open: any ball cannot be contained inside the line

but the open square is

[sullen lynx]

yeah i see

thank you very much for your help

[pearl holly]

the simplicies [v_0, ..., v_n] and [v_n, ..., v_0] are the same, right?

[orchid forge]

orientation?

[pearl holly]

oh yeah right

[plain raven]

what context are you talking about

like i half know but

i'm confused. do hatcher's delta complexes have orientable cells

they do

obviously

something's throwing me off. i gotta reread that section to see how he does it

[proud ridge]

Show that the set ${(x_1,x_2,x_3,x_4)}\in \mathbb{R}^{4}:\text{rank}\left(\begin{pmatrix}x_1 & x_2\x_3&x_4\end{pmatrix}\right)=1$ is a 3-dimensional submanifold of $\mathbb{R}^4$. How could I go about even visualizing this? My first thought is its all 2 dimensional subspaces of $\mathbb{R}^4$, but is that correct?

[gentle osprey]

deathcode

[plain raven]

hmm not all two-dimensional subspaces.

[proud ridge]

In terms of the actual geoemtry part, I can probably figure out some submersion from $\mathbb{R}^4 \rightarrow \mathbb{R}^3$, but I'm unsure

[gentle osprey]

deathcode

[proud ridge]

my reason for thinking 2d subspaces is bc the rank being one implies $x_1=\lambda x_2, x_3=\lambda x_4$

[gentle osprey]

deathcode

[proud ridge]

loosely

bc there is the case where either column is 0

[orchid forge]

since they said simplices, i assumed it was the standard (n+1)-simplex with labelled vertices, and orientation induced by the labeling

but wasn't sure if orientation mattered to them, hence the question mark

[orchid forge]

the rank being 1 implies that every vector is an eigenvector for some nonzero eigenvalue

wait err

[proud ridge]

I thought it just meant the basis vectors lie on the same span, i.e. coplaner

[orchid forge]

the image of the basis vectors are multiples of each other

the image of everything is a multiple of each other

[proud ridge]

oh

[orchid forge]

this is wrong btw

just so we're clear

[proud ridge]

Maybe trying to reason about this geometrically is the wrong approach, and I'd be better off proving its a manifold without understanding the geometry

Bc at the end of the day, its the set of all points such that $\mu_1 x_1 = \lambda_1 x_2, \mu_2 x_3 = \lambda_2 x_2$

[gentle osprey]

deathcode

[proud ridge]

Which amounts to the basis vectors lying on the same span, which means they're lines, I think?

[orchid forge]

i think you had the right idea initially. the matrix sends (1,0) --> (x1, x3) and (0,1) --> (x2, x4), and (0,1) = \lambda (1,0) so x2 = \lambda x1 and x4 = \lambda x3

[proud ridge]

I used a second constant, because its possible x_1, x_3 = 0, etc

[pearl holly]

I'm reading the proof that the cup product is commutative "up to sign change". Here Hatcher defines a singular n-simplex from another one by just reversing the order of the verticies and I didn't think through and thought that they were the same lol

[plain raven]

this is a truly important question

because like

there are permutations of the vertices of a simplex, that extend to linear automorphisms of the simplex

and precomposing a singular simplex with that linear automorphism gives a different simplex

so like

you need to know whether [v0....vn] and [v_n....v_0] are the same simplex with different signs, or whether one is given from the other by precomposing with some automorphism of the base simplex

in which case the two singular simplices would be totally unrelated in the free group

[pearl holly]

ah yeah right I see

so when I'm at it I will quickly ask this: let $\epsilon_n = (-1)^{n(n+1)/2}$ and $\rho: C_n(X) \to C_n(X)$ be defined as $\rho(\sigma) = \epsilon_n \bar{\sigma}$ where $\bar{\sigma}$ is the singular $n$-simplex obtained by preceding $\sigma$ by the linear homeomorphism of $[v_0, \ldots, v_n]$ reversing the order of the verticies. Hatcher will prove that $\rho$ is a chain map and that it is chain homotopic to the identity and so induces the identity on cohomology. He now writes "from this the theorem quickly follows", the theorem being that $\alpha \smile \beta = (-1)^{kl} \beta \smile \alpha$ where $\alpha \in H^k(X; R)$ and $\beta \in H^l(X; R)$. I don't see how it follows immediately tho

[gentle osprey]

Tokidoki ✓

[pearl holly]

so does $\overline{\alpha \smile \beta} = \beta \smile \alpha$ and does $\epsilon_{k+l} = (-1)^{kl}$?

[gentle osprey]

Tokidoki ✓

[plain raven]

i think he explains that in the next two sentences

[pearl holly]

oooohh

lmao

[plain raven]

this is why i'm always on call to provide my vital assistance

[pearl holly]

yeah okay then I get it lmao. I thought that it was up to the reader to verify that. Thank you so much!

[plain raven]

i think the whole permutation is just a slight subtle bug in his definition. you can easily fix it just by modding out the chain complex groups by the relation that two simplices are opposite iff they differ by an odd transposition of vertices

it gives you the same theory

this is a theorem

not obvious maybe lol

[gritty widget]

The symmetric algebra on a vector space V

This is just formal combinations of elements in V?

And we make

Two formal combinations equal if

They only differ by commutative and associative “replacements”

also this seems wrong to me?

I understand how we need char 2 for (2) to imply (1)

but i dont see how all of this is equivalent to the graded commutative condition

if w has even degree then we should have

w \wedge v= v\wedge w

but (2) is saying that its always anti commutative?

[plain raven]

no, that's not how graded-commutative stuff works here.

this law only applies to elements of degree 1

the law is extrapolated to higher degree terms by applying it iteratively, i.e. permuting the elements one transposition at a time

in the exterior algebra every element of degree p can be written as a sum of products of p elements of degree 1

so specifying this law for elements degree 1 automatically implies the usual graded commutative law for higher degree terms

[plain raven]

yeah. if you choose a basis for V, then the symmetric algebra is isomorphic to the polynomial ring with variables drawn from the basis and coefficients in R

[gritty widget]

Thank you

[orchid forge]

$(v + w) \wedge (v + w) = v \wedge v + v \wedge w + w \wedge v + w \wedge w$. If $v^2 = 0$, then this gives us $0 = v \wedge w + w \wedge v$. If $\wedge$ anti-commutes, setting $w = -v$ gives $0 = v^2 + v^2 = 2v^2$. This implies $v^2 = 0$ provided the characteristic is not 2

[gentle osprey]

Kogasa

[orchid forge]

fuck

i put wedges instead of +'s

and now it's too late to fix it

just ignore this

$(v + w) \wedge (v + w) = v \wedge v + v \wedge w + w \wedge v + w \wedge w$. If $v^2 = 0$, then this gives us $0 = v \wedge w + w \wedge v$. If $\wedge$ anti-commutes, setting $w = -v$ gives $0 = v^2 + v^2 = 2v^2$. This implies $v^2 = 0$ provided the characteristic is not 2

[gentle osprey]

Kogasa

[digital peak]

[plain raven]

Yeah. ok so. A subset of C^2 is called algebraic if it's in the image of the functor V(-)

so the algebraic subsets of C^2 form one half of this equivalence

the other half is kinda interesting. so here's a condition on the ideal I that is clearly necessary.

If f^n is in I for some natural number n, then f is in I.

Why is this necessary? Because if f^n(x,y) =0 for some pair (x,y) in C^2, then f(x,y) =0, because a complex number to the nth power is zero iff that number is zero

[digital peak]

right so if f^n in J then f in I(V(J))

or whatever

[plain raven]

Right. Yeah.

So, a famous theorem of Hilbert established that this condition is also sufficient. Very cool result

so there's a contravariant equivalence of posets between radical ideals in I and algebraic sets in C^2

[digital peak]

called?

[plain raven]

uh the Nullstellensatz, theorem of zeroes.

[digital peak]

[plain raven]

but it's often cited in a few different forms depending on the commutative-algebraic guise most convenient for it to appear under

so let's pick a point (a,b) in C^2

this is obviously algebraic because it's defined by the system of equations f(x,y) = x- a and g(x,y) = y-b

so it's V( (f,g) ) for that choice of f,g

conversely it's not too hard to see that (f,g) is a maximal ideal. like, there's a ring homomorphism C[x,y] -> C which sends x to a and y to b and sends all complex numbers to themselves

and by looking at the ring homomorphism laws and the fact that polynomials are built up by addition and multiplication

you can prove that this sends the function t(x,y) to t(a,b)

in particular the kernel of this homomorphism is exactly the set of polynomials which vanish at (a,b)

that means that C is isomorphic to C[x,y]/ I((a,b))

which proves that I((a,b)) is a maximal ideal.

