Mathematics - 2021-02-13 (page 1 of 2)

Thorgott

00:00

happy birthday! @Ted

BigSocks

eyy happy birthday @TedShifrin hope it's gone well

Ted Shifrin

Impeachment trial. Scintillating!

But tanx.

Clarinetist

Happy birthday @Ted

BigSocks

@TedShifrin Jimminy Crickets, what a blast!

\

Ted Shifrin

Thanks, Clarinet.

Clarinetist

00:14

Here's a question for you all: suppose all I know is $\cos(x + y)$ and $\cos(2x + y)$. Would it be possible for me to calculate $\sin(x+y)\sin(2x+y)$?

(assume $x$ and $y$ can be any real number)

Ted Shifrin

Easy up to sign.

Clarinetist

Yeah, it's the "up to sign" part that gets me.

Ted Shifrin

Why are you doing this?

Clarinetist

It's an intermediate step in attempting to show that $X_n = \Re{e^{i(nx + y)}}$ for each integer $n \geq 1$ are not independent random variables.

where $x, y$ are independent uniform in $[0, 2\pi]$

Amin Idelhaj

Happy birthday Ted!

Clarinetist

00:36

Never mind, the answer is no

There is an aspect to this that I was missing

Ted Shifrin

Thanks, Demonark.

Clarinetist

Not a fan of the weather on weekends lately.

We're expecting -20F (without wind chill)

Rithaniel

01:05

Ah jeez, that's cold

vzn

02:37

in theory salon, 52 secs ago, by vzn

Microsoft’s Big Win in Quantum Computing Was an ‘Error’ After All / wired

BigSocks

03:20

anybody here mess with knots? was wondering how in general one could tell if a knot diagram was that of a ribbon knot

like does it have to contain the singularity for ribbon knots? I guess otherwise it would kind of trivially be a ribbon so not really a ribbon knot... idk

robjohn

@TedShifrin so you're now even more ancient? Congratulations!

copper.hat

@TedShifrin Happy Birthday! Today is my daughter's birthday as well!

Mike Miller

@BigSocks There are certain specific diagrams which are obviously ribbon but there's certainly no algorithm to determine if a knot is ribbon from the diagram

BigSocks

Right I was looking at 8,20 on the knot atlas and they have the ribbon diagram right there

but I was wondering how you could tell by looking

katlas.org/wiki/8_20

I thought maybe it had to come off as a link of 2 components or something

maybe they haven't done the saddle move? idk I am a bit confused

Rithaniel

03:40

$\mathcal{I}$ usually denotes the set of all intervals in measure theory, right?

Or am I mis-remembering?

Mike Miller

@BigSocks In the ribbon diagram think of your immersed disc as a long strip (ribbon). The two ends are at the bottom-left and top-right. You first fill up a disc at the bottom left which thins out into a strip, which loops right, twists, loops through the first part of your strip on the inside, and moves right again before expanding out into the end of the ribbon

BigSocks

ok yeah I can see that I guess. I can see how the big hoops on either side have the intersections that go away in 4 d too. Do ribbon diagrams need those though?

Mike Miller

I don't follow what you mean

A ribbon knot is a slice knot, because you can push the ribbon in on one side near the self-intersections to avoid self-intersection

So yes you need to be able to do that

BigSocks

Ribbon knot

In the mathematical area of knot theory, a ribbon knot is a knot that bounds a self-intersecting disk with only ribbon singularities. Intuitively, this kind of singularity can be formed by cutting a slit in the disk and passing another part of the disk through the slit. More precisely, this type of singularity is a closed arc consisting of intersection points of the disk with itself, such that the preimage of this arc consists of two arcs in the disc, one completely in the interior of the disk and the other having its two endpoints on the disk boundary. == Morse-theoretic formulation == A slice...

like here, you can push through those 2 self intersections in 4 dimensions bc it's slice.

like you said, exactly

but could you say a knot with ribbon diagram with no self intersection when "filled out" constitutes a ribbon knot?

lol the first sentence there answers the question nvm

I should probably read as well as draw

oh no wait, it says only ribbon singularities

which lends itself to the interpretation that it could have 0 ribbon singularities (and 0 other non-ribbon singularities) so it "only" has ribbon singularities (none)

idk actually- what do you think?

