point-set-topology

[plain raven]

Natural transformation of sheaves, still don't need the space.

Similarly to do change-of-basis stuff, looking at how sheaves are transported along a continuous map f : X-> Y, you don't really need the continuous function f. What you need is the preimage map f^{-1}: T(Y) -> T(X). Then the direct image functor f_* is just given by precomposing with the map f^{-1} : T(Y) -> T(X)

And as long as you have maps g : T(Y) -> T(X) that are well-enough behaved (preserve finite intersections, preserve unions/suprema, etc.) you'll get a good change of basis functor

And you could just define inverse as the left adjoint to this precomposition functor, assuming it exists

So it's a nice general setting for thinking about sheaf theory.

Anyway there are some problems here that arise

like

What is the general class of posets we want to consider here? I can answer that, we'll say posets with finite meets and arbitrary joins. And maps between them preserving those finite meets and arbitrary joins.
Ok. Does every map g : T(Y) -> T(X) preserving finite meets and arbitrary joins arise from a continuous map f : X-> Y? If f exists, is it unique?
Does every poset with finite meets and arbitrary joins arise from looking at the lattice of open sets of a topological space?

And so on.

Ok.

Definition

A locale is a poset with finite meets and arbitrary joins

A morphism of locales is an order homomorphism of posets preserving finite meets and arbitrary joins

Denote by Loc the category of locales

Then there is a contravariant functor T(-) from Top to Loc^op

sending a topological space to its lattice of opens.

Theorem: T(-) has a right adjoint. neat!

I'll describe the construction of this right adjoint. Remember that we said earlier that points of a space determine completely prime filters in the lattice of opens. The right adjoint we want to construct now will associate to each locale A a space X whose underlying set of points are precisely the set of completely prime filters in A, which we think of as the filters containing the corresponding points of X.

The topology on X is not hard to describe, it should have as its open sets exactly the elements of A. If x is a point in X, then by definition x is just some filter F_x \subset A. If U is an element of A, we'll say that U contains x iff U is an element of F_x. Thus the open sets are just those of the form { x\in X | U\in F_x} for all U in A.

I won't describe in too much detail how it acts on morphisms. It's an easy exercise to show that any map g : A -> B which preserves arbitrary unions and finite meets will have the property that g^{-1} of a completely prime filter F\subset B will be again a completely prime filter in B. Then you have to show it's continuous.

Ok

Cool.

So

let's call this right adjoint Sp(-) for "space".

There is a separation axiom for spaces, called being "sober". A space X is said to be sober if every irreducible closed subset F has a generic point, i.e. a point x\in F whose closure is F itself. (x is necessarily unique if it exists)

It's not hard to show that X is sober iff every completely prime filter on the lattice of opens T(X) corresponds to a unique point in X. Thus the set of elements of X is in bijection with the set of elements of Sp(T(X)). Indeed, one can show that the unit of the adjunction X -> Sp(T(X)) is not just a bijection but a homeomorphism if X is sober.

Moreover, this is clearly iff, as the unit won't be a bijection unless X is sober.

Furthermore, it's a proposition that the functor Sp(-) sends all locales into the full subcategory of sober spaces.

Conversely you can identify a certain distinguished full subcategory of the category of locales - let's call these the spatial locales - such that the functor T(-) carries every space to a spatial locale, and every spatial locale is isomorphic to a locale in the image of T.

Again, it's a proposition that if A is itself a spatial locale, then the counit A->T(Sp(A)) is again an isomorphism of locales.

To summarize: There is a contravariant adjunction between topological spaces and locales. This restricts to an equivalence of categories between the category of sober spaces and spatial locales.

This situation is much like the situation in sheaf theory where one has an adjunction between presheaves over a fixed space X, and bundles fibered over X, which restricts to an equivalence of categories between sheaves over X and etale spaces (local homeomorphisms p :E -> X)

In the situation with sheaf theory, the forgetful functor from sheaves to presheaves has a left adjoint, which is given by passing from a presheaf P to the etale space of the presheaf and taking global sections. This gives the sheafification of P.

The same situation is repeated here.

The forgetful functor (inclusion) from sober spaces to all topological spaces has a left adjoint. The soberification of a space X is given by forming the lattice T(X) of all open subsets of X, and then considering the completely prime filters on the lattice of opens.

You are adjoining all the points you "want" to be there based on consideration of the lattice of opens.

The universal map X -> Sp(T(X)) will be injective iff X satisfies the T0 separation axiom.

Alright. I have given this long winded rant.

What is the point of all this

what the hell does this have to do with your question

The answer is this: the functor t in your question is exactly the soberification functor. Well, that's not exactly true because a variety is more than a topological space, it's a topological space equipped with a sheaf of regular functions. But if P is any sheaf of rings on a space X, and Sp(T(X)) is its soberification, then we can equip Sp(T(X)) with the structure of a ringed space in an obvious way - by taking the direct image of P along the unit of the adjunction, X -> Sp(T(X)). However, on the level of topological spaces, the scheme associated to a maxspec variety is exactly its soberification.

As you say, it is a matter of proving that these functions must map closed points to closed points. I don't think that this is a question about topological spaces and soberification tbh so i'm sorry for making you read all that if you were just looking for an answer to your question.

Now, I actually think this is an extremely interesting fact. I can't fully justify this by pointing to any concrete result. But to me this feels like almost a sheaf-theoretic motivation for preferring schemes over varieties which helps us go beyond the common sense thing "oh well arbitrary ring maps don't necessarily send maximal ideals to maximal ideals"

Why do I say this is a sheaf theoretic motivation? Because the structure sheaf only knows and cares about the lattice of open sets of the topological space. For the sake of sheaf theory, we can take a "stalk" (a filtered colimit) among any filter of opens that looks like it might correspond to a point, say, every completely prime filter. So saying that every completely prime filter in the lattice of opens necessarily corresponds to a unique point can be thought of as saying: we want to assert as an axiom that for every filter of opens we might reasonably take a "stalk" at (a filtered colimit), that point really exists, and we can take the stalk of the sheaf at that point. All the stalks for the sheaf that might reasonably exist, do.

The soberification functor lets us force all of our varieties to satisfy this property by replacing them with schemes which do have this property.

Or perhaps we can say this: Some of the arguments in the old days spoke about these fictitious 'generic points' somewhere in the space, but nowhere in particular, a point p such that every property A which held almost everywhere held at p. Sometimes it was important that such a point p actually exist, and Krull and Noether devoted time to proving existence results for sufficiently generic points. Other times the points didn't exist actually exist at all, and you could prove that there wasn't any point which satisfied those properties. The arguments weren't really about the point at all. The arguments were pre-sheaf theoretic language, but in modern sheaf theoretic language, they were really taking place in the lattice of opens in some completely prime filter in the lattice of opens. Grothendieck's schemes with the honest-to-god generic points which existed in the space let us formalize these arguments by talking about that generic point, but it's my belief that many of these arguments could be carried out perfectly rigorously for maxspec varieties simply by translating everything about the generic point into an argument which takes place more explicitly in the sheaf of functions. Perhaps in the case where the field k you are working over is not ACF, there are morphisms T(Y) -> T(X) which don't correspond to maps of varieties X-> Y but are still perfectly adequate for taking the direct image / inverse image, and so in the pre sheaf-theoretic language (as opposed to presheaf theoretic language) you would have to talk about "functions" X -> Y carrying "generic points" to "generic points"

I should admit that the stuff in the last paragraph about the history is... conjectural. There's just too much goddamn literature out there and I haven't had time to really study any of the classical generic point arguments and get a sense as to what is going on. Too much to read, too little time

One of the big insights from topos theory is that arguments in the sheaf of functions over a space have a natural forcing logic which parallels the one used in set theory to speak about generic filters. Something I'd like to do eventually is go through and see if it's possible to write down explicit translation of these arguments into a forcing argument in the rings of regular functions of the structure sheaf over a variety. Ingo Blechschmidt's thesis is one of the only resources I know of on this kind of stuff

[pearl holly]

[swift fjord]

"I don't want to give a massive monologue" gives a massive monologue

[cedar pebble]

i don't want to give a massive monologue but like
Chapter 1, section 1: frames and locales

[plain raven]

I recently started taking an ADHD drug and I have not yet adjusted to it.

[empty grove]

Yooo this is the stuff nikita has been telling us all along

[plain raven]

If the ADHD drug causes me to waste like 1.5 hours writing up a long rant that no one asked for and no one is going to read then perhaps it is not going to be super helpful for treating my productivity related dysfunctions. Hmm

[empty grove]

I've read half of it so far, very cool. I didn't get why a generic point of an irreducible closed subset is unique though

[plain raven]

oh, yeah. i mean i guess it's not obvious but it's also not a very hard exercise.

[empty grove]

oh ok I'll try that

[plain raven]

Is it? thinking

half an hour later no, it's obvious.

[empty grove]

[plain raven]

Oh, wait. maybe this is only true for T0 spaces.

all spaces are T0 tho

[empty grove]

By point that is its own closure you mean some x such that cl({x}) = {x} right?

Oh wait

[plain raven]

when did i say point which is its own closure

[empty grove]

Fuck

Yeah so that's why I was finding it so hard

I thought I was almost done when I got to the point where you said here's what this has to do with your question

Finished reading it now, it was very nice are these generic points something like prime ideals in spec? Since max ideals correspond to actual points in traditional varieties, and non maximal primes are sort of everywhere where they have 0s? I think vakil mentioned something like this but I haven't read it properly yet

[gritty widget]

Is terminology correct

Suppose there is some top space X

not an abitrary top space, something with lots of nice properties like a manifold with a lie group action

we are concerned with calculating the cohomology of X

suppose instead i can make a differential graded algebra A

such that the cohomology of (A,d) is the same as the cohomology of X

(A,d) being the chain complex oyu get from the dga

is it right to call A an algebraic model of X?

would it be right to call the derham comlex an algebraic model of a manifold

[empty grove]

1 step ahead of you 😎

Wait, on the reading part

Not the starting new ADHD drug part

[gritty widget]

Is there anyway we can go about defining this differential intrinsically

Like without reference to a basis

[abstract pagoda]

tbh idk what that is? wedge product g^V tensor S(g^V)

what class?

