#point-set-topology | Mathematics | Page 195

tight agate Jan 12, 2021, 6:42 AM

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noice

elder yew Jan 12, 2021, 6:42 AM

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AG? RG?

sleek thicket Jan 12, 2021, 6:42 AM

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that's my next app actually

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analysis!

elder yew Jan 12, 2021, 6:42 AM

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Oh rly

tight agate Jan 12, 2021, 6:42 AM

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wait UCLA REU apps started?

sleek thicket Jan 12, 2021, 6:42 AM

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it's gangbo's program also

elder yew Jan 12, 2021, 6:42 AM

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Ah. Gangbo is an interesting guy

sleek thicket Jan 12, 2021, 6:42 AM

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I believe it's open

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https://ww3.math.ucla.edu/david-harold-blackwell-summer-research-institute/

UCLA Mathematics

The David Harold Blackwell Summer Research Institute - UCLA Mathema...

David Harold Blackwell Summer Research Institute A collaboration between the UC Berkeley Electrical Engineering and Computer Sciences (EECS) department, the UCLA Math department and the Stanford bioengineering department. Lawrence Berkeley Nat’l Lab – Roy Kaltschmidt, photographer about the program David Harold Blackwell (1919-2010) David Harold...

elder yew Jan 12, 2021, 6:42 AM

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Super friendly

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He's a great mentor

tight agate Jan 12, 2021, 6:42 AM

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I applied sometime late march or early april last year lol

elder yew Jan 12, 2021, 6:42 AM

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Not the best as a professor, but super chill

sleek thicket Jan 12, 2021, 6:42 AM

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I would be applying to hopefully work on "Calculus of Variations" I think

elder yew Jan 12, 2021, 6:43 AM

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He has a thick north african accent

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Sounds like the warlords in documentaries

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I know it's immature, but I couldn't stop laughing

tight agate Jan 12, 2021, 6:44 AM

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sleek thicket https://ww3.math.ucla.edu/david-harold-blackwell-summer-research-institute/

this seems to be some special program

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the usual UCLA REU apps open in march I think

sleek thicket Jan 12, 2021, 6:45 AM

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this? http://reu.dimacs.rutgers.edu/

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wait thats rutgers lmao

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why is it on the ucla sitr

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yeah idk I just put everything that seemed vaguely interesting on a big spreadsheet

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and am applying to all of them

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2 down ^_^

tight agate Jan 12, 2021, 6:46 AM

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are you applying to chicago

sleek thicket Jan 12, 2021, 6:46 AM

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yup

tight agate Jan 12, 2021, 6:46 AM

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one of my friends went there last year

sleek thicket Jan 12, 2021, 6:47 AM

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didn't get in last year but I think my app is stronger now

tight agate Jan 12, 2021, 6:47 AM

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he said it was fun

sleek thicket Jan 12, 2021, 6:47 AM

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yeah it seems really cool

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otoh I would like to do something more researchy

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but I wouldn't hate u chicago's program either lol

tight agate Jan 12, 2021, 6:47 AM

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I dont want to do researchy stuff

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I'd prefer reading a shit ton and doing an expository paper of some kind

pastel linden Jan 12, 2021, 6:48 AM

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is it usually people at a knowledge level comparable to you all that apply to REUs

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this makes me think applying to some is a waste of time

sleek thicket Jan 12, 2021, 6:49 AM

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nope

elder yew Jan 12, 2021, 6:49 AM

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You should apply bacono - they factor in potential and who your letter writers are

sleek thicket Jan 12, 2021, 6:49 AM

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there are a couple things to consider

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several people in my reu last year had only done like, an intro proofs class

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or like linear algebra

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they weren't on my project because that project requires more mathematical knoweledge/experience

tight agate Jan 12, 2021, 6:50 AM

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what was your reu on shamrock?

sleek thicket Jan 12, 2021, 6:50 AM

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but similarly I didn't apply to the other projects because I wanted something more advanced

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so people self select

pastel linden Jan 12, 2021, 6:50 AM

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that makes me feel better

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very good to know

sleek thicket Jan 12, 2021, 6:51 AM

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tight agate what was your reu on shamrock?

which of these do you have knowledge about?
(1) braid groups
(2) braided monoidal categories
(3) operads
(4) hopf algebra
(5) quantum groups

tight agate Jan 12, 2021, 6:51 AM

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none

sleek thicket Jan 12, 2021, 6:51 AM

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oh okay sick

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sorry Im not trying to flex or anything

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I just like, want to know where to target my explanation

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okay so you have knots, yeah?

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this is at its heart a knot theory project

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knots are not nice algebraically

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you can't like, compose two knots

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also it's sort of arbitrary to restrict to embeddings of just one circle, so instead lets look at links, which are like knots with multiple components

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following so far?

tight agate Jan 12, 2021, 6:52 AM

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yup

sleek thicket Jan 12, 2021, 6:53 AM

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well there is a closely related object which is nice algebraically

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braids

📎 5_Strand_Braiding_Technique.png

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idk what the arrows are I stole this from wikipedia

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a braid on n strands is like, you fix two sets of n points and then let n disjoint copies of the unit interval connect them

elder yew Jan 12, 2021, 6:53 AM

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They're describing how to overl-lap the braids

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to get your braid relation

sleek thicket Jan 12, 2021, 6:54 AM

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cool i ddin't really look at it lol

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(the embeddings are subject to some dumb regularity conditions or whatever so that they're not super wild. just imagine physical strands connecting n points to n more points)

tight agate Jan 12, 2021, 6:54 AM

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sleek thicket a braid on n strands is like, you fix two sets of n points and then let n disjoi...

oh nice, so do you compose in the obvious way?

sleek thicket Jan 12, 2021, 6:54 AM

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yup!

tight agate Jan 12, 2021, 6:54 AM

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cool!

sleek thicket Jan 12, 2021, 6:54 AM

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here's some examples from wiki

📎 unknown.png

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and you can like, twist and untwist

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if you think about two strands

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crossing them one way and then the other

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and stacking those on top of eachother

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you can untwist to get the identity

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so the inverse is like, reverse all crossings and reverse the ordering

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so it's geometrically kind of clear that like, the nth braid groups is generated by the operations of crossing the ith and (i+1)th strand

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with some relations that you have to stare at but can convince yourself make sense

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📎 3aad386475c4da2dba588817665b1d79b977d1b6.png

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so the last one holds for i, j not adjacent

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crossing far away strands commutes

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and then the first one is a weird kind of twist

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also more examples

📎 Braids_1000.png

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and here's the braid relation

📎 braid_3_artin_all.png

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you just sort of pass the middle strand through

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so the second important thing

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you can close up a braid to get a link

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just connect the top and bottom dots

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yeah?

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alright so

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this is our connection between geometry and algebra

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the equivalence relation "closes to the same links" can be described as the reflexive transitive closure of some simple things called markov moves

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which I won't go through

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but in principle you can determine that like, if you have an invariant of braids you can check that it descends to links

tight agate Jan 12, 2021, 7:00 AM

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hmm ok

sleek thicket Jan 12, 2021, 7:00 AM

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just by checking it's invariant under these three moves

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okay so

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new perspective

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stop caring about knots

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care about braids

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(and then eventually bring it all back to braids)

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we're going to look at systems of representations $B_n \to \mathrm{Aut}(X_n)$

tight agate Jan 12, 2021, 7:01 AM

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sleek thicket stop caring about knots

no problem, never did

gentle ospreyBOT Jan 12, 2021, 7:01 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:01 AM

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lol

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alright so

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there are two paths

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are you a noncommutative algebraist or a category theorist at heart?

tight agate Jan 12, 2021, 7:02 AM

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uuh

sleek thicket Jan 12, 2021, 7:02 AM

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everyone is one of the two

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this is a theorem

tight agate Jan 12, 2021, 7:02 AM

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aight lets go cats for now

sleek thicket Jan 12, 2021, 7:02 AM

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maybe "representation theorist" instead of noncommutative algebraist

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okay cool

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that one is a nicer story

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even though i personally prefer/worked on the other

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so here is an idea

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hmm lemme think about how to phrase this

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do you know the definition of a monoidal category?

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symmetric monoidal?

tight agate Jan 12, 2021, 7:03 AM

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yup

sleek thicket Jan 12, 2021, 7:03 AM

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alright sick, so say $(\mathscr{C}, \otimes)$ is a symmetric monoidal category

gentle ospreyBOT Jan 12, 2021, 7:03 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:03 AM

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then for any $x \in \mathscr{C}$ we get a sequence of representations $S_n \to \mathrm{Aut}(x^{\otimes n})$

gentle ospreyBOT Jan 12, 2021, 7:03 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:04 AM

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do you see how this works?

tight agate Jan 12, 2021, 7:04 AM

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yup

sleek thicket Jan 12, 2021, 7:04 AM

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right so this also plays nice with the tensor product of maps

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and a certain operation on Sn

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which i think of as the direct sum

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if you have a permutation in Sn and one in Sm you get one in S_{n+m}

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do the first permutation on the first n things you're acting on and the second on the last n

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yeah?

tight agate Jan 12, 2021, 7:05 AM

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yup

sleek thicket Jan 12, 2021, 7:05 AM

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this is not a group homomorphism, but it's something for sure

tight agate Jan 12, 2021, 7:05 AM

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seems to be some kind of "join"

sleek thicket Jan 12, 2021, 7:05 AM

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so this system of representations turns "join" into "tensor"

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does that make sense?

tight agate Jan 12, 2021, 7:06 AM

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hmm ok

sleek thicket Jan 12, 2021, 7:06 AM

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if you think about the like, n = 4 case

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and S2

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this is just saying

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take two permutations in S2

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then we have $x \otimes x \otimes x \otimes x$

gentle ospreyBOT Jan 12, 2021, 7:06 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:06 AM

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the join will do the first permutation on the first $x \otimes x$ and the second permutation on the second $x \otimes x$

gentle ospreyBOT Jan 12, 2021, 7:06 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:06 AM

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right so

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there's some niceness to this representation

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this isn't like, super relevant I think

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but I just want to point out we have nice systems of representations

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okay so

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slogan: braids are permutations with more memory

tight agate Jan 12, 2021, 7:07 AM

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makes sense

sleek thicket Jan 12, 2021, 7:08 AM

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right

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just trace the permutation of the dots

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and this gives a homomorphism Bn -> Sn

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also we have a sort of join on braids too, right?

tight agate Jan 12, 2021, 7:08 AM

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right

sleek thicket Jan 12, 2021, 7:08 AM

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literally just stack them next to one another horizontally

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okay so we want a different flavor of monoidal categories

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instead of being able to act on tensor products by permutations, we want to act on them by braids

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and then permute according to the underlying permutation of the braid

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so the definition is what I said, plus some coherence conditions on the isomorphisms

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https://en.wikipedia.org/wiki/Braided_monoidal_category

Braided monoidal category

In mathematics, a commutativity constraint

    γ
  

{\displaystyle \gamma }
on a monoidal category 


  
    
      
        C
      
    
  

{\displaystyle {\mathcal {C}}}

is a choice of isomorphism

      γ
      
        A

...

