« first day (5204 days earlier)      last day (112 days later) » 

Jakobian
Jakobian
00:02
@DannyuNDos Isn't arctan such homeomorphism
@DannyuNDos they're not uniformly isomorphic, sure
Dannyu NDos
Dannyu NDos
Yeah
Jakobian
Jakobian
Because uniformly continuous functions should preserve such things as boundedness
So uniformly continuous surjection $(0, 1)\to\mathbb{R}$ cannot exist
well, totally bounded at least
I think bounded is not something they necessarily preserve
Dannyu NDos
Dannyu NDos
Yeah; the notion of standard bounded metrics ruin boundedness of functions.
Jakobian
Jakobian
Ah, right. The metric spaces $(X, d)$ and $(X, d\land 1)$ are uniformly isomorphic
If $X$ is totally bounded and $f:X\to Y$ is uniformly continuous then there is an extension of $f$ to $\tilde f:\tilde X \to \tilde Y$ where $\tilde X$ is compact, so $f(X)$ is totally bounded
Where $X, Y$ are uniform spaces and $\tilde X, \tilde Y$ are their completions
Dannyu NDos
Dannyu NDos
00:24
Never knew the notion of total boundedness extends to an arbitrary uniform space; TIL, thanks.
Jakobian
Jakobian
Yes. A uniform space is totally bounded iff its completion is compact as well
Equivalently, $X$ is totally bounded if for every compatible with its uniformity pseudometric $d$, every $d$-discrete set is finite
leslie townes
leslie townes
i demand to know how this generalizes to bornological spaces
Jakobian
Jakobian
00:40
No clue. Never looked at bornological spaces
leslie townes
leslie townes
nobody has, its some thing from the mid 20th century that people thought would help do functional analysis but doesn't
Jakobian
Jakobian
@leslietownes I've walked 11 kms at a steady pace today
~7 miles
leslie townes
leslie townes
in comfortable shoes, i hope. that is approaching a distance where you might injure yourself otherwise
Jakobian
Jakobian
They're my old shoes, so I suppose
leslie townes
leslie townes
when i was in high school i walked home every day, which was about 4 miles. i plan on bothering my daughter about this when she is old enough to appreciate '4 miles'
Jakobian
Jakobian
00:49
In what sense appreciate them?
leslie townes
leslie townes
like, to understand what that is as a distance. her conception of distance and time is relatively undeveloped. her conception of "an hour" is about 20 minutes right now.
i also did this with a fully loaded backpack. i invented the rucking workout
Jakobian
Jakobian
I've used to sit under my house door for hours. I know patience
leslie townes
leslie townes
her elementary school and likely high school are both about is 2.5 miles away and we drive her there, and because of traffic safety i would not dream of letting her walk that route. our main street that is just nearby, a pedestrian gets killed on it every 4-6 months. it is not a nice place. and that's before you cross a highway.
Jakobian
Jakobian
I've walked maybe 2 miles to my school
leslie townes
leslie townes
so i'm rehearsing my speech about how in my day we walked 4 miles home from school and were happy to do it
Jakobian
Jakobian
00:55
@leslietownes that sounds very american
leslie townes
leslie townes
yes, very much so
which is weird, because most of the places i've lived have been very walkable, but very much not this one
Sine of the Time
Sine of the Time
@leslietownes parents describing how they went to school:
leslie townes
leslie townes
sine: uphill both ways!
