Mathematics - 2016-01-16 (page 2 of 3)

Krijn

17:01

@iwriteonbananas Elliptic curves, algebraic curves, some homology and cohomology and some algebraic number theory

Although, at the moment, mostly cohomology

iwriteonbananas

@Krijn Cool, cohomology is fun.

r9m

@Superstar I did a rage quit on the series :( not gonna look at it again until I cool down :P

Mike Miller

@iwriteonbananas: It's just what one needs to do to have a derivative

Superstar Monica

@r9m Hey, nice to see you around again! :D You're very powerful at calculation stuff, don't worry about that. Anybody can fail for a while. It's better to put it away for a while, and then return to it.

iwriteonbananas

@MikeMiller right

Mike Miller

17:12

so for instance on the level of functions $\nabla f$ is the Riemannian gradient, meaning you identify the 1-form $df$ with a vector field using the metric; This is not a terribly exciting result but it does tell us that this is basically judt differentiation

iwriteonbananas

What's the Riemannian gradient?

Mike Miller

Now suppose you'd like to take a second derivative. On the form side you might want to say "Well I'll just take $d\omega$. But when you're doing that to forms of positive rank this "more or less obviously" isn't what you want, since a function $f$ could be really wild ($e^z$, say) and its second derivarive $ddf$ should not really be zero.

Superstar Monica

@r9m have you seen my answer above? Initially after I posted it I received 3 downvotes. Then I thought to make a great favour to my downvoters: to also give them the possibility to take a bounty of 100 points from me.

r9m

@SuperstarMonica okay .. I got the idea .. but can't figure out if I am leaving a term somewhere that's creating all the mess .. I'll install mathematica sometime and recheck the whole calculation again :P

Mike Miller

@iwriteonbananas: Do you know how to identify 1-forms with vector fields using the metric?

iwriteonbananas

17:15

@MikeMiller Yeah, generally $TE\cong T^*E$ if $E$ is a bundle admitting a metric, right?

Overflowh

I'm sorry, I have a question.
If the geometrical interpretation of the derivative of a single (real) variable function in a given point $P$ is the angular coefficient of the tangent in $P$, what is the geometrical meaning/interpretation of the derivative of a two (real) variables function? (give both functions being real functions)

Mike Miller

@iwriteonbananas Yes (which is every bundle you'll naturally encounter), but the isomorphism depends on the metric.

iwriteonbananas

Right

Mike Miller

$\nabla f$ just means $(df)^\sharp$ or whatever symbol it is that means I pass to the vector field it corresponds to.

Superstar Monica

:26883681 It was about downvotes on purpose. Now I want them to teach me better solutions and then give them bounties, but it seems they delay with brilliant solutions.

I'm disappointed.

Mike Miller

17:17

Note BTW that if you, like me, prefer forms to vector fields, you could instead consider the LC connection as something that takes sections $T^*M \to T^*M \otimes T^*M$.

iwriteonbananas

@MikeMiller Ok,that's good to know

Superstar Monica

Where are my downvoters to give them bounties? That's the question!

:D

Mike Miller

Now that is the second derivative we want. The reason we would get $ddf=0$ is because we skew-symmetrized in the definition of $d$, and mixed partials are equal, so everything died

iwriteonbananas

@MikeMiller Shouldn't it be $\Gamma (T^*M)\to \Gamma (T^*M \otimes TM)$?

Mike Miller

But $\nabla df$ is really rather rarely zero.

r9m

17:19

@SuperstarMonica sigh.. there's nothing new with these scenarios .. has happened to you in the past too .. why waste time (reputations too? :P )

Mike Miller

@iwriteonbananas: Yiur second term is wrong. A connection is a map $\Gamma(E) \to \Gamma(E \otimes T^*M)$.

iwriteonbananas

Right, whoops

Superstar Monica

@r9m I'm annoyed since this happened after I posted my solution in chat. I shouldn't be annoyed, you're right. My philosophy is that if anybody wants to give me lessons in math to do it, I'm ready to learn, I prefer lessons more than downvotes. :D

user303542

I agree @r9m

iwriteonbananas

Any particular reason you prefer forms to vector fields, by the way?

r9m

17:22

@SuperstarMonica :-) .. more like lessons in 'how to please the spoon feeding league' :P

Superstar Monica

@r9m :-)))

Mike Miller

@iwriteonbananas: Now let me start working abstractly for a second. Given a connection $\nabla_E$ on $E$ on a Riemannian manifold, you get higher connections $E \otimes T^*M \to E \otimes T^*M \otimes T^*M$, eg, given by seconding $\sigma \otimes \omega \to \nabla_E \sigma \otimes \omega + \sigma \otimes \nabla_g \omega$

This means I can make sense of "$\nabla^2 \sigma$"

iwriteonbananas

Ok, I haven't seen that before

Mike Miller

You might have seen $d_A$ - I want to talk about that in a second

r9m

@Khallil yo!

