random cohomology for quantum nerds

Balarka Sen

02:07

@loch You want to tell me about that stuff?

Daminark

Hmm, how far does the LES of a fibration go? I don't have much time to pursue this because I've gotta eat, charge my phone, and then study a bit for Galois (watch as I ignore that and just do smth else...) but does it go down to, say, π_0? Where exactness means pointed sets?

Leaky Nun

hi @BalarkaSen

Balarka Sen

Ey

@Daminark Yep

Daminark

Okay so I think that gives you fundamental group of RP^n

Balarka Sen

Running it over the double cover of RP^n, yeah

Daminark

02:10

Sick, okay building the inventory of known homotopy groups bit by bit

Balarka Sen

It's a special case of, if $G$ acts freely on Hausdorff space $X$ then $\pi_1 X/G \cong G$, which you also get from following the fibration LES.

MatheinBoulomenos

Oh one thing I wondered about that are the maps with pi_0 group homomorphisms if you put in topological groups and assume your fibration sequence consists of continuous group homorphisms?

Balarka Sen

But it follows from some covering space theory too

@MatheinBoulomenos I think so, yeah

MatheinBoulomenos

I mean it's clear for the maps from pi_0 to pi_0, I'm just wondering about the connecting map from pi_1 to pi_0

Mike Miller

pi_1 to pi_0, and yes, the point being that the product in pi_1 can be computed using the group structure

Balarka Sen

02:13

So if you have $H \to G \to G/H$ a fibration then the connecting map $\pi_1(G/H) \to \pi_0 H$ comes from the connecting map $G/H \to BH$ at the space level, no?

That connects (smirk) with what I was talking about earlier

MatheinBoulomenos

@MikeMiller ah, this is an application of Eckmann-Hilton, right?

Mike Miller

yup

thought about naming it but forgot the name

Balarka Sen

02:32

Huh, so the Euler class is nothing but pulling the Thom class in $H^n(E, E - 0)$ back to $H^n(X)$ by the inclusion of $(X, \emptyset) \hookrightarrow (E, E - 0)$ as the 0-section.

Hm, but the Thom class in $H^n_{\text{vertical sup}}(E)$ is Poincare dual to $0 \subset E$.

Balarka Sen

02:45

It is unclear to me why the $\iota^*\Phi$ with $\Phi$ being the Thom class is then going to be Poincare dual to the zero locus of a generic section of the bundle.

I'll think about it; gonna transport to somewhere else in a few minutes

loch

08:56

I think this whole thing about a complex oriented generalised cohomology theory is really just abstracting the notion of Thom class you find in (co)homology. More specifically, a complex oriented generalised cohomology theory $\mathcal{E}$ is a multiplicative cohomology theory (i.e a cohomology theory (but might differ from the singular case if you're not requiring the dimension axiom) with products) where you have a notion of Thom class (for complex vector bundles)

i.e. For every complex bundle $E\rightarrow B$, there is an element $\mathcal{E}^{2n}(E,E-0)$ which when restricted to fibre …

Which is natural in the sense that it is compatible with pullbacks and direct sums

loch

09:09

And I think the upshot is that the definitions and arguments you see when talking about Thom classes. For example you have Thom isomorphism and your Gysin maps, i.e. in the singular case when you have a submanifold $Z$ of codimension $c$ in $M$ (with appropriate orientations etc. - for the source I'm reading everything are complex manifolds) , you have maps $H^{i-2c}(Z) \rightarrow H^i(M)$ obtained by Excision and tubular neighbourhoods etc.,

which allows you to define the fundamental class of $Z$ to be the image of $1$ under $H^0(Z) \rightarrow H^{2c}(M)$ (which is poincare dual to the fu…

Balarka Sen

14:15

@loch Huh I see

What spectra are these complex oriented generalized cohomology theories represented by?

loch

14:50

@BalarkaSen
I don't know a lot about spectra (i.e. I basically only know their definition) - but I think in general a multiplicative cohomology theory is represented by a ring spectrum - and a complex oriented generalized cohomology theory is represented by a (commutative, associative) ring structum equipped with a complex orientation for the tautological bundle over $\mathbb{C} \mathbb{P}^{\infty}$ (in the sense as above, basically)

I'm not sure if there's more to say about this in the general context.. but specifically for $K$-theory because of Bott periodicity, the corresponding spectra…

Balarka Sen

Aha

Thanks, that's very interesting

Mike Miller

18:11

@Balarka I think it is quite difficult to understand such spectra.