[digital peak]

wait what

[plain raven]

i guess in the end that's not really what i wanted to prove.

let me back up lol

sorry what are you gonna say

[digital peak]

it's an ideal, where'd the maximality come from?

[plain raven]

Oh, C is a field and it's a known characterization of maximal ideals that I is maximal iff the quotient ring R/I is a field

[digital peak]

[plain raven]

give me a second to think about this

it's not too hard why that's true

[digital peak]

no like, it's probably a thing that I should know

[plain raven]

Oh right

If R/I is a field

then it has no non-trivial ideals other than the zero ideal

if it has some nontrivial ideal I' then no element of I' can be invertible

because the product of any x with its hypothesized inverse x^-1 would also be in the ideal

and then xx^-1 = 1 would also be in the deal

thus every element of R would be

so this proves that the only ideals of a field are the zero ideal and the entire ring R

[digital peak]

the way I look at it, all ideals of R/I are of the form J/I where J is an ideal of R superset to I

[plain raven]

yeah.

[digital peak]

if R/I is a field then J/I = R/I or I/I

i.e. I is maximal

[plain raven]

Right.

[digital peak]

good refresher lmao

[plain raven]

Cool

Ok. i think like

all i should have said earlier was that it's obvious that the ideal C[x,y]/(x-a, y-b) is isomorphic to C. Because quotienting out by the ideal forces x-a =0 and y-b =0 in the new ring, i.e. x=a and y=b

so the quotient takes the two variables and replaces them with elements of C

it gives a map C[x,y] -> C

which sends x to a and y to b

that proves that (x-a,y-b) is maximal, because it's the kernel of this homomorphism

does that make sense?

[digital peak]

sure

[plain raven]

So, since (x-a,y-b) \subset I(V(x-a,y-b)), and it's maximal, the equality is strict.

Question: are there any maximal ideals in C[x,y] that don't arise from points in this way?

is every maximal ideal generated by (x-a,y-b) for some a,b?

[digital peak]

unless I(V(x-a, y-b)) = C[x, y]

but that's fortunatley not the case

[plain raven]

Yeah. Right. We know that not every function vanishes on V(x-a,y-b) = (a,b). for example no nonzero constant function

So the answer to this question turns out to be again, no!

all maximal ideals in C[x,y] correspond to points!

this is another corollary of hilberts' nullstellensatz

so maybe the first topology that comes to mind to put on C^2 is just like, the usual one on the underlying metric space R^4. however in algebraic geometry all of our functions are polynomials and shit so there aren't that many functions we want to be continuous

and uhh idk if i have a great excuse for this topology but all you really need is a topology which is fine enough that the functions you work with are continuous

so since we have only polynomial functions, and later rational functions

we can take like a very coarse topology

here we're doing to define the Zariski topology on C^2 by taking the closed sets to be exactly the algebraic ones

it's not too hard to show that this is indeed a topology

[digital peak]

ok

though

[plain raven]

?

[digital peak]

that makes very few functions C^2 -> ... continuous

[plain raven]

yeah but if your whole category is built out of similar spaces to this one then it doesn't matter. like, if you just stuck to C^n and closed subspaces of C^n and those were the objects of your category

[digital peak]

then that's not really motivating

[plain raven]

it's algebraic geometry lol

the field is about the study of polynomial curves

and stuff like that

[digital peak]

if you consider Top(X, Y) then coarser Y means more maps, and coarser X means fewer maps

both coarser means???

[plain raven]

yeah. ok i getcha. my point is: you're right that there are not many continuous maps from X to Y. That's fine because we're not really interested in all continuous maps from X to Y. we are not talking about a full subcategory of Top here. It's like, you want smooth functions between smooth manifolds to be continuous, definitely, because topological arguments are important! but you're not interested in arbitrary continuous functions between smooth manifolds.

so like, we only care about polynomial maps from X to Y anyway, so it's not a concern of mine that there's not many more continuous maps than just those

[digital peak]

but if you make Y coarse as well then it kinda balances out

so it's not clear what you've done

[plain raven]

Maybe. Haha

i really could not tell you what the set of continuous functions X-> Y looks like outside of the polynomial functions. no clue

[digital peak]

if all spaces are discrete, then all functions are continuous

on the other side of the spectrum, if all spaces are codiscrete, then again all functions are continuous

[plain raven]

actually some of these spaces are pretty weird. I think the expected AG spaces are discrete, like a finite union of points.

[digital peak]

(insert bell curve meme)

[plain raven]

You don't really run into infinite unions of points because you couldn't define them by polynomials nicely

btw the only algebraic sets in C (as a one-dimensional C-Vector space) are the finite sets

[digital peak]

that's FTA

[plain raven]

so it has the cofinite topology for us

yeah

exactly

so there are lots of continous functions C ->C

does that help

as long as every fiber is finite right

[digital peak]

there's something implicit going on here

a leap from vanishing sets to closed sets

I guess a vanishing set has to be closed

we're saying nothing else is closed?

[plain raven]

yeah

[digital peak]

and that's somehow enough to make polynomials continuous even outside their vanishing sets

[plain raven]

yeah.

[digital peak]

that's peculiar

[plain raven]

uhh ok let me think about this.

so let f(x,y) be a function from C^2 -> C

take a closed set in C. that's just a finite union of points by FTA

so it suffices to show that the preimage of a singleton is closed

[digital peak]

it's obvious in the C case

[plain raven]

and then do induction on the number of elementsd.

[digital peak]

I'm trying to generalize

[plain raven]

yeah, for function vectors C^n -> C^m of the form (f1, f2. .. fm) each one in n variables

i guess it suffices to consider closed sets of the form V( (f) ) where f is just a single element. Because if I is an arbitrary ideal then V(I) is the intersection of all the sets V( (f ) ) for each f in I

and taking preimages plays nicely with taking (even infinitary) intersections

so i think we should just be able to argue that the preimage of V(f) is closed and then the general case follows by taking intersections

but i think that now (f1... fm)^{-1} (V(f ) ) is exactly V( (f(f1, f2,... fm)) )

[digital peak]

nah

[plain raven]

it's the vanishing set of the composition

[digital peak]

the thing I'm looking for is

wait what am I looking for

not sure what the extent of our spaces is

but suppose f : C^n -> C^m and x in C^n

now suppose f(x) in A, where A is a vanishing set of some p : C^m -> C

then what we want is p(f(x) - f(.))

[plain raven]

yeah. i mean i can sketch a reasonable category for you for the sake of argument, it's the category of all C^n for n in N and closed subspaces of these

[digital peak]

which is a polynomial on C^n

whose vanishing set is f^-1(A)

so f is continuous at x, for all x

ok that seems somewhat reasonable

[plain raven]

ok. cool

[digital peak]

it works because you use f to relate polynomials in cod(f) to those in dom(f)

which only works if f is polynomial itself

[plain raven]

I guess that's it.

Yeah

[digital peak]

well, not "only"

but yeah

[plain raven]

so yeah this is this standard topology. so one thing that i can point out here is that like, you can transfer this topology from C^2 to the set of maximal ideals of C[x,y] by the bijection we established earlier

so that they're homeomorphic as topological spaces by definition

[digital peak]

hold up

[plain raven]

sure

[digital peak]

ok

[plain raven]

and uh you can translate the definition of closed sets we gave into a condition on when a set of maximal ideals is closed

so, say I is some ideal.

Then V(I) is a closed set in C^2 by the definition of the topology

(a,b) is in V(I) iff every function in I vanishes on the pair (a,b). In other words, iff every function on I belongs to that maximal ideal m_{(a,b)} of all functions that vanish on (a,b).

It's about the inclusion reversing correspondence i guess

[digital peak]

sure I guess

[plain raven]

(a,b) \in V(I) iff {(a,b)} \subset V(I) iff I({(a,b)}) \superset I

[digital peak]

I superset I

[plain raven]

sorry.

