random cohomology for quantum nerds

12:50 AM

rip with a capital b

Daminark

I'll just black box this for now and figure it out later

Now coboundaries

1:12 AM

Alrighty. First G-P exercise.
A smooth function on $\Bbb R^k \subset \Bbb R^l$ is a function $f : \Bbb R^k \to \Bbb R^n$ such that for every point $x \in \Bbb R^k$, there is an open set $U \subset \Bbb R^l$ and a smooth function $F : U \to \Bbb R^m$ such that $F|_{U \cap \Bbb R^k} = f$. But $F$ restricted in this way is essentially just the first $k$ coordinates of $F$, and this is clearly smooth in the usual sense in $\Bbb R^k$, as it has continuous partial derivatives of all orders.

Conversely, suppose $f$ is a smooth map on $\Bbb R^k$ itself. Then the map $f \times \text{id} : \Bbb R^k \times \Bbb R^{l-k} \to \Bbb R^l$ is smooth (again, having continuous partial derivatives of all orders), and so for each point $x \in \Bbb R^k$, the whole space constitutes a nbhd on which you have a smooth function that restricts to $f$. So $f$ is smooth on $\Bbb R^k$ considered as a subset of $\Bbb R^l$.

I feel like I might be missing something somewhere here.

Alex

@BalarkaSen Do you know much about triangulated categories?

2 seems easier. Let $f : X \to \Bbb R^m$ be smooth, and let $x \in Z$. There is a nbhd $U$ at $x$ such that a smooth map $F : U \to \Bbb R^k$ exists where $F|_{U \cap X} = f_U$. But then $F|_{U \cap Z} = f_{U \cap Z}$, so that $f|_Z$ is a smooth map.

lots of typos in that but I think the sense of it is clear

1:45 AM

3 is giving me some trouble

I'm not sure how to "glue" the information of the smooth maps

Alex

Which text @Fargle?

Guillemin-Pollack.

I think I have something here tho

The problem asks to show that for $X,Y,Z$ subsets of Euclidean spaces, if $f : X\to Y$ and $g:Y\to Z$ are smooth, then their composition is smooth

So let $x \in X$; there is a nbhd $U$ such that there's a smooth function $F : U \to \Bbb R^m$ (the ambient space of $Y$) yadda yadda.

There is also a nbhd $V$ of $f(x)$ such that there's a smooth function $G : V \to \Bbb R^l$ (the ambient space of $Z$) yadda yadda yadda.

$f$ is smooth, hence continuous--that means $f^{-1}(V)$ is open, and $F|_{U \cap f^{-1}(V)}$ is smooth. Furthermore, this restriction can now be composed with $G$ to give a nbhd $U \cap f^{-1}(V)$ of $x$ such that there's a smooth map $G \circ F|_{U \cap f^{-1}(V)}$ for which $(G \circ F|_{stuff})|_{stuff \cap X} = (g \circ f)|_{stuff \cap X}$.

3:43 AM

Alright, number 4:
(a) The function is clearly smooth on its domain. Its inverse is $y \mapsto \frac{ay}{a^2 + \|y\|^2}, also clearly smooth. Bada bing bada boom.

(b) Given $x \in X$, $U$ nbhd of $x$ gets mapped by $\phi$ to $V$. There is a ball $B$ in $V$ containing $f(x)$ since $V$ is open, and this pulls back to an open set containing $x$--since $\phi$ is a diffeomorphism, we conclude that $\phi^{-1}(B)$, a nbhd of $x$, is diffeomorphic to a ball $B$, which is diffeomorphic to $\Bbb R^k$.

Oops, didn't even check the TeX. That's $y \mapsto \frac{ay}{a^2 + \|y\|^2}$

@Alex Vaguely

9:37 AM

@Alessadro u done w exam

I have an algebraic number theory presentation to give next Friday

that'll also be my last exam

ahh

cool

(I should also finish writing my thesis at some point)

is your thesis on lagic

yeah, model theory to be precise

9:44 AM

very cool

I want to focus on model theory and set theory during my master

I know nothing about that particular discipline of mathematics

I should, someday, learn something about it

LoL looks like the modules handbook for next year was written by a mathematician

LOL

How is it going with interviews and stuff by the way? When will you know where you're going to study next year?