[gritty widget]

Sorry I hope this doesn’t come across as rude I’m just so curious

Did you ask this just out of curiosity

Or

So that if you know what these are you will then be able to answer the question

This was mainly funny to me because I always picture people as being their profile pictures

So it felt like

Asking a question to a group of mathematicans

And then a wild otter swims up out of nowhere

‘sorry mate you’ll have to explain that again to me’

[gritty widget]

This also makes @plain ravens messages much more interesting

Absolutely wicked smart dog

nine9s is an otter, clerk is a dog, daminark is a sloth, mniip a cat, and i a hyena

we're all animals that learned how to use computers

The sunglasses throw me off for you

im just that cool

[empty grove]

Buff hyena with sunglasses

I can imagine the broad shoulders right where the pfp ends

[gritty widget]

unfortunately i am not buff

[surreal estuary]

I don't know any of that, but if you have a question you should just ask rather than asking to ask

[spring moat]

idk

[swift fjord]

Just ask

[orchid forge]

I look exactly like my pfp irl

I also do not know

[gritty widget]

my screen is not high res enough that your pfp looks like squidward

is shamrock still around

Can someone help me with the exterior algebra again

seemingly

the exterior algebra on V* should be

$\mathbb R \oplus \mathbb R[\theta]$

[gentle osprey]

lime_soup

[gritty widget]

oh no its on only

$\mathbb R \oplus \mathbb R\theta$

[gentle osprey]

lime_soup

[spring moat]

@gritty widget I guess i haven't learned it correctly.

can anyone explain Grothendieck's finiteness theorem? specifically how it relates to this version of Euler's characteristic?

Specifically exactly how he used Chow's lemma which proves proper morphism is = to projective morphism?

and how a surjective morphism induces an isomorphism.

how does f : X' > X go to f^-1(U)≈U?

[gritty widget]

This is confusing me

W(g) is the tensor of the exterior algebra on the dual of g

With the symmetric algebra on the dual of g

It make sense that the symmetric algebra is given by R[u]

But I’m confused why they have only given the degree 0 and degree 1 elements in the exterior algebra

[orchid forge]

it is the exterior algebra of a 1-dimensional vector space

[plain raven]

yeah. nonmaximal primes are thought of as being generic points. the justification for this is that like, every ideal gives rise to a "variety" in the spec, which is formally the set of all prime ideals containing it. so a prime ideal which is not maximal represents both a variety (the set of all maximal primes containing it) and a point itself, the "generic point" which lives somewhere in the variety but "nowhere in particular" - formally what we mean by this is that every open dense set which meets the variety anywhere at all contains the generic point. (In the zariski topology every nonempty open set is dense, but i like to throw the word 'dense' around anyway to stress the point that we're interested in properties that hold 'almost everywhere'. generic point arguments are something like arguments by appeal to baire category in analysis where you take a countable family of properties which hold "almost everywhere" i.e. on an open dense set, and show that there's at least one point in the intersection for which all those properties hold)

[plain raven]

i don't know exactly what the weil algebra is but it looks a lot like a koszul complex? i can't tell exactly because i can't read the notation but i recently did find a very nice coordinate free definitiion of the koszul complex

[plain raven]

tell me if you find an answer tho. i love collecting coordinate free definitions of complexes >:)

also i think there is an exercise on this in Cartan and Eilenberg's book on homological algebra?

exercise 14 in the chapter on lie algebras

page 287

what source is this? what are you reading

[orchid forge]

dudes be like: "i love coordinate free stuff"

then they're like "just consider the free product of all possible systems of coordinates"

[plain raven]

[orchid forge]

i ain't saying it's a bad thing, because it isn't

but there is a part of my brain that perceives some absurdity here

[wary furnace]

does anyone know any books, pdfs or other kinds of resources with a bunch of tasks and exercises? (for topology ofc)

[orchid forge]

at what level? most textbooks will have exercises

like an introductory general topology book with good exercises?

[wary furnace]

yeah exactly

[orchid forge]

munkres is the standard reference for that, but i'm sure some people here like other ones too

[wary furnace]

my main problem is that the one I'm using rn has very little exercises, it's like "prove this simple fact and then let's move on"

[orchid forge]

what are you using?

[wary furnace]

Topology through Inquiry

Michael Starbird and Francis Su

[orchid forge]

chapter 1 is about cardinality

chapter 7 is about continuous functions

what a very strange book this seems to be

[wary furnace]

all of that is the prereq

[orchid forge]

which chapters are you looking at in particular?

all the point-set stuff in here is covered in munkres and those exercises should be good

[wary furnace]

Chapters 2.1 and onwards

[orchid forge]

but i would recommend looking elsewhere for manifolds + homology

yeah i would say munkres for chapters 2 through 9

[wary furnace]

hmm alright I'll look it up

thanks for the advice

[orchid forge]

just bear in mind the order of topics will be different

[gritty widget]

It’s exactly the koszul complex of a Lie algebra

[gritty widget]

Yes but

Why does it have no terms above degree one

[orchid forge]

what would a degree 2 term look like?

[gritty widget]

Why is it $R\oplus R\theta$ not $R\oplus R[\theta]$

[gentle osprey]

lime_soup

[gritty widget]

Thank you

God

Damn

That’s obvious

So if V is and n dim vector space

The exterior algebra dies at

Terms of degree n

[orchid forge]

yes

and the nth exterior power is really just a 1-dimensional vector space

consisting of the multiples of (the wedge product of) any given basis

[gritty widget]

Someone recently described to me

That the exterior algebra properly captures the idea of

Squaring infinitesimals is zero

[orchid forge]

that's not the intuition i have for the exterior algebra. i would say that concept corresponds more directly to the ring of dual numbers

but it's possible?

I am pretty sure I wrote a clerkpost about what i think the exterior algebra is geometrically once

[plain raven]

Tbf the exterior algebra of k as a k vector space is exactly the ring of dual numbers over k

[orchid forge]

yeah but that's kind of a special case

[plain raven]

yeah

[orchid forge]

the exterior algebra over a vector space is basically what you get when you want to capture a notion of volume which transforms linearly with respect to linear transformations

[gritty widget]

What are dual numbers

[orchid forge]

$R[x] / (x^2)$

[gentle osprey]

Kogasa

[gritty widget]

So we have something that square to zero

Is the term dual used in the sense of

We now have “two numbers”

Or that these are some “dual” of regular numbers

[orchid forge]

i'm not actually sure what dual is supposed to mean here, i would guess it's either in the first sense or in that x^2 = 0

they come up pretty naturally when you try to define a notion of intersection for algebraic varieties

like in $\mathbb R$, the origin corresponds uniquely to the maximal ideal $(x)$ in the sense that ${0}$ is the vanishing set of the maximal ideal, but there are subideals $(x^n)$ which satisfy the same property, so $\mathbb R[x] / (x^2)$ has the same underlying set but with a richer algebraic structure. under the appropriate definitions and associations, that structure can be thought of as a 1-dimensional (co)tangent space at a point

[gentle osprey]

Kogasa

[orchid forge]

for example, the intersection of the line $y=0$ with the circle $x^2 + (y-1)^2 = 1$ is just a point (0,0), but after translating "sets of points" into "algebraic varieties," the notion of "intersection" turns into a tensor product of coordinate rings, and the intersection of these things as varieties is precisely the ring of dual numbers $\mathbb R[x]/(x^2)$, indicating a "multiplicity 2" intersection

[gentle osprey]

Kogasa

[orchid forge]

when you define the cotangent space abstractly for varieties, you also consider something of the form $m / (m^2)$, where $m$ is the ring of (regular) functions vanishing at a point. the elements of this quotient are functions whose zeroth order is specified (it is 0 at x), and whose higher-than-first order terms are quotiented out. so it's just the "first order data of regular functions at x," which captures the tangent space idea kind of intuitively

[gentle osprey]

Kogasa

[orchid forge]

but this isn't the post i wanted to write

[plain raven]

my speculation is that maybe the "dual" relates to the geometric interpretation you're talking about? tangent vectors operate on functions, so a dual number thought of as a tangent vector on the line is kind of dual to the ring of functions in that sense

(??)

[orchid forge]

yeah that might be right

that thing i defined $m / m^2$ is really the cotangent space, so we need to take the dual to get proper "tangent vectors"

[gentle osprey]

Kogasa

[orchid forge]

Let $V_k(v_1, \dots, v_k)$ be the function which sends a collection of $k$ vectors to the $k$-dimensional volume of their convex hull. For the standard basis $e_i \in \mathbb R^k$, we have $V_k(e_1, \dots, e_k) = 1$. Then $V_k(\alpha_1 e_1, \dots, \alpha_k e_k) = \prod_{i=1}^k \alpha_i = \prod_{i=1}^k \alpha_i V_k(e_1, \dots, e_k)$. It follows that, if one of the $\alpha_i = 0$, then the volume is $0$. So $V_2(v_1, v_1) = 0$. This, together with linearity, implies that $V_k$ is alternating: if $\mathbf{v} = (v_1, \dots, v_k)$ and $\sigma \in S_k$, then $V_k(\sigma \mathbf{v}) = -1^{\mathrm{sgn} \sigma} V_k(\mathbf{v}).

