aloha

hey

ok, let's do this

so what I was thinking of is the Thom-Pontryagin correspondence. the classical version of this says that there's an isomorphism between the framed Bordism groups and the stable homotopy groups

what's the framed bordism group?

consider the set of submanifolds of dim $k$, $M \subset S^n$, equipped with a trivialization of their normal bundle. we say they're framed bordant if there's a cobordism between them in $S^n \times I$ with a trivialization of the normal bundle of the cobordism $W \subset S^n \times I$

ok

addition is given by disjoint union, let's say

call this $\Omega^n_{k,fr}$. I claim this is isomorphic to $\pi_n(S^k)$.

ok, that's a cool theorem. but it surprises me that this thing is ever computable.

given a smooth map $f: S^n \to S^k$, a regular value $f^{-1}(x)$ is a smooth submanifold - but better; the map $f$ induces an isomorphism $N(f^{-1}(0)) \to N(x) = T_x(S^k)$

so regular values come equipped with trivializations of the normal bundle. you can see where this is going; pick a smooth homotopy such that $x$ is still a regular value of $f_t: S^n \times I$, and eventually see that the framed cobordism class is independent of the choice of regular value, or of map $f$

similarly one can show every framed submanifold arises as a regular value.

so, why is this computable? let's start with dim 0 - immediate corollary is that $\pi_0^s = \Bbb Z$ (framed 0-manifolds are the same as points with an orientation.)

true, not a bad start

framed 1-manifolds. first, this is a disjoint union of circles; given two framed circles, there is a unique framing on a pair of pants extending those two framings. so every pair of circles is bordant to a single framed circle.

sorry, what I wrote up there was garbage; it's $\Omega^n_{n-k, fr}$ that's the same as the relevant homotopy group

ok

now there are two (stable) framings on a circle: if you pick one, the others differ by an element of $\pi_1 GL(n,\Bbb R) = \Bbb Z/2$

(one framing differs from another by 'rotating' the normal spaces as you go around the circle)

right, ok

i'm maybe forgetting how to show that they're not framed bordant :p give me a minute

sure

oh, I'm silly. so the point is that one of those framings is trivial - the one that arises as the boundary of the disc

and the only thing we have to do is show that one of them is not trivial

yeah fuck it :p

can you repeat what exactly you're trying to show?

I want to show that $\pi^1_s = \Bbb Z/2$ by showing that the two framed circles, which every framed 1-manifold is bordant to, are not framed bordant

ok

here's someone who can actually do this

haha

it's possible to extend this to all the way up to $\pi^3_s = \Bbb Z/24$; in that case you play the same trick (everything is cobordant to a framed $S^3$, so we just need to understand its framings), you do something cute to show that the framing it gets from being a Lie group is 24-torsion, and you do something I don't know how to do to show that it's $\Bbb Z/24$

I'm fine at showing that these groups are quotients of the things they actually are, but I always forget how to actually show that elements are nonzero...

what issue does one encounter for $\pi_s^4$?

meh

it's no longer true that everything is framed bordant to a framed sphere or something like that

right, ok

you had to do surgery before to simplify things, and in all of those cases you could extend the framings across the surgeries

just because certain homotopy groups were zero in small degrees

and that just stops being true

sorry if this was a bit underwhelming

mhm ok

no problem, it's not

actually i was completely unaware of this alternative method of computation before

let me tell you how to do it with leray-serre

let's start by looking at, say, $S^3$. let's choose a postnikov tower $\cdots X_5\to X_4\to X_3=K(\Bbb Z,3)$

I always forget how these run - there's a map $X_n \to S^3$ that's $n$-connected?

the basic idea now is to look at the fibrations in that tower and to compute as many (co)homology groups of the spaces $X_k$ as we can

the map is $S^3\to X_n$ and that's $n$-ctd.

ok

the Whitehead tower is where I kill from below, and the Postnikov tower is where I've killed from above

and the spaces $X_k$ have same homotopy groups as $S^3$ up to degree $k$

all higher ones are zero

yes

gotcha gotcha

so the first fibration in this tower is $$K(\pi_4S^3,4)\to X_4\to K(\Bbb Z,3)$$

we know what the $\Bbb Z_2$ cohomology of the base space is, by Serre's theorem

using some Bockstein-trickery, we can compute the $\Bbb Z$-cohomology of it, at least in low dimensions

did you figure out how that works?

yeah i did, at least i think so

cool

let's accept it for now

and consider the leray-serre SS of that fibration, with $\Bbb Z$ coefficients

ok

We know a few cohomology groups of $K(\pi_4S^3,4)$...in degree $5$ it is exactly $\pi_4S^3$

Oh I see, we know $X_4$'s cohomology agrees with $S^3$ up to degree 4 and we want to use that to break the previous?

