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12:00 AM
So look it up! Take the definition of the direct limit, which you can surely find, first figure out what it does in simple examples, then write it out for this instance and try to see what it does here
I once took two weeks to understand a single sentence in a paper, it happens.
 
I'll try to read the referenced Whitehead paper, I guess...
he just starts hammering away with CW complex stuff without defining a CW complex...
@ACuriousMind Do you know if this heavy-duty homotopy theory stuff shows up again later
so, what I understand now is this
we have a family of continuous maps $f_i:M^{a_i}\to K_i$, which are in fact homotopy equivalences
we then take $K=\sqcup K_i$ with the topology in which each $f_i$ is continuous
er
No...
@ACuriousMind What space actually gets this topology
$K=\sqcup K_i$ gets the direct limit topology, which is the topology s.t. each $f_i:M^{a_i}\to K$ is continuous?
then, by some theorem of disjoint unions, there is a unique continuous map $f:\sqcup M^{a_i}\to K$ that restricts to $f_i$ on each $M^{a_i}$
Hmm, i think I can buy that
 
12:50 AM
@ACuriousMind So...why is any manifold dominated by a CW-complex
Milnor says you embed it in some $\Bbb R^n$ and then retract a tubular neighborhood?
 
1:09 AM
How do I TeX this?
@ACuriousMind Does a homotopy equivalence induce isomorphisms on the direct limit of homotopy groups in all dimensions?
@ACuriousMind So the directed set here is the sequence $a_1<a_2<\cdots$
And the groups are $\pi_n(K_i)$?
So what are the homomorphisms $f_{ji}:\pi_n(K_i)\to\pi_n(K_j)$?
I don't know how this is a direct system of groups
@ACuriousMind First, let's go back to the general case
$D$ is a directed set and $G_\alpha$ are abelian groups for $\alpha\in D$. I think the homomorphisms $f_{\beta\alpha}:G_\alpha\to G_\beta$ tell us how they "relate" to each other
the condition $f_{\gamma\beta}\circ f_{\beta\alpha}=f_{\gamma\alpha}$ basically tells us that if we know how $G_\alpha$ relates to $G_\beta$ and how $G_\beta$ relates to $G_\gamma$, then we know how $G_\alpha$ relates to $G_\gamma$
Now, we have the equivalence relation $f_{\beta\alpha}(g)\sim g$ $\forall g\in G_\alpha$ $\forall \beta>a$.
right, so what this does is it tells us that the $g'(:= f_{\beta\alpha}(g))$ in $G_\beta$ related to $g\in G_\alpha$ is basically "the same"
and that all $g'$s in all the "higher" groups are basically "the same"
So, this fixes a relationship between all the groups and between all the elements in all the groups
Seems reasonable.
well, not all
the maps need not be surjective
ah, but they tell you how the elements in each group are related to the one right above, and so on
Now, we define $\varinjlim G_\alpha=(\bigoplus G_\alpha)/\sim$.
So, we put together all the groups and identify the elements which we said are "the same"
@ACuriousMind Ok. So are the homomorphisms in the Milnor case just maps induced on the homotopy groups by the inclusion maps?
Do I sound insane right now
@ACuriousMind So, we have these nested sets $K_1\subset K_2\subset\cdots$
Now, we have to take the union $\bigcup K_i$, presumably
Now, what is $\pi_n\left(\bigcup K_i\right)$
 
1:34 AM
Anything fun happening in this room?
 
@ACuriousMind I conjecture that $$\pi_n\left(\bigcup K_i\right)=\varinjlim \pi_n(K_i)$$
where $f_{ji}:K_i\hookrightarrow K_j$.
@ACuriousMind Things that are unclear: the existence of the direct limit topology
Let $g_i:M^i\to K_i$ be the maps in the figure above.
Then we can take $g_i:M^i\to \bigcup K_i$ and that makes perfect sense.
@Mikhail Perhaps. I'm trying to learn algebraic topology.
 
