not an O module
#point-set-topology
tight agate
wanton timber
That sounds likely to me, as well
tough imp
that's what I was thinking
wanton timber
I have a suspicion that when X is a scheme and F a sheaf of quasi-coherent modules, j^* and j^-1 agree; that j^-1(O_X) = O_X_P. So it may have been a notational oversight because it's irrelevant in all intended applications.
tight agate
Z/2 is a quasicoherent sheaf over spec Z
if you pullback (of sheaves) along the inclusion of 2 into spec Z, then you get (Z/2)_2
oh wait nvm
wanton timber
Yeah, I think Spec(R)_P = Spec(R_P) and j^-1(O_R)=O_R_P
wanton timber
I agree, but in the problem, P was a closed point. Definitely very special situation
tight agate
wait a sec , we're saying for any R module M, and for any prime ideal p, M_p iso M_p \otimes k(p)
where k(p) is the residue field at p
wanton timber
Are you sure? I think we're saying that M_p iso M_p \otimes R_p
tight agate
no, we're trying to say j^-1 M = M_p, is the same as j*M = M_p \otimes k(p)
this shouldnt be true
lemme look up the definition of pullback
gimme a sec
tight agate
Are you sure? I think we're saying that M_p iso M_p \otimes R_p
this is definitely false, take M = Z/2, R = Z, p = 3
wanton timber
I'm looking at the definition on pg 109 of Hartshorne, where f=j, Y=Spec(R) and X=Spec(R_P).
Then I think j^-1(O_X)=R_P, O_X=R_P
And f^-1 M = M_P
tight agate
yes, that's the inverse image sheaf
for the pullback you need to tensor with the structure sheaf of the domain
in our case it is the residue field at the point
wanton timber
That's where I disagree. I think the domain is spec(R_P)
Spec(R_P) is the space with the topology of X_P as defined in the problem
As far as I can tell, anyway
wanton timber
Totally fair
meager python
I see that hartshorne often writes F|Y := i^-1F instead of F|Y := i^*F
wanton timber
From the way he defines cohomology, that doesn't surprise me (as it's done as sheafs of ab not qcoh)
meager python
So it looks like he means j^-1 above
wanton timber
Yeah, but I think it agrees with j^* when X is a scheme, P a closed point, and F quasi-coherent
desert bloom
Heya
I have a question regarding the shape operator
Is there any intuitive way to think about why the Gaussian curvature is the determinant of the shape operator?
and why the curvature directions will be its eigenvectors?
<@&286206848099549185>
desert bloom
Anyone?
desert bloom
For me its defined as the derivative of the normal map
oh I think I get it now
I need to think through stuff more
gritty widget
at least in the 2d case, the eigenvalues of the shape operator S are given by the min and max of <S(x),x> over |x| = 1 (because the shape operator is self-adjoint), and that's the second fundamental form at x (i think?)
maybe that's one way you can approach this
just throwing an idea out there
willow spear
can soemone explain to me what teh diffreence bwteen "continuous deformation", "homeomorphism" and "Homotpy"
or are there similarities?
feel free to ping me when you answer
👍
yes it does thanks so much
so "continuous deformation" is like a general term?
bijection is the same thing as continuous right?
cloud owl
def not
willow spear
whats teh difference?
cloud owl
bijection is just something else entirely, right?
set of natural numbers is bijective with rational numbers?
i don't know topology
willow spear
ok
gritty widget
they're not at all the same
you can have a discontinuous bijective function, and you can have a continuous not-bijective function
maybe one abstract way to look at it is that bijectivity is a property of a set or the functions defined thereupon, but that continuity of a function depends on extra structure on the set (a topology)
cloud owl
not even that, surely
bijectivity is a property of a function, not sure how a set can be intrinsically bijective
gritty widget
or the functions defined thereupon
cloud owl
yeah
gritty widget
i'm being vague because i feel like there is something precise to be said there, but i don't know how to put it into words
maybe i am talking out of my ass, who knows 
cloud owl
ok so is a bijective function not just a function with an inverse, both injective and surjective
ie. f: X->Y
no y in Y is given by more than one f(x), injectivity
every y in Y is given by at least one f(x), surjectivity
gritty widget
i see
modern trail
coarse kestrel
fading vale
fading vale
oh oops im silly
this is like
the disjoint union of all the X^k and we are identifying points in X^k with points in X^{k-1}?
rugged quiver
Can you help me find CW complex with following cellullar chain complex?
$0\to...\to 0\to\mathbb{Z}^r=C_2 \to \mathbb{Z}^k=C_1 \to \mathbb{Z}=C_0$
gentle ospreyBOT
Gigrise
marsh forge
@fading vale you’re identifying any tuple containing the base point with the base point
fading vale
yeah okay i think that makes sense
wait hmm
@marsh forge but if you had (x1, ..., e, ...., xk) and (x1', ..., e, ..., xk') wouldnt those be identified with (x1, ..., xk) and (x1', ..., xk')?
i thought if it has the base point you just delete it you dont identify the entire thing with the base point
ur not
marsh forge
take r spheres w basepoint, k circles w basepoint, and one free point
fading vale
thats correct
marsh forge
glue all those points together
this should work fine
oh wait
maybe you are right sloth
yes okay okay
J is the space of all k-tuples without basepoint and the basepoint
fading vale
cool
marsh forge
oh and maybe no basepoint
hard to tell what hatcher means here
Because either e by itself is excluded
or we identify e with the empty tuple
which is weird
fading vale
mhm
he said this connects to loop spaces in 4J so i assume there is a base point of some kind
marsh forge
okay so he itends for J_2(X)=XxX/~ and under this relationship we still want a point (e,e) so i think we want a representative for the basepoint
fading vale
yea that makes sense
this section is kinda wack
but i guess its cool
a lot of it feels like set up for ch 4 stuff which is ok ig
fading vale
something about creating a space where J(S^n) is like
it has a cohomology ring thats "almost Z[x]" with coeff in Z
and is exactly Q[x] with coeff in Q
idk why we care about this though lmfao
marsh forge
i see
fading vale
is there any actual motivation behind the realization problem
fading vale
or finding a space whose cohomology ring is Z[x] or Q[x]
marsh forge
realization problems are always interesting to me
fading vale
its interesting i think
im just not sure how this specific case of it is important beyond this
marsh forge
For example it's interesting that you don't easily get analogues of EM-spaces for rings and coho
fading vale
or maybe its not and its just for its own sake
marsh forge
oh i mean hatcher is just giving an example i assume
marsh forge
you mean RP?