9:54 AM

I have the interview on 21st

I think I will know by a few days after that if I'm going to get in

What about the other place (or places? I remember only two) you applied to?

I applied to two, I got in one of them, of which I will be taking the interview

I didn't get in the other

Ah, I see, that sucks

Eh its ok

I don't really know much about it but the Indian admissions sound crazy

9:58 AM

It has been driving me quite insane for the last few months. But I have sort of stopped caring now

Whatever shall happen will happen

10:10 AM

I see, good luck!

thanks

12:53 PM

OK, I'm going to rant about function spaces and technicalities about the topology on function spaces and how classical result follows from such conversations

spicy

The only reasonable function spaces are Banach, all else is garbage

That is correct, they will be Banach.

Good, so it's not too bad

1:08 PM

$M^m, N^n$ be smooth manifolds. Denote $C^k(M, N)$ to be the space of $C^k$-functions from $M$ to $N$ and $C^\infty(M, N) = \bigcap C^k(M, N)$ consists of smooth maps from $M$ to $N$ (so they are $C^k$ for all $k = 1, 2, \cdots$).
Denote by $\mathscr{G}(M, N) = \bigcap \mathscr{G}_p(M, N)$ to be the space of germs of functions, where $\mathscr{G}_p(M, N) := \{f \in C^\infty(U, N) : p \subset U \subset M \text{is open}\}/\sim$ where $f \sim g$ if say $f$, $g$ are defined on some neighborhoods $U$ and $V$ of $p$, and $f|_W = g|_W$ for some smaller open set $W \subset U \cap V$ containing $p$.

Now we can start putting topologies on these objects.

The topology on $J^k(M, N)$ comes from pulling back the topology (and also the smooth structure) by the local trivializations of $p^k$, where $J^k(\Bbb R^m, \Bbb R^n)$, which is a space of tuples of polynomials, is given the Euclidean topology on the coefficients of the polynomials.

There is a natural morphism $C^k(M, N) \to J^k(M, N)$ by sending a $C^k$ function $f : M \to N$ to it's $k$-th order Taylor polynomial in the $k$-jet space, which let us denote as $J^k f$. This is the "k-jet prolongation/extension of $f$"

We define the weak Whitney topology on $C^k(M, N)$ by setting the basic open sets up as $\mathcal{V}(K, U) = \{f \in C^\infty(M, N) : J^k f(K) \subset U\}$ for any compact set $K \subset M$ and open subset $U \subset J^k(M, N)$.

If $M$ is noncompact, a better choice is the strong Whitney topology, where the basic open sets are $\mathcal{V}(\{K_i\}, \{U_i\}) = \{f \in C^\infty(M, N) : J^k f(K_i) \subset U_i\, \forall i\}$ where $\{K_i\}$ is a family of compact subsets of $M$ and $\{U_i\}$ is a locally finite atlas on $J^k(M, N)$.

If $M$ and $N$ are closed manifolds, then as proved in this paper, $C^k(M, N)$ under the (weak or strong, they agree if $M$ is compact) Whitney topology forms a Banach manifold.

1:28 PM

In the immortal words of one of my analysis professors, "That's good. draws smiley face on the board We like that."

The topology on $C^\infty(M, N)$ is given by the "limit topology" coming from the $C^k$ Whitney topologies. Namely, we have inclusion maps $C^\infty(M, N) \hookrightarrow C^k(M, N)$ for all $k$. Give $C^\infty(M, N)$ the topology given by union of the topologies of $C^k(M, N)$ under these inclusion maps.

Alternatively just add "for all $k$" in each of the definitions above.

Here's a classical result in differential topology. First a definition: If $M, N$ are two closed smooth manifolds and $W \subset N$ is a submanifold then a smooth map $f : M \to N$ is said to be transverse to $W$ if $df_p(T_p M) + T_{f(p)} W = T_{f(p)} N$ for all $p \in M$. (If $f(p)$ lies outside of $W$, then this is vacuously satisfied). We say $f \pitchfork W$.

Theorem: If $f : M \to N$ is so that $f \pitchfork W$, then for any smooth homotopy $h_t : M \to N$ of $f$ for $t \in [0, 1]$, there is an $\epsilon$ such that for all $t < \epsilon$, $h_t \pitchfork W$. Transverse maps are stable under perturbation.

