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7:00 PM
First step is to understand explicitly why that is $\Bbb{RP}^3$
If you can do that you can do it for all the rest
 
yeah because $[-w:-x:-y:-z] = [w:x:y:z]$ right
let's say $\Bbb RP^2$ is $\{ - \mid w=0 \}$
 
@LeakyNun This is important to keep in mind when you work with sheaves more in the future (e.g. if one day you decide to live in some derived category of sheaves)
 
@loch cool
 
so far I see a bunch of symbols, I do not know how that is a justification for the cell structure yet
 
I'm not sure about living in some category
 
7:03 PM
life is hard enough in the society of the spectacle, i can't begin to imagine the society of the natural transformation
3
 
$\Bbb{RP}^3 = \Bbb R^3 \sqcup \Bbb{RP}^2$ where $\Bbb R^3$ is $[1:x:y:z]$ and $\Bbb{RP}^2$ is $[0:x:y:z]$
 
Hi @Ted
 
hey @Ted
the people just keep coming!
 
Hi, a @Balarka, @Leaky, @MikeM
I've been busy moving and unpacking. Ugh.
 
@BalarkaSen oh lens spaces right
 
7:04 PM
my trick was not to bring anything
 
I have two backpacks everywhere I travel
 
That's a trick I also used when finishing grad school, but after that it was hopeless. Although most of the furniture I bought during my postdoc is now long gone.
 
my trick was not to travel
 
Hi @Ted
 
That's all I need
 
7:05 PM
Hi @Mathein
 
@TedShifrin I am not planning to move many more times
 
@BalarkaSen then what Wiki says doesn’t make sense to me. Unless they forgot to say negative curvature or something
 
@RyanUnger yeah it's nonsense
 
Well, @MikeM, I have now lived in 3 places in 4 years in SD. I hope never to move again. In the first years after Ph.D. (36 years) I lived in 3 places total.
 
7:06 PM
let's say I map $B^3 \to \Bbb{RP}^3$ by $(x,y,z) \mapsto \left[1 : \dfrac1{1-x} : \dfrac1{1-y} : \dfrac1{1-z} \right]$
is this a good map?
 
I don't know why people think Wiki is the Bible of mathematics. I've found lots of wrong stuff.
 
God Leaky
 
What is a "good map"?
 
he's trying to write down the cell decomp of $\Bbb{RP}^n$
 
$\Bbb{RP}^3$ is $S^3/\Bbb Z_2$. You should be able to say why it's $D^3$ attached to $\Bbb{RP}^2$ by the double cover from there.
 
7:07 PM
the particular mechanisms of that are a mystery
@TedShifrin probably because it's free and easy to reach, nothing more
 
oh ok that's better
$S^3/C_2$
 
I don't like your map at all, @Leaky. The boundary of the ball should be double-covering the $\Bbb RP^2$ at infinity
 
$S^3/C_2 = (D^2 + D^2 + S^2)/C_2 = D^2 + S^2/C^2 = D_2 + \Bbb{RP}^2$
there you go @BalarkaSen
 
WTF
 
Oh god.
 
7:09 PM
how about two diffeomorphic space forms are also isometric. This seems elementary
 
lol
 
That is a theorem with a name, @Ryan. Minding's Theorem, maybe?
 
$D^2$ and $D^2$ are the north and south "hemispheres"
$S^2$ is the "equator"
amirite
oh I messed up the dimensions didn't I
 
I prefer east and west
$D^3$'s, yeah
 
$S^3/C_2 = (D^3 + D^3 + S^2)/C_2 = D^3 + S^2/C_2 = D^3 + \Bbb{RP}^2$
 
7:10 PM
@TedShifrin i think this is de rham
 
With a prime meridian instead of an equator?
 
Hmm, really?
 
i've heard that name attached before
presumably the answer is in wolf
 
I wish I had kept Wolf's book.
 
@BalarkaSen and that's why it is the double cover
lol
 
7:11 PM
@LeakyNun this is neither topology, algebra, nor mathematics
 
LOL
It's arithmetic!
 
