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12:02 AM
@copper.hat What occasions that remark?
 
12:15 AM
User 123. I ended up blocking them because I couldn't take it anymore. I observed you being very patient yesterday.
Half of the registers were down at Trader Joe's which blocked aisles and significantly reduced throughput. Decided I didn't need the 40 min. wait right now.
 
0
Q: Order of a Module

user193319Here is the problem I am working on: If $M$ is a finitely generated torsion module over a PID $R$ such that there exists $m \in M$ with $ann(m) = (r)$, where $R \ni r = order(M)$, what can you say about the invariant factors? What exactly is $order(M)$?

@robjohn But those aren't matrices over $\Bbb{Q}$.
 
 
1 hour later…
1:26 AM
@TedShifrin They are not singing dogs ;-)
Now we are going to pick up food.
 
1:37 AM
OK, @robjohn, I recant.
 
2:10 AM
1
Q: Is AlphaFold just making a good estimate of the protein structure?

AaronIn the news, DeepMind's AlphaFold is said to have solved the protein folding problem using neural networks, but isn't this a problem only optimised quantum computers can solve? To my limited understating, the issue is that there are too many variables (atomic forces) to consider when simulating h...

 
2:30 AM
@user193319 If you don't allow matrices over $\mathbb{C}$, then you can't always find a Jordan form.
 
2:44 AM
@robjohn I know, and I'm trying to argue that that example is one which doesn't have a Jordan form over $\Bbb{Q}$. I'm just not sure about my argument; I think it's correct, but something seems off...
 
3:38 AM
@user193319 Are the eigenvalues in $\Bbb Q$?
 
9
Q: Introduction to protein folding for mathematicians

user37344My background is mostly in (applied) math with healthy doses of physics and computer science. Are there any good introductions to protein folding and its challenges for someone with that kind of quantitative background but very little knowledge of biology, chemistry, etcetera?

 
4:00 AM
hi chat
I proved a fine theorem recently
*derived
here it is:
let $p_n$ be the $n$th prime. Then $$p_n\le\frac{p_x}{x\log x}n\log n\quad\forall n\ge x$$ for any natural number $x$
This gives very wonderful bounds for $n$
for example $$n\log n<p_n\le 1.120904141401622842917499039n\log n\quad\forall n\ge 1000000$$
and 1.120904141401622842917499039 is very close to 1 so this helps to approximate large primes
can't give a proof here
I have many more bounds for functions of primes
my question: is this bound known already?
 
4:36 AM
this is a 3d one
bye I have to go
 
5:33 AM
 
123
6:11 AM
Hi All. Good Morning..
 
6:24 AM
@LeonhardEuler so $\frac{p_n}{n\log(n)}$ is decreasing?
 
6:55 AM
@robjohn it tends to 1, and is decreasing
 
Have you tried plotting it? I think you'll find out it is not monotonically decreasing.
 
7:13 AM
Ok thanks
Where are the maximum values?
Do you know anything about them?
For example, the maximums of prime gaps occur at multiples of 6
 
Hey guys I need some tips on how to properly write down my math equations
I am usually confused when to subscript or spell out a certain variable
 
8:22 AM
Hey I derived this identity:
 
 
1 hour later…
9:27 AM
According to Stirling, $\frac{x!e^xx^{{\tiny\frac12}-x}}{\sqrt{2\pi}}\sim x+\frac1{12}$
so, $p_x\sim\left(x+\frac1{12}\right)\log(x)$
any idea about the error?
actually, $x\log(x)$ seems closer
 
@robjohn So here is a better one:
ah I don't have
but I have derived many asymptotics
even for prime gaps
 
 
2 hours later…
11:21 AM
@robjohn could you post a link please?
 
12:12 PM
@TedShifrin Over $\Bbb{Q}$ the only eigenvalue is $0$. Over $\Bbb{C}$, we also have $\pm i$
 
12:28 PM
Anyone know of any rings $R$ with uncountably many primes all of the form $nu$, $n\in\mathbb{Z}$ and $u\in U(R)$? You'd need uncountably many units, clearly, but I don't know a good example of this offhand
 
12:44 PM
Power series ring over Z in uncountably many variables? Each p(1+x_i) is prime
 
12:57 PM
@skullpatrol there
 
1:13 PM
thanks
 
Hey
 
I'm reading An Introduction to Systems Programming, by Leland Beck
It's pretty difficult book
struck in first chapter itself
SIC/SE Model Machines are very hard tho!
 
