They're both very friendly books I just think that Lee focuses more on the important topics for people who are going into math at large
#point-set-topology
honest narwhal
honest narwhal
So yeah basic analysis and algebra would be good then Lee topological manifolds
slow urchin
Thanks Sloth King,
I'm going to need to.read this as a refreshers as soon as I finish Tao.
graceful sequoia
Was studying differential geometry and came to vector fields. They are defined as functions that assign to every point "p" in R^n a tangent vector v_p to R^n at p. How can we write this in functional notation? Such as F: A ---> B?
gritty widget
the best way to do so is with the tangent bundle
gritty widget
sections are vector fields
"sections of the tangent bundle" is basically the best way to talk about vector fields that i know of
(i hope i didnt make any errors above)
it might feel a little complicated but it's actually a pretty nice construction (the tangent bundle) cause it also lets you talk about differential forms and stuff in a neat way
graceful sequoia
Thnks a lot!
gritty widget
it's through the tangent bundle that all this stuff (vector fields, covector fields, forms, etc.) gets generalized to smooth manifolds, so it's important to learn this if you are looking to go further in geometry (at least thats what the first half of my differential geometry course this summer has taught me)
graceful sequoia
I just understood the first picture, manifolds and everything else is too complex for me, haven't studied that yet
This answered my question, thanks a lot.
by tangent bundle, you do mean the tangent space yes?
gritty widget
yeah, it's fine to put off manifolds until later and to just start in R^n
no, the tangent bundle
graceful sequoia
Hm, are they different?
gritty widget
the union of all the tangent spaces (a bundle of tangent spaces!)
graceful sequoia
Oh
so each point in R^n gets assigned a tangent vector v_p, and that v_p is in the tangent space of R^n at p. How does tangent bundle come into play here?
gritty widget
the tangent bundle is the collection of all of the tangent vectors to every point
honestly if you're starting with R^n you don't need to worry about it
marsh forge
The tangent bundle at T_p (the vectors tangent to p) parameterizes all possible vectors you could choose
a section is a simultaneous selection of one such vector for each point
gritty widget
i only brought up the tangent bundle because it's a nice (and "the") way to tal about vector fields
ill let max j take it from here
marsh forge
Normally you put conditions on this choice (differentiable, smooth, continuous)
graceful sequoia
So every point has a tangent vector, and if we have the collection of all these tangent vectors, we have a tangent bundle?
I'm sorry if I sound stupid but it is a difficult topic for me atm, I just started it
marsh forge
Well, every point has in general many tangent vectors
I think you're looking at this the wrong way
graceful sequoia
Yes, but we are assigning just one?
marsh forge
The tangent bundle is an object worthy of its own study
graceful sequoia
It hasn't been taught to me yet so I do want to stay away from it for now..
marsh forge
and it tells us about all the vectors tangent to a manifold (e.g. R^n) at any given point
As a side effect, this lets us defne vector fields using the tangent bundle
but the tangent bundle isn't really defined using vector fields as such
However, your definition works just fne
we select a vector for each point
and we want these vectors to vary nicely i.e. if a point p is close to a point q, we don't want abrupt changes between v_p and v_q, intuitively
graceful sequoia
I see, thanks a lot.
wdym by the abrupt change? I mean if the points are different, they'll have different vectors assigned to them
marsh forge
Uh let me get a picture
graceful sequoia
Btw, is it possible to have same vectors assigned to them?
graceful sequoia
Ah.
marsh forge
anyway do you see how the vectors 'turn' smoothly here?
graceful sequoia
Yes.
graceful sequoia
Why?
marsh forge
Well it depends on your definition of a vector field, but almost always we make this part of the definition
Do you have a definition from your class or book?
They should state formally what I am describing intuitively
graceful sequoia
Definition of tangent vectors?
Because we haven't studied tangent bundle yet
sorry I replied late I was doing some irl work
Just that the tangent vectors to point p are the vectors with point of application "p" and all the vectors then make up the tangent space. Nothing about what you're saying atm, like the directions etc
Oh you mean vector field. The definition of vector field that my book has that it takes a point from R^3 and it assigns to it a vector.
That's pretty it I believe
But when I was asking my question, I wanted it to be in general, so R^n
and then wanted a functional notation for it, and someone else answered
nimble jolt
@graceful sequoia It is more than notation that is different. When you are talking about tangent vectors to R^n, you have the convenient/boring property that every one of your tangent spaces can be identified in a very natural way.
Which means if R^n is your manifold, your set of all tangent vectors could be written as R^n x R^n, and this coincides with the tangent bundle mentioned by others above.
graceful sequoia
I see, well thank you, I think I'll leave this for now. It is getting too complex
nimble jolt
On the other hand, if you are working on some arbitrary manifold, eg a 2-sphere, then each tangent space will be a copy of R^2, but there will not be a way to canonically identify these copies of R^2
Okay, I won't go further then. Just to sum up though, in general a tangent bundle of a manifold will only "locally" look like a product of a piece of your manifold with a copy of R^n, and there need not be a way to globally glue these together into a consistent product.
graceful sequoia
Any recommendations on a differential geometry book for independent study? Currently our uni has barret o neils book but imo it is not suitable for independent study
I could not find one in #books-old or maybe I didn't look thoroughly
viscid shadow
heey
not sure if this is appropriate for #questions, because it seems rather complex and particular
let's say the sides of the big rhombi are 1
what other info could one find about anything else ?
viscid shadow
i think you need to have more stuff defined to actually solve
honest narwhal
@viscid shadow this would go in #geometry-and-trigonometry rather than here
viscid shadow
oops
forest rover
dw
sleek thicket
I have decided to try and stumble through Hartshorne again
just going to spam here while working on problems
okay, so why do irreducible closed subsets have generic points? Lets consider the affine case first. If Z is an irreducible closed subset of Spec A, we can write Z = V(J) for some radical ideal J. I think irreducibility of Z will imply J is prime. If J is prime, it's easily a generic point of Z, since cl({J}) = V(I({J}) = V(J).
It doesn't feel like Z being irreducible is enough for J to be prime though... I know that if ab = J for ideals a, b, then a = J or b = J, but that says nothing about when ab <= J
okay so conversely if ζ is a generic point for Z, then V(ζ) = Z, so I think J has to be prime for this result to be true?