BigSocks

04:09

nvm I just understood how to draw them better

Leonhard Euler

Knot theory

Not theory

No theory

Ted Shifrin

04:31

@copper.hat I knew she had good taste!

copper.hat

My present for her was a large box of chocolate and a fire extinguisher :-).

The latter is a joke about my concern for my offspring's safety.

Certainly one of the happiest days in my life :-).

anakhro

04:48

luv u all

Ted Shifrin

@copper.hat Sometimes I worry about you!

copper.hat

You would have a lot of company there :-).

Fitz Watson

05:09

@robjohn Yes. Is there a possible combination?

Semiclassical

05:44

so, here's something mathjax-weirdness i noticed on an old question

5

Q: Characterising spaces of linear recurrent sequences.

Let $K$ be a field and $\def\N{\mathbf N}K^\N$ the infinite dimensional space of all sequences of elements of$~K$. Any linear recurrence relation of order $d$ with constant coefficients $$ a_{i+d} = c_0a_i+c_1a_{i+1}+\cdots c_{d-1}a_{i+d-1} \qquad\text{for all $i\in\N$,} $$ defines a $d$-dimens...

Question statement: $K^\N$ renders.
Answer statement: $K^\N$ doesn't render

copper.hat

the answer is missing a def

edit the q to see.

porridgemathematics

hi, in this question, I can see why $\Phi$ is surjective, but I don't really get why the kernel of $\Phi$$ is the normal closure of $$\bigcup \delta_{i}$$ math.stackexchange.com/questions/959180/… , could anyone help take a look?

*why the kernel of $\Phi$ is the normal closure of $\bigcup \Delta_{i}$

I can see the normal closure is in the kernel, but its the other inclusion that i sort of feel must be true, but cant put a rigorous argument towards

robjohn

06:27

@Semiclassical that's because \N is defined in the question, but not in the answer.

Semiclassical

ahah

copper.hat

that was what i meant :-)

Semiclassical

gotcha

i always forget that you can use \def in mathjax

copper.hat

definitely

porridgemathematics

06:48

hm, I managed to prove it via doing a reduction on the image of $$\Phi$$ on a reduced word, but its via a messy case analysis

is there a slicker way?

Vishesh Mangla

07:19

general formula for union of set 6k + 3, 6k - 4, 6k - 1, 6k?

copper.hat

what does that mean?

Vishesh Mangla

like 6k+3 is a set of numbers {9 , 15, .... } , 6k-4 { 2, 8.....}

so union of all sets

these are the cosets of Z6

copper.hat

$6\mathbb{Z}+\{0,2,3,5\}$ ?

or $\mathbb{Z} \setminus ( 6 \mathbb{Z} + \{1,4\})$.

Vishesh Mangla

07:40

just a sec, those are just sets, k is immaterial

you can say 6k+3 , 6t - 4, 6s-1, 6f . k is just an iterator i.e., k = 1, 2, 3, . I just mean to find a formula for nth element of the union set

copper.hat

ok, you asked for a formula for the union of the sets, now you want something different?

Here: $1_{6\mathbb{Z}+\{0,2,3,5\}}$.

Vishesh Mangla

no I meant this only from the start. Sorry my math notation knowledge is too bad. Now I 'm feeling it would had been good if I had focus on notations since everyone can understtand them equally

ok, let me tell what I m trying to do. I m actually stuck

so in collatz sequence, every sequence ends at a power of 2. Once you get a power of 2 you get 1 directly

so, 3k+1 = 2^s , k = 2^s - 1 / 3 , 2^s = 1 mod(3) implies s is 2k

now we travel one step back , this number above which was odd and to which to applied 3 k + 1, it would have come from a number 3k'+1 = 2^m * (4^s - 1)/ 3

you can strip off the 2^m to get the k which would give 2^s

k' = 2^(m+2s) - 2^m - 3 / 9 , clearly k' is integer. Using mod arithmetic , I got s = 2 , m = union ( 6k+3, 6k-1, 6k - 4, 6k)

first of all how can s be fixed?

secondly how can m be written as a function of whole number?

