Couldn't find a member matching gm to est!
#point-set-topology
gentle ospreyBOT
pearl holly
okay wait lemme google
gentle ospreyBOT
No results found at:
https://ncatlab.org/nlab/search?query=3pm+gmt
abstract pagoda
today I hesrd about this
pearl holly
lmao why is this so hard 
gentle ospreyBOT
pearl holly
ye 10 am
pearl holly
hmm well maybe we can just talk and stuff after your class?
idk
we will see I guess
swift fjord
Toki muthafuckin doki
gritty widget
Groups act on spaces
Do groupoids act on spaces?
I know the defintion of a groupoid in terms of
A groupoid is a category where all the morphisms are isos
And I can see this generalized a group because a group is
One object
And only isos
Partial function being
I cannot multiple everything
Okay for example
S^2 has an action of S^1
And also has an action of Z/2Z
Is there a way two thing about
Ah yes
We can make the cat with two objects
By it isn’t connected
You mean the actions aren’t
So we can just take the two groups and create their product category
Yeah
Yeah yeah okay
I though this but then doubted myself
So then
The groupoid will be connected when there is an interplay between the actions
Object in the groupoid X
Yeah okay
Sorry
Yeah
The object x in some groupoid
Not the whole groupoid
I’m trying to think of some example of an interplay of actions
I was going to ask
Is there then for any space T
Some universal groupoid
That has all the actions one could put on T
orchid forge
You can always just take a bunch of groups and call it a groupoid
orchid forge
Also, you know a group action is just a homomorphism G --> End(X)
As in, a nice (homomorphism) way to define, for each g in G, the endomorphism "multiplication by g : X --> X"
So you're looking for two groups G, H acting on X that don't "commute," so like g(h(x)) != h(g(x))
For example, let X be an n-gon, and fix an axis of reflection
The reflection generates a group of size 2
gritty widget
Ah yes thank you!
orchid forge
And there's also a rotation by 360/n degrees that generates Z/nZ
orchid forge
We have an action of Z/2Z and an action of Z/nZ, and yes exactly
gritty widget
Ah thank you you very much
orchid forge
This defines an action of the semidirect product of Z/2Z and Z/nZ over that nontrivial morphism
This semidirect product is just the dihedral group D_(2n)
gritty widget
Is there any other simple example that isn’t
Just a group
As in with this example we could squish these two points together
(They are isomorphic I think)
orchid forge
Squish how?
gritty widget
We could have the same information in
A one object category
Oh
By definition
All the objects have to be isomorphism
Wait so
Every connected groupoid
Is just a group
orchid forge
If by connected you mean like, the underlying diagram of the category is connected as a graph, yes
Eh, actually
It can be a collection of isomorphic groups
The skeleton is a group
For example the fundamental groupoid of a path connected space is made up of all the groups $\pi_1(X, x)$ for all $x \in X$
gritty widget
Ah yeah so
gentle ospreyBOT
Kogasa
gritty widget
If we had multiple copies of the same group
gritty widget
We might get some problem with having multiply identity elements?
orchid forge
Possibly, you can have issues anywhere isomorphism isn't the same as equality
empty grove
If by connected you mean like, the underlying diagram of the category is connect...
That is how it's defined yeah
gritty widget
So the main thing is really that
empty grove
Is just a group
It's equivalent to a group
gritty widget
Groupoids are a nicer way of organizing things ?
empty grove
But there are reasons that you'd not want to treat it as such
orchid forge
Nicer way of organizing groupoids
gritty widget
(In the contexts that they are used)
orchid forge
The fundamental groupoid is not always so simple
empty grove
Yeah the fundamental groupoid of the circle is just Z at each object connect by all these isomorphisms
If you just treat it as a group you lose information
gritty widget
And now tell me this
empty grove
Can't use van kampen for example
orchid forge
And you can sometimes express theorems about "the" fundamental group in terms of the groupoid and now they work on non path connected spaces
orchid forge
Van Kampen is an example, you can tweak it to work with the fundamental groupoid
gritty widget
Take the “homotopy”? Category
empty grove
Van kampen for groupoids is more powerful even with connected spaces
orchid forge
gritty widget
And make all isomorphic objects equal
Sorry very rude of me I have to run off
I’ll read back over this tomorrow
Thanks !
orchid forge
You mean the skeleton category, and that's essentially what you do when you refer to "the" fundamental group of a (path connected) space
Bye 
orchid forge
Van kampen for groupoids is more powerful even with connected spaces
I mean one of the conditions for the classical version is the intersection is path connected, using the groupoid relaxes this
But of course this makes it more powerful for (covers of) even path connected spaces
You can probably strengthen it more tbh, any time you're secretly looking at a diagram in Set there's hella theorems
SvK/Meyer Vietoris, while obviously related directly through the abelianization functor, can also both be derived from finding a suitable functor from Set, and applying it to the diagram defining "union" (or intersection)
For SvK it's a bit tricky, you need to figure out how to translate the set theoretic conditions into a quotient of Top
But you can phrase lots of "union" and "intersection" flavored stuff like this
Brb I'm gonna go look for who asked
limpid leaf
theres some post on mathflow but i dont want to try it cuz it looks cancer
quasi forum
So unless my intuition is wrong, shouldn't this end up being the entire space?
If not, how do I need to change my thought process.
orchid forge
can you explain your thought process?
quasi forum
Well, I guess for starters, for any x in R^w, x,0,0,0,... is a sequence that is in R^inf
But since that's true for any x, wouldn't R^w=R^inf?
Clearly my thinking is flawed somewhere, I am just not sure how exactly
orchid forge
Well, I guess for starters, for any x in R^w, x,0,0,0,... is a sequence that is...
x is an infinite sequence right?
quasi forum
Yea, that is eventually 0
orchid forge
in R^\omega we don't necessarily need that to be the case
it's just an infinite sequence of real numbers
quasi forum
How is what you said different from what I said? 🤔
orchid forge
you said x is eventually 0
i said it is not necessarily eventually 0
because it is an arbitrary element of R^\omega, not R^\infty
quasi forum
But isn't that how R^inf is defined?
orchid forge
x is an arbitrary element of R^\omega, not R^\infty
quasi forum
"Consists of all sequences that are eventually 0"
orchid forge
$\mathbb{R}^\omega$, not $\mathbb{R}^\infty$
gentle ospreyBOT
Kogasa
quasi forum
Yes, but the sequence x,0,0,0... is certainly in R^inf
orchid forge
x is an infinite sequence
quasi forum
For any x in R^omega
quasi forum
Oh, there must be some confusion then. x is just some element in R omega and x,0,0,0... is the sequence I am constructing
It satisfies the definition of R^inf
quasi forum
What about this sequence is problematic? I fail to understand
orchid forge
x is an infinite sequence of real numbers
you've written "(x_0, x_1, x_2, ...), 0, 0, 0"
quasi forum
I agree with this
orchid forge
backwards
quasi forum
Ah yes
orchid forge
R^\infty is not "sequences of elements of R^\omega"
it is a subset of R^\omega, consisting of sequences of real numbers
so like (1, 2, 3, 0, 0, 0, ...) would be in R^\infty
quasi forum
It just says consists of all sequences, not consists of all sequences of real numbers.
orchid forge
it says it's a subset of R^\omega
quasi forum
But the R^inf you are talking about is far more interesting
orchid forge
I am talking about exactly what they've defined in your book
quasi forum
Okay sure
Hmmm, so how might one go about this completely different subset than that one I was thinking of.