From what I can see you're trying to learn latex
 
I know latex just fine
@ACuriousMind Now, we need a topology on $K:=\bigcup K_i$ such that each $g_i$ is continuous.
Hmm, is that supposed to be a disjoint union?
Well no probably not, we take the union and then topologize it.
But then my above claim does not make sense.
 
I'm trying to derive some boring sampling criteria for measuring diffusion :-)
 
Presumably, we can take a topology on $K$ in which all the $K_i$ plus $\emptyset$ and $K$ are open.
Then we need to further refine the topology such that each $g_i:M^i\to K_i\subset K$ is continuous in the $K$ topology
How do we know such a topology exists?
Well, direct limit has to do with abelian groups, and I see no connection as of right now.
@Mikhail I've just spent an hour rambling
I need to eat something
 
1:45 AM
like a "sock"?
 
no
my advisor gave me a book that's way above my level
I don't want to give up
 
Sounds like fun
 
but I also don't want to read a 400 page book on algebraic topology
in 4 days
 
As we know books are for reference
typically
 
this is a learning book
 
1:48 AM
Does it say "lemma"?
 
there are lemmata in it
 
2:09 AM
0
Q: Why was this Very Low Quality flag declined?

Emilio PisantyI have always felt that the criteria for the use of the Very Low Quality flag are fuzzy, confusing, and ultimately wrong, and I recently had a VLQ flag declined that I would like some clarifications on. The answer in question is this one, which I'll quote as an image in case it is deleted. ...

 
user54412
@HDE226868 Did you know you have a doppelganger?
 
> I hail from Alberta, Canada also known as the Texas of Canada.
CANADIAN CONFIRMED
 
What's wrong with that?
 
Did you calculate that limit yet?
 
@bolbteppa What limit?
 
2:16 AM
The limit in the definition of the derivative of a product of things
If I said a "direct limit" was a formal power series, would you go insane?
 
@bolbteppa The product rule?
@bolbteppa No.
@ACuriousMind Do we even have to prove that the direct limit topology exists?
or do we just define it as the topology with that property
but then how do we know there is such a topology
Ahhhhh
 
Using
$\frac{\partial }{\partial x}f(x)g(x) = \lim_{h \rightarrow 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x+h)g(x+h) - f(x)g(x+h) + f(x)g(x+h)- f(x)g(x)}{h} = \lim_{h \rightarrow 0} \frac{[f(x+h) - f(x)]g(x+h) + f(x)[g(x+h)- g(x)]}{h} = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}g(x+h) + \lim_{h \rightarrow 0} f(x)\frac{g(x+h)- g(x)}{h} = f'(x)g(x) + f(x)g'(x)$
we have
$\frac{\partial }{\partial x} \vec{A} = \frac{\partial }{\partial x} [A^i(x,y)\vec{e}_i(x,y)] = \lim_{h \rightarrow 0} \frac{A^i(x+h,y)\vec{e}_i(x+h,y) - A^i(x,y)\vec{e}_i(x,y)}{h} = \lim_{h \
qed
 
2:33 AM
is that on a manifold?
 
If you want to apply this simple idea on a manifold you need to invent some definitions to do that, you could do it a few ways
 
I like how you think you know more than me, but ok
 
But on the plane, or in $\mathbb{R}^n$, nothing wrong with this
 
well now that I know this wasn't on a manifold, I buy it from the beginning
@bolbteppa sure
don't know why you've been stressing about it
 
After all, the theory of manifolds is all about reducing things to $\mathbb{R}^n$
haha
I just think that's cool tbh
 
2:38 AM
but putting it on a manifold is pretty involved
and nonunique
 
I remember learning all the definitions in Lee and using other books and just memorizing it all and wondering how in the world that really related to calculus and the disconnect just became infuriating
 
it's always been obvious to me
 
Do you want to summarize every section of GP for me? :(
 
Not particularly
But if you want to know something specific I can help
 
Basically wtf is going on in chapters 2 & 3, let me dumb them down as best I can
In elementary calculus, you can define a curve as the intersection of two surfaces, as far as I can see it, transversality is really just generalizing that baby idea
 