fading vale
yea
u get like
extra generators with coeff in Z
its cursed for odd
i mean not that cursed but like
marsh forge
anything not simple is cursed
fading vale
i dont even remember the structure
its like
Z[a, b]/(a^k+1, 2a, ab)
ummm
b^2?
for RP^2k+1
gritty widget
Alright
placid hearth
I thought #geometry-and-trigonometry was just high school stuff
placid hearth
Oh sorry I didn’t see lol
willow spear
can someone explain to me differenbce between algbebraic geomrtry, hyperbolic geomtry and topology
do they have things in common
?
ivory dragon
hyperbolic geometry studies hyperbolic spaces, which are a specific type of noneuclidean space
algebraic geometry studies varieties (and later schemes), which are (roughly speaking) the geometric structures formed by the solution sets to polynomials
some varieties are themselves hyperbolic, in particular many quotients involving the cone
though its perhaps better to view hyperbolic geometry as a specific case of riemannian geometry
(which would fall more naturally under differential geometry)
topology is an entirely different field, although there are a ton of connections between topology and geometry (and between topology and the rest of mathematics, really)
defining topology in a satisfying way is difficult, but at least as far as its geometric connections go, you can consider it as the study of the "distance-free" or "qualitative" properties of spaces.
in any case, topological definitions (in particular, the notions of a topological space and topological manifold) form the basis of basically every modern geometric definition
add a metric on and you get distances (of a sort) and hence calculus and whatnot, and can start doing "real" geometry.
wicked obsidian
can someone explain to me differenbce between algbebraic geomrtry, hyperbolic ge...
Topology is "less restrictive".
A cube is not the same as a sphere in geometry.
But they are homeomorphic. Which means they are the same as far as topology is involved.
You might have heard of TOPOPHOBIA. Fear of holes. Topoç means hole. Topology and homeomorphisms is all about holes 🕳. A torus has a hole. A double torus has 2 holes a triple torus has 3 holes, so none of those are homeomorphic.
ivory dragon
first off
topophobia is not a fear of holes
perhaps you mean trypophobia
but theres no relation
second off, the "hole" stuff of topology makes up like
1-2 lectures in an algebraic topology course
even if it makes up 99% of youtube videos about it
willow spear
fax
fading vale
me computing H^*(RP^infty; Z2): learning a lot about the holes right now
ivory dragon
please try and know what you're talking about before you say it so confidently
cursive flume
guys how can one define a hilbert space over a manifold?
ivory dragon
what do you mean by "define [...] over"?
cursive flume
so in classical mechanics we have the phase space as the cotangent bundle,or alternatively a section in gamma(tm). and we have a canonical isomorphism between the cotangent and the tangent bundles given by the poisson rbacket. I would like to somehow construct a similar point of view for QM/QFT
like how can I define a hilbert space or "hilbert bundle" if smth like that exists over a manifold?
the manifold is r^4 with a pseudo riemannian metric
how can i construct a hilbert space structure anyhow on top of that structure?is there any way?
or can one define a manifold over a hilbert space?(meaning let's say I have a hilbert space and I want to equip it with manifold structure)
cursive flume
is it possible to take other chart than the global one?
fading vale
i imagine itd depend on the space
cursive flume
in QFT we usually have infinite dimensional hilbert spaces
arent they all isomorphic to l^2? or is it jsut for some special cases of inf dim hilbert spaces? (i havent done proof rigorously)
fading vale
i havent done much with this stuff but that seems very unlikely to me?
what about L^2?
cursive flume
i mean L^2 should be the only inf dim hilbert space, or well, up to an isomorphism
and then can one take another chart than the global one on L^2?
fading vale
ah yeah you're right there is only one up to isomorphism
ok im not gonna risk giving any more wrong info 🤔
little hemlock
what is Delta^{n-1} supposed to mean here?
something homeomorphic to a disk, but idk that notation
ah, thanks
marsh forge
does anyone have an example of a jacobson space with non-jacobson subspace
tight agate
Spec Z is jacobson right?
I think the subspace (0), (2) is not jacobson
yeah (2) is the only closed point of that subspace, but the closed subset {(0), (2)} is not the closure of {(0), (2)} intersect {(2)} = closure of {(2)} = (2)
(topologically it is the same as spec(Z localized at 2), which is not jacobson as Z localized at 2 is not a jacobson ring)
the affine line /C also works I think
take the subspace consisting of the generic point and some closed point
thin bramble
I need a question course advice. So these two courses I will be taking next semester. However, I never took a geometry in high school. Should I be worried or do some reading before I take these classes. semester starts on Febuary 1, 2021.
**Transformation Geometry
**Classical theorems of Menelaus, Ceva, Desargues, and Pappus. Isometries, similarities, and affine transformations for Euclidean geometry.
**Introduction to Differential Geometry
**Differential geometry of curves and surfaces in Euclidean space, frames, isometries, geodesics, curvature, and the Gauss-Bonnet theorem.
Do at somecases but not really
river granite
I don't think either will assume a ton of HS geometry -- the first should take either an axiomatic or group-theoretic approach, and the second might at most assume you had calc 3
river granite
if you want book recs., I liked George Martin's "Transformation Geometry: An Introduction to Symmetry" (it has a late chapter on these classical theorems), Do Carmo's classic curves and surfaces book and apparently Dover also has a nice book by Guggenheimer. But I'd look at the syllabus in both your courses and see if I can grab those
thin bramble
@river granite George Martin's "Transformation Geometry: An Introduction to Symmetry" is the book we will be using for the first class. Not sure about the other course
river granite
that's great, I thought it was really well written
Do Carmo is one of the most usual ones for curves and surfaces, idk if any other is usually used (in English-speaking courses at least)
tight agate
are you asking if idempotent matrices are dense in rank k nxn matrices?
okay, well that's not true
take n = k
what's an oblique projection map?