Here's a proof sketch using these technical topology on function spaces business. Consider the subspace $\mathscr{R} \subset J^1(M, N)$ of $1$-jets $(x, f(x), L(x))$ where $f : U_x \to N$ is a function defined near some chart around $x \in M$ and $L\in \text{Hom}(T_x M, T_{f(x)} N)$ is a linear map, such that this 1-jet is transverse to $W$, ie, $L(T_x M) + T_{f(x)} W = T_{f(x)} N$, then

$\mathscr{R}$ is a closed subset of $J^1(M, N)$. This is easy to see; say $(x_i, f_i(x_i), L_i(x_i))$ is a sequence of 1-jets in $\mathscr{R}$ converging to $(x, f(x), L(x))$. Then $L_i(T_{x_i} M) + T_{f_i(x_i)} W = T_{f_i(x_i)} N$ for all $i$.

But by choosing a smooth path $\gamma$ containing $x_i$ for all $i > n$ for some large enough $n$, and trivializing $TM|_{\gamma}$, we can realize this situation as a sequence of subspaces $(P_i, Q_i)$ of $\Bbb R^n$ such that $P_i \pitchfork Q_i$ for all $i$ and $P_i \to P$ and $Q_i \to Q$, but that of course implies $P \pitchfork Q$ because transversality of subspaces of a vector space is clearly stable.

That wasn't correct, I got confuzzled. The complement $\mathscr{R}^c$ is a closed subset of $J^1(M, N)$, consisting of 1-jets $(x, f(x), L(x))$ such that $L(T_x M) + T_{f(x)} W \neq T_{f(x)} N$. Replace all the $=$'s in the last three messages by $\neq$. It is easy to construct a sequence of transverse subspaces of $\Bbb R^3$ which converges to non-transverse subspaces (take a 2-plane and a line in $\Bbb R^3$, and at each stage reduce the "angle" between them to $1/n$)

Non-transversality is a closed condition on linear subspaces.

1:51 PM

what's the idea behind transversality

To clear the confusion: I'll state it. The complement $\mathscr{R}^c$ consisting of 1-jets not transverse to $W$ is a closed subset of $J^1(M, N)$.

@Fargle Guillemin-Pollack. It's all there :)

:^|

nah you right

Anyhow, so $\mathscr{R}$ is an open subset of $J^1(M, N)$. But that means the set $\mathscr{T} = \{f \in C^\infty(M, N) : J^1 f(M) \subset \mathscr{R}\}$ is an open subset of $C^\infty(M, N)$, as the weak $C^\infty$ topology consists of union of weak $C^k$ topologies for all $1 \leq k \leq \infty$. But $\mathscr{T}$ is exactly the set of smooth functions $f : M \to N$ such that $f \pitchfork W$, by definition of $\mathscr{R}$.

So the set of smooth functions transverse to $W$, which let's call $C^\infty_{\pitchfork W}(M, N) \subset C^\infty(M, N)$ is an open subset. By the paper sited above, $C^\infty(M, N)$ is a Banach manifold, hence for any $f \in C^\infty_{\pitchfork W}(M, N)$ we can choose a Banach chart $\mathscr{U} \subset C^\infty_{\pitchfork W}(M, N)$ containing $f$.

If $h_t$ is a smooth homotopy of $h_0 = f$, then there exists an $\epsilon > 0$ such that for all $t < \epsilon$, $h_t$ are all contained inside the Banach chart $\mathscr{U}$. Therefore $h_t$ for all $t < \epsilon$ are also transverse to $W$, as desired.

@BalarkaSen Correction: No, some $k$ where $1 \leq k \leq \infty$. It's the union of all the topologies.

@MikeMiller Do you want to check if this proof of the stability theorem works :3

@Fargle I was writing so was trying to hold off an exposition on transversality. Linearly, it's easy to describe: If $V_1, V_2 \subset W$ are two subspaces of a vector space, they are said to be transverse if $V_1 + V_2 = W$. Simple as that.

2:07 PM

I see

If a thing cuts another thing by slicing

Not tangentially

For manifolds, it's an easy follow-up of the linear definition. $N_1, N_2 \subset M$ two submanifolds are transverse if for any $p \in N_1 \cap N_2$, $T_p N_1 + T_p N_2 = T_p M$.

