random cohomology for quantum nerds - 2020-04-22 (page 1 of 2)

Alessandro Codenotti

09:08

0

Typically in categories of algebraic objects (I will use $\mathsf{Grp}$ as an example), every object is the direct limit of its finitely generated subobjects. Based on this observation I was wondering how general this phenomenon is, and it turns out that the answer is locally finitely presentable...

I asked on main

Alessandro Codenotti

10:47

@Balarka lmao so I'm having a geometric measure theory class on zoom and the professor just said "let's see this in detail even if we're doing an overview, every lecture should have a proof, even better if it's a proof by Gromov"

Balarka Sen

Yea boi

Alessandro Codenotti

Is there any field Gromov didn't do work in? Apparently he did stuff on the isoperimetric problem too

Balarka Sen

Yeah that's actually famously Gromov's research work

He revolutionized isoperimetric ideas

There's a paper by Larry Guth on waist inequalities check it out

Balarka Sen

11:11

wrong chat

Alessandro Codenotti

11:24

Oh no he pulled out a paper copy of Federer

Balarka Sen

Gone

Alessandro Codenotti

15:07

@Balarka I asked one single category theory question on main and I got invited in the comments to a conference of Italian category theorists, I'm starting to think they're a massonic lodge or something

Balarka Sen

Oh shit what the fuck

Alessandro Codenotti

math.stackexchange.com/questions/3637955/… third comment lol

Mike Miller

ask elsewhere

Balarka Sen

lmfao

Alessandro Codenotti

washington?

Balarka Sen

15:09

Mike isn't in washington

Mike Miller

Oh you can delete

Forgot

Make me mod again so I can see deleteds

Balarka Sen

yeah 1 sec

Feeds

Balarka Sen has added Mike Miller to the list of this room's owners.

Mike Miller

lmao

Alessandro Codenotti

lol

Alright enough gossip, time to go back to math

Balarka Sen

15:15

this is why i want to be in academia

just for the gossip

Alessandro Codenotti

You should have studied law or economics

Balarka Sen

its more fun to pick on math nerds

Alessandro Codenotti

They're always ready to backstab each other with no shame as soon as they need to elect the head of the department or any other official role

Balarka Sen

because everyone here is dumb

Mike Miller

@AlessandroCodenotti mathoverflow.net/q/358219/40804

Why isn't the first sentence obviously false

Alessandro Codenotti

15:16

Meanwhile in the math department there was a single candidate because nobody wants to deal with bureacracy and they all want to just do math

Mike Miller

$(\ell^1/V)^* \subset \ell^1 = \ell^\infty$, so that the dual space of any quotient of $\ell^1$ is separable, whereas $\ell^\infty$ has non-separable dual

Ah dumb

ell^inf is not separable

Alessandro Codenotti

yeah, also I don't see why it's an isometry, but the part of the argument sketched in the question seems like an extremely reasonable approach

Mike Miller

Yeah, the proof seems fine

This reminds me of the amusing fact that given a Frechet space $V$ one can find a continuous map $\ell^1 \to V$ with dense image

Alessandro Codenotti

What's a Frechet space again?

All the functional analysis I learned was over Banach spaces at least

Mike Miller

Inverse limit of Banach. Topology defined by countable family of seminorms

$C^\infty([0,1])$, or $C^0(\Bbb R)$ are good examples

In the former the norms are the $C^k$ norms for all $k$; on the latter the norms are $C^0([-n,n])$

Alessandro Codenotti

15:22

I see

Mike Miller

They are much more poorly behaved, unfortunately

Oh

Separable Frechet space

Obviously the conclusion implies separabe

Anyway, denote the seminorms by $\rho_n$, and pick a dense set $x_n \in V$. Define constants $C_n$ so that $C_n \geq \sum_{i,j \leq n} \rho_i(x_j).$

Then consider the space $\ell^1(C_n)$ of sequences so that $(C_n a_n)$ is $\ell^1$