@skullpatrol was that what you wanted? or did you want a specific link?
 
@RewCie ask here
 
1:21 PM
@skullpatrol Thanks Pal. Helpful!
 
@robjohn I thought the webinar was on YouTube
 
I 💚 JAVA!
 
@RewCie you'll have to wait for professor Buffy
 
@skullpatrol Ah, you wanted the link to the webinar...
 
@skullpatrol Which room? here or The Classroom?
 
1:27 PM
@RewCie yes
 
Oh I see him
@skullpatrol Should I tag him?
or wait?
 
tag
 
@robjohn Your cat ran across the keyboard
 
@robjohn yes! Thank you, sir :-)
 
Yesterday I mistakenly uninstalled NetworkManager and wpasupplicant from Debian Buster...
Had to reinstall everything
4 hours wasted!
 
1:29 PM
:-/
 
💔
 
@MikeMiller Danke schön. That's a good one.
Whenever I think about this stuff, I start wondering if what I'm looking at is isomorphic to anything that you'd see in the hyperreals or surreals or anything like that. I'm still super-unfamiliar with anything in nonstandard analysis
 
Sir is online now
 
@Rithaniel I never do the nonstandard stuff
 
Yeah, I don't think anybody does, at least officially
But, that's the crux of the wondering. Maybe when we look at this stuff, we're working with something from that field without realizing it
But then again, it's really just an abstract thought about labels
 
1:48 PM
hi chat
 
Hi
 
Hey Leonhard, check out that one
 
how can I prove this:
Aaaaa @Rithaniel how you got that
 
I unfortunately didn't write down the formula
 
I shouldn't have told the name of that software
 
1:52 PM
But I'll see if I can recreate it
 
secret came out :P
 
Tag professor Buffy with your book recommendation @RewCie
 
Why not? I could have easily made my own fractal rendered
Well, maybe not easily, but the information is out there
 
@skullpatrol Oh okay..
 
I was just joking @Rithaniel
 
1:53 PM
Gortcha
 
this one is z^2+sin(c)
okay could anyone solve my problem?
okay a book arrived today
 
@LeonhardEuler sho us
 
dickson's history of theory of numbers vol. 1
uh it's bought rather than arrived
here's the cover:
bought that book because of apostol's book
 
Wow! Cover is cool!
 
apostol's book said:
Note. Every serious student of number theory should become acquainted
with Dickson's three-volume History of the Theory of Numbers [13], and
LeVeque's six-volume Reviews in Number Theory [45]. Dickson's History
gives an encyclopedic account of the entire literature of number theory up
until 1918. LeVeque's volumes reproduce all the reviews in Volumes 1-44 of
Mathematical Reviews (1940-1972) which bear directly on questions commonly regarded as part of number theory. These two valuable collections
2
 
2:05 PM
coolio
 
that " Every serious student of number theory" line caught my attention
 
yup serious
 
Serious
 
serious
 
60 courses in college... 4 years...
I'm damn nervous!
They have called us on 9th of Dec.
 
2:10 PM
@RewCie prof Buffy has answered
 
@skullpatrol I see... He gave a books on OS
 
yup
 
@skullpatrol I already had both... I'll later read Minix and Linux Dev ones...
 
60 courses =4 years
1 year=15 courses
1 course$\approx$24 days
that was my speed of learning calc 1
Okay that book looks fine
any serious number theory student here?
 
@LeonhardEuler Yep. Sed. One course = 4 books (atleast), one book 1200 pages :-(
During my JEE time, I used to read hardcore mathematical books, around 120-125 pages a day with 300+ problems a day!
 
2:14 PM
@RewCie ok
 
$\sqrt{-1}\,8\,\sum\,\pi$
decipher this...
 
@LeonhardEuler I ate some pie....
 
Xenioux!
 
if an odd perfect number exists, it must be of the form 12k+1 or 36k+9
proof?
 
2:22 PM
Proof has been left as an exercise for the reader
2
 
I can find one but the main problem is that mathematical associations require login (which requires institution name) but I am an independent researcher
The proof is trivial, easy to see, a corollary of Theorem 1, and left as an exercise to the reader.
The thing to be proved: RH
Theorem 1: 2+2=4
the truth of all books
 
@LeonhardEuler I'm glad Euler is not alive today here in this chatroom...
 
I am Euler
I am alive
 
Nah, the big Euler
 
2:29 PM
Not big Euler, not small Euler, I am just Euler
the same Euler who liked to prove everything
Ah don't say that it kills me
 
Euler is pronounced as /'oI'lr/
 
@skullpatrol ty
 
Note that the first one is Capital I and second one is small l
font! damn font!
 