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we have an isomorphism $\sigma : A \otimes B \to B \otimes A$

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called the "braiding"

gentle ospreyBOT Jan 12, 2021, 7:11 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:11 AM

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but unlike in the symmetric case, $\sigma^2 \neq id$

gentle ospreyBOT Jan 12, 2021, 7:11 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:11 AM

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(i mean it can, every symmetric monoidal cat is braided too, but in general this equality fails)

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and then you have ugly axioms like the pentagon axiom (edit: but for the braiding)

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making sure coherence theorems work

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lol

tight agate Jan 12, 2021, 7:12 AM

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aight

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dont need to see those

sleek thicket Jan 12, 2021, 7:12 AM

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yeah haha

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but they ensure that

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this one family of isos

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gives a system of braid group actions

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like, $\sigma$ represents the generator of $B_2$

gentle ospreyBOT Jan 12, 2021, 7:12 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:12 AM

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the thing where you flip one strand over the other

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and then all braids decompose as compositions of braids where you just flip one strand over an adjacent one

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and a braid like that is just a trivial braid joined with $\sigma$ joined with another trivial braid

gentle ospreyBOT Jan 12, 2021, 7:13 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:13 AM

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pictures are useful here lol

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if you look at the left braid here, that's $(\sigma \oplus e)(e \oplus \sigma)(\sigma \oplus e)$

📎 braid_3_artin_all.png

gentle ospreyBOT Jan 12, 2021, 7:14 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:15 AM

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so basically the point is all that matters is this generator $\sigma \in B_2$, all other braids are built up from that, so for a braided monoidal category as long as we have this one braided we can act by the braid groups

gentle ospreyBOT Jan 12, 2021, 7:15 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:15 AM

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akin to how the single flip AxB -> BxA in a symmetric monoidal cat gives you an action of all the Sn on length n products

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okay cool

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so in particular this kind of category produces loads of braid group representations

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yeah?

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pick any object and look at the automorphism groups of its tensor powers

tight agate Jan 12, 2021, 7:17 AM

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👍

sleek thicket Jan 12, 2021, 7:17 AM

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Alright so

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How do I even know these things exist?

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Like, braided monoidal categories

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Well it comes down to that braid relation $(\sigma \oplus e) (e\oplus \sigma) (\sigma \oplus e) = (e\oplus \sigma) (\sigma \oplus e)(e\oplus \sigma)$

gentle ospreyBOT Jan 12, 2021, 7:18 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:19 AM

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completely unrelated to all of this (in the sense of like, how things developed historically) physicists were interested in solution to the "yang-baxter equation"

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these are matrices $R \in \mathrm{End}(V \otimes V)$ such that $(R \otimes I)(I \otimes R)(R \otimes I) = (I \otimes R)(R \otimes I)(I \otimes R)$

gentle ospreyBOT Jan 12, 2021, 7:20 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:20 AM

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wait a minute...

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....

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...

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those are the same equation!

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so people had already come up with these things called quantum groups

tight agate Jan 12, 2021, 7:21 AM

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hmm

sleek thicket Jan 12, 2021, 7:22 AM

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quantum groups (or "quasi-triangular hopf algebras") are certain kinds of hopf algebras $H$ along with a special element $R \in H \otimes_k H$, satisfying axioms which make multiplication by $R$ on $V \otimes_k V$ a solution to the yang-baxter equation for any representation $V$ of $H$

gentle ospreyBOT Jan 12, 2021, 7:22 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:23 AM

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(the element R satisfies some weird properties that you stare at for a couple days and think really hard about and then realize are a really clever way to state that it gives solutions to the YBE)

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so if you have a quasi-triangular hopf algebra, its category of representations is braided monoidal, with braiding $A \otimes_k B \to B \otimes_k A$ given by "multiply by $R$ and then use the usual swap iso $A \otimes_k B \to B \otimes_k A$ in $k-Vect$"

gentle ospreyBOT Jan 12, 2021, 7:24 AM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 12, 2021, 7:24 AM

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okay so i feel like I am brushing a lot of details under the rug here

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but it's really not useful to talk about what quasi triangular hopf algebras are lol

elder yew Jan 12, 2021, 7:24 AM

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That's quite a lot

sleek thicket Jan 12, 2021, 7:25 AM

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they are things that had already been extensively studied and their category of representations is braided monoidal

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is the point

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so now we have a pipeline

tight agate Jan 12, 2021, 7:25 AM

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ok

sleek thicket Jan 12, 2021, 7:25 AM

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quasi-triangular hopf algebras -> braided monoidal categories -> braid group representations -?> knot invariants

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thats a lot isn't it

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but wait

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what's that?

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you think it's too simple?

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not enough category theory for you?

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oh I should note at this point, none of this is my work

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this is all already known

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so here's where the idea for my project starts off

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what if instead of regular knots we consider *singular* knots

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where we allow self intersections

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and these have a notion of "singular braids" analogous to braids, where the strands are allowed to intersect and come apart

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so literally just like, let the strands collide

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does that make sense?

tight agate Jan 12, 2021, 7:27 AM

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uuuh sorta

sleek thicket Jan 12, 2021, 7:27 AM

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you can still concatenate singular braids

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but once you've created a singularity you can never pull it apart

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so we no longer have groups, only singular braid monoids

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examples of singular knots

📎 unknown.png

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monoid presentation of the singular braid monoid

📎 unknown.png

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we still have those usual twists si

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and their inverses

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but now we also have collisions/self intersections/singularities ti

tight agate Jan 12, 2021, 7:29 AM

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👍

sleek thicket Jan 12, 2021, 7:29 AM

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so once again these look scary

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but like

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most of those are not actually weird

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first half is commutativity stuff

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"far away things don't matter"

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(3) is interesting and geometric, it says you can twist a singular braid to move a crossing above a singularity

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hard to show you without a physical model lol

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(4) is the braid relation and then two singular versions

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this but the top left and bottom right crossings are now singular

📎 braid_3_artin_all.png

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you just sort of twist to move it up

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(and the mirrored version)

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anyways, if we want to find singular knot invariants we might start with singular braid monoid representations

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oh but wait! we saw earlier that those naturally come from braided monoidal categories

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(oh also thing I forgot earlier: all of the basic knot invariants you'd learn in a first course on knot theory arise from quantum groups/braid group representations in the way I described earlier)

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so what's the analogue of a braided monoidal category for the singular braid monoid?

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oh well, clearly what we need to answer this question is

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O P E R A D S

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see, (strict) braided monoidal categories are algebra over the fundamental groupoid of the little 2-disks operas in Cat

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(general braided monoidal categories are of course a modified weak version of all of this)

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(also that fundamental groupoid is pretty easy to write down explicitly and you don't need to know about the little 2-disks operad or anything, Im just meming)

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so the moral is there's a nice operad people already care a lot about

elder yew Jan 12, 2021, 7:34 AM

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Oh my god sham

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You're really letting 'em have it

sleek thicket Jan 12, 2021, 7:34 AM

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and it somehow parameterizes braided monoidal categories

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okay moonbears keep in mind

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i had to fucking generalize all of this

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like

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this was my first week or two of summer

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learning ALL OF THIS

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and I haven't even stated our goal yet!

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okay so back on track

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the idea of the project is to find an analogue of the paranthesized braid operad (the thing I mentioned above) for singular braid monoids, so that it's algebras give a notion of singularly braided monoidal categories, then look for the right singular version of a quasi-triangular hopf algebra so we can find explicit singular braid monoid representations/singular knot invariants

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it took me like a week to understand what the project was about lmfao

elder yew Jan 12, 2021, 7:37 AM

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I think you secretly really love LDT sham

sleek thicket Jan 12, 2021, 7:37 AM

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lol

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well the thing is like

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it's an algebra/category theory project for which I can draw pretty pictures

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and reason with braids

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that's like the perfect thing

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abstract nonsense+geometry (actual geometry, not AG)

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sorry for the long explanation brofib

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but there is a lot of setup required to explain the project lol

tight agate Jan 12, 2021, 7:38 AM

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lol it was fun listening

sleek thicket Jan 12, 2021, 7:38 AM

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ng is also into this stuff

tight agate Jan 12, 2021, 7:38 AM

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ng?

sleek thicket Jan 12, 2021, 7:38 AM

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thing i am thinking of

tight agate Jan 12, 2021, 7:38 AM

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ah okay

sleek thicket Jan 12, 2021, 7:38 AM

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ngroupoid, sorry

elder yew Jan 12, 2021, 7:39 AM

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Braid relations are interesting

sleek thicket Jan 12, 2021, 7:39 AM

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anyways, turns out an operad/notion of singularly braided monoidal category can't exist

elder yew Jan 12, 2021, 7:39 AM

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They describe dehn twists!

sleek thicket Jan 12, 2021, 7:39 AM

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for subtle reasons

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but we found an appropriate version of this stuff

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a "PROB"

cedar pebble Jan 12, 2021, 7:39 AM

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ah just catching up, I was taking a nap monkey

sleek thicket Jan 12, 2021, 7:40 AM

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and I ended up working out some of the hopf algebra generalization stuff

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someone asked what my reu was

sleek thicket Jan 12, 2021, 7:40 AM

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sleek thicket which of these do you have knowledge about? (1) braid groups (2) braided monoida...

and I needed to explain this

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oh yeah you asked what came of our reserach

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https://www.jonathanbeardsley.com/PROB_of_Singular_Braids_annotated.pdf

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here's the talk jonathan gave

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paper is still WIP

cedar pebble Jan 12, 2021, 7:41 AM

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cool I'll check it out

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oh btw, did you end up going to Olivia's talk on this cactus operad stuff?

sleek thicket Jan 12, 2021, 7:41 AM

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oh no I missed it!