Sine of the Time
Sine of the Time
user image
Jakobian
Jakobian
Ankles still intact 👍
I injured my achilles once. I could only jump
leslie townes
leslie townes
01:02
my wife and i went to london once and walked everywhere, like 6-12 miles a day for several days, and by the end of it we basically couldn't move due to tendon stuff. london is a really walkable city though. would do it again
i'm told that they have some kind of underground rail system that people can also use to get from place to place
Jakobian
Jakobian
@leslietownes pretty sure all major cities do. Same with Warsaw
leslie townes
leslie townes
jakobian, did you know that los angeles has a light rail system and that it doesn't (yet) go to the airport :)
until very recently it didn't even come close. they will probably have something working by the olympics
pre WW2 we had a train system that connected literally everything. i could have walked from my current house to a train stop and gone basically anywhere in metro LA. the land that those train tracks used to go over is now a long awkwardly shaped park and we have an interstate highway instead
i can get on the 405 and be nowhere in 2 hours
 
3 hours later…
Xander Henderson
Xander Henderson
04:36
@leslietownes You misspelled "interstate parking lot system".
leslie townes
leslie townes
:(
 
2 hours later…
Soumik Mukherjee
Soumik Mukherjee
06:49
I prefer walking over cycling.
 
3 hours later…
Sine of the Time
Sine of the Time
09:43
@SoumikMukherjee cycling>walking
Soumik Mukherjee
Soumik Mukherjee
10:01
>:(
Sine of the Time
Sine of the Time
10:13
<:)
think_meaning_builds
think_meaning_builds
10:36
>8(
Sine of the Time
Sine of the Time
chat is sleeping
think_meaning_builds
think_meaning_builds
yup, gotta get its 8 hours
Ben Steffan
Ben Steffan
rise and shine
Sine of the Time
Sine of the Time
Rise and sine
psie
psie
11:18
@Thorgott I was wondering, you said that $x$ is contained in exactly one $Q_i$, are you sure about that? If I draw a cube in the plane, I see that a point may land on the boundary of some cube, but these cubes may share their boundary with other cubes. I would like to conclude something like $$\left|\sum_1^{N} |\det D_{x_j}G|\chi_{Q_j}(x)-|\det D_xG|\right|=\left||\det D_{x_j}G|-|\det D_xG|\right|<\epsilon,$$which is only possible if $x$ lands in exactly one $Q_j$.
No matter how I change the partition, the $x$ will remain on the boundary of several cubes.
I could probably make the partition coarser instead of finer, which would maybe ensure that $x$ only lands on the boundary of one cube (or no boundary at all).
But the problem then is that $x$ may not be a distance $\delta$ away from the center of some cube...
I'm stuck.
Thorgott
Thorgott
11:48
sorry, I was thinking the cubes would be half-open or something
hmm, but I don't think the claim is correct without such a stipulation?
Pizza
Pizza
Hi
Sine of the Time
Sine of the Time
hi
Pizza
Pizza
@SineoftheTime Anyway I'm still waiting for the exam results, I don't know why the professor is taking so long
Sine of the Time
Sine of the Time
more than one week is strange
Pizza
Pizza
It usually took 3 to 5 days
@SineoftheTime Yes, maybe this weekend he'll publish them
Sine of the Time
Sine of the Time
12:03
yeah
nickbros123
nickbros123
Im trying to come up with an example for a non-abelean group such that $\{g \in G: |g|<\infty\}$ is actually not a subgroup. Is there something like dihedral group of order infinity xD
ok now that I think about it $symm(\mathbb{N})$ might work
Pizza
Pizza
@SineoftheTime anyway I was studying algebra and geometry theory, what do you recommend I do better?
That is, I saw that there are many propositions and demonstrations on these
Sine of the Time
Sine of the Time
@Pizza what do you mean?
Pizza
Pizza
I mean I saw that some people only did the "named" topics
Sine of the Time
Sine of the Time
I don't understand the question
is the exam written+oral?
nickbros123
nickbros123
12:07
@nickbros123 I think $f=(12)(34)(56) \cdots$ and $g=(23)(45)(67)\cdots$ might do the trick
or not
Pizza
Pizza
@SineoftheTime Yes the oral is apart
Sine of the Time
Sine of the Time
Did you already do the written part?
