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12:01 AM
Oh that reminds me I never got around to the "geometric/thin triangle C'(1/6) implies hyperbolic" explanation (I was going to write a blog post or something but I think there was a sort of subtle part I didn't really finish) @BalarkaSen
I'd love to understand that
Do write a blogpost!
Here was the idea, take a large geodesic triangle(say each side is 100 times the length of the longest relation)
The triangle describes a van Kampen diagram that you can fill in with relations
So you can apply Greenlinger lemma/Dehn's algorithm
Mhm, ok
well you will have to "remove a corner cell" since that interacts with both sides of the triangle, and you won't be able to do the shortening procedure on a geodesic
This should give a new geodesic triangle with shorter perimeter
That's an interesting way to intuit Greenlinger lemma, as a "shortening procedure on a geodesic". Basically if a reduced word (or it's cyclic conjugates) has a long relation then it's trivial
12:10 AM
repeat until the corner cell is touching more than one cell. At this point that should give you a coarse center which is close to the other side of the triangle. This part needs some work
really all you need is that this can be done till at least on side is < 100* longest relation (or something like that
I vaguely follow
Basically a geodesic triangle should be "1-relation" thick everywhere except the center where the geodesics diverge, so the shortening procedure on a corner should bring you closer to this place.
hi, a @Balarka
Hi @Ted!
12:26 AM
@mathsresearcher So how do you use the Mean Value Theorem to prove that if $h'=0$ on an interval and $h$ is continuous (on the closed interval), then $h$ is constant?
I found it strange that Dehn basically discovered geometric group theory, but for whatever reason the direction went more combinatorial and the more geometric aspects where not really dug up and expanded on until Gromov
Gromov says in his Abel prize interview that everything was known and was obvious, he just wrote it down lol
ahhh I'm just gonna wing this one :/
freaking out for no reason i guess nvm
12:45 AM
it's the most craziest time of the yearrrrrrr
There is some sort of truth in that though. If Dehn went in a somewhat different direction I could imagine a world where Hyperbolic Groups was written by Dehn in 1915 (time to write the nerdiest alt-history book) although lacking the important 3-manifold motivation happening in the 70's and 80's @BalarkaSen
I'd watch that movie
Boo @Fargle
Hopefully it has a time travel plot about saving Galois
12:49 AM
@PaulPlummer that'd be a twist
Aaah @Ted
yeah I guess Gromov was really trying to prove hyperbolization with his perspective more than renovating combinatorial group theory
I make soaps as a hobby
I'm working through a proof and the following claim is made: if $T$ is a normal operator, then $||T^2v|| = ||TT^*v||$. I've tried like mad to prove this, but I haven't had much luck.
Isn't there something specific assumed about that $v$, @user193319?
12:54 AM
I guess that'd mean that you'd have $\langle T^2 v,T^2 v\rangle = \langle TT^* v,TT^* v\rangle$
@TedShifrin It's assumed that $v \in \ker T^2$.
You shouldn't leave out stuff.
I was about to pick a $v$ that was in $\ker T^*$ but not in $\ker T$.
Sorry, I thought that the norm equality was something that held for any $v$.
12:56 AM
Did the text say that?
No, but it didn't say anything to the contrary.
Dehn publishes Hyperbolic groups... topologists compute some fundamental groups 'wow a lot of these are hyperbolic I wonder if they have constant curvature metrics' ....Thurston is born years later and becomes a lawyer @BalarkaSen
Okay so I'm trying to prove that for commutative rings $A, B$, given an epimorphism $\phi : A \to B$ that the induced map $\phi^* : \operatorname{Spec}(B) \to \operatorname{Spec}(A)$ is a homeomorphism of $\operatorname{Spec}(B)$ onto the closed subset $V(\operatorname{ker}(\phi))$ of $X$. I managed to show that $\phi^*$ yields a continuous bijection onto $V(\operatorname{ker}(\phi))$. To show that $\left(\phi^*\right)^{-1}$ is continuous I claimed that
for any $f \in B$ we have $\phi^*(\overline{V(\{f\})} = V(\overline{\phi^{-1}(f)})$, but I'm having some trouble proving that $\phi^*(\overline{V(\{f\})} \supseteq V(\overline{\phi^{-1}(f)})$
Oh, wait. The assumption that $v \in \ker T^2$ comes after that equality...Here's the relevant sentence: "Since $T$ is normal then $||T v|| = ||T^*v||$ and thus $||T^2v|| = ||T T^*v||$, so I don't think the author is assuming $v \in \ker T^2$.