Oh hm

exactly!

so we know the cohomology groups on the x-axis in low degrees and we know what happens on the y-axis up to degree 5

and we know that the SS converges to the cohomology of $X_4$, which coincides with that of $S^3$ up to deg 4

sorry, why is it $\pi_4 S^3$ in degree 5?

why not degree 4?

$\pi_4S^3$ is finite, then we apply the UCT

that was phrased poorly

icic

$H_4(K(\pi_4S^3))=\pi_4S^3$, which is finite. Then we can apply the UCT to figure out that $H^4(K(\pi_4S^3))=0$ and $H^5K(\pi_4S^3))=\pi_4S^3$

So, from convergence of the SS and from what we know about $H^*(X_4)$, we get that the differential $d_6:\pi_4S^3\to E_2^{6,0}=\Bbb Z_2$ is an iso

that's in the stable range, so $\pi_1^s=\Bbb Z_2$

sorry, I need a second

no problem

ok, cohomology arrows go left $n$, up $(n-1)$

no, they go right+down :P

fucking god damnit

i always get that wrong too

ok, but why do we know that's an iso? it seems like you can only say something about that if you know something about the 5th cohomology of $X_4$

and I don't think I do

Actually, you do. $H_i(X_4)=H_i(S^3)$ for $i\leq 5$

why?

I know the map is an iso on $\pi_n$ for $n \leq 4$, which shows that it has the same homology in those degrees as well... is the point that $H_5(S^3) \to H_5(X_4)$ is surjective or something?

cuz it seems to me the exact sequence I get is $H_5(S^3) \to H_5(X_4) \to H_5(X_4, S^3) = \pi_5(X_4, S^3)$, which I don't know is zero

$\pi_5(X_4,S^3)$ is still zero

oh, 4-connected means the map one level up is surjective.

yeah

and this is obviously true in the base case of $S^3 \to K(\Bbb Z,3)$ because you surject onto zero :p

my bad. thanks

ok, I'm happy.

np, i was having the same issues (and just got confused again :P)

so that's the computation of the 1st stable homotopy group

so we've computed $\pi^1_s$. I take it 2,3 go the same way, but that beyond that we run into trouble because we usually don't have spectral sequences simple enough to have just one arrow that tells us $\pi_9 S^6$ or w/e?

yeah, and we also know less and less about the cohomology of the spaces in our fibrations. so the fibrations in higher degrees look like $K(\pi_s^k, n+k)\to X_{n+k}\to X_{n+k-1}$ and it becomes increasingly difficult to say something about cohomology of the spaces involved

yeah, you'd need to be good at Leray-Serre for the things before it to start with

I see

exactly

so neither of us know a way to do $\pi^4_s$ :(

we need to learn that. probably the adams spectral sequence gets the job done

what's that

i think i read that all group up to $\pi_s^{60}$ or something are known lol

i think thats using complicated techniques though

some old spectral sequence, i dont know shit about it

just hear people talking about it and saying how great it is

lol I looked at it and now I'm spooked.

is this your personal taste in stuff? you really like the computations and spectral sequences?

yeah i think they're quite fun. do you like em?

I respect them, but they're a little too spooky for me

you're a down to earth guy

the most badass thing I've seen recently is this

arxiv.org/abs/1601.02184

(in the area of spectral computations I mean)

that paper does look badass

do you know the J-homomorphism?

wow

yeah, i know the J-hom

to show that the smooth Poincare conjecture in dimensions bigger than... let's say 126 to be safe, it suffices to show that the J-hom is not surjective in most degrees

I think you need to show that for $n \equiv 2 \pmod 4$

wow

I forget all of the details here though

the J-homomorphism is that map $\pi_i(O)\to \pi_i^s$ right?

and $\pi_i(O)$ is known

yeah

that's a superb answer, i remember reading parts of it

the point is that there are three points of complication in understanding exotic spheres: 1) this "theta_{bp}" subgroup (solved by Milnor and Kervaire), 2) the Kervaire invariant (solved by Hill-Hopkins-Ravenel 2009? except for in ONE specific dimension) 3) the image of the J (solved by bananaman, 2017)

i dont really understand how a lot of these calculations work but i do find them fascinating

haha, i wish...

so there's definitely still a lot of very calculationally-minded people working in these fields, I think; a lot of the tools are becoming more abstract though

like tmf is somehow a cohomology theory coming from the stack of elliptic curves over schemes and like :I

what does tmf stand for?

topological modular forms

it's a cohomology theory that can be used to show that $\pi^{125}/J$ is not zero i guess

how would you describe your personal taste? why dont you do more of this calculations stuff if you find it fascinating?

ah, ok lol

algebra scares me. i don't really think about this stuff :) the past couple days i've been looking at like some of the homotopy theorist's models of equivariant homology, but that's just because i want to build a chain complex for equivariant homology

and i'm trying to understand how to do that

algebra scares me too. ok, i see

i have a sort of understanding of how spectral sequences work but hardly a good one

same here. still trying to learn.