Thanks Mike
 
how about $S^3 = \partial D^4 = \partial(D^2 \times D^2) = \partial D^2 \times D^2 \cup D^2 \times \partial D^2 = S^1 \times D^2 \cup D^2 \times S^1$
what is this
 
Very cool, @MikeM.
 
If you find the paper let me know, i do not know the proof
 
7:13 PM
I think I know the paper but it’s in French. It was not obviously related to this
 
It seems to me something like this is in Spivak volume 4 or something.
 
@MikeMiller this is a compact way to pack a proof :P
 
this might be it
hmm, Gilkey cites Wolf for the second theorem but not this one
 
I don't see it in there, @Ryan.
 
@BalarkaSen and then the construction inductively continues, so we get $\Bbb{RP}^\infty$
 
7:18 PM
That paper seems more focused on simplicial/cellular structure.
But I haven't looked carefully.
 
@TedShifrin me neither but the French was the main barrier
 
@LeakyNun Essentially, yes
 
That's the paper that Fernando Marques cites
Though the actual result wasn't clear from what he wrote
 
Well, my French was essentially as good as my English, so that's not the obstruction.
 
and of course there's no page or theoem number given
 
7:19 PM
@BalarkaSen oh and Li Peng died today
 
I would need to take a few minutes to read more carefully and now it's lunchtime.
 
@LeakyNun Oh.
 
@BalarkaSen now... let's do $K(\Bbb Z,2)$ lol
 
Strangely coincidental for him to die when another Tiananmen is on the verge of conception.
 
so we start with $S^2 = e^0 \cup e^2$
@BalarkaSen hmm
$\pi_3(S^2)$...
 
7:21 PM
Indeed.
 
$S^1 \to S^3 \to S^2$
$0 \to \Bbb Z \to \pi_3(S^2) \to 0$
$\pi_3(S^2) = \Bbb Z$ generated by the Hopf map
 
Yup.
 
no I do not want to attach $e^4$ to $e^0 \cup e^2$ using the Hopf map
 
You should!
 
Does one really need spherical space form for this to be true? I guess a scaled torus rules it out for the flat case.
 
7:22 PM
It's also known as $\Bbb{CP}^2$.
 
right
the general Hopf fibration goes $S^1 \to \Bbb{CP}^{2n+1} \to \Bbb{CP}^{2n}$ right
no
 
No.
$S^{2n+1}$ in the middle.
 
oh
 
$\Bbb{CP}^n$ in the last
 
oh right, $S^2 = \Bbb{CP}^1$
$S^1 \to S^{2n+1} \to \Bbb{CP}^n$
 
7:24 PM
I mean just think of $\Bbb{CP}^n$ as $\Bbb C^{n+1} \setminus 0$ modulo $\Bbb C \setminus \{0\}$ then restrict to a sphere around $0$.
 
cool
 
oh wow there are also isospectral but not isometric lens spaces
 
Damn!
 
of course Gilkey's other book isn't on libgen
 
there should be explicit flat examples in dimension 3 too but i don't know what the spectrum of lens spaces is in 3d
 
7:30 PM
$\Bbb{CP}^2 = (\Bbb C^3-0)/(\Bbb C-0) = (\Bbb C^3-\Bbb C^2+\Bbb C^2-0)/(\Bbb C-0) = (\Bbb C^2 \times (\Bbb C-0) +(\Bbb C^2-0))/(\Bbb C-0)$
 
Man I'm not going to look at these horrifying 'formulas' anymore
 
$= \Bbb C^2 + (\Bbb C^2-0)/(\Bbb C-0) = \Bbb C^2 + \Bbb{CP}^1$
 
They're hurting my eyes but more importantly my soul
 
I don't have one so it's alright for me
@BalarkaSen so yeah the attaching map is $(\Bbb C^2 - 0) \to \Bbb{CP}^1$, i.e. $S^3 \to \Bbb{CP}^1$
so inductively blah blah blah $K(\Bbb Z,2) = \Bbb{CP}^\infty$
8 mins ago, by Leaky Nun
$S^1 \to S^{2n+1} \to \Bbb{CP}^n$
using this fibration
 
Right.
 