I think elliptic cohomology theories are examples, associated to each elliptic curve, and this thing tmf is somehow supposed to be the "derived sections of this sheaf of Spectra over M_ell"

And nobody knows wtf the deal with that is

MatheinBoulomenos

I like elliptic curves

Mike Miller

they're nice

MatheinBoulomenos

@MikeMiller is this for ellitpic curves over general fields or over $\Bbb C$ or number fields?

Mike Miller

18:28

c

MatheinBoulomenos

i c

Mike Miller

remember, they were talking about being complex oriented above

so if an elliptic curve shows up somehow, it seems natural it be complex

MatheinBoulomenos

@MikeMiller apparently complex oriented cohomology theories are related for formal groups

I studied a lot of formal groups this semester (not over $\Bbb C$ though), so it seems cool they show up in topology

Mike Miller

1-dimensional formal groups, but yeah

MatheinBoulomenos

I only worked with 1-dimensional formal groups, too

you can use them to prove local class field theory

Mike Miller

18:33

there's a perspective I never got into but is popular nowadays that a lot of homotopy theory is dictated by formal algebraic geometry

there's a book in preprint by eric petersen

MatheinBoulomenos

afaik (but I don't know much) formal groups and elliptic curves are something that more arithmetically inclined algebraic geometers would care about

So I'm a bit surprised they show up in topology

but pleasantly surprised :)

Mike Miller

well, homotopy theory

Mike Miller

18:58

Codimemsion 1 hypersurfaces of a manifold with H^1(X;Z/2) are orientable.

Proof

1) Given a space $Y$, and a line bundle $\xi$ over $Y$, the pullback $\pi^* \xi\big|_{\xi \setminus 0}$ is trivializable away from the zero section. Write elements of $\xi$ as $v$ where $\pi(v) = p$; write elements of the pullback as $(v, w)$ where $w \in \xi_{\pi(v)}$. Then there is a trivailization $(v, w) \mapsto (v, c)$ where $c$ is the unique real number with $cv = w$ (which always exists).

2) Given a codim 1 submanifold $Y$ of a manifold $X$, the normal bundle is a line bundle over $Y$. It extends to a tubular neighborhood by pullback; by the above we may trivialize it over the tubular neighborhood, and thus extend it globally (by the trivial bundle everywhere else).

This is the topological construction of "line bundle associated to a divisor".

MatheinBoulomenos

ah

Mike Miller

Now H^1(X; Z/2) = iso classes of real line bundles over X, so det(TX) is trivial, and X is orientable; but so is the line bundle we just constructed. That restricted to the normal line bundle on Y itself, so the normal bundle is trivial

And hence Y is orientable

MatheinBoulomenos

I only knew you could do that in algebraic geometry

Mike Miller

you definitely get a cooler construction in algebraic geometry (you get variation as you vary the divisor)

But the fact that such a construction exists in topology is clear from a few basic things: 1) line bundles are determined by their class w_1 in H^1(X;Z/2); 2) a codimension 1 submanifold gives a codimension 1 homology class; 3) Poincare duality

the class w_1 is Poincare dual to the zeroes of a generic section of the line bundle, and this is the statement you're used to where a divisor arises as the zero set of a section of the line bundle

MatheinBoulomenos

hmm, I'm unsure whether I should learn universal bundles or simplicial sets first

Mike Miller

19:05

Tell me the things that have excited you most in the last two weeks

MatheinBoulomenos

I found the proof that if the characteristic of a local field is not p, then the p-completion of the absolute Galois group is topologically finitely generated really cool

Mike Miller

Hmm, tell me strictly the non-algebra things because those won't help me so much

Also, are you done with school for a while?