I did not realize sequences were elements of R^omega, so that Is where my mistake was
So if we were to look one dimension at a time, so we restrict the N where it is eventually 0, then this consists of an N-1 dimensional plane
Or an N-1 dimensional slice of R^omega if that is preferred
orchid forge
that N would depend on the particular sequence
you want to find the sequences which are always "epsilon-close" to something in R^\infty
so that they are contained in every open neighborhood of R^\infty
quasi forum
So I guess the question of interest is what happens if we examine an element x which is not in R^inf
orchid forge
the question is when such an x is "always epsilon-close to something in R^\infty"
quasi forum
Yes, for sure
Ah, but in order for this to happen for some (x_1,....,), then d(x_n,0)<epsilon for all n
Well, at least for some tail of the sequence
But if that holds, then x must be in R^inf, since this needs to happen for every epsilon ball.
So I believe the answer is that the closure of R^inf is itself
orchid forge
not exactly, but along the right lines
quasi forum
Where is my mistake
orchid forge
But if that holds, then x must be in R^inf, since this needs to happen for every epsilon ball.
quasi forum
Ah, but in order for this to happen for some (x_1,....,), then d(x_n,0)<epsilon ...
Do we agree w/ this?
Again, for some tail
quasi forum
So I guess I'm struggling a small bit.
I am understanding the conclusion needs to be the set of all sequences that converge to 0.
But I am not sure how to show a rho ball around (x_n)->0 meets R^inf
orchid forge
you need to use the definition of a sequence converging to 0
quasi forum
I guess my issue is we are taking the Supremum of the distance of each coordinate
Well, as long as it's smaller than 1
Oh, I see. For each N, (x_1,...,x_n,...)'s the rho ball of radius epsilon will meet (x_1,...,x_(N-1),0,0...)
quasi forum
Ty for the help. If I realized that elements in R^omega were sequences from the beginning, this probably would have been a lot smoother
fading vale
Oh i did this exercise
I dont think theres a nice easy solution
u just have to cooompute
limpid leaf
how is this a good problem
orchid forge
because you know you're about to use it for some good shit
orchid forge
have you defined "chain homotopy" yet
orchid forge
it's not complicated, but you will use this problem for singular homology if/when you talk about it
limpid leaf
we are doing alg top for the entire month of november but i dont think chain homotopy is on the list
orchid forge
if homology is on there, then you will
limpid leaf
Fr Nov 5 Homotopy equivalence and algebraic topology Munkres 9.51, 9.58.
We Nov 10 Path components Munkres 2.25.
Fr Nov 12 The fundamental group Munkres 9.52.
We Nov 17 The fundamental group of the circle Munkres 9.53, 9.54.
Fr Nov 19 Brouwer fixed point theorem Munkres 9.55, 9.57.
We Nov 24 Fundamental theorem of algebra Munkres 9.56.
Fr Nov 26 Seifert-van Kampen and some knot theory Munkres 9.70.
is not
orchid forge
F
bronze lake
yooo seifert-van kampen and some knot theory is hype tho
limpid leaf
tsomamerman js gna win a fields medal u know
abstract pagoda
jesse is a butcher?
abstract pagoda
have you defined "chain homotopy" yet
what is this
orchid forge
given two chain complexes, if you can find a map h between them that anti-commutes with the differential, it induces the 0 map on homology
orchid forge
the differential of the chain complex
gritty widget
abstract pagoda
chain complex is a tuple containing a sequence of maps and objects C_n
orchid forge
either way, it's just a map of chain complexes that is trivial with respect to (co)homology
abstract pagoda
oh boundary
orchid forge
the maps of a chain complex are what i meant by differential
but differential is probably for cochain complex
abstract pagoda
oh you were talking about cohomology?
orchid forge
i'm talking about both, or either
abstract pagoda
im gonna need big help understanding parts of homology
So what are C_n called?
each C_n
orchid forge
1-(co)chains
abstract pagoda
a cochain group 1?
orchid forge
or the group of 1-(co)chains
abstract pagoda
ok why are they called chains? and what does each contain?
i barely remember
but for singular
orchid forge
would probably help to fix a (co)homology theory
abstract pagoda
they contain singular simplcies
orchid forge
for simplicial homology, an n-chain is just a formal sum of n-simplices
with coefficients in Z or Z/2Z or whatever
orchid forge
the boundary is an alternating sum
abstract pagoda
fuckkk
oh wait
i remember this
the n chain is a formal sum of eini
ei ni
where ei is in Z and ni is a sigma?
orchid forge
ye
abstract pagoda
but uh
i need to break this down further
why is ei Z and ni sigma and why do we define those to be elements of a n-chain
what does it represent is a better question ig
orchid forge
i would recommend reading about simplicial homology
abstract pagoda
ok yes ill have to reeeed
i skimmed and attended lectures
but reading probably helps solidify
i do have questions about a lot of things
like for defining a standard n simplex why do wr need orientation on the verticies
orchid forge
if you don't pick an orientation the definition of the boundary doesn't make sense
the fact that the simplices are oriented, and these orientation descends to a coherent orientation of the faces, is what makes the boundary operation (and homology groups) meaningful
so you can do stuff like, given a line a --> b --> c, the boundary is +c - a, and it's made up of components a --> b and b --> c, with boundaries +b - a and +c - b, and their sum is +c - a
abstract pagoda
i know how to compute boundary too
yeah
wait
the ith boundary is sum of sigmas restricted to i’th point removed from a simplex
alternating*
I need to read you are right
but reading feels like a rush compared to before
I feel really pressured to get it done super quickly
honest terrace
@dull goblet
honest terrace
So we say for a topological space $X$ that it is Hausdorff if for any $x, y \in X$, we can find open sets $U, V$ such that $x \in U, y \in V$ and $U \cap V = \emptyset$
gentle ospreyBOT
Shika
honest terrace
And a topological space $X$ is said to be compact if for any open covering $(O_i){i \in I}$, we can extract a finite subfamily $O{i_1}, \cdots, O_{i_k}$ that is also an open cover
gentle ospreyBOT
Shika
dull goblet
didnt you do that exercise about proving that spec A is compact a few days ago
quasi compact
lol
honest terrace
Where a family of subsets of a topological space is an open cover if each of the subset is open and that the union of the family is the whole space
(basically that you're covering every point of your space with open sets)
empty grove
right so quasi compact is usually just compact
honest terrace
(disclaimer: the defn of a compact set without any example and a few theorems about them usually makes no intuitive sense 🐒)
empty grove
in topology, we say compact and hausdorff separately
empty grove
because of the french, in commie alg we say quasi compact when compact but not hausdorff
and compact means compact + hausdorff
honest terrace
🥖
dull goblet
and compact means compact + hausdorff
oh lol ok
so prime spectrum is already compact from what i did before
just gotta prove haussdorff
honest terrace
It usually isn't hausdorff, if you equip it with zariski topology 🤔
empty grove
ye
honest terrace
Oh, ok, so yeah I can see why that makes it hausdorff I think 
gritty widget
Guys what are the prerequisites for topology?