Hmm I have to use radicalness somehow, since the ab = J => a = J or b = J thing works even if J isn't radical
can I do galaxy brain things with integral=reduced + irreducible?
I have a closed immersion Spec A/J -> Spec A which gives a homeomorphism between Spec A/J and V(J), so Spec A/J is irreducible. It's reduced since J is a radical ideal (reduced iff reduced at stalks, and the localization of a reduced ring at a prime is reduced)
so Spec A/J is an integral scheme, and in particular its global sections A/J are integral, so J is prime
This is a terrible proof
@tough imp u up?
tough imp
Yeah I am
tough imp
Sure
sleek thicket
With J radical
I want to say J is prime
My proof is that with the reduced induced closed subscheme structure, V(J) is integral, so its ring of global sections A/J is an integral domain
Since it's reduced+irreducible
this feels bad lol
There's gotta be a better proof
tough imp
So the way I did this right
There exists a minimal prime P containing J, so we get that V(J) < V(P), but V(P) = {P} closure so that V(P) < V(J)
tough imp
You can do Zorn’s
If you can show minimal@primes exist you can show they exist for primes containing an ideal
sleek thicket
Hmm okay I'm more okay with my proof then
Oh right
Just take a minimal prime of A/J
derp okay
But still, if P contains J then we'd have V(P) contained in Z, right?
tough imp
Yeah so they’re equal
Although I’m thinking about this for a second
And now I forget how I get that inclusion you pointed out
sleek thicket
But why do we get V(P) >= V(J)?
tough imp
I might have fucked this up lmao
sleek thicket
Hartshorne proves integral=reduced+irreducible after this problem
But I don't really care
ohhh
tough imp
Not containing
sleek thicket
So you do need a new Zorn
tough imp
Yeah but it should work all the same
sleek thicket
Sure
marsh forge
cursed notation for the stable homotopy category of G-Spectra with respect to a universe U
$\tilde{h}\mathcal{GSU}$
gentle ospreyBOT
MaxJ:
sleek thicket
yeah?
sleek thicket
ugh is that the adjoint
tough imp
Yes
sleek thicket
If so fuck off
tough imp
Then said I had strong reason to believe it’s true
I have since done the entire thing
And ffs
Honestly the time I got like 75% of the way done
I felt more convinced it’s true/ felt like I saw why its true
More than I do now after having actually done it all
sleek thicket
lol
sleek thicket
I mean the important thing for actually working with it is that I know how to translate functions back and forth
tough imp
I did literally all of it except this one last part without reference to elements
Only in the last bit I just could not fucking do it with diagrams
I was honestly disheartened
Then remembered I know what an element of the f^-1 sheaf is
By like, pretty much the same construction as the stalks
sleek thicket
lol
tough imp
I gotta say tho, there’s a problem about an inverse limit sheaf I think
sleek thicket
sometimes elements...are good
tough imp
And like
My proof was just limits commute 😎
Then for the direct limit you have to sheafifify
And so I showed the presheaf one is the direct limit in presheaves
Then
Sheafification is adjoint to forgetful
😎
Honestly after getting used to sheaves and stuff
sleek thicket
remember when you used to get mad at me for saying words like functor and natural transformation
Good times
tough imp
The hardest problem there was that adjoint and the fucking problem about the flasque sheaves lol
sleek thicket
Ugh I should go back to section 1 shouldn't I :/
tough imp
Honestly
sleek thicket
The flasque stuff at least
tough imp
Not that crucial IMO
It’ll show up eventually
Honestly there’s a lot of stuff in II that are important
sleek thicket
Sure, I'll wait until sheaves of modules then
tough imp
I think like II.2.18 is the one
sleek thicket
I'm probably going to give up on this in a week or w/e
But it's good to keep me occupied right now
tough imp
Relating properties of affine scheme morphisms to the one on rings
That ended up being clutch
sleek thicket
Oh yeah I really want to do that one
tough imp
And the one about the generalized open sets
tough imp
Generalized distinguished opens sorry
X_f for a global section f
The set of x such that f_x not in m_x
sleek thicket
Ah yeah
tough imp
Doing that
sleek thicket
2.16?
tough imp
Yeah
That and me showing isos of schemes respect distinguished opens
I.e. they send a distinguished open to a distinguished open
tough imp
And it’s the one it should be, like X_f goes to Y_phi^#^-1(f)
That made me finally accept the identifications we do of affine open subsets
To their actual schemes that is “physically” SpecA
And like the simultaneous distinguished opens thing you know?
sleek thicket
Yeah
tough imp
Can cover the intersection of two open affines
tough imp
Pretty easy using the language of the generalized distinguished opens
sleek thicket
time to break out vakil
I just feel like starting from mostly scratch
and doing more problems
tough imp
Namely that X_f \cap Spec A = D(f)
Yeah
I did that too
And it was a great help IMO
sleek thicket
alright I got the reduction to affine done
Problem 2.9 complete
Probably not doing 2.10 or 2.11 lol
Maybe I should do 2.11?
The residue fields are the finite fields of characteristic p. How many is combinatorics
Ig I could look at the factors of x^(p^n) - x and set up a recurrence relation? Sounds dumb
ughhh do I really have to do the gluing lemma
I probably have to do the gluing lemma
Ughhhhh
alright the dream is dead
spark skiff
I need help T-T.... I cant compute using light cones 😦
marsh forge
You might get more help in #old-network on the physics server
tough imp
@sleek thicket apparently counting it used mobius inversion
I looked it up since I had literally no idea wtf to do for that
And was like ¯_(ツ)_/¯
Also I started and should finish 2.11
This is the first time I’ve had to like glue shit so I was really hazy on how it worked
This sucked mega hard when I proved the scheme fiber and set theory fiber are homeomorphic since I had to dive into the construction of the fiber product as this glued piece of shit
But if you can glue manifolds together in a similar way I imagine it’ll be less mysterious to you what’s going on
sleek thicket
You absolutely can't
I think
manifolds are way harder to glue
You get singularities
But schemes are just like "eh, who cares about singularities"
Yeah think about two lines
And just two intervals on them
those don't glue to a manifold
anyways I did a problem on the fiber product with Jessie a while ago and the construction finally made sense
marsh forge
this is more due to the fact that manifolds don't play nicely categorically
like if you ever considered a category of manifolds
you get literally nothing nice lol
sleek thicket
yeah, don't glue = don't have pushouts
marsh forge
they aren't closed under anything
manifolds were the historic humans first version of CW complexes
Really Top is just a model category for hTop, the actually useful category
tough imp
I really really love proofs which are show existence and uniqueness
And to show existence you need to glue stuff locally, and in order to do that you need to know if something exists it is unique
I smile every time it happens
They seem to occur in AG a lot
marsh forge
it happens in AT too
although a lot of times something ends up only being unique up to homotopy (homology)
but then you remember you're only working up to homotopy!