The x-axis and the y-axis in $\Bbb R^2$ are transverse, but $y = x^2$ and the x-axis are not.

oh that makes perfect sense

nice example

Yup it's a very intuitive thing

seems like it extends easily enough too--e.g. $S^2$ and a coordinate axis are transverse, but $S^2$ and a tangent line are not

Careful

2:15 PM

oh

yeah

I see the problem

What is the problem? State it.

Alright I still believe the first part of the sentence I said

The problem is that you haven't stated where they are transverse inside of

The normal line plus the tangent plane make 3-space

ah right

they're transverse as submanifolds of $\Bbb R^3$

Correct, they are transverse in $\Bbb R^3$

Not in $\Bbb R^4$ though

2:18 PM

right

plane plus line $\neq$ 4 dim

2 + 1 $\neq$ 4 quick maffs

that property is a uckas

lmao

alright, so under your latter formulation, it looks like transverse manifolds have to add up to at least the dimension of the space

Indeed

2:21 PM

I guess my question now is, where is that sum of tangent planes taking place? Inside of $\Bbb R^n$, the ambient space of the manifold $M$?

No, inside $T_p M$.

$N_1, N_2$ are submanifolds of $M$

Oh right right

If $p \in N_1 \cap N_2$, then $T_p N_1, T_p N_2 \subset T_p M$ are subspaces

So for example, two (non-parallel) planes are transverse in $\Bbb R^3$ as well

Yes.

2:22 PM

I just wanted to make sure there wasn't a more rigid condition there I didn't see

"Transversality of manifolds" is a meaningless term. Transversality is an adjective of submanifolds (well, you can slightly generalize it to maps and submanifolds, but whatever)

In the same way you wouldn't call a group normal

Yep

Here's the magic of transversality. If $N_1, N_2 \subset M$ are transverse submanifolds, $N_1 \cap N_2$ is also a submanifold of $M$, with $\text{codim} \, N_1 \cap N_2 = \text{codim}\, N_1 + \text{codim} \, N_2$.

So what does it mean to say a map is transverse to a submanifold--how does that translate into this definition?

Oh I buy that

@Fargle There's no hurry to learn that stuff.

2:25 PM

Fair enough

It's all in G-P, you can read at your leisure. Really one should understand it for submanifolds first.

(that probably sounds a bit dejected but nah srsly, fair enough--I trust your judgment)

Yeah I just don't want to rewrite Guillemin-Pollack in this chat while you already have the book, is my standpoint :P

I think I've done enough of the first section's exercises to feel justified in moving on

Better to talk about ideas togather

2:26 PM

yee

Non-transverse submanifolds can have totally bad intersection

You can find two submanifolds of $\Bbb R^2$ intersecting at a Cantor set.

>:c

@BalarkaSen what

that's just disgusting

Even a fat Cantor set

2:28 PM

intervals are just cantor sets in progress

That's the kind of bullshit you expect from real analysis, not geometry!

That's why geometers and topologists invented transversality

Transverse intersection is always nice

You go think about non-transverse nonsense, analysts

Sounds like a great solution to me

I really liked doing the exercise for stereographic projection

@Fargle Did you work out the bump function?

That's probably all you need to do from section 1

2:30 PM

oh I need to do that?

shit

Yeah those will be essential throughout GP

I worked out that the $e$ thing is smooth--that's a classic example of a smooth non-analytic function

Yup

but that functional equation scared the bejesus out of me

Good work

It looks scary, but isn't afair

3:00 PM

Okay I've finally convinced myself it's zero outside $(a,b)$

Nice

3:11 PM

but why the gosh dang heck do it gotta be positive

oh is there a typo in my edition

oh, yep, just found Ted's errata

Check here

snip

yeah so it had $g(x) = f(x-a)g(b-x)$

and, like, no

I'm not doing that

but $g(x) := f(x-a)f(b-x)$? ez

Alright, so a function that's $1$ in the ball of radius $a$, $0$ outside the ball of radius $b$, and in between in the in between: Let $g^*(x) = f(\|x\| - a)f(b - \|x\|)$. Then define $h$ as before but as a quotient of line integrals of $g^*$.