okay
chrome dew
@tight agate why did that guy delete his messages in shame
chrome dew
same lmao
river granite
it left me intrigued as well lmao
digital glacier
Because the ${\epsilon}_i$ form a basis for the 1-forms $ T^{*}M$
gentle ospreyBOT
E8
digital glacier
$\Omega$ is an element of $\Lambda ^n T^{*}M$ by definition.
gentle ospreyBOT
E8
quartz rover
how do you compute the coefficients of the first and second fundamental forms
of a surface
and is there a general method for calculating the gaussian curvature of any hypersurface?
digital glacier
how do you compute the coefficients of the first and second fundamental forms
you should be able to find that in any standard text book
quartz rover
i'll skip ahead in the one i'm in
daring bluff
How do you compute the maximal flow generated by a vector field? I've got a vector field given by X in the picture. Any help appreciated.
daring bluff
<@&286206848099549185>
summer jolt
Hi, I came across this question but I'm struggling to understand the solution
Why does the calculation proves that phi_t is a 1-parameter group of diffeomorphisms?
gritty widget
because that's what it means for phi_t(g) to be an integral curve of A
something something uniqueness maybe
elder yew
I think all you have to do is differentiability
cuz you already know the other properties of the exponential map
summer jolt
Oh I see! So the equation reads y'(t) = the value of vector field at t. Am I right?
gritty widget
if y(t) = phi_t(g), yes. that's what it means for y(t) to be the integral curve of A (starting at g)
lie groups ☺️
they're pretty neat
summer jolt
@gritty widget thanks that makes sense. This is for a Differentiable Manifolds class (we only briefly touched on Lie Groups)
little hemlock
is this just because you have more maps getting identified to the same morphism of A under T? So in a sense the association of morphisms in Top with morphisms in A is weaker?
marsh forge
I dont really think this is super formal
I think thats one takeaway for sure
the idea i guess is that Top->A loses information
so if the problem can be translated then it willl be simpler
sometimes there are statements in Top that cant be answered in A
for example, no homotopy invariant can tell me whether im looking at a point or a disk
fading vale
yeah the loss of information is how i think about it too
but honestly i think its just better to imagine working in hTop most of the time
marsh forge
which one lol
fading vale
naive i guess
marsh forge
in this context you might as well work up to whe
fading vale
need to see CW approximations soon
nimble flower
If i want to show a set is closed, what can i do? I can show the complement is open, i can show the closure of the set equals itself - anything else?
if i have a homeomorphism, i can show that the set maps to a closed set under the homeomorphism
if the space is hausdorff, i can show the set is compact
sleek thicket
How do you compute the maximal flow generated by a vector field? I've got a vect...
You solve an ODE in coordinates
Suppose you have some point p = (θ0, φ0) and you want an integral curve γ with γ(0) = p
We can write γ(t) = (θ(t), φ(t))
nimble flower
i can show the set contains all its limit points
i can show the set contains its boundary
sleek thicket
the condition that γ be an integral curve says γ'(t) = X_γ(t), so θ'(t) = 0 and φ'(t) = 1/(sin(θ(t))
Then θ must be constantly θ0 and φ'(t) = 1/sin(θ0) is constant, so φ(t) = t//sin(θ0) + φ0
This means the integral curve starting at p = (θ0, φ0) is γ(t) = (θ0, t/sin(θ0) + φ0), and so the flow is F(p, t) = (θ, t/sin(θ) + φ)
This isn't actually well defined if sin(θ0) = 0, but your vector field isn't either so 🤷
nimble flower
From Munkres. How does this show that X - Y is open?
I get the proof - just dont see why that shows what it says it shows
yes
ah, we needed to make sure the open sets containing the points in X-Y were contained inside X-Y
gotcha, thank you 🙂
nimble flower
Suppose X' is the 1 point compactification of X and and C is compact in X. then it is also compact in X', right? since the embedding of X into X' is a cts map?
elder yew
Compact subsets of a topology space are compact with respect to the subspace topology
topological space
nimble flower
i thought the ambient topology was not relevant to compactness
ie. the image of a compact set under a cts map will always be compact
nimble jolt
that is correct @nimble flower
tough imp
@nimble flower the point he's making is that you need to consider what topology C has inside of X'
When we say C is compact in X it's compact wrt the topology we give it as a subspace of X
theoretically its topology as a subset of X' could be different
but it turns out the way X embeds into X', X's normal topology coincides with the subsapce topology it has in X'
so everything works out
This says C's subset topology in X and in X' are the same
nimble flower
okay that makes sense
then would it be true that if C is compact in X', not containing the added point, then C is compact in X
nimble flower
hmm, the only issue is the open sets that cover C could contain the added point *
but i guess when we take the preimage of those sets, we get back the original sets minus *, which are open in X
and hence we get a finite subcover from that?
tough imp
It’s equivalent to say that C is compact in the sense that open sets of X which cover it has a finite su cover (this covering mean it’s contained in the union)
And that open sets in C (under the subspace topology) which cover it have a finite subcover (in the sense that they union to C)
So it’s all internal to C’s topology as a subspace of X’ and X, and those agree
nimble flower
And that open sets in C (under the subspace topology) which cover it have a fini...
they could union to more than C
C just needs to be contained in their union, right
tough imp
No
That’s the point I made
The first message is the sense of open sets of C which cover it aka C could be contained
The latter says open sets OF C which are subsets of C
These are the form C intersect U with U open in the amebient space
The point is that we can talk about compactness of C in X completely in terms of C’s subspace topology
Is C a compact topological space via the subspace topology
So we don’t need to worry about open sets of X’ containing infinity, we only care about C’s topology as a subspace of X’
Which is the same as it’s topology as a subspace of X
Hence C is compact in X as well
nimble flower
hmm
okay i see what ur saying, i think
now this doesn't work when C contains infinity
tough imp
But then it isn't a subset of X
so it doesn't make sense to consider if its compact in X
nimble flower
yeah that's true
but we can still consider C - infinity and ask if that is compact
tough imp
I think it's false
let X be non-compact
then take C = X'
C is compact
but C minus infinity = X is not
the nbd of infinity just sucks up so much stuff
nimble flower
yeah that's true
paper crag
What's the most beautiful way for proving that inversions are conformal mappings and that circles remain circles under inversions ?
paper crag
Also, any book recommendation for studying advanced geometry for complex analysis, topology and other college courses at grad and undergrad level ?
paper crag
please be polite
daring bluff
This isn't actually well defined if `sin(θ0) = 0`, but your vector field isn't e...
thank you! Why isn't the vector field well defined?
summer jolt
Hey, how do I compute the derivative for a mapping from R^n to R given by
where b is a symmetric constant matrix
gritty widget
product rule and then maybe some playing around with the indices (try differentiating with respect to a fixed x^k first)
remember the derivative of x^i with respect to x^k is delta^i_k, which might help you simplify
sleek thicket
thank you! Why isn't the vector field well defined?