[digital peak]

sounds adjointy

[plain raven]

yeah that's what i'm using i think

{(a,b)} \subset V(I) iff I \subset I(a,b)

[digital peak]

yeah this is the Hom = Hom version

[plain raven]

it's just taking transposes i think

ok cool

yeah so we arrive at

the definition of the topology on maximal ideals

which is that the closed sets are of the form V(I) = { m | I \subset m }

hopefully that provides a little intuition for the spec construction

what this is is the subspace of Spec(C[x,y]) consisting of only the maximal ideals. the prime ideals are a bit harder to get intuition for, for the sake of a lot of things you can dismiss them as a gadget needed to make the construction work right when we try and expand this to arbitrary rings

one other thing

if f : C[x,y] -> C[x,y,z] is a ring homomorphism

say it sends x to some f1 and y to some f2

ok let me say this.

A basic concept of geometry is that if you have a good map of spaces f : X -> Y

and you look at the ring of functions on Y

say from Y into some fixed ring R

[digital peak]

sheaves 🥴

[plain raven]

then you should be able to pull back the function along f and get a function X -> R

so like

if I have a polynomial function f : C^3 -> C^2

then this should give a homomorphism C[x,y] -> C[x,y,z] by precomposition

[digital peak]

and that describes all polynomial functions

[plain raven]

conversely if you have any homomorphism C[x,y] -> C[x,y,z], say it sends x to some f(x,y,z) and y to some g(x,y,z), then you can think of these as what you get from projecting a map C^3-> C^2 onto the two coordinates of C^2

[digital peak]

that's really weird

[plain raven]

so you can associate to this the polynomial function <f(x,y,z), g(x,y,z)> : C^3 -> C^2

yeah it's pretty weird i agree

making sure this shit all works right takes some work

hmm

one slight generalization i should mention is that our techniques are easily adaptable to closed subspaces of C^n

if A is an algebraic set in C^n

[digital peak]

closed as in zariski-closed

[plain raven]

yeah.

[digital peak]

that did not immediately click

[plain raven]

then you can just define the ring of functions on A to be the quotient of C[x1... xn] by the ideal I(A) of all those functions which vanish everywhere on

so you're saying, yes give me the usual polynomial functions but identify them if they agree everywhere on A

[digital peak]

wait

[plain raven]

sure

[digital peak]

somehow I missed the part where we went from C to C^n

on C the topology is cofinite

[plain raven]

Yeah

[digital peak]

but on C^n it's more weird

[plain raven]

yeah it has lots of closed sets

so what are you concerned about in this setting

[digital peak]

the fact that I missed the transition where we're no longer talking about cofinite sets

[plain raven]

yeah so an example of this would be like

the algebraic subset V(y-x^2) in C^2

the ideal is generated by a single polynomial, y-x^2

and the closed set is the parabola y=x^2

[digital peak]

wouldn't it be the whole riemann surface

[plain raven]

this is how you should think of a typical closed set, it's some kind of algebraic curve or surface or shape. a hyperboloid, an ellipsoid, yadda yadda

yeah i'm just visualizing this in R^2 because i can't see for shit in C^2

[digital peak]

but ok

[plain raven]

we still say 'curve' tho, even tho it's a surface, because it's 1-dimensional over C

[orchid forge]

it's just a torus minus a point

[plain raven]

😮

[orchid forge]

wait no, a sphere minus a point

[digital peak]

so a point then?

based path induction

[plain raven]

ok so a minute ago i was explaining how we could define the ring of functions on an algebraic subset A in C^n as just the ring C[x1 ,... xn] modded out by the ideal I(A)

and you can still talk about polynomial functions from one closed subset A of C^n to another closed subset B of C^m using the background coordinate systems

and again this induces a ring homomorphism (C[y1....ym]/I(B)) -> (C[x1...xn]/I(A))

and vice versa

[digital peak]

I'll need to think about that correspondence a bit

[plain raven]

and the same logic as before works out but you can do it

so to sum up this correspondence

what are the properties of this ring C[x1...xn]/I(A)

well

I(A) was a radical ideal, remember

this was hilbert's nullstellensatz

[digital peak]

radical?

[plain raven]

oh sorry it satisfies f^n in I for some integer n implies f in I

[digital peak]

o

[plain raven]

and like

it's not hard to see that when you quotient out by this ideal you get a ring with the property

f^n = 0 -> f =0

so introduce a name for this, we'll call it "reduced"

so a ring is reduced if the zero ideal is radical

too much jargon smh

smh

and closed subsets of a space, we called them algebraic sets, that name, considered as spaces in their own right they're sometimes referred to as affine varieties

here affine basically means that it's embedded in C^n

in contrast to like a more general object which is glued together out of coordinate charts which are locally in C^n

[digital peak]

not related to the other affine whatsoever?

[plain raven]

not that i know of.

[digital peak]

[plain raven]

like how a manifold is glued together out of open subsets of R^n

a manifold which could be embedded in R^n or which is embedded in R^n might be thought of as affine but like

[digital peak]

flashback to covariant vector vs covariant functor

[plain raven]

Hahahaha

yeah

a more general manifold which is just thought of as being glued together out of open subsets is not affine

[digital peak]

why do names exist, we should refer to properties by the merkle hash of their definitions

[plain raven]

lmao

uh

ok what are the other interesting properties of those rings

they're finitely generated as algebras by the variables x1...xn

that's pretty much it

so

what we have is that

there is an equivalence of categories between

[digital peak]

"those rings" --- reduced?

[plain raven]

• the category of affine varieties and polynomial maps between them
• the (opposite) category of finitely generated, reduced C-algebras and homomorphisms between them

yeah reduced

[digital peak]

C-algebras as in algebras over C the field?

[plain raven]

yeah

so this basic duality between the algebraic side and the geometric side is the starting point for a lot of interesting algebraic geometry

this is kind of where it starts

[digital peak]

hmmmmm

I think I kinda see

this is like the upgraded version of the toy C^k -> C^m vs C[x1...xm] -> C[x1...xk]

[plain raven]

yeah. not too far upgraded, all we did is quotient out by some ideals and you have to check that that works

which isn't too bad

just a standard algebra exercise

annoying

[digital peak]

so I'm assuming there's things like "manifold" "varieties"?

[plain raven]

it also suggests why you might want to try and take this galaxy approach you were talking about in your class of trying to search for geometric intuition in CRing^op, because you know in this nice extremely small subcategory of CRing (finitely generated reduced algebras over C) you do have geometric meaning for the op category

yeah.

so the most basic example is the projective line, the set of lines in C^2 that pass through the origin

a line through the origin can be specified by giving a pair of points on it, say (x,y)

we usually denote [x:y]

but like, [x:y] denotes the same line as [2x:2y] and [-0.4x : -0.4y]. any scalar multiple of the same vector determines the same line.

[digital peak]

yup that's projective line

[plain raven]

it's convenient to fix a unique representative by scaling so that the first coordinate is 1

[digital peak]

so far looks like a circle

[plain raven]

then your coordinates are of the form [1:y/x]

so like

if you just set Z = y/x

you see that the whole thing is controlled by a single one-dimensional parameter away from the point [0:1]

so intuitively you think of this as having a copy of C^1 embedded within it

which is the whole space except for one point

[digital peak]

oh wait C not R

so a sphere then

[plain raven]

yeah so this is a sphere

the Riemann sphere

conversely you could do the same thing with the other parameter.

so yeah this is two affine varieties (two affine lines) glued together

along the map, say, z \mapsto 1/z'

which is defined on both lines away from the origin

this is where sheaves actually do have to really come in because

before this i was talking about like, understanding a subspace of C^n by looking at the global functions on it.

but there are no interesting global functions on the Riemann sphere, except for the constant ones

[digital peak]

in a whatever manifold, we want the chart transition maps to be whatever

[plain raven]

Yeah.