MatheinBoulomenos

wait, non-algebra things exist?

not really

Mike Miller

You're asking for epsilon-guidance about non-algebra, so I want epsilon more input :)

I have no idea how your school system works

MatheinBoulomenos

the summer semester I'm currently in lasts until end of july, so two more months roughly

Mike Miller

19:10

I see, you actually do stuff during the summer semester

MatheinBoulomenos

But I finished my seminar talks for this semester, so I have a little more time than previously

Mike Miller

How do you feel about spectral sequences?

MatheinBoulomenos

they're a bit scary, but I like them in principle. I want to learn to work with them better

I saw how you can get them from double complexes or filtered complexes

not sure if that's sufficient for topology

Mike Miller

I only ever use them from filtered complexes, it's not sufficient for the homotopists but it's sufficient for me

I really like multicomplexes, which are like double complexes except there are more than just d_0 and d_1, and in fact d_i which are (i, 1-i)-graded

I really like learning-via-project (aka, having a hard paper to read that has a bunch of different input you would learn on the way), because if you're excited about a project it gets you motivated to learn the input

MatheinBoulomenos

I'm not sure if that counts as non-algebra but I was excited by the realization that discrete valuations on the field of meromorphic functions on a compact Riemann surface that are trivial on the constant functions correspond to points on the compact Riemann surface

Mike Miller

19:14

I'm trying to think of a project related to the things you suggested above

MatheinBoulomenos

I'm still working on the details, but I think you can use that to reconstruct the surface from the meromorphic functions

back when I took algebraic topology, Brown representability was the result I found the most exciting

Mike Miller

@MatheinBoulomenos So I think you would like either of Quillen's great papers on group cohomology: either "cohomology and K-theory of the general linear group over a finite field", or "the spectrum of an equivariant cohomology ring"

MatheinBoulomenos

wow, both sound really cool

Mike Miller

In the first he calculates H^*(GL_n F_q; F_ell) (ell prime to q), and as a corollary gets a calculation of K(F_q)

MatheinBoulomenos

maybe the first one a little more

I know roughly how k-theory works, but I don't know any actualy details

Mike Miller

19:17

It uses complex K-theory, Brauer lifting, and an Eilenberg-Moore (homotopy pullback) spectral sequence

MatheinBoulomenos

so a lot to work on

Mike Miller

It turns out that F_q-representations are a lot like C-representations with eigenvalues in qth roots of unity

MatheinBoulomenos

I heard that the K-groups of finite fields have been computed

Mike Miller

Brauer Lifting is probably the thing he explains worst

MatheinBoulomenos

But now how that went

Mike Miller

19:18

Yup, this is how

MatheinBoulomenos

hmm, there's a section in Serre on that

Mike Miller

In the second paper, he considers the diagram of elementary abelian subgroups and the conjugation maps / actions between/on them

of a finite group G

And proves that the map $H^*(G; \Bbb F_p) \to \lim H^*(A; \Bbb F_p)$ is nearly an isomorphism, the limit taken over that diagram

It is an isomorphism "modulo finite p-groups", I believe, or maybe modulo p-nilpotence

MatheinBoulomenos

Learning K-theory would be beneficial for some number-theoretic stuff I want to learn at some point

Mike Miller

What kind of K-theory?

MatheinBoulomenos

algebraic K-theory

Mike Miller

19:21

2 hard

Quillen uses his definition via the plus-construction, if you've heard of it

BTW, here is a nice spectral sequence argument - with geometric input - that gives you a result on the cohomology of finite groups

MatheinBoulomenos

is that the Q-consruction where you have triangles with an admissable epi and an admissable mono as objects?

Mike Miller

No, totally different

MatheinBoulomenos

then I haven't heard of it

Mike Miller

Given a space $X$ with perfect normal subgroup $G \subset \pi_1(X)$, the plus construction spits out a space with $G$ killed in the fundamental group but no change in homology

It's a stupid trick: kill off the loops in G with discs, since the group was perfect they're zero in homology; they add a bunch of free 2-homology. Kill off the 2-homology with 3-discs. That is the plus construction

$GL(R) = \text{colim } GL_n(R)$ has subgroup consisting of the elementary matrices (generated by the matrices inducing elementary row operations), which is perfect (those matrices are commutators of other such matrices), so $E(R) \subset GL(R)$. Take $BGL(R)^+$. Then $K_i(R) = \pi_i BGL(R)^+$, when $i > 0$

$K_0(R) := K_0(R)$ lol

(do it as grothendieck group of projectives)

MatheinBoulomenos

wait does this imply that the quotient by a perfect subgroup has the same cohomology groups?