@PaulPlummer lmfao
12:59 AM
Am I on the right track?
Also the overlines denote complements not closures
OK, @user193319. Now we can get it. Compute $\|Tv\|^2 = \langle Tv,Tv\rangle$. How can you use normality?
I actually just figured it out.
OK, good.
1:50 AM
I figured my stuff out too. can't believe I went into panic mode :/
I need chocolate
2:47 AM
@Perturbative that's not true. consider $\phi : \Bbb Z \to \Bbb Z[\frac12]$ with trivial kernel, but the image of $\phi^\ast$ is $\operatorname{Spec}(\Bbb Z) \setminus V(2)$
Ahh okay thanks @LeakyNun
3:02 AM
Simpler question, let $A$ be a non-zero ring. Show that the set of prime ideals of $A$ has a minimal element. I thought that a proof of this would just be an application of Zorn's Lemma by making $\operatorname{Spec}(A)$ into a poset $(\operatorname{Spec}(A), <)$ by defining $\mathfrak{p} < \mathfrak{q}$ if and only if $\mathfrak{q} \subseteq \mathfrak{p}$.
Consider any chain $ ... < \mathfrak{p} < \mathfrak{q} < \mathfrak{k}$ since $\mathfrak{k}$ is contained in a maximal ideal of $A$ it follows that this chain has a lower bound, and so by Zorn's Lemma we conclude that $(\operatorname{Spec}(A), <)$ has at least one minimal element.
Ahh okay nevermind I just wrote garbage by misinterpreting the question
2 hours later…
4:43 AM
@MatheinBoulomenos @loch What's the best way to see that $H^n(G; M)$ for a finite group $G$ is annihilated by $|G|$?
I can prove it explicitly by an averaging argument on the crossed homomorphisms for $n = 1$. But in general this is not obvious to me.
@BalarkaSen There are maps going both ways between $C_*(BG;M)$ and $C_*(EG;M)$, where the first is defined with local coefficients and the latter is not
One is the usual projection
The other is the "transfer", sending a chain downstairs to its $|G|$ lifts
The composites are both multiplication by $|G|$
True. That's how Bockstein arises. Oh I see.
Oh I guess this is precisely why the averaging argument worked. I think I can translate this to algebra
$H^0(\{e\}; M) \to H^0(G; M)$ is defined by $m \mapsto \sum_{g \in G} g \cdot m$ on degree zero that's the transfer map.
Also, this shows that if $|G|$ is invertible in $M$, then $H^n(G,M) = 0$ (or rather, $M^G$ concentrated in degree 0)
4:59 AM
Right, that's what I was trying to work towards (I wanted to prove normal abelian subgroups of a group with order and index coprime have a complement. Boils down to $H^2(G/A; A) = 0$ for $|A|$ and $|G/A|$ coprime)
The topological intuition behind the fact that $A \to G \to G/A$ splits iff a corresponding class in $H^2(G/A; A)$ is $0$ I believe is that it gives me a fibration $BA \to BG \to B(G/A)$ and obstruction to getting a section over the 2-skeleton lies in $H^2(B(G/A); \pi_1(BA))$, where the coefficient is somehow twisted, instead.
A presentation complex level section gives me a group level section, I think
The twist in the coefficient is not clear to me. Perhaps the point is that we don't just want a topological section, but it preserves the additional $G$-structure on the bundle
I'll think about this by myself though, so don't tell me!
Hi guys, if given $(X,d)$ MS, $a \sim b$ means $\exists A \in X$ such that $A$ connected and $\{a,b\} \in A$, how can we prove that the equivalence classes $C_a = \{b \in X : b \sim a\}$ is closed?