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I think I meant to go there

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but forgot

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oh also did you see this https://www.ias.edu/math/events/quantum-groups-seminar ?

Institute for Advanced Study

Quantum Groups Seminar

cedar pebble Jan 12, 2021, 7:42 AM

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Likewise. I'm really interested in what she's working on (we had quite a few conversations about this early on while she has been working on this)

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oh no I did not see this!

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incidentally there are a good handful of quantum groups people at my university 🙃

sleek thicket Jan 12, 2021, 7:42 AM

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She's super close to the stuff I was doing this summer but also has left the uw lmfao

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so

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I could've worked on stuff with her maybe

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if not for that

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sadge

cedar pebble Jan 12, 2021, 7:43 AM

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WAAAAH

sleek thicket Jan 12, 2021, 7:44 AM

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oh right I just remember what I was thinking before brofib's question

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"I should get dinner"

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hmmm

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Mistakes were made

cedar pebble Jan 12, 2021, 7:44 AM

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screaming

tight agate Jan 12, 2021, 7:45 AM

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I think there's a couple of grad students over here who work on quantum groups stuff

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I know that my freshman year algebra TA does for sure

cedar pebble Jan 12, 2021, 7:46 AM

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I'm taking homological algebra this semester from one of the quantum groups/geometric rep theory people here

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excited for that

sleek thicket Jan 12, 2021, 7:46 AM

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ooh

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I'm hoping one of the groups in my class presents on hopf algebra coho

cedar pebble Jan 12, 2021, 7:46 AM

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time to actually learn spectral sequences properly 😛

tight agate Jan 12, 2021, 7:46 AM

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everyone's doing homological algebra lmao

sleek thicket Jan 12, 2021, 7:46 AM

#

yeah haha

cedar pebble Jan 12, 2021, 7:47 AM

#

I use spectral sequences all the time I just

#

happen to only work with cases where I never have to think about what the differentials are doing

#

if I have to actually think about what the differentials are doing then uhhh

#

monkey

tight agate Jan 12, 2021, 7:47 AM

#

ah geometers

cedar pebble Jan 12, 2021, 7:48 AM

#

spectral sequence that immediately degenerates on the E_2 page
haha yea I know how to use spectral sequences

#

or like spectral sequences but there's only like one column of nontrivial entries 😛

sleek thicket Jan 12, 2021, 7:48 AM

#

I haven't learned about spectral sequences, they haven't come up for me yet. I'm a very concretely focused mathematician

#

alright now back to this thing about 2 functors

tight agate Jan 12, 2021, 7:49 AM

#

sleek thicket I haven't learned about spectral sequences, they haven't come up for me yet. I'm...

how did you prove the snake lemma then thonk

cedar pebble Jan 12, 2021, 7:49 AM

#

lol you don't need a spectral sequence for that tfw

tight agate Jan 12, 2021, 7:49 AM

#

(it's a joke)

cedar pebble Jan 12, 2021, 7:50 AM

#

WAAAAH

sleek thicket Jan 12, 2021, 7:50 AM

#

lmfao

cedar pebble Jan 12, 2021, 7:50 AM

#

sorry I'm tired as hell

sleek thicket Jan 12, 2021, 7:50 AM

#

We did it the standard way dude

cedar pebble Jan 12, 2021, 7:50 AM

#

WAAAAH

sleek thicket Jan 12, 2021, 7:50 AM

#

We watched "It's My Time"

cedar pebble Jan 12, 2021, 7:50 AM

#

kek

#

beat me to it

sleek thicket Jan 12, 2021, 7:50 AM

#

No lie that's what we did today in my class

#

3rd class seeing that clip lmfao

tight agate Jan 12, 2021, 7:52 AM

#

ye im super excited to get better with spectral sequences too

#

just learned how the adams SS is set up a couple of days ago

cedar pebble Jan 12, 2021, 7:52 AM

#

aaaa

#

that's a thing I used to know

#

what an utterly horrifying thing to try to compute

#

someday I want to learn how to actually compute things in algebraic K-theory

tight agate Jan 12, 2021, 7:54 AM

#

someday I will learn what algebraic k theory is about

sleek thicket Jan 12, 2021, 7:54 AM

#

hmm okay it is Tortellini Time

cedar pebble Jan 12, 2021, 7:54 AM

#

or pigs in a blanket?

sleek thicket Jan 12, 2021, 7:54 AM

#

more important than algebraic k theory

#

lol

cedar pebble Jan 12, 2021, 7:54 AM

#

kek

sleek thicket Jan 12, 2021, 7:54 AM

#

Oven is borked

cedar pebble Jan 12, 2021, 7:54 AM

#

WAAAAH

sleek thicket Jan 12, 2021, 7:55 AM

#

we need to get it fixed

tight agate Jan 12, 2021, 7:55 AM

#

the prof teaching my hom alg class does algebraic k theory

sleek thicket Jan 12, 2021, 7:55 AM

#

Ooh

#

Do you know the story of why weibel wrote his homological algebra book?

tight agate Jan 12, 2021, 7:55 AM

#

no

sleek thicket Jan 12, 2021, 7:55 AM

#

Well he kept hearing this rumor

#

Like at conferences

#

That Weibel was writing a book on algebraic k theory

#

Like people asked him how it was going and when it would be finished

#

and he at no point had intended to write a book on algebraic k theory

#

But (I assume) felt peer pressured

cedar pebble Jan 12, 2021, 7:56 AM

#

the fact that K_{2n-1}(Z) has rank 1 for n≥3 odd and rank 0 otherwise has something to do with the conjectured transcendence properties of the odd zeta values \zeta(n)

sleek thicket Jan 12, 2021, 7:56 AM

#

However he felt like before he could do that, he needed to write a book on homological algebra

#

Since there weren't any good enough ones

#

And then after that he eventually wrote an algebraic k theory book

#

Imo a hilarious story

tight agate Jan 12, 2021, 7:57 AM

#

I might get to start reading some algebraic k theory next quarter hype

sleek thicket Jan 12, 2021, 7:57 AM

#

Hype!

cedar pebble Jan 12, 2021, 7:57 AM

#

nice!

tight agate Jan 12, 2021, 7:58 AM

#

some people might present some of Quillen's foundational stuff in my topology class

sleek thicket Jan 12, 2021, 7:58 AM

#

oooo

tight agate Jan 12, 2021, 7:58 AM

#

which hopefully should give me the background required

sleek thicket Jan 12, 2021, 7:58 AM

#

I am hoping my bundles class talks about k theory but it doesn't seem like it will :(

cedar pebble Jan 12, 2021, 7:59 AM

#

Even after seeing the foundational stuff for like the tenth time I have never walked away from that stuff thinking "great, now I know how to compute this!"

tight agate Jan 12, 2021, 7:59 AM

#

sleek thicket I am hoping my bundles class talks about k theory but it doesn't seem like it wi...

hatcher's book is good

sleek thicket Jan 12, 2021, 8:00 AM

#

yee, it's just one of those things I don't really have the time for

cedar pebble Jan 12, 2021, 8:00 AM

#

yea it's stupidly hard screaming

sleek thicket Jan 12, 2021, 8:00 AM

#

This was supposed to be a lighter quarter, and yet so far...

#

my analysis class is very very very stupid

cedar pebble Jan 12, 2021, 8:01 AM

#

btw Shamrock given your stuff about knot invariants do you know much about the Kontsevich integral of knots?

sleek thicket Jan 12, 2021, 8:01 AM

#

no, sorry

cedar pebble Jan 12, 2021, 8:02 AM

#

Ah it's super cool and I've had quite a few project ideas related to it for a while that I probably will never get around to

#

it's conjecturally a complete knot invariant 👀

#

zoomEyes

sleek thicket Jan 12, 2021, 8:03 AM

#

:PogChamp:

cedar pebble Jan 12, 2021, 8:03 AM

#

it's also EXTREMELY related to the whole Drinfeld associators/parenthesized braids/parenthesized chord diagrams picture

sleek thicket Jan 12, 2021, 8:03 AM

#

yeah I didn't learn the knot theory stuff very well

#

because I was trying to learn everything else and actually do new stuff lol

cedar pebble Jan 12, 2021, 8:04 AM

#

Also is extremely related to quantum invariants, you can read every quantum knot invariant off the Kontsevich integral if you set things up carefully

sleek thicket Jan 12, 2021, 8:04 AM

#

ooh

cedar pebble Jan 12, 2021, 8:04 AM

#

(in fact you can read literally any Vassiliev invariant off the Kontsevich integral, which is related to the whole singular knots picture)

sleek thicket Jan 12, 2021, 8:05 AM

#

Yeah, interesting

sleek thicket Jan 12, 2021, 8:06 AM

#

sleek thicket yeah I didn't learn the knot theory stuff very well

also related, Jonathan is many things but he is not a knot theorist

#

Which I mean, he was very open about in our project

cedar pebble Jan 12, 2021, 8:06 AM

#

I'm sad he deleted his twitter 😦

sleek thicket Jan 12, 2021, 8:06 AM

#

So if I had to ask knot theory questions I would have to go to another mentor

#

Yeah

cedar pebble Jan 12, 2021, 8:07 AM

#

I mean I have him on facebook and I get why he deleted twitter

#

but I miss having him around there

#

WAAAAH

sleek thicket Jan 12, 2021, 8:08 AM

#

I should hopefully be meeting to talk about simplicial stuff and dold kan

#

In the near future

cedar pebble Jan 12, 2021, 8:08 AM

#

nice smol_nozoomi

sleek thicket Jan 12, 2021, 8:13 AM

#

sleek thicket also related, Jonathan is many things but he is not a knot theorist

oh lol this reminds me, we had to define the notion of a natural transformation for our grad student mentor