Pizza
Pizza
no, but I wanted to do the oral part because I more or less already know how to do the written things
Because maybe it ends up that the oral exam is a few days later and I won't have time to do everything
Sine of the Time
Sine of the Time
propositions and theorems in linear algebra are not hard
what are you asking precisely?
Soumik Mukherjee
Soumik Mukherjee
@nickbros123 try matrix groups
nickbros123
nickbros123
12:14
GL_n(R) ?
Pizza
Pizza
for example there is a demonstration on : S is a basis;
2 S is maximal with respect to the property of being independent; So (1) <-> (2)
is this important?
Sine of the Time
Sine of the Time
yes
Pizza
Pizza
Oh okok, then I'll try to do everything with a demomstration well
What were you asked, if you remember?
Sine of the Time
Sine of the Time
in my exam?
Pizza
Pizza
I mean, what did they ask you in the algebra and geometry oral exam?
@SineoftheTime yes
Steinitz's theorem it is almost always asked
Sine of the Time
Sine of the Time
12:21
@Pizza what's that?
Pizza
Pizza
it.m.wikipedia.org/wiki/Lemma_di_Steinitz
Ben Steffan
Ben Steffan
that's Steinitz' lemma, not Steinitz' theorem :)
Sine of the Time
Sine of the Time
@Pizza Teorema di struttura per operatori isometrici di uno spazio Euclideo
Pizza
Pizza
@BenSteffan In my slides It says theorem :'(
Sine of the Time
Sine of the Time
The the professor asked other random questions
Ben Steffan
Ben Steffan
12:24
Steinitz' theorem also exists but is something else
Sine of the Time
Sine of the Time
@Pizza I know it, but I've never heard that name
Pizza
Pizza
@BenSteffan If I write Steinitz theorem that Wikipedia link comes up
Ben Steffan
Ben Steffan
uhh
Steinitz's theorem
In polyhedral combinatorics, a branch of mathematics, Steinitz's theorem is a characterization of the undirected graphs formed by the edges and vertices of three-dimensional convex polyhedra: they are exactly the 3-vertex-connected planar graphs. That is, every convex polyhedron forms a 3-connected planar graph, and every 3-connected planar graph can be represented as the graph of a convex polyhedron. For this reason, the 3-connected planar graphs are also known as polyhedral graphs. This result provides a classification theorem for the three-dimensional convex polyhedra, something that is not...
Pizza
Pizza
Given a vector space $V$ and two sets of vectors $S = \{ u_1, u_2, \dots, u_m \}$ and $T = \{ v_1, v_2, \dots, v_n \}$. If the vectors in $S$ are linearly independent and each of them depends on the vectors in $T$, then the cardinality of $S$ is less than or equal to the cardinality of $T$, that is, $m \leq n$.
Ben Steffan
Ben Steffan
that's the lemma, yes
Pizza
Pizza
12:29
to me its wrote Steinitz's theorem and this
@BenSteffan Ah then I think it's a mistake
Ben Steffan
Ben Steffan
maybe italian uses a different convention
but in english it's like this :)
nickbros123
nickbros123
@SoumikMukherjee testing out elements in GL_2(R), seems like im just throwing darts lol. trying out A^2=1 elements and seeing where it lands
but also turns out $D_{\infty}$ is actually a thing
Pizza
Pizza
@BenSteffan another prof calls it lemma, though
psie
psie
@Thorgott I guess we could just ignore the boundary since it has Lebesgue measure $0$. Folland is saying $\sum_1^{N} |\det D_{x_j}G|m(Q_j)$ is the integral of $\sum_1^{N} |\det D_{x_j}G|\chi_{Q_j}(x)$, where $Q_j$ are cubes that partition $Q$ and have disjoint interior. However, it is also the integral of $\sum_1^{N} |\det D_{x_j}G|\chi_{P_j}(x)$, where $P_j$ are cubes with the boundary removed.