empty grove
set theory, but real analysis and metric spaces are extremely helpful
gritty widget
just the concept of a metric space?
empty grove
and by set theory I mean naive set theory, you should know how unions intersections complements inclusions inverse images etc work
yeah you can make do without metric spaces, the benefit is that results and examples of metric spaces motivate topology
gritty widget
which introductory topology book uses just set theory and introduces real anal concepts?
empty grove
Munkres is standard for topology but I don't think it does too much real analysis
idk any topology books that start with real ana
gritty widget
okay thanks
gritty widget
Munkres is standard for topology but I don't think it does too much real analysi...
does it have problems at the end of a chapter?
empty grove
yes
gritty widget
just "topology munkres solutions"?
empty grove
yeah
coarse night
I prefer topology by simmons (topology and modern analysis)
flint cove
in topology, we say compact and hausdorff separately
when even the name satisfies the separability axiom
gritty widget
Can any algebraic geometry people describe to me the pictures in their head that they see for schemes and affine schemes
So in diff geom I know my pace is locally modeled by Euclidean space
So I can imagine globally I may have some potatoe blobule
But locally I know all it’s properties
I’m really not sure how to do picture this in alg geom
Even if we step back and go to “classical” varieties
Where we have the zero set of some polynomial
This doesn’t feel like something that is made by glueing together smaller pieces
Or at least
I don’t know what the property of coming from the zero of a polynomial looks locally
modest agate
I'm currently doing my first assignment on general topology, but I'm stuck at the second problem. Could I get a hint to get me started?
Let $(X,\tau_X),(Y,\tau_Y)$ be topological spaces. Let $f:X\to Y$ be a continuous map, with image $Z=f(X)\subset Y$. Factor $f=i\circ p$ where $p$ is the surjective mapping $p:X\to Z$ and $i:Z\to Y$ the inclusion map. Let $\tau_1$ denote the quotient topology on $Z$ induced by $p$, and let $\tau_2$ denote the subspace topology on $Z$ inherited from $Y$. How can I show that $\tau_1$ is finer or equal to $\tau_2$?
gentle ospreyBOT
notsushY
modest agate
empty grove
The identity set map j is continuous 
Equality is not true, hint is T_1 is the finest topology such that p is continuous (definition of/theorem about quotient topology) ||but p remains continuous if you make T_1 coarser||
long hornet
The zero set of z^2 + w^2 - 1 in C^2, thought of as R^4, is a 2-manifold, right?
What is it like?
I'd also like to know whether the general case is easy. I mean suppose we have an entire map f : R^n --> R. If I'm not mistaken, we can extend this to a map F from C^n to C. Then if 0 is a regular value of f, then it is also a regular value of F, and we can get an analytic manifold (?) of complex dimension n, so of real dimension 2n. How is it related to f^-1(0)?
orchid forge
The zero set of z^2 + w^2 - 1 in C^2, thought of as R^4, is a 2-manifold, right?
Consider the projectivization of $z^2 + w^2 - 1$ living in $\mathbb{CP}^2$, $z^2 + w^2 - \zeta^2$
gentle ospreyBOT
Kogasa
orchid forge
each element $[z, w, \zeta]$ of the zero set with $\zeta \ne 0$ corresponds to a unique element $(z/\zeta, w/\zeta)$ of your zero set, but you have more points at $\zeta = 0$: the equation becomes $z^2 + w^2 = 0$, which means $(z/w)^2 = -1$, $z/w = \pm i$. this gives two points of $\mathbb{CP}^2$, $[i, 1, 0]$ and $[-i, 1, 0]$.
gentle ospreyBOT
Kogasa
orchid forge
so your zero set, plus two extra points, forms a closed subset of the compact space $\mathbb{CP}^2$
gentle ospreyBOT
Kogasa
long hornet
so your zero set, plus two extra points, forms a closed subset of the compact sp...
I see!
Any idea how does it look like?
orchid forge
I'm pretty sure it can only be one of a small number of things
but I forget exactly what it is
abstract pagoda
Equality is not true, hint is T_1 is the finest topology such that p is continuo...
equality true if f injective
wow im late
thats a cool topology question
empty grove
No, you can have S = {1,2} with the discrete top mapping via identity to S with the indiscrete top
Then the quotient top is also discrete
subspace top is indiscrete
abstract pagoda
oh yea im forgetting its a map between topologies
i was just saying that f = i o p is true if f injective
not that the topologies are equivalent
empty grove
wait thats always true isnt it
long hornet
I'm pretty sure it can only be one of a small number of things
Well, is it a submanifold?
empty grove
orchid forge
which i think implies it's a disk?
wait no it's a sphere 
so it should just be a sphere minus two points
it might be more helpful to just construct the riemann surface explicitly
pearl holly
ok very quick question: Hom(Z³, Z) is just Z³ right?
I just need this to be verified lmao
in case I do something horribly wrong
tight agate
yes
pearl holly
ok thank you so much! 
orchid forge
Well, is it a submanifold?
https://math.stackexchange.com/questions/1552627/riemann-surface-of-w2-sqrt1-z2 
you can see the "disk with two points removed" from here too but this is more "complex analysis"-y
orchid forge
I'd also like to know whether the general case is easy. I mean suppose we have a...
how did you mean to extend the map to C^n? you would need twice as many variables in general, although polynomials on R^n do extend to polynomials in C^n
tough imp
ok very quick question: Hom(Z³, Z) is just Z³ right?