2:47 AM
Oh god, why would you define a curve like that
 
I know
 
No, transversality is not that
transversality is when two submanifolds intersect in a way that is "nice", in the following sense: their tangent spaces add up to the whole tangent space of the big manifold
 
Let me say it in another way: another way of asking when the intersection of two surfaces forms a curve is asking when two submanifolds form a submanifold, another way of thinking about forming a curve on a surface is to intersect it with another surface as a way to impose this constraint, according to this pdf math.toronto.edu/mgualt/MAT1300/week4.pdf those are the motivations for transversality, the exact same ones from that baby calculus idea
 
Yes, there is a theorem on when the intersection is a submanifold
It's precisely when the intersection is transversal
 
Just as when the intersections of planes/surfaces forms a line/curve
Even more proof is that other differential topology concepts are motivated similarly like submersions and immersions coming from the idea of implicit $F(x,y) = 0$ or parametric $z = f(x(u,v),y(u,v))$ surfaces respectively
Ok so that's not totally insane, moving on: Is intersection theory a way to count the number of solutions to $F(x) = y$, i.e. $F(x_1,\dots,x_n) = (y_1,\dots,y_n)$? If so, is degree mod 2 about whether it has an odd or even number of solutions? If so, why the **** does anybody care?
 
3:00 AM
No, intersection theory is a way of assigning a topological/homotopy invariant to intersections
 
If transversality is about forming a curve via intersection of surfaces, then intersectionality is about ??? a) defining "intersection" formally? b) counting the no. of times the curve you created intersects itself? c) the number of solutions to an equation for some reason?
 
Intersection is defined formally
It's just $X\cap Z$ :P
The really important thing about intersection is this: the intersection number depends only on the homotopy class of the map
and using intersection theory you can define powerful invariants like the Euler characteristic and the Lefschetz number
 
Sure, but you can also define those ideas using grothendieck groups, so there's something underlying it that makes it tick (not sure about Lefschetz)
 
and using intersection theory you can prove Jordan-Brouwer, Borsuk-Ulam, compute the homotopy of spheres, Hopf theorem, etc.
@bolbteppa you do it "by hand"
using the smooth structure
Most things in "elementary" differential topology can be done with algebraic topology
 
I guess the thinking might go like this: Just as transversality came from the idea of using intersections of surfaces to generate a curve, what happens when you then intersesct two curves generates by the intersection of surfaces (in a Russian doll turtles-all-the-way-down iterative kind of way)?
But generalized to submanifolds
 
3:08 AM
The intersection of two curves is not interesting at all
Their intersection number will be 0
(in 3 or more dimensions)
 
Then since you're on a manifold where you want to find things that remain invariant under any change of coordinates, you search for invariants like that intersection number etc...
Hmm
Maybe not
Maybe that is a consequence
 
@bolbteppa any manifolds with noncomplementary dimension have 0 intersection number
 
Yeah
Ahh ok
 
@bolbteppa i.e. if X,Y in Z and dim X + dim Y < dim Z
 
Before thinking further, since GP have tubular neighborhoods in an exercise above the mod 2 section:
 
3:13 AM
you can always perturb $X$ away from $Y$ a small amount and they will no longer intersect
 
Here is the insane thinking I found behind tubular neighbourhoods so far:
Just as the implicit function theorem starts from an open set and finds a subset which maps onto the image, tubular neighbourhoods arise by starting from the image, inverting and trying to construct the open set you started with in the IFT
So if you were just playing games with the IFT and wanted to reverse the logic, you'd end up with tubular neighbourhoods
 
it's not that simple
what you do is you have a manifold $X\hookrightarrow \Bbb R^n$
and the tubular neighborhood is a part of the normal bundle $NX\to X$
yes, the IFT is used
but it's more like you can "fatten" $X$ inside of $\Bbb R^n$
and it's still a manifold
if $X$ is compact, you can make it a "constant" fattening
if it's not, it will generally be a variable one
 