[digital peak]

what happens here

[plain raven]

uh, locally rational transition maps are the basic morphisms

[digital peak]

[plain raven]

a map f : X -> Y is said to be regular if X admits an open cover where f can be described as g/h where g and h are polynomial functions

[digital peak]

why rational

[plain raven]

idk it's a slightly more general class of ''algebraic '' functions than the polynomials

like arguably rational functions are still algebraic

at least on their domain of definition

[digital peak]

intuitively I would want to adjoin inverses of polynomials

not reciprocals

[plain raven]

wait like inverses in terms of function composition

[digital peak]

yeah

[plain raven]

mm that's pretty weird

[digital peak]

because a transition map is f . g^-1

[plain raven]

ok. let me think about that

[digital peak]

I don't think there's anything useful if you require transition maps to be just polynomial

but rational seems like a super arbitrary cutoff

[plain raven]

ok maybe this is pushing it but

say i have the affine line

it's just a line, it's C

[digital peak]

wait

is the inverse of a rational function even always rational?

[plain raven]

i don't think that these should have inverses necessarily

[digital peak]

well suppose you have charts f and g

you transition from f to g via g . f^-1

you have to transition backwards too, using f . g^-1

[plain raven]

i guess i would say no. i can't think of any reason why it should be. so you'd need the extra assumption that the maps in both directions were rational

[digital peak]

sounds very restrictive

not sure what the word is in english, but az+b/cz+d are such

[plain raven]

my memory is a bit foggy wrt these questions you're asking

[digital peak]

anyway I wanted to bring up that I guess x is a polynomial of y and x/y

[plain raven]

yeah so this works like, if you pass from one dimensions to 2

the hyperbola y=1/x is the set of solutions to the curve yx-1

this is a closed (algebraic) subset of C^2

and at each point on it, y=1/x and so you can use the variable y as you would 1/x

so you can sometimes make rational functions polynomial by passing to a different context

(localizing)

alright i think i gotta walk away i'm getting tired

my eyes are shutting lol

[digital peak]

but that would mean cbrt(x) is also

[plain raven]

i think i'm going to go to bed early.

Uh can you explain that

i don't know why you'd say that

[digital peak]

y = cbrt(x) is the same curve as y^3 = x

[plain raven]

hm

[digital peak]

anyway thanks for all this

[plain raven]

np!

yeah idk how to answer that question. i must be sweeping something under the rug about why this works in some cases and not in others lol

it has something to do with the fact that x=0 is a closed subset of the affine line

and so when you remove that point you get an open subset

the function 1/x is defined on an open subset, like it belongs to the sheaf of locally rational functions

idk haha

alright goodnight

gotta go

[shy moss]

a noetherian space is reducible?

[tough imp]

Not necessarily

Spec Z for example is irreducible

Meanwhile Spec Z x Z is Noetherian but is reducible

[honest narwhal]

So this maybe goes here

Let's do some differential operator stuff

[gentle osprey]

Sloth King Daminark

Sloth King Daminark

[honest narwhal]

:/

[orchid forge]

[gritty widget]

[digital wraith]

I'm having trouble seeing why the fundamental group of the Klein bottle isn't just Z

I'm trying to do it via Van Kampen

[orchid forge]

with what open cover?

[digital wraith]

I was trying a few different ones

But now I see neither works

[orchid forge]

[digital wraith]

My initial thought was to decompose it into two mobius bands but that didn't seem to work

But apparently someone online was saying you can?

[orchid forge]

that should be fine

[digital wraith]

How do you draw the mobius bands though

Or wait is it just upper left and bottom right squares

If you use the definition of the Klein bottle as a quotient of [0,1] x [0,1]

[orchid forge]

not sure what you mean by upper left and bottom right squares

but i would also do it with the fundamental polygon

[digital wraith]

Don't know what that is

[orchid forge]

the presentation of the klein bottle in terms of a labeled square

with orientations

[digital wraith]

No that's what I mean

[orchid forge]

yeah, i'm saying that's a good way to do it

[digital wraith]

Square where you identify the top and bottom edges and the the left and right edges with a twist

I'm saying if you break up the square into a grid of 4 smaller squares

And then take the subspace that's two diagonally adjacent squares

That's a mobius strip

[orchid forge]

i don't think it is, since removing a point from that would make it disconnected

keep in mind what the fundamental polygon of a mobius strip looks like

and try to draw a line through the klein bottle that produces two mobius strips

[digital wraith]

*Deformation retract of a mobius band

[orchid forge]

in case you're still looking at this

try cutting like this

[orchid forge]

actually you probably want to make a vertical cut

[coral pawn]

Is the identity point on an elliptic curve always of degree one ?

[gentle osprey]

Phi

[minor hornet]

Does it make sense to talk about a metric space whose topology isn't the metric topology?

eg. (Reals, Discrete topology) with the euclidean metric on R
My gut tells me no because you'll have two distinct notions of open sets, continuity etc.

Corollary: If I'm told I have a set X equipped with a metric d, is it safe to assume we are dealing with the metric topology?

[hollow harbor]

you're right that it doesn't make a ton of sense

it is safe to assume you mean the metric topology

however, there are contexts in which you will talk about different topologies on a space

this becomes really important when you start talking about topologies on spaces of functions.

like we can talk about functions converging pointwise, or converging uniformly, or converging in L^p norm, or converging weakly in L^p, etc, and all of these are sequence convergence for different topologies on the same space of functions.

usually your space of functions comes with a particular metric, but you still talk about other kinds of convergence which are related to the metric somehow (but not exactly it)
typically you'll talk about norm/strong convergence (the standard kind), or weak convergence, and then sometimes pointwise convergence.

[minor hornet]

interesting. so we'd get like a countable sequence of topologies for the same function space and metric?

[hollow harbor]

i don't know about a countable sequence

(there are definitely situations in which you might consider such a thing)

here's a way to think about it: you might be familiar with "the weakest topology generated by a collection of functions"
this says, you have a set X, and you have functions f_n from X to some topological spaces Y_n. the question is, what's the least amount of open sets you can put in X so that all the f_n's end up being continuous?
basically, you preimage all the open sets in each Y_n by each f_n, and then you take the topology generated by that.
when the f_n's are projections, this is the product topology.
but in a function space, we can think of functionals (functions which take functions as input and return numbers, e.g. the evaluation functional at y which takes any function and tells you its value at the fixed point y, or say the integration functional which tells you the integral of your function over its domain) and then we can use the weakest topology on your space of functions which is generate by a collection of functionals.

so if you pick an increasing sequence of functionals, then you get stronger and stronger topologies as you add more and more of them.

but usually we just look at the "weak topology" which is the weakest topology which takes ALL the linear functionals which used to be continuous for the metric topology, and keeps them ALL continuous (this usually isn't the metric topology anymore, usually we can throw some information out - it's the same in finite dimensions but function spaces are infinite dimensional usually!).

so for example, the sequence (sin(nx)) in the space L^2(0, pi) does not converge to anything in the metric topology. the wiggles just keep getting steeper and steeper, what would it even converge to?

well, in the weak topology on L^2, somehow the wiggles cancel out and this sequence converges to 0.

(in a weaker topology, it's easier for things to converge)

the point being, integral of sin^2(nx) from 0 to pi is always pi/2, but integral sin(nx)f(x) from 0 to pi for any FIXED f in L^2 goes to 0. you can see this for differentiable functions by integrating by parts, and then you can approximate functions in L^2 by differentiable functions to see it always works.

this is kind of strange example, but it turns out to be really important

since in analysis we get a lot of things which oscillate like crazy, and sometimes we want to make sense of them, but we can't in the metric topology. however, we might be able to say something about their behavior in the weak topology. and even though it's harder to prove theorems and do things with weak limits, it's a lot better than if we didn't have a limit at all (which we usually don't if we stick to the metric topology)

many times you actually also "upgrade" convergence from weak to strong. sometimes you know a theorem which tells you that something has a weak limit, and then you do some work to show that the sequence actually converges in the metric topology to the weak limit.

weak limits are super powerful because in infinite dimensional spaces, closed, bounded balls are basically never compact for the metric topology. but they are compact in the weak topology! and compactness lets us claim that bounded sequences have (weakly) convergent subsequences.

so it's totally essential in functional analysis / harmonic analysis to be able to talk about multiple topologies on a metric space. but usually we just stick to the strong (metric) topology and the weak topology.

[lunar yoke]

Thanks for the writeup ryc

I had a first course in functional analysis like a year ago, but it seems i forgot most of it

[hollow harbor]

an important point which i glossed over here is "how in the hell do you find all the continuous linear functionals on a function space"

and the good thing is for all the common spaces these are well known

except for L^infinity lol, that one is a mess

[lunar yoke]

i remember for just continuous functions the dual space was some weird space consisting of measures, right?