Here's my attempt:

Suppose $C_a$ is not closed, so there is a limit point $x$ not in $C_a$. The closure $\overline{C_a}$ is a closed set and contains $x$. If $C_a$ is a connected set then so is $\overline{C_a}$, then $C_a$ cannot be the set that contains all $b \in X : b \sim a$?
I guess that banks on the fact that $\overline{C_a}$ is also connected, but I think intuitively that makes sense
1 hour later…
6:17 AM
Do we always use the word closure under some operation? Is there any operation involved while we are defining closure of a graph ? What is the motivation of defining closure of a graph in the way it is defined?
Usually, I have used the word closure as the set we get after adding element in a set to make that set self content under some operation. For example, closure of $\mathbb{Z}^*$ under multiplication is $\mathbb{Q}^*$.
@taritgoswami I am not sure how closure of $\mathbb{Z}^*$ under multiplication should be anything other than $\mathbb{Z}^*$ itself. So I am not really following what you are asking.
@TobiasKildetoft Sorry, I mean we need to add inverses of elements to make $\mathbb{Z}^*$ closed under multiplication.
Don't we call it closure?
No, we don't need to add inverses to make it closed under multiplication
@TobiasKildetoft Ooh sorry. I mean type of suppose the set $\{1,2,3\}$ , the closure under $+$ will be $\mathbb{N}$
Here we are saying closure under addition . Like that is there any operation involved in closure of graph?
@MikeMiller I think explicitly, at the level of the bar resolution, the transfer map can be defined as follows. Let $H \leq G$ be a subgroup; every $G$-module $M$ can be thought as an $H$-module as well. Define $\tau : C^1(H; M) \to C^1(G; M)$ as follows: let $f \in C^1(H; M)$ be a function $f : H \to M$. Note that $G$ is partitioned by $n=|G : H|$-many cosets, let's say $x_1 H, x_2 H, \cdots, x_n H$ of $H$ in $G$.
6:29 AM
I am not sure how the closure of a graph is defined in the first place
For any $g \in G$, you can choose translates $g_1 g, \cdots, g_n g$ of $g$ such that each $g_i g$ belongs to the $i$-th coset of $H$ in $G$, i.e, $g_i g H = x_i H$. Then define $(\tau f)(g) = \sum_{i = 1}^n x_i \cdot f(x_i^{-1} g_i g)$.
I think $\tau$ is well defined upto $1$-coboundaries, i.e., independent of the representatives $x_1, \cdots, x_n$ of the cosets
This should give the homomorphism $\tau : H^1(H; M) \to H^1(G; M)$. The higher generalizations shouldn't be hard to write down from this either
Hi @BalarkaSen I hope you are well.
(Note that $g_i$ is determined by $x_i$ for each $1 \leq i \leq n$.)
Hi @JasperLoy. I'm well, thank you
@taritgoswami Closure or closed can be used in many different mathematical contexts. There need not be any operation.
In fact, it can even be used in many different non-mathematical contexts, eg emotional closure.
@JasperLoy Ok. Can you tell me what motivates the definition of closure of a graph ?
6:45 AM
:0 nerds
@Daminark Hello Amin.
per Tobias's remark above, it's not even clear what definition of "graph closure" you have in mind
can't expect to know how a definition is motivated without having the definition
If you are @Semiclassical then I am Bimodern, LOL.
You guys have no Christmas hat.
7:08 AM
@JasperLoy You are right. I need to fix that
8:06 AM
@Isa Why translate using emojis? For fun and profit: Unicode string benchmarking. Universal Declaration of Human Rights is the most translated document in the world, which makes it an ideal corpus (semantically equivalent body of text in plethora of languages and scripts). Emoji is a fast growing language independent script and testing the string processing performance on this subset is increasingly important.
@Isa The translation is just a procrastination task I enjoy doing. I'm going for expressionistic translation: trying to invoke the ideas represented in Article 1. Using only Emojis proved lacking for expressing the relations, so I've expanded by Unicode palette used to include Math Symbols, too. 🤰 is a pregnant woman. I've used the 🤰➡️👶 triplet to invoke the idea of being born.
@Isa After few iterations I got this:

∀👤 ∈ 👥,🤰➡️👶: 👤 ∈ 🗽
👤 ≡ 👤: {🤝,📜}
👥 = {👤 | 🧠, 👼🤔👿}
👤∼👤: 👬👻
@Isa … which hopefully invokes the ideas from Article 1:
All human beings are born free
and equal in dignity and rights.