The function with the $P_j$'s converges uniformly to $|\det D_xG|$ as I tried to show, though I'm not sure about the other function with the $Q_j$'s.
psie
psie
12:54
I take my statement back...$\sum_1^{N} |\det D_{x_j}G|\chi_{P_j}(x)$ does not converge uniformly on $Q$ to $|\det D_xG|$, probably only on the interior of $Q$...hmm.
Maybe I need to change my argument entirely.
Soumik Mukherjee
Soumik Mukherjee
@nickbros123 try A=([0 1][1 0]), B=([0 2][1/2 0])
 
1 hour later…
Jakobian
Jakobian
14:10
Does anyone else find it annoying that sometimes you need to catch up to massive amounts of context to answer someone here. Not directed towards anyone, it just takes so much time to read up on it all
I'm basically complaining that I need to read
@psie I don't get it. What is $\Omega$? An open set right? Then why are we decomposing an open set into a finite number of compact sets?
ah no sorry, we are decomposing a fixed compact cube $Q$
If $f_n$ are continuous, and converge pointwise to continuous $f$, all on a compact set, then this convergence doesn't have to be uniform. So you have to do some special argument like Leslie is suggesting
Ryder Rude
Ryder Rude
14:25
hi
Jakobian
Jakobian
Sure, lets assume $|x-y|\leq \delta \implies ||\det D_x G| - |\det D_y G||\leq \varepsilon$
nickbros123
nickbros123
Can anyone provide an example of an isometry from X which is a non compact metric space to itself that is non surjective? The unit vectors in $\ell_{\infty}$ is one example, are there others?
By shift operator
Jakobian
Jakobian
@nickbros123 come again?
oh you mean $X$ is a non-compact metric space?
Certainly, plenty examples exist
For instance if $X$ is an infinite discrete metric space, with the standard metric, and $X\to X$ is any injection, then its an isometry
and since there exist non-surjective injections from an infinite set to itself, we have an example for every cardinal number
If you like, you can also, for example, consider shift-to-right operator on $\ell^p$ for any $0 < p\leq \infty$
Jakobian
Jakobian
14:48
@RyderRude hi
 
1 hour later…
psie
psie
15:52
@Jakobian Indeed, we have a fixed cube and we decompose it into subcubes. Then we derive the inequality $$m(G(Q))\leq (1+\epsilon)\sum_1^N |\det D_{x_j}G|m(Q_j),$$which only holds for $Q_j$ having a certain side length $\delta$ small. Then it is claimed the function whose integral is the sum above converges uniformly to $|\det D_xG|$ on $Q$ as $\delta\to0$.
user image
Consider the cube I've drawn here, and the circle there represents a point on the boundary of a cube. I feel like my argument here fails for that point.
This is just an illustration of what I'm trying to convey. It lies on the boundary of several cubes and we might get contribution of several terms from the sum.
psie
psie
16:15
user image
Thorgott
Thorgott
@psie this works if the $P_j$ are half-cubes
nickbros123
nickbros123
@Jakobian thanks 👍. This was asked today in an analysis quiz in class, I couldn't figure it out in the quiz. After some thinking I thought of ell_infty. Sad life.
Jakobian
Jakobian
16:37
@psie yeah you should go for what Thorgott is suggesting i.e. the cubes can be taken to be of the form $[a_1, b_1)\times ...\times [a_n, b_n)$
psie
psie
ok 👍 no uniform convergence on $Q$ then, as Folland claims...
it doesn't matter probably
or maybe we do get uniform convergence on all of $Q$, I'm fried
Jakobian
Jakobian
@psie no, the subcubes need to definitely be taken to be disjoint
which is always possible, so its not a big deal
psie
psie
ok
Jakobian
Jakobian
which page is it
psie
psie
page 75
Jakobian
Jakobian
16:50
which version
psie
psie
2nd
Jakobian
Jakobian
its not in the errata
> Additional corrections will be gratefully received at [email protected] .