and u smile
sleek thicket
shows up a lot in manifold land, except you get to skip uniqueness because partitions of unity are OP
imagine having to make sure things are equal on overlaps LMAO
marsh forge
imagine even believing charts exist
sleek thicket
ahh good point, you probably need to use choice somewhere
Sorry for disrespecting your anti-choice beliefs max
sleek thicket
does anyone want to explain Proj to me
The structure sheaf is weird
I feel like it's the sheafification of some simple presheaf but I'm not sure what that should be
O(U) = degree 0 elements in the localization of S with respect to T = { homogeneous f in S : f not in p for any p in U }
maybe?
Something similar happens with the structure sheaf on Spec
@tough imp does this look right to you?
tough imp
Honestly, I have no idea what it might be the sheafification of
I related it to the structure sheaf on Spec and saw how it looked like that
and by that time I was actually really comfortable working with the structure sheaf on Spec
@sleek thicket
sleek thicket
I mean, I'm not not comfortable with it
tough imp
Proj, Spec, both?
sleek thicket
But I felt like I understood it better when I figured out it was the sheafification
Structure sheaf on Spec
tough imp
Ah yeah so I didn't really look at it that way
sleek thicket
and it was even more clear when max talked about it as a sheaf on a basis
tough imp
I remember you mentioned it in the winter
but after having worked with it for a while I got reaaaaaally used to working with it as functions such that...
I'm getting to the point Sandor talked about where functions like that seem exactly like tuples
sleek thicket
...
tough imp
but I am talking about
FEELINGS
Yeah I know I did
because I didn't feel it
and I asked Sandor if it's okay to just pretend when I make the free group or whatever I treat it as tuples even in the uncountable case in my head
and he actually sort of said you can it's just hella annoying
but he said to him they just seem like literally the same thing at this point
sleek thicket
okay so you can just do the same sheaf on a base thing
Which is nice
degree zero elements of S_f
marsh forge
The only thing woker than sheafification is spectrification
Bc the model cat of omega spectra is not closed under a bunch of stuff
So whenever we do something not closed
We take spectrification
The issue is that all the experts know when we do or dont do this
So they omit the notation
sleek thicket
Lmao
loud scarab
How is the 2nd condition for basis verified here in the definition of a subbasis? The 2nd condition demands that if x is in the intersection of two basis elements then there exists a basis element such that it contains x and contained in the intersection, but here the proof only shows that B1 intersected B2 is itself B3?
gritty widget
this looks like its out of munkres so they're probably referencing this when they say B_3
loud scarab
I need to show there is a basis element contained in B1 intersected B2
gentle ospreyBOT
slimvesus:
loud scarab
but the last line says that B1 intersected B2 is itself a basis element
gentle ospreyBOT
slimvesus:
loud scarab
so the subset symbol here
means subset or equal?
well doesnt that make the 2nd condition trivial
the condition says there has to be an element contained in the intersection of two elements that contains x
but this will be true for any topology right?
if B3 is itself B1 intersected B2
Well what does the 2nd condition really demand
Ohh
Got it
the renaming of B1 intersected B2 to B3 made it a bit misleading
thank you
Yes
its the 2nd section
Gonna hopefully figure it out
in any case, thank you 🙂
dense sundial
I know very little differential geometry, and am curious about calculating things like mean/gaussian curvature of a surface. I found a million sources online for the case when the surface is embedded in 3D. But what if it's embedded in 4D or something? Is there a good resource/book I could jump right in to a calculation like that?
honest narwhal
@dense sundial Riemannian geometry (e.g. in Do Carmo) studies similar ideas for arbitrarily k-manifolds in R^n
gentle ospreyBOT
slimvesus:
honest narwhal
Tbh in hindsight it wasn't a relevant assumption I just knee jerk stick my manifolds in R^N when I don't wanna think about it
But yeah basically the idea is that dF(x,v) = (F(x),dF_x(v))
And at least in the case where our manifolds are in R^N, but really more generally when you have local coordinates, that's what it looks like in coordinates
sleek thicket
@honest narwhal I'm gonna explain Proj to you
tough imp
Ping me when you want to talk about b) cuz I'm gonna be coding in the meantime
honest narwhal
So like, in general for rings
tough imp
If you even do
sleek thicket
So define D+(f) = { p in Proj S : f not in p }
@tough imp sure
for f a positive degree homogenous element
honest narwhal
Let's say A is a ring, S multiplicative subset. Then prime ideals of A dodging S are the same thing as prime ideals of S^{-1}A right?
sleek thicket
Right
honest narwhal
That should just be taking the preimage
sleek thicket
This is how it works for Spec
Like D(f) being iso to Spec A_f
the issue is that D+(f) isn't iso to Spec S_f
It's iso to Spec S_(f)
Which is the quotients p/q in S_f where p and q are homogenous of the same degree
This should be familiar from variety land
This is what regular functions on open subsets of P^n look like, right?
honest narwhal
Yeah
Is S_(f) just notation here or is this a localization? (f) isn't automatically a prime ideal is it?