They are endowed with reason and conscience
and should act towards one another in a spirit of brotherhood.
@Isa I was hoping that all you native math speakers would help me to tune this, to not include some majorly incorrect use of notation, that would be too distracting. But logical rigor and excessive formalism is not my goal here.
3 hours later…
11:40 AM
@TobiasKildetoft I got mine for I don't even know what.
@Palimondo If you want to get a serious answer to this question, I think it is better to post on the main site. This is a question on formal logic more than the English language, and native speakers of English who also study math may not know much formal logic, even though they should know informal logic.
2 hours later…
1:14 PM
@PaulPlummer do you have a reference for a proof that $\Bbb Z$ is amenable that constructs a left-invariant mean rather than a Følner sequence?
1:59 PM
2:21 PM
2:36 PM
$\text { Let } D \text { and } R \text { be integral domains, and } \theta : D \rightarrow R \text { be a ring homomorphism such that } \theta \left( 1 _ { D } \right) \neq 0 _ { R }$
$\text { Prove that } \theta \left( 1 _ { D } \right) = 1 _ { R }$
Im not sure how to proceed with this proof
@SharathZotis Hint: The image of an idempotent is an idempotent
What exactly is idempotent I am not too familiar with that term
$x$ is idempotent if $x^2 = x$
I see. So since $1_D *1_D = 1_D$
$1_D$ is idempotent?
2:43 PM
By def of homomorphism $\theta(1_D*1_D) = \theta(1_D)*\theta(1_D)$
3:01 PM
How to proceed from here?
hi chat
Hey Semiclassical
3:21 PM
$\text { Let } D \text { and } R \text { be integral domains, and } \theta : D \rightarrow R \text { be a ring homomorphism such that } \theta \left( 1 _ { D } \right) \neq 0 _ { R }$
$\text { Prove that } \theta \left( 1 _ { D } \right) = 1 _ { R }$
How do I proceed without knowing the fact that image of idempotent is an idempotent?
3:41 PM
tfw you think you've come to a final endpoint for a problem...and then realize you missed some crucial components, and you don't know how to incorporate them without a lot of work
(polytopes, why do you torment me)
hey hey hey
@AlessandroCodenotti I don't have a reference that I know of. Wikipedia has a proof. Note that the equivalence of the definitions is not true in ZF (to get from one to the other you use things like nonprincipal ultrafilters). Maybe it is worth looking in Geometric Group Theory by Kapovich and Drutu. They have a section on amenability, but I havn't really looked at it
Hello, guys. How do I know in terms of orientability (both of $M$ and $\Sigma$) when a embedded submanifold $\Sigma\subset M$ admits a unit normal globally defined vector field?
@PaulPlummer yeah, I'm interested in how much choice is needed since we're using the left invariant mean definition in class
just confirmed by an explicit example that my previous approach isn't sufficient.
3:56 PM
@AndersonFelipeViveiros $\Sigma$ is codimension $1$ in $M$?
@BalarkaSen Yes, sure. Hypersuface
Right, so by orientability, $TM$ and $T\Sigma$ are both orientable vector bundles over $M$ and $\Sigma$ respectively
Pull $TM$ back to $\Sigma$ to get $TM|_\Sigma$. This is an orientable bundle over $\Sigma$.
Take the orthocomplement of $T\Sigma$ in $TM|_\Sigma$ (maybe after giving it a Riemannian metric). That's an orientable line bundle on $\Sigma$ embedded in $TM|_{\Sigma}$.
Prove that orientable line bundles are trivial
Then take a section; that gives a normal field
Ok, thank you! I'll try this
(To briefly sum up my howl: I assumed it would be sufficient to take a high dimensional-simplex, map it into 6D space, then restrict to the intersection of three planes to get a 3D set. Turns out I need to take the intersection of a lot of planes with my high-dimensional simplex, and then map this to 3D.)
@BalarkaSen So, if $M$ and $\Sigma$ are both orientable, the answer is yes, there is such vector field.