Mike Miller

19:29

No. $BG^+$ is not a $BG'$

MatheinBoulomenos

oh of course

Mike Miller

This construction messes really bad with homotopy groups

Unrelated: This is what I wanted to show earlier: If G, G' are finite groups, and f: G -> G' is an isomorphism in cohomology with integer coefficients, then it is an isomorphism of groups; in particular the only acyclic finite group is trivial.

Proof: 1) first, let's see f must be injective; consider i: G'' = ker(f) -> G

MatheinBoulomenos

I have no idea why is the same as the homotopy groups of the geometric realization of the nerve of the Q-construction applied to the category of finitely generated projective $R$-modules

Mike Miller

of course not, it is quite mysterious

Neither do I

The K-theorists do

Proof: 1) first, let's see f must be injective; consider i: G'' = ker(f) -> G. Clearly the map on cohomology from G'' to G is zero, but the map on cohomology from G' to G is an isomorphism, so the map in cohomology from G to G'' is zero. But I believe a map that induces zero in cohomology is actually zero

I am forgetting how this step goes :(

I will assume it and move on

MatheinBoulomenos

okay so I need to learn modular representation theory, this Eilenberg-Moore spectral sequence, complex K theory and I guess this plus construction

Mike Miller

19:35

I just told you the plus construction, I think black box it and don't get too hung up on it

You will learn Eilenberg-Moore in practice. I suggest starting with the Brauer theory

MatheinBoulomenos

okay

Mike Miller

I learned it from Benson and I hated it so I suggest finding a better book

2) Now that f is injective, pick a faithful unitary representation G' -> U(n); of course f: G -> G' -> U(n) is also injective. Consider the spectral sequence H^*(G'; H^*(U(n))) => H^*(U(n)/G'), and similarly with $G$; there are maps between them

Because $U(n)$ is connected and $G$ (resp $G'$) act by translation, the action of $G$ (G') on $U(n)$ is trivial on cohomology

In particular the E^1 pages are $H^*(G') \otimes H^*(U(n))$

We see by assumption that $f$ induces an isomorphism on the $E^1$ page. This means it must induce an isomorphism on the $E^\infty$ page

But the map $U(n)/G \to U(n)/G'$ is a covering map of degree $[G' : G]$, and thus induces mutliplication by that integer on top cohomology. Because this map is an isomorphism, $[G' : G] = 1$

QED

This proof is due to Culler and I like it a lot

MatheinBoulomenos

wow

Mike Miller

I am just having trouble reproving the injectivity part, that an injective map inducing zero on cohomology is zero

Maybe it's about a clever restriction to cyclic subgroups somehow?

a similar spectral sequence is also the usual proof (Venkov's) that group cohomology of a finite group is a finitely generated ring: A faithful U(n)-rep induces a module structure of group cohomology over $H^*(BU(n)) = k[c_1, \cdots, c_n]$. Consider the SS corresponding to the fibration U(n)/G -> BG -> BU(n) to get an SS of $k[c_1, \cdots, c_n]$-modules $$H^*(U(n)/G) \otimes H^*(BU(n)) \implies H^*(BG)$$

U(n)/G is a compact oriented manifold and so it has finitely generated cohomology

Thus $H^*(BG)$ is finitely generated over the ring of Chern classes. Taking those finite generators and the restrictions of the Chern classes to BG you get generators of the whole ring

@MatheinBoulomenos I think the theorem I need for the injectivity part is that the induced map by restriction to a cyclic subgroup $H^*(G) \to H^*(C_p)$ is nonzero

But I don't remember how to prove that

ams.org/journals/proc/1960-011-06/S0002-9939-1960-0124050-2/…

It's another spectral sequence argument

And again you have learned that unchecked, I will just talk arbitrarily long

MatheinBoulomenos

19:57

I don't mind

I don't really follow though

but it sounds interesting

Mike Miller

Yeah, sorry about that

MatheinBoulomenos

I also didn't really work with maps induced on cohomology from group homomorphisms, even. (of course I know how you get them, that's just abstract nonsense), for what we did in ANT, restriction, corestriction and inflation for subgroups was enough so far