7:35 PM
now let's do $K(C_3,1)$...
 
No, let's stop lol
 
why
 
I mean you can do it. I have to do other stuff lol
 
$e^0 \cup e^1 \cup e^2$ with $\phi^2 : S^1 \to S^1 : z \mapsto z^3$
I've never seen $K(C_3,1)$ though
now what on earth is this space
 
It's not something you can recognize unlike previous examples
Clearly not a manifold, for one
 
7:38 PM
oh no
how do I compute $\pi_2(e^0 \cup e^1 \cup e^2)$ then
I should probably find a triple cover
that would be the universal cover
 
@MikeMiller Igor Belegradek has a question on MO about this that I'm sure you saw
 
It didn't seem to give anything I didn't find beforehand in that Gilkey ref
 
He says it's the paper I linked.
 
ah I missed that comment
 
The paper is completely unreadable to me
 
7:46 PM
I feel like a triple cover would be $e^0 \cup e^1 \cup e^2_1 \cup e^2_2 \cup e^2_3$
 
I've suffered through French papers but this one doesn't have any equations
 
wait that's just inserting a disc inside a sphere
hmm, does that work
 
Sounds right, yes.
 
$(r,\theta,i) \mapsto (r,\theta+120^\circ,i+1)$?
 
The Cayley complex of $\Bbb Z/3$ is the $3$-cycle, and you attach three disks to it. $\Bbb Z/3$ acts by rotations on the circle, while simultaneously permuting between the three disks.
It is an $S^2$ with a disk glued along the great circle
 
7:50 PM
the what complex
see my horrendous formulas work
ok so now I need to compute the $\pi_2$ of this horrendous thing
hey this looks like $S^2 \vee S^2$
I just pinch the middle disc to a point
 
It is, which is why I think you're screwed.
 
how do you compute $\pi_2(S^2 \vee S^2)$?
 
Hurewicz
 
how do you compute $H_2(S^2 \vee S^2)$?
oh, MV
what's screwed about it then
it's just $H_2(S^2)^2$ right
which is $\Bbb Z^2$
 
Ok, now attach the 3-cells.
 
7:54 PM
oh I need two 3-cells
 
$e^0 \cup e^1 \cup e^2 \cup e^3_1 \cup e^3_2$
eh... where do I attach them lol
 
There are two spheres in the universal cover, given by hemisphere+meridian disk
 
ok I'm visualizing $e^0 \cup e^1 \cup e^2$ as a quotient of the closed unit disc
 
Attach two D^3's there, and squash by the covering map at the boundary
 
7:55 PM
we need $\phi^3_i : S^2 \to e^0 \cup e^1 \cup e^2$
 
"Just when I thought I was out, they pull me back in" - balarka sen
 
LOL
 
i'm hungry
wait so it's a quotient of $\overline{B^3}$? @BalarkaSen
since adding the two D^3 to the sphere+disc thingy gives you B^3-closed
 
This is going to be a hopeless cell structure. There should be a better cell structure on $K(\Bbb Z_3, 1)$. You can realize it as direct limit of lens spaces, I think.
 
$S^\infty \subset \Bbb C^\infty$ is contractible
 
7:59 PM
Right, better to think about that
Act on each coordinate by rotation
 
$C_3$ doesn't even act on $S^2$
 
You're acting on $S^{2\infty - 1}$
Big difference
 
oh
 
But I think you're still screwed if you want to figure out the cell structure.
 
$S^\infty$ is the unit sphere in $\Bbb C^\infty$?
 