honest narwhal
I guess the idea might be something like
(f) isn't automatically a homogeneous ideal so you don't get the nice grading on S_f
So you find some subring which does still preserve gradings maybe? Hmm
sleek thicket
(f) is a homogenous ideal
honest narwhal
Idk I'm shooting at the dark here
sleek thicket
Since f is homogeneous
sleek thicket
I mean I think I get why it's worth looking at S_(f)
In analogy with varieties
I just don't understand how the primes correspond to homogenous primes of S not containing f
(and not containing S_+)
honest narwhal
Hmm
sleek thicket
Okay so Hartshorne says that you take a prime p of S
You extend it along S -> S_f
Then you contract along S_(f) -> S_f
So there's like a span or smth
It's not even clear this preserves primality
honest narwhal
Wait what's S_(f) -> S?
honest narwhal
Ah gotcha
sleek thicket
S_(f) is a subring of S_f
honest narwhal
Well if you give me a prime ideal in S that doesn't contain f
You push it across to S_f, by the general Spec business it's still prime
Preimage of prime ideal is prime
honest narwhal
Fucking amazing
sleek thicket
anyways I think that makes sense now
Extending along S -> S_f preserves primality
Since we don't contain f
Okay this is good enough for me right now
I'm not gonna worry about why it's a bijection yet
honest narwhal
I actually wanna think about this hmm
sleek thicket
I'm not vibing with it yet
honest narwhal
I have it
Okay so the map from D+(f) to Spec S_f
The problem here is it can't be surjective likely
If it weren't injective then that'd get fucked even harder once we go to S_(f)
tough imp
I actually wish I wrote up my work in understanding Proj fuck maybe I scrawled some stuff in my notebook but oof
honest narwhal
So basically we have some prime ideals of S_f which are no bueno
sleek thicket
It's not surjective since we're ignoring all the nonhomogenous primes
it's injective by properties of localization
D+(f) is a subset of D(f) in Spec S
honest narwhal
Oh wait so in that case are prime ideals of S_(f) just gonna be the homogeneous ones in S_f or something?
sleek thicket
Hmm maybe?
So S_f is Z graded right
nah that doesn't sound right to me
Idk
I mean your thing doesn't sound right
It is Z graded
honest narwhal
But lemme still try
sleek thicket
weird stuff
Okay so first off, why is injective?
Once we contract down to Spec S_(f)
suppose $(p S_f) \cap S_{(f)} = (q S_f) \cap S_{(f)}$
For p, q in D+(f)
gentle ospreyBOT
shamrock:
sleek thicket
Right
But I don't see how to get that
How to get pS_f = qS_f I mean
Oh I see
I think it's easy
So what does the intersection with S_f look like?
It has elements like g/f^k
honest narwhal
So just to recall, S_(f) is the set of x/f^n in S_f such that deg(x) = n deg(f) right?
sleek thicket
Right
so if g/f^k is in p S_f
then g has to be in p S_f
by primality
Right?
and then actually g is in p
I think?
honest narwhal
So first part I buy because 1/f^k couldn't be in pS_f. Now if g is in pS_f, then g is a sum of things in the image of p
g = g_1/f^{k_1} + ... + g_n/f^{k_n}?
And then clear denominators
So yeah I buy that g is therefore in p
Because of primality again and f isn't in p
Sorry I'm not used to this so I'm trying to go line by line to be sure
sleek thicket
Yeah no worries
I find it easy to assume too much about localization
So it's easy to screw me up
But p S_f is stuff of the form g/f^k for g in p I think
which is 👌
honest narwhal
And I guess we can split g into homogeneous parts
sleek thicket
Okay so injectivity follows from the fact that the denominators of stuff in S_(f) is super constrained?
Hmm no I'm being too sloppy
I think I need to do this later, sorry
Too sleepy
honest narwhal
Just to be sure let's say p \ne q. Let's say g is in p but not q, in fact I think we can choose g such that deg(f) divides deg(g)?
That would do it
I think
honest narwhal
Because then g/(the right power of f) would be in (pS_f)\cap S_(f)
sleek thicket
So g^(j) not in q for some j
so wlog g homogenous
And this is where we're using homogeneity of p and q
and then we can look at g^n/f^m
sleek thicket
yee
yeah you can't state deg(g) the way you were
I guess you can define it to be the sup of all homogeneous degrees which aren't zero
sleek thicket
right I'm just saying you need to take g homogeneous
honest narwhal
Oh yeah
sleek thicket
So you need to say that if p ≠ q there's a homogenous element in one but not the other
honest narwhal
But yeah okay that should do injectivity for us
sleek thicket
Okay so for surjectivity, if p is a prime of S_(f) let q be generated by all numerators of stuff in p
this is automatically homogenous
Why doesn't it contain f?
Since it's a homogenous ideal, it suffices to check primality on homogenous ideal
And I think you can go up to p
with homogeneous elements
Okay so lemma
If x in S is homogenous, x in q iff x/f^n in q for some n
Backwards is definitional
If x in q, then x = g1 + … + gn
For gi such that gi/f^ni in p for some ni
does that make sense so far?
By definition of S_(f), each gi has to be homogenous too
and then they all have to be homogenous of the same degree
err, maybe some cancelation occurs
Like maybe g1 + g2 = 0
sleek thicket
right
honest narwhal
S_(f) is (S_f)_0 I think
honest narwhal
Ah that's coo
sleek thicket
yee
Graded stuff seems cool
anyways my lemma
suppose x is homogenous and x in q
Then x = g1 + … + gn for gi numerators of stuff in p
By definition of q
Each gi is homogenous
By looking at the deg x'th homogenous part of the right hand side, we can assume wlog that each gi is homogenous of the same degree as x
Does that make sense?
honest narwhal
So I'll read in a moment but since I may go to sleep soon I fucked around a bit on Stacks
And there is a relevant lemma if you wanna see
sleek thicket
Sure
honest narwhal
honest narwhal
Woo
tough imp
Das coo
Honestly the entire problem is pretty much just "prove something about graded rings"
Which is why I managed to do it lol
sleek thicket
yeah lol
so far I haven't had to think about schemes much
Every problem has been immediately reducible to algebra
Section three though...
sleek thicket
maybe I'm just not doing the hard exercises yet
tough imp
Oh, the latter ones involve thinking about schemes a bit more
sleek thicket
I mean I skipped the gluing one
sleek thicket
I think I worked out gluing of schemes in painful detail months ago
Oh actually I did use scheme stuff
tough imp
Did you do the uhhh
sleek thicket
uhhhhh
tough imp
Nvm
I was thinking of some other cool thing
but it was just the cool way to do that problem LOL
sleek thicket
The generic point one?