Mike Miller

Hmm, interesting

MatheinBoulomenos

and coinflation of course

though you can't quite patch together inflation and coinflation in Tate cohomology afaik, that's why usually just use inflation for positive degrees

Mike Miller

Yeah, I don't know how to do those things in Tate

Which might be why I think of everything from homomorphisms

MatheinBoulomenos

20:03

The inflation-restriction sequence shows up at some critical point

we mentioned that you can get that from a spectral sequence, but we proved it directly

Mike Miller

I believe it, some edge map

MatheinBoulomenos

I guess it comes from the Grothendieck spectral sequence? Since when you have a normal subgroup N in G, then the functor taking invariants with respect to G is the same as first taking invariants with respect to N and then with respect to G/N

Mike Miller

I feel that this is probably more down to earth than a composition of derived functors spectral sequence

I'm pretty sure you're just describing Serre for BH -> BG -> BK

MatheinBoulomenos

hmm, I see

this is probably really simple, but why does a short exact sequence of groups induce a fibration of classifying spaces?

Mike Miller

You think of them all as EG x EK mod the natural actions

Hmmm

Something close to that

That doesn't quite work

Sorry here it is

I'm being bad

EG is equivariantly natural under group homomorphisms (perhaps by viewing it as a bar construction). So we have an equivariant map EG -> EK, which by a path space construction we may assume is a fibration. Quotienting by everything in sight, we have a map BG -> BK. The fiber is the quotient of the fiber of our original map by H. But because this fibranion is an equivalence the fiber is contractible, and H must act freely as it is a subgrojp of G. So the fiber is EH.

Therefore our fibration BG -> BK, induced by the group homomorphism, has fiber BH.

MatheinBoulomenos

20:27

What is EG?

Mike Miller

Hm? I am confused as to how you define BG!!

EG is a contractible space on which G acts freely and properly

MatheinBoulomenos

Ah

I just assume that G is discrete and take an Eilenberg-Maclane space

I haven't looked at the definition of BG in general

Mike Miller

EG is then the universal cover of course

MatheinBoulomenos

I see

Mike Miller

Do you want to talk this out? I bet you could come to a good understanding of the universal bundle

MatheinBoulomenos

20:30

yes, please

if you have time

Mike Miller

Let G be a (reasonable) topological group. Let EG be a space which is contractible and so that G acts freely and properly, and in particular there is a principal G-bundle EG -> EG/G =: BG

MatheinBoulomenos

that's the universal bundle and I guess the contractibility gives us some lifting by obstruction theory?

Mike Miller

(principal G-bundle means fiber bundle whose fibers are copies of G, and transition functions are left translation by elements of g on the fibers

To understand why we would ever define things this way let's remember the defining property of BG: it carries a universal bundle, which every G-bundle should be the pullback of

That's a good guess. Let's prove it

If I had a space X, equipped with a principal G-bundle P, what would it mean to demonstrate that bundle as a pullback of the universal bundle?

(We think of P as carrying a G-action, simple and transitive on each fiber, so X = P/G. For G discrete these are the 'regular G-covers')

(For a vector bundle, say, we let G = O(n) and the corresponding principle bundle is the space of framings of each fiber)

MatheinBoulomenos

hmm

Mike Miller

Mainly this is a convenient way of talking about vector bundles and covering spaces in the same breath but it also turns out to be sufficient to talk about fiber bundles in generslb

MatheinBoulomenos

20:40

Somehow we should get a map to BG from P

but I don't see it

I mean a map from X to BG, but somehow it should be induced by P

Mike Miller

Go backwards: if P is the pullback of the tautological bundle by a map f: X -> BG, what data does that give you in terms of P?

MatheinBoulomenos

okay I need to draw some diagrams

Okay, this is cheating: does the functor that sends each reasonable space to the set of principle G-bundles over it the conditions for Brown representability?

Mike Miller

Yup, but it's not what I'm looking for :P

Do you want the answer so we can continue?