Since the IFT is used to construct manifolds, and since it uses a subset of an open set to do that, and by Whitney's theorem every manifold is a submanifold of some $\mathbb{R}^n$, is the fact you can fatten not because you had the room to fatten from the fact you took a subset of an open set to construct that manifold?
 
well of course you use an open set somewhere
I'm not sure what you mean
 
Alright, I will think about that one
So with transverality, two manifolds that are transversal don't have to be of complementary dimension in general? But in the case they are you get 0-dimensional intersections leading to the next section of GP?
Hmm
Using the wiki
In mathematics, transversality is a notion that describes how spaces can intersect; transversality can be seen as the "opposite" of tangency, and plays a role in general position. It formalizes the idea of a generic intersection in differential topology. It is defined by considering the linearizations of the intersecting spaces at the points of intersection. == Definition == Two submanifolds of a given finite-dimensional smooth manifold are said to intersect transversally if at every point of intersection, their separate tangent spaces at that point together generate the tangent space of...
Dimensionality issue
The intersection of two planes in space is 4 dimensional, space is 3 dimensional, so you'll get a manifold of dimension 4 - 3 = 1, a curve
Awesome
 
3:32 AM
Correct
 
Intersection of a plane and a line is 3 dimensional, complementary dimensions, 0-dimensional set of points
What is the meaning of the intersection of two lines in space, why is that not a good thing
 
because you can perturb one away from the other
it's unstable
if two surfaces intersect, you can move each one a little bit
no matter how you do it, the intersection will still be a line
 
Oh good point
"If one were asked to give Sherlock Holmes a clue as to how he might best succeed in unlocking the secrets of the smooth Universe he inhabits, one might do worse than observe that most pairs of lines in the plane meet once, and most pairs of lines in space do not meet at all"
 
good quote
@bolbteppa that's in section 1.6 in GP
 
Ahh cool
This book is full of goodness
Even the comment about measurement is stunning
Ok so you have transversality of submanifolds, in the case of complementary dimensions their intersection is a bunch of points, why consider "odd or even", and then when orientation comes in do we go further
I guess on a basic level the oddness or evenness will remain invariant under perturbations
 
3:47 AM
yes
 
"Physically, stability means that transversal maps are actually observable" what lunatics
Jesus
In linear algebra, isn't solving a system of linear equations about finding the points or lines of intersection, is this not a generalization to non-linear equations having a unique solution & just saying that we can approximate the solutions???
When you solve a system of linear equations, i.e. intersection of hyperplanes, and get a line or a plane or something, you have transversality, but when they intersect in a point, you get a unique solution
 
@bolbteppa ah, I might have something interesting
let $A$ be a linear operator and let $V\subset\Bbb R^n$ be a linear subspace
then $A\pitchfork V$ means $A(\Bbb R^n)+V=\Bbb R^n$
so...$A\pitchfork \{0\}$ means $A$ is a vector space isomorphism
 
4:03 AM
Ah I think that makes sense yeah
Will come back to it but yeah that is a good idea
 
@bolbteppa also, a submersion is transversal to any submanifold
this is so important
 
4:15 AM
I just can't think of a children's analogue of why the mod 2 number of solutions of a system of equations remains invariant under perturbations, the pictures make it seem clear enough
One idea might be in solving polynomials, there is this crazy thing where you add a parameter to different terms in the polynomial and you can lose roots
In $ax^2 + bx + c = 0$, if you add a parameter $\varepsilon$ to $c$ you still have two roots by sending it to zero or whatever, but if you do that to $a$ you lose a root
and the quadratic formula goes to infinity
If it's a parabola, I guess it intersects the x axis either 0 or 2 times for transversal intersections by varying it on the $c$ term (i.e. shifting it up or down), but id what it means to play with it on the $a$ term
Did I just find a counter-example? :p
 
Uh...let me see
Well, let me try to explain something
if you have a double root, then it's not a transversal intersection (do you see why?)
 