Because you're dividing by something which might be zero
shut moat
could be easier to just differentiate it in the form $x^T [b]x$
gentle ospreyBOT
~S^1
chrome dew
nah, easier the way it is, it's pretty explicit
$$\pdv{x^k} (b_{ij}x^ix^j) = b_{ij} \pdv{x^i}{x^k}x^j + b_{ij}x^i\pdv{x^j}{x^k} = b_{ij} \delta^i_k x^j + b_{ij} x^i\delta^j_k$$
gentle ospreyBOT
Merosity
chrome dew
$$=b_{kj}x^j + b_{ik}x^i = b_{ki}x^i + b_{ki}x^i = 2b_{ik}x^i$$
gritty widget
nooo nooo noooooooooo there's more than one index here i can't handle it
gentle ospreyBOT
Merosity
chrome dew
tbh it's not even necessary for b_{ij} to be symmetric, it's just a convenience since the result ends up symmetrizing itself anyways
and it might as well be from the start anyways, so w/e
shut moat
alternatively just $$(x+h)^T B(x+h) - x^TBx = x^T Bx + x^T Bh + h^TBx + h^TBh$$ $$= x^T Bh + h^TBx + o(h) = x^T Bh + x^T B^T h +o(h)= x^T(B+B^T)h$$ so the derivative is $x^T(B+B^T)$
idk, indices are gross lol
gentle ospreyBOT
~S^1
shut moat
but ig it's a taste thing 
chrome dew
lol I guess, I feel like there's too much worrying about row vs column stuff for me
gritty widget
pushing indices around is fun
frosty sundial
@gritty widget but I thought you liked RG?
gritty widget
"pushing indices around is fun" wasn't sarcasm 
elder yew
I opened a book and say
saw
$$ \mathrm{d}s^2 = \frac{\mathrm{d}x^2 + \mathrm{d}y^2}{y} $$
gentle ospreyBOT
MoonBears-C-
elder yew
It said this
"dx^2 + dy^2 represents the usual metric in the plane"
And I was like
You mean
(x-x_0)^2 + (y-y_0)^2
???
I am still confused as to what it means
Fkin book
shut moat
presumably it means the metric tensor doesn't it 
not a metric in the sense of a metric space
chrome dew
(x-x_0)^2 + (y-y_0)^2
that can't be a metric in general since d(x,x)=0 is a requirement for a metric. you can think of every point of space having its own separate vector space with the metric tensor dependent on this point, telling you how to measure vectors there at that specific point
shut moat
ngl I think the notation already implies it's a metric tensor
gritty widget
How would one start solving 7.5?
I have I Don't understand how to compute the beta polynomial of the mirrored trefoil, I missed a few classes due to health issues
Oh sorry I thought I wrote I haven't
I vaguely do yes
Oh thank you very much
@molten mica
unique juniper
Im not sure if this is where to ask this question but I am thinking of signing up for a differential geometry course. However, I enjoy the much more applied mathematics courses. So stuff like analysis I find very difficult. Is this course like a differential equations course or is it more of a proof based course if anyone knows
tough imp
Kind of depends as far as I know
It might be like a lot of calculus
Vector calc stuff
Or it could be much more theoretical
I’d ask people at your school who’ve taken it or even shoot an email to the person teaching it
sleek thicket
if you post the syllabus we could probably tell you
sleek thicket
but id expect it to be proof based (even if it's not super duper theoretical)
tough imp
I imagine it’ll be a proof based course either way
unique juniper
Okay thank you guys. I'll email the prof tomorrow its a little late now
tough imp
👍
You could just send it now
And they’ll reply when they see it
It’s not like they’ll get a notification, unless you don’t want to be viewed as a degen for being up too late hahaha
unique juniper
Yeah I just dont like sending emails very late. I turned in my numerical homework at like 3 am every time when it was due at 8 am haha. Thank you for the suggestions! Have a good night
rotund thicket
what i do is schedule an email send
for like some weird time like 6:47am or smth like that
elder yew
I send it at 4 am just to let them know that they cause me to suffer
They includes the voices
digital glacier
riemannian metric <:catThink:526983414226747393>
HyperKähler metric 
Kähler metric 
Riemannian metric 
river granite
Okay thank you guys. I'll email the prof tomorrow its a little late now
most email providers (particularly GMail) let you schedule emails to send them at a set hour/date btw
gentle ospreyBOT
Divya Ranjan
feral dragon
**Divya Ranjan**
Does this proof of the identity make sense, I mean it's supposed to be this way of course, but i tried it myself so just wanna check
uncut surge
I think this looks okay, I don't think you need to a separate case for n = 0, though. One thing you might want to think about for a moment is why the calculation in coordinates is really sufficient, since, of course, this only shows that d(\phi^* \omega) = \phi^* d \omega on this coordinate patch
@feral dragon
feral dragon
I think this looks okay, I don't think you need to a separate case for n = 0, th...
Umhmm....but it's going to be the same with any other patch I guess, but yeah let me think why just the calculation in co-ordinates would be sufficient
uncut surge
Yeah, that's true, but sometimes if you do calculations in local coordinates, it's not sure that you can glue all the local pieces together again to global objects
But here, you don't need to worry about that anyway: Both d(phi \omega) and \phi (d \omega) are globally defined forms, and you've shown that they're equal at every point, so the forms are equal 🤷♂️ It's just good to keep that in mind, because it's not always so straightforward @feral dragon
tough imp
Please forgive for the crosspost (in the algebra and geometry channel), but I think it’s justified. A lot of people here have heard about the moduli and stacks course that Alper is running this winter. Here is a link to the course website, in particular note at the bottom he’s included a way to contact him if you would like to informally join. Many people have asked me about this, so here’s the info: https://sites.math.washington.edu/~jarod/math582C.html
cursive flume
Can someone give me any hint/intuition what topology and smooth atlas to equip u(1) or gl(n) with to make them manifolds?