MatheinBoulomenos

no I'm still thinking

Mike Miller

K, take your time :)

MatheinBoulomenos

20:57

I think I have some parts of a useful diagram

$\require{AMScd} \begin{CD}
P @>>> P \times EG @>>> EG\\
@VVV @VVV @VVV\\
X @>>> ? @>>> BG
\end{CD}$

nothing interesting happened so far

Can we find a map from a space to $BG$ such that the pullback of $EG \to BG$ is $P \times EG$?

Mike Miller

That's a cool diagram, but the middle arrow is a trap

What you were looking for was the map $P \to EG$ (we already knew the three other arrows of the pullback square); the requirement on this map is that it is $G$-equivariant (so a map of principal $G$-bundles)

Such a map induces a map on the quotient, $P/G \to EG/G$, aka $X \to BG$

MatheinBoulomenos

ah true

Mike Miller

And pulling back the fibration EG -> BG (with fiber G) along a map $X \to BG$ gets us a map P -> X with fiber G; because EG's fibers had a G-action, the pullback does too, by acting on EG

MatheinBoulomenos

wait, what's the map $P \to EG$? The map in the above diagram is just constant

Mike Miller

Really? That is quite confusing

MatheinBoulomenos

21:03

if we take the composition $P \to P \times EG \to EG$

Mike Miller

what is your first map?

P x *? That seems like the wrong thing to consider, as the map is not G-equivariant

MatheinBoulomenos

oh no I guess you can make it $G$-equivariant

Mike Miller

What I mean is that when you have the arrow X -> BG and EG -> BG we can take the pullback, right? And the pullback should have a map to X and a map to EG

And we should be able to identify the fibers of the map to X with the fibers of the map EG -> BG

Remember the way us boring people think of pullbacks: in terms of sets, this is $\{(x, e) \mid f(x) = \pi(e)\}$, and those properties are easier to check in terms of this

(at least for me)

MatheinBoulomenos

So fix a point in $EG$, then we can take a constant map $P/G=X \to EG$ and because our action is simple and transitive, this should extend to a $G$-equivariant map $P \to EG$

Mike Miller

How did you get a map from X to EG?

MatheinBoulomenos

21:08

a constant map

Mike Miller

Then P = X x G

Since you're going to identify it with the pullback of the trivial bundle over a point

MatheinBoulomenos

hmm

Mike Miller

So you totally do get an equivariant map, but for boring reasons

My claim is that the statement "P -> X is a pullback of the universal bundle" is the same data as a proper G-equivariant map P -> EG (where you may ignore 'proper' when the group is compact), which covers a map X -> BG; this is the map we identify it as the pullback over

MatheinBoulomenos

okay, but how do we get that map?

Mike Miller

You mean when you already have a map X -> BG?

Or are you asking how to show that everything is a pullback of the universal bundle (and hence it deserves that name)?

MatheinBoulomenos

21:11

No, I understand how you take pullbacks. But how does the bundle P->X determine a map X -> BG?

yes

Mike Miller

That's the fun part! :)

Have you ever seen a proof of Whitehead's theorem?

MatheinBoulomenos

we haven't used the fact that EG is contractible so far

no

Mike Miller

In that case you probably haven't seen the following proof idea before, so I think I'll just walk through it instead of hinting it, if that's ok

MatheinBoulomenos

sure

Mike Miller

Assume X is a CW complex for convenience of discussion. (The result in the end should be true for any paracompact space, maybe with local decentness assumptions, I don't remember)

We have a bundle P -> X. I am going to inductively construct an equivariant map P -> EG, inducting on the skeleta of X.

$P^{(0)} \to X^{(0)}$ is just a discrete set of copies of $G$ (mapping to a set of points). It is clear how to come up with a map $P^{(0)} \to EG$: pick basepoints in each copy of $G$, and a point on $EG$ for each of them to map to

MatheinBoulomenos

21:16

okay

Mike Miller

Inductively we have defined the map on $P^{(k-1)}$. Let $D^n \to X^{(n)}$ be the characteristic map of an n-cell. We may pull back the bundle $P$ over this characteristic map; on $D^n$ the bundle is trivial (as $D^n$ is contractible). We already have a defined equivariant map on the boundary $G \times S^{n-1} \to EG$

So trace out $\{1\} \times S^{n-1}$ in $EG$. Because $EG$ is contractible, that extends to a map from the disc, and hence to an equivariant map from $G \times D^n \to EG$. Because this extended the map that already existed, we have extended the map on $P^{(n-1)}$ to (the portion of $P^{(n)}$ including the preimage of that cell).