8:01 PM
@BalarkaSen no way
lens spaces have famously nice cell structures
 
oh my god Wolf even proves the Sylow theorems
this book has everything
 
in quite literally the same way that cyclic groups have famously nice free resolutions
 
(except for the result I'm looking for)
 
@RyanUnger what's the title?
 
spaces of constant curvature
 
8:02 PM
how is that related to Sylow theorems
@MikeMiller @BalarkaSen oh right $S^1 \to S^1 \to K(C_3,1)$
 
you want to find p groups in your fundamental group, I'm guessing
 
@MikeMiller It's hard for me to visualize what's happening in higher dimensions but a'ight
 
@BalarkaSen If you can see it in dimension 3 I don't think anything really changes. In the end there's still just one cell in each dimension
 
ok so $S^\infty = S^1 \cup S^3 \cup S^5 \cup \cdots$
 
jesus christ
 
8:04 PM
@MikeMiller Dude I was writing a long ass question for you
 
I think he left
are we just taking the limit of the Hopf fibrations
and restricting it to $C_3 \le S^1$
help I can't visualize $S^3 \subset \Bbb C^2$
 
@MikeMiller Can you remind me the Morse theory argument for $H_*(\Omega S^n)$? This is what I vaguely remember: I take the energy functional on $\Omega S^n$ which is a Morse functional in an appropriate sense, whose critical points are closed geodesics at a fixed point on $S^n$. These would be a great circle wrapping around $m$ times.
The index is the multiplicity of the conjugate point antipodal to the basepoint, which is $n - 1$ (number of orthogonal directions to the geodesic, in which if I perturb I hit that point again). That gives a total index of $m(n-1)$. This is means $\Omega S^n$
 
I'll just think of it as $(a_1, a_2, \cdots)$ with finite support with $\sum |a_i|^2 = 1$
 
It seems like you've already written the whole story!
 
@MikeMiller Igor says Ricci flow is not a "simple" proof in 3D. I fail to see how Ricci flow could be used to prove this.
The metrics are assumed to have constant curvature so they move by scaling.
 
8:16 PM
I had some additional questions that popped up: At the level of the 1-st approximation, we have a $2n$-complex with a $D^{2n}$ attached to $S^n$ by $S^{2n-1} \to S^n$. What is this space? Knowing the algebra structure gives me the Hopf invariant for this map
I replaced $n-1$ by $n$ because whim
 
The claim is that if $g_1,g_2$ are metrics on $S^n/\Gamma$, both with constant curvature 1, then there's a diff $\varphi$ of $S^n/\Gamma$ such that $g_1=\varphi^*g_2$, right?
 
@BalarkaSen first approx. of what?
 
The cell structure on $\Omega S^n$ I just wrote down
When $n$ is odd, $H^*(\Omega S^n)$ is the divided polynomial algebra on a generator of degree $n-1$, let's call it $\alpha$. Then $\alpha^2/2$ is the generator of $H^{2(n-1)}$.
So it's Hopf invariant is $2$, I guess.
 
@BalarkaSen I'm not sure why there should be a good name for it
I'm not remembering anything more than you do about the Morse theory
 
For even $n$, $H^*(\Omega S^n)$ is a tensor product of an exterior algebra on a generator of degree $n-1$ and a generator of degree $2(n-1)$, say $\alpha$ and $\beta$. Then $\alpha^2 = 0$. That means the Hopf invariant is zero, yeah?
Does that mean the map is nullhomotopic
Or the complex is $S^{2n-2} \vee S^{n-1}$
 
8:22 PM
The Hopf invariant only detects (some part of) the free part of $\pi_{2n-1}(S^n)$, and that is only infinite when $n$ is even
 
Ah, I see
Right, good point
 
It's a homomorphism to $\Bbb Z$, right? But there's no reason to believe that this tells us the rest
 
I wasn't sure to what degree the Hopf invariant determines the free part. Is it an isomorphism to $\Bbb Z$ (kills the torsion) when $n$ is even?
 
It does completely tell us how to build $S^n \otimes \Bbb Q$ in general - it's either $K(2n+1,\Bbb Q)$ or a fiber bundle $K(4n+1,\Bbb Q) \to S^n \otimes \Bbb Q \to K(2n,\Bbb Q)$
 
Right.
 