Balarka Sen

Cute

Mike Miller

Then the map $\ell^1(C_n) \to V$ just sends $(a_1, \cdots)$ to $\sum a_i x_i$; the finite sums are Cauchy sequences in each seminorm and hence this is a convergent sequence in $V$

And of course it contains all of the $x_i$

Alessandro Codenotti

Neat

Balarka Sen

15:39

$f(t)= \|J(t)\|^2$ implies $f''(t) = 2\|J'\|^2 + 2J \cdot J''$ and $J'' + 2R(J, d/dt)d/dt = 0$

So $f''(t)/2 = \|J'\|^2 - 2R(J, d/dt)d/dt \cdot J$

So if sectional curvature is negative, the norm squared of Jacobi fields are convex

That's the proof of Cartan Hadamard iirc

right because $J(0) = 0$ and convex functions cannot dip to $0$ after starting at $0$

Cool

Mike Miller

@BalarkaSen Lol very clever

Balarka Sen

:P

Mike Miller

It's sufficient that the sectional curvature be nonpositive so that $f'' \geq 0$, and thus that $f$ is bounded below by a convex function

Balarka Sen

True, good point

Mike Miller

@BalarkaSen I guess I don't get the argument --- how does this give you the conclusion that the exponential map is a diffeomorphism when $\pi_1 M = 0$?

There's some lemma you're using that I don't know

Balarka Sen

15:52

Ah so what I'm using is if Jacobi fields are always nonvanishing there are no conjugate points; those are points where a bunch of nearby geodesics clutter and meet - these correspond to singularities of $d\exp_p$

So that implies $\exp_p$ is a global diffeomorphism

If $(d\exp_p)_v(w) = 0$ then the Jacobi field in the direction of $w$ along the geodesic emanating from $p$ with direction $v$ must have a point where it vanishes, yeah?

Because take the variation $F(t, s) = \exp_p(t(v + sw))$

Then $J(t) = d/ds |_{s = 0} F(t, s) = (d\exp_p)_{tv}(tw)$

$J(1) = 0$ as required

I suppose if you live in the manifold and $p$ is a photoluminescent source, light rays are emanating from $p$ and taking geodesic ray paths, so you'll see some interference happening at these conjugate points. If you're looking along a geodesic ray that contains a conjugate point, you'll see these as dark spots, I suppose

The convexity condition says there are no dark spots wherever you look at $p$ from, which is essentially the same as saying $\exp_p$ is a covering map

but whatever I'm nerding out

Mike Miller

16:07

@BalarkaSen OK.

Got it

Thugh your expression for J(t) has no s in it lol

Balarka Sen

It shouldn't right?

$s$ is the direction of the variation; $t$ is the direction of the geod ray

Alessandro Codenotti

How do people come up with the right matrices to show that $SL_2(\Bbb Z)$ has a free subgroup by ping pong?

Balarka Sen

I think it becomes natural if you look at the action on H^2

Alessandro Codenotti

Uhm the proofs I've seen usually have it act on $\Bbb R^2$

Balarka Sen

Yeah you can do it by acting on H^2 as well, let me recall how this goes

1 sec

Alessandro Codenotti

16:14

I actually like to have it act on R^2 since I'm not familiar with H^2 though

Balarka Sen

You wanted an answer to your question though right? :P

Alessandro Codenotti

fair enough

Ah wait

Turns out I'm unable to draw what the subset of R^2 given by |x|>|y| looks like

Ok it's clear to me how to do this now

Balarka Sen

Okay, so let $\text{SL}_2(\Bbb Z)$ act on the upper half plane $\Bbb H^2 = \{z \in \Bbb C : \Im(z) > 0\}$ by $z \mapsto (az + b)/(cz + d)$. Then $z \mapsto z + 1$ is translation and $z \mapsto 1/z$ is inversion about the unit semicircle, $z \mapsto -z$ is inversion about the infinite radius semicircle $\Re(z) = 0$