Yeah
Solving a quadratic equation is tantamount to determining the intersection numbers of the submanifolds $y = 0$ and $y = ax^2 + bx + c$ or something right? Homotopic deformations of $y_{\varepsilon} = ax^2 + bx + \varepsilon c$ offer an example of the theorem, but for $y_{\varepsilon} = \varepsilon ax^2 + bx + c$ we go from an odd to an even number of intersections no?
 
no
it's much simpler
what is the tangent space to the parabola at the double root point
 
Yeah but that case doesn't matter
If you have a parabola with two real roots, and then you perturb the highest term to zero, it will turn into a line passing through the x-axis without ever having become non-transversal no?
 
4:31 AM
uhhh
what
the perturbations are small
arbitrarily large ones can mess stuff up, yeah
you could perturb two transversal surfaces if you try hard enough
but any small perturbation will not matter
 
Ok cool will come back to this too :p
Alright thanks for the help
 
np
 
vzn
5:02 AM
↑ GR+QM still incompatible :|
 
 
2 hours later…
user116211
7:06 AM
@PhysicsMeta Absurd and a ridiculous request ;((
 
user116211
Though sometimes Lumo becomes snarky, that doesn't mean he has to leave ;/
 
user116211
WTH is that rubbish talk - OP didn't get @dmckee's advice in any way.
 
0
Q: Pull and push problem

Santhosh B R I have a question about my Physics Stack Exchange post: PULLING AND PUSHING TWO OBJECTS THROUGH RIGID MEMBER IN BETWEEN THE OBJECTS

 
user116211
7:21 AM
@PhysicsMeta These newbies have compelled me in hating the CAPS ;(((
 
8:22 AM
@vzn : not a great article. "However, quantum spin is, confusingly, not related to the rotation of a body. Indeed, if you calculate how fast an electron needs to rotate in order to generate its spin angular momentum, you'll come up with a ridiculous number (especially if you take the idea of the electron being a point particle seriously)."
 
 
2 hours later…
10:31 AM
@vzn Because we refuse to believe that these two theories are separate
What if they are really separate? Similar in the way some might think about the consistent null results in magnetic monopoles, are there any test that instead of showing that they are interrelated, show that they are separate instead?
 
10:52 AM
@0celo7 In this case, the topology on $K$ is just the topology such that a set is open if and only if there exists a $K_i$ in which it is open (cf. final topology).
 
11:35 AM
@ACuriousMind
So I'm studying for my exams, and I've bumped into this small question
I learned how monopoles (in d=4) and instantons can both be viewed as solutions to $F_{MN}=\pm \star F_{MN}$, but I'm not exactly sure what the relation between the two is. I think monopoles are supposed to be finite energy solutions, while instantons have finite Euclidean action.
However, my notes say something like "they differ by boundary conditions at infinity", which leaves me (??)'ing
 
@Danu I think that's just another way to say "finite energy" vs. "finite Euclidean action". The monopoles must approach pure gauge, i.e. no field, at spatial infinity for thie magnetic field to fall off like that of a monopole should, while the instanton usually approaches pure gauge at temporal infinity to be viewed as the "mediator" between to vacuum configurations.
 
Hmkay.
And those 4d Euclidean instantons correspond to what in the language of solitons?
 
What is the "language of solitons"?
 
I remember something like $D+1$-dim soliton corresponding to $D$-dim instanton in Euclidean time... or the other way around.
 
I have never seen a more precise definition of "soliton" than "some solution that carries topological charge/is topologically stable", so I can't comment on that
 
11:49 AM
OK.
To me, a soliton is the following: A finite energy, (top.) stable solution that admits a particle-like interpretation in the sense of having correct behavior under boosts and admitting asymptotic scattering (in- or out-)states.
 
Well, by that definition I don't see how e.g. the BRST instanton is not a soliton
 
I don't know what the BRST instanton is.
 