@LeonhardEuler np, pal
 
I am neither called Wheeler nor Youler; I am eeler
eel (fish)+er
Leonhard fisher
so I am fisher
Leonhard fisher
 
2:33 PM
You seem to be a lot of things at once.
 
I am perfect, I am prime, I am even
I am better than that little boy Gauss who solved 1+...+100 and thought that he was a superhero
 
I think Apostol was not talking about this serious student...
 
lol
 
I am better than fermat, who had to make an excuse because he couldn't prove a simple theorem
I am better than pythagoras; he did nothing
I am supreme
What I am is the opposite of all this
 
2:37 PM
ok ok calm down pal
:-)
 
start reading your new book!
 
Okay now let me discuss actual math
@skullpatrol I have some questions so I will probably stay here till I get the answers
 
Here's the proof to your theorem
 
There are more
does anyone know the reason why maximums of prime gaps occur at multiples at 6?
This math is beyond the era of Euler
 
2:41 PM
@skullpatrol Okay pal, I'll have to go.... reading :-)
Bye everyone :-)
 
cya
:-)
 
@skullpatrol following your advice, I too am going
 
later, pal
 
 
1 hour later…
4:08 PM
Why does the multiplicity of a holomorphic map not depend on the charts on that point of the riemann surface.
 
multiplicity as in least non-vanishing coefficient of a taylor expansion in coordinates?
 
4:23 PM
If $M$ is a simple $R$-module, why does $J(R)$, the Jacobson radical, act trivially on $M$? Clearly $J(R)M$ is a submodule, but why can't it be that $J(R)M = M$?
 
what's your definition of the Jacobson radical
 
@Thorgott It's the order of the zero when I write the taylor expansion in coordinates
So if you have $f : C \to \Sigma$, where $\Sigma$ is the Riemann sphere, the you have given coordinates on $C$ centered at $p$, $f = a_nz^n + O(z^n+1)$ , this $n$ is said to be the multiplicity. So locally any holomorphic $f$ looks like $z \to z^n$
 
If you know that such a map will look like $z\mapsto z^n$ in suitably chosen charts, then coordinate-independence is clear, cause it can't look like $z\mapsto z^n$ and $z\mapsto z^m$ for $n\neq m$ simultaneously
but you can also see this directly: a change of centered coordinate charts is a biholomorphic map mapping zero to zero, so looks like $\sum_{n\ge1}a_nz^n$ with $a_1\neq0$, so if you plug that power series into the other power series, the lowest non-zero term (which is the order of the zero) won't change
 
123
Hi All..
 
4:39 PM
@Thorgott J(R) equals the intersection of all maximal right ideals of the ring
 
Right! Nice
 
Or r is in J(R) if 1-rx is invertible for every x.
Perhaps we can show it is contained in the annihilator ideal.....
 
every simple right $R$-module has the form $R/\mathfrak{m}$ for some maximal right ideal $\mathfrak{m}$
 
Oh, so that's pretty much the proof. I have to check if that theorem is in my book.
Actually, maybe it is easy to prove.
 
For the physical interpretation of a holomorphic function (satisfying cauchy-riemann) as a fluid flow: do I understand correctly that the real part describes how much the fluid is "dragged" into a point, in such a way that the the fluid flows into the direction of steepest ascent of the real part (the gradient), satisfying the laplace equation, so that what flows out is equal to what flows in
And that the imaginary part gives the amount of fluid that flows through a streamline, meaning the line that we get when we follow the gradient of the real part
physics wikipedia really doesn't satisfy me as much as maths wikipedia, meh
 
4:47 PM
@user2103480 I'm not thinking very hard about this but let me rephrase CR in a useful way
A holomorphic function is a vector field <u, v> so that the two "reflected vector fields" <v, u> and <u, -v> are irrotational
So your job is to understand the reflected vector fields in terms of the first
 
Let $f : M \to N$ be a smooth map and $p\in N$ be a regular value of $f$. Then is there a relation between $\text{PD}_M(f^{-1}(p))$ and $f^*(\text{PD}_N(p))$.
 
irrotational does mean what exactly intuitively? I know it means that the curl is zero, but that's just the formalisation
 
Obviously, $M$, $N$ are compact and orientable.
 