Consider $z \mapsto -1/z$ and $z \mapsto 1/(z + 1)$

Alessandro Codenotti

Hmmm

Balarka Sen

Do I remember how to do this

Lol

Mike Miller

16:25

@BalarkaSen It should if you're taking the derivative with respect to s

The derivative w/r/t s of a function independently of s is famously 0 :p

I was being an ass since I didn't read one line up though

It's clear where the s goes now

Balarka Sen

@MikeMiller Uh? $F(t, s)$ depends on $s$, it's $\exp_p(t(v + sw))$. The derivative at $s = 0$ does't

Mike Miller

lol

I'll see myself out then

(of course you're right)

Balarka Sen

i frequently make calculation mistake so i wouldnt be surprised if i was wrong tbh

glad im not

@Alessandro I do believe that if you let $A$ to be everything in $\Bbb H^2$ outside of the unit semicircle and $B$ to be everything inside, then $z \mapsto -1/z$ takes $A$ inside $B$ and $z \mapsto 1/(z + 1)$ takes $B$ inside $A$

Does that suffice for ping-pong? Should right?

Alessandro Codenotti

You want that iterating the maps also swaps them

Wait let me write that properly

If your maps are $f_1$ and $f_2$ we need $f_1^n(A)\subseteq B$ and $f_2^n(B)\subseteq A$ for all $n$

Balarka Sen

Ahh, yikes.

Wait but how does that make sense if $f_1$ has finite order

You'll end up proving $A \subseteq B$

You mean non-identity elements, got it

Alessandro Codenotti

16:38

There's some restrictions, I think the least that one can ask is for one element to have order $\geq 3$ and one order $\geq 2$

Balarka Sen

Ok, pass to $\text{PSL}_2(\Bbb Z)$. My $z \mapsto -1/z$ has order $2$ whereas $z \mapsto -1/(z + 1)$ has order $3$.

@Alessandro haha nice

exactly what I want

Alessandro Codenotti

And then you look at the actions of the subgroups they generate minus the identity

@BalarkaSen lol beautiful

math.la.asu.edu/~paupert/CookPingPongLemma.pdf yeah that's how it's written there for example

Balarka Sen

Yeah I am pretty sure if $f_1$ and $f_2$ are those maps respectively, $f_1(A) \subseteq B$ and $f_2^i(B) \subseteq A$ where $i = 1, 2$

So this should tell you $\text{PSL}_2(\Bbb Z) \cong \Bbb Z_2 * \Bbb Z_3$

Alessandro Codenotti

@BalarkaSen Right

Balarka Sen

Is this enough for your purposes? These guys contain free groups by a covering space argument, and $\text{SL}_2(\Bbb Z)$ is a central $\Bbb Z_2$-extension, so that should also be fine.

Alessandro Codenotti

16:40

Neat

yeah, actually this whole thing is a very roundabout proof, what I want is to show that free groups are residually finite, and apparently this is done by embedding $F_2$ in $SL_2(\Bbb Z)$ and using the fact that $GL_2(\Bbb Z)$ is residually finite and that subgroups of residually finite groups are residually finite

Balarka Sen

ahhh

Alessandro Codenotti

(and then there's a small argument to deal with infinitely generated free groups)

Balarka Sen

Surprising there's no direct argument. Normal subgroups of $F_2$ are automatically finite index, so you just need to find, for every element of $F_2$, a normal subgroup that avoids that element

Alessandro Codenotti

There might be a direct argument, but that's how it's proved in the book I'm reading

Balarka Sen

Garbage. Finite rank normal subgroups of $F_2$ are finite index

Ignore me

@Alessandro Pretty cool fact

Alessandro Codenotti

16:49

38

Q: Why are free groups residually finite?

Why is it that every nontrivial word in a free group (it's easy to reduce to the case of, say, two generators) has a nontrivial image in some finite group? Equivalently, why is the natural map from a group to its profinite completion injective if the group is free? Apparently, this follows from...

gr.group-theory topological-groups profinite-groups

There's a direct proof there even though all the indices are confusing to read and I'm still deciphering the permutation

Balarka Sen

Ugh!