[hollow harbor]

yes, if you look at a locally compact hausdorff space, then the dual of the continuous functions vanishing at infinity is the space of radon measures (with bounded variation)

so if you just look at a compact space, this is the dual of continuous functions on your space

[lunar yoke]

oh yeah it hink we only had it for compact sets in R^n, nice that this extends to that generality

[hollow harbor]

radon = compact sets have finite measure, open sets approximate measurable sets above, compact sets approximate open sets below.

[lunar yoke]

lebesgue is radon then right?

[hollow harbor]

(for positive radon measures. for negative ones, just subtract 2 positive ones)

[hollow harbor]

yes

[plain raven]

this is kinda wild. i've never heard this.

is this definitional

[hollow harbor]

because bounded sets have finite measure, we can use open intervals as our covering intervals for any set, and for open sets we can just approximate each interval in the open set from the inside.

[plain raven]

or a consequence of some cool theorem

[hollow harbor]

well, this is kind of a funny point

many people would define radon measure to be "positive linear functional on C_0(X) for X locally compact hausdorff"

many other people would define it to be "measure which is outer regular, and inner regular on open sets"

so "definitional" depends on which kind of person you are

[plain raven]

I'm a complete asshole

which one should I pick

[hollow harbor]

but e.g. the way I learned it was, this is literally the definition, and then we proved properties about radon measures culminating in "they are actually measures, and they have these 3 properties, and those 3 properties also give you a continuous linear functional on C_0(X)"

it's also funny that people call it the "riesz representation theorem" when there is another "riesz representation theorem" that is also about the dual of a large class of spaces!

[plain raven]

yeah i've been confused about this in the past

[hollow harbor]

which is why people sometimes call it Riesz-Markov-Kakutani representation, but no one actually calls it that

[gritty widget]

what do we mean when we say some object

corepresents something

is i saying that Hom(object, -)

is isomorphic to some F(-)?

[plain raven]

uh sure

then (object) corepresents F

one example would be like, if F is the forgetful functor from groups to sets

then the group Z of integers corepresents F, as maps from Z into a group G are in one to one correspondence with elements of G

[uncut surge]

this map is not well-defined right

if k = 1 then its image changes when i replace f_1 with f_1 + constant

y dis published

[cedar pebble]

no this should be fine I think

forgetting f_0 this is just the usual anti-symmetrization map (without the 1/k!)

[uncut surge]

wait really how

the f_0 is still there, right

[cedar pebble]

sure, but that factors out

[uncut surge]

why does this not depend on the choice of f_1

[cedar pebble]

oh I see the issue

I think this should be like

df_{\sigma(1)}\otimes...\otimes df_{\sigma(k)}

where this lands in \Omega^1(X)^{\otimes k}

though the original map should be well defined up to exact forms I think

[uncut surge]

yeah that map would have a better chance

that's the full statement

so they're claiming this is a splitting to the HKR map or something

[cedar pebble]

source?

[uncut surge]

https://arxiv.org/abs/0912.3729 i think they're being silly

i don't think this is largely relevant for what they're doing, but the morphism only goes one way i'm breddi sure

[cedar pebble]

hmmm

maybe the point is that \Omega*(X) is being considered with the zero boundary map?

this definitely wouldn't make sense for \Omega*(X) with the usual de Rham differential I agree

but yea this seems fishy

[uncut surge]

ye i think i'll just be wary with this one

i wonder if it's good etiquette to write an e-mail about something like this

i kinda have a feeling i'd just get a "lol true but idc" back

can't be the first person to notice

[cedar pebble]

I'd write an email yea

[feral dragon]

Write an email and see if the author responds with an interest in it. I think they would.

[cedar pebble]

I mean if this proposition is correct it's probably something really basic but the author just doesn't do a good job dispelling any confusion like this

although I would maybe try to search for similar instances in the literature first

[uncut surge]

ye, appreciated!

[gritty widget]

thanks

[west pawn]

I was told to ask this in here, as opposed to in Math Help.

What's the difference between an open neighbourhood, and a basic open neighbourhood? Google is being very unhelpful with this...

[lunar yoke]

dont think i've ever heard this terminology, but I would guess a basic open neighborhood is an element in the neighborhood basis of the point

[cedar pebble]

^

yup, elements of a neighborhood basis are called basic opens

[west pawn]

And more terminology I don't recall...

Wikipedia has:

The set of all open neighbourhoods at a point forms a neighbourhood basis at that point.

So which'd imply they're the same then, no?

[cedar pebble]

no, typically you choose a smaller neighborhood basis

[empty grove]

if you don't know what a neighbourhood basis is then maybe whatever source you are referring to just means elements of a basis instead of a neighbourhood basis

basic open set refers to elements of the basis. You fix a basis before you can talk about basic open sets

[west pawn]

Ok, thankyou. I think that's what it's referring to, which helps a lot.

[pearl holly]

Okay I have a few questions on this proof. So first, why is there a big i+1+n-j exponent in the second sum instead of just (-1)^j? And in the next area marked by red, why can we just replace i by i-1? Don’t we then include a nonzero term? And in the last area marked by red, why does the theorem follow when A is non empty?

I just realized that there’s not enough info in that pic

This is the stuff before the first pic I sent

And also why does epsilon_n[w_n,...,w_0]-[v_0,..v_n] represent p(sigma) - sigma? Isn’t the simplex with w_i different from v_i? In the sum they are different but now they seem to be equal wtf

[orchid forge]

Notice the order of vertices, v0 ... vi wn ... wi. Here, w_j is the (n-j)th element after w_n, which is the (i+1)th element. Hence you're deleting the (i+1+n-j) element

[gritty widget]

god that looks messy

[pearl holly]

Ye right that makes sense. Thank you!

[orchid forge]

For the second question, you're just reindexing. The sum is over i < n, so for all i except the initial one (0 or 1?) there is a corresponding i-1 term, which cancels with the i term of the first sum

Then they state the remaining non-cancelled elements

[pearl holly]

Right but if I have a sum over i < n and I replace i with i-1 then I sum over i-1 < n and so I include the term when n=i and that term is nonzero right?

[orchid forge]

No, you don't want to actually change the sum. It would be the sum over i-1<n-1

He's just saying the i-1 term of the second sum is the negative of the i term of the first sum, for each i where this makes sense (0 <= i-1 < n-1)

[pearl holly]

Okay so if I do this calcelation, won’t I still have a term left? The term being when i = n?

Fuck this proof man

Okay I need to go, my train will arrive soon but I will review all of this later. Thank you so much!

[orchid forge]

The i=n term of the first sum is cancels with the i=(n-1) term of the second

[pearl holly]

Okay right I see. Thank you!

[wise panther]

If $X$ is contractible complex (CW Complex), show that there exists a function $F:X\times X \to X$ such that $F(x,x) = x$ and $F(x,y)=F(y,x)$. Does anyone have any insight into this?

[gentle osprey]

Pan

[orchid forge]

Any kind of function? Or a continuous one? A homotopy?

[wise panther]

F should be continuous

In the very simple case of X = [0,1] you can do it by (x,y)->(x+y)/2

But I have no idea how to generalize this

Also unsure what about CW complexes is needed that general contractable spaces don't have

[orchid forge]

here's an idea that might work

instead of going directly from (x,y) to (y,x), we just want to find some path that switches x and y, and look at where it intersects the diagonal

[wise panther]

But there isn't a unique way to do that

[orchid forge]

uhh

i wanted to just write enough of my idea to be sure that it probably works

but i keep writing and not being sure, so i'm either wrong or going to end up writing out the entire problem

I want to say that the fact that X is a CW complex lets you consider the CW-pair $(X \times X, \Delta)$ where $\Delta$ is the diagonal, and use your theorems about CW-pairs / "good pairs"

[gentle osprey]

Kogasa

[orchid forge]

In particular, when the subcomplex is contractible.

[orchid forge]

to be clear i don't think the map is supposed to be unique

[orchid forge]

i'm pretty sure my solution works, but it's not very short

[orchid forge]

I guess the idea is: if you draw X x X as a square, and indicate the diagonal as a... diagonal, you get two triangular regions, and you should try to retract one triangle onto the diagonal, then do that same retraction backwards on the other triangle. that traces out a continuous family of paths (x,y) --> (y,x) through the diagonal

[wise panther]

Thanks @orchid forge , I'll meditate on this and see if it goes anywhere

[trail tiger]

What does restricted mean

Restricted tensor algebra

[orchid forge]

verboten

[trail tiger]

[loud scarab]

Why does Munkres say 'we show that f(x)>0' here?