Since the interiors of the cells are disjoint (and that's the only place we change the definition of the existing map) we can do this for all of those cells to construct an extension to $P^{(n)}$

MatheinBoulomenos

The part with "Because $EG$ is contractible" is where obstruction theory comes in, right?

Mike Miller

Exactly

Well, I mean, to clarify

I literally used $\pi_{n-1} EG = 0$ to extend a map from $S^{n-1}$ to a map from $D^n$

That is not obstruction theory so much as the definition of homotopy groups

MatheinBoulomenos

ah I see

Mike Miller

One could also do this by way of obstruction theory, but if you chased this through, you would find it's just hiding the argument we just did behind the language of obstruction cocycles

MatheinBoulomenos

21:24

okay

Mike Miller

I guess my punchline is that "Everything gets pulled back from the universal bundle" is very explicitly the statement that EG is contractible

Via the proof above

If EG was not contractible you could come up with a G-bundle over a sphere that was not represented

MatheinBoulomenos

okay so this way we get an equivariant map and because you pulled everything back from $X$, this should cover a map $X \to BG$ if you just mod out $G$

Mike Miller

Yup yup

This also explains the uniqueness, as the same argument shows that any two EGs are equivariantly homotopy equivalent

But I suppose the universality of the universal bundle does that too

MatheinBoulomenos

Yoneda basically

Since we know that $\operatorname{Hom}(-,BG)$ and $\operatorname{Hom}(-,(BG)')$ must represent the same functor, we get that $BG$ is unique up to homotopy, but then for two different universal bundles $EG \to BG$ and $(EG)' \to BG$, both must correspond to the identity in $\operatorname{Hom}(BG,BG)$

Mike Miller

21:41

Yup

So I really liked your point about Brown representability earlier, let's use that

GBun(X) = [X, Z] for some connected space Z

Now here is a fact. When X is a suspension (abuse notation and rewrite it as $\Sigma X$), $GBun(\Sigma X) = [X, G]$. Proof: This is the "clutching bundle" construction. Trivialize the bundle over the two cones; the transition function on the equator X is called the "clutching function" and its homotopy class is what we care about

GBun(Sigma X) has a product coming from the coproduct on Sigma X (you can "pinch it" at the equator to get a map to Sigma X vee Sigma X)

And this product is identified as the product on $[X, G]$ coming from multiplication in $G$ (you "concatenate" transition functions by multiplying them, is more or less what this says)

Now we see $[X, G] = [\Sigma X, Z] = [X, \Omega Z]$ as groups. This implies that $G \simeq \Omega Z$ as topological groups (there is a zig-zag of homomorphisms, each of which is a homotopy equivalence)

When you have a topological group, you can take its classifying space, and the result of all of this is that $BG = B \Omega Z = Z$

Thus just knowing Brown representability (for both set-valued functors and group-valued functors) gets you that the representing space for G-bundles is BG

MatheinBoulomenos

really cool

So this explains why we call $BG$ the "delooping" of $G$

Mike Miller

Indeed. It is worth noting that given a space X, you cannot usually pick out a unique space "BX" that loops to X. The point being that you get an equivalence Omega BX = X, and Omega BX has extra structure - the concatenation of loops gives us a product structure on X

But this is what you need. When you have a group (the fully homotopical thing would be an A_infty-space), you can indeed deloop

One option is to geometrically create BG = EG/G as before

The more functorial option is to discover the bar construction for topological groups, which outputs a space B(*,G, *) = BG

MatheinBoulomenos

21:56

hmm, I know the Bar resolution

Mike Miller

It's precisely that fella

The best way to understand bar resolutions in general is via simplicial sets

If M is a left G-space, N a right G-space, B(M, G, N) is the realization of a simplicial space whose n-simplices are M x G^n x N. This means you put it together by gluing together copies of M x G^n x N x Delta^n by certain rules on the boundary of the simplices Delta^n

The bar construction is used and talked about extensively in Emily Riehl's book on abstract homotopy theory

it's often the explicit way to write a homotopy quotient, or other homotopy colimits