[gentle osprey]

Kogasa
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[orchid forge]

oh my god

We now ask, given a linear transformation $f : V \to V$, how does $f$ act on volumes? Well, by linearity, we just need to check a basis. If we can find an eigenbasis for $f$, then the volume would just change by... the product of the eigenvalues. After some checking, we find that $f$ acts on "ordered sets of n vectors" by multiplication by $\mathrm{det} f$

[gentle osprey]

Kogasa

[orchid forge]

If you have some collection of $k$ vectors, with $0 \le k \le n$, then there is a similar functorial interpretation, in terms of restrictions of $f$ to certain subspaces

[gentle osprey]

Kogasa

[orchid forge]

the exterior algebra of $V$ is just what you get by making this precise and asking "what is the natural way to define objects, so that the maps between them are precisely the volume-scaling maps induced by linear maps"

with some thought it's not hard to believe that this would just be tuples of vectors, whose order is considered up to a sign

so, modulo details, this says that the exterior algebra is correctly understood as consisting of "oriented volumes"

at least, this makes sense when you're talking about a relatively concrete vector space, like R^n or the tangent space to a manifold

[tough imp]

So maps to P^n are determined by a line bundle L and n+1 elements y_0,…y_n which generate L

However we also have copies of A^n inside P^n given by saying that x_i ≠ 0

I’m trying to see what these affine spaces, for simplicity let’s say x_0≠ 0 corresponds to in terms of this description of maps into P^n

My point is, I feel as though maps into the space where x_0≠0 corresponds to a line bundle L with n + 1 elements y_0,…,y_n such that y_0 actually generates L

Is that right?

[finite heath]

may i ask a question -- i dont wanna interrupt chmonkey

[tough imp]

Go ahead @finite heath

[gritty widget]

Sorry I fell asleep before getting to reply last night

Thanks very much that helped tie a lot of ideas together

[uncut surge]

Just to be sure: If we have two topological spaces $X,Y$, and a cover $\mathcal{W}$ of the product space $X \times Y$, then there does not necessarily exist a refinement $\tilde{\mathcal{W}}$ of $\mathcal{W}$ coming from the product of two covers $\mathcal{U}$ and $\mathcal{V}$, i.e. $\tilde{\mathcal{W}} = {U \times V : U \in \mathcal{U},V \in \mathcal{V}}$. As a consequence, if we consider the collection of all open covers of $X \times Y$ as a directed system w.r.t. refinement, then the collection of product covers of the above shape is not cofinal in this system

[gentle osprey]

Lartomato

[uncut surge]

My thought is something like filling out (0,1) \times (0,1) with smaller and smaller balls towards the edges and corners, so that any cover coming from the product of two covers becomes too rigid

Just writing this down because this sucks ass

no wait maybe this isn't true at all -- gonna have a thonk

[gritty widget]

I would have guessed there is such a refinement

Just because the converse of tychonoffs theorem is true

I know this is not super related but if you could make a “bad”cover of the product like this then I wouldn’t expect the converse of tychonofff to be true

[uncut surge]

What's the converse of Tychonoff precisely?

[empty grove]

probably that if a product is compact then each factor must be

[uncut surge]

'cus the counterexample I'm thinking of is filling out an open rectangle in this way, so that the width of the interior rectangles becomes smaller and smaller as you go further up. You can make this into an open cover, so that if you take any open interval (a,b) on the x-axis, there will be an arbitrarily small interval (x,y) on the y-axis so that (a,b) \times (x,y) will not be contained in any of these small rectangles

shitty pic is shitty but I think this idea makes it so that no product cover can be finer than this one

[uncut surge]

o thanks

ye maybe thinking about tychonoff helps

ye but the product topology allows you to do more, since you can use unions of cartesian products to create things which aren't cartesian products, so this doesn't automatically falsify tychonoff or so lol

[gritty widget]

Oh

Maybe this is stupid way to make an example

This is an open cover of the product

But

It does not come from a product of open covers

And

There is no refinement

I remember someone once told me

Any non theorem in topology is is a non theorem because of finite topological spaces

[uncut surge]

hahaha nice one

Ye alright so I can be pretty sure that this statement is garbage

okay i sent the spooky german professor who made that claim in his paper an e-mail now

god, i wish this didn't happen all the time

[true robin]

I might be misunderstanding the example here, but isn't the product space supposed to have 9 elements?

[gritty widget]

Oh shit yeah

But think it’s still okay

[empty grove]

The product of 2 discrete spaces is discrete right? Then can't you refine any cover to be a cover by singletons?

And a cover by singletons is a product of covers

[true robin]

a refinement here probably means a subcover of the open cover (atleast that is what I'm guessing) in that case it should be easy to find counter examples. I think you are thinking of a cover such that each element is a subset of an element of the original cover

[empty grove]

Refinement means that every element in the cover is contained in some element of the original cover

Wait what

Is that not what's being used here

[true robin]

yeah, so if the original thing doesn't have singletons, how are you getting a subcover that has singletons

[empty grove]

Wait are you sure they're using the subcover definition

[true robin]

yeah

because otherwise the product top has a basis of elements of the form he asked for

so I assumed it meant subcover

[empty grove]

Not necessarily

They're taking product of covers

[true robin]

I see

yeah then what you are saying makes much more sense

[empty grove]

Like in this the U and V are not just any open sets in X and Y

[uncut surge]

Oh yeah actually that thought is way more useful

actually nvm

Yeah I've been thinking about moldilocks' refinement definition

does it become trivial then

But yeah okay I get it now, this won't fail in the discrete setting because you can always take an open cover by singletons

[true robin]

yeah I think your original idea should work: we take an open cover of R^2 by saying that (x,y) has to sit in a open ball of radius 1/max(|x|,|y|) around it (or 1 if both co-ordinates are zero). then taking any open covers U and V of R, if 1 belongs to some U_i in U, we can find an open interval of size 1/N around 1 inside U_i. I'm pretty sure U_ixV_i cannot be contained in an element of the cover containing the point (1,2N) for any V_i. let V_i be any open set containing 2N in V. the open sets in our original cover that contain (1,2N) cannot have radius greater than 1/N as then (1,2N) will we a distance 1/N away from some point whose maximum coordinate is <=N which is impossible. so that means (U_i)x(V_i) cannot be a subset of this open cover

[gritty widget]

opps lmao

[uncut surge]

thanks

what joy it is to find extremely critical errors like that in 60 year old german papers with a bunch of citations

[gritty widget]

now

does it actually end up being a critical error

[true robin]

oh was this claim in some paper? what was it about?

[gritty widget]

or is the general claim right

you just need to do some extra steps

[uncut surge]

the claim is very very implicit, but they're basically saying that you can calculate cech cohomology of a product of sheaves by products of open covers on the base spaces

so essentially a proof of a very general künneth formula

and this works in a lot of settings, in the algebraic geometry setting you can take covers by open affine sets, and their products are open and affine again, so then there's some Leray Theorem telling you that you get sheaf/cech cohomology from just looking at these covers

but in this paper, they're saying "alright and now we're taking the direct limit over the refinements of open covers", but they're just taking the direct limit over the above product covers

https://link.springer.com/article/10.1007%2FBF01111357 dis, for anyone who's overly interested

i think the author's still alive, i sent them an e-mail, but idk if anything's gonna come from this

[true robin]

I see, that is quite interesting. Is the main result in any trouble because of this or is it just a minor correction?

[uncut surge]

i honestly don't see how to fix it, but i'm just a lowly phd nerd

i feel like this happens once every couple months in my area of mathematics and it's super frustrating; there's not even a standard kind of venue I can go to

like, if the author was dead, who would i even ask? where would i even complain to? if it ended up being a legitimate issue with the paper, would it even get retracted? somehow i don't think so

i mean if this ends up being an issue, i'm certainly not the first person to notice it in all these years, but what does that imply, that no one else cared enough? oh man

and then you end up sending e-mails about this kind of stuff, get an e-mail back saying like
"Dear Lartomato,

Yes, that is true. Very interesting. Let me think about this."

and you never hear back from them

i am hurt, mathematics has hurt me, i will never recover

[gritty widget]

I don’t get why this should be 0

W(g) is the weil algebra, or the koszul complex on g

so really $\theta=\alpha \otimes 1$ where $\alpha$ is the basis on $\mathfrak{g^*}$

[gentle osprey]

lime_soup

[gritty widget]

so $\delta \theta(X)= u(X)=1\otimes \alpha(X)$

[gentle osprey]

lime_soup

[gritty widget]

alpha(X) should just be some real number

are real numbers zero in the symmetric algebr?

i don't think thats true

so alpha lives in S^1(g^*)

so alpha is a polynomial of degree 1

are the degree zero elements of S(g^*) always 0?

[uncut surge]

I think the interpretation $\delta \theta(X) = (\delta \theta)(X) = u(X)$ is wrong; rather, $\delta$ acts on the real number $\theta(X)$

But the (anti-)derivation \delta is necessarily zero on real numbers, so it vanishes there

@gritty widget does that halp

[gentle osprey]

Lartomato

[gritty widget]

Ahhh

That makes sense

So

In general for cartans formula

The d iota _x (whatever)

Is really

d(iota_x(whatever))

[wise ruin]

Simple question, by why is the (ball subset of cube) condition required for cubes to be basis of R^n? Shouldn’t the (cube subset of ball) condition be enough since then, for all points in open sets U, there exists an open cube in an open ball in U that contains p, satisfying the condition of a basis?