psie
psie
yeah, I checked the errata too
Jakobian
Jakobian
wait a second
Folland says its the integral that converges uniformly
ah no he isn't, sorry
yeah sure its something you can bother Folland about I would say
psie
psie
well, it's the highlight of the section
Jakobian
Jakobian
16:55
the convergence is uniform on a subset $A$ of $Q$ with $\lambda(Q\setminus A) = 0$ that's for sure
so its not like the argument is wrong
the "edges" of the cubes are the problem
psie
psie
@psie yeah, we get the last inequality here either way, but it's the statement about the uniform convergence of the function on $Q$ which is...questionable
Jakobian
Jakobian
you can do what Thorgott said about considering half-cubes, or just accept the uniform convergence isn't on the whole of $Q$, but on $Q$ except for a set of measure zero
@psie its not questionable, its just false
psie
psie
then I should probably email him :D
@Jakobian if we partition the cube into half-cubes though, the union of the half-cubes won't be the entire cube, or?
Jakobian
Jakobian
@psie not all of them have to be half-cubes
psie
psie
true
Thorgott
Thorgott
17:05
all you have to do is partition the cube into subsets with diameter that can be bounded by $\delta$ in some way
psie
psie
17:16
ok, thanks for this idea 👍 I'm getting comfortable with this argument.
Pizza
Pizza
17:53
Hi
Sine of the Time
Sine of the Time
18:10
@Pizza hi
Pizza
Pizza
@SineoftheTime i have a news
Sine of the Time
Sine of the Time
good or bad?
Pizza
Pizza
the results are out
I got 26
Sine of the Time
Sine of the Time
@Pizza 😬
@Pizza respect
the max was 27?
Pizza
Pizza
yes I think between 26 and 27
:)
Sine of the Time
Sine of the Time
18:16
well done!
Pizza
Pizza
all thanks to your help!
Ben Steffan
Ben Steffan
👏
psie
psie
Pizza is delivering :) (instead of delivering Pizza)
Vladimir Lysikov
Vladimir Lysikov
Congrats
Sine of the Time
Sine of the Time
@Pizza nah, that's not true. Yes, I've helped you, but you've been on those exercises for months so it's your merit
Binky
Binky
18:19
🎉🎉🎉
Pizza
Pizza
Thank you all !
@SineoftheTime yes but it's also partly your merit
Sine of the Time
Sine of the Time
I'll wait for my paycheck then🙃
Binky
Binky
Does anyone here play Pokémon?
@SineoftheTime are you free ?
Sine of the Time
Sine of the Time
@Binky yes
Binky
Binky
@SineoftheTime I thought you were paid
Sine of the Time
Sine of the Time
18:28
@SineoftheTime I was joking
Binky
Binky
Anyway, I bought all the books you told me to buy
ModularMindset
ModularMindset
Where might I look to figure out how to put a global metric on a stratified space?
Sine of the Time
Sine of the Time
@Binky I did not tell you to buy anything
I told you what are the most used books
Uni books cost a fortune
ModularMindset
ModularMindset
The singularities look like $\sqrt{x}$
as $x \to 0$
Binky
Binky
Yes, I meant that I bought the most used ones
Sine of the Time
Sine of the Time
18:33
@Binky did you find all of them?
Ben Steffan
Ben Steffan
spending money on math books? in this economy?