@user2103480 Bro draw a picture
What's different about <x,y> or <x,-y> from <-y, x>
(Or appeal to Green's theorem)
Irrotational means the line integral along every tiny loop is zero
AKA the vector field does not rotate around any point
 
whats PD
 
5:02 PM
Poincare dual. If S is a k-submanifold of M. PD_M(S) is the unique (n-k)-form $\eta$ such that for any $\omega \in H^{k}(M)$, $\int_S \omega|_S = \int_M \omega \wedge \eta$.
(such a thing exists because Poincare duality)
*unique upto cohomology
 
Bruh
This guy writes $PD_N(p)$ instead of `pairing with the fundamental class'
Same thing
One lives in homology and the other is cohomology but in top degree universal coefficients canonically identifies them
So whatever
 
@MikeMiller there are enough pictures on wiki; I'm pretty dense when it comes to diff geo and most things related to differential forms. Same holds for complex analysis, for probably very similar reasons. That's why I asked for someone to weigh in on my above interpretation
 
Yeah but I didn't want to think about it
Just gave you what I have
 
fair fair
 
@feynhat These are probably equal
I think I'd need more time than I have right now to prove it from first principles.
Maybe look in Bredon, but I can't promise it's in there
It's definitely true and the proof is "clear to me" but chasing through details seems to be a serious headache
 
5:23 PM
Does it involve some heavy result?
 
I dunno what a heavy result is
It should reduce to knowing that cup product is poincare dual to intersection product
and knowing how to make sense of intersection product of homology classes
Probably everything is more easily phrased in terms of a slant product or something but I never think about that stuff
And that fact comes from Thom iso in the end
 
What exactly does "pullback of a module" mean in the following: "Since the Jacobson radical
acts by zero on an irreducible module we see that any irreducible $U_n(\Bbb{C})$−module
is a pullback of irreducible $\Bbb{C} \oplus ... \oplus \Bbb{C}$−module."
Note: $U_n(\Bbb{C})$ is the ring of all $n \times n$ upper triangular matrices over $\Bbb{C}$.
And we are considering an irreducible (simple) module over $U_n(\Bbb{C})$. I am trying to show they are all 1 dimensional.
 
EM4
Contour integral of tan(z) , C = |z| = $\frac{3\pi}{4}$
would the poles be pi/2 and -pi/2
then use cauchy or what?
 
around the circle $|z|=\frac{3\pi}4$?
What are the residues of $\tan(z)$ at $\pm\frac\pi2$?
 
5:38 PM
@user193319 it's saying the same thing as what Lukas told you a couple days ago: a simple $R$-module is the same thing as a simple $R/J(R)$-module
 
I don't even know what is homology. I am doing forms and integration.
 
$U_n(\mathbb{C})/J(U_n(\mathbb{C}))\cong\mathbb{C}\oplus\cdots\oplus\mathbb{C}$
pullbacks just restriction of scalars, ig
 
Fine, then prove it using forms and integration
 
EM4
@robjohn , yes around $|z|=\frac{3\pi}4$.
 
6:20 PM
@feynhat: If $N$ is $n$-dimensional, take an $n$ form supported in a neighborhood of $p$ (with nonzero integral, of course) over which $f$ looks like a local product. It will boil down to Fubini.
 
6:30 PM
(My English sentence structure in that last penultimate sentence is just awful. But I think it's clear what I meant.)
 
 
1 hour later…
7:33 PM
Raghvan Narasimhan YVES complex analysis
is a beautiful book
 
8:17 PM
Does anybody know a decent resource to read about directional (Frechet) derivatives of maps like $f \colon x \mapsto Ax$ where we use coordinates instead of doing things generally?
 
Why would you use coordinates?
 
Maybe I worded my question poorly
I found a question on main and it confused me a lot
0
Q: Jacobian of trace of matrix product

tphillipsI would like to compute the following Jacobian with respect to $x$: $$\frac{\partial trace[A(x)A(x)^{T}]}{\partial x}$$ where $A(x) \in \mathbb{R}^{n\times m}$, $x \in \mathbb{R}^{m\times 1}$. The result should be a vector in $\mathbb{R}^{m\times 1}$. So far I have something like: $$\frac{\partia...

 
8:32 PM
First of all, you wrote $Ax$, not $A(x)$. Totally different animals.
 