4:04 PM
If $M$ is not orientable, then...?
Or the other possibilites
Not true :) Take $\Bbb{RP}^1 \subset \Bbb{RP}^2$
So it might be true or not
Or it will be always false?
4:06 PM
It can be true. Take connected sum of $\Bbb{RP}^2$ with a torus, then take the meridian circle of the torus.
That has a normal field.
Ok. So the answer is only conclusive when we have both of them orientable. Alright
So, what field of math do higher dimensional simplex fall? (or simplex in general, because I'm not really familiar with the topic)
broadly speaking, linear algebra
a 1-simplex is an interval, a 2-simplex is a triangle, a 3-simplex is a tetrahedron etc
and you can think of that as being generated by convex combinations of the vertices
(Intuitively: Suppose you've got a set of outcomes labelled by real vectors and some probability distribution on those outcomes. Then the average value of that vector is going to be somewhere inside the convex hull of those vectors.)
Okay, seems easy enough.
So an n-simplex is the simplest geometric structure you can create in n-dimensional space which is still necessarily n-dimensional?
a somewhat annoying part is that an n-simplex has n+1 vertices
which means I have to be careful not to use the wrong number
In the present case, I've got a whole bunch of outcomes (e.g. in one case I've got 14 different possible outcomes) which translates into a high-dimensional simplex
4:22 PM
Yeah, just need to think of the edges being intervals and an interval having 2 endpoints.
But I only want to consider those probability distributions which satisfy certain constraints
In my small cases (where there's not as many outcomes) I can get away with doing it in a simpler way, where I only worry about making sure the variances of those vectors are right
but I'm now realizing that this won't be sufficient if I'm doing larger cases
guys if I have a questtion
Do you potentially get vectors leaving the bounds of the simplex?
if i know that the series of an converges and bn converges
can I say that max(an,bn) is an+bn
is less than*
or equal to
so by the comparison test
so the series of max(an,bn) converges as well?
@Rithaniel that's not so much an issue, at least not with the code I'm running
(thank goodness for smart people who have already worked on this)
4:28 PM
max(an,bn) is less than or equal to an+bn
Depends on the sign of $a_n$ and $b_n$, maths. Think about if $a_n = -b_n$
oh and we assume that an and bn are non negative
The point is more that I assumed I could map a linear mapping on my simplex and apply the constraints to that
Then yeah, you should be okay.
the image of the simplex you get is complicated, but the constraints are simple
4:29 PM
whew, thanks rithaniels
i'm realizing, though, that I need to do the constraints first (and there's a lot of constraints)
which means I'm going to end up with some horrible high-dimensional polytope
things won't be so bad once I manage to get that, since doing linear maps on polytopes is pretty easy
but it's going to require some significant reworking of my approach
and high-dimensional here means dozens of dimensions. in my previous attempt, I "only" had to worry about a polytope in 6D space :)
Oh jeez
the funny thing is that the linear mapping I get now will actually be nicer, in a certain respect
I've only dipped a toe or two into thinking about 6D or 7D space. I imagine working with 81D or 121D space would be effectively the same, just more of it, but still, that seems daunting.
@Rithaniel definitely
I'm effectively trading the weirdness of the mapping for weirdness of the preimage
4:38 PM
I only know 3D space because that's what I seem to live in. =)
3+1 :P
Or perhaps 3.5, since one dimension has only one apparent direction.
But I don't know the implication of Poincare Conjecture on the universe shape.
@Rithaniel nah. rays are still 1D
4:40 PM
though there is always is something strange about viewing time as a dimension along with space, precisely because you can't move back-and-forth in time
(philosophically, anyways. mathematically there's nothing particularly weird about minkowski spacetime)
Well, space-time has 4 dimensions, and space has 3, so I don't see why we should bring in time when we talk about space only.
But I don't know any physics. I am only a banana.
the real question is whether it ultimately makes sense to talk about space only
@JasperLoy trying to decide whether that would make this book perfect for you or exactly not: amazon.com/Bananaworld-Mechanics-Primates-Jeffrey-Bub/dp/…
Well, when we talk about 3-space we live in, we are not even talking about the theoretical physicist's spacetime. We are just talking about the layman world he lives in and applying some math to it.