on a sidenote, springer has a 40% sale on hardcovers going on and I'm tempted, for the first time in a while
Sine of the Time
Sine of the Time
Lately, I'm downloading and printing instead of buying
I don't know if it's morally acceptable
Ben Steffan
Ben Steffan
it is morally accepted
I recommend looking into how the business model of academic publishers works if you have some time, your moral qualms will disappear :)
Binky
Binky
@SineoftheTime yes
Sine of the Time
Sine of the Time
I don't know how it works in Germany, but in Italy it's hard to find a math book less that 50-60 euros
Ben Steffan
Ben Steffan
18:36
in particular note that authors generally don't get paid when you buy their books
at least not with the big academic publishers like springer
ModularMindset
ModularMindset
I have purchased 2 math books in my career
Ben Steffan
Ben Steffan
@SineoftheTime it's the same here
Sine of the Time
Sine of the Time
I printed a 400 pages book, payed less than 10 euros
Ben Steffan
Ben Steffan
in fact, the more mathematics you learn the more expensive the books get, as a rule of thumb
there are books I'm interested in that cost well beyond 100 euros
Sine of the Time
Sine of the Time
yeah, that sucks
Ben Steffan
Ben Steffan
18:38
see also global.oup.com/academic/product/…;
ModularMindset
ModularMindset
Any good books on algebraic topology (in general, and in particular covering stratified spaces)?
Ben Steffan
Ben Steffan
I don't know a single book covering stratified spaces
Hatcher is pretty much everyone's baseline nowadays, owing at least part to the fact that it's available online for free from the author's webpage. tom Dieck's text is also popular and probably better in certain aspects, but I haven't used it much and can't give much of an opinion. May's "A Concise Course in Algebraic Topology" is fantastic, but it's, well, concise and not meant as an introductory textbook per se, more as a reference.
It's follow up work "More Concise Algebraic Topology" is also great and very useful as a reference; the material it covers is not collected in any single other book as far as I know.
Spanier is old but has some merits. Whitehead's "Elements of Homotopy Theory" is also sometimes referenced and contains some useful things that you can't necessarily find in other texts.
Soumik Mukherjee
Soumik Mukherjee
@BenSteffan I recently bought a springer book with 61% discount
Ben Steffan
Ben Steffan
The amount and extent of sales and vouchers that springer offers should tell you something about the margins in their business model
Soumik Mukherjee
Soumik Mukherjee
Yeah
Ben Steffan
Ben Steffan
18:47
I get some form of sales promotion or voucher by mail every other week or so, usually at least 30% discount
Soumik Mukherjee
Soumik Mukherjee
Ooh
Ben Steffan
Ben Steffan
"huh, wonder how they can afford to do this"
:)
Soumik Mukherjee
Soumik Mukherjee
@SineoftheTime see this king walk
Sine of the Time
Sine of the Time
@SoumikMukherjee circumnavigation of the white pawns
played a couple of bullets today, I hung my queen like 3 times
Soumik Mukherjee
Soumik Mukherjee
It's not a bullet otherwise
Sine of the Time
Sine of the Time
18:54
I was tired
and I need a back massage
:(
Soumik Mukherjee
Soumik Mukherjee
I offered draw in a totally lost position and my opponent accepted. It was like -8.5 or something 😂
Thorgott
Thorgott
@BenSteffan I can vouch for tom Dieck, also Bredon's Topology and Geometry (though not strictly algebraic topology), those are two of my favorite books
Sine of the Time
Sine of the Time
@SoumikMukherjee menace
Thorgott
Thorgott
Whitehead is also very respectable, but it's purpose is mostly as a reference work imo
insane amount of knowledge in that book, though
like ancient sorcery levels type of stuff
you open that book randomly and find a section computing the nilpotency class of the group $[X,G]$ for $X$ a CW-complex and $G$ an $H$-group
Soumik Mukherjee
Soumik Mukherjee
@SineoftheTime here
Sine of the Time
Sine of the Time
18:59
You're in zugzwang
Soumik Mukherjee
Soumik Mukherjee
Yup
Pizza
Pizza
Anyway, now there should be the oral on November 7th, but I'm undecided whether to do it or not.
Soumik Mukherjee
Soumik Mukherjee
Funny how I avoided 3fold only to blunder the game
@Pizza optional exam? Is that real?
Pizza
Pizza
I mean, in the end it's in 5 days, but I should also do the other subjects that I have behind, I don't know what to do :(
@SoumikMukherjee The theoretical questions were in the written exam, so the oral exam is not mandatory.