Isn't the metric always > 0 anyways?

This is the definition of d(x,A)

[ivory dragon]

≥ 0

munkres is showing its nonzero

[loud scarab]

Thank you!

[gritty widget]

Parallelgrom ABCD whos side AB is 2x bigger than BC. M is the center of side CD, determine the angle AMB

[tough imp]

[gritty widget]

Got it

[raw sedge]

d(x, x) = 0, but if x =/= y then d(x, y) > 0

[fathom cave]

dang, ur doing cohomology 😮 @pearl holly
I should stop slacking off lol

[gritty widget]

toki hijo de puta doki.....

[bleak path]

Re: geometry, can I ask if it's commonly understood that a sphere curve is defined as as a curve x(s) lying on the surface of a sphere? Or is it something that is unique to the course I'm taking

[gritty widget]

that's what id think it is

[gritty widget]

i think for how short it would take to clarify the definition

i would just quickly state the definition

I am slow and if i heard sphere curve i wouldn't know whether you meant a curve taking values in a sphere

or a curve from sphere

the second doesn't really make sense

but i'd still be confused

always good to think of us small brains when you present stuff

[bleak path]

I would, but just asking because I had a discussion with another friend taking it in a different school and he didn't have the same impression that i did

[gritty widget]

as a rule of thumb

i think it never hurts too be too precise when defining something

but you can be too precise when presenting proofs

there are technical lemmas that are more believble than their proof is illuminting

[tawny jewel]

Hey, so I have an exercise I'd like to figure out, "Which capital letters of the Roman alphabet are homeomorphic? Ex-
plain."

So I have no idea how to visualize and notice homeomorphisms just by the shape

And I'd like to learn how

[gritty widget]

read the first few pages of hatcher

[tawny jewel]

Alright, brb!

[gritty widget]

sorry thats definetly way more than you need but

i was mainly hoping the pictures might spell it out

[tawny jewel]

Seems a bit outreaching for me 😦 I'm only in calculus 3 class right now

[bleak path]

My friend was discussing with me about a problem I have for an assignment. Could anyone give a hint how to prove the following?

"The curvature of a sphere curve is always greater than or equal to the inverse of its radius"

$\kappa(x) \geq R^{-1}$

[gentle osprey]

wOne

[gritty widget]

are u sure

[bleak path]

I can transcribe the question if you want

"A sphere curve lies on the surface of a sphere in euclidean space and can be described as |x(s) - p|^2 = R^2; the sphere has radius R and is centered at p.

Prove that the curvature of a sphere curve k \geq R^{-1}"

[bleak path]

To be honest it is not intuitive but I can loosely see it

Of course I'm not sure if I'm simply seeing things that aren't there, given that it was in the question and it implied that it was true

[gritty widget]

oh, i misread what you wrote, sorry

[bleak path]

cheers, no worries

[gritty widget]

okay well basically

and are they actually asking for homemorphic or homotopy equivalent

because there could be someting slightly subtle going on with homemorphic

T and L are homotopy equivalent

but if they "thin lines" i don't think they are homemorphic

anyway so homotopy equivalent

two letters are homotopy equivalent if they have the same number of holes

if two letters are homeomorphic they are homotopy equivalent

but you can be homotopy equivalent and not be homomorphic

[tawny jewel]

Yeah, homeomorphism can bend, twist, stretch, and wrinkle

But I'm kinda not very happy with that XD

[gritty widget]

so if i was doing this question i would first group together all the homotopy equivalent letters

QAORDP

these ll have one hole

shrink all the legs

[tawny jewel]

Yeah

It kinda gives me the answer

But not the intuition

Like

[bleak path]

Another interpretation might be (speaking very very informally even though this is loosely accurate) 2 spaces are homeomorphic if you can deform one to become another. So in this context I'm going to assume a looser definition of homeomorphic (since you're calc 3 and not topology).

Example: C and I are homeomorphic because you can bend I to form C, or straighten C to get I. E and T are homeomorphic because you can rotate the T counterclockwise to get the tail pointing to the right, and bend the edges of the T to get the top and bottom bits of E. O and D are homeomorphic because you can flatten one side of O to get D. Check the font to be more exact.

[gritty widget]

wait are you guys in the same class?

[tawny jewel]

Hm I see, so by applying these transformations

I'm just composing well-known homeomorphisms

[bleak path]

If we are, I don't know him

I doubt so, since I'm not taking calc 3

[tawny jewel]

Hm I don't know how you got that impression @gritty widget but we're not 🤣

[gritty widget]

sorry i think the "neon" pfps made me mix up who sent what message

and then it seemed like you were asking the same question

[tawny jewel]

Oh 🤣 well nvm, seems like I got the intuition for homeomorphisms

But the problem is now uninteresting to me

thanks

[bleak path]

Well I'd still like help with my problem, if anyone knows where to start that would be appreciated 🙂

[obtuse meteor]

Geometry

[gritty widget]

geometry

[pearl holly]

geometry

[cedar pebble]

Oh god I hate this

The torus is not supposed to be hyperbolic

[obtuse meteor]

Clearly you’re not ascended

[gritty widget]

i am blue now

this brings me great joy

[sweet wing]

[bleak path]

I've tried comparing it to the Freser-Serret Frame equations and trying to make sense of it, but I haven't made much headway there, maybe I'm missing something from it

How can we tell? I can see how it would lead to knowing some properties about the curve and it's derivative but I'm not sure how we can establish that it's 1 everywhere

Sorry, it's not a priori to me - is that just an intrinsic property of all spheres of R^n?

what do you mean by norm R, don't fully get you

[hollow harbor]

Spheres are the sets of points which are a certain distance from some point

So any point x that lies on the sphere of radius R about the point p satisfies ||x - p|| = R

This is what it means to be a sphere

(todd male is saying that if you want, you can translate your curve by -p so that it lies on a sphere of radius R centered at 0. then we would just have ||x|| = R. But you don't need to do this).

[bleak path]

I understand this, but I don't understand how this translates to the curve has norm 1 everywhere, or norm R (presumably for a sphere of radius R?)

[hollow harbor]

What specifically is it that you understand in what I said

If gamma is a curve such that gamma(t) is on the sphere of radius R centered at p for all t, what this means is that, for all t, ||gamma(t) - p|| = R

(by definition of what it means to be on a sphere)

[bleak path]

I understand that that is the definition of what it means to be a sphere

Maybe I'm missing something about the meaning of norm?

As far as I know from my course the norm to gamma(t) would be the normal vector at any point

Have I missed something fundamental?

[hollow harbor]

Oh

No

These are different uses of the word "norm"

[bleak path]

Would this be what you mean by norm in this context?

[hollow harbor]

"norm" usually means "distance from zero", it's the usual ||.|| square root of sum of squares thing. Norm ≠ normal. The latter is the vector you mean.

Sorry, that was probably confusing.

[bleak path]

Thank you for clarifying, informal usage in my earlier weeks definitely didn't help

[hollow harbor]

The words "normal/norm" are criminally overused in math for tons and tons of different things lol

[bleak path]

I noticed lol, along with proper

[hollow harbor]

Normal subgroup, normal operator, normal vector, normalized, norm, etc

[bleak path]

Definitely many more examples

Sorry, just trying to read through what was earlier mentioned with this new meaning of norm

So here, and later when he mentioned Norm R, this is saying (in other words) that every point on the curve has a distance R from the origin, correct?

I'm afraid I must ask a few more fundamental questions here - these were not covered very well in a previous course - how exactly do you differentiate <gamma(t), gamma(t)> = R^2? I would presume implicitly with reference to t, but I haven't seen this notation with angle brackets yet. I know a bilinear map uses this notation, but is this notation in the context of geometry referring to a bilinear map as well? And if so, how would you differentiate it?