tough imp
Oof yeah lmao
sleek thicket
Also there's no way I'm going to keep doing this
tough imp
Lmao I like when I read a proof
sleek thicket
Maybe if I actually get covid
tough imp
and it's like "isos on stalks lol"
sleek thicket
good shit
tough imp
Like literally I have this thing thats like
homeo is immediate, note these sets cover X
sheaf maps are isos on stalks
and that's my solution
sleek thicket
Lol
sleek thicket
oh lol yeah
I think I see it
it's obviously an iso on stalks
And also a homeomorphism
so yes
lol
sleek thicket
l o c a l l y
tough imp
Honestly sheaves op
imagine showing presheaves are isomorphic
pffffffft
Can't even just go to stalks, nerds
sleek thicket
I mean here's another proof that should work
the gluing lemma for scheme morphisms holds
Apply that to the inverse maps
tough imp
That is what my proof was basically
but showing the local isos go global is like haha stalks
sleek thicket
But the stalk thing is different
You don't really even refer to the inverse maps at all
tough imp
How do you show gluing isos together is an iso globally for sheaves?
w/o just hopping to the level of stalks
sleek thicket
I'm saying that if you have an open cover {Ui} of Y, and scheme morphisms fi : Ui -> X which agree on Ui cap Uj, you get a scheme morphism Y -> X
tough imp
Oh yeah I get that
sleek thicket
apply this lemma to the inverses of the maps f^(-1)(Ui) -> Ui
tough imp
Ohhhhhhh
sleek thicket
and you get a global inverse
sleek thicket
this is how you prove e.g. a bijective local diffeomorphism is a diffeomorphism
Or even that a bijective local homeomorphism is a homeomorphism
At least it is in my brain
tough imp
Ah in my brain that isn't how I thought of it
But yeah that makes sense
Yo did you see my thing in #groups-rings-fields
sleek thicket
no ill look
tough imp
so the map f*:SpecB -> Spec A being surjective is equivalent to f: A -> B having the going up condition
but apparently it's also equivalent to this other thing
shut moat
Not sure if this is the right place, but I've been trying to find the gaussian curvature of an ellipsoid in cartesian coordinates using its taylor polynomial
but idk why I'm failing so hard 😂
From using coordinates, the gaussian curvature is $$K = \frac{a_{2,0} a_{0, 2} - a_{1,1}^2}{(1 + a_1^2 + a_2^2)^2}$$
gentle ospreyBOT
ApproximatelyASphere:
shut moat
where a1 and a2 are the linear coefficients of the taylor polynomial and all that
Except the expressions for these things get absurdly large
$a_1 = \frac{cu}{a^2}\sqrt{1-\frac{u^2}{a^2}-\frac{v^2}{b^2}} + \frac{u}{2a^2} - \frac{u^3}{2a^4} - \frac{uv^2}{2a^2b^2}$
gentle ospreyBOT
ApproximatelyASphere:
shut moat
$a_2 = \frac{cv}{b^2}\sqrt{1-\frac{u^2}{a^2}-\frac{v^2}{b^2}} + \frac{v}{2b^2} - \frac{v^3}{2b^4} - \frac{u^2v}{2a^2b^2}$
gentle ospreyBOT
ApproximatelyASphere:
shut moat
and then squaring it and inserting it into the above expression does not give anything remotely pretty, and mathematica isn't able to simplify it
am I approaching this wrong?
marsh forge
Can you take a screenshot
willow spear
why do you need to make it look it different tho
limpid mural
Loring Tu in his book $\textit{Differential Geometry Connections, Curvature and Characteristic classes}$ defines the directional derivative of on a submanifold of $\mathbb{R}^n$ in this way: $\forall p \in M$ if $X_p$ is a tangent vector at $Y$ and $Y = b^i \partial_i$ is a vector field tangent to M in $\mathbb{R}^n$ then the directional derivative $$D_{X_p}Y = X_p(b^i)\partial_i.$$ My question is: why is it defined in this way? Why don't we derive along X the vector basis $\partial_i$ too?
gentle ospreyBOT
emme:
limpid mural
and a vector field along M in $\mathbb{R}^n$ is a vector field defined on $R^n$ such that $\forall p \in M$ you have $X_p \in T_pM \subset T_p\mathbb{R}^n$
gentle ospreyBOT
emme:
limpid mural
I think this is the key to understand the definition, probably he's reading $X$ in the "canonical" basis of $\mathbb{R}^n$ so it's possible for the directional derivative of $Y$ along $X$ to not be tangent to $M$
gentle ospreyBOT
emme:
loud scarab
With this theorem in hand, how is it possible that the ordered square is not convex in R2? Isn't every interval connecting two points in the square contained in the square?
uncut surge
google says dis
wanton marsh
then the square is not convex in R²
uncut surge
Yeah, if you have for example the point (0,0) and the point (1,0), the dictionary-ordered interval between the two contains all elements (a,b) with a between 0 and 1 and all real numbers b, I think. Something along that line, depending on how you define your order
In any case, some intervals between elements will be unbounded, so they cannot be contained in this bounded square
loud scarab
Oh I'm sorry for not defining convex set
But I got it, it's because of the dictionary order that makes the intervals between elements unbounded, thank you
gentle ospreyBOT
slimvesus:
nimble jolt
@gritty widget If F was just required to be smooth you could take F constant. This choice of F makes both compositions trivially smooth for completely arbitrary G.
are you talking about a or b?
That equality looks a bit nonsensical to me, the things on each side are restrictions of different functions to different sets.