you've got me beat

especially once you learn this adams stuff

man that 61-sphere paper is gangster

still scrolling through it

yeah... it looks like the technology is not that much bigger than what's in Hatcher's book right?

there's that plus the EHP sequenece?

indeed

do you know characteristic classes?

unfortunately not but i wanna learn about them at some point

ok, so what I proposed reading before was this

math.cornell.edu/~hatcher/Papers/MW.pdf

I want to give a little background about why one should care

i know the euler class

ok sure

given any type of "bundle" - real vector bundle, oriented real vector bundle, whatever - one would be interested in understanding how to tell them apart (or see that they're nontrivial)

sure, that's a natural question

real vector bundles of rank $n$, say, are "classified" by a space $BGL(n) = BO(n)$, meaning that there is a bijection between iso classes of bundles over $X$ and homotopy classes of maps $[X, BO(n)]$

so if we understood $H^*(BO(n))$, we would have a way of telling apart these bundles. precisely, if $\alpha \in H^k(BO(n);R)$, we can define $\alpha(E) = E^*\alpha \in H^k(X;R)$

ok. sorry, why is BGL(n)=BO(n)?

whenever you have a group homomorphism that's a homotopy equivalence, it induces a homotopy equivalence of the classifying spaces

ah, i see. gram-schmidt or something...

yeah, so because O(n) -> GL(n) is a homotopy equivalence (by Gram-Schmidt), pass to the map on classifying spaces, and it's also a homotopy equivalence. which is not trivial, but not too hard after building the whole machine

i only pass to O(n) here because it's compact and that's nice

ok

anyway, we actually do have computations, to a degree. $H^*(BO(n);\Bbb Z/2) = \Bbb Z/2[w_1, \dots, w_n]$, where there are no relations and $|w_k| = k$

yeah, i saw those before

those are the Stiefel-Whitney classes. they're relevant for other reasons too; eg $E$ is orientable iff $w_1(E) = 0$

we also have $H^*(BU(n);\Bbb Z) = \Bbb Z[c_1, \dots, c_n]$, where $|c_k| = 2k$; these are the Chern classes of complex vector bundles

oh, that's neat.

now these aren't good enough to show that vector bundles are trivial, because there are such things as stably trivial bundles; eg the tangent bundle of $S^2$ is not trivial (Hairy ball theorem), but sum it with the normal bundle (which is trivial), and you get a trivial bundle

right

by certain properties of most of these characteristic classes, you see that $w_i(TS^2) = 0$ automatically

I think but forget that it's actually a theorem that the Chern classes of a complex vector bundle all vanish iff it's stably trivial

maybe not tho idk

ok sure

anyway, this thought process extends to all kinds of bundles. $G$-bundles instead of vector bundles: look at the cohomology of $H^*(BG)$!

now let's push this even further: can we study fiber bundles with fiber $M$?

sweet. we know something about $H^*(BG)$ from spectral sequences

and I claim yes; I claim this is actually the same thing as studying principal $\text{Diff}(M)$-bundles (or if you want the transition maps to only be homeomorphisms, $\text{Homeo}(M)$-bundles)

so what does this lead us to? it leads us to studying $H^*(B\text{Diff}(M))$.

right

in 1D we understand this completely. the inclusion $O(2) \hookrightarrow \text{Diff}(S^1)$ is a homotopy equivalence, so $BO(2) \hookrightarrow B\text{Diff}(S^1)$ is a h.eq

and we actually see that every smooth circle bundle is equivalent to one (and only one) with linear transition functions

hm, ok

in 2D we actually still have a very good understanding. $O(3) \hookrightarrow \text{Diff}(S^2)$ is a h.eq; $GL_2(\Bbb Z) \times S^1 \times S^1 \hookrightarrow \text{Diff}(T^2)$ is a h.eq; and for surfaces of genus greater than 1, the identity component $\text{Diff}_0(\Sigma_g)$ is contractible

wow, ok

so if we define $\text{MCG}(\Sigma_g) = \pi_0 \text{Diff}(\Sigma_g)$, we see that to understand $\Sigma_g$-bundles, it suffices to understand $B\text{MCG}(\Sigma_g)$.

actually, what do you mean by the identity component of $\text{Diff}_0(\Sigma_g)$?

I mean the identity component (which I call Diff_0) of Diff

which is a topological group; you can think of the topology as coming from the metric $d(f,g) = \max d(f(x),g(x))$

ok

anyway, the MTW theorem calculates $H_k(B\text{MCG}(\Sigma_g))$ in a "stable range" - for $k < g/2$, maybe, I forget precisely what the range is

looks like it might be $k \leq 2g/3$

amazing. is it with spectral sequences?