4:44 PM
Handing a book for primates to a banana seems sorta wrong
And who knows? All this spacetime thing is just a theory which might be outdated one day.
Of course the curvature of spacetime can explain things like gravity currently, but it is still just a theory.
Now, generally higher dimensional space is used to discuss things with lots of free variables. A 3D graph can only really handle three variable, unless you somehow encode multiple variables into one, in which case the graph might not be able to accurate represent the data.
@Rithaniel yeah
@Semiclassical Instead you need to hand a banana to a primate.
at the end of the day, I end up taking my complicated high-dimensional polytope and making a certain linear transformation to 3D space
at which point I can easily make a visualization
but the preimage? very much not visualizable
4:46 PM
Makes me think about the visualization of p-adic space.
Are you in an academic job now @semi?
@Rithaniel yuck
Spacetime is way cooler if it is emergent from something more formless
@Secret This is a very deep remark...
Lemme see if I can find an image
4:47 PM
@JasperLoy I'm not sure that distinction is ultimately possible. I mean, it's fair to say that in most everyday experiences the notion of space and time being related is pretty irrelevant
newtonian physics has space and time, but they really don't have much to do with one another
on the other hand, there are experiments you can do where the relationship between space and time from special relativity is borne out
In other news, I like things to come in threes, space and time makes spacetime, but it will also be cooler if there is some third something that has the same significance as space and time
@Secret Well, let me add one for you. Your experience of life can be a function of (space, time, mind).
So I think special relativity is very much 'experiential'. it's just not involving everyday experiences
(by contrast, the 'experiential' aspect of quantum mechanics is waaaay more confusing)
@JasperLoy I'm in between things tbh
To me, quantum mechanics is less confusing than GR
@Secret my sense is that, mathematically, QM is simpler than GR
but philosophically, GR is not nearly as strange as QM
4:50 PM
I didn't like the phrase in one of the Transformers movies where some actress said "we need to move from fourier analysis to quantum mechanics". It just sounds deep without any meaning, and gives the impression that the latter is more advanced and more difficult.
yeah, that's a pretty dumb phrase
I don't know which movie exactly now. I stopped watching the series after the first one or two.
yeah that's one reason I found quantum mechanics more tractable, because I can always go back to the maths to find what it is trying to tell me, but for GR, even just trying to check something the maths is way harder to do it on the back of the envelope
"we need to move from quantum mechanics to quantum field theory" would be more appropriate
@Secret That's why I don't do math on envelopes.
4:52 PM
or even "from wave physics to quantum physics"
@Semiclassical I hope you do get a professorship somewhere someday, if that is what you want.
in The h Bar, Mar 23 at 8:54, by Secret
GR is harder than quanutm IMO. cause it f888 up the intuitive notion of distance
tbh I don't know what I want
in The h Bar, Mar 23 at 8:59, by Secret
GR, on the other hand, you can no longer draw diagrams without massively distorting what the model is describing
I can't really see myself in academia
but I still have the research itch
4:53 PM
@Semiclassical Me too, other than wanting to get well. =)
though I suspect I might be able to overcome this now that I have a more firm grasp of the concept of spacelessness
@Semiclassical I don't like the publish or perish thing, but that is just the way many systems are now.
So, couldn't find the image I was thinking of, but, 3-adic numbers are often visualized with the Sierpinski triangle, apparently.
@Rithaniel I think 2-adic numbers can be visualized with the Cantor set?
@Secret Although you can't grasp spacelessness itself. =)
4:56 PM
That would make sense.
with the digits of your 2-adic number telling you which third of the interval your point is in at each step of the construction
@JasperLoy of course, else I am not a human being and has no physical body
Actually, that's a cool way to bridge two seemingly unrelated concepts.
@Secret Well, maybe in some beliefs, even the body exists only in the mind.
e.g. any 2-adic number ending in 001 would be somewhere in (0,1) -> (2/3,1) -> (2/3,7/9) -> (2/3,19/27)
no idea how the 2-adic norm would work there tho
4:59 PM
Did you guys know that wombat faeces being cube shaped has got to do with the shape of their intestines and is a matter of fluid dynamics?
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