Sine of the Time
Sine of the Time
@SoumikMukherjee yes, that happens to me all the times
Soumik Mukherjee
Soumik Mukherjee
19:01
Ooh
Sine of the Time
Sine of the Time
@Pizza how does it work? The professor can give you a lower mark or you start from $\ge 26$?
Pizza
Pizza
@SineoftheTime Those who wish to improve (or worsen :-)) the grade obtained (for a maximum of 20%), will be able to take the oral exam on both the theoretical and practical topics of the course.
It's written like this
Sine of the Time
Sine of the Time
makes sense
it's a bit strange, but I assume the professor wants to put emphasis on the written part
Ben Steffan
Ben Steffan
@Thorgott ah, I feel a bit more ambivalent about Bredon
I actually own a physical copy, but I rarely use it
it's not a book I go to first or second
although it's fine
yeah, I wouldn't read Whitehead per se, nor, say Spanier, even though that text can also be useful
How do you feel about Aguilar-Gitler-Prieto? :)
Pizza
Pizza
@SineoftheTime Yes, I don't know what to do :(, obviously because it is written that the vote can also get worse
otherwise obviously I would have tried
Ben Steffan
Ben Steffan
19:11
or about Fomenko-Fuchs?
Thorgott
Thorgott
I also own a copy of Bredon, it's admittedly not primarily an algebraic topology textbook, but I still think it has arguably the best treatment of singular cup and cap products among the standard references, also a very clear exposition on Eilenberg-MacLane spaces and obstruction theory imo
@BenSteffan never read it
Ben Steffan
Ben Steffan
@Thorgott it's a funny book, of sort
they define $H_n(X) := \pi_n(SP(X))$
Sine of the Time
Sine of the Time
@Pizza that's up to; it depends your preparation
I can't help you
But if you feel insecure, I'd not risk
Thorgott
Thorgott
@BenSteffan a very workable definition, I'm sure
Ben Steffan
Ben Steffan
they seem to make it work, anyhow
I'm at a loss why you would decide to do it that way, but
Thorgott
Thorgott
19:14
@BenSteffan I once spent multiple hours discussing with a friend what the homology of a space in one of that books pictures is
Ben Steffan
Ben Steffan
I'm baffled that Fomenko-Fuchs got published in the state it's in
Thorgott
Thorgott
it has some good stuff on spectral sequences, but I've admittedly never read as much of it as I should
Ben Steffan
Ben Steffan
there's so many typos
Thorgott
Thorgott
same goes for Mosher-Tangora
Ben Steffan
Ben Steffan
Mosher-Tangora I should look at sometime
ModularMindset
ModularMindset
19:16
Is SP a configuration space of some kind?
Thorgott
Thorgott
@BenSteffan typos are everywhere in books that aren't like 3rd edition+, I feel
Lurie also has so many
Ben Steffan
Ben Steffan
@ModularMindset in a sense
it's the infinite symmetric product
you can find it e.g. in Hatcher :)
@Thorgott HTT pales in comparison to Fomenko-Fuchs
HA has more, but then again HA is not a published work
(in terms of typos)
Thorgott
Thorgott
I see lol
Ben Steffan
Ben Steffan
but the pictures are very nice :)
Thorgott
Thorgott
indeed
a friend told me recently he had some drugs stashed away in his copy of Fomenko-Fuchs and I suggested he might comprehend the pictures if he takes them
Ben Steffan
Ben Steffan
19:21
lol
Vladimir Lysikov
Vladimir Lysikov
Oh
Algebraic topology
I have a question
I work in applied algebra/theoretical computer science
Never learned algebraic topology properly, but kind of remember definitions of the cohomology of a complex, de Rham cohomology, and how to compute simple cohomologies using Mayers-Vietoris
Also know Hilbert polynomials for projective varieties
And I guess I want to know if there is something there which will help with questions I am interested in
For example, "can this orbit closure appear in the boundary of this another orbit"
Is there something I can try to read or the only way is to find an algebraic topologist who will listen to me and talk to him?