Hom is additive so direct sums in the first component become direct sums on the outside (direct sum’s universal property is just so that this is so)
Then if you know Hom_Z(Z,G) = G
You get your answer
tough imp
quasi forum
Oh man, my head's spinning on this one (6 part (a))
orchid forge
the distance from x to y in the uniform metric is the supremum of the distances of each component right?
quasi forum
That, but we are finding the supremum of the bounded metric. So min{1, d(x,y)}
orchid forge
so if $\mathbf{x}, \mathbf{y} \in \mathbb{R}^\omega$ such that $\sup_n \bar \rho(x_n, y_n) < \epsilon$, that means there is a real number $\delta := \sup_n \bar\rho(x_n, y_n)$ so that all the components are $\delta$-close
gentle ospreyBOT
Kogasa
quasi forum
Yes, I agree with that
orchid forge
what do you think about the claim $(0, 1/2, 3/4, 7/8, 15/16, ...) \in B_{\bar \rho}(0, 1)$?
gentle ospreyBOT
Kogasa
quasi forum
Oh, the supremum of those distances is 1, so it is not in that ball
orchid forge
drew
quasi forum
drew?
orchid forge
true
quasi forum
So there is the big issue. If we take an element two elements x,y so that the supremum to be equal to epsilon, than y would be in U(x,e) but not the ball.
orchid forge
if the supremum equals epsilon, it's not in the ball
quasi forum
To be more precise, the supremum equal to epsilon when the maximum is undefined
orchid forge
remember the ball is the set of points with distance less than epsilon
oh
nevermind sorry you are right
if the supremum is epsilon but all the individual distances are strictly less than epsilon, it's in $U(\mathbf{x},\epsilon)$
gentle ospreyBOT
Kogasa
quasi forum
Thats a better way to say it
So, if i'm not mistaken, U(x,e) is the closure of the rho epsilon ball around x
orchid forge
I think that's true
quasi forum
So to show that this set is not open, we take a point y, along the boundary of U(x,e) show we cannot find an open neighborhood of ycontained in U(x,e)
orchid forge
seems reasonable
bold minnow
Oh man, my head's spinning on this one (6 part (a))
This is munkres right?
bold minnow
Ok nice, I'm currently going through it and I recognized the type/paper look/notation
Admittedly i wasnt sure lol
quasi forum
Definitely seems like a good resource. I've had to scratch my brain a few million times
I am trying to use the triangle inequality here, but I seem to be a little stuck.
orchid forge
so by assumption we can make $\bar \rho(x_n, y_n)$ as close as we want to $\epsilon$ by picking large enough $n$
gentle ospreyBOT
Kogasa
orchid forge
if $z$ is in the $\delta$-ball around $y$, can we make $\bar\rho(z, x) \ge \epsilon$?
gentle ospreyBOT
Kogasa
orchid forge
the triangle inequality says that the biggest we can possibly make that distance is $\bar\rho(x,y) + \bar\rho(y,z) = \epsilon + \bar\rho(y,z)$ which is certainly greater than epsilon
gentle ospreyBOT
Kogasa
orchid forge
but can you actually achieve that distance? how should you pick $z$ to be as far from $x$ as possible
gentle ospreyBOT
Kogasa
quasi forum
Well, on a metric like the norms (just as a visual aid), we can make e+rho(y,z) be the length of the line segment
But of course, we aren't in the norms, so it isn't that obvious
orchid forge
you mean like, take a line segment going from x to y
then just follow that line outwards
for some distance delta
that is a point with distance delta from y, which has maximal distance from x
that is the idea, for any $0 < \delta' < \delta$ we can pick $z \in B_{\bar \rho}(y, \delta)$ having distance exactly $\bar\rho(x, y) + \bar\rho(y,z) = \epsilon + \delta'$
gentle ospreyBOT
Kogasa
orchid forge
now by assumption, eventually $|y_n - x_n| > \epsilon - \delta'$
gentle ospreyBOT
Kogasa
quasi forum
Does it suffice to show any ball centered at y is not contained in U(x,e)?
orchid forge
yes
quasi forum
Oh really? Why don't I have to choose an arbitrary ball that has y?
quasi forum
**Kogasa**
Yea. But here you chose a ball specifically centered at y
gentle ospreyBOT
Kogasa
orchid forge
so it suffices to show this
quasi forum
Oh okay, that's good to know
orchid forge
you should think about how to prove that
there is a geometric picture here though
your "boundary element" y is just a sequence of reals in (x-epsilon, x+epsilon) that converges to one of the endpoints
a delta-neighborhood of y consists of all the sequences that are always close to y
but for any delta, if y_n --> x + epsilon, then (y_n + \delta/2) --> x + epsilon + delta/2
quasi forum
Yea, I can agree to that
orchid forge
that's a delta-close sequence to y which is not contained in (x - epsilon, x + epsilon), and delta was arbitrary, so there is no open neighborhood of y contained in B(x, epsilon)
quasi forum
Where did the delta/2 come from?
orchid forge
y is converging to x + epsilon from the left, and i wanted to make it leave the interval, so i shifted it to the right a bit
by a distance smaller than delta
if y --> x - epsilon then i would shift it to the left by delta/2
to be more precise, since x is a sequence it's not necessarily the case that y is a convergent sequence at all
but (y_n - x_n) does converge to 0
so really, $y \in \partial U(x, \epsilon) \leftrightarrow$ the sequence $(y_n - x_n) \in (-\epsilon, \epsilon)$ converges to $\pm \epsilon$
gentle ospreyBOT
Kogasa
orchid forge
and $z \in U_{\bar \rho}(y, \delta) \leftrightarrow$ the sequence $(z_n - y_n) \in [-\delta', \delta'] \subset (-\delta, \delta)$
quasi forum
A lot's happening right now. I'm gonna need a minute to process this
gentle ospreyBOT
Kogasa
shy moss
Let $X=V(S) $ be an algebraic variety. Suppose the coordinate ring $A(X)$ is a field. Then for all $f \in K[x_1,...,x_n]$ exist $g \in K[x_1,...,x_n]$ such that $(f+I(X))(g+I(X))=1+I(X)$, this implies that $f(x)g(x)=1$ for all $x \in X$. Now let $p \in S$ and $a \in X=V(S)$ then $p(a)=0$ but $p(a)$ is also unit. Are there something wrong with this?
gentle ospreyBOT
Or x1
shy moss
This implies that X is a point?
orchid forge
yes
more directly, remember the coordinate ring is a quotient of a polynomial ring
when a quotient is a field, the thing you're quotienting by is a maximal ideal
and maximal ideals correspond to points
shy moss
and maximal ideals correspond to points
why?
orchid forge
think first about the case where K is algebraically closed
you just identify the point p with the ideal of polynomials vanishing at p
all of them have a factor of (x-p), which generates a maximal ideal, so we have identified p with the ideal (x-p)
if k is not algebraically closed, then the maximal ideals of the polynomial ring aren't necessarily of the form (x-a)
orchid forge
i guess in your proof it was necessary to state that f and g are both nonzero (or that I(X) is proper)
long hornet
how did you mean to extend the map to C^n? you would need twice as many variable...