It has a well-defined "center" which you may interpret as the particle, and I'm pretty sure you can scatter these things
@Danu The general solution to the anti-self-dual equations on $\mathbb{R}^4$
 
Well, the solitons are also supposed to be located in a Lorentzian space
 
11:54 AM
Hence somethingsomething imaginary time
So maybe the following
a $D$ (always spatial)-dim instanton should be a $D-1$ spatial-dim soliton in imaginary time
No, sigh
Let's see
 
Ahhhhh
You mean like the 4D solitonic Polyakov-'t Hooft solution looks like a 3D magnetic monopole/instanton
 
probably
:D
So is $D$ supposed to be spatial dimensions here or not?
For you
Because the $4d$ Polyakov-'t Hooft monopole is indeed an example of a $D=3$ (spatial dimension) soliton for me
 
Yes, there is a correspondence between instanton/monopoles in D Euclidean dimensions and solitonic solutions in D+1 dimension
 
So how does this correspond to an instanton?
 
@Danu The field of P-t' H monopole in 5D (which is really a soliton in the 4D theory) is exactly the same as that of a BPST instanton in 4D, I think. Here's a paper about it.
 
12:03 PM
I can tell you a bit more about what I've seen: We have this $4d$ P-'t H monopole, and you can write it in a $5d$ form by just defining $A_4=\phi$. Is it correct to say that, in doing so, we switch from Yang-Mills-Higgs to pure Yang-Mills by adding a dimension, hence translating a $4d$ soliton/monopole into a $5d$ (pure gauge theory) instanton?
But I still didn't go to Euclidean signature, lol.
 
@Danu I The correspondence is the other way around: You start from the Lorentzian monopole solution and find that it is parametrized by the same type of moduli as the generic instanton solution in Euclidean space one dimension lower - the paper I linked is neither long nor hard.
The 4D -> 5D trick for the P-'t H monopole is for solving the equation, but the 4D YMH equations do not become the 5D instanton equations.
At least, I don't think they do
 
@ACuriousMind The correspondence is supposed to come purely from the moduli space?
So, in temporal gauge, $A_0=0$ and in the $5d$ language for the monopole (YMH equations) we can just ignore the 0-component, so we effecitvely get some $4d$ equations.
These are the same as the $4d$ instanton equations
And that makes a lot of sense
 
12:18 PM
Yes
 
Because the $A_0$ component is exactly the time-like part
So we get a $4d$ Euclidean thing, from a $5d$ Lorentzian thing
 
Yeah, right, abuse your mod power to hide your confusion :D
 
<3
Didn't wanna drag innocent bystanders down with me
 
Keep telling yourself that.
 
Serve & protect, my child.
 
12:45 PM
Where do we go now?
 
2:03 PM
@ACuriousMind So now, to come back to the boundary conditions. In what sense do instantons interpolate between ground states?
 
@Danu See this answer of mine. The instanton is itself an (Euclidean) ground state, and it interpolates between spatial slices of pure gauge configurations that differ in their topological charge exactly by the instanton number.
There's another way to see the significance of instantons even if we want to just do the perturbative path integral around a flat, uncharged configuration:
Locality/Cluster decomposition/whatever locality principle you like best roughly says that the path integral should factor as the integral over field configurations of pieces of the spacetime. Now, instantons usually have anti-instantons, meaning that even if we start from a uncharged flat configuration, we can put an instanton on one half of our space and an anti-instanton on another half of the space and then this is a globally uncharged configuration that is included in the integral
But as soon as we want to factor this to compute it more easily, it has to run over instanton configurations on the subspaces
Witten says this somewhere far more eloquently :P
 
Hmmkay...
You misspelled principal as principle, btw ;) I corrected the typo for you (as well as $\pi_3(S^3)$, which should've been $\pi_3(G)$, I guess you have $SU(2)$ in mind...)
 
@Danu I knew about the former typo, but it didn't warrant bumping the post to me
 
@ACuriousMind I don't agree with the anti-bumping principle :P
 
@Danu I personally hate it when I see activity on an interesting question just to find out someone corrected a trivial typo
 
2:18 PM
I never look at that tab.
 
I don't use any other tab, except the featured and newest tabs occasionally
 
I only use the "newest" tab.
 