8:24 PM
No, this is that theorem of Adams called "hopf invariant one theorem" which says that there is only a map of spheres $S^{4n+1} \to S^{2n}$ which has Hopf invariant one iff $n = 0, 1, 2, 4$
Most people get more excited about the stuff about (nonassociative) skew fields
 
Oh right I vaguely recalled but forgot this
 
I do remember that James proved some nice things about $\Omega \Sigma X$ in general. If $\tilde H_*(X;R)$ is free I think $$H_*(\Omega \Sigma X;R) = \bigoplus_{n \geq 0} \tilde H_*(X;R)^{\otimes n},$$ with algebra structure just "stick some tensors next to each other"
And if you suspend once more there's no interesting data at all
 
Wow nice
I remember that. $\Sigma^2 \Bbb{CP}^2$ and $\Sigma^2(S^2 \vee S^4)$ are homotopy equivalent, right?
 
$$\Sigma \Omega \Sigma X \sim \bigvee \Sigma(X^{\wedge n})$$
@BalarkaSen I dunno let's check
 
Suspension does weird things. Sullivan pointed it out below my answer once.
Maybe only $\Sigma \Bbb{CP}^2$ and $\Sigma(S^2 \vee S^4)$
 
8:28 PM
$\Bbb{CP}^2 = D^4 \cup_{\eta} S^2$, where $\eta: S^3 \to S^2$ is the Hopf map. $\eta$ suspends to become the non-trivial generator of $\pi_1^s$ so this space does not stably split, so no
 
Ah, so you have to suspend twice?
 
$\pi_1^s$ is stable bro
it doesn't matter if you suspend more
those aren't equivalent
 
Oh ok. Hm.
 
The cohomological proof is that $\text{Sq}^2 \neq 0$
which you can see with Wu classes if you want
 
I see. When you suspend your CW complex you're also suspending the attaching maps. Somehow I never noticed that.
And saying a map of spheres suspends to something nullhomotopic is demanding it becomes trivial in the stable homotopy group
This is fun.
 
8:32 PM
I think $\Bbb{CP}^\infty$ does not split stably into any wedge summands. I think you get various steenrod operations connecting all of the nonzero cohomology groups
 
But if attach by the square of the Hopf map to get $D^4 \cup_{\eta^2} S^2$ and then suspend it, it does become $\Sigma(S^2 \vee S^4)$, right?
$\pi_1^s$ is 2-torsion
 
Yup
 
3
Q: Isometries between spherical space forms

TotoroLet $S^n/\Gamma_i\,(i=1,2)$ be a $n$-dimensional spherical space form, where $\Gamma_i \subset SO(n+1)$ is a finite subgroup acting freely on $S^n$. Suppose $S^n/\Gamma_1$ is diffeomorphic to $S^n/\Gamma_2$, can we show they are isometric?

9 years later, he still doesn't know. Huh.
 
@MikeMiller Nice, I like this a lot
 
Unlike $\Bbb{CP}^k$ though we learned from the James stuff that $$\Sigma \Omega S^3 \simeq S^3 \vee S^5 \vee S^7 \vee \cdots$$
 
8:36 PM
I have heard the Steenrod squares are stable analogues of cup-squared, with which I detect CP^2 and S^4 v S^2 are non-equivalent in the first place, but don't know how to compute.
 
And indeed as you say, the Hopf invariant of the first stage of this guy is 2 and so it should stably split as $S^3 \vee S^5$
 
@MikeMiller Ah, ok, so at least that map $S^3 \to S^2$ in $\Omega S^3$ is some even power of the Hopf map
Square, actually
The Hopf invariant $\pi_3 S^2 \to \Bbb Z$ is an isomorphism
 
You use $\text{Sq}^1 = \beta$ (bockstein), you use various relations to compute from the things you know, and you use that they're natural under pullback
 