[bleak helm]

You need both to prove that they generate the same topology

[uncut surge]

The "Therefore" conclusion is, a priori, weaker than what you're showing there, I guess

[wise ruin]

Doesn’t generate just mean that it’s a basis? I thought that was just some equivalent terminology

[bleak helm]

You are correct, I was overthinking the requirements

[uncut surge]

The "ball subset of cube" condition is basically just a very prototypical version of saying "the open cubes are open sets within the standard topology", I guess; because of course, to have a basis of a topology, you need to know that all the sets in your basis are open

'cus of course all the singletons lie in every ball as well, but no open ball lies in them – so yes, to an extent, you need both statements to know that these constitute a basis

[bleak helm]

Yes, thank you, I knew something was off

[wise ruin]

That’s kinda what I was thinking as well, thanks for the solidification

[uncut surge]

Same! Glad that's figured out

[uncut surge]

@true robin the fucking madman fixed the theorem in an article 3 years later

"Ein Satz über topologische Urbildgarben", which has THREE citations, whereas the original paper has like fifty

now my e-mail is pointless, my soul is healed, my crops are watered and spring is coming

i despise the fact that academic mathematics is just a constant emotional rollercoaster

oh but the original theorem is still wrong

the rollercoaster continues i suppose

NOOOOOOOOO WHY DOES EVERYONE MAKE THIS MISTAKE

THEY ARE NOT COFINALI

[meager python]

Very stupid question. Let x* be a geometric point and let FEt_S be the category of finite etale morphisms over S. Let F(X) = Hom_S(k(x*), X) and G(X) = X x_S Spec k(x*). Why are these isomorphic?

[gritty widget]

Does anyone know where to/ how to

Find a translation of Ellie cartans thesis

[gritty widget]

I'm trying to understand this.

This is the set up.
I have $H^(X)$ and $H^(Y)$ where $X$ is a topological space, and $Y$ is some diffenerential graded algebra, so $H^(X)$ is the regular cohomology and $H^(Y)$ is the cohomology of the complex you get from the differential graded algebra.

Now $X$ is actually an infinite dimensional manifold. I can approximate $X$ by a finite dimensional manifolds $X_n$. Doing this one can show that there are isomorphisms $H^n(X)\cong H^n(Y)$ for every $n$.

The next step is I want to show that we actually have an isomorphism as cohomology rings. The text states: "The approximtions $X_n$ of $X$ form a direct system of topological spaces. The cohomology functor induces an inverse system of graded algebras. By the universal property of the inverse limit, there is a ring morphism from $H*(X)$ to the inverse limit of the $H^*(X_n)$. For given $q$ if $n$ is sufficently large $H^q(X_n)$ stablizes to $H^q(X)$, so the ring morphsim to the inverse limit is really an isomorphism.

[gentle osprey]

lime_soup

[gritty widget]

so i a direct system of top spaces is a sequence of inclusions

I don't understand how

the stablizing induces a ring isomorphism

it doesn't seem to tell you anything about the product structure

[orchid forge]

The stabilization $H^q(X_n) \to H^q(X)$ says that the inverse limit is isomorphic to $H(X)$ at each degree. The product structure is fixed too: for each a,b having degree n, m there is an N so that the cohomologies in degrees n, m, and n+m all stabilize

[gentle osprey]

Kogasa

[gritty widget]

Hmm

[finite heath]

A = [0, 00)

int(A) = (0, 00)

cl(A) = [0, 00]

on R

right?

oops i didnt mean to interrupt!

[gritty widget]

Suppose N is such that H^k(X_n) is isomorphic to H^k(X) for

I in degrees n m and n+m

Does this mean that the product of a b for X_N will be the same as for X?

[orchid forge]

by assumption $H^p(X_N) = H^p(X)$ for $p \in {n, m, n+m}$, so i'm not sure how it could be otherwise

[gentle osprey]

Kogasa

[gritty widget]

Just because you can have an isomorphism in all cohomology groups

That isn’t an isomorphism of rings

[orchid forge]

it's a direct system of graded algebras that's stabilizing

[gritty widget]

Ah

[orchid forge]

so presumably the product is preserved too

[gritty widget]

And then the morphism from the universal property will be a graded

Yes thank you

[orchid forge]

the way it's stated, the universal property is just a ring morphism

but i think you can easily show that the product is also preserved in the way i said

pick N large enough that all the relevant degrees stabilize, then project from the inverse limit onto H(X_N) = H(X)

actually wait, it's even simpler

or it may not be, i g2g rip

yeah nevermind, you can split the map $H(X) \to \varprojlim H(X_n)$ by sending $\alpha \to \pi_N \alpha$ for large enough $N$ that $H(X_N) = H(X)$. this is well defined, and a ring homomorphism for the reason i said, if $\alpha, \beta \in \varprojlim H(X_n)$ in degrees n and m, pick $N$ large enough for degrees n, m, n+m to stabilize, then $\pi_N(\alpha \cdot \beta) = \pi_N \alpha \cdot \pi_N \beta = \alpha \cdot \beta \in H(X)$

[gentle osprey]

Kogasa

[orchid forge]

err, by that last part i meant $\pi_N \circ \phi$ instead of $\pi_N$, and $\alpha, \beta \in H(X)$

[gentle osprey]

Kogasa

[crimson path]

A singleton set is path-connected, right?
f : [0,1] ---> {a}
would just be a constant map and thus continuous, no?

[crimson path]

@gritty widget thank you!

[bleak path]

Hi, have a very simple question that I can't definitively give an answer to:

Is (0, 1] homeomorphic to [0, 1)? What about (0, 1) and (0, 1]? And what about (0, 1] and [0, 1]?

Intuitively I think the answer is yes to all 3 but I cannot think of a way to definitively say either way. Some part of me thinks all 3 might be no because (for example with (0, 1) and (0, 1]) you must remove a point to get from the latter to the former, and that means they are not homeomorphic, but isn't this point just at the end since they are both connected?

[wanton marsh]

I'm not sure what you mean but, since you can remove a point from (0,1] to get (0,1), if you assume that (0,1] is homeomorphic to (0,1), then you deduce that you can remove a point from (0,1) to get something homeomorphic to (0,1)

[bleak helm]

(0,1] and [0,1) are homeomorphic with homeomorphism ||f(x) = 1-x||. The others are not.

[wary furnace]

quick question, is the discrete topology on X just the power set of X? (nvm I figured it out already, it is)

[rancid umbra]

R and (0,1) are homeomorphic but you have to remove uncountably many points from R to get to (0,1). just a heads up, that’s not what a homeomorphism detects in general

[bleak path]

Well thanks for the clarification then. The homeomorphism for the first question helped quite a bit

Actually while you're here, do you have an intuitive explanation of why removing a single point from S^2 gets you R^2?

My lecturer mentioned that but I don't see a "stretching and bending" way of getting from one to the other

[tough imp]

Just like

Remove the top

And then like

Lol I’m about to say the same thing

But

There’s a hole in the top now

Okay new idea

You have a playdoh S^2 on the table

You cut a little hole in the top

Stick your hands in and like pull the hole… out and down?

[bleak path]

like stretch the hole out?

[tough imp]

You should be able to flatten it out into a disk

Yeah

And also like drag it down

What used to be the inside of the hole at the top

Like the edge

Is now the outer edge of the disk

On your table

[bleak path]

right I can see that now

[tough imp]

That’s R^2

When you stretch

Like uh

Assume this disk is now the unit disk

You can send the point at angle theta and radius r

To the point at angle theta

And radius 1/(1-r)

I think

And that fills out all of R^2

And well the origin just stays fixed

[bleak path]

ahhh using polar is a good way to visualize it

I only considered cartesian and it felt like the "edges" were joining up which we don't really want

[tough imp]

You can do it like

Explicitly

Using the projection map

Like it’s called uh

Fuck what is it called

[honest terrace]

stereographic projection

no ?

[tough imp]

Yeah

That

[bleak path]

ah

henry segerman

[tough imp]

😐

Maybe

Idk

[bleak path]

binged a bit of his youtube videos and I saw it there

[tough imp]

You take a point on the sphere

[honest terrace]

here's a picture from wikipedia that gives a picture of what happens with S^1 and R

[tough imp]

Draw a line from the top of the sphere through that other point

[bleak path]

https://www.youtube.com/watch?v=VX-0Laeczgk

[honest terrace]

When you remove the north pole of S^1

[tough imp]

And then send it to where it intersects the plane

This defines a homeomorphism from S^2\top to R^2

[honest terrace]

you can send the remaining points in a continuous way to R

yeah

chmonkey already explained it

I'm too slow sorry lol

[tough imp]

To show its continuous with continuous inverse you can write it all out in coordinates

[bleak path]

your way feels like a different way of explaining it though, that helped

[tough imp]

It ends up being like rational functions in the coordinates

With maybe sqrts

[bleak path]

although I might discover in the future that its actually the same way

[honest terrace]

I think the explicit map is in Lee's ITM

let me check

[tough imp]

Yeah

It is

It’s also in like any complex analysis book

[bleak path]

the reason im asking the R^2 S^2 bit is because I have an assignment question, which is "Is S^1 homeomorphic to S^2? Prove why or why not" and I tried the "adding a point bit"

So my logic is that S^1 - a point = R^1, and similarly S^2 - a point = R^2, and so since R^n is not homeomorphic to R^m for n != m, S^1 and S^2 are not homeomorphic

[honest terrace]

Nvm it isn't in Lee's ITM, there's only map from B^n to R^n, no S^n to R^n

😔

[bleak path]

but it also feels not rigid to me

[coral pivot]

If you want something a bit more direct

[honest terrace]

https://math.stackexchange.com/questions/1122931/showing-that-stereographic-projection-is-a-homeomorphism

[tough imp]

How much topology do you know?