#

Who was great at knot theory/low dim top

#

but uh, that made things awkward considering how categorical our project was

digital glacier Jan 12, 2021, 8:34 AM

#

Ooo thanks! Will I need to understand perfectoid stuff for the scholze?

cedar pebble Jan 12, 2021, 8:35 AM

#

The biggest thing to understand right now is condensed mathematics

#

This is how Scholze sidesteps some of the foundational issues trying to correctly define the derived category of the moduli of G-bundles on the Fargues Fontaine curve

#

Perfectoid space stuff is very necessary yes, as is the more general notion of diamonds

#

It seems like the main technical inputs are like

etale cohomology of perfectoid spaces and diamonds and Artin stacks in the sense of diamonds
The Fargues-Fontaine curve and the moduli of G-bundles on the Fargues-Fontaine curve, and the geometrization of p-adic Hodge theory
condensed mathematics, especially solid modules

#

(Condensed mathematics has a lot of appeal that is independent of all of this and is probably the easiest of these three items to learn: for instance condensed sets are in a lot of ways preferable to working with topological spaces)

digital glacier Jan 12, 2021, 8:40 AM

#

"Indeed, unlike in classical topologies such as the Zariski or ´etale topology, sheaves on the pro-´etale
site of a point are not merely sets. What seemed like a bug of the pro-´etale site is actually a feature
for us here, as Clausen explained to the lecturer."

#

I already want to die haha

cedar pebble Jan 12, 2021, 8:41 AM

#

They are a little funny to set up

#

The point is that they give you more or less the correct framework for doing algebra in a topological setting

#

E.g. the category of topological Abelian groups or the subcategory of locally compact Abelian groups are very poorly behaved and aren’t well suited for homological algebra and other categorical constructions

#

Condensed Abelian groups form a much nicer category that faithfully includes the classical picture

digital glacier Jan 12, 2021, 8:45 AM

#

"If S is a profinite set, then all cohomology
groups vanish for i > 0, and H0
sheaf(S, Z) are the continuous maps S → Z, while H0
sing(S, Z) are all
maps S → Z"

#

scratches head Well there's nothing left to study...

#

"Mumford writes in Curves and their Jacobians: “[Algebraic geometry] seems to have acquired
the reputation of being esoteric, exclusive, and very abstract, with adherents who are secretly
plotting to take over all the rest of mathematics. In one respect this last point is accurate.”
For some reason, this secret plot has so far stopped short of taking over analysis. The goal
of this course is to launch a new attack, turning functional analysis into a branch of commutative
algebra, and various types of analytic geometry (like manifolds) into algebraic geometry. Whether
this will make these subjects equally esoteric will be left to the reader’s judgement."
-Scholze

#

@honest narwhal Looks like we can merge a few channels catshrug

#

@cedar pebble do you work on the langlands or geometric langlands?

cedar pebble Jan 12, 2021, 9:01 AM

#

Just classical Langlands stuff for number fields

digital glacier Jan 12, 2021, 9:15 AM

#

Oh right. Need someone I can discuss the new stuff with.

#

The Arinkin et al paper is a lot more familiar in its language. I can read this paper at least.

#

Clausen and Scholze... I don't know if I can handle it

#

Have you read Arinkin et al?

gritty widget Jan 12, 2021, 10:01 AM

#

my Prof said this holds if phi is diffeo

📎 unknown.png

#

and thus we can have a notion of 'pullback of a vector field'

#

here omega is a vector field

#

how can this be true?i'm a bit confused

uncut surge Jan 12, 2021, 10:03 AM

#

Denoting a vector field by omega is some sicko stuff, but other than that, this seems güd I believe

gritty widget Jan 12, 2021, 10:04 AM

#

but how can we pullback vector fields

#

i thought we can only pullback forms

#

prof said its crucial that phi is diffeo

#

but i cant see why would that give a pullback of a vector field

#

think about why pullback for vector fields doesn't make sense usually and then think about what a diffeo is

uncut surge Jan 12, 2021, 10:04 AM

#

Yes, it's very important that you have a smooth inverse map

#

Philosophically, this is fine, because if you have a diffeomorphism between two manifolds, this basically means that the two manifolds and their respective smooth structures are "the same", so everything you can do on one manifold you should be able to do on the other

gritty widget Jan 12, 2021, 10:06 AM

#

so if I had a diffeomorphism could I push forward forms too?

uncut surge Jan 12, 2021, 10:06 AM

#

And, I mean, your screenshot from up there is just a definition, anyway, right? So there's nothing to check in terms of "whether it's true or not"

gritty widget Jan 12, 2021, 10:06 AM

#

similarly?

uncut surge Jan 12, 2021, 10:06 AM

#

I'd say so, ye

gritty widget Jan 12, 2021, 10:06 AM

#

well

#

think about why the pullback of a vector field isn't well defined to begin with

#

then you can see why you need a diffeo

uncut surge Jan 12, 2021, 10:08 AM

#

ye that sounds like a healthy exercise

gritty widget Jan 12, 2021, 10:08 AM

#

my problem is i've never seen/heard this

#

i only heard definition of pullback for forms

#

i can't imagine what would change if i try to do it for vector fields

#

like it's not even defined

uncut surge Jan 12, 2021, 10:08 AM

#

Yeah you never talk about this because the pullback of forms and the pushforward of vector fields are the only canonical operations; i.e. they are the only ones which are always okay to do

gritty widget Jan 12, 2021, 10:08 AM

#

no that thing you posted was your definition

#

see what happens when you remove the condition that phi is a diffeo

#

then it's not invertible

#

and i can't go to RHS

#

yes well what happens if \phi issn't injective

#

you could get multiple vector fields on your domain manifold for the same point

#

and if it isn't surjective then you don't even have a pullback for at least one point in the target

#

yes,makes sense

#

and so the map has to be smooth and bijective for the pullback of vector fields to be well defined

#

which means it needs to be a diffeo

meager python Jan 12, 2021, 11:31 AM

#

@tough imp am I just dreaming or were there notes for the stacks course that were much longer than the current version?

summer jolt Jan 12, 2021, 5:21 PM

#

Can someone explain why the highlighted part is true?

📎 unknown.png

astral cedar Jan 12, 2021, 5:23 PM

#

Isn’t it precisely what it says in the formula

#

If you discard the christoffel part

#

?

gentle ospreyBOT Jan 12, 2021, 5:32 PM

#

TTerra

📎 222878770862882816.png

gritty widget Jan 12, 2021, 5:32 PM

#

petTheCat

#

(and on X at p petTheCat )

summer jolt Jan 12, 2021, 6:54 PM

#

gentle osprey **TTerra**

Thanks that makes sense!

tough imp Jan 12, 2021, 7:06 PM

#

meager python <@146419489083228160> am I just dreaming or were there notes for the stacks cour...

Yeah they used to be longer

sonic hill Jan 13, 2021, 12:31 AM

#

For pullback of vector fields it’s only necessary for F to be a local diffeomorphism, if I am not wrong? So neither injectivity nor surjectivity are strictly required

cedar pebble Jan 13, 2021, 12:50 AM

#

To pull back vector fields it's enough to have a local diffeomorphism

cursive flume Jan 13, 2021, 12:54 AM

#

is someone familiar with rep theory

📎 unknown.png

#

im missing a crucial point

#

this

#

so my main trial is that i have SU(2) and SO(3) as groups, I try to do rep of this by using their lie algebras,which are isomorphic,i.e. su(2) iso so(3). I know how to do rep theory for so(3) now I can use the fact that the rep from the lie algebra level can be lifted to the covering lie group ( I can't prove this so help would be appreciated)

#

but unfortunately the covering lie group is SU(2) not SO(3) and I would need to make the transition

cedar pebble Jan 13, 2021, 1:10 AM

#

what is the representation D^(j) here?

cursive flume Jan 13, 2021, 1:12 AM

#

📎 unknown.png

cedar pebble Jan 13, 2021, 1:14 AM

#

okay sure SU(2) is simply connected so representations of su(2) lift to representations of SU(2) like this

cursive flume Jan 13, 2021, 1:17 AM

#

yes

#

but how to go from SU(2) to SO(3)?

#

why does it work only for integer j

cedar pebble Jan 13, 2021, 1:19 AM

#

if you have half integer spin then D^(j)(2π) is minus the identity

#

so this can't form a representation of SO(3): rotation through an angle 2π has to be sent to the identity

#

you still get a projective representation in this way

cursive flume Jan 13, 2021, 1:21 AM

#

i can't really see this

#

why would this not form a representation of SO(3)?

cedar pebble Jan 13, 2021, 1:23 AM

#

I mean you're forced to map the identity to the identity in such a representation

#

half integer spin would be forcing you to map minus the identity to the identity, which you can't do unless you pass to the corresponding projective representation

digital glacier Jan 13, 2021, 1:46 AM

#

@cedar pebble Oh, scholze has a really nice lecture series on condensed maths on youtube. Though at 20 hours... it could not be said to be condensed. 😛

gritty widget Jan 13, 2021, 1:46 AM

#

20 hours is just under a day, get watching

#

you can do them all in a sitting with a coffee or two!

cedar pebble Jan 13, 2021, 1:47 AM

#

yea it's a good lecture series!

digital glacier Jan 13, 2021, 1:49 AM

#

You are trying to scramble my brain, TTera 😮

#

Maybe if I leave it on when I go to bed, his soothing german accent can lull me to sleep and then I can learn by osmosis and have downloaded everything when I wake up? hmmm

#

Oh, he basically covers basic AG at the start. Perhaps people could watch this and skip hartshorne in future? catshrug

tough imp Jan 13, 2021, 2:22 AM

#

If people can skip Hartshorne by watching only a few hours of Scholze I will destroy the video as I annoy handle others spending < hundreds of hours on Hartshorne and learning AG

sleek thicket Jan 13, 2021, 2:25 AM

#

annoy handle

cedar pebble Jan 13, 2021, 2:58 AM

#

(they can't, they are totally orthogonal topics)

elder yew Jan 13, 2021, 3:05 AM

#

I think going through the standard grad books/courses is good to give a broad and deep understanding where a lot of things come from

#

But in research you can kinda just pick a branch in a tree to jump up to

#

And hope the branch will support you

#

Without going up the trunk and exploring branches that will support you

digital glacier Jan 13, 2021, 3:24 AM

#

I say this with some degree of cheek. After spending all the time on grad studies, it's easy to watch a category theory explanation of something and go, "this is so clear and concise, why wasn't it explained like that the first time?". But back in my grad student days, an explanation with "left adjoint" and "right derived functor" would have been opaque.

willow spear Jan 13, 2021, 5:07 AM

#

To leanr Algebraic topolgoy what pre-reps are required?