I have them as sets and i have an operation and i'd like to show they're lie groups
tough imp
One thing I forgot to mention: he said that he will record lectures and post them to the website as well. I don’t believe that’s stated on the course page.
sleek thicket
Can someone give me any hint/intuition what topology and smooth atlas to equip u...
GL(n, R) is an open subset of the space of n×n matrices, and the space of n×n matrices can be thought of as a R^(n^2) so it naturally has a smooth manifold structure
also be careful with your capitals here, gl(n, R) does *not* mean GL(n, R), it means the lie algebra of GL(n, R)
cursive flume
Is there a very explicit proof of your statement?
Iike to actually inherit the structure from it
sleek thicket
which part specifically?
any open subset of a smooth manifold inherits a smooth structure & topology
sleek thicket
take your atlas on $U \subseteq M$ to be all sets in the atlas on $M$ contained in $U$
gentle ospreyBOT
shamifold
sleek thicket
and yes the subspace topology
I'm assuming that we're looking at maximal atlases here
otherwise take your atlas to be $U \cap V$ for $V$ an element of the atlas on $M$
gentle ospreyBOT
shamifold
cursive flume
GL(n, R) is an open subset of the space of n×n matrices, and the space of n×n ma...
Why can we think of it as r^n^2
sleek thicket
just lay out all the elements of the matrix in a row
so in the 2d case $\begin{bmatrix} a & b \ c & d\end{bmatrix} \approx (a, b, c, d)$
gentle ospreyBOT
shamifold
sleek thicket
And GL(n, R) is an open subset since it's the set where a polynomial in the matrix entries is nonzero (the determinant)
also it turns out any vector space, like M(n, R), has an intrinsic smooth structure and topology, but imo it's easier to think of it as just M(n, R) ≈ R^(n^2) for now
cursive flume
also it turns out any vector space, like M(n, R), has an intrinsic smooth struct...
Can you pls elaborate on first
Intrinsic structure
sleek thicket
well I'd prefer to make sure you understand the stuff we were just talking about first
like, does the smooth manifold structure on GL(n, R) make sense now?
cursive flume
I got everything except the open set part
Why would it be open subset if det is not zero
Why would the det be the restriction in the topology
sleek thicket
If X is a topological space and f : X -> R a continuous function then { x in X : f(x) ≠ 0 } is open
cursive flume
Is r^n^2 equipped with standard top?
sleek thicket
yes
the reason for this is because that set is equal to f^-1(R\{0}), and R\{0} is open and continuous functions pull back open sets to open sets
cursive flume
If X is a topological space and f : X -> R a continuous function then { x in X :...
Right
sleek thicket
Since the determinant is a polynomial, it's a continuous function M(n, R) -> R
so the set where it's nonzero (ie GL(n, R)) is open
cursive flume
I understood this but gotta go board on flight home
Can we discuss the intrinsic topology later?
cursive flume
Yes
sleek thicket
So if you have a finite dimensional normed vector space, you can define limits
cursive flume
But on a vector space i dont have apriori an inner product
sleek thicket
Yeah, that's a good point
We'll get back to that
so if you have a fd normes space you can define limits and stuff
and make sense of open sets
(because a norm gives a metric)
cursive flume
I'll write in 2h when i get home to see where we'd get the norm from
sleek thicket
I'm going to be asleep
But have a nice flight :)
Maybe here's a better explanation: any choice of ordered basis gives coordinates on V, ie it makes V into a chart via the map V -> R^n sending a vector to its coefficients with respect to this basis. The transition maps R^n -> R^n are like change of basis maps, and in particular they're linear, so smooth. Since we defined this atlas in terms of all bases for V (instead of some basis) I like to think of this as intrinsic, although others might disagree
wicked obsidian
How can C be open and closed and
1/n (where n is an integer and n > 1) be neither?
bleak helm
Do you mean as subsets of $\bC$ ? Because {1/n} is closed as a subset of R or C
gentle ospreyBOT
Lunasong
red garden
number 2
what i said was suppose [A_I] be an open cover for Z in X now suppose [B_I] is an open cover for Z in Y --> B_k = Z intersects N where N is open in Y --> Y is a subspace of X --> B_k = Z intersects X intersects F where F is open in X
idk how to continue
would i just find the subcover of [B_I] and say each set is a subset of [A_I] ?
im getting confusedd and it sounds easy
red garden
ping if help
woeful oasis
@red garden You should start with an arbitrary cover of Z by open subsets of Y.
It looks like you're starting off with a cover of open subsets of X first, so try the other way around.
meager python
What are the logical dependencies in Hartshorne to read chapter III?
tough imp
You really only need up till II.5 to make good sense of it
That’s what I’m rolling with and up till now it’s been readable. this is possibly bound to change, but you don’t need a crazy amount of scheme theory to do cohomology
There’s an example in III.4 that uses differentials but you can just skip that
frosty sundial
this isn't a full answer but I think you have to be careful with how cup products interact with the boundary map in the mayer vietoris sequence
even when that map is an isomorphism, as you kind of noted the cup products land in different places
my instinct says that MV might be useful here but I'm not sure in exactly which wya
tough imp
Is anyone able to verify this map is a group homomorphism? Technically the set of torsors modulo isomorphism isn’t really a group, it’s in bijection with H^1(X,O_X) via some map defined two sections earlier
This is mainly like “has anyone verified this before, and if so, how”
tight agate
i think you can do it using cocycles
in fact that's how i would imagine the identification between pic and H1
yeah it should work with cocycles
take a line bundle
write down the data in terms of cocycles
BAM
it's an element of H1
translate the tensor product of line bundles to the language of cocycles
and you're done
tough imp
blech, I wanted to avoid cocycles lmao
that's like just doing it in terms of the Cech cohomology essentially right?