Oh Stallings has a covering space proof let's see

I was wondering if I can find a finite covering of $S^1 \vee S^1$ that misses a path

Beautiful!!!

Alessandro Codenotti

@BalarkaSen That's done in another answer to the above question

Balarka Sen

Yeah I was reading that

Alessandro Codenotti

At the end they use the fact that every finite index subgroup contains a finite index normal subgroup just in case you haven't seen this before (I hadn't until yesterday)

Balarka Sen

Yeah, that's taking the permutation representation and then it's kernel

It's a nice group theory fact

I remember being shocked to hear $A_\infty$ has no finite index subgroups

'cuz it's infinite simple

Alessandro Codenotti

17:00

That's a fact I've heard but I don't know how to prove

There's a simple full classification of the subgroups of $S_\kappa$ for all $\kappa$ iirc

Balarka Sen

Yikes

@AlessandroCodenotti I think you just intersect it with the finite levels $A_n \subset A_\infty$; in each of that it's normal so full $A_n$

Alessandro Codenotti

mathoverflow.net/questions/12943/… first answer

normal subgroups though, of course there's plenty of messed up subgroups

Mike Miller

lol

Balarka Sen

You should absolutely know who HG is

Alessandro Codenotti

Looong entered the chat

I'm betting on Balarka being flagged for saying shitless

See you in a few days

Balarka Sen

17:11

Lololol

feynhat

Hello

Balarka Sen

Hi!

Let $M$ be a smooth $n$-manifold. For any smooth submanifold of dimensional $k$, $Z \subset M$, you can extract a natural homology class out of it by triangulating it and letting that be your singular cycle $[Z] \in H_k(M; \Bbb Z_2)$. This is called the fundamental class. It's important that the value group is $\Bbb Z_2$ and not eg $\Bbb Z$ because of orientability issues

Purely topologically, this is pushing forward the orientation class $[Z] \in H_k(Z; \Bbb Z_2)$ by $H_k(Z; \Bbb Z_2) \to H_k(M; \Bbb Z_2)$ induced by inclusion

Let $Z_1, Z_2 \subset M$ be $k$- and $l$-dimensional submanifolds respectively. It's curious to ask what $\text{PD}(\text{PD}[Z_1] \smile \text{PD}[Z_2]) \in H_{n - (k+l)}(M; \Bbb Z_2)$ is. Is it also fundamental class of some submanifold?

Do you have a guess

(Just want to make sure if you know the story already)

feynhat

$Z_1 \cap Z_2$?

Balarka Sen

After you make $Z_1$ and $Z_2$ transverse to each other, that's right.

So it's interesting to ask if the cup product can be defined purely in homology, as a pairing $H_k(M; \Bbb Z_2) \times H_l(M; \Bbb Z_2) \to H_{n - (k+l)}(M; \Bbb Z_2)$. If you can prove it's a nondegenerate pairing for $l = n - k$, this will give you an isomorphism $H_k(M; \Bbb Z_2) \cong H_{n-k}(M; \Bbb Z_2)^*$, which is the mod 2 homology Poincare duality

I need to get dinner but I will come back and rant more

feynhat

Sure. I have a few questions though.

Balarka Sen

17:25

Yeah ask

feynhat

@BalarkaSen What did you mean by 'extracting a natural homology class'?

Balarka Sen

As in any submanifold gives a homology class

Gotta run. I'll come back and talk more

feynhat

@BalarkaSen Okay, so you'll triangulate $Z$ in a way that boundaries of the triangles cancel?

Mike Miller

17:57

@feynhat Yes, or just use the abstract fact that oriented manifolds have fundamental classes, which you can push forward.

feynhat

@MikeMiller Compact, right?

Mike Miller

Yeah, sorry. Compact (without boundary).

I think of a cycle (simplicial, singular, or otherwise) as more or less being a manifold up to codim 2 singularities, and a boundary is the boundary of something which is a manifold-with-boundary up to codim 2 singularities. The idea is you can paste the simplices or whatever into a manifold --- the "closed" condition means that you can pair off the sides of your simplices (that is, for each face of a simplex, we choose exactly one face of a different simplex to glue it to).