#

is it just Point-Set and Group Theory?

#

is there anything else?

ivory dragon Jan 13, 2021, 5:11 AM

#

yeah basically, but knowing more algebra will make it easier

elder yew Jan 13, 2021, 5:12 AM

#

@tough imp did you actually do most of Hartshorne's exercises

tough imp Jan 13, 2021, 5:13 AM

#

I've done exactly 1/2 of the ones from chapters II and III

elder yew Jan 13, 2021, 5:13 AM

#

That is impressive

tough imp Jan 13, 2021, 5:19 AM

#

I finished on Dec 26th and haven't done one since

#

kekw

willow spear Jan 13, 2021, 5:24 AM

#

ivory dragon yeah basically, but knowing more algebra will make it easier

What else besides GT in algebra is good to know?

sleek thicket Jan 13, 2021, 5:46 AM

#

sanity check

#

Let $E, F$ be vector bundles. Do we have $E^* \otimes F^* \cong (E \otimes F)^*$?

gentle ospreyBOT Jan 13, 2021, 5:47 AM

#

shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 13, 2021, 5:47 AM

#

Pretty sure this is true...

#

You have a natural iso of this kind for vector spaces, right? So just apply that on fibers

#

also do we have an adjunction of the tensor/hom bundles? Feels like we should since they're all defined fiberwise...

tight agate Jan 13, 2021, 6:09 AM

#

they work for locally free sheaves

#

so should work for bundles

tough imp Jan 13, 2021, 6:14 AM

#

Define globally check locally

#

¯_(ツ)_/¯

#

For locally free sheaves I found the hardest part is figuring out the way to define it globally and once you can do that you go local and it's just a linear algebra statement that's obvious

sleek thicket Jan 13, 2021, 6:19 AM

#

yeah I think I see how you can do this

#

It's like

#

Define the map on fibers

#

via the result for fd vector spaces

#

Then by naturality, the local coordinate representation should just be the thing applied to that local coordinate representation

#

idk lol

#

Still do not see this isomorphism of bundles...

#

Ugh I hate this problem it feels like there's some really beautiful geometry that I'm just not seeing

#

Maybe I need to read about symplectic stuff

sleek thicket Jan 13, 2021, 6:43 AM

#

alright so

#

if I have an oriented real bundle $E$ over a paracompact $X$ of rank $2$

gentle ospreyBOT Jan 13, 2021, 6:44 AM

#

shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 13, 2021, 6:45 AM

#

There's a way to find a complex line bundle $E'$ on $X$ whose underlying real bundle is $E$

gentle ospreyBOT Jan 13, 2021, 6:45 AM

#

shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 13, 2021, 6:45 AM

#

I see how to construct it, but it's bad and doesn't seem at all canonical

#

by Gram Schmidt you can cover by oriented orthonormal frames

#

Then transition functions land in SO(2) ≈ U(1) <= GL(1, C)

meager python Jan 13, 2021, 6:47 AM

#

Why is May’s book bad @cedar pebble?

sleek thicket Jan 13, 2021, 6:48 AM

#

hello tolaria

meager python Jan 13, 2021, 6:48 AM

#

Hi

sleek thicket Jan 13, 2021, 7:01 AM

#

Yeah this confuses me

#

Like

#

The complex multiplication depends on the choice of frame, eight?

#

*right

#

Ooh maybe orientedness saves us here....

#

Right okay you can like, scale and rotate

#

Rotation is well defined independent of basis

#

right??

tight agate Jan 13, 2021, 7:07 AM

#

You're given an orientation of each fiber

#

which is an equivalence class of bases

#

the complex multiplication you define will not depend on the choice of rep from that equivalence class

#

and then the fact that all the transitions are in GL+ ensure that whatever you defined on trivilaizing open sets will be compatible on intersections

#

all the choices are made for you when you assume the bundle is oriented

#

I think

sleek thicket Jan 13, 2021, 7:24 AM

#

Right, this is my thinking too

#

the orientation on each fiber gives us a way to rotate vectors without a choice of basis

#

and so you can define complex multiplication

#

I think the extra confusion I was having is that Lee chooses a fiber metric and says all the transitions lie in SO(2)

#

and then SO(2) is the circle

#

But adding the fiber metric is actually adding even more apparent noncanonicity

#

When really there's a high level description of the complex bundle

#

alright so what I believe is

#

hmm

#

hmmmmm

#

I'm no longer certain this works

#

Orientability gives us that all the stuff has positive determinant, sure

#

But rotations do actually preserve distance, which is not invariant...

#

so I think maybe you do need to start with an oriented bundle which has a metric on it

#

Ugh maybe this isn't true? I'm no longer sure

sleek thicket Jan 13, 2021, 8:15 AM

#

ooh but my actual problem involves looking at this for S^2 = CP^1

#

Where I have a very nice metric (round metric or fubini study, pretty sure these are the same?)

#

Okay yes so I still need some noncanonical choice of fiber metric I think

#

But given that and an orientation you get a complex multiplication

sleek thicket Jan 13, 2021, 8:49 AM

#

aha, so I think this is actually really intuitive for the sphere

#

rotation is given by the right hand rule

#

With axis the normal vector

heady grove Jan 13, 2021, 6:39 PM

#

Any ideas on how to do part b

📎 unknown.png

fading vale Jan 13, 2021, 7:30 PM

#

hm if we have f: X -> Y, g: Y -> X a htpy equivalence, A subset X, B subset Y, and f | A: A -> B, g | B: B -> A

#

is f | A, g | B a htyp equiv from A to B

#

mhm yea pretty sure thats true

sleek thicket Jan 13, 2021, 7:43 PM

#

Don't you need something like thy homotopy extension property?

#

yeah moth what about like, X = R^2, Y = pt, A = S^1, B = pt

#

f squashes everything to a point and g includes the pt into the circle

#

@fading vale

fading vale Jan 13, 2021, 8:09 PM

#

yea ur right

#

in this context it was fibers so it works out

#

but def not in general lol

#

otherwise everything would be htpyically trivial by embedding into its cone

#

ugh this construction is really confusing

#

📎 20210113_152501.jpg

#

the section about homotopies in Top^K or Top_B being normal homotopies with H_t morphisms

#

tammo tom dieck

#

the H_t morphisms bit makes sense when im thinking about homotopies between maps f, g : X -> Y

#

say with i : K -> X, j : K -> Y

#

but when i try to think about it in the context of homotopy equivalences it starts being pretty confusing to me

#

like if i have f: X -> Y, g : Y -> X we need fg simeq id_X and gf simeq id_Y

#

but i dont understand how the homotopy from fg to id_X works because i dont get what spaces under the K H_t between them are supposed to be morphisms between

#

sigh

#

like the problem im having makes sense right

#

cuz we have fg: X -> X, gf : Y -> Y, ok sure

#

where does the K come in though?

#

i guess fg is a morphism from i : K -> X to itself and likewise the identity so is that what the point is here??

#

that kind of makes it work out because when verifying the deformation retract stuff you would get H_t(k) = H_t(i(k')) = i(k') = k so the htpy is rel K...

heady grove Jan 13, 2021, 10:35 PM

#

Any ideas for part b

📎 unknown.png

frigid patrol Jan 14, 2021, 12:25 AM

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Can you imbed a bouquet of 3 circles in a torus?

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yes

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📎 unknown.png

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arrows are the circles, (identify opposite sides of a square to get a torus as usual)

elder yew Jan 14, 2021, 2:43 AM

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Can you find a curve on the torus such that if you cut the torus along that curve you end up with two manifolds whose connect sum is the torus?

river granite Jan 14, 2021, 2:49 AM

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nice cup

frigid patrol Jan 14, 2021, 3:14 AM

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the only way two manifolds have connect sum to be the torus is if one of them in a torus and the other is a sphere

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this is a corollary of the classification of surfaces

elder yew Jan 14, 2021, 3:17 AM

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without using the classification of surfaces

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C'mon

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You're ruining the fun

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Try drawing a curve

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why does it always fail?

tight agate Jan 14, 2021, 3:33 AM

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elder yew Can you find a curve on the torus such that if you cut the torus along that curv...

but the torus is a curve moonbears

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this post was made by the complex geometry gang

woeful oasis Jan 14, 2021, 3:33 AM

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that was pure genus

elder yew Jan 14, 2021, 3:57 AM

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fjdka;

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Yeah, can you prove it for all ways that you can draw a disk?

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I mean draw a simple closed curve?