Or I guess alternatively just compose $h$ with the magnitude function.

Oh no that gives me something inside out

oops

oh I guess I can just do some $1 - h$ business

3:45 PM

oh snap, tangent space is well-defined

@Balarka In class we defined the tangent space of an abstract smooth manifold at a point $p$ as the space of derivations on the germs of smooth functions at $p$, but I find this definition a bit unsatisfactory because it's using the smooth structure while it looks like a $C^1$ manifold should be enough to define tangent spaces

It doesn't!

How is the tangent space of a $C^k$ manifold, $1\leq k<\infty$ defined?

You can't do the same derivations definition

You'll get an infinite dimensional vector space if you do that for $C^k$ with $k < \infty$

@Alessandro Let me recall the point. So the definition is that, the tangent space of a $C^\infty$ manifold $M$ at $p$ is the space of linear functionals $\mathcal{G}^\infty_p(M) \to \Bbb R$ on the space of smooth germs at $p$, which satisfy the Leibniz rule (these are derivations), yes?

@BalarkaSen I'm not even sure how do that for $C^k$ manifolds, "germs of smooth functions at $p$" doesn't seem to work at all here

@BalarkaSen Right

3:57 PM

You can look at germs of $C^k$ functions at $p$.

So $A_k = \text{Der}(\mathcal{G}_p^k(M), \Bbb R)$

If $k = \infty$, we know $A_\infty = T_p M$.

Right

So the problem, if I recall correctly, is that $A_k$ is infinite dimensional for $k < \infty$.

Let's see. I think the issue stems from the fact that $\mathcal{G}^k_p(M)$ is not a Noetherian ring for finite $k$.

Yeah, of course it isn't.

Hmmmm, I'm looking at the proof that the derivations in $A_\infty$ induced by a chart in $p$ are a basis that we did in the $C^\infty$ case and I can't pinpoint where smoothness is used and being $C^k$ isn't enough

You still get an injective map $T_p M \to A_k$ by using a tangent vector as a differential operator, like if $X \in T_p M$ then the differential operator is $X(f) = D_X f$, directional derivative of a germ at $p$ along $X$.

But this would not be surjective

How do you prove surjectivity? That every derivation appears as a tangent vector?

You need the soupled up Taylor expansion for that (every smooth germ $f$ can be written as, in local coordinates $(x_1, \cdots, x_n$), $f = f(0) + \sum_{i = 1}^n x_i g_i$ for smooth germs $g_i$), and for $k$ finite you reduce the regularity of $g_i$ to $C^{k-1}$, which fuckducks the proof

That's what we did, but it seems to me that $g_i$ is still $C^k$, wait a minute I'll check it again

4:09 PM

$g_i$ are the partial derivatives of $f$

Ah, right

ok I'm convinced now

So the key point is $\infty - 1 = \infty$

:3

quick maths

Now here is the real point, which is really algebraic of nature

$\infty = \infty + 196883$

deep fact about monster group or something

4:11 PM

So we need a completely different construction in the $k<\infty$ case?

$\mathcal{G}^k_p(M)$ are really in bijection with $k$-order Taylor polynomials with remainder in $x_1, \cdots, x_n$ at $p$ for some choice of local coordinates $(x_1, \cdots, x_n)$ at $p$. That remainder term, however, can be totally irregular and is an element of $\mathcal{G}^0_p(M)$, the space of continuous germs at $p$.

So there is this kind of decomposition "$\mathcal{G}^k_p(M) = P(k, n) \bigoplus \mathcal{G}_p^0(M)$" (I don't know if this is the correct algebraic description)

And when you pass to derivations, intuitively, you get a contribution from $\text{Der}(\mathcal{G}^0_p(M), \Bbb R)$ which is a massive ass space

There are no $C^0$ tangent spaces

I see, that makes sense

That's why $C^k$ derivations fail. The remainder term is bollocks

@AlessandroCodenotti Yeah, but the cool thing is that if a manifold has a $C^k$ structure for $0 < k < \infty$, it has a $C^\infty$ structure

That's a bit annoying

@BalarkaSen what

Yup

4:18 PM

I kinda have to run away very soon but I'll think about what a smooth structure on the unit cube looks like <.<

Where very soon is right now! Bye and thanks for your help!