[coral pivot]

Remove two points from the circle

[tough imp]

You can argue this a million ways

[honest terrace]

Yeah, removing two points is probably the simplest way

[tough imp]

Yeh

[honest terrace]

Removing two points from S^1 gives you a non-connected space while removing two points from S^2 gives you something that is still connected

[bleak path]

feels like my answer is too handwavy for where I am at the course

[tough imp]

Can you see how S^2 minus two points and S^1 minus two points

Differ in a crucial way

[bleak path]

ah lol I dont know why I didnt consider another point

[bleak path]

I am currently approaching the midway point of a first course in general topology

[tough imp]

Okay then yeah

[bleak path]

thats the hard bit lol

[coral pivot]

The idea is kinda the same, like circle minus a point is R and then you remove another point to get the usual arguement

[honest terrace]

yeah well

ti's literally the same

[tough imp]

I think removing two points is easier than my idea anyway

Yeah

[honest terrace]

because S^1 - 1 point is R and same for R^2 minus 1 point

[bleak path]

yeah, that one just clicks with me

[coral pivot]

To be fair I think that would be circular

All proofs I have seen of that Todd require you to do it on spheres first

[tough imp]

S^2 minus a point is R? Neat!

[honest terrace]

lol

[bleak path]

Actually I know I can use the minus points argument for sure, it's in one of the answers demonstrated a few weeks ago

[tough imp]

Yeah I was gonna suggest you look at the fundamental group lmfao

[bleak path]

That sounds like it belongs in algebra

[tough imp]

No this is topology

It’s about closed loops in your space

[honest terrace]

This is a bit of an overkill here, isn't it ?

[bleak path]

🤔

[tough imp]

The idea is you can make a loop in S^1

That you can’t like

Squish down to a point

[coral pivot]

The way you show the spheres are not the same is usually using like homology for instance

[tough imp]

Cuz S^1 Hs a hole

In S^2 you can just like

Squeeze the loop

[honest terrace]

I mean, if he's working with "low dim things", it's probably because he doesn't have the tools yet to prove this kind of things

[tough imp]

Until it’s a point

[bleak path]

Algebraic topology is after general topology for me, is that normal?

[tough imp]

Yes

[coral pivot]

So it’s a valid arguement that shows the power of these tools

[bleak path]

coolio

[tough imp]

That’s why this proof is bad

[honest terrace]

Oh sry I missed wOne brought it up, nvm then

[tough imp]

The invariance of dimension things is easy in the n = 1 m > 1 case tho

[bleak path]

There's something about being new to a topic and answering some bits excessively

[tough imp]

Because of precisely this argument

Remove a point

[honest terrace]

but yeah, basically fundamental group memery is just "to each top space, you associate some group that preserves some info about the top space"

[tough imp]

Yeah

[honest terrace]

And computing this group for two given top spaces gives you a way to discriminate non homeomorphic spaces

because homeomorphic spaces have the same associated group

[bleak path]

This is with regards to R^n != R^m?

[orchid forge]

the first thing in algebraic topology is to state this and be like "we'll prove this in a bit"

[honest terrace]

lol

[bleak path]

Is there a symbol I can use to show non-homeomorphy without latex, or is != fine

[orchid forge]

state the invariance of dimension theorem

[tough imp]

Like you could do

[honest terrace]

words ?

[tough imp]

$\not\cong$

[gentle osprey]

Chmonkey

[honest terrace]

so ugly lol

[bleak path]

words yeah but if there's a symbol that would be good too

point taken though

[honest terrace]

Just write it down with words, there's no nice and standard symbol for that

[tough imp]

I just

Provided one

SMFH

[honest terrace]

I mean, I think he just meant $\neq$

[gentle osprey]

Shika

[honest terrace]

I said nice

[bleak path]

I do just mean neq

[tough imp]

[bleak path]

but there's no neq symbol and most people do associate the two

[tough imp]

Yeah there is

$\neq$

[honest terrace]

but yeah anyway in this case what you want is definitely not \neq

[gentle osprey]

Chmonkey

[bleak path]

there's no neq symbol outside latex

[honest terrace]

because equality isn't equality up to homeomorphism

[tough imp]

≠

What’s this then

[coral pivot]

I disagree homeomorphism is equality

[bleak path]

I said without latex because I was wondering what you guys would use in casual conversation around here

[tough imp]

Soooooo bad

[coral pivot]

If the category ur working in is clear this is fine smh

[bleak path]

copied 🙂

[tough imp]

No bad

[honest terrace]

I mean, I agree, univalence ftw, but pedagogically here it's just bad lol

[tough imp]

Unit disk = R^2

[coral pivot]

Yes

[tough imp]

🤡

[coral pivot]

These are literally the same

[tough imp]

🤡

[honest terrace]

what's the difference ?

[coral pivot]

No difference

[honest terrace]

I don't know Todd, ask chmonkey

[tough imp]

S^2 projecting onto R^2 is surjevtive

🤡

[bleak path]

in the world where the complete world is the unit disk?

[tough imp]

Unit disk = R^2 after all!

[coral pivot]

I mean you can compose the projection with the homeo

[bleak path]

lol I see why it can be confusing now

[coral pivot]

Smh

[honest terrace]

Unit disk = Unit disk 🤡

[tough imp]

My point is topology is based on sets and it’s meaningful to deal with actual sets

So abusing = like that is badge

[honest terrace]

pointless topology ftw

[bleak path]

I know enough to not be confused

[coral pivot]

I personally think it’s fine pedagogy, like morally these shouldn’t be considered different if ur only working in top

[tough imp]

But working in Top

[honest terrace]

Yeah I'm mostly trolling here, ignore me

[bleak path]

thanks for the tips though 🙂

[tough imp]

Involves set functions

Which where they can matter if they’re surjective or not

Unless you abstract away the points

[bleak path]

Learning about "clopen" for the first time was pretty funny

[tough imp]

You care about points

Lol

[coral pivot]

I don’t.

[tough imp]

You’re bad

[honest terrace]

I think making an explicit distinction between equality and homeomorphic is important pedagogically

[bleak path]

reminded me of linear algebra, where they defined "linear independence" as "a span of vectors that is not linearly dependent" or something like that

[tough imp]

Like = for canonical iso is admissible

But for just any iso is

Incredibly brain worms

[coral pivot]

Unit interval has two points, the boundary and the interior

[orchid forge]

this greatly simplifies the theory of groups

[tough imp]

This again matters though!!!

You can have isomorphic subgroups

Where one is normal and the other isn’t

[orchid forge]

i'm am yolking

[tough imp]

Tfw

[bleak path]

I have done a basic group theory course but also have 100% not applied it in the context of topology

[honest terrace]

thanks captain obvious

[tough imp]

[coral pivot]

Todd group theorists

[tough imp]

Can we bring topology-and-geometry back to its actual purpose

Talking about anime

[honest terrace]

probably a reference to a group is a one object category

My short term memory is.. wait what was I talking about again ?

[tough imp]

You were talking about how you were gonna pay me back the 20 bucks for pizza

[honest terrace]

Oh yeah right

Is paypal okay ?

[tough imp]

Sure

[bleak path]

you are homeomorphic to a goldfish through the homeomorphism f: shika -> goldfish taking memory as your input

[tough imp]

[e-mail]

Thxthxthx

[honest terrace]

what

please

what

[tough imp]

Doesn’t that give you a ring

Lmao

[honest terrace]

this

[tough imp]

Like a Boolean ring or some shit

[coral pivot]

It’s z mod 2z to the n

Or smth

[honest terrace]

is any group iso to a subgroup of such a group ?

[coral pivot]

No this group only had elements of order 2

[bleak path]

alright I'm going to bounce for now, thank you guys as usual for helping

[tough imp]

Yeah this is called Cauliflower’s theorem

Hmm

Maybe I should’ve said Kale’s theorem

That sounds more like Cayley

[honest terrace]

You mean Yoneda ?

(just preshoting John)

[orchid forge]

is this where i post the uh

[honest terrace]

Don't share such a pic in a public channel please 😔

Installing latex takes so much time wtf

Last time I did it wasn't so long

[orchid forge]

you ever update TeXLive and it's 6000 packages and takes an hour

[honest terrace]

it's 4500 packages in this case, and it already took me more than two hours

[wary furnace]

I'm trying to prove that this thing over here (#A is just the size of a set) is a topology on R. It's quite a trivial task but I'm stuck at proving the arbitrary union point. I know that if you union intervals with a finite number of holes all together you'll end up with an interval with a finite number of holes as well (or just R), but how do I write it down rigorously?

[gentle osprey]

c squared

[wary furnace]

sneaky de-morgan's law

[rancid umbra]

big hints: ||any subset of a countable set is countable and A intersect B is a subset of both A and B||

[wary furnace]

thanks for the hints and the explanation

[wise walrus]

Any idea how I could take the rationals, and find an equivalence relation on the rationals that forms a quotient topology homeomorphic to the topological space on a two element set {x,y}?

My first thought was to take an equivalence relation that sets all even rational equivalent and all odd rationals equivalent, that's a clear two equivalence classes. But I don't quite know how that would transfer to {x,y} through homeomorphism. Maybe I don't understand quotient topologies enough

would the topology on Q/~ then be {nullset, {[a],[b]}, {[a]}, {[b]}} with a an even rational and b an odd rational? That is clearly a homeomorphism to a topology on a two-element set

[fair idol]

I'm trying to learn about embedded vs immersed submaniolds and I don't quite get the distinction between the two. I hear that immersed submaniolds might not necessarily have the same topology as the big space but what does that mean?

Any idea where I can read more about this?