Because the interior of a simplex is a manifold, and our glued-up space is a manifold at the codim 1 simplices (because we just glued two of them along their faces, which gives a manifold). We can say nothing about what the space looks like in codim 2.

Try this as an exercise. Start with a 1-cycle in a space. (The space X we're taking chains in is totally irrelevant.) Show that there is a way to glue these by their boundary points so that the glued up thing still comes equipped with a map to X, but so that it's a 1-manifold without boundary, aka, a bunch of circles.

(This circle of ideas leads to --- among other things --- a geometric proof that $\pi_1^{ab} = H_1$. This might be the proof in Hatcher?)

feynhat

18:16

@MikeMiller I am not sure I follow what you meant by 'a way to glue these by their boundary points'.

Balarka Sen

@feynhat I am using $\Bbb Z_2$ coefficients so actually you don't have to bother with cancellation

The cancellation (with $\Bbb Z$ coefficients) happens only if your manifold is orientable. Try it with $\Bbb{RP}^2$!

Sorry for disappearing I started playing Resident Evil 4 again

feynhat

@BalarkaSen oh wow $\mathbb{RP}^2$ was a good exercise.

No matter how I choose orientations on the square, they will not be compatible after the identifications.

Balarka Sen

That's right. So orientability can be understood as obstruction to a certain chain being a cycle.

Alessandro Codenotti

@BalarkaSen Oh wow I remember playing that when it came out on the PS2, ages ago

Hm actually it came out in 2005 when I was 10, so it was probably a couple of years later rather than when it came out

Mike Miller

@feynhat I'm sorry, I don't really know what else to say. A 1-chain is a formal sum of bunch of maps from [0,1], and you can glue spaces along subspaces. If it's a cycle you get a certain relation between the boundary values of these maps from intervals.

Balarka Sen

18:31

@Alessandro Oh nice

It's a fun game

Alessandro Codenotti

I remember being very irritated that you can't move and aim at the same time

Balarka Sen

Hahaha yeah

Alessandro Codenotti

Apart from that it was a great game for its time

But it came out around the same time as shadow of the colossus, which is still my personal favourite as far as games are concerned by far

Balarka Sen

@MikeMiller I am forgetting the obstruction story. The triangulation defines a $\Bbb Z$-valued $n$-chain and it's boundary is an $(n-1)$-cycle. Upto homology it should give a well-defined class in $H_{n-1}(M; \Bbb Z)$ which vanishes when you go mod $2$. What is this class again?

Surely it's PD would give me something like $w_1$ but I am blanking out on how to see that

I have never played SoC

feynhat

@MikeMiller Okay. Suppose $\sum c_i \sigma_i$ is a singular 1-cycle. So, $\partial(\sum c_i\sigma_i) = 0$ or $\sum c_i (\sigma_i(1) - \sigma_i(0)) = 0$. Is this the relation you're referring to.

Alessandro Codenotti

18:36

It's a great game, they also made a remastered HD version which has basically no changes from the original one apart from the much improved graphics

Balarka Sen

Nah I must be confused. If you do it with the triangulation on $\Bbb{RP}^2$ the boundary of the chain is $``2\Bbb{RP}^1"$. It doesn't give anything meaningful in homology.

My bad

Mike Miller

Yeah that relation

@BalarkaSen You've just shown the existence of a codim 1 2-torsion class, right?

UCT should take that class to the mod 2 fundamental class

SotC is solid

Relatively short which is a plus for me nowadays

You know nowadays we can do this stuff on zoom where I can write on a whiteboard too.

Balarka Sen

That would be awesome lol

@MikeMiller Ah fair enough got it

Mike Miller

I should really write a lesson plan but I could be on whiteboard for a bit

Alessandro Codenotti

@MikeMiller Did you play the last guardian too?