I see, yikes
 
You use that the Wu classes have $\text{Sq}(\nu) = w$ and that on $H^{\dim M - k}(M;\Bbb F_2)$ we have $\nu \smile x = \text{Sq}^k x$
 
8:39 PM
This is where things start becoming muddy for me. I have no conceptual understanding of characteristic classes with which I can fact-check everytime I compute something
I'm horrible at formulas
 
So if you know something about the SW classes you can often pull that back to learn something about the Wu classes and then learn something about the Steenrod squares
The conceptual understanding is just that they're the stable cohomology operations on $H\Bbb F_2$. You won't get the insight you want
To compute them is a computation in homotopy theory and such is hard
 
Is it possible to learn to use Steenrod squares as a one-day trip on a rainy day?
If so, would you recommend some book?
 
Hatcher has a nice brief section
 
Excellent, thanks. I might look into it.
 
I think the usual reference is Mosher-Tangora
but that's a book
 
8:43 PM
Noted as well
 
They'll calculate $H^*(K(\Bbb Z/2, n); \Bbb F_2)$ in there
Maybe integrally as well though I am less sure of that, that sounds hard
 
I saw the calculation in Hatcher's SSAT but got scared off
 
i never read it
 
9:31 PM
@BalarkaSen can $H_{p+2}(X^p,X^{p-1})$ be non-zero for a CW-complex X?
it's just $H_{p+2}(X^p/X^{p-1})$ right
which should be zero right
 
Yes, $X^p/X^{p-1}$ is a wedge of $p$-spheres
That $H_i(X^p, X^{p-1})$ survives at only $i = p$ is sort of the algebraic reason why cellular homology works.
The general statement is that if you have a chain complex $C$ and a filtration $F_p C$ of $C$ such that $F_p C/F_{p-1} C$ has homology only at $p$-th grade, then $H_\ast(F_\ast C/F_{\ast-1} C)$ make a chain complex whose homology computes to $H(C)$
 
oh no I misunderstood something
big oof
oh and they say that Mayer-Vietoris can be generalized to open cover by artbrarily many opens
and it becomes a spectral sequence
but does this only work if your thing is locally contractible?
 
If the cover is good, then the nerve of the cover is homotopy equivalent to the space. The filtration of the nerve by the $n$-skeleta gives a spectral sequence converging to the homology of the full nerve, which is homology of the space.
That's your Mayer-Vietoris spectral sequence
$E^0_{p, q}$ is like $\bigoplus_{|I| = p} C_q(U_I)$ where $U_I = U_{i_1} \cap \cdots \cap U_{i_p}$ if $I = (i_1, \cdots, i_p)$.
 
I heard about extracting exact sequences from spectral sequences; have you heard of such thing?
 
Yes.
 
9:44 PM
what is it?
 
Think of a spectral sequence which has two consecutive rows in $E^2$, hence in particular degenerates at $E^3$.
Give me an exact sequence relating $E^2$ terms with $E^\infty$ terms
 
I don't know it
 
skip 1-4
 
Unfortunately I haven't read all the papers in the world
I am not going to read it, I have other things to do lol
 
9:51 PM
ok
 
It's not worth reading all the internal mechanics of a spectral sequence unless you're working with some really technical spectral sequence or something
It's a very simple idea; you have a filtration and you get a spectral sequence. Computing is the hard part
So just learn to compute with the simplest spectral sequence in the world; the Serre spectral sequence
 
Serre Grothendieck
 
Grothendieck unfortunately did not prove the finiteness theorem
Serre did
 
$E^2_{p,0} \to E^2_{p-2,1}$ right
 
9:54 PM
$0 \to E^3_{p,0} \to E^2_{p,0} \to E^2_{p-2,1} \to E^3_{p-2,1} \to 0$ right
 
Right, and there are exact sequences $0 \to E^3_{n-1,1} \to H_n \to E^3_{n, 0} \to 0$ if you're converging to $H_n$, because $E^3 = E^\infty$ and that's what associated complex of a filtration means
Splice your thing and mine togather and you get a long exact sequence
Er
 

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