[rancid umbra]

whats an even rational and an odd rational

[wise walrus]

... does that not work

[tough imp]

What is an even rational and what’s an odd rational lol

[rancid umbra]

lol i could see even or odd integers working

[orchid forge]

(the image of) an embedded submanifold should have the subspace topology

An easy way this can fail is if you bring the ends of an open interval together

like send (0,1) to a circular arc tangent to the x axis at the origin and [1,2) to the interval [0,1). This is a loop-de-loop looking thing and every small neighborhood of 0 is homeomorphic to a cross

Or in general send (0,1) to a curve passing through 0

The immersion kind of blends together open sets by sending distant points close to each other

[plain raven]

Folks i spent all day today thinking about chain complexes and boy are my arms tired

[rancid umbra]

since the concept of even and odd rational numbers is not a thing (google confirmed)lol, why not just say p ~ q iff sign(p) = sign(q), where sign(0), instead of normally being set to 0, is instead set to 1

[plain raven]

@gritty widget i spent some time today thinking about your question about a nice abstract characterization of the weil algebra. I think i have the idea but i didn't get time to do the computations and check whether i'm correct

I'll tell you my conjecture tho. it should not be hard to check, i just haven't really had a chance to write it down?

let C denote the cochain complex concentrated in degrees 1 and 2, with C1 = R and C2 = R and the only nontrivial differential C1 -> C2 is the identity

[orchid forge]

If you want {a} and {b} to have open preimage, and {a,b} has preimage Q, then f^(-1)(a) \cup f^(-1)(b) should be a separation of Q in the subspace topology right?

[plain raven]

hmm idk. i don't really want to say this because it might be total nonsense. but like, the category of cochain complexes has all the structure you need in order to talk about the exterior algebra: you form 1, C, C\otimes C, C\otimes C\otimes C..... and take the direct sum of these. then quotient out C^{\otimes n} by the alternating action of the symmetric group, and there's your "Exterior algebra". actually maybe it wouldn't be the alternating action

i think if you set C1 = V, C2= V it gives the koszul algebra generated by V?

hmm. maybe it's the symmetric algebra rather than the exterior algebra, being careful about signs...

[plain raven]

one simple counterexample would be like, imagine a map from R^1 into R^2 that's in the shape of a fishhook. It starts out going from left to right, say it sends (-\infty, 0] to the negative x-axis

then as it goes from 0 to positive infinty, the curve loops around back on itself, making a 270 degree counterclockwise turn, and starts converging toward the origin (0,0), getting slower and slower as it gets closer and closer so it never quite reaches it

then the map R -> R^2 is injective but it doesn't carry the subspace topology

it's an immersion but not an embedding

i think more important examples of immersions that aren't embeddings come up in the theory of foliations

essentially like

idk say you have a vector field X on R^2

and for each point you have a solution

which takes the form of a curve f: R -> R^2 such that the standard unit tangent vector of R at p maps to the vector X(f(p)) for each point p\in R

well under good conditions you have a solution passing through every point

so you have a family of parallel curves side by side, which partitions the manifold up into one dimensional curves

in the study of this sort of thing you often run into one dimensional submanifolds which are not embeddings because the different slices get too close to each other. so like, if you took the countable coproduct of copies of R, and fix an enumeration {q_n} of the natural numbers, and mapped \coprod_{n\in N} R into R^2 by a map that sends (n, x) to (q_n,x) in Q\times R \subset R^2, this would not be a subspace embedding because the slices would be too close together.

to see how this comes up in the study of vector fields, suppose you have some vector field on the torus and you have found a solution, at some point p, this solution is defined for all x in R and the curve just loops around the torus over and over again in a spiral shape, so that in the end the graph of the solution is dense in the torus

the solution would be an immersed submanifold which is not dense

[finite heath]

hey @orchid forge

can i share my proof with u

[fair idol]

Ahhh okay so the figure eight and some dense curve over a torus are the immersive but not embedded submanifold examples in John lee. I guess I didn't know what subspace topology was but now I have a better understanding

It's like with that curve in R^2 you described where the open balls in R^2 don't work but the open intervals in R are okay. I think it's something like that.

[jagged pivot]

For something like $$conv{ 0, e_1, e_2,...,e_d },$$ how would I prove that any two vertices share an edge?

[gentle osprey]

beeswax

[jagged pivot]

I can see it in 2d and 3d, but I'm not sure how I would prove it for all dimensions

[uncut surge]

What's your definition of "sharing an edge"? If you simply mean that the straight line between them is contained in the set, that's essentially due to the definition of the convex hull

[jagged pivot]

I mean like, let's say in R^2, we have conv{0,(1,0),(0,1)}. All of the points essentially connect to eachother

Or the standard simplex in R^3, which has 4 points. If you take any one of the points, it would connect to the other 3 points

So in R^d, we would have d+1 points and if you take any of the d points, it should connect to the other d points

[uncut surge]

Yeah, and it always does if you write the simplex as a convex hull of these points, because by definition of the convex hull, it is a convex set; so the whole line $t * e_i + (1 - t) * e_j$ for $0 \leq t \leq 1$ is contained in the set for all indices $i,j$, which is exactly the edge they share

[gentle osprey]

Lartomato

[uncut surge]

Continuous functions from the interval [0,1] to the real or complex numbers

[gritty widget]

are you saying something like

abstractly

the differential should be able to just be defined in terms of the differential from the cochain C?

[gritty widget]

Why can any compact group G be embedded in O(n) for some large n?

[uncut surge]

Compact Lie group? Because then you can argue with the fact that G acts by isometries on its Lie algebra, where the Lie algebra is equipped with the negative of the Killing form, making it into a Euclidean space

oh wait maybe this only works for compact simple things

[solar shore]

In topology (or maybe in math in general? not sure) the word imbedding - why does it have i as the first letter instead of e? Is it just an archaic spelling of the word embedding? Or did mathematicians modify the word on purpose to mean something different than usual?

[gritty widget]

oh but then we should be able to do something like

the representation is a sum of irredcuicle representations

and what you say should work on each irreducible rep?

[bleak helm]

I usually see embedding. Might just be a different spelling used by some.

[wise sigil]

have you poked around with sequences?

you gotta find a cauchy sequence of elements in V that doesnt converge in V

imagine what a cauchy sequence X_n would look like

the norm forces a "total maximum error" on the terms of the sequences X_m and X_m' for m, m' large

its hard to put into words

does that make sense?

So you would need the sequences X_m to change little by little when m is large

for example

X_1 = (1, 0,0 ...)
X_2 = (1, 1/2,0, 0 ...)
X_3= (1, 1/2, 1/4, 0, 0, ...) etc

as u go along, the changes get should get smaller and smaller

[bleak helm]

What would be the sequence 1/2^(n-1)?

There isn't one sequence here. You have a sequence of sequences.

And none of the sequences is equal to 1/2^(n-1) for all n

x_i is the sequence 1/2^(n-1) for n <= i and 0 for n > i

Brian gave you a sequence (sorry for interrupting, im bored lol), so prove it is cauchy but not convergent

[empty grove]

Boredom is not an excuse to interrupt Luna

Sorry for interrupting im bored lol

[wise sigil]

Excuse me moldilocks

jk its cool luna idrk how to explain well

[bleak helm]

Same tho

[bleak helm]

Mold star

[marble socket]

can someone help me with this? It's a homework problem, but the course has been too fast for me to digest everything as soon I see it. I proved that closed points in t(V) are precisely singletons {p}. Needed to use nullstellensatz for the <= direction in (a), (b) i think follows from the fact that when you look at the corresponding local maps of stalks, and then the residue field for f(P) will map injectively to that of P which is k, forcing the residue field to be k itself. For injectivity of (c) i think we can recover f from t(f) as {f(p)} = t(f)({p}). For surjectivity, enough to restrict our attention to affine varieties. if V and W have coordinate rings A and B, then we know that t(V) and spec A are isomorphic are schemes, so any map from t(V) --> t(W) will give a unique map spec A --> spec B, and looking the global section will give a ring map B --> A, this will give us a map from V --> W which sends a point p to the point q such that the m_q is inverse image of m_p under B --> A. but this is precisely the map we get from restricting the original map t(V) --> t(W) to the closed points, as {p} corresponds to the maximal ideal m_p in spec A. So it suffices to show that scheme maps between spec A and spec B correspond precisely with ring map from B --> A, am I right?
would need help in filling in details, i'm not sure about what i'm saying completely.

[plain raven]

yeah. The category of chain complexes itself has certain built in constructions that eat chain complexes or pairs of chain complexes and returns chain complexes. the tensor product is one of these, so is the direct sum / product...

[plain raven]

oh boy do i have a rant for you

[marble socket]

[plain raven]

someone just asked this question like yesterday or two days ago in this channel lol if you scroll up.
but uhh hold on i have to read your question. it seems like you have the basic ideas you need to solve it already i'm still trying to see what you need exactly

[marble socket]

yea it was Moldi, we're taking the same course lol

[empty grove]

We are spending all our time bitching about the bad pace in our DMs

[plain raven]

hmm it does sound like the right idea. certainly scheme maps Spec A -> Spec B are in bijection with ring maps B -> A, this is like, the whole point of the Spec construction so it would be quite alarming indeed if this were not true. definitely something you want proved and in your notes anyway regardless whether you need it for this construction. and if they're fibered over k then the maps between them are exactly the k-algebra maps

if you're worried about details maybe just try proving some theorems about how t behaves with respect to like, open covers

like
just thinking in terms of Top here.
let t be the functor which eats a space X and constructs the following space t(X).

1. The underlying set of t(X) is exactly the set of irreducible closed subsets of X. Write x_F for the point in t(X) corresponding to the irreducible closed set F.
2. The topology T(t(X)) is in bijection with the topology T(X), that is, for each open set U \subset X there is a corresponding open subset U' \subset t(X), and this is a bijective correspondence.
3. We define U' to be the set of all points x_F in t(X) such that F\cap U is nonempty in X.