Ew
Why would I want to see all the questions, but not the answers coming in?
 
(very rarely the featured, too, and the hot in order to see why crappy questions got so many upvotes haha)
@ACuriousMind I don't know!
I just started out like this because I was here to answer questions.
Now this is how I roll :P
 
haha
 
2:23 PM
I'll give the active tab a try.
 
@ACuriousMind did you read my wall
I don't think the thing in Milnor actually requires category theory
 
Of course it does not
 
You said it does...
 
@ACuriousMind By the way, I'm now in a section of my notes where it is argued that you don't need a cylinder-type deal to see this interpolation stuff.
 
@0celo7 No, I said that "direct limit" is an algebraic/categorial construction. That you can brute-force define the object $K$ that is constructed here without using the general concept of a direct limit may happen.
 
2:35 PM
@ACuriousMind So, the map $g$ is just the $g_i$s pasted together, right?
 
In fact, most of the classical big theorems on algebraic topology can be rephrased in categorical language
 
Nothing fancy about it
 
vzn
hi dudes trying to follow some of the above re tricky soliton correspondences & note the july 2016 paper ref (!) :) btw 't hooft cited above has some "very unconventional" ideas lately (a lot of papers on cellular automata etc), wonder what ppl think of those
 
@ACuriousMind Is my intuition on direct limits correct?
It doesn't seem that strange to me
 
@0celo7 Yes
 
2:37 PM
@vzn They are completely mislead is what 99% of people think of those.
 
vzn
@Danu yeah ok got that impression myself (about the crowd judgement that is). so he seems to have some einstein-like qualities/ tendencies it appears aka unorthodox etc
 
@ACuriousMind btw that comment I left is genuine, always wondered about it
 
@0celo7 Yeah, sounds correct
@0celo7 I said don't ask about the $\pi$! :P
 
@vzn Einstein was never very unorthodox.
 
@ACuriousMind actually, I think you need a direct limit notion
because what is $\pi_n(K)$
 
2:40 PM
@ChrisWhite It's kind of hard to be active on Worldbuilding and not be aware of TrEs-2b
 
you need to show that $g$ induces an isomorphism of $\pi_n(M)\to \pi_n(K)$ for all $n$, I think
 
vzn
@Danu think he was, he just didnt publish much after becoming that way largely the 2nd ~½ of his life. it is portrayed more by mainstream physics history as "quixotic" like (again parallels to 't hooft)
 
@0celo7 Ah, you're right. But you'd first have to show that the homotopy group functor preserves direct limits ;)
 
fuck
can you give a reference?
 
2:43 PM
@0celo7 I don't even know whether that is true
 
@ACuriousMind Also would you mind telling me exactly what needs to be shown?
Milnor skips so many steps in that proof it's insane
And I don't know why the manifold is dominated by a CW complex either
@Danu you're free to chime in too
 
vzn
@Secret @JohnDuffield figured someone would ding me over the popsci article. ok, its not a great treatment, the tone is rather facetious, but think its better than nothing, & it seems to contain no outright errors, and one can always ref the underlying high quality paper contents/ result instead (and not a lot of sympathy for those who criticise it merely on the arstechnica summary). anyway top theorists agree there must be some way to mesh QM+GR, its esp critical wrt black hole theory etc...
 
@0celo7 You need to show that $g$ induces isomorphisms on homotopy. It should follow from all the $g_i$ doing that, so I think you need to show that two maps $S^n\to K$ that are homotopic in $K$ already are homotopic in some $K_i$.
 
@ACuriousMind What exactly is $M$ and $K$?
Disjoint unions, regular unions, ??
 
@0celo7 Uh, $M$ is just your manifold, isn't it?
 
2:57 PM
@ACuriousMind Sure
 
@vzn I am not criticisng anything about that finding, I am just wondering whether there are groups who have considered the two theories might be independent framework of each other. From what you said, do you mean the observation of black holes rules out this possibility and that quantum gravity has to exist in some form?
And suppose such possibility is not ruled out, how it would be experimentally tested?
 
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