2^even number mod(9) forms a cycle

Fitz Watson

08:29

@robjohn Does there exist any such combination?

Could you please give me a hint to look into?

robjohn

09:12

@FitzWatson Consider the functions $f_n(x)=\frac{1-\cos^{2n}(x)}{\sin^2(x)}$

Fitz Watson

09:39

Doesn't this become undefined at 0 ?

fido9dido

guys I learned at university in computer animation class quaternions, but I forgot it, I still have the lect notes but it's terrible, anyone knows good material to learn it from? another thing that confuses me according to my lect notes the algorithm to convert quaternion into rotational matrix is [(1-2y^2-2z^2) (2xy-2sz) (2xz+2sy) etc]and this guy on youtube youtu.be/3Ki14CsP_9k?t=183 shows it another way so i am confused

Fitz Watson

@robjohn Doesn't this become undefined at 0 ?

robjohn

think

what is the limit as $x\to0$?

robjohn

10:04

what does it tend to away from $0$? as $n\to\infty$

user 85795

10:22

17 hours ago, by user 85795

@FitzWatson Please explain why you believe it's undefined at 0?

or at least try your best to :-)

user15072279

Why is it -1 here ?

robjohn

@user15072279 Because $\sum\limits_{k=\color{#C00}{0}}^n\binom{n}{k}=2^n$

@user85795 well, technically, it is undefined at $0$, but extend...

It is important to be able to tell when it is important to worry about functions which have isolated, removable singularities, and when it is not.

@user85795 once he gets past his worries about the removable singularity, he can get to the business of finding the sum of nice functions that sum to $\csc^2(x)$

user 85795

Indeed sir, and that is why I asked him to try to formulate an explanation of "why" :-)

robjohn

10:38

Well, $\frac00$ is an issue, but noting that the limit as $x\to0$ is defined is the point here.

49 mins ago, by robjohn

what is the limit as $x\to0$?

user 85795

@FitzWatson^ study the diagram carefully and think about its meaning

robjohn

It's like shrink-wrapping $\csc^2(x)$

user 85795

nice way of putting it, sir

Shrink wrap

Shrink wrap, also shrink film, is a material made up of polymer plastic film. When heat is applied, it shrinks tightly over whatever it is covering. Heat can be applied with a handheld heat gun (electric or gas), or the product and film can pass through a heat tunnel on a conveyor. == Composition == The most commonly used shrink wrap is polyolefin. It is available in a variety of thicknesses, clarities, strengths and shrink ratios. The two primary films can be either crosslinked, or non crosslinked. Other shrink films include PVC, Polyethylene, Polypropylene, and several other composition...

robjohn

Shrink-wrap can be used as a verb

user 85795

10:54

I was trying to provide some familiarity with the noun first :P

robjohn

11:06

I guess shrink wrapping was a common term when I was growing up so I assumed everyone knew it. Good to have a sanity check ;-)

Of course, the person asking about the sum left while we were discussing it.

Hopefully, they figured it out and they get their extra credit.

user 85795

yup

maths student

11:26

what is complement of language (a+b)* aba (a+b)* ?

Mike Miller

@BigSocks Yes, if you bound an embedded disc, you're also ribbon. But the only knot that bounds a disc in S^3 is the unknot...

maths student

@MikeMiller any idea about this ?'

love_sodam

Is there anyone know the latex code of this?