Mike Miller

18:46

No, though I figure it's good

Balarka Sen

The math camp organizers wanted the instructors to get a drawing pad but unfortunately couldn't deliver it in time to anyone except people in US

Hopefully I'll get it next time

Seems useful

Mike Miller

@BalarkaSen lol yeah no surprise

Alessandro Codenotti

I liked it, but you're pretty much forced to proceed linearly in the story which is a pity, the setting is visually stunning so I was expecting it to be a roam freely type of thing as in SotC

Mike Miller

That doesn't surprise me, a lot of those really pretty games aren't robust enough to be pretty from all angles

feynhat

@MikeMiller Are you expecting me to say that some function is constant some some set, and you mod out by that set, and you get that the image of the function under this quotient is a bunch of circles? But I don't see how that relation helps me conclude this.

Mike Miller

18:52

I am really confused about what is unclear. Assume that all of the $c_i = 1$ (either work over $\Bbb F$ or work over $\Bbb Z$, where a singular 1-simplex is iirc identified with - its orientation reversal). Then the relation $\sum \sigma_i(1) - \sigma_i(0)$ means that for each $\sigma_i$, there is some $j(i)$ so that $\sigma_i(1) = \sigma_{j(i)}(0)$.

Take the quotient space of $\sqcup_i [0,1]_i$ (subscript to denote which copy) by gluing $1_i$ to $0_{j(i)}$.

Each boundary point is identified to exactly one other boundary point so the quotient is a manifold.

Because $\sigma_i(1) = \sigma_{j(i)}(0)$ these descend to a map $\sigma$ from the quotient.

I'm sure this would all be much clearer with pictures

That very first sentence was a bit unfair. Sorry.

feynhat

Are you invoking freeness of singular chain groups, to conclude the existence of $j(i)$ for each $i$? That is a $\sigma_i(1)$ cannot cancel another $\sigma_{i'}(1)$.

Mike Miller

@feynhat Fair point. I was really thinking of the integral case, where finite sums of "1" are not zero.

But still. Start with $\sigma_1(1)$. There is some $j$ so that for $t$ equal to one of $0$ or $1$ we have $\sigma_{j}(t) = \sigma_1(1)$, where here $(j,t) \neq (1,1)$.

Glue together $\sigma_1$ to $\sigma_j$ by gluing $t_j$ to $1_1$.

You now have fewer 1-simplices in your cycle, and you check that it remains a cycle. (In this process you may have glued one endpoint of $\sigma_1$ to the other.)

feynhat

I think you need some sign there?

Mike Miller

Sorry for the 0/1 carelessness.

I'm working over $\Bbb F_2$ so I don't need to deal with that junk

If you were working over $\Bbb Z$, and you had oriented the simplices so that all of the $c_i = 1$ (none are negative), then you always glue some $1_i$ to $0_{j(i)}$ --- this order for the gluing is what guarantees the glued-up thing has a canonical orientation

(And given that I made an error, my first line was much more than just a bit unfair.)

feynhat

19:10

@MikeMiller Okay. This makes sense.

Mike Miller

Here is the general statement. Let $\sigma$ be a $k$-cycle over $\Bbb F$, represented as $\sum_i \sigma_i$, where $\sigma_i: \Delta^k \to X$ is some map. Then there is a k-dim simplicial complex $S$, so that each (k-1)-simplex in $S$ appears as a face of exactly two k-simplices, and there is a simplicial map $p: \sqcup_i \Delta^k \to S$ which glues along those (k-1)-faces.

Then there is a map $\sigma_S: S \to X$ so that $\sigma_S p = \sqcup_i \sigma_i$.

$S$ is called a pseudomanifold, or manifold up to codimension 2.

If we work over $\Bbb Z$, then further $S$ is oriented, in the sense that each top simplex is oriented and the boundary-gluing is compatible with orientation (that is, the boundary orientation on any (k-1)-face is different depending on which k-simplex you induce it from).

There is a similar statement for chains which are boundaries (they arise as the boundary of some pseudomanifold-with-boundary, oriented if you so desire).

If you care I can explain why this proves $\pi_1^{ab} = H_1$.

feynhat

Yeah sure.