I was meaning to use CA stuff. If we have a function which is analytic on all R, we can analytically continue it to C, right?
(I know very little CA)
orchid forge
you can, provided that there is a taylor series that converges everywhere
like 1/(1 + x^2) is analytic everywhere but the taylor series at x=0 only converges in |x| < 1
but there is no hope of a nice relationship like you said
because there's no relationship between the real roots of a given analytic function and its complex roots
if you restrict to a class of functions that have such a relationship then you might be able to say something
long hornet
like 1/(1 + x^2) is analytic everywhere but the taylor series at x=0 only conver...
Oh, right. What about the multivariate case?
orchid forge
i'm supposed to be leading a calc review session on zoom rn and they didn't make me a host so i can't start the meeting
orchid forge
Oh, right. What about the multivariate case?
multiple complex variables makes things a bit tricky
orchid forge
1 variable complex functions be like
"i'm 0 on a single convergent sequence? then i'm 0 everywhere, fucker"
multiple variables don't give a single FUCK unless you're in codimension 1
hollow harbor
First of all, If x is in cl(A) and y is in bd(B), why is (x, y) in bd(A x B)?
(x, y) in cl(A x B) because x is in cl(A) and y is in cl(B), and therefore (x, y) in cl(A) x cl(B) = cl(A x B).
(x, y) in cl((A x B)^c) because x is in cl(A) and y is in cl(B^c), so (x, y) is in cl(A x B^c) which is a subset of cl((A x B)^c). (note that A x B^c is disjoint from A x B to see this).
so therefore (x, y) in bd(A x B), and cl(A) x bd(B) is a subset of bd(A x B). By a symmetric argument, bd(A) x cl(B) is a subset of bd(A x B).
this is one direction of the proof.
for the other direction, we need to show that any point in bd(A x B) is in either cl(A) x bd(B) or bd(A) x cl(B).
limpid leaf
ryc this is graded homework 
hollow harbor
well being in bd(A x B) means you're in cl(A x B), so you're in cl(A) x cl(B), and therefore we just need to check that EITHER x in bd(A) or y in bd(B).
tough imp
what are you proving anyway
limpid leaf
he's showing this
limpid leaf
yeah
tough imp
Is X also second countable?
limpid leaf
no
limpid leaf
the assignment was pretty good honestly
tough imp
so fucking rip
limpid leaf
this question just sucks
anyways i still havent solved the fucking cube question
🧊
hollow harbor
We know (x, y) in bd(A x B), so (x, y) in cl((A x B)^c). We can write (A x B)^c as a union, A^c x Y union A x B^c. Check this. If (x, y) in cl(A^c x Y), then x in cl(A^c), but we know x in cl(A), so then x in bd(A).
If (x, y) in cl(A x B^c), then we know y in cl(B^c), but we also know y in cl(B), so y in bd(B).
therefore bd(A x B) is a subset of (bd(A) x cl(B)) union (cl(A) x bd(B))
end proof
orchid forge
this makes the right side not symmetric in X and Y
the left side is also not symmetric in A and B
tough imp
wut
orchid forge
A x B is not the same as B x A
tough imp
I mean
sure but they did the < direction already
so if the right is symmetric in A and B
wait
omega think
maybe ur right
orchid forge
i mean why do you say second countability implies the rhs is not symmetric in A and B
tough imp
lol
no you're right
I was thinking if A and B are both subsets of a second countable space
you'd only need to show that one of bd(A) x cl(B) < bd(A x B)
but you're right the right hand side isn't symmetric
so it don't work
:(
or something like that idfk
I just want to always prove something once then get the others 4 free
orchid forge
well
you kind of can
we know that $\partial(A \times B) \subset cl(A \times B) = cl(A) \times cl(B)$, so automatically both coordinates are in $cl(A) \times cl(B)$
gentle ospreyBOT
Kogasa
orchid forge
so the equation says the boundary of $(A \times B)$ is all the points $(a,b)$ where $a \in \partial A$ or $b \in \partial B$
gentle ospreyBOT
Kogasa
orchid forge
if not, then both $a$ and $b$ are in the interior of their respective sets... so $(a,b) \in int(A \times B)$
gentle ospreyBOT
Kogasa
orchid forge
it was done before you proved it too
chmonkey just wanted to prove it in one step, which is possible
hollow harbor
thats nice
orchid forge
u ok?
hollow harbor
im sorry, im just mad at jesse for asking me to do this problem for him if it was already done lol
tough imp
Wut
hollow harbor
if there is actually a way to do it in one step that is nice is what i meant
limpid leaf
im sorry, im just mad at jesse for asking me to do this problem for him if it wa...
ive never proved this
tough imp
Copium
orchid forge
you told me to stop talking about a problem because you solved it
a problem out of an ancient textbook that every math student ever has or will read
you can just google a proof
gritty widget
coral pawn
How would you phrase the solution to this?
For example if I write $\frac{\partial f}{\partial r}|_{r_0,\theta_0}$, does it mean I am evaluating the derivative at the point $(r_0\cos{\theta_0},r_0\sin{\theta_0}$ or at the point $(r_0,\theta_0)$?
gentle ospreyBOT
Finitely Many Bananas
coral pawn
The notation is kinda sucky and I don't wanna lose points because of it
orchid forge
I would read that as evaluating at the point (r_0, \theta_0) in polar coordinates
which is (r_0 cos(\theta_0), r_0 sin(\theta_0)) in cartesian coordinates
i would phrase the solution by stating x = r cos(theta), y = r sin(theta), compute $\frac{\partial x}{\partial \theta}$ and $\frac{\partial y}{\partial \theta}$, then use the chain rule to write $\frac{\partial}{\partial \theta} = \frac{\partial x}{\partial \theta} \frac{\partial}{\partial x} + \frac{\partial y}{\partial \theta} \frac{\partial}{\partial y}$
gentle ospreyBOT
Kogasa
coral pawn
Can you read through my solution?
orchid forge
maybe
coral pawn
coral pawn
maybe <:Hmmm:768218192975101982>
How does it look?
orchid forge
correct but maybe could be more concise
coral pawn
Maybe so
coral pawn
Can someone verify if my solution for this is correct? It seems to follow from the definitions pretty much
orchid forge
Assuming proposition 2 is applied correctly it seems fine
long hornet
so it should just be a sphere minus two points <:Hmmm:768218192975101982>
Cool! Incidentally, that's homotopy equivalent to S^1, the set of real zeros. Coincidence?
I think so
orchid forge
i dunno about homotopy equivalence, but you can see the circle inside it by gluing two slit CP^1's to form the riemann surface explicitly
the two removed points are the points at infinity in each copy of CP^1
and the set of real zeros is a circle around the middle
long hornet
Fair enough
long hornet
i dunno about homotopy equivalence, but you can see the circle inside it by glui...