MatheinBoulomenos

hmm I see

thanks a lot for this! This was quite insightful

@MikeMiller Are monoidal categories and monads used in modern homotopy theory? I think they are quite nice from an abstract point of view, but I haven't seen them get used much. (Although faithfully flat descent in alg geo follows from Beck's monadicity theorem which is pretty sick)

Mike Miller

22:14

I'm certain so but that's where you get into the homotopy theory that is both too hard for me and usually further from my interest

MatheinBoulomenos

I see

Mike Miller

I have definitely seen Peter May write things that pass through monads, in particular about operads

But the most complicated operads I have ever needed are A_infty = Associative operad

which is the simplest of them all

MatheinBoulomenos

I think monads are quite natural/cool when you think about them as extra nice adjunctions

which I guess doesn't fully cover all monads

Mike Miller

And might cover why they don't excite me! :D

I am glad you liked the discussion, I am always glad to --- especially as it is a tool for procrastination

MatheinBoulomenos

idk, I think the fact that a monadic adjunction $C \to D$ implies that you can view $C$ just as objects in $D$ endowed with a special kind of extra structure (defined internally in $D$, via the monad) seems pretty cool

and it seems like a natural question: what kind of minimal data from the adjunction (suppose it's nice enough) do we need to reconstruct $C$ from $D$ and the extra data?

but that's probably just me being an algebraist

Mike Miller

22:23

You start by understanding and appreciating adjunctions

I still never have

MatheinBoulomenos

yeah I guess if you don't like adjunctions, it's hard to like monads

Mike Miller

yeah, that's all

If they work for you I'm gald

MatheinBoulomenos

I guess you could think of them as just a kind of representability condition where you can naturally vary an additional parameter

So, say instead of just taking one functor $F:C \to \mathbf{Set}$ and asking if you can find an object $X \in C$ such that $F \cong Hom(X,-)$, you ask simultanously for every set Y if you can represent the functor $ A \mapsto Hom(Y,F(A))$ by some object in $C$. If you can do that for every $Y$, you of course want this to depend functorially on $Y$

Mike Miller

I suppose that makes sense

I think with any categorical notion, I would need to feel it in the math I'm actually doing - and see that it really is just this basic level of abstraction - to be comfortable with it

(And it should be frequent enough that there's value to the change, you feel?)

MatheinBoulomenos

And then you see that unlike representable functors which make sense for concrete categories (or enriched categories I guess) this actually makes sense without the functors going to some "concrete/enriching/etc. category"

we used them in our undergrad algebra course actually

Yeah it's hard to appreciate something if you don't use it

I don't think that's even something related to abstractness

Adjunctions are really something I just see everywhere since I understood them

Whenever you use that a map of sets from a basis of a vector space uniquely extends to a linear map, you're actually using some free-forgetful adjunction

Mike Miller

22:45

Sure, there are sometimes obvious ones

MatheinBoulomenos

In general I don't think of adjunctions as a deep concept, I think they're rather shallow, as the free-forgetful examples show (okay I guess the "free" functor from topological spaces to compact Hausdorff spaces is a bit more interesting), but it's just a kind of pheonomenon that happens often enough that it's convenient to have some language for it

I guess you could say that about a lot of categorical stuff

like the only reason we care for limits and colimits are because they come up so often

If you want to be really vague then you can just say that adjunctions formalize the situation that even if a functor is not an equivalence, there might still be some "universal" way to go in the other direction

Mike Miller

The difference perhaps being that I have found use infrequently for limits and colimits but not so much for adjunctions - perhaps because my use of limits is so infrequent

MatheinBoulomenos

I guess they are more common in algebra

of course the Smash-Hom adjunction is quite important

or the exponential law for compactly generated Hausdorff spaces

which is left adjoint for the Hom with other variance

Mike Miller

23:03

Yeah

I'm sure you would like the way the homotopists think about things

I don't think it's for me though

MatheinBoulomenos

fair enough

Transcript for

♦ $Mathematics$ random cohomology for quantum nerds

Transcript for

♦ random cohomology for quantum nerds

♦ $Mathematics$ random cohomology for quantum nerds