Mike Miller

1) For connected CW $X$, the forgetful map gives an isomorphism $p: \pi_1^{ab}(X,x) \cong \Omega^{SO}_1(X)$.

The right thing is oriented bordism of 1-manifolds in X.

Proof: $p$ is surjective, because given any finite collection of loops in $X$, these are homotopic to a collection which always sends $\gamma(0) = x$, and then bordant (using a pair-of-pants type thing) to a map from a single loop based at $x$. Injectivity is more subtle.

feynhat

oh wow. I don't know what you're talking about.

Mike Miller

19:25

I guess I should stop.

feynhat

Whatever happened to showing that if two paths are homotopic, they are homologous if thought of as simplices?

Balarka Sen

That's more or less clear. I want you to write down a proof for me.

Hint: Triangulate the homotopy square

feynhat

Okay. Let $H$ be the homotopy between $f$ and $g$. Let $\Delta_1$ be the triangle that lies above the $x = t$ line and $\Delta_2$ be the other one.

Let $\sigma_i = H|_{\Delta_i}$. Then, I claim, $\partial(\sigma_1 + \sigma_2) = f - g$.

Oh wait, I didn't give orientations.

Orient the top edge from $(1, 0)$ to $(1, 1)$, and move along clockwise to orient all the edges

For $\Delta_1$ orient the diagonal from $(0, 0)$ to $(1, 1)$, and in the opposite direction for $\Delta_2$.

Now, $\partial \sigma_1 = f - H|_D + f(0)$ and $\partial \sigma_2 = g - H|_D + g(1)$

D is the diagonal

Okay I needed $\sigma_1 - \sigma_2$.

Balarka Sen

19:49

This seems like too much symbolic work for something direct, no? Take the square $[0, 1] \times [0, 1]$ where $[0, 1]$ is oriented standardly. $H$ is a map from this square to $X$, restriction of $H$ to the top is $f$, bottom is $g$

Dissect the square along the diagonal going from $(0, 0)$ to $(1, 1)$. Restrict $H$ to one triangle, that's $\sigma_1$. Other is $\sigma_2$

feynhat

Yeah right. I was orienting each triangle separately. Didn't realize oriented $[0, 1]$ will induce an orientation on $[0, 1] \times [0, 1]$.

Balarka Sen

OK, this finishes the proof.

So you get a map $\pi_1(X) \to H_1(X)$, by sending a loop to the homology class of the $1$-simplex given by the loop.

By abstract nonsense, it gives a map $\pi_1(X)^{ab} \to H_1(X)$ because we're mapping to an abelian group

feynhat

Yes, any map from a group G to an abelian group factors through the abelianization of G.

Balarka Sen

Correct, @feynhat

The question is why is this injective/surjective?

Which one do you want to see first

feynhat

Injective?

Balarka Sen

20:04

Okay.

Suppose a loop $\gamma$ in $X$ is nullhomologous.

Then there is some $2$-chain $\xi$ such that $\partial \xi = \gamma$ (I am being fidgety with notation but this shouldn't be unclear)

According to what Mike said, $\xi$ can be realized as a map $\xi : K \to X$ from a $2$-dimensional simplicial complex $K$ "with boundary" $\partial K$ (consisting of open edges) such that $\xi | \partial K = \gamma$

Is that pictorially clear?

feynhat

No. $\xi$ is not a cycle.

Balarka Sen

Mike said a version of the thing for cycles is true for chains which realize boundaries.

But we can go over a sketch of a construction if you want

feynhat

@BalarkaSen Yes, please. Do it for a 2-chain please. I don't think I'll understand the general construction.

Balarka Sen

Yes, OK

Roughly speaking, if $\partial \xi^2 = \zeta^1$, then you can expand out $\xi^2$ as a linear combination of $2$-simplices, and then take the boundary of each of them.

Their formal sum is a massive expression, and then they cancel out in a way that gives $\zeta^1$

feynhat

@BalarkaSen Is this something other than 'a chain is linear combination of simplices'? Am I missing something?