tight agate
yes
tough imp
I figured that there should be some way to verify this is a group homomorphism without going to cocycles, because I find it very odd that it would be presented at this point in time (before anything Cech is introduced) if it required going to cocycles
tight agate
what is your definition of Pic
tough imp
Invertible sheaves modulo isomorphism
The main issue is that 1: I don't really see what torsor the tensor product of two sheaves goes to
it isn't going to be the product of the two torsors you'd get separately
secondly, the map from torsors to H^1 is super opaque
I don't see how it really relates to the original torsor very much, you just kind of make an exact sequence, extend it, then take the image of 1 via some boundary map
eh actually I see how maybe you could get it by going to opens
you'd have to go local enough such that your torsor has sections, and you can track that through the diagram
any element of the torsor gets mapped to 1 inside the Z-sheaf on that open
so you can glue the images of those to figure out what 1 was sent to in terms of your torosr
but that's still pretty fucking opaque
tight agate
okay, I don't really follow everything you're saying, but the connection between vector bundles and torsors isn't bad
tough imp
Right, sorry this is about the diagram at the bottom of this https://stacks.math.columbia.edu/tag/02FN
I'm not sure if you're familiar with how you get a section of H^1 via a torsor
gah fuck, I have ideas now and it all involves just going to a trivializing cover of the line bundles
it seems kind of hellish hahaha
tight agate
maps to H1 (derived functor version) in general will be opaque as the only way to map to it is either by sticking it in an exact sequence or via its universal prop.
tough imp
yeah
tight agate
as that is how it's defined, it's the only way you have access to the object
tough imp
the issue is we have to verify that this composition of the map Pic -> Torsor/~ -> H^1 is a group hom
and Torsor/~ is just a set lol
so that's why I'm so confused precisely because it just says "we omit the proof it's a homomorphism"
but that's precisely the hardest part of this lmao
tight agate
Torsor/ ~ should have a nice group structure
tough imp
Nah
I mean... not that I can tell
direct product doesn't work
the action won't be transitive anymore
you can give it a group structure via the bijection with H^1 but I'm pretty sure that's just a super arbitrary operation
tight agate
I mean, the functor that assigns all G-torsors over a space is representable by a topological group
tough imp
like it doesn't line up with any natural operations on the torsors
tight agate
the so called 'BG' spaces
tough imp
tight agate
so it definitely has a group structure on it without looking at H1
tough imp
rip
Well, if I want to conclude the homomorphism by making Torsor/~ a group it will have to be the one you would induce by the bijection
so I guess I can try to think harder about that and see wtf happens
The construction of the map in the other direction constructs all the torsors as a subsheaf of an injective sheaf I with H a subsheaf of I
so there's nice represntatives of them all which might behave nicely
but if you're given arbitrary torsors you'd need to know which subsheaf of I it's isomorphic to in order to use that
I think...?
tight agate
okay, I do not have all the details in my head right now, but the correspondence is (vector bundles) <--> (cocycles) <---> Principal G-Bundles <---> Torsors
in general (outside the context of AG)
honest narwhal
Pfffffffffft
tough imp
Hey Dami
honest narwhal
I literally go to profs' houses on holidays to ask math questions
honest narwhal
"Yay it's the new yea... Oh who's there?"
"Yo Scholze mah boi what's a perfectoid space again?"
tight agate
lemme try to figure out a way to put a product on it directly
tight agate
does just taking the fiber product work?
tough imp
fiber product as...
sheaves
hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
as sheaves over what?
tight agate
oh wait, I'm probably confused by looking at the topoligical case
because in the topology sense, you can view the whole thing as some space sitting over your base
and the space above has a G action
that is free and transitive over each fiber
tough imp
ah
tight agate
so an example of a GLn torsor would be, take a manifold, look at the frame bundle of the tangent bundle
tough imp
I literally don't know what any of that means besides manifold and GLn
Anyway, I'm going to just move on from this for th emoment
I have been thinking about this for hours
and need to do other things with my life
tight agate
okay, but cocycles are your friend
tough imp
tight agate
pretty useful way to write down maps to cohomology groups
I have seen it used in ~4 instances
so yes, very useful
crap
it might be 3 instances
ffs
well 3 ~ 4 so that's okay
meager python
@tough imp doesn’t it follow from L \otimes Hom(L, O_X) \cong O_X
meager python
How is the group structure on Pic defined?
I don’t know this at all but since both structures are given by tensor products isn’t the morphism some form of evaluation map?
L^* is Hom(L, O_X) or?
tough imp
It’s by tensor product
But we straight up don’t have a group structure on the torsors I think. Or at least not one that’s obvious to me
And H^1 is just a cohomology group, it’s operation is abstract.
tough imp
Right, that's what I'm talking about
The torsors don't have an operation
the torsors are a sheaf of sets
we know that the torsors up to isomorphism are in bijection with a group, H^1
what it means for that to be a homomorphism is really that the compositon of that with the map torsors -> H^1 is a group hom
unless you just make torsors a group by stealing the operation on H^1 via the bijection
but that's literally the same thing as knowing that the composition is a group hom
meager python
So you have L |-> isom(O, L) in one direction and T |-> T \otimes_O* O in the other
willow spear
what is the link between topolgy and metirc spaces?
sweet wing
thonkzoom
sharp yoke
that was a joke right
coarse kestrel
all topological spaces are metrizable
wait fuck I can't tell if he's joking or not
sleek thicket
All topological spaces are either cringe or metrizable
chrome dew
all posets have a maximal element
gritty widget
all manifolds are smooth
All functions are continuous 🙂
all modules are vector spaces
coarse kestrel
All knots are rational knots
fading vale
my favorite vector space is Z
sharp yoke
silence monkeys
ivory dragon
@gritty widget classifies the finite simple groups
and also, by corollary, all the other groups
sure, but have you written your proof up in Lean?
elder yew
All curves are essential, simple, and closed
shut moat
Analysis is applied triangle inequality
velvet finch
you're applied triangle inequality
feral dragon
Everything is a matrix.
gritty widget
Every object is a set
uncut surge
every diagram is commutative
cursive flume
is the covariant derivative of a function equal to its partial derivative?
and only differs if we consider tensors?
i'm a bit confused by this claim
gentle ospreyBOT
slimvesus
cursive flume
i know this
but i'm kinda super confused of this statement
the wavefunction is not a function,cause it would be,nabla acting on it would yield the partial derivative
but if we look at the wavefunction as a section of a complex line bundle,we can make sense of the extra terms
chrome dew
is the covariant derivative of a function equal to its partial derivative?
they coincide if the metric tensor in your coordinate system at that point is the identity matrix
steep chasm
Und which conditions is the curvature 2-form of a principal G-bundle given by the exterior covariant derivative of the connection 1-form
Or when is not the case
summer jolt
Hey, I was trying to prove this exercise
In the solution to prove that the union of opens is open the following remark was given
Can someone explain how is one allowed to "pull out" the union outside the chart map?