Ah yea I was hoping reading up on $Ax$ would give me an idea on how to tackle $A(x)$
 
I would approach that problem by differentiating the matrix function $f(A) = \text{tr}(AA^\top)$ . And I would do that with the chain rule.
No. $Ax$ is a simple linear function. The derivative of a linear function is itself.
 
how, in general, does one determine if an $n$-manifold without boundary is the boundary of an $(n+1)$-manifold with boundary
 
Ah, cobordism groups.
I'm sure @MikeM can give you a nice lecture on this. It's a beautiful topic.
Take a look at Hirzebruch's Topological Methods in Algebraic Geometry. This might also be in Milnor's Characteristic Classes.
 
0
Q: Applying Pompeiu formula for large disc R

user3046168 Let $g(z)$ be a continuously differentiable function on the complex plane that is zero outside of som compact set. Show that $$g(x)=-\frac{1}{\pi}\iint_\mathbb{C}\frac{\partial g}{\partial \overline{z}}\frac{1}{z-w}dx\,dy,\quad w\in \mathbb{C}.$$ Remark. If we integrate this formally by parts...

 
8:40 PM
ah, this is cobordism theory, I see
I just read that every closed 7-manifold is the boundary of an 8-manifold and this seems really surprising to me
 
I guess the easiest example, if I remember right, is that $\Bbb CP^2$ cannot be a boundary. Characteristic classes tell you that. Or I misremember.
 
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.”
cute. Does anybody know what he meant?
 
LOL, not exactly. Of course, he ultimately switched to applied math stuff.
 
bummer, I thought there was an exciting conspiracy
 
If someone explained it to me, I would probably say, "Oh," but I don't know off-hand. Maybe the Grothendieck-Bourbaki school ... I dunno.
 
8:48 PM
Oh yea the trace is linear so $\text{d}(\text{tr} \circ f)(x)v = \left( \text{d}\text{tr}(f(x)) \circ \text{d}f(x) \right)v$
 
Right, @Khallil. Now what's the derivative of $AA^\top$? Best way is to compute $d(AA^\top)(B)$.
 
is RP^2 a boundary?
 
Hmm, non-orientable.
I've forgotten the non-oriented cobordism groups.
Oh, duh. $\Bbb CP^2$ cannot be a boundary because it has $p_1\ne 0$.
 
Is there a neat way to write out $\text{d}(AA^{\top})(B) = AB^{\top} + BA^{\top}$, @TedShifrin ?
(I imagine commutators might come into a nice expression but not sure)
 
Just write the directional derivative as a limit definition ... piece of cake.
 
8:56 PM
Oh sorry I meant re-writing $AB^{\top} + BA^{\top}$ in a neater way
(already worked it out with the directional derivative defn)
 
I don't even know: is the Möbius strip a boundary?
 
It has boundary (or is non-compact).
No, no nicer way to write it, @Khallil, but when you use tr you can play a little.
 
Hi all. Things are looking bad for my PhD. In less than a week, I might be moved down to the Master's degree. (I've struggled with my mental health these last few years, so my life has not been conducive to studying.) I feel like a failure and I don't think I'll ever get on to another PhD programme.
 
I think Stiefel-Whitney class will tell us that $\Bbb RP^2$ cannot be a boundary and actually generates the non-orientable cobordism group in dimension 2.
The question, Shaun, ultimately is, might you be happier and healthier not trying for the Ph.D. For a lot of people, I know the answer turned out to be yes.
OK, it's late for lunch for this bonzo. Back later.
 
Thanks for the help as always, @TedShifrin
Enjoy your lunch <3
 
9:05 PM
Maybe, @TedShifrin. I don't know. I'd like to be an academic, so, as far as I'm aware, a PhD is required. I enjoy studying. I don't think I'd be happy if I just tried the once.
Enjoy your lunch!
 
9:18 PM
@AlessandroCodenotti I just spent more than an hour trying to prove that a strongly continuous semigroup is continuous in the operator-norm just to learn that its false in general
I thought that the strong operator topology is just the usual norm topology
whack
 
@user2103480 Wait isn't it?
Anyway the thing you wanted to prove should be equivalent to being a uniformly continuous semigroup
 
In functional analysis, a branch of mathematics, the strong operator topology, often abbreviated SOT, is the locally convex topology on the set of bounded operators on a Hilbert space H induced by the seminorms of the form T ↦ ‖ T x ‖ {\displaystyle T\mapsto \|Tx\|} , as x varies in H. Equivalently, it is the coarsest topology such that, for each fixed x in H, the evaluation map ( T , x ) ↦ T x {\displaystyle (T,x)\mapsto...
The SOT is stronger than the weak operator topology and weaker than the norm topology.
@AlessandroCodenotti yeah which implies that the domain of the generator is the whole space, I've since learned
 
Ah ok
But then the uniformly continuous thing should be equivalent to continuity wrt the norm topology I think? Maybe I'm getting confused
Hi @Balarka
 
@AlessandroCodenotti yes yes this is the case I think
 
9:33 PM
unif. cont. implies continuous wrt the norm topology pretty much by definition
 
Howdy, confused demonic @Alessandro.
 