Balarka Sen

20:18

No so far this is straightforward. I'm thinking of writing a proper argument for the construction of $K$

The idea is the following. You take these simplices $\Delta_i$ in $\xi$, and keep track of when the boundary of $\Delta_i$ cancels with that of $\Delta_j$ when applying $\partial$. They must always cancel in pairs

Whenever a pair cancels, glue $\Delta_i$ and $\Delta_j$ togather along these faces.

OK, so this is what it is

$\xi^2$ is a formal expression of the form $\sum \epsilon_i \Delta_i$ where $\epsilon_i = \pm 1$, allowing repetition of simplices - I consider repeated simplices as distinct geometric simplices.

feynhat

> They must always cancel in pairs

What guarantees this?

Can a face of $\Delta_i$ cancel another of its face? (If some simplex in $\xi$ is not a honest 2-simplex but a degenerate one).

Balarka Sen

Apply $\partial$ to $\sum \epsilon_i \Delta_i$ and write it out again as a formal sum of all the faces of $\Delta_i$ with $\pm 1$ coefficients

This is exactly $\zeta$, but in the algebraic expression for $\zeta \in C_1(X)$, some of these faces of the simplices of $\xi$ do not appear, right?

So their coefficients must have summed up to $0$

@feynhat And this happens because whenever a sum of a bunch of $1$'s and $-1$'s is $0$, you can always pair the $1$'s and $-1$'s togather

@feynhat Yep.

It's all a little fidgety to write down, but tell me if the general idea is clear

feynhat

I understand how to glue the simplices together when restriction of $\xi$ on a face of $\Delta_i$ is canceled by $\xi$ restricted to a face of some other $\Delta_j$.

Balarka Sen

Wait hold your horses

There is an issue with degenerate simplices we'll deal with it later

First choose pairs of cancelling signs - it's important to fix a choice

OK, here is how you fix degenerate simplices. I claim there is a choice of cancelling sign such that when you glue, a simplex isn't glued to itself

Example time

Let's say $\Delta_1, \Delta_2, \Delta_3$ are three simplices in the whole decomposition, middle one is degenerate

They all have positive signs

Ah, but I can just have a single simplex mapping to a cone

Fine, it will be a delta complex

@feynhat Yeah I guess you can't fix the degenerate crap. Choose cancelling pairs of signs, and glue accordingly, do nothing to faces which do not cancel with anything.

You'll get a delta complex $K$ this way

Are you convinced lol

feynhat

20:42

34 mins ago, by Balarka Sen

According to what Mike said, $\xi$ can be realized as a map $\xi : K \to X$ from a $2$-dimensional simplicial complex $K$ "with boundary" $\partial K$ (consisting of open edges) such that $\xi | \partial K = \gamma$

We are here ^ right?

Balarka Sen

Right

$\partial K$ is union of the open edges

My next claim is for the exact same reason Mike said $K$ is a manifold with boundary. Because interiors of faces are obv manifold points, points on the edges are manifold points if they glue to some other edge (they always glue in pairs by construct) and on the open edges they are bd points of the mfld

feynhat

Sure.

Balarka Sen

OK!

So I have a map $F : M \to X$ from a compact manifold with boundary such that $\partial M = S^1$ and $F | \partial M = \gamma$, my loop

By classification, $M = \Sigma_g \setminus D^2$ where $D^2$ is a small disk on it.

No way

But I want you to tell me what the element in the boundary of $M$ as a loop in $\pi_1(\Sigma_g \setminus D^2)$ looks like

feynhat

element in the boundary of $M$?

@BalarkaSen $S^1$ that bounds that hole in $\Sigma_g\setminus D^2$.

Balarka Sen

the boundary is a circle

its a loop

the loop gives an element in the loop group

aka the fundamental group

feynhat

20:59

I dont know.

Okay wait.

Can I deform it to wedge sum of circles?

Transcript for

♦ $Mathematics$ random cohomology for quantum nerds

Transcript for

♦ random cohomology for quantum nerds

♦ $Mathematics$ random cohomology for quantum nerds