What property allows for that?
tepid depot
@summer jolt how does your book define open?
U is open if each point has a coordinate neighborhood contained in U?
tepid depot
What property allows for that?
phi is a diffeomorphism (in particular phi is bijective)
summer jolt
A subset A is open if A phi(A intersected U) is open in Rn. Where (U, phi) is a chart
obtuse meteor
@trail apex
so like
There are two routes to this
I can talk about homology and maybe cohomology if I remember how it goes
or I can talk about de rham cohomology
which I should be learning formally next semester but which I can broadly describe
homology I can basically tell you the complete story
cohomology I always forget bc I'm bad at math
de rham cohomology I can tell you the full story if you blackbox what the cotangent bundle is
trail apex
@obtuse meteor sure
trail apex
Homology
obtuse meteor
cool
so first lets denote the standard n-simplex by Δ^n
now let's look at a topological space X
trail apex
What's an n-simplex
obtuse meteor
we call an n-simplex in X is a continuous map Δ^n -> X
the standard n-simplex is like
okay
Δ^0 is a point
Δ^1 is a line
Δ^2 is a triangle
Δ^3 is a tetrahedron
trail apex
Ah
obtuse meteor
something like this
trail apex
So it's the n+1 regular polygon?
obtuse meteor
hrm something like that I guess
but you want it like filled in
so you want the interior of the tetrahedron too
trail apex
Can you be a bit more precise Idk if n simplex is just the equivalence class or the shape
obtuse meteor
in analytic terms it's like the space {(x1, ..., x_n | x_1, ..., x_n >= 0 and x_1 + ... + x_n <= 1}
and that's Δ^n iirc
let me replace something because I'm dumb
there we go
trail apex
How does it describe a triangle for n = 2
obtuse meteor
obtuse meteor
ok so
an n-simplex in X is a continuous map Δ^n -> X
we define the group of singular chains in X to be the free abelian group generated by all n-simplices in X
this is denoted S_n(X)
trail apex
Isn't S_n the symmetric group
trail apex
Ok
obtuse meteor
anyway
now we want to construct a map which goes from S_(n + 1)(X) -> S_n(X)
which we'll call the boundary map
this is motivated by the fact that given any simplex Δ^(n + 1) you can look at its boundary which will be a bunch of n-simplices unioned together
obtuse meteor
intuitively you view a + b for like 1-simplexes a and b as following the line a then the line b or something of this sort
and you can build chains this way
trail apex
Ok
obtuse meteor
like if you link up a bunch of 1-simplices you can form chains
and in calculus you integrate over these bois
now
we want the kinda like oriented boundary of Δ^(n + 1)
as a motivating example
if we have a 1-simplex phi : Δ^1 -> X
this is a line
and it's boundary should be phi(1) - phi(0)
endpoint - initial point
think fundamental theorem of calculus vibes
(or Stoke's theorem if you're huge brain)
trail apex
this is a line
What's the line?
obtuse meteor
Δ^1 is a standard line
and this is like a curve in X
not really a line
a continuous curve
trail apex
Yeah well you didn't say that phi was continuous
obtuse meteor
some topological space we're going to look at homology of
not necessarily
not all topological spaces are metrizable
trail apex
Ok
trail apex
An n+1 simplex?
obtuse meteor
sorry
brian dum
you take the sum over i from 0 to n of (-1)^i * f restricted to the i-th face Δ^(n + 1)
but this doesn't quite work
because the i-th face of Δ^(n + 1) is not the standard n-simplex
trail apex
What's a face of a simplex
obtuse meteor
for a triangle it's like the lines which form its boundary
and for a tetrahedron its the triangles forming its boundary
trail apex
Ok
obtuse meteor
so you construct a nice embedding e_i : Δ^n -> Δ^(n + 1)
which embeds in the i-th face
trail apex
Yep cool
obtuse meteor
and then you tke sum over i from 0 to n of (-1)^i * (f . e_i)
which will be an n-chain in X
extending by linearity we get a map S_(n + 1)(X) -> S_n(X)
we'll call it d_n
you can prove by a bunch of bullshit
that the kernel of d_n contains the image of d_(n + 1)
it's basically a bunch of shitty computation with simplices
but the intuitive idea is that if you have a loop then endpoint - initial point = 0
and in higher dimensions you should have similar phenomena
trail apex
Why the (-1)^i
obtuse meteor
you want orientation
to get endpoint - initial point for curves
or something of this sort right
obtuse meteor
that the kernel of d_n contains the image of d\_(n + 1)
it's basically to guarantee that this works
obtuse meteor
call the kernel of d_n the n-cycles in X and the image of d_(n + 1) the n-boundaries in X
then the n-th homology group H_n(X) = n-cycles mod out by n-boundaries
trail apex
Ah
obtuse meteor
intuitively
if you have a cycle that's not a boundary of something
then there's a hole in the space where the cycle should be able to be filled in
and this hole should be n-dimensional
this means that the dimension of H_n(X) gives you the number of n-dimensional holes
at least in the absence of torsion this is a reasonable definiton
(torsion is when H_n(X) is not a nice free group that's like Z, Z^2, Z^3, etc)
obtuse meteor
so like
H_n(X) in a lot of cases
will be a freely generated abelian group
so like Z is the free abelian group on 1 generator
Z^2 is the free abelian group on 2 generators
etc.