This is what the schemification functor does on the level of spaces and then the sheaf of regular functions on t(V) is just the direct image of the sheaf of regular functions on V

so like

I only mention this because if you're worried about stuff like 'just prove it on affines'

i would suggest that you try proving that if X is the union of a family of open sets {U_i}

then t(X) is the union of t(U_i)

Maybe we want to say something stronger, that t(X) is the direct limit of the t(U_i) by a similar diagram? t preserves these certain kinds of colimits?

Sorry. i'm just being chatty. The technical details are still a bitch regardless of my handwaving but it does sound like you have the right idea. I don't see anything you said that's wildly wrong. if you have some criticism of a certain point in your reasoning that you're worried about, point it out

[marble socket]

right, i was mostly worried about the gluing part with affine stuff. and don't know much about regular functions. from what i remember a map between two affine varieties V and W thinking extrinsically was like a restriction of polynomial map by choosing some representative in the map of K-alg of their corresponding coordinate rings.

[plain raven]

Are you following hartshorne?

This is a hartshorne exercise

so i assumed you had some experience with regular functions

gtg bbl

[marble socket]

our prof might be following it... but this was an homework problem so they also referred to Hartshorne

[empty grove]

So they started with sheaves, went to schemes, then did one class on varieties where they defined all those affine, projective and the quasi versions and defined this functor in the problem

I think regular functions were defined

But really no level of familiarity beyond definitions

[marble socket]

[slender plank]

u€L(E) u continuous <=>ker(u) closed 🔐

Can't figure out how to do the reciprocal

[tough imp]

Can you solve this via an equalizer diagram?

I’m pretty sure that Hom(X,Y) is the equalizer of a diagram of the following form

Hmmm

Nvm

Well…

Hmmmmmmmmmmmmmmm

There’s a dependence on the map f

Which I don’t like

I mean maybe this still works lol

But if you map into the product of all Hom(V,U_i) over all V open in X

Then this into like Hom(V\cap V’,U_i)

My idea is that a map f:X -> Y is equivalent to maps

f_i:f^-1(U_i) -> U_i which agree on the intersection

And then from this I want to reduce to the affine case, or at the very least the target is affine, as then we know the result I think (the bijection from variety maps to scheme maps)

And then it should just be a formal consequence that the equalizers are in bijection

[dim radish]

What is the intuition/idea when letting fibers of the covering projection act on the fundamental group

[hearty copper]

as someone who is very interested in category theory, and has come to understand that topology is almost a prerequisite, does anyone have any good introductory topology books?

I have been reading "Math 131: Introduction to Topology" by Denis Auroux, is that one good?

[empty grove]

Standard recommendation

Don't think it's that important for cat theory tho

[hearty copper]

hmmm maybe, i know it isnt strictly required but every time Riehl says "ok, for a grounded example of the usefulness of categories, take a look at this functor from topological spaces to groups" or stuff about homotopy :PPP

[empty grove]

Riehl needs to learn more math

[hearty copper]

thank you!

[empty grove]

Yeah it's an important source of examples

But

[hearty copper]

what does

mean

[empty grove]

The fact that you don't just understand it is very

[hearty copper]

:'( ime beginner hobbyist 😭

[obtuse meteor]

I'd recommend https://topology.mitpress.mit.edu/ actually

[hearty copper]

oh awesome, it seems to teach the two fields together from the get-go?

[obtuse meteor]

Yes to a degree

[gritty widget]

what's wrong with munkres then learning category theory from someone like Borcherds?

or hatcher even

[wary furnace]

Riehl moment

[swift fjord]

Hatcher continuously dodges taking a categorical approach

[obtuse meteor]

Topology good

[obtuse meteor]

Yeah I think Hatcher isn’t really appropriate for an intro to modern alg top

It’s good as an intro to alg top

For someone who’s not going to like work in alg too

[swift fjord]

I'd disagree on that but it really depends on the reader

but it definitely isn't good if you wanna see things in full generality from the start

[obtuse meteor]

I really think suspension/loop space Adjoint should be like, one of the first things talked about bc I’m weird like that

[orchid forge]

I need to learn stable homotopy theory

At least some basics

Is there a good book

I'm reading about some spectrum level knot homologies and so far all I know about spectra is stuff that has been scribbled on a blackboard or hurriedly read off nlab

[plain raven]

ok you realize what this sounds like right

[plain raven]

i'm clueless about spectra also but i was going through switzer's book on algebraic topology recently and it seems like he does all the standard AT stuff from the POV of spectra

[gritty widget]

what

[plain raven]

was thinking it might be cool to check it out if i can. (i was consulting switzer because it has a chapter all about spectral sequences and its applications)

[orchid forge]

nice nice

are you still looking for a sheaf buddy for that one book

[plain raven]

Manzareh the word "Category" is first mentioned in Hatcher in section 2.3. It is not a categorical book at all. If there were any less category theory in Hatcher, it would signify some weird opposition to category theory.

[gritty widget]

i came in thinking we were talking about topology

[orchid forge]

clerk this book is probably exactly what i'm looking for

[plain raven]

Ok, I see. Yeah I think foobles is trying to learn category theory . I was just thinking that a book that goes through the fundamental group and like, the entire construction of singular homology and its basic properties without mentioning the words "category" "functor" and "natural transformation" is not a good place to learn about categories in topology

[swift fjord]

Yea you wanna read dieck for that

Or that categorypilled book fiona is reading

Topology: a categorical approach

[plain raven]

Yeah Kogasa i'm still interested i need to get my priorities straight and decide whether I should pull the trigger on committing to it... I definitely want to, I was spending a lot of time trying to badger people earlier about it.
anyway i'm home visiting my family right now so i have some time to relax unwind and figure out my schedule.

The only other problem is that i'm a complete idiot and am bad at scheduling and i never do what i say i'm going to do which is why i'm looking for a partner lol

someone who can bug me and be like "yo weren't we gonna meet and read this?"

i started it, i'm roughly through the first 50 pages

[orchid forge]

maybe it would be best if i started reading too and just bugged you about it

"bugged" as in talked, not as in "hey you need to read remember"

i too am very good at saying i'm going to do things then simply not

[plain raven]

ok i worked it out to make sure i was correct. this is a bit abstract so i hope you'll forgive the generality.
Let A be an Abelian category. Say A has infinite direct sums. Suppose that A has a "tensor product", here I mean a symmetric monoidal product, and let's say it distributes nicely over direct sums. In this category using the notions of tensor product and colimit you can carry out the construction of the symmetric algebra for any object X: letting 1 denote the unit of the monoidal product, we can form the sequence 1, X, X\otimes X, X\otimes X\otimes X, and so on. Now, there is an obvious action of the symmetric group S_n on X^{\otimes n}. Write Y_n for the quotient of X^{\otimes n} by this action. (The coequalizer of all the automorphisms in the group action.) Then the direct sum 1 \oplus Y_1 \oplus Y_2 \oplus ... is called the "symmetric algebra on X". I haven't checked this but presumably this should be a commutative monoid in A and it should have the universal property that every map from X into a commutative monoid M extends uniquely to a monoid homomorphism from the symmetric algebra to M.

now the category of cochain complexes is indeed like, rich enough so that it has all these properties. there is a standard tensor product of cochain complexes which you can look up, there is a standard symmetry isomorphism, it has infinite direct sums and all this shit

The Koszul algebra is just the symmetric algebra generated by the cochain complex 0 -> V -> V -> 0 concentrated in dimensions 1 and 2.

and yes its differential is automatically given, at all points during this construction we work purely within the category of chain complexes

ok kogasa sure let's do it. idk. you want to agree to talk about this, say, next week on this date

not that we have to schedule a meeting but like

check in october 20th

i'll start rereading / reviewing the stuff i've read

[orchid forge]

sure, i'll see what i can do with the first 50 pages

are there exercises?

[plain raven]

i don't think so. i don't do exercises usually i just work out the proofs in detail to my satisfaction. the books i read usually have pretty terse proofs because idk i guess i hate myself, so there's enough exercises there for me lol

"It is left to the reader" has about 15-20 hits on ctrl-f lmao

[orchid forge]

that's fine with me

my now-cancelled algebra class frees up a good reading hour every other day and it seems reasonable to fill that with cool topology flavored algebra

[plain raven]

what was the class on

[orchid forge]

commutative algebra

apparently it might not be cancelled if we can find a time to move it to? i dunno it's a shitshow

[gritty widget]

hello, I was working on this question since yesterday, and I determined the nature of the opens in R/N but I'm stuck in the unmetrizability

I was wondering, can't I say that all metric on R/N, if it exists will induce a topology were d(0,1) is closed in R/N, but the reverse image of d(0,1) by the projection from R is an open

sorry for the size of the screen

[plain raven]

ooh weird

cool

What is d(0,1)

i am wondering if you have tried figuring anything about it. Like is this space compact? is it hausdorff?

[gritty widget]

distance between 0 and 1, in R/N it's 0 so the (0,d(0,1)) ball is closed in R/N

that's not the case in R

sorry for my english or my french notations

[gritty widget]

can't figure it out if it's hausdorff, looks like yes

[plain raven]

Yeah. Ok. I don't think your argument works. The quotient map R -> R/N doesn't have to preserve the distance between points

[gritty widget]

but it's continue so the reverse image of a closed should be closed isnt it ?

[plain raven]

Yes.

distance between 0 and 1, in R/N it's 0 so the (0,d(0,1)) ball is closed in R/N
This is what I have a problem with. Write f : R ->R/N for the quotient map. There is not necessarily any interesting relationship between d(x,y) and d(f(x),f(y)) other than the relationship imposed by continuity.

Sorry I am not explaining this well.