[bleak path]

I'm not given an explicit expression for a curve, and this is probably because we want the equation to work for every sphere curve. Is there an expression or special term in geometry for this case? i.e. the inner product of a curve with itself

I can see how the fact that the curvature is a constant means it's part of a sphere

[remote beacon]

How do I prove SO(3) is diffeomorphic to RP^3

[gentle osprey]

Todd Male

Todd Male

[bleak path]

Thank you for this, I will spend some time thinking about it. I greatly appreciate the help of both you and rew york city.

[flint cove]

wait I thought spheres were just glorified suspensions

[obtuse meteor]

Based

[dreamy smelt]

I am like

freaking out right now

how is the interval we choose (-1, 1)

This thing is nonzero outside of (-1, 1), for example f(-5)=1-(-5)^2\neq0

?????????????

why is the support such a small part of R

[empty grove]

Why is that x sideways

😵‍💫

[dreamy smelt]

LMAO

Wikipedia moment

[empty grove]

|-5| ≥ 1, so it's 0 there by definition

[dreamy smelt]

Oh 🤦‍♂️

thanks

My mind was just deleting the | | when I was checking points x<0 with full confusion lol

[gritty widget]

I want to ask something. The lesson name is analytic geometry. Our teacher tried to explain something but I got nothing seriously. I really don't know the topic but he mentioned "affine frame". I really don't understand anything even if I listen to the records again and again. Can someone show me a way to get this topic like for example you must know derivative to understand integral and limits to understand derivatives right? I need someone to tell me the topic sorting to study respectively something like that

[coral pivot]

can you give more details on what the topics are?

[gritty widget]

I don't even know the topics, but was the first stuff our prof started to tell, affine space or affine frame something like this

I should study all the required topics before the deadline of the homework

[empty grove]

I'm guessing it's affine space

[gritty widget]

okay sir what do I need to study before I get to understand affine space?

[empty grove]

Do you know anything about vector spaces?

[gritty widget]

too little

[empty grove]

That is usually the starting point, but you can study affine spaces directly as long as you have intuition for what vectors are

[ivory dragon]

what type of course is this

like whats the course name

[gritty widget]

analytic geometry

university second

[empty grove]

Affine spaces are sets in which you can draw a vector from any one point to another, and vector addition makes sense: vector from a to c is the sum of vectors from a to b and from b to c

This is imprecise but should probably be enough for the course if it was only mentioned in passing, but idk

[gritty widget]

knowing vector spaces is enough for the course?

[empty grove]

I wouldn't know, depends on if it was only mentioned as a side note or as a prerequisite

Or as something to be covered later

[gritty widget]

shall I translate the homework question and paste here?

[empty grove]

Yeah

[gritty widget]

but after I pee, I'll tag you, hold on a few mins

[tough imp]

World’s longest pee

[gritty widget]

I'm back, latex took time sorry

Consider $({\mathbb{R}}^2, {\mathbb{R}}^2)$ affine space. If $\phi :{\mathbb{R}}^2 \cross {\mathbb{R}}^2 \longrightarrow {\mathbb{R}}^2, (p,q) \longrightarrow \phi (p,q) = (q_1-p_1, q_2-p_2)$, then, is ${p_0(1, 3), p_1(-1, 2), p_2(3, 2)}$ an affine frame on ${\mathbb{R}}^2$ that is starting from $p_0$ If so, find the coordinates according to the this $(3, 5)$ frame.

[gentle osprey]

junait

[gritty widget]

.

it's a miracle for me to translate this into english

does it make sense

[empty grove]

oh I have not seen affine frame before let me see what that means

but the first part makes sense

Affine space (R^2, R^2) means that the underlying set of the space is R^2, and the vectors you draw from any one point to another also come from R^2

yeah it makes sense

https://encyclopediaofmath.org/wiki/Affine_coordinate_frame

so you have to show that in R^2, p1 and p2 are linearly independent

but this seems like the kind of stuff that would have a lot of different but equivalent definitions (there are like 3 different mainstream definitions of affine spaces ) so maybe check with your instructor if this is what they mean

[gritty widget]

do you mean the question is not clear enough in terms of the definition of affine space

[empty grove]

question is clear

but the way you write the answer will depend on what definitions your course is using

all the definitions are equivalent so they would fundamentally all be the same

but your instructor might have a specific definition in mind so better to use that

[gritty widget]

I got it, so what should I study, linear algebra?

[empty grove]

you can try reading about affine spaces directly if you know what the vector spaces R^n look like, you shouldn't need to go too deep into linear algebra

but you should know what linearly independent means, along with maybe a few other basic definitions

maybe look them up as you encounter them

[gritty widget]

okay thank you sir

I appreciate it

[maiden oracle]

i know that homotopy is what makes the fundamental group a useful object to study but just theoretically couldn't you also define a group structure on paths just by allowing to reparametrize the paths and not complete homotopy of paths, since the only aspect where homotopy is really needed is to show associativity. or am i missing something?

[empty grove]

yeah you could do it

[maiden oracle]

k good to know

[plain raven]

but would paths be invertible then

because no matter how fast you traverse a path from a to b and back

it's not the constant path

[empty grove]

oh true

@maiden oracle mb it doesn't work

[abstract pagoda]

whats an example of non abelian fundemental group

[hollow harbor]

Pi_1 of a figure 8

If you go around one loop, then around the other, then backwards around the first, then backwards around the other, this is not the identity

In fact, pi_1 of a figure 8 is the free group on two generators, which is very nonabelian

[stone cipher]

Hi. Guys,.... I know that every subspace of a separable metric space is separable. Does that hold for any topological space too?

thx

[finite heath]

Cl(A) = A U A', so while limit points are not necessarily contained in A, they must be contained in the boundary of A?

[bleak path]

After a day of thinking and playing with equations I've solved it. Thanks again to you and @hollow harbor for having the patience to explain it to me.

[hollow harbor]

Well done! These questions are tricky

[pearl holly]

so cohomology over a ring R is also a R-module?

is that true?

ye okay and since they are rings and R-modules they are R-algebras by definition right?

or is that not the definition?

oh okay I see

my module and algebra stuff are super rusty and it feels like Hatcher will be doing some stuff with algebras and modules

[tough imp]

Isn’t your cohomology

Defined to be quotients of things that are R-modules

That gives you the R-module structure

[pearl holly]

I don't even know anymore

[tough imp]

And for a definition of R-algebra, when R is commutative it’s just a ring map R -> S which lands in the center of S

You define r•s to be like f(r)s

What do you mean by cohomology over R then?

[tough imp]

When R isn’t commutative algebras make less and less sense so idk

[pearl holly]

I mean the homology of the of the chain complex I get by dualizing C_n(X) to Hom(C_n(X), R)

[tough imp]

Yeah

[pearl holly]

oooh okay I see I see

[tough imp]

It’s a module by letting r•f be defined like pointwise

Like uh

(r•f)(x)

= r•f(x)

And well addition is pointwise

You’ve probably seen this before and I’ve talked about it before maybe to someone else but

Well hold up

[tough imp]

What category is this Hom in?

Are these like abelian group Homs?

Slim it literally matters

If you say restrict to ring maps between something

[pearl holly]

bruh I don't even know

[tough imp]

It isn’t a module

Yeah so I think it’ll be linear maps

[pearl holly]

ye

[tough imp]

Like Z-linear

And in that case you’re fine

Since f + g is still linear

As is rf

Why this matters is if you take ring maps

And define f + g to be like done pointwise

This isn’t a ring map

I mean C_n here isn’t a ring so you’re okay

But you have to worry about closure under addition and scalar mult

[pearl holly]

ye okay I see. So let me try to summarize this: since Hom(C_n(X), R) are R-modules, the kernels and the images are too. Quotients are modules too so cohomology is a R-module. The direct sum of R-modules is a R-module and so the cohomology ring is a R-module?

[tough imp]

Right

It’s an algebra I think

Or a graded ring probably

Like you’d need a way to multiply an element of a certain cohomology with one in a different one

Which… idk probably has some weird ass procedure to do

Cup product?

Idfk

[pearl holly]

ye Hatcher says that it's a graded ring but maybe if you had R as a field instead then it would become an algebra?

[tough imp]

No it’s an algebra anyway

Like

Idk what’s in the degree 0 part

But that should be a ring

With a map from R

So you make the graded ring an R-algebra via the map sending an element in R to something homogeneous in degree 0

Probably

Like what is in degree 0?

[pearl holly]

wtf is degree 0?