but in bad cases H_n(X) is not freely generated
for example for the real projective plane
there's an n so that H_n(X) = Z/2Z
in algebra the elements of finite order of a group are called torsion
and it means that the holes have some weird property
which means like
you can have a loop in your space
that when you go around it once it's not nullhomotopic
but when you go around it twice it is nullhomotopic
which is weird
and kinda cursed
pseudo crane
what is the identity in S_n(X)
trail apex
You mean Z is a monogene
obtuse meteor
idk what that means but maybe
trail apex
Z/2Z is also a monogene with 1 as a generator
trail apex
It's cyclic
obtuse meteor
Z/2Z and Z are different
in that Z/2Z has torsion
and Z doesn't
which is what that comment was trying to capture lol
trail apex
Z/2Z is cyclic because it's a finite monogene
obtuse meteor
yeah it is cyclic
but that's not what I was trying to talk about in the comment ^^
trail apex
I don't get what you mean by torsion
obtuse meteor
it's just the elements of a group that have finite order
finite order means z^n = 0 for some z in your group and n a natural
trail apex
What about the nullohomotopic thing
obtuse meteor
well in abelian notation this should be n * z = 0
nullhomotopic is like using the fundamental group perspective on homology
we want to envision like
if we have a loop a
then a + a is going around a twice
so if we have an a such that a + a = 0
this means that going around a twice is the same as staying where we are
(up to homotopy)
trail apex
Ok
pseudo crane
so it's just the empty word
yes but to what mapping does the empty word correspond. the trivial map?
sleek thicket
S_n(X) is the free abelian group on the set of maps, not just the set of maps
obtuse meteor
so since H_1(RP2) = Z / 2Z
sleek thicket
S_n(X) is the free abelian group on the set of maps, not just the set of maps
the set of maps does not have an intrinsic group structure to it, so we formally add inverses and formally add things
obtuse meteor
that means there's a loop x : Delta^1 -> X so that x + x = 0
obtuse meteor
the set of maps does not have an intrinsic group structure to it, so we formally...
this is what I consider "the cheating at the heart of algebraic topology"
sleek thicket
omegalul
pseudo crane
oh ok
obtuse meteor
"it just works" (tm)
sleek thicket
pls attribute things correctly faye, it's "Noether's cheat"
obtuse meteor
🧠
sleek thicket
haha wow you have numbers? what if those were G R O U P S
obtuse meteor
GROUP
GROMP
the fact that homology actually captures holes is a fucking miracle lmao
at least imo
sleek thicket
is it alright if I ask a question in here? not sure whether you're done explaining
obtuse meteor
sure in some sense it's definitional
but like if you look at it for curves and surfaces we can see
it does a damn good job
and it's spooky
sleek thicket
lol
sleek thicket
I think graphs were what made me believe
there's one exercise in weibel iirc
which is like
simplicial homology of a graph counts holes
obtuse meteor
graphs are nice yeah
because like there it's like
you can make a reasonable definition
lol
I haven't done much exercises with homology and stuff besides poking around
hence why I'm taking an AT course next semester
to make me actually internalize it lol
sleek thicket
nice!! I'm taking homological algebra and am really looking forward to it
obtuse meteor
I think that's a reasonable intuition to have of it
n-chains are a very formal object
which makes them hard to think about in a nice way
at least imo
sleek thicket
this is why cohomology is more natural, since cochains are just functions assigning a number to each singular simplex in X. no formal addition needed!
im not actually sure how sarcastic this is lol
pseudo crane
wait are S_n constructed the same for simplicial cohomology
obtuse meteor
also it's kinda a lie lol sham
sleek thicket
wait are S_n constructed the same for simplicial cohomology
you look at group homomorphisms of S_n into Z
obtuse meteor
bc like if you have a cochain that immediately gives a map S_n(X) -> M
sleek thicket
and then dualize the boundary maps
sleek thicket
right
obtuse meteor
by free universal prop
and for the boundaries you like
definitely need to at least think in terms of the free thing
imo
sleek thicket
yeah haha the coboundary map looks a little sus without thinking about chains
okay big question time
So I proved Lemma 4.17.2 here https://stacks.math.columbia.edu/tag/09WN
and I want to use it to deduce $\mathcal{O}{X,x} = \mathrm{colim}{D(f) \ni x} \mathcal{O}_X(D(f))$ for $X$ an affine schemem. I don't see why the inclusion $\mathsf{Dis}(X)^{op} \to \mathsf{Open}(X)^{op}$ would be cofinal though, since it seems to fail condition (2) of definition 4.17.1 (where $\mathsf{Dis}(X)$ is the poset of distinguished opens). If I'm dualizing everything correctly then condition 2 says that for any distinguished nbhds D(f), D(g) of x contained in U we can find a common upper bound D(h) which is still contained in U, and I don't see why this is true (it's actually like a chain of upper bounds but still). Unions of distinguished opens aren't distinguished so...???
sleek thicket
lol
obtuse meteor
let me read
gentle ospreyBOT
shamifold
sleek thicket
it comes up a lot in AG and I had the intuition but never actually proved the genreal lemma lol
sleek thicket
also the stacks project definition seems stronger than the case of posets given on wikipedia https://en.wikipedia.org/wiki/Cofinal_(mathematics)
obtuse meteor
huh I see
I'm kinda not enough energy to answer this in full
but I'll think about it
obtuse meteor
sham it'll reduce because all diagrams commute in posets right?
so (2) should be automatic?
unless I'm brain dead
sleek thicket
right so commutativity isn't the issue
sleek thicket
yup
obtuse meteor
rip
sleek thicket
as far as I can see
obtuse meteor
that is what one might call
sleek thicket
lemma see if I can find a counterexample
obtuse meteor
an L
sleek thicket
lmao
sleek thicket
right
sleek thicket
exactly
obtuse meteor
mhm
wait
fuck
I hate this
so iirc in a poset
ok no this is fine
there's like a thing where arbitrary meets or joins implies you have arbitrary joins or meets respectively
and this is like
so fucked up to me
anyway
sleek thicket
lol
obtuse meteor
can you just take U <- D(1) -> V?
sleek thicket
you don't have binary meets in this poset though
nope, D(1) won't be contained in U still
err like
okay so we want an upper bound which is still contained in this y element
yeah?