Or at least code to upload in MSE

user 85795

Nov 29 '20 at 3:02, by skillpatrol

A new commutative diagram editor for anyone who wants it.

robjohn

11:54

@user85795: actually, if one simply computes $\sum\limits_{k=0}^\infty\cos^{2k}(x)$ using the geometric formula one gets the same result as computing $\sum\limits_{k=1}^\infty\frac{\cos^{2k-2}(x)-\cos^{2k}(x)}{\sin^2(x)}$

user193319

Let $G$ be a countable discrete group. Does anyone know where I can find a proof that $(\ell^{\infty}(G))^*$ is isometrically isomorphic to $\ell^1(G)$?

Alessandro Codenotti

12:34

@user193319 That seems false, are you sure you have the dual on the right space?

Krijn

@love_sodam you should try q.uiver.app

porridgemathematics

13:24

hi, I get that in this imgur.com/a/POetAoX , the homeomorphism from the second figure to the first involves just straightening out the loop boundary of the second figure, but im wondering what exactly is going on here, shouldn't this only be a homeomorphism up to identifying the vertices of the 'new edge' in the first figure?

oh, nvm, the vertices of the new edge in the first figure are identified by the arrows anyways, i just wasnt seeing it

user15072279

13:55

@robjohn thnx for previous answer for my Q

robjohn

14:24

@user15072279 anytime.

@FitzWatson: did you get your sum of nice functions?

Leaky Nun

@user85795 thanks

Mike Miller

15:48

@AlessandroCodenotti If $\kappa, \mu$ are cardinals write $\kappa_{co-\mu}$ for the underlying set of $\kappa$ equipped with the co-$\mu$ topology (the closed sets are $X$ and its subsets of cardinality at most $\mu$). Are there necessary and sufficient conditions on a space $X$ so that any continuous map to $X \to \kappa_{co-\mu}$ is constant?

Clearly $X$ must be connected

I was thinking about this because as an exercise one can prove that if $X$ is connected has a countable dense subset of cardinality $\mu$ then any continuous map $f: X \to \kappa_{co-\mu}$ is constant

Call that $\mu$-separable

The same result follows if you assume $X$ is connected and locally $\mu$-separable

Is that necessary too?

Nah it's gotta be more than that it's already weird for the finite complement topology

BigSocks

@MikeMiller ah, excellent explanation right there

Alessandro Codenotti

This looks awful

More Anonymous

16:03

Any differential geometers in the chat?

Alessandro Codenotti

There could also be some annoying dependence on $\kappa$. There are only constant functions from $[0,1]$ to a countable set with the cofinite topology, but there are nonconstant ones into sets of cardinality $\geq \mathfrak c$ with the cofinite topology and I have no idea what happens for sets in between

Thorgott

that moment when even Alessandro calls your topology question awful

Leaky Nun

is it a topology question though

Mike Miller

Yeah I'm asking him to turn it into a set theory question

Leaky Nun

the identity function [0,1] -> [0,1] is continuous right, if the second [0,1] is equipped with the cofinite topology

Alessandro Codenotti

16:14

sure

any injection is

Mike Miller

Can we at least characterize when you have maps to $\kappa_{cof}$?

Alessandro Codenotti

Hmm if $X$ is $T_1$ and $|X|\leq\kappa$ then there are nonconstant continuous functions $X\to\kappa_{cof}$ but that's trivial (any injection works)

Simone

17:35

Can I put a text behind spoilers in Stackexchange?

Thorgott

yes

Simone

How?

Thorgott

forgot the code, but google is your friend

Simone

I use Duckduckgo, sorry

user2103480

@AlessandroCodenotti Is the reduced product topology for a nonprincipal ultrafilter equal to the quotient topology on the ultraproduct

Simone

17:38

:unoreverse:

Thorgott

lol

Alessandro Codenotti

@user2103480 what's the reduced product topology? And how do you take ultraproducts of topological spaces?