What would the "complex n-sphere" look like? It's not obvious to me how this method generalizes.
fading vale
probably the one point compactification of C^n
honest terrace
It's this just S^{2n} ? 🤔
honest terrace
Oh yeah ofc sry
fading vale
orchid forge
physicists need to stop pretending "tensor is what transforms like a tensor" is a rigorous definition
these motherfuckers define manifold as "a locally euclidean space" and never talk about it again i have seen their fucking books
long hornet
probably the one point compactification of C^n
It's not compact. Also, the "complex 1-sphere" turned out to be a 2-sphere punctured twice.
orchid forge
when you say "complex sphere"
do you mean the zero set of the polynomial x_1^2 + ... + x_n^2 = 1
over the complex numbers?
long hornet
Yes, except in C^n.
long hornet
do you mean the zero set of the polynomial x_1^2 + ... + x_n^2 = 1
That's the "complex (n-1)-sphere"
orchid forge
that's not usually what we mean by "complex sphere," but i see what you're saying
long hornet
Oh, I see
orchid forge
well, you can try constructing riemann surfaces again explicitly, or you can projectivize it and look at it inside CP^n
but now you'll have higher genus
for n=3 it'll be a closed genus 1 riemann surface
orchid forge
orientability
long hornet
Orientability?
long hornet
orientability
Aha
orchid forge
we usually say euler characteristic instead of genus as well, but it's mostly the same information
long hornet
I see
long hornet
for n=3 it'll be a closed genus 1 riemann surface
It would be nice if it deformation retracts to the 2-sphere
orchid forge
wait no sorry, it's always genus 0
i dunno why i thought it was a degree 3 polynomial for n=3
so it should always compactify to a sphere
orchid forge
it has dimension (n-1)
complex dimension
and yes it's a submanifold of CP^n
the projectivization is x_0^2 + ... + x_n^2 - z^2 = 0, and the extra points you get in compactification come from solutions to x_0^2 + ... + x_n^2 = 0
fading vale
It's not compact. Also, the "complex 1-sphere" turned out to be a 2-sphere punct...
What isn't?
long hornet
The set of complex solutions of z^2 + w^2 = 1
long hornet
the projectivization is x_0^2 + ... + x_n^2 - z^2 = 0, and the extra points you ...
No idea how to visualize that, but the idea is that the complex sphere is then the solutions of the projectivization minus the extra points?
long hornet
But thats not the standard definition of a "complex 1-sphere"
Yeah, I didn't know that. I will denote it by G^n.
orchid forge
an issue now is that you're not dealing with surfaces
but codimension 1 (hyper)surfaces
long hornet
I'd prefer a non-algebraic approach
I mean not relying on the fact that S^n is a variety
But as you said, it doesn't seem easy or even possible
fading vale
GAGA moment 
long hornet
I mean not relying on the fact that S^n is a variety
I thought I could mess with the imaginary parts to make G^n retract onto S^n, but realized it was dumb
fading vale
You can probably translate the necessary AG statements into things about the complex geometry but
Dunno enough about the higher dim things to say anything
long hornet
I thought you could only translate "high-level stuff"
orchid forge
If AG can't describe basic analytic geometry we would probably have issues
orchid forge
Same
fading vale
Relatable content
orchid forge
To describe the zero set to x_0^2 + ... + x_n^2 in CP^(n-1): one of the x_i must be nonzero, say x_0 so divide through and we get 1 + (x_1/x_0)^2 + ... + (x_n/x_0)^2 = 0. Multiply x_j/x_0 by i to get (i x_1/x_0)^2 + ... + (i x_n/x_0)^2 = 1. This is a transformed copy of G^(n-1). So the zero set is the union of n copies of G^(n-1). But each two of these intersect at a copy of G^(n-2) and so on
For n=2, the zero set is made of two transformations of x^2 = 1, namely [x, iy] and [ix, y] = [x, -iy] intersecting at a point (the origin), although this isn't actually an intersection in CP^1
pearl holly
Okay so I'm supposed to calculate the cohomology groups of the Klein bottle with Z coefficients directly from the definition. I already did H^0 and H^1 but now I'm stuck at calculating H^2.
So I'm interested in the image of the map $\partial^: \text{Hom}(\langle a, b, c \rangle, \mathbb{Z}) \to \text{Hom}(\langle U, L \rangle, \mathbb{Z})$. I know that $\partial^(\varphi) = \varphi(\partial)$ and now $\varphi(\partial(U)) = \varphi(a) +\varphi(b) - \varphi(c)$ and $\varphi(\partial(L)) = \varphi(b) - \varphi(a) - \varphi(c)$. But now what?
gentle ospreyBOT
Tokidoki ✓
orchid forge
What is the kernel? So which linear combinations of U and L equal 0?
pearl holly
I got the kernel to be Z
orchid forge
Explicitly?
pearl holly
ye
orchid forge
I mean what is the generator
pearl holly
well I don't know I just used like a lot of presentations
orchid forge
You gotta keep track of the generators
Otherwise how will you compute quotients
A ~ Z and B ~ Z doesn't let you compute A/B
pearl holly
oh no
so if I have that A is isomorphic to Z^2 and that B is isomorphic to Z, then A/B is not Z?
orchid forge
So like H^0 is generated by the homology class <v>
Well we care about the kernel because it's part of the definition of the homology groups
pearl holly
shouldn't it be the image?
orchid forge
No, homology groups are "closed chains modulo boundary," closed chains are the kernel and boundary is the image
pearl holly
ye but I'm doing cohomology
orchid forge
It's the same
pearl holly
so I have 0 ---> <v> ---> <a, b> ---> <U, L> ---> 0 where by <a, b> I actually mean its dual group. So H^2 will be <U, L>/image of the third map or am I completely insane?
orchid forge
The cohomology of $\partial^$ is the kernel of $\partial^$ modulo the image
gentle ospreyBOT
Kogasa
pearl holly
ye and I have the kernel since the last map is 0 so now I want the image
orchid forge
Yes
Reeee
I have a class starting rn
I can explain in an hour if nobody else does
pearl holly
okay great! Thank you so much!
pearl holly
ye I think that we talked about this before moldi
orchid forge
You gotta look at how these homomorphisms are related
empty grove
orchid forge
a+b-c = (a-b+c) + 2(b-c)
So f(U) - f(L) = 2(b-c)
We're modding out by this relation
pearl holly
ah yeah right okay I see now. I will keep this in mind!
pearl holly
so if I have that A is isomorphic to Z^2 and that B is isomorphic to Z, then A/B...
but now I have to redo the earlier parts because of this
it's so easy for me to miss this
it feels so true
actually never mind, I never used that fact
idk why I mentioned that wtf
shy moss
let Y an affine variety, if i have an element f in the coordinate ring A(Y), what means that f(x)=0?