You will need to use invertibility at some point to peel off the F.
sleek thicket
The F|U(U) \cong F(U) thing doesn't really make sense
You should just say "... such that F restricts to a diffeomorphism U -> F(U)"
the equality G = (G ° Fhat) ° Fhat¯¹ is also a little ehhh
I normally wouldn't say anything
But the point of this problem is to be careful about domains
Yeah
Haha sure :P
He's actually on discord
On my department's lounge server
My 78 year old freshman analysis professor followed me on Twitter earlier
dim meadow
lmao
sleek thicket
Also is there any specific notational trouble you're having? I could try to help
Oof
dim meadow
I don't think lee's notation is bad, I think manifold notation is necessarily very dense
dim meadow
lol
sleek thicket
That's in the intro to ism I think
small obsidian
I do like Tu's book as well, but lee takes the cake in many places, and is a more full book
sleek thicket
What's your previous background in math?
small obsidian
Lee gets thicc in some places
elder yew
78 year old analysis prof at UW? Is that Marshall?
small obsidian
Tu has a nice consistency, but likes to hide important facts in proofs
sleek thicket
nah, morrow. I don't think he's published in decades lol
small obsidian
I've not heard that one! Lol. I wouldn't know, personally
sleek thicket
huh, can't relate
We skipped three chapters over two books
homology in ITM, de rham's theorem in ISM, and symplectic stuff in ISM
small obsidian
I definitely don't know all of Lee. That tubular stuff just was like "Nop"
sleek thicket
I was deeply confused about that stuff
Decided I'd just skim
not a big deal
Then my homework that week was all about slightly improving the results
small obsidian
"I really don't care" kind of chapter. I could be wrong and maybe could revisit
sleek thicket
And modifying the proofs
Nah it's really good stuff
Tubular neighborhoods are super important. Idk about how big of a deal it is to know the proofs though
small obsidian
Okay I guess I'll see what all the fuss is about haha
elder yew
Tubular neighborhoods is used a lot in Low Dimensional Methods
I don't know if it's popular in higher dimensions
sleek thicket
I'm pulling up the third quarter schedule
elder yew
But it was the pathway to more sophisticated techniques
sleek thicket
Then 3 and a half weeks to get up through 12
So 10.5 weeks I guess, over two quarters
yeah that's too much
our pace was brutal
and it was about a chapter a week, sometimes more
Why do you need to?
you won't get 75% in 6 weeks
small obsidian
What's a good book for after Lee?
sleek thicket
I would recommend against this plan. Just read Lee during your manifolds course and ignore the prof
small obsidian
Not that I'm necessarily ready for that but I at least want to see
sleek thicket
I mean that sucks
But you can't learn all of this in 6 weeks
You'd find it hard in 12 weeks without peers or a teacher
elder yew
You use them in the proof of Waldhausen's
small obsidian
I see it as possible
sleek thicket
It's a ton of material
small obsidian
But I don't know diff geo
elder yew
Waldhausen says that the Heegaard Splittings of S^3 are unique up to isotopy
Yes!
The proof, at least the one I learned, uses tubular neighborhoods
TO push things around, slice things up
Ok, so I gotta go read it haha
But they're used in the very definition of heegaard splittings
I'm trying to find my PDF of Schulten's three-manifolds
Ehh it's something I should know
I did a whole year on this stuff, so I should be able to answer basic questions
So Heegaard splittings are used to decompose 3-manifolds into two pieces, which are called handle-bodies
To construct the handle-bodies you rely on tubular neighborhoods
This was at the very end of my 3-manifolds class
So it's a little hazy
In general, it just seems like a way of classifying 3-manifolds into their handle-body decomposition
Since you can tell 3-manifolds apart using various properties
The handle-bodies have an explicit construction that's easy to check for topological/geometrical properties
And then you can piece it back together to say something about the original 3-manifold
I wouldn't be surprised if you could
I am by no means a MCG expert
However the natural connection between these areas, namely the
study of the mapping class group Aut(M, Σ) of a Heegaard splitting
Σ, has not been systematically investigated. One key to understanding
Aut(M, Σ) is the observation that an element of the kernel of the map
from Aut(M, Σ) to Aut(M) corresponds to a non-trivial loop of embeddings of the Heegaard surface. In fact, for a hyperbolic 3-manifold the
space of embedded surfaces isotopic to Σ is a classifying space for the
kernel [18]. Just as strongly irreducible Heegaard surfaces define minimal surfaces of index one, such loops of embeddings should determine
minimal surfaces of index two. See Bachman [2] and Hass, Thompson,
Thurston [12] for further developments of similar ideas.
- From that paper
So it seems like a very promising research area
This is a very interesting quesiton
That sounds about right
There's a book
"A Primer on MCG"
One of my profs colleagues read that thing in his first year of grad school
He said it was one of the best decisions he ever made
You just gave me a rabbit hole to chase down Jan
I hope you know the rest of my summer is ruined
I'm planning on putting up a math blog soonish
small obsidian
MCG?
elder yew
Mapping Class Group
elder yew
This is the only way someone can convince me algebra is cool
The MCG of the sphere is trivial, for the Torus it's SL_2(Z)
I think of it having "one way to orient yourself on the sphere, two ways of orienting yourself on the torus"
But again, I'm just some lazy student
So don't believe anything I say
We ran a grad student seminar for half a semester on MCG and pair of pants decompositions
Costco pizza, soda, lots of fun
If I get into Cal, I'm probably doing LDT
but I won't get into Cal
sleek thicket
Manifolds with corners are an infinity category?
We only talked about them extremely briefly to prove homotopy invariance of integrating closed forms with stokes
ahh okay sure
elder yew
I'm gonna pull a reverse Larry Guth: do a PhD in Analysis, switch to LDT, and not be a great mathematician
(He is a great mathematician, did LDT, and went to Analysis)
sleek thicket
my plan is to learn all the categorical nonsense that I can and then move into pdes instead
galaxy brain
elder yew
My only goal is to be the best-god damned prof I can be
And try research
I remember my revelation at 8 years old "Professors get paid to sit around and think without manual labor? Sign me up!"
Anyways, I need to learn more topology stuff
You can study Diff. Geom. without going through Lee's topological manifolds
What text does the diff geo class take?
If it's spivak you can probably jump right in
I never really used any Lee, but you can get pretty far without reading Lee
I skipped into Riemannian Geom.