Thorgott

@user2103480 did I miss the announcement that today is the horrible topology question day

user2103480

@Thorgott yes

@AlessandroCodenotti mscs.mu.edu/~paulb/Paper/ult.pdf

Alessandro Codenotti

math.stackexchange.com/questions/4023082/… well since it's horrible topology question day here is my horrible topology question

Thorgott

17:40

damn, I need to come up with an entry

user2103480

thats the reduced product topology, and for a nonprincipal ultrafilter I would just define the ultraproduct as a topological space via the quotient

BigSocks

@Thorgott lmao

user2103480

quotient on the product topology

BigSocks

considerably evil to just link a paper as an answer to a question of clarification

user2103480

the definition is literally on the start of page 2

BigSocks

17:43

cool looking paper tho

user2103480

This reduced business seems way too nice to be equal to the sledgehammer-type quotient topology lol

Alessandro Codenotti

I think this reduced thingy is finer than the quotient from the product topology because you can have products of infinitely many open sets, as long as their indices are in the filter

BigSocks

$f_i$ is I guess a cts function in $X_i$?

user2103480

https://mathoverflow.net/questions/201289/which-compact-topological-spaces-are-homeomorphic-to-their-ultrapower

I'm not sure whether they use the quotient topology here, but then Eric Wofsey's comment would answer it: "Maybe I'm understanding your definitions wrong, but doesn't any nonprincipal ultrapower in the category of spaces have the indiscrete topology?"

the construction is so abstruse, it's trivial

Alessandro Codenotti

Oh no, right, it's the other way around because the ultrafilter contains no finite sets

user2103480

17:47

@BigSocks not continuous, any tuple

Alessandro Codenotti

That's a weird construction

user2103480

We just take the set-theoretic product and then quotient it by the equivalence relation

@AlessandroCodenotti I know, it's horrible topology question day

BigSocks

I don't understand- I said $f_i$. that shouldn't be a tuple? asking what is one element of one coordinate of the domain of the ultraproduct

should be a function, and I was guessing cts since it's somehow relevant to the space $X_i$. I could see how maybe an equivalence class of the $f_i$ would be a tuple maybe

user2103480

I meant it like this: f is just any function from I to the union of X_i and f_i is just any element of X_i

BigSocks

I mean it seems pretty standard to me, this construction. But most interesting topological properties are not first order I don't think, so Los can't give you those

Alessandro Codenotti

17:51

@BigSocks topological spaces are not first order structures to being with

(ok some people did study model theory of topological spaces using two sorted languages and other painful tricks, but it doesn't work super well)

BigSocks

yeah, that too. seems pretty rough to work with these from that pov

user2103480

@BigSocks finding interesting topology was never the goal, the goal was horrible topology

but I guess thats not pretty horrible, its trivial

BigSocks

large kek

@user2103480 this tho

user2103480

the reduced product seems horrible enough though

since it is not trivial and preserves properties

@BalarkaSen youtube.com/watch?v=mKr7ThMib2w

wtf that muppet film is actually pretty dark. is that supposed to be children's entertainment

Mike Miller

That's a pretty direct reference to an old movie which is now part of standard Americana, "It's A Wonderful Life"

user2103480

18:04

And of course to Dickens, but still

BigSocks

@user2103480 wondering about example 1.3

user2103480

@BigSocks what about that

BigSocks

seems tough to show. idk what could go wrong though

user2103480

I would need to verify myself the claims there, but the construction sounds reasonable

Then we're both on the same level

BigSocks

yeah the construction seems really reasonable. just that big properties like that... I don't think about those

user2103480

18:10

I really have no more knowledge about these things than you. Okay, maybe a bit, depending on your exposure to ultraproducts

BigSocks

I've dabbled, but yeah, not really an expert

user2103480

But I just thought about ultraproduct topologies for the lulz, and have no experience with those whatsoever

now please provide your own horrible topology question

BigSocks

oh ok. I still think they are pretty cool. just quantifying over functions... hmm

user2103480

They are neaaat

BigSocks

@user2103480 hmmm

user2103480

18:13

But I got scared of ultrapowers after having a closer look at a few pages of chang & keislers model theory book