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BOO

tight agate Jan 14, 2021, 4:12 AM

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?

elder yew Jan 14, 2021, 4:13 AM

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You can start by trying to draw curves

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and see what it property it violates

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There's only a few ways you can draw curves

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Up to isotopy

tight agate Jan 14, 2021, 4:13 AM

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well yeah, they're all in the same homology class as one of the standard generators or contractible

elder yew Jan 14, 2021, 4:14 AM

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So lame

tight agate Jan 14, 2021, 4:14 AM

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lame because I used the word homology class?

elder yew Jan 14, 2021, 4:14 AM

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yA

tight agate Jan 14, 2021, 4:14 AM

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You can say the same thing without using that word

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ye

elder yew Jan 14, 2021, 4:15 AM

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yA

tough imp Jan 14, 2021, 4:16 AM

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Proofs using homology is better because then you’re using algebra

sweet wing Jan 14, 2021, 4:32 AM

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tough imp Proofs using homology is better because then you’re using algebra

agree

frigid patrol Jan 14, 2021, 4:34 AM

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i hate algebra

sleek thicket Jan 14, 2021, 4:38 AM

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it's okay, lots of mathematicians live a full life while hating algebra

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being wrong is not a crime

hazy stratus Jan 14, 2021, 5:32 PM

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Are single points 0-dimensional submanifolds of R? e.g. take the set {-1,1}, it's the image of phi: {-1,1}->R, phi(x)=x. If phi is an embedding then {-1,1} is a submanifold...is phi an embedding?

small phoenix Jan 14, 2021, 6:02 PM

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yes.

hazy stratus Jan 14, 2021, 6:19 PM

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Ok, thank you! Is that because phi: {-1,1}->phi({-1,1}) is a homeomorphism? I'm trying to sort out all of the terms

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And there should be a diffeomorphism such that {-1,1} map to a point on the axis, in this case one needs two diffeomorphisms, one for each point? Or is there a way to define single diffeo that fits the definition

sleek thicket Jan 14, 2021, 7:00 PM

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I think it would be useful if you shared your understanding/definition of what an embedding and a manifold are

cedar pebble Jan 14, 2021, 7:45 PM

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@sleek thicket did you ever work out the 2-category meme 😰

sleek thicket Jan 14, 2021, 7:45 PM

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hoping to get it finished over the weekend

cedar pebble Jan 14, 2021, 7:45 PM

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hyper

sleek thicket Jan 14, 2021, 7:45 PM

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twitter voted 26 to 25 so I have to now

cedar pebble Jan 14, 2021, 7:45 PM

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oh nice

sleek thicket Jan 14, 2021, 7:45 PM

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I can post here or smth, will probably put it up on my website

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actually I have an AG-related bundles question you might be able to help with, but I think @hazy stratus was asking a question

hazy stratus Jan 14, 2021, 7:46 PM

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We defined a manifold by 3 different characterizations, I was trying to match {-1,1} being a manifold to two of them

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(sorry for the late answer)

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So the first one was that M is a manifold if all points in M have an open neighborhood V s.t V is the image of an embedding

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An embedding being an injective immersion that's also a homeo

sleek thicket Jan 14, 2021, 7:48 PM

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is a manifold an abstract topological space for you or a certain kind of subset of R^n for some n?

hazy stratus Jan 14, 2021, 7:48 PM

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hazy stratus Are single points 0-dimensional submanifolds of R? e.g. take the set {-1,1}, it'...

That was this

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Oh hm

sleek thicket Jan 14, 2021, 7:48 PM

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From what you're saying I'm assuming M is a priori contained in R^n

hazy stratus Jan 14, 2021, 7:48 PM

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I'd say the second one

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Yes

sleek thicket Jan 14, 2021, 7:49 PM

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since you need to make sense of immersion/derivatives in the definition

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okay cool

hazy stratus Jan 14, 2021, 7:49 PM

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It doesn't have to be n dimensional though

sleek thicket Jan 14, 2021, 7:49 PM

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well my n there could be anything

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I'm not saying n = dim M

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just that it sits inside some euclidean space

hazy stratus Jan 14, 2021, 7:50 PM

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Oh I meant it does't have to have the "full" dimension

hazy stratus Jan 14, 2021, 7:50 PM

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sleek thicket since you need to make sense of immersion/derivatives in the definition

Yeah the thing about the derivative having to be continuous as well isn't really intuitive to me

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That it has to be differentiable, sure

sleek thicket Jan 14, 2021, 7:51 PM

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have you seen the example with corners of a square?

hazy stratus Jan 14, 2021, 7:51 PM

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But why continuous

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Hm no

sleek thicket Jan 14, 2021, 7:51 PM

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err wait sorry that's for why we require immersion

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I was thinking about derivatives not having to vanish, sorry

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you can find a C^1 function R -> R^2 whose image is square (so not "smooth"), but its derivative must vanish

hazy stratus Jan 14, 2021, 7:52 PM

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Ah yeah that was one of the 3 definitions we had

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That one I haven't understood entirely though

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Why does the image have to be square?

sleek thicket Jan 14, 2021, 7:53 PM

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nah this is just an example of how things can go badly if you weaken the definition

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if we don't require that there's an immersion (just that there's a C^1 map which is a homeomorphism onto the image) then a square is a manifold too, which is intuitively bad

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since it has sharp corners

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I'm trying to think of a convincing reason for why we don't want all differentiable maps

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at a high level, discontinuous derivatives can be really weird and bad

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but it would be nice to say something more concrete

hazy stratus Jan 14, 2021, 7:55 PM

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sleek thicket if we don't require that there's an immersion (just that there's a C^1 map which...

Because the derivative on the corners isn't well defined?

sleek thicket Jan 14, 2021, 7:55 PM

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sort of, yeah. Manifolds should look like smooth curves or surfaces or whatever

hazy stratus Jan 14, 2021, 7:55 PM

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Like the derivative of abs in 0

sleek thicket Jan 14, 2021, 7:56 PM

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but by making the derivative of your parameterization go to 0 as you approach the corner you can still get a C^1 map

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think about like, moving along the top of the square

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and slowing down as you approach the corner

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slower and slower

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so that your velocity reaches 0 when you hit the corner

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then you turn around and speed up from there

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we can ensure that the derivative does actually exist

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since it's 0 as you approach from both sides

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but we've created a corner

hazy stratus Jan 14, 2021, 7:57 PM

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Is there a contradiction there?

sleek thicket Jan 14, 2021, 7:57 PM

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no, sorry. I don't think I'm explaining this well

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your definition includes that the parameterization be an immersion

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yeah?

hazy stratus Jan 14, 2021, 7:57 PM

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Yep

sleek thicket Jan 14, 2021, 7:58 PM

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I'm trying to provide intuition for why we want that (ie, why we want to make sure the derivative is nonzero)

hazy stratus Jan 14, 2021, 7:58 PM

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In fact

sleek thicket Jan 14, 2021, 7:58 PM

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because you can have a smooth parameterization of a curve which doesn't look smooth

hazy stratus Jan 14, 2021, 7:58 PM

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Ohhhh

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That makes sense

sleek thicket Jan 14, 2021, 7:58 PM

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also I think my reason for why manifolds should have C^1 parameterizations is so that we can use the inverse/implicit function theorems. these are like the most important tools in smooth manifold theory

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in fact I bet if you post your definitions then the IFT's will be how you go between them

hazy stratus Jan 14, 2021, 8:00 PM

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Right, I was trying to get a diffeo from half of a circle to an interval and tried to check if it would work by calculating the jacobian

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By the IFT, the function is only invertible if the jacobian is invertible if I'm not wrong

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Much thanks for the intuition🍀

sleek thicket Jan 14, 2021, 8:51 PM

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oh yeah okay I had a question so

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I am trying to understand the twisting sheaf

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just in the simplest case

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so we have $X = \mathbb{P}^1_\C$

gentle ospreyBOT Jan 14, 2021, 8:52 PM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 14, 2021, 8:52 PM

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we can defined this as $X = \mathrm{Proj} \C[x,y]$

gentle ospreyBOT Jan 14, 2021, 8:52 PM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 14, 2021, 8:53 PM

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we can cover by two "distinguished opens" $D(x), D(y)$, both of which look like $\mathbb{A}^1_\C$

gentle ospreyBOT Jan 14, 2021, 8:53 PM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 14, 2021, 8:53 PM

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with coordinate y/x on the first and x/y on the second

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so we can define the twisting sheaf on this base

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$\mathcal{F} = \mathcal{O}(1)$ is defined by $\mathcal{F}(D(x)) = (S_x)_1$ and $\mathcal{F}(D(y)) = (S_y)_1$

gentle ospreyBOT Jan 14, 2021, 8:56 PM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 14, 2021, 8:56 PM

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does this seem right?

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we're basically like

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hmm

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so what would this look like if I were actually interested in bundles on the manifold CP^1 lol

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$C^\infty(\C\mathbb{P}^1)$ is some weird ring

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is it graded? probably not lol!

gentle ospreyBOT Jan 14, 2021, 9:03 PM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 14, 2021, 9:04 PM

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so we have an injection into smooth functions on C^2{0}

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as the homogeneous ones

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err no

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not homogeneous, just invariant under scaling

tight agate Jan 14, 2021, 9:04 PM

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it might be easier to try to visualize O(-1) first

sleek thicket Jan 14, 2021, 9:04 PM

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well yeah O(-1) is what I actually care about

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it's the tautological bundle

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I'm trying to understand why

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so like, my understanding is that in the algebraic case we the tautological bundle equals O(-1)

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I'm trying to figure out if we can make sense of O(n) in the smooth case

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via a twisting kind of definition

tight agate Jan 14, 2021, 9:06 PM

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you mean you want to try do define it in terms of graded modules?

sleek thicket Jan 14, 2021, 9:06 PM

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Right

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I could define O(-n) as the nth tensor power of the tautological bundle

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and then O(n) as the dual of O(-n)

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but I want to understand the global sections of these bundles/what happens when you tensor them with another bundle

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so that's not really giving me anything new

tight agate Jan 14, 2021, 9:08 PM

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you can define it explicitly in terms of transition functions

wanton marsh Jan 14, 2021, 9:08 PM

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aren't half of those global sections empty

sleek thicket Jan 14, 2021, 9:08 PM

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is this different from defining it in terms of the tautological bundle @tight agate?