empty grove
Coordinate ring is k[x1,...,xn]/I(Y) right?
shy moss
yes
empty grove
f = [p(x1,...,xn)] for some polynomial p, and f(x) means p(x) (you can check this is well defined)
The idea of A(Y) is that it is the ring of polynomial functions on Y
But different polynomials can define the same function
That's why you quotient by I(Y), then each element corresponds to a unique polynomial function and vice versa
shy moss
yes
sleek thicket
okay it seems to be unoccupied
fading vale
@sleek thicket you've seen like ramification in riemann surfaces right
sleek thicket
nope. my understanding of ramification is a point which splits into too many points (or too few? like z -> z^2) and stops something from being a covering map
and also this is what primes splitting looks like in ANT
fading vale
thats all u need really
Yes thats it exactly
wait
uhh ill be back in a few minutes
or less
sleek thicket
lol
fading vale
ok back
So basically if i have an map p: Y -> X of curves then being unramified in the algebraic sense at p of pB splitting into primes if and only if B/pB is an etale k(p)-algebra
sleek thicket
I am going to need further elaboration here
First, are X = Spec A and Y = Spec B affine?
second I'm assuming the map p and the prime p are distinct
fading vale
they should be affine normal but projective also works
Sorry this is kinda a mess cause i just had to change classes
sleek thicket
lol
fading vale
yes uh lets do like f: Y -> X instead
basically the idea is that over C the closed points form Riemann surfaces, and f will define a holomorphism of Riemann surfaces
and if we have a closed point p in X then the splitting of its extension in Y is going to look like ramification in the sense that like
if p splits into q_1^e_1 ... q_n^e_n in the coordinate ring of Y
sleek thicket
so we're taking some point p in X, this is a maximal ideal m of A
fading vale
Yep
fading vale
Yeah
sleek thicket
You're talking about the pushforward factoring as a product of primes
fading vale
Right
fading vale
itll look like q_1^e_1 ... q_n e^n right, with the q_i prime
fading vale
when the e_i are all 1 and it splits into distinct primes
fading vale
um wdym
sleek thicket
itll look like q_1^e_1 ... q_n e^n right, with the q_i prime
Here
Should it be q_n^e_n?
fading vale
Oh yeah lol
sleek thicket
okay so
Suppose the pushforward is a product of primes
Now further assume it's a product of distinct primes
fading vale
product of primes is automatic in the normal case btw
because we're in dim 1 so thatll be dedekind
sleek thicket
So is being a product of distinct primes unramified?
fading vale
Right
sleek thicket
Cool
fading vale
and this is equivalent to B/pB being a finite etale algebra over the residue field of p
where B = coord ring of Y
sleek thicket
So we have A -> B
Oh p is maximal
I see where the algebra structures comes from
κ(p) is just A/p
sleek thicket
So what does it mean for an algebra to be etale?
sleek thicket
I was told I would be etale pilled moth
fading vale
lolz
sleek thicket
Product as in direct product?
fading vale
yep
fading vale
finite product of finite separable extensions
sleek thicket
it's the crt
fading vale
right
sleek thicket
And then you just need like
A theorem about comparing direct product decompositions
Which should be easy
fading vale
yeah its not a hard proof
so like the idea here is that like, if i have a proper holomorphism of riemann surfaces
its gonna be ramified at only finitely many points
and everywhere else itll be a cover
sleek thicket
Why does it only ramify at finitely many points?
fading vale
complex analysis reasons, its just that the branch points end up being finite discrete and stuff
(sry for the delay im technically in class )
sleek thicket
That makes sense
fading vale
i dont rly remember the complex analysis details tbh xd
sleek thicket
This fits with my understanding of complex analysis
fading vale
Poggies
sleek thicket
We're looking for points where it's not a local homeomorphism right?
fading vale
Yep
sleek thicket
If the derivative is nonzero then it's gonna be locally invertible
(this is all happening in charts)
IVT stuff
fading vale
Yes this sounds right
so like if we have a finite separable morphism Y -> X of affine curves (X integral blah blah)
its gonna end up being etale on a cofinie subset of X
sleek thicket
Cool
fading vale
so heres the cool part
u know how pi_1 can be thought of as like, sets with actions of pi_1 on it correspond to covers?
fading vale
you can get a group pi_1^et that will do the same thing, but with etale covers instead of topological ones
sleek thicket
π1(X) is like the automorphism group of the identity functor of π1(X)-set right?
In the classical case
Smth like that?
fading vale
I think a fiber functor
fading vale
what ends up happening is that like
each etale cover of X corresponds to a finite extension of the function field
just bc all finite maps above X will
(it will induce an injection on coord rings and stuff)
so like, if i take the compositum of all these extensions
its galois group will be the etale fundamental group
sleek thicket
neat
fading vale
and basically when u work over C u get this really nice relationship where like, if i pick some finite discrete subset S of X
then the profinite completion of pi_1(X - S, x) regarding X as a riemann surface will be the etale fundamental group
fading vale
yep
fading vale
i know! its weird
orchid forge
ah yeah right okay I see now. I will keep this in mind! <:catthumbsup:6145401887...
Did we end up getting it
fading vale
Linear algebra and vector spaces
sleek thicket
linear algebra and vector spaces
it's worse than you think
this problem set is so long and so boring https://courses.cs.washington.edu/courses/cse446/21au/assignments/hw0.pdf
fading vale
sleek thicket
now you see why im trying to learn measure theoretic probability
i can technically say it's for my ML class
and it is
so much more interesting
plain raven
oh, they're the zeros of the derivative?
this should be right. for every map of riemann surfaces f : X -> Y, given x in X there are coordinate charts U around x and V around y such that f is just f(z) = z^k for some k, f : U -> V. this fact really helps me to sort out my intuition about theorems about maps of riemann surfaces
sleek thicket
makes sense
plain raven
it's an extremely nice result
pearl holly
Did we end up getting it <:Hmmm:768218192975101982>
yeee boiii thank you for the help!
sleek thicket
i am taking a complex geometry class next quarter which should be fun
plain raven
so it's ramified iff k is not one
sleek thicket
good to know i remember enough complex analysis to muddle my way around
plain raven
you should check out forster. i'm loving it. and he does a bunch of proofs with sheaf cohomology which has helped me to gain intuition for what sheaf cohomology is trying to do. i mean i think i get it but like, certainly divisors are easier to understand in the CA setting than just talking about abstract nonsense about schemes
i can't remember. did you say you were a masochist?
no nvm i'm thinking of chmonkey
my bad
sleek thicket
lol
that sounds fun, I'll check it out
i expect the class to be about as far from AG as a complex geometry class can get
based on the professor (lee)
Lectures on Riemann Surfaces by forster?
plain raven
yeah that's what i mean
also like, this isn't super surprising but complex analysis does require some analysis. obviously this is kind of just 'wtf' but for me it helped me to link the analysis to the AG. for example the theorem that every every compact riemann surface has finite genus is a consequence of some results of analysis, some very technical arguments that ultimately appeal to the Banach open mapping theorem
maybe you'd be able to get away with a completely different argument style
orchid forge
you should check out forster. i'm loving it. and he does a bunch of proofs with ...