BigSocks

twitter.com/agolian/status/1360269132822835201

which one of these is slice (if any)?

that, I think, is pretty horrible

@user2103480 I should really dust off that book and read it man...

user2103480

@BigSocks if you've got like 5 years to spare

BigSocks

actually

user2103480

do all the exercises and then face shelah as the final boss

I'm pretty sure nobody has beaten him yet and this is why he is immortal

BigSocks

lmao. as you keep fighting, he just writes loads of papers and crushes you under them

user2103480

18:16

looool

@BigSocks I thought "this looks like a topology analogue of the four colour theorem"

then looked at the abstract

"by computer experimentation"

nice

Semiclassical

Was there a chat room for “help me find a question I can’t track down”

Mike Miller

The comments below point out that Nathan Dunfield & Sherry Gong calculated that these aren't even topologically slice

Semiclassical

I thought there was but I’m not seeing it

user2103480

@MikeMiller so this doesn't really tell us much?

BigSocks

@user2103480 lol I guess contemporary knot theory is kind of dependent on computers too

Mike Miller

18:23

@user2103480 These are 21 knots for which all standard obstructions to smooth sliceness specifically vanish but still aren't smoothly slice. That's all you've learned from this.

BigSocks

@MikeMiller but that was 16/23?

hmm so the ones with trivial Alexander poly. should be discounted? why include them?

Mike Miller

Oh my bad.

So there's some number more they haven't discounted.

BigSocks

right I guess there is still a chance and the problem is in a sense easier

Mike Miller

It's still worth doing sometimes. There is an older saga where some people put up a similar list of candidates for exotic 4-spheres though they could only manage a computer calculation showing one is standard. The next month someone showed by calculation that all of their examples were standard. The month after that someone wrote up a paper (now refuted --- there was a tiny point that didn't work and in fact the result was wrong) saying that the entire approach couldn't work.

user2103480

What's worth doing?

BigSocks

18:26

That's so nuts... can't really wrap my head around the difficulty surrounding making an exotic 4-sphere. But the history is cool

user2103480

Dunno what to substitute there

BigSocks

my guess is "trying to come up with an exotic 4-sphere/ show someone else's guess is not an exotic 4-sphere"

user2103480

everything's fair in love and exotic 4-sphere theory

Thorgott

exotic structures are way too wild

Mike Miller

@user2103480 The computational stuff giving guesses at possible exotic things

It leads to new theory as people try to (dis)prove that they work

user2103480

18:33

IIRC 19th century mathematicians had their personal human calculators so that they could find patterns

so I don't see anything wrong with computer experimentation

Alessandro Codenotti

Mayer (of the Mayer-Vietoris sequence) was known as Einstein's calculator apparently

user2103480

except that it kills valuable jobs!

so sad human calculators had no union smh

Alessandro Codenotti

So when are you HoTT people going to leave all mathematicians jobless with authomatic proof writers?

user2103480

are you really including me in HoTT people just because I'm one of about 50 people in the world that took two courses on it

BigSocks

I wonder if it will be HoTT people/theorem-prover-using people or AI that just reads loads of math and generates proofs to then put into theorem provers

geocalc33

18:38

what is HoTT

Thorgott

HoTT "people"

Astyx

HoTT type theory

user2103480

I feel like I just learned graph theory on loads of bull steroids

Alessandro Codenotti

@Thorgott HoTT robotic overlords

Zuckenberg is a type theorist confirmed

user2103480

Lizard mathematics

BigSocks

18:40

link.springer.com/chapter/10.1007/978-3-030-53518-6_1

every day is a race vs google to see who writes my thesis first

user2103480

Hrmph, @MikeMiller, whats the nice MV decomposition for computing homology of genus g orientable/non-orientable surfaces

Thorgott

"Senator, that is an excellent question. So, a homotopy type is..."