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and also zef im working on a smooth manifold

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the global sections capture all of the data of the sheaf

tight agate Jan 14, 2021, 9:10 PM

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sleek thicket is this different from defining it in terms of the tautological bundle <@!351398...

idk how different do you want it to be, but it isnt just 'take tensor powers'

sleek thicket Jan 14, 2021, 9:10 PM

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I guess I don't see what transition functions you're thinking of

tight agate Jan 14, 2021, 9:11 PM

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I mean define it to be trivial over each of the standard opens

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and then for the m'th twisting sheaf, the transitions are given by something like (x_j/x_i)^m

sleek thicket Jan 14, 2021, 9:12 PM

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hmm okay I think I see what you mean

sleek thicket Jan 14, 2021, 9:12 PM

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gentle osprey **shamroc$\\overline{k}$**

this thing I said earlier

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but in the smooth case

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that might work

tight agate Jan 14, 2021, 9:12 PM

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I havent really thought about the smooth case so idk

sleek thicket Jan 14, 2021, 9:13 PM

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I guess what I want to do is understand $\Gamma(E \otimes_{\C} L)$ in terms of $\Gamma(E)$ for a complex vector bundle $E$

gentle ospreyBOT Jan 14, 2021, 9:13 PM

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shamroc$\overline{k}$

📎 228562950703546368.png

sleek thicket Jan 14, 2021, 9:13 PM

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where L is the tautological line bundle on CP^1

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and I thought thinking of it as the twisting sheaf might help me do that

tight agate Jan 14, 2021, 9:13 PM

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but if you use the transition function definition for the holomorphic case, then the fact that O(-1) and the tautological bundle are the same is pretty straightforward to derive

sleek thicket Jan 14, 2021, 9:13 PM

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hm okay

tight agate Jan 14, 2021, 9:17 PM

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another way to identify O(-1) and the tautological bundle is using blow-ups

sleek thicket Jan 14, 2021, 9:17 PM

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i will take it one AG concept I barely understand at a time

astral cedar Jan 14, 2021, 11:31 PM

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Btw you can also define (if I recall correctly) O(m) as the quotient $\frac{(\mathbb{C}^{n+1} - {0})\times \mathbb{C}}{\mathbb{C}^*}$

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Where C* act on the first n+1 components via multiplication for t and on the last via multiplication for t^m

gentle ospreyBOT Jan 14, 2021, 11:35 PM

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alehmannbis

📎 769628190926176266.png

astral cedar Jan 14, 2021, 11:35 PM

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This makes it sort of easier to see how the section should work

sleek thicket Jan 14, 2021, 11:40 PM

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This makes sense

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Well, something like this makes sense

tight agate Jan 14, 2021, 11:41 PM

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mildly related to that construction: the complement of the zero section of O(-1) can be identified with C^n - {0}

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on P^n

sleek thicket Jan 14, 2021, 11:42 PM

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ah yeah I have that as a homework problem

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(in the smooth case)

tight agate Jan 14, 2021, 11:42 PM

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ye it's also an exercise in huybrechts

fathom cave Jan 15, 2021, 4:18 AM

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So an exercise has asked to show the skew-symmetric part $$S_{ij}=\frac{\partial T_i}{\partial Z^j}-\frac{\partial T_j}{\partial Z
^i}$$ of the variant $\frac{\partial Ti}{\partial Z^j}$ is a tensor. Transforming, we get
$$S_{i'j'}=S_{ij}J^i_{i'}J^j_{j'}+T_i\frac{\partial^2 Z^i}{\partial Z^{i'} \partial Z^{j'}}-T_j\frac{\partial^2 Z^j}{\partial Z^{i'}\partial Z^{j'}}$$
But i am not exactly sure what the problem is asking me. Is it asking me to show the first part in the above pde transforms tensorically?

gentle ospreyBOT Jan 15, 2021, 4:23 AM

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Zero0

📎 522733361031151616.png

hidden bear Jan 15, 2021, 4:24 AM

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Might be a silly question but what would be the symmetry group of a parallelapiped in R3 with L=/=W=/=H

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in R2 i know you can rotate by 180

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and by 0 of course, so im guessing in R2 itd be order2

chrome dew Jan 15, 2021, 4:36 AM

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📎 Screenshot_29.png

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@fathom cave

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notice you can relabel j to i

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that's it

fathom cave Jan 15, 2021, 4:38 AM

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ooooo right right Ty @chrome dew

chrome dew Jan 15, 2021, 4:39 AM

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yup you're welcome

bright acorn Jan 15, 2021, 11:27 AM

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Any examples of topological spaces that naturally arise in some fields of math but are not Hausdorff?

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Or being even more strict, any examples of important topological spaces that do not attend any separabiliy axioms nor countability axioms?

tough imp Jan 15, 2021, 11:39 AM

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The zariski topology on an affine scheme is usually not even T1

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It only satisfies the incredibly weak T0 axiom, in fact schemes exist where the closure of a single point is the entire scheme

bright acorn Jan 15, 2021, 11:42 AM

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That's some crazy stuff, man, I really need to study some algebraic geometry.

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There's some interesting stuff going on.

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I wonder why this is so.

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Anyway, any easy examples of algebraic or projective varieties with these properties?

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I've never studied about schemes yet

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But I think that it is somehow a generalization of the usual algebraic sets and projective varieties.

meager python Jan 15, 2021, 12:51 PM

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There are schemes without closed points too

shy moss Jan 15, 2021, 3:52 PM

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What is an example of non-path conected space?

gritty widget Jan 15, 2021, 3:52 PM

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your favorite non-connected space

shy moss Jan 15, 2021, 3:56 PM

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geometrically, What means the fundamental group is trival?

gritty widget Jan 15, 2021, 3:57 PM

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no holes

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petTheCat

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honkhonk

wicked trout Jan 15, 2021, 3:58 PM

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Hi folks - I'm trying currently to construct an orientable 3-manifold (probably with torus boundary components) with the same fundamental group as the Klein bottle.

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i.e. the Klein bottle cross the interval is no good

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Is there a way using the orientable double cover maybe?

river granite Jan 15, 2021, 4:05 PM

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shy moss What is an example of non-path conected space?

if you're interested here's a bunch of connected spaces that are not path connected https://topology.jdabbs.com/theorems/T000040

shy moss Jan 15, 2021, 4:09 PM

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thanks

wicked trout Jan 15, 2021, 4:16 PM

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wicked trout Hi folks - I'm trying currently to construct an orientable 3-manifold (probably ...

<@&286206848099549185> sorry for the ping but could use a hand with this one

marsh forge Jan 15, 2021, 5:32 PM

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@wicked trout

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nvm this is wrong in general

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📎 image0.png

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this seems to prove its still not possible tho

wicked trout Jan 15, 2021, 5:59 PM

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@marsh forge thats a v interesting result u have there

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basically im looking at (generalized) Heegaard diagrams for 3-manifolds with Baumslag-Solitar fundamental groups

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so if you take a genus 2 surface and draw a curve representing the element bab^{-1}a that defines a manifold w the fundamental group of the Klein bottle

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but i want to identify what it is so im going backwards

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trying to take the Klein bottle and do something to it to make it an orientable 3-manifold

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im thinking maybe a twisted line bundle??? who knows

finite meteor Jan 15, 2021, 8:17 PM

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Am I correct in saying that open balls cannot be complete metric spaces?

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ty

tacit stratus Jan 15, 2021, 8:47 PM

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wicked trout im thinking maybe a twisted line bundle??? who knows

How about this: The klien bottle is a fiber bundle of a circle over a circle, given as the mapping torus of the orientation reversing diffeomorphism of a circle. Instead, realize it as a bundle over a mobius strip

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so, itd be the mapping tours of the diffeo S1xR reversing both the S1 factor and the R factor. Since it reverses both, it should preserve the volume form on S1xR, and thus be orientable

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more concretely, it'd be $[0,1] \times \mathbb{C}^*, (0,z) \sim (1,z^{-1})$

gentle ospreyBOT Jan 15, 2021, 8:51 PM

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eeliot

📎 125777519717777408.png

tacit stratus Jan 15, 2021, 8:51 PM

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I think that should work?

wicked trout Jan 15, 2021, 10:27 PM

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@tacit stratus that sounds like what i was going for!! ill look over it tomorrow:) thanks!!

crimson imp Jan 15, 2021, 10:31 PM

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Hey guys, can someone help me with this exercise?

📎 image0.jpg

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I was able to prove the first implication, but I don’t know how to use the compact and ENR hypothesis

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Here ENR spaces are the ones such that there exists $r:V\to Y$ where V is a neighborhood of Y

gentle ospreyBOT Jan 15, 2021, 10:33 PM

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pmorelli

📎 339788063062360064.png

crimson imp Jan 15, 2021, 10:38 PM

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The book which I’m studying assume V and Y as subsets of R^n on the def of ENR. I don’t know if this would help on the solution

finite meteor Jan 15, 2021, 11:18 PM

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I am a little confused as to why a closed subspace of a complete metric space must be complete.

My understanding is the counter-example below is wrong based on the above, but I am having a hard time understanding why.

Lets say we have a metric space A {x in R: 0 <= x <= 5}
and a subspace B {x in Q: 0 <= x <= 5}

In my mind, B is closed, but it has a Cauchy sequence that converges to pi, which is in A but not in B.

Our book's definition of complete space is: every Cauchy sequence in a metric space X converges.

I don't see how the A/B example would break that. A sequence in B (subspace of A) might not converge to a value in B, but it does to one in A. B would not be complete as it doesn't have Pi.

gritty widget Jan 15, 2021, 11:19 PM

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B aint closed

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complement is irrationals in [0, 5], those dont contain an interval

finite meteor Jan 15, 2021, 11:20 PM

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ahh tyvm

gritty widget Jan 15, 2021, 11:20 PM

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np

astral cedar Jan 16, 2021, 1:23 AM

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@crimson imp i would think this has something to do with the compact-open topology?

bright acorn Jan 16, 2021, 1:42 AM

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river granite if you're interested here's a bunch of connected spaces that are not path connec...

This site is all about giving examples of topological spaces satisfying certain properties?

gritty widget Jan 16, 2021, 1:42 AM

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it's like an online version of the counterexamples book

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and more usable

bright acorn Jan 16, 2021, 1:43 AM

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I've never even heard of that book before.