I have absolutely no idea how the proof goes

oh ok

so the MTW theorem, at least stably, helps us understand two things: 1) characteristic classes of surface bundles. 2) how characteristic classes of surface bundles change when you "stabilize" by connected summing on an extra handle, say

I think that's real exciting, and that's my pitch

homological stability has become a whole thing now and they have proofs of similar theorems in many dimensions and contexts

i'm sold on this.

btw since I love diff groups, here are some other theorems. Hatcher's theorem: $\text{Diff}(S^3) \simeq O(4)$; some extensions to $\text{Diff}(L(p,q)) \simeq \text{Isom}(L(p,q))$ (where you give $L(p,q)$ the metric it inherits from $S^3$ with the round metric); there's Smale's general conjecture that for 3-manifolds with constant curvature 1, $\text{Diff}(M) \simeq \text{Isom}(M)$

hatcher also proved $\text{Diff}(S^2 \times S^1) \simeq O(2) \times O(3) \times \Omega SO(3)$

lol that's incredible

(the first term acts by rotating the circle, the second by rotating the sphere, the third by rotating the sphere as you loop around the circle)

that's a nice interpretation

what else have I got... for a hyperbolic manifold of any dimension bigger than 2, $\text{Diff}(M) \simeq \text{Isom}(M)$ - but note that isometry groups of hyperbolic manifolds are finite, discrete, and equal to $\text{Out}(\pi_1)$

and there's generally a pretty good but not complete understanding of diffeomorphism groups of 3-manifolds. also, we have that $\text{Diff}(M) \hookrightarrow \text{Homeo}(M)$ is a homotopy equivalence when $M$ is of dimension at most 3

you might hope from this that $\text{Diff}(S^k) \simeq O(k+1)$ for all $k$. this is known to be false for every $k \geq 5$; in fact, infinitely many of the relative homotopy groups $\pi_n(\text{Diff}(S^k),O(k+1)) \neq 0$. I was in a seminar once reading stuff about how this was done and gave up because it was too crazy for me

that sounds fun. how hard are these diff groups to compute?

lol

S^1 is a great exercise for someone motivated, S^2 is a hard theorem of Smale's which you usually do by solving a certain PDE ("isothermal coordinates") or sometimes by invoking dynamical systems theory, S^3 is a hard theorem of Hatcher that nobody understands

the higher-dim stuff is computed using a big machine called Waldhausen's algebraic K-theory of spaces. that's what the seminar was about. but I didn't end up getting to the point where I could understand how the computations were carried out.

:D i see

but - if you believe the hard theorem of Smale's - the rest of the computations in 2D are done in that note by Hatcher I suggested readig

sweet. i'm down to try reading that thing but by the looks of it, it'll be very difficult for me to say the least

I haven't taken too hard a look, but I think it's worth a try

do you have an idea of how to go about reading that?

BTW in dimension 4 Homeo and Diff start becoming crazy different; Danny Ruberman found examples of 4-manifolds $M$ s.t. $\pi_0 \text{Diff}(M) \to \pi_0 \text{Homeo}(M)$ has infinitely generated kernel. I think that all of the $\pi_k$ should be really different; a personal pipe dream of mine is to find an $M$ such that I can find an element in the kernel of $\pi_1$. but enough bout that garbage.

I think we should start with sections 1 and 2 and then regroup, maybe

haha :D

ok, let's try it

the general way I've done reading seminars with people in person is that we would meet once a week or so to talk about what we've read and ask each other questions about things we were confused about

might be a good plan here, tho the first two sections seem pretty introductory, so we could regroup before next week

the only thing in that paper I know are the contents of appendices A and B

ok, sounds good

im gonna print the first two sections and start reading tomorrow. i'll probably be very slow.

since you asked what my taste was, here's a paper from last year i liked a lot :p file:///home/chronos/u-91a17ee88eea85960d482994e487cd6bc4ae8a5a/Downloads/jsharp‌​-paper.pdf

oh.

that's not going to link you to anything.

lol

math.harvard.edu/~kronheim/jsharp-paper.pdf

that's fine, we both have other things we have to do

we should star this room so its easy to get back here

everything in that paper seems totally foreign to me. :/ kk, i'll star this room.

it's more differential geometry than much else

see eg prop 3.1

man it's unbelievable how much more you know than every other phd student at my university...

appreciate the compliment but i don't really

it'll be meaningful if i can ever turn something i know into a paper :p

i am quite optimistic about that...

i need to cook dinner

maybe. my optimism depends on the day

I'll be here

my optimism depends on all kinds of shit. it's like a leaf in the wind. :P

room topic changed to stuff: homological stability and other garbage (no tags)

yeah I get that.

im off to shower+bed. see you later dude.

night. isn't it eRly?