Sheaf cohomology measures obstructions to coherently gluing data 
orchid forge
I didn't see it this way until I looked at that application paper
About sheaves and data fusion problems
uncut surge
Sheaf cohomology measures obstructions to coherently gluing data <a:catHyperJAM:...
i sincerely don't know how people teach sheaf cohomology and don't teach people this intuition
how can that even be done
orchid forge
But it is straightforward to interpret what I said precisely as the definition of the cohomology group
orchid forge
how can that even be done
I have no idea how people teach sheaf cohomology
uncut surge
ye me neither
orchid forge
The two times it's been introduced in classes we basically stated the definition then did computations
uncut surge
i mean maybe i'm just missing a crucial alternative interpretation of sheaf cohomology
orchid forge
So I'm glad I learned this shit prior
At least the basic ideas
I think it's because sheaf cohomology was only mentioned in passing in those classes though
uncut surge
maybe the gamer way would be to just teach people cech cohomology
and then say "okay and sheaf cohomology is then the abstract limit of this garbage but just think about cech"
orchid forge
That is what we did in the first class lmao
uncut surge
vewwy gud uwu
orchid forge
Except we defined sheaf cohomology, then only ever did computations with cech
Professor was feeling based that day
uncut surge
ye i guess that's the only way to do any explicit calculations anyway
uuuh unless you're doing more abstract things ofc
orchid forge
He stated some more abstract theorems and examples but didn't really expect us to do much with it ourselves
uncut surge
i've used it a bunch in my research
okay nobody cares but i'm too giddy not to say it
you can calculate lie algebra cohomology of vector fields on a smooth manifold by doing it on smol open neighbourhoods (basically R^n) and then gluing things together via cech/sheaf cohomology
this is what people call gelfand-fuks cohomology and nobody understands it
you don't have to care but i did this and i'm very happy and it made me really appreciate cech/sheaf cohomology
orchid forge
This sounds cool
You should always be able to glue local cohomologies together like that right?
uncut surge
this is actually quite interesting, some cohomologies behave well with respect to this gluing and some become really nasty
orchid forge
But getting something nice would intuitively require a very particularly well behaved local cohomology
Like I would think this would usually be hard to compute
uncut surge
lie algebra cohomology of vector fields is actually rather nasty; but if you look at something like hochschild cohomology of smooth functions, that is a very local thing
i don't quite get why, but i'm actually quite in love with that whole cohomology business
i want to be the one person in the world who can tell you something about every cohomology theory
-- in any case, i think the right intuition is that cohomological things should generally not behave very nicely with respect to localization; after all, you can't see, say, the simplicial cohomology of a sphere by just looking at its small open sets; you somehow need global data
orchid forge
Yes
uncut surge
but by combining your cohomology with cech/sheaf cohomology, you can capture "how badly" the localization process fails
uncut surge
the sad part is that this will generally only give you something awful like a spectral sequence -- but better than nothing
that's oké, bye! thanks for listening to my cohomology stories
plain raven
yo lartomato
if you're interested in this stuff i have a question for you.
this is really bugging me
do you know of any nice abstract or categorical constructions of the de Rham complex? i'm talking about like, the free widget generated by a blah blah blah in a symmetric monoidal category
like
if you wanted to rigorously and precisely define the de Rham complex (say in the category of sheaves of real vector spaces) what is the most efficient way to describe this
i think of free constructions as kind of 'induction principles', because they give you laws by which you can 'introduce' a diagram of a certain shape. i'm looking for an 'induction' principle which would allow me to construct the de Rham complex.
closely related to this is V(g), the Chevalley-Eilenberg complex associated to a Lie algebra
i think an answer for the case of V(g) could be easily translated into a case for the other
i came up with a satisfying answer in the case of the Koszul complex associated to a module equipped with a finite number of pairwise commuting endomorphisms. And the chevalley-eilenberg complex has the Koszul complex as the special case where g is an Abelian lie algebra
from googling around i've seen that you can think of the chevalley-eilenberg complex as a kind of deformed Koszul complex, which seems to make sense, but i have no idea how to use this information to give a nice general construction of the complex
something something Lie operads
uncut surge
that's an interesting question and i'm not sure if i have a satisfying answer, but i'll thonk about it for a moment
plain raven
yeah it seems hard to me, it took me forever to come up with the answer for the Koszul case, but i also really know almost nothing about Lie algebras.
uncut surge
what is a fun construction of the de Rham complex is that it equals the continuous hochschild cohomology of the algebra of smooth functions lol
but that's not quite what you're asking i believe
uncut surge
i think of free constructions as kind of 'induction principles', because they gi...
i'm not sure i understand this line, do you have an example of such an induction principle?
plain raven
well, it might, but is there a general construction for a 'bar resolution' type thing for hochschild co/homology of smooth functions which gives you the de Rham complex as a special case?
Yeah i'll tell you what i figured out for Koszul complexes.
let's work in like, the setting of a symmetric monoidal category.
i think that's really all i need here.
Let V and M be two objects in the category.
Suppose V acts on M by means of a morphism $h : V\otimes M\to M$.
gentle ospreyBOT
diligentClerk
plain raven
Suppose this action is "commutative" in the sense that this map iterated twice, $V\otimes V\otimes M \xrightarrow{id \otimes h}V\otimes M\xrightarrow{h}M$ commutes with the symmetry isomorphism $V\otimes V\otimes M\cong V\otimes V\otimes M\to V\otimes M\to M$.
gentle ospreyBOT
diligentClerk
plain raven
That's really all I want to assume here. It's a pretty simple setup.
uncut surge
ye i can see how this is koszuly
plain raven
The case of this which is relevant in the definition of the Koszul complex is that like, $V$ would be the free $k$-module on symbols $x_1,\dots,x_n$, and each of the $x_i$ has an endomorphism of $M$ associated to it, and these commute pairwise. The action is extended to arbitrary elements of $V$ by linearity.
gentle ospreyBOT
diligentClerk
plain raven
So, the setup I have described freely generates a big monoidal subcategory of the monoidal category by taking the closure under the unit object $1$, the tensor product, adding in the symmetry isomorphisms, associativity isomorphisms... everything that can be built out of monoidal category operations (composition, identity, tensoring) and the distinguished constants $V, M$ and the morphisms i've explicitly suggested
gentle ospreyBOT
diligentClerk