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Is there something similar to other fields of math?

gritty widget Jan 16, 2021, 1:43 AM

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https://en.wikipedia.org/wiki/Counterexamples_in_Topology

bright acorn Jan 16, 2021, 1:43 AM

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Like

gritty widget Jan 16, 2021, 1:43 AM

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for analysis, as well

bright acorn Jan 16, 2021, 1:43 AM

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A site which gives lots of counterxamples to stuff concerning manifolds?

gritty widget Jan 16, 2021, 1:44 AM

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if you find one, let me know petTheCat

bright acorn Jan 16, 2021, 1:46 AM

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Well

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In fact

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A while ago I've found this site here

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http://www.map.mpim-bonn.mpg.de/2-manifolds

2-manifolds

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The manifold atlas project

gritty widget Jan 16, 2021, 1:47 AM

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interesting

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bookmarked, thanks

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petTheCat honkhonk duck

bright acorn Jan 16, 2021, 1:47 AM

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Take a look at it, idk how much of a use it would have to me at the moment.

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So that's why I asked you if you had any other references.

gritty widget Jan 16, 2021, 1:47 AM

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only the copy of lee on my desk

crimson imp Jan 16, 2021, 2:41 AM

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astral cedar <@339788063062360064> i would think this has something to do with the compact-op...

I don't know, the only thing I manage to prove, which is a consequence of triangle inequality is that d(f(x),g(x))≤n\epsilon

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This n may depend on epsilon, I'm trying to use compacity argument to prove that it doesn't, but I don't know where this would take me

fathom cave Jan 16, 2021, 4:29 AM

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Does chirstofel's symbol transform tensorically under linear transformation? I got this does and just making sure

gritty widget Jan 16, 2021, 4:43 AM

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no, i don't think they do. here's an excerpt from lee's riemannian manifolds book

📎 unknown.png

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if i'm misunderstanding what you mean by "transform tensorically" lemme know

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i just have a hunch this is what you're looking for petTheCat

chrome dew Jan 16, 2021, 4:48 AM

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if by linear transformation he means A has constant entries, then its derivative is 0 and so you don't have that second term, so it would transform as a tensor

gritty widget Jan 16, 2021, 4:53 AM

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petTheCat

fathom cave Jan 16, 2021, 4:53 AM

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ya that merosity

chrome dew Jan 16, 2021, 4:54 AM

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😎

fathom cave Jan 16, 2021, 4:54 AM

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haha

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and what is that book TTerra

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i dont think there is a word 'tensorically' i just made tht up 🤣

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well u get the idea

gritty widget Jan 16, 2021, 4:58 AM

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introduction to riemannian manifolds

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by john m lee

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very good book

shut moat Jan 16, 2021, 5:01 AM

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is there an intuitive reason for why this was defined like this? nervousSweat

📎 Screen_Shot_2021-01-16_at_12.01.14_AM.png

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it seems reasonable that the result is a k+l form but I can't really picture this

gritty widget Jan 16, 2021, 5:06 AM

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maybe bad answer: because then dx^1 wedge ... wedge dx^n is the volume form (up to a scalar multiple)

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although i wouldn't be surprised if that's not a bad answer, since a lot of things relating to DG seem to be defined just for the sake of getting a certain property lol

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that's just something i pulled out my ass on the spot, btw petTheCat

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good question

shut moat Jan 16, 2021, 5:08 AM

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I hope there's more to it than that, but I do agree that thinking of it in the context of elementary forms is easier

gritty widget Jan 16, 2021, 5:08 AM

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there might be some stupid algebraic way to look at the wedge product that characterizes it as something that's completely natural to consider, idk

shut moat Jan 16, 2021, 5:09 AM

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damn there's a stackexchange post for everything

jagged ocean Jan 16, 2021, 5:09 AM

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this could be helpful or not, or useful or not, but I always go back to the idea that forms are merely alternating tensors of covariant rank, and in the same way we have the tensor products to make higher order tensors, we want an alternating (or antisymmetric) product to make higher order alternating covariant tensors. the wedge product should be the antisymmetrization of the tensor product.

gritty widget Jan 16, 2021, 5:10 AM

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god fucking fuck of course there's a goddamn algebraic natural isomorphism bs to it pepega

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fking algebraists

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i do like the "wedge product is the antisymmetrization of the tensor product" that craft suggested

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since it's just Alt(phi tensor omega), ya?

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maybe you can show that as an exercise, ~circle petTheCat

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(maybe, up to a scalar multiple, though)

jagged ocean Jan 16, 2021, 5:14 AM

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yeah so in warner they construct the exterior algebra as equivalence classes of covariant tensors modulo the ideal generated by stuff of the form (v tensor v), the stuff that definitely needs to be trivial in the exterior algebra. from here, you can just say that the wedge product of two forms, or two equivalence classes, is just the equivalence class of the tensor product of any pair of representative samples from each of the two original equiv classes

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(not to imply that warner's construction is unique, just I like that reference in particular...)

shut moat Jan 16, 2021, 5:21 AM

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That makes sense, ty!

gritty widget Jan 16, 2021, 5:27 AM

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warner monkaS

shut moat Jan 16, 2021, 5:30 AM

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Just to clarify about the argument that uses
$$\bigwedge^k V^* \equiv \qty(\bigwedge^k V)^*$$
it identifies the two vector spaces via the isomorphism $$\phi_1\wedge \phi_2\wedge...\wedge \phi_k \mapsto \qty(v_1\wedge v_2\wedge ... \wedge v_k \mapsto \det(\phi_i(v_j)))$$
So the wedge product of forms is equivalent to sortof just concatenating them into the matrix?

gentle ospreyBOT Jan 16, 2021, 5:30 AM

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~S^1

📎 593959711020154891.png

obtuse meteor Jan 16, 2021, 9:41 PM

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So I remember doing this exercise in an algebraic topology book once but I can't find the reference

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so I'm crowdsourcing it

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As I recall there was some construction of a ring associated to a topological space that preserved all the information of the space. I think up to homeomorphism but maybe up to homotopy

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the point of the exercise was that certain algebraic constructions for topological spaces are too fine-grained. If you recover all the information then these things are like impossible to compute and don't allow you to solve any problems easier

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but I can't remember the book or the construction

astral cedar Jan 16, 2021, 9:45 PM

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Maybe the fundamental infinity groupoid?

little hemlock Jan 16, 2021, 9:45 PM

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i know what you are talking about

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📎 unknown.png

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@obtuse meteor you are talking about this, right?

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this is from page 13 of rotman's intro to AT

obtuse meteor Jan 16, 2021, 9:54 PM

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thanks!

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I think this is it

little hemlock Jan 16, 2021, 9:54 PM

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npnp

obtuse meteor Jan 16, 2021, 9:54 PM

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yeah I knew it was a simple construction

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but my brain was just gone

sleek thicket Jan 16, 2021, 10:09 PM

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The same is true for smooth manifolds, btw

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Taking the ring of smooth functions is a full and faithful functor

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and actually if you think of the manifold as a locally ringed space, the (co?)unit M -> Spec C^infty(M) of the spec/global functions adjunction is a topological embedding

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and when M is compact (iirc?) the image is exactly the maximal ideals

tidal cedar Jan 16, 2021, 10:11 PM

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Isn't there also an equivalence with C^infty modules and bundles over the manifold or similar?

sleek thicket Jan 16, 2021, 10:12 PM

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¯\_(ツ)_/¯

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there's serre swan

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but your thing sounds more general

gritty widget Jan 16, 2021, 10:12 PM

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please remove your category theory from my calculus spaces petTheCat

tidal cedar Jan 16, 2021, 10:17 PM

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Oh yeah it was Serre-Swan

marsh forge Jan 16, 2021, 10:23 PM

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that theorem is also the starting point of Top K-Theory -> Alg K-Theory

obtuse meteor Jan 16, 2021, 11:08 PM

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@sleek thicket yeah I'm familiar with this idea. It's very nice !!!

gentle ospreyBOT Jan 16, 2021, 11:08 PM

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Cohomologay

📎 131041620131840000.png

sleek thicket Jan 16, 2021, 11:09 PM

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oh hey my name thing actually works

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i tried it earlier and it didn't lmao

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and yeah I think it's sweet!

obtuse meteor Jan 16, 2021, 11:11 PM

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I hate your name thing tbh

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I want to be able to mention you without texit going "brrrrrrrrrrrrrrrrrrrrrrrrrrrrrr LaTeX"

marsh forge Jan 16, 2021, 11:17 PM

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ban latex usernames 2021

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i tried in ivory-tower

twin nebula Jan 16, 2021, 11:18 PM

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Hello everyone could someone spare 2 mins to help my dumb ass with a question thats probs very easy for you to spot the answer to

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Do i send it here? Its not exactly related to topology, although i wish i knew more about thee subject to be able to oost questions here about that

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My question was about this equality

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📎 unknown-3.png

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And if it would hold true from -inf to 0

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Instead of 0 to inf

obtuse meteor Jan 16, 2021, 11:25 PM

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hmm

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IDK where the best place to ask this would be

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problably #multivariable-calculus or #calculus

This equality seems to come from integration by parts + gaussian integral. It will definitely hold from -inf to 0 because the functions are even, aka f(x) = f(-x)

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topology/geometry isn't closely related to this question :)

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so let's move to #calculus

twin nebula Jan 16, 2021, 11:28 PM

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Sorry to ask this but would you mind helping me understand cause in my head it seems like in 0 to inf u sub in inf first and thrn 0 which woukd =0

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But for -inf to 0 ud sub 0 first which =0 and then -inf

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And since when u sub in its [sub a] - [sub b]

obtuse meteor Jan 16, 2021, 11:29 PM

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let's move to #calculus and we can talk about it there

twin nebula Jan 16, 2021, 11:29 PM

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Wouldnt thr result be muktiolied by -1?

obtuse meteor Jan 16, 2021, 11:29 PM

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we try not to clog up these channels with this stuff :)

twin nebula Jan 16, 2021, 11:29 PM

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Sure

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Sorry

obtuse meteor Jan 16, 2021, 11:30 PM

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it's no problem!

summer jolt Jan 16, 2021, 11:55 PM

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Hey, I was reading topology and groupoids by Ronald Brown and came across this characterisation of neighbourhoods

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📎 unknown.png