Ben Steffan
Ben Steffan
this is a bit too vague to be answerable I'm afraid
it depends on what questions you're interested in
but right off the bat "orbit closure" does not sound promising
neither does "projective variety," really
are you sure you don't want an algebraic geometer?
Vladimir Lysikov
Vladimir Lysikov
19:38
I know an algebraic geometer I can talk to :)
But I was always wondering if we can actually do something with cohomologies there
Basically because I don't understand them
Ben Steffan
Ben Steffan
if you give more details on the problem then the answer is maybe
Vladimir Lysikov
Vladimir Lysikov
The general problem is we have two homogeneous polynomials in $S^d V^*$ or two tensors in $V_1 \otimes V_2 \otimes \dots \otimes V_d$ and want to prove that one is not in the orbit closure of another
wrt $GL(V)$ in the first case or $GL(V_1) \times \dots \times GL(V_d)$ in the second
I can give you some specific tensors but not sure if this is helpful
Ben Steffan
Ben Steffan
so essentially you have two points in a space, a group action, and you're asking whether the closure of the orbit of one of the points contains the other or not
this does not strike me as a problem where cohomology will help
on a basic level we study group actions and their quotients, but generally what we want to do only works if the group action is suitably nice
Vladimir Lysikov
Vladimir Lysikov
Ok
Good to have an optinion of someone who knows more than me
ModularMindset
ModularMindset
20:37
user image
Does what I've written here seem clear?
user image
$\mathcal I$ is essentially a type of stratified space of dimension 2
Nevermind - there's too much prerequisite material in the previous pages that it may not make a lot of sense standing alone
Vladimir Lysikov
Vladimir Lysikov
21:05
It seems strange to me that $\tau(x_1, x_2, x_3) = (x_1, \frac12-x_3,x_2)$ is called a "rotation about $x_1 = \frac12$ axis"
$x_1 = \frac12$ is called a plane before, not an axis
and the transformation seems nothing to do with the set defined by $x_1 = \frac12$
Is the formula correct?
ModularMindset
ModularMindset
probably need to revise that - you're right @VladimirLysikov
ModularMindset
ModularMindset
21:23
the axis of rotation here should be orthogonal or normal to the slicing plane $\Sigma$
psie
psie
user image
Here's an excerpt from the proof of the change of variables formula in Folland's book, where by assumption $G$ is a $C^1$ diffeo with domain $\Omega$:
> Next, let $W_K=\Omega\cap\{x:|x|<K\text{ and } |\det D_xG|<K\}$. If $E$ is a Borel subset of $W_K$, by Theorem 2.40 there is a decreasing sequence of open sets $U_j\subset W_{K+1}$ such that $E\subset \bigcap_1^\infty U_j$ and $m(\bigcap_1^\infty U_j\setminus E)=0$.
For reference, see Theorem 2.40 above. I don't understand how he is using Theorem 2.40 here. Moreover, I don't understand the role of $W_K$ and in particular, why $U_j\subset W_{K+1}$. It'd be awesome if someone could clarify these points.
psie
psie
21:44
I think $|x|$ above is a typo for $\|x\|$, since we are in $\mathbb R^n$ and he actually introduced the notation $\|x\|$ for $x\in \mathbb R^n$ at the very beginning of the proof. But $|x|$ is how it's written in the book.
Sine of the Time
Sine of the Time
$|x|$ is usually used for the euclidean norm
psie
psie
yeah, that could be the case, true
 
2 hours later…
Sine of the Time
Sine of the Time
23:31
Making a Manual Transmission Pen
►
Popped on my yt @Xander

« first day (5204 days earlier)      last day (112 days later) »