[tough imp]

The cohomology rinf

Is like

A big ole direct sum

Yeah?

[pearl holly]

yeee

[tough imp]

Indexed by n

[pearl holly]

oh so H^0?

[tough imp]

Right

So that’s probably easier written a different way

Like is it just R?

[pearl holly]

no it turns out that H^0(X) is iso to Hom(H_0(X), R) by the universal coefficient thing

[tough imp]

Okay

Is H_0(X) a uh

Free Z-module?

Or something

Or hmmm idk there’s not an obvious ring map here

Unless it’s literally just Z

It’s definitely an R-module

And it’s also a graded ring I think

So like it should be an algebra

But idk how cuz idk AT

[pearl holly]

H_0(X) is generated by the path connected components in your space so I guess that it is a free Z module?

[tough imp]

Hmmmm

Well… that makes it iso to R^(+alpha)

For some alpha

But there’s not like a canonical map from R to that (unless alpha = 1)

Besides maybe the diagonal

But if alpha isn’t finite that isn’t a valid map

[pearl holly]

ye right I see

[tough imp]

Idk I think it’s probably an R-algebra

Anyway in terms of if R is a field

Actually irrelevant

[pearl holly]

okay I see

[tough imp]

Algebras are just ring maps

So it really doesn’t matter if the source is a field or not

I guess the only thing it simplifies is

Actually no

This isn’t even true so

Yeah it doesn’t simplify anything

[pearl holly]

yeah okay I see. Thank you so much for making stuff a lot clearer!

btw would anyone recommend some good source that talks about the cohomology ring? It feels like I don't fully understand the stuff that Hatcher is talking about anymore

I tried searching stuff up on youtube but I can only find videos that only mention it quickly

[gritty widget]

Hello I was minding how this conclusion is made, cant figure it out

[surreal estuary]

Is there an actual explanation for what the pushforward and pullback are in differential geometry other than the abstract definitions?
And are there any resources that make an actual attempt to explain them or give actual concrete non-trivial examples?

[gritty widget]

perhaps helpful https://math.stackexchange.com/questions/194147/intuition-about-pullbacks-in-differential-geometry

[surreal estuary]

I saw that but I have trouble following stackexchange since the asnwers tend to use weird notations and concepts I've never seen

they also tend to jump around instead of working through an entire example from start to finish

[surreal estuary]

yeah I looked at the top answer and it just immediately states

the push-forward amounts to changing coordinates for a vector field (viewed as a derivation) for example: for cartesians verses polars in
without giving any reason why I should believe this or how to show it

[finite heath]

Can anyone help me with this problem:

[tough imp]

Let a be an arbitrary point, then you want to know that for any open set eventually (x_n) is inside that open set for all n sufficiently big

So let U be an open set containing a

It’s the complement of a finite set

So now you just gotta go high enough so that all the elements of that set no longer appear in x_n

That happens cuz pigeonhole principle

[finite heath]

Ah, I am a tad confused by this, sir

[tough imp]

Think about it, if I just tell you then you won’t learn

[finite heath]

U is infinite, but aren't sequences infinite too?

[fair idol]

Is there a straightforward method for computing a local inverse for a smooth mapping between smooth manifolds?

Like say I had a simple map between R^2 and R^2

[bronze lake]

In general local inverses don't necessarily exist depending on the injectivity of the smooth map

[fair idol]

That's true, I should have prefaced this question by the assumption of the inverse function theorem

[gentle osprey]

fajitas

[bronze lake]

I might be talking about of my ass, but the inverse function theorem says that a function is invertible at a neighborhood of a point if the jacobian matrix is invertible, so perhaps something taking the inverse of the jacobian map to get a linear approximation in the reverse direct? idk

[fair idol]

Hmmmmmmm that's not a bad idea 💡

[gritty widget]

I was working on pressley's Elementary Diff Geometry book, and the last part of question 2.3.4 has exploded into a wall of trig functions, where you show Viviani's curve has the property that was proven

Viviani's curve:
[\gamma(t)=(\cos^2t-\frac{1}{2},\sin t\cos t,\sin t)]

[gentle osprey]

Manzareh

[gritty widget]

and it's exploded completely in my solution and I'm wondering if I missed something

showing that it was spherical was easy, but the torsion, curvature ratio looked terirble

[wise ruin]

In Loring Tu’s Introduction to Manifolds, he sets about to prove this proposition. In proving the leftward implication, he says that τ contains the empty set, but if we define B to be just the set S, the two conditions hold and τ={S} which does not contain the empty set. Am I missing something?

[orchid forge]

the "empty union" is the empty set

this is technically speaking a definition / convention, not something which should necessarily be obvious

[wise ruin]

Ok thank you!

[abstract pagoda]

is baycentric division only brought up for excision theorem?

[gritty widget]

bumping myself because i'm shameless

[orchid forge]

i'll be honest i tried doing it and shit got out of hand pretty quickly

and i got bored

[gritty widget]

same

i even tried following my proof for the forward direction after showing that its spherical and i'm still handling ugly trig trees

[empty grove]

I think faithful is pretty obvious and I'm stuck on proving that this is full. I know that V embeds into t(V) as the subspace of all closed points. I think I should prove this: closed points of t(V) must map to closed points of t(W) so that the restriction to V → W makes sense, and this restriction uniquely determines the whole map. But idk how to even start either of these parts

I have no intuition for maps of schemes in general I feel

[stone barn]

hey guys, i was assigned this just now and the question is very vague, does this sound like a problem that should be modelled with graph/network theory to you guys?

i've heard of the travelling salesman problem but i haven't been taught it yet, and i just googled it and it looks applicable to this

[empty grove]

This isn't the right place, you should ask in a CS server probably (see #old-network)

[hollow harbor]

hints: there are many possible approaches

This is also a throwback to the horrific final project I had in my advanced algorithms course

That we actually didn't do terribly on

But also not super well

[plain raven]

yo this is really interesting

@empty grove
Warning The following rant is not directly relevant to your question. here is what I have to say which is relevant:

I think the answer to your question definitely needs some commutative algebra. I can't remember if you need the full power of hilbert's nullstellensatz over an algebraically closed field... I believe this might hold for spaces equipped with a sheaf of k-algebras for k an arbitrary field and maps of spaces together with k-algebra maps of the associated structure sheaf. i think you want to prove that like, on the level of stalks, using the fact that k-algebra map of local k-algebras necessarily induces an isomorphism between the residue fields over the two points, that closed points map to closed points. you might need some kind of characterization of the closed points in terms of what the residue field looks like at closed vs nonclosed points. something like the residue field at a closed point is just k and at a nonclosed point it must be more funky than that.? I'm making shit up but this is buried in my AG notes so ping me if you want me to look it up.

But rereading the question it does say assume k=ACF. i think you should really just try and prove what's purely a theorem of commutative algebra, that the preimage of a maximal ideal among a k-algebra map is a maximal ideal (k an ACF, all rings being finitely generated, maybe reduced, whatever)

anyway, rant time

this is actually part of a more general result about a certain class of topological spaces.

i don't want to give a massive monologue but like, basically there's a certain class of spaces that can be uniquely reconstructed from their lattice of opens regarded as a poset

Write T(X) for the lattice of opens of X.

Each point x in X determines a set of opens in T(X), the ones that contain it.

This set has some interesting properties. It's a filter - it's closed upward. It's closed under taking finite intersection.

Also, if {U_\alpha} is a family of open subsets of X, none of which contain x, then neither does their union contain x. So the filter of opens containing x has the additional order-theoretic property that you can't get into it from outside by taking suprema - if A \subset T(X) is disjoint from the filter F_x of opens containing x, then the supremum of A is not in F_x either.

We say that F_x is completely prime.

For the name convention, I don't really know what to tell you other than that if you view the meet as a kind of multiplication, then it makes sense to call an order-theoretic ideal "prime" if the meet a ^ b is in I iff a or b is in I. Dualizing this definition by flipping everything upside down gives a criterion for a filter to be prime - the join of a and b is in F iff either a, b is in F. Then "completely prime" is the natural generalization from finite to infinite subsets.

Anyway

The study of the lattice of opens is pretty interesting, particularly from the view of sheaf theory, because to define a sheaf on a space, you don't actually need the space per se, you need the lattice of opens. The gluing condition can be fixed by replacing the word "union" with "supremum"