I just read spivak for diff. geom on my own
Oh ok
No, when I took Riemannian Geometry at LA from Petersen he kinda just let me in
We used a combination of Petersen and some other obscure text
I had taken a Calculus on Manifolds class at my Community College, which got up to Riemannian manifolds, and covariant derivatives
And I was able to TA for that class twice before I took Riemannian Geometry
So I informed Petersen of this, and he decided I was prepared
I think you'll probably be fine. Auditing is a good idea
Enroll only if you think you have the time to commit to the class, and are reasonable in your ability to do well
But I think you can get the main ideas through auditing
In my seven years in university, there isn't a single term I didn't skip a pre-requisite in some form or another
Y'know? You just take courses and pick it up as you go
Sometimes ppl get overly caught up in "proving everything in the book"
Well the book exists for a reason!
For you to read it. Don't waste your life re-doing results that are already done
Get through it, go do something you find interesting
You'd be surprised at how much you can contribute to research with the proper mentor
Are you a junior/senior
IDK, I really dislike intro-grad courses
Have you had an intro to manifolds?
Spivak's Calculus on Manifolds is much shorter
I'm a huge fan of that text, really gets you up to speed faster
Topology and Manifolds
LOL. I think either way you'll be fine
It's not like Riemannian will be incomprehensible to you
There might be some technical details that escape you ~ find a grad student
Make friends, buy them beer
I'm serious haha
They'll help you out! When I did grad complex my senior year, all the undergrads got together to do the problem sets
There was four of us. Lots of fun
You'll be surprised how much you can contribute to the conversation
I was by far the least prepared in the four undergrads, least background
But I was able to identify issues/geometric concerns quicker than my classmates
Yeah, I just finished my MS
I'm applying to PhDs eventually ~ depending upon how my living situation stabilizes. I have a wife so it's trying to solve a two-body problem
No she's going to CS, a big tech company
I'm gonna be taking some courses/maybe some side research. I'm applying to teach at CC
I'm split between Topology and Analysis
Yeah that's the issue. I have to find a PhD program where she finds work as well
The maximum time off I'm taking is 3 years
the minimum is a year. So it'll probably be 2 years off
My first love was topology, I got stockholmed into analysis
Yeah the current goal is every school on my list has something to offer topologically or analytically + tech industry nearby
elder yew
So Maggie Miller is a co-author of a co-author of my prof
I'll take a look at what they do. It's largely Floer Homology and other things that I know nothing of
dawn sable
I have a question about the definition of topological dimension in Hartshorne.
So for topological spaces being Noetherian, we require them to satisfy the DCC, since an ascending chain of ideals in A[x1,...,xn] corresponds to a descending chain of closed sets in A^n.
But then why do we have the dimension of a topological space being the longest chain of closed subsets Z_0 ⊆ ... ⊆ Z_n, the same direction as Krull dimension for rings? One of the exercises is to find an infinite dimensional Noetherian topological space, which seems much simpler relative to finding an infinite dimensional Noetherian ring, since all you have to do is find an infinite ascending chain of closed subsets.
sleek thicket
The chain Z0 <=… <= Zn isn't really directed
Since it's finite
You could just add easily write it Zn <=… <= Z0
also it's very hard to write down an infinite dimensional non noetherian ring
You can get much easier examples than that for this problem
Also your definition of dimension is wrong
it should by irreducible closed subsets
otherwise any space with infinitely many closed points would be infinite dimensional
Have you thought about what the topological dimension "looks like" geometrically? Like if you take a hyperbolic paraboloid z = x^2 + y^2, is it intuitive why this is 2 dimensional?
dawn sable
right right they have to be irreducible
I got the answer to the exercise fairly easy and was wondering why the definition was worded they way it is
It makes total sense for finite dimensions
sleek thicket
What was your solution?
dawn sable
I had a sort of contrived one, but the idea was similar to the topology on N where the closed sets are {1,...,n}
just not as clean
sleek thicket
hmm it seems like you won't have any irreducible subsets
Other than points
Also maybe here's a way to see why dimension is defined the way it is: dim Z(J) = dim k[x1,...,xn]/J. Do you see why?
dawn sable
Aren't they irreducible because if {1,...,n} is the union of {1,...,m1} and {1,...,m2} either n=m1 or n=m2?
I'm still a bit confused on the dimensions of varieties and was trying to grasp the dimension of general topological spaces first
sleek thicket
Oh lol I misread what you said
You're totally right
I thought you just had any finite set
dawn sable
I guess mainly I'm just confused why you don't reverse the direction of the inclusions because it matters in the infinite case?
sleek thicket
yeah that works
You do reverse the dimensions of the inclusions
But you're only every looking at finite chains
The dimension is the sup over all lengths of finite chains, right?
sleek thicket
Yeah, ascending vs descending only matters in the case of an infinite chain
dawn sable
That helps
At first I was worried I'd have to do some awful construction like the infinite dimensional Noetherian ring
Haha
dawn sable
Yeah
sleek thicket
Here's a related problem
Find a non noetherian ring A where Spec A is a noetherian topological space
sweet wing
somewhat long geometry problem coming lol:
does anyone have a good way of seeing why mobius transforms on the riemann sphere $=\mathbb S^2$ are the conformal transforms of $\mathbf B^3$
ig that like we can show $\text{M"ob}\mathbf B^n\cong\text{M"ob}\mathbb S^{n-1}$ by looking at circles orthogonal to $\partial\mathbf B^n\cong\mathbb S^{n-1}$ and then we have one of Liouville's rigidity stuff that says all conformal transforms on $\mathbb R^3$ are just M"obius but feels kinda unsatisfying.