I guess the question is whether there are semigroups that are continuous wrt the strong operator topology but not the norm one then
hi @Ted, analysis often has that effect on me
 
Logic always has it on me.
 
I don't do much logic either, it's too dry
 
Wait. You've forsaken?
 
9:36 PM
I think I use logic in a much stricter sense than you do, I consider it as a different thing compared to set theory or model theory
 
Oh. I really liked model theory when I took my one grad logic course.
 
Hi everyone
 
Model theory is very neat
 
You are picky as I am when people refer to everything with a manifold in it as differential geometry. It is utter balderdash.
Heya a @Balarka!
 
The things I'm working on for my PhD are a very interesting mix of topological dynamics, descriptive set theory and model theory
 
9:38 PM
@AlessandroCodenotti grrrrr
 
Balarka: You can talk to Thor about cobordism :)
 
Was there a question unanswered
I know very little about cobordisms
 
I attended a couple of interesting seminars on hyperbolic groups earlier this week, Balarka
 
Balarka: What's the one-line proof that if $w_2\ne 0$, then a surface cannot be a boundary?
 
waves angrily with his arms and mumbles something about modal logic formalizing theological arguments
 
9:39 PM
I am like Hippa's meme. I've forgotten everything.
 
Double the 3-manifold it's the boundary of, I think.
 
How does that show $w_2=0$? You know that it's just $2$-torsion to start with.
 
Suppose $S = \partial M$. Consider $M \cup_{\partial M} M$. This is a closed $3$-manifold, so has $0$ Euler characteristic modulo $2$. That means $2\chi(M) - \chi(S) = 0$ modulo $2$... so $\chi(S) = 0$ modulo 2.
 
Oh, I see. Very cool.
 
@AlessandroCodenotti Nice. What seminars?
 
9:43 PM
I remembered about Pontryagin numbers for oriented cobordism, but I don't remember those proofs either.
 
@TedShifrin I suppose if $W$ is a cobordism between $M$ and $N$, then the tangent bundle on $W$ restricts to $TM \oplus \varepsilon$ and $TN \oplus \varepsilon$ on the two boundary components $M$ and $N$ of $W$ respectively. So by naturality and stability of characteristic classes, $TM$ and $TN$ must have the same char classes.
Characteristic numbers, rather.
 
Right, I get it.
 
I wish I knew how the various converse of such theorems are proved
Thom's theorem(s)
 
I guess if $M$ bounds $X$, then Stokes's Theorem says that any Pontryagin number of $M$ has to vanish (because of your argument on restricting the tangent bundle of $X$).
I was reviewing that a little. It's all classifying space stuff you love.
 
That makes sense
 
9:48 PM
Hirzebruch gives a two-sentence summary.
 
@TedShifrin what's the issue with being non-compact
 
We only do cobordism in the compact setting, @Thor. How could a Möbius strip be a boundary?
Boundaries of manifolds with boundary have no boundary.
 
for all I know (read: nothing), the Möbius strip could be the boundary of a non-compact manifold
open Möbius strip, that is
 
Take cone on the open Mobius strip, and throw away the vertex?
Anything is a boundary if you allow noncompact crap
Take cone on it and throw away the vertex
 
Right. Balarka beat me to the punch.
 
9:52 PM
I guess taking cone and throwing away vertex is really my idiosyncratic way of saying $M \times [0, \infty)$
 
@Balarka:
 
Oh thanks! That looks surprisingly readable. Maybe I can try to fill in the details
 
right...
 
@BalarkaSen Here in Muenster. First one was about the classification of different types of actions an hyperbolic group can have, the latter was about showing that every hypernolic group admits an affine isometric action on $\ell^p$ for some $p$
 
so I guess restricting to boundaries of compacta is the only sensible thing
 
9:55 PM
@Alessandro Ah ok interesting
 
@Thor: Yup, but good to think that through!
And Balarka's proof that $\Bbb RP^2$ cannot be a boundary is elementary (Euler characteristic argument).
 
Yeah. I remember being very surprised that $\Bbb{RP}^2 \# \Bbb{RP}^2$ is nonetheless boundary of something.
 
Mike would know in a millisecond.
 
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