[gritty widget]

no it's ok I think I get it

[plain raven]

The preimage of the (f0,d(f0,f1)) ball should not be the (0,d(0,1)) ball

[gritty widget]

sad

[plain raven]

I think that's what i'm saying

[gritty widget]

so that doesn't work

yes

so my teacher said using the first countability axiom

I suppose to prove the non convergence of a countable family of open balls

[plain raven]

nonconvergence?

[gritty widget]

nvm I just have to disprove the first countability axiom in R/N

or maybe I can show R/N is not Hausdorff

Is a solution to say that R/N is not Hausdorff since we can't separate the integers ?

[orchid forge]

The integers are separated, except the nonnegative ones which are all the same point (in the quotient)

[gritty widget]

yeah by integers I meant the N elements

so if their are in the same point that means there are no opens that contains one without the second, in other words R/N is not Hausdorff

[plain raven]

No, they are not distinct

they have to be distinct points

[gritty widget]

ok so in a quotient topology all elements with the same class are not distinguishable ok

[plain raven]

The elements of the quotient space are the equivalence classes themselves. We just refer to them by choosing representatives

[gritty widget]

so like in algebra with quotient ring

[plain raven]

yes!

[gritty widget]

geometry and topology

so cool

[coarse kestrel]

So I’m reading Lee’s introduction to topological manifolds, and there are just many random conditions that imply each other, like second countable locally compact hausdorff implies paracompact, but first countable and compactly generated hausdorff implies every proper map to this space is closed. So notice how like randomly one theorem needs first countable and second one needs second countable

Is there a way to like systematically remember them, like an overall picture or something

Or I just have to like remember these case by case

Or are these just technical details that don’t matter at all in real practice

Btw ch4 is fun but pain to remember everything

[jagged pivot]

For the n-dimensional cube, is there a general form for how many facets and vertices we have?

Would 2n faces and 2^n vertices generalize for the nth dimension?

[coarse kestrel]

I feel like the number if $m$ dimensional faces will be $2^{n-m}\binom{n}{m}$

[gentle osprey]

Whoever

[coarse kestrel]

Because the m dimensional faces are where you choose n-m of the coordinates to be 1 or -1, and let all other m coordinates vary freely

@jagged pivot so in this case the face and the vertices you’re talking about are the m=n-1 and m=0 faces

Which is consistent with the formula

[jagged pivot]

oh

[gritty widget]

there are many nice things for drawing commutative diagrams

is there any nice thing for making spectral sequeucnes

[drifting sundial]

I went thru the chapter recently and decided to skim through the proper map stuff for now

I figured I'll probably forget most of the stuff anyway when I reach the applications so might as well revisit it then

[coarse kestrel]

That’s fair

It seemed like a pretty useful thing though tbh

[gritty widget]

Trivial homology groups need not imply that the cohomology groups are trivial?

does anyone have an example to this

[pearl holly]

wait is that true?

like if ALL homology groups of a space are trivial then the cohomology groups are also trivial but I guess that you're asking something different

[gritty widget]

by trivial homology i mean

homology of a point

so the zeroth homology need not be the trivial group

or sorry

its true that H_k(X)=0 does not imply H^k(X)=0

[uncut surge]

I guess if you have coefficients in a field this doesn't happen, right? Due to the universal coefficient theorem

[pearl holly]

Ye I think so too

[uncut surge]

In any case I imagine that theorem is what you should be thinking about if you want counterexamples or so

They definitely exist in other coefficients since that theorem depends on H_{k-1} which you can relatively freely choose

[vocal anchor]

If $q\colon B \to C$ is a quotient map, why doesn't $id \times q\colon A \times B \to A \times C$ have to be a quotient map? I thought it was like this: The open sets of the product are the products of the open sets, so if $(id \times q)^{-1} (U \times V)$ is open, $q^{-1}(V)$ would be open, and thus $V$ and $U \times V$ would be open.
But that's apparently wrong?

[gentle osprey]

expectTheUnexpected

[bronze lake]

I'm pretty sure you're right, what makes you believe it's wrong?

[vocal anchor]

https://encyclopediaofmath.org/wiki/Quotient_mapping

At the end of the third paragraph:
The Cartesian product of a quotient mapping and the identity mapping need not be a quotient mapping

No reference is provided.

[bronze lake]

[vocal anchor]

But my "proof" would also show that the product of quotient maps is a quotient map I think, and there's a counterexample somewhere on math.SE

[bronze lake]

oooh wait

yeah munkres chapter 22 has an example

it looks like the requirement is that A is locally compact

https://math.stackexchange.com/questions/31697/when-is-the-product-of-two-quotient-maps-a-quotient-map

[vocal anchor]

Thanks, now I have to understand this, seems non trivial. however, another question: where is the mistake in my reasoning above? It's got to be "(id x q)^-1(U x V) open implies q^-1(V) open", right?

[bronze lake]

Yeah, that's specifically what you have to show

oh wait

you want to prove that (id x q)^-1(U x V) is open implies U x V is open for the product to be a quotient map

ooh I think I see what the problem is

[bronze lake]

the trouble comes from the fact that an open set in $A\times C$ isn't necessarily of the form $U\times V$ for $U\subset A$, $V\subset C$. It can be written as a (not necessarily finite) union of sets of that form, though

[gentle osprey]

cgodfrey

[bronze lake]

the product of open sets in each space form a basis for the product topology, but not every open set can be written that way

like, if you have the open set $(0,1)\times (0,1) \cup (0.5, 1.5)\times (0.5, 1.5)\subset\mathbb{R}^2$, it's definitely open as the union of open sets, but it can't be written as $U\times V$, $U, V\subset\mathbb{R}$

[gentle osprey]

cgodfrey

[vocal anchor]

I see, and locally compact means that every open set looks like a finite union of sets of the desired form

[bronze lake]

I imagine so

I didn't look into the correct proof close enough to see where it's used

[pearl holly]

why is the exterior algebra a graded ring?

why would one even define a exterior algebra? What is its "purpose"?

why am I even posting this here lmao

[marble socket]

cuz you can take something in the k-th wedge and something in the l-th, and so something to get something in the (k+l)-th wedge

cuz it's cute and fun. idk, only use i've seen is for differential forms... you get to precisely define what dx and dy mean and how does dxdy relate to it

[pearl holly]

oh that's sounds kind of cool tbh

[pearl holly]

what do you mean by wedge? Just "multiplication"?

[marble socket]

yea the wedge product...

the usual dxdy is independent of orientation...

we get do define dxΛdy which also captures orientation

dx Λ dy = - dy Λ dx

[pearl holly]

ye I don't know what that is lmao. THis is the definition I have: $\bigwedge_R[\alpha_1, \cdots, \alpha_n]$ over a commutative ring $R$ with identity is the free $R$ module with basis the finite products $\alpha_{i_1} \cdots \alpha_{i_k}, i_1 < \cdots i_k$ with associative, distributive multiplication defined by the rules $\alpha_i \alpha_j = - \alpha_j \alpha_i$ for $i \neq j$ and $\alpha^2_i = 0$

[bronze lake]

you also get some properties that fall out really nicely with wedge product

[pearl holly]

and I don't see how I can "decompose" the exterior algebra so that it becomes a direct sum of other stuff

[gentle osprey]

Tokidoki ✓

[pearl holly]

I don't know the command for the fancy ^ lmao

[marble socket]

$\wedge R^{*}(M) := \bigoplus{n \ge 0} \wedge^n_R(M)$

[gentle osprey]

det

[marble socket]

this is the definition i've seen

M is any R-module

[pearl holly]

oh lmao

wait what does \wedge^n mean?

[marble socket]

so have you seen tensor products?

[pearl holly]

oof no I haven't

[marble socket]

oh okie... so maybe think of it like this... the n-th wedge of M consists of objects that are linear combination of (m1 Λ m2 ... Λ mn) think of Λ as a billinear product which is alternating... so a Λ a = 0 which implies a Λ b = - b Λ a

[pearl holly]

oh okay I see I see

[marble socket]

so there is a natural way to take something in Λ^n and something in Λ^m to get something in Λ^(m+n)

these m and n are the grading

[gentle osprey]

det

[pearl holly]

okay so let's say that this alpha that you wrote above is $x_1 \wedge \cdots \wedge x_n$ and say that the beta is $x'_1, \wedge \cdots, \wedge x'_m$. Then $\alpha \wedge \beta = x_1 \wedge \cdots \wedge x_n \wedge x'_1, \wedge \cdots, \wedge x'_m$ and this is in $\wedge^{n+m}$?

[gentle osprey]

Tokidoki ✓

[marble socket]

yep

that's how we define it on "pure wedges" i.e. if alpha and beta have single terms... now just extend via linearity

[pearl holly]

yea okay I see now. This kind of reminds me of the cross product

like the relation that a ^ b = -b ^ a and that a ^ a = 0

[marble socket]

they sure are related 😛

[pearl holly]

oof okay okay

Okay thank you so much for the help! I get it now!

[gritty widget]

[marble socket]

[empty grove]

tokology and dokeometry

so cool

[pearl holly]

[jagged pivot]

I have been looking at this exercise for hours, and I haven't made any significant dent. Is someone able to help with this?

I did some examples in R2 and R3, but I'm still not sure how to prove this

[novel acorn]

Toki doing diff geo now?

I've been thinking of jumping into Lee
Although I've kinda gotten hooked to physics rn

[gritty widget]

I have a space $X_n$, I know that $H_k(X_n)=0$ for $0<k\leq n$ (in particular $H_0(X_n)\neq 0$. Can I conclude that $H^k(X_n)=0$ for $0<k\leq n$

[gentle osprey]

lime_soup

[gritty widget]

I can see that use the universal coefficent theorem it should work out

but it seems there should be a "slicker" answer