Compare this to saying we can somewhat naturally extend mobius transforms on $\mathbb C\cup{\infty}$ to the upper half space $\mathbb H^3$ of $\mathbb R^3$ and give it the metric $\mathrm ds^2=\frac{|\mathrm dx|^2}{\Im x}$ and we got the isometries of the hyperbolic space, feels like there should be something equally nice for the ball model with $\mathrm ds^2=\frac{4|\mathrm dx|^2}{\left(1-|x|^2\right)^2}$
gentle ospreyBOT
ariana:
sleek thicket
It's a good problem grub
this is why noetherian scheme ≠ noetherian underlying topological space lol
sweet wing
counterexamples in algebra book when
sleek thicket
That's just anything by Nagata lol
Haha thanks
They're wearing masks with twin fantasy on them :)
sweet wing
as in?
sleek thicket
yeah which thing max?
marsh forge
ah ok
okay maybe someone here can point out why im an idiot
or ill figure it out as i type
ok so we have a fibration
SU(n-1)->SU(n)->S^{2n-1}
Serre SS gives us
$E_2^{p,q}=H^p(S^{2n-1},H^{q}(SU(n-1))$
gentle ospreyBOT
MaxJ:
marsh forge
Assume the inductive hypothesis and lets say for simplicity we are working over $\bR$
gentle ospreyBOT
MaxJ:
marsh forge
The only way $E_2$ can be nontrivial is if $p=0,2n-1$ and $q=0$ or $q$ is an odd less than $2n-3$
gentle ospreyBOT
MaxJ:
marsh forge
So our E_2 page is basically just two vertical rows at p=0 and p=2n-1
marking 0 and every odd number up to 2n-3
in particular this is just a copy of $H^*(S^3)$ at each such q
gentle ospreyBOT
MaxJ:
marsh forge
Now the differentials are only nontrivial if we have something of the form
$E_r^{0,q}\to E_r^{2n-1,q-2n+2}$
gentle ospreyBOT
MaxJ:
marsh forge
But! the issue is that since $2n-3<2n-1$ our $q$s will be negative before we get far enough
gentle ospreyBOT
MaxJ:
marsh forge
Hence all the differentials are trivial
the problem is that unless i am high (i am not)
This implies that there should be a generator for $H^{2n}$ given by $E_{\infty}^{2n-1,1}$
gentle ospreyBOT
MaxJ:
marsh forge
This is known to be false
So clearly what we want is for $E^{2n-1,q}_\infty$ to be trivial for $q>0$
gentle ospreyBOT
MaxJ:
marsh forge
But I don't see why this would happen, they are not the target of any differentials
In particular we still want it to be nontrivial for q=0
Just as a sanity check
$E^{2n-1,2k+1}=H^{2n-1}(S^{2n-1},H^{2k+1}(SU(n-1))$
gentle ospreyBOT
MaxJ:
marsh forge
Which should be isomorphic to just $H^{2k+1}(SU(n-1))$
gentle ospreyBOT
MaxJ:
marsh forge
which should be just a copy of $\bR$
gentle ospreyBOT
MaxJ:
marsh forge
(i should add that these computations are wrong but i was not wrong in the way i thought i was which was the point)
quiet pilot
I have a question about pointwise continuity
We define $$f: X \rightarrow Y$$ to be continuous at $$x\in X$$ if for all neighbourhoods V (i.e open sets containing f(x)) of f(x) there exists a neighbourhood U of x such that $$f(U) \subset V$$
gentle ospreyBOT
AoiKunie:
quiet pilot
I initially thought that this was equivalent to $$f^{-1}(V)$$ to be open for all neighbourhoods V of f(x) but this seems to not be the case. This statement obviously implies the definition but am I correct in concluding that the definition does not imply this?
gentle ospreyBOT
AoiKunie:
quiet pilot
Thanks a lot!
cobalt lily
sorry for likely wrong channel - "ρ must be a homomorphism, i.e. ρ(gh) = ρ(g)ρ(h)" - is this legit?
why a homomorphism and not an isomorphism?
i guess i am having trouble understanding the difference. i have referred to google before posting this - i am just not a very mathematically inclined (or particularly bright) individual
serene geode
In general, a homomorphism between structures of the same type (rings, groups, whatever) is a function that respects the underlying structure. So in your example, there's is some kind of product that is preserved by ρ, in the sense that you can do the product before of after applying ρ and it's the same. An isomorphism is a bijective homomorphism. Isomorphisms are cool beacuse they tell you both structures are "essentially the same". In the question you posted, context isn't clear, so maybe for whatever you want, ρ might need to just be a homomorphism, or it might need to be an actual isomorphism.
cobalt lily
you can do the product before or after applying ρ and it's the same.
this was a lightbulb moment! i was just thinking about it wrong. thanks so much!
green flame
Someone told me this is part of topology, does that para make sense to any of you guys?
marsh forge
Yes
green flame
Can you break it down for me? Or maybe suggest me topics I should read about, this is from a fluid mechanics book.
wanton marsh
that's a whole chapter of point set topology
ivory dragon
i mean
studying them is a whole chapter but i dont think the definitions are too bad
intro analysis couses cover them in like a day
is there a specific definition you're not able to grasp?
wanton marsh
it's a lot to take in at once like this
green flame
Oh well I just can't make sense of whatever is in the selected text. Which book in your opinion would be easy to follow for someone like me?
I tried looking for a decent video/lecture on YouTube but got india spammed by set theory tutorials and lectures.
Oo I found this, pretty good explanation https://www.youtube.com/playlist?list=PLgAugiET8rrJEL_3aovzCWK2LBSwvPsFd
marsh forge
If you want the shortest possible intro
allen hatcher has point set topology notes
look at those
willow spear
Wait guys an open set in topology refers to the eleemnts contained ina topology right?
like all elemenst of a topology, tau, are open sets?
serene geode
Yeah, you usually define a "topology in a set" as the "set of open subsets"
marsh forge
I say usual because you can technically do it three ways
and the statement is only correct in one of them
sleek thicket
what are your three ways
Imo the most galaxy brain way is with a closure operator
Imagine having more than one axiom for topological spaces LMAO
couldn't be me
marsh forge
oh
i was referring to the three ways in munkres
i only remember two of them lmfao
The first two are right
i dont remember the third bc i stopped reading
sleek thicket
A topological space is a pair $(X, \mathbf{c})$, where $\mathbf{c} : 2^X \to 2^X$, satisfying
$\displaystyle A\cup \mathbf {c} (A)\cup \mathbf {c} (\mathbf {c} (B))=\mathbf {c} (A\cup B)\setminus \mathbf {c} (\varnothing )$
for any $A, B \subseteq X$
Oof it got cut off
fuck
marsh forge
A topological space is a simplicial set such that all horns can be filled
sleek thicket
Lmao
sleek thicket
Alright
dim meadow
Gross
gentle ospreyBOT
shamrock:
sleek thicket
Yeah