9

user2103480

do I just cut a disk out and work with the polygons

Astyx

@Thorgott lmao

user2103480

@Thorgott dying

But man, type theory as a categorical framework is really really nice

Couldn't get too warm with kan-complexes since I lack cognitive ability, but the general theory via presheafs is very very nice

geocalc33

18:44

@user2103480‎ "There are two ways to do great mathematics. The first is to be smarter than everybody else. The second way is to be stupider than everybody else -- but persistent.”

Balarka Sen

@MikeMiller We should make a combinatorial exotic R^4.

Thorgott

@user2103480 can you abelianize $\pi_1$?

user2103480

@geocalc33 there are two ways to build a tunnel

one is using advanced technology and statics skills

the other is running against the wall head first until you're done

Balarka Sen

Find a notion of hyperbolic triangulations of R^4. This will involve angle defects at the vertices. Prove that they are all simplicially standard.

Thorgott

ah, unnecessary I guess, just observe that it retracts onto a wedge of circles upon puncturing

user2103480

18:46

@Thorgott so cut out the disk it is

Mike Miller

@user2103480 Polygonal repreesntation

Or using $\Sigma_g = \Sigma_{g-1} \# T^2 = \Sigma_{g-1,1} \sqcup_{S^1} \Sigma_{1,1}$

The latter requires you understand what's going on tho

user2103480

@MikeMiller meaning "understand why the last isomorphism holds"?

Balarka Sen

Fix an end $\xi$ of a tree $T$. Given any vertex $v$ of $T$, there is a unique ray joining $v$ to the end $\xi$. Call the immediate successor of $v$ along this end "the $\xi$-parent of $v$". Call the other neighbors as the $\xi$-children of $v$. This gives a height function $h : T \to \Bbb Z$ on the tree, unique upto translations on the codomain, where if $w$ is the parent of $v$, $h(w) = h(v) + 1$, and if $w$ is a child of $v$, $h(w) = h(v) - 1$.

geocalc33

has anybody here taken a course on Several complex variables?

user2103480

@MikeMiller Sounds reasonable though. In the context of MV this seems like a good decomposition for connected sums in general, for inductive proofs

Mike Miller

18:51

No understanding homology of simple surfaces with boundary like $\Sigma_{1,1}$ and understanding what their relative homology is etc

yes I use this example when teaching

Balarka Sen

This is the analogue of the following, I think: If $\xi$ is an ideal point of $\Bbb H^2$, there's a natural height function $h : \Bbb H^2 \to \Bbb R$, unique upto translations in the codomain, such that the gradient field of $h$ is the horocycle flow:

Flow along the red lines towards the end $\xi$.

user2103480

@MikeMiller any other simple examples like this I can work on?

Mike Miller

RP^n and CP^n after doing all surfaces

You could look up Dehn surgery and prove things about that

user2103480

Ok. Also via MV? Because for these, our main tool was cellular homology

Mike Miller

MV = cellular homology

Prove those via MV and you'll see that this is true

user2103480

18:55

dang, I get what you mean

the boundary is an extra cell and such

(e.g. in connected sums)

Mike Miller

Yeah exactly

user2103480

thanks! will work on this

Mike Miller

Yeah if you get this for RP^n (start with RP^2 and RP^3 for ease of visualization) you'll see how my statement makes sense

user2103480

I have to explicitly work with the MV maps right

Mike Miller

In one case yes

Balarka Sen

18:57

MV = homology is a sheaf
Cellular homology = run the sheaf on the cells

Thorgott

compute homology of lens spaces

Mike Miller

Note that MV is also basically the same as H(X, A) = H(X/A)

@BalarkaSen h o m o t o p y s h e a f

Balarka Sen

yes

homotopy sheaves in the homotopy category of chain complexes

user2103480

@BalarkaSen do you have an nlab article for additional unclarity

Balarka Sen

my spirit animals

dont listen to me

im being facetious

Thorgott

18:59

orly

Balarka Sen

im sick of helping people every day to compute homology of RP^n

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$Mathematics$ Mathematics