Khallil

17:24

Great to hear from you, @r9m!

We haven't been on together for a while!

How are you?

iwriteonbananas

(haven't seen $d_A$ either :P)

Mike Miller

Anyway, if I have a metric on $E$ and $M$, all of these tensor bundles have a metric, so I can obviously write down $\|\sigma \otimes \omega\|_{C^0}$ - just the max of the fiberwise norms of the sections

iwriteonbananas

sure

r9m

@Khallil good and alive! :) How're things going?

Balarka Sen

@Krijn I am now.

iwriteonbananas

17:25

Why the $C^0$ index?

Mike Miller

That's the name of the norm

Khallil

Not too bad. Just been chilling out, @r9m. Been working but not hard :-)

Mike Miller

Like in $\Bbb R^n$, $\|f\|_{C^0}$ is the sup norm

iwriteonbananas

Oh, i've never seen that...only $\|f\|_\infty$

r9m

@Khallil (y) awesome! :)

Khallil

17:26

Are you in your second semester now, @r9m?

r9m

@Khallil I completed my UG :-)

Mike Miller

@iwriteonbananas: Different contexts. If your functions are continuous, then in the $C^0$ norm limits are the same thing as uniform limits of continuous functions

Khallil

Ah wicked! Are you a working man now or postgrad, @r9m?

Mike Miller

so it's a banach space in its own right

but you have a different sort of limit in $L^\infty$

now I finally know how to define $\|\sigma\|_{C^k}$. It's $$\|\sigma\|_{C^0} + \|\nabla \sigma\|_{C^0} + \dots + \|\nabla^k \sigma\|_{C^0}.$$ This is the notion of $C^k$ norm I've been after this whole time.

r9m

@Khallil waiting for application dates for a masters program again :P (failed in a course in the last UG sem and had to repeat that entire semester)

Mike Miller

17:29

That's what I like about connections. They give me a way, on a manifold or even on a bundle on a manifold, to define the $C^k$ norm. Of course there are a lot more selling points but this is my starter package.

Khallil

Sounds awesome. Wasn't the master in-built into the UG, @r9m?

iwriteonbananas

@MikeMiller Cool, that's quite interesting

Krijn

@BalarkaSen I was wondering, when two spaces have the same homology groups, what does this say about the spaces? What about cohomology for $G = \mathbb{Z}$ or even any $G$?

Mike Miller

Now I'll stop after another minute, but there's another way to extend $\nabla$ to higher things

r9m

@Khallil nope .. 3 years UG course ..

Balarka Sen

17:30

@Krijn It doesn't say much.

iwriteonbananas

I'm all ears.

Mike Miller

You could have written $E \otimes T^*M$ as $\Omega^1(E)$; the usual name is "1-forms with values in $E$"

Krijn

What are easy examples with the same (co)homology groups?

Khallil

Ah I see. Is it uncommon where you are to do a 4 year UG/masters, @r9m?

r9m

@Khallil nope .. but my college didn't have a program like that ..

Balarka Sen

17:31

@Krijn $S^2 \vee S^4$ and $\Bbb {CP}^2$ works.

Mike Miller

then $E \otimes \Lambda^k T^*M$ is called $\Omega^k(E)$, and rewriting my original connection as $d_A: \Omega^0(E) \to \Omega^1(E)$, I can easily extend this to higher forms by demanding the Leibniz rules hold: if $\omega \in \Omega^k(M)$, $\sigma \in \Gamma(E)$, then I demand

Balarka Sen

However, iirc, it is true that if X, Y are simply connected and f : X --> Y is a map inducing isom on all H_n's, then X and Y are homotopy equivalent. I think this was a corollary of Whitehead's theorem.

X, Y CW complex.

iwriteonbananas

@BalarkaSen No, it's not true

Ok, now it is. Fun counterexample: The double comb space has vanishing homology, cohomology and homotopy groups in positive degrees, but it is not contractible.

Mike Miller

$d_A(\omega \otimes \sigma) = d\omega \otimes \sigma + (-1)^k \omega \wedge d_A \sigma$

iwriteonbananas

@MikeMiller Right

Mike Miller

17:35

Just like $d$ is a skew-symmetrized derivative, this is a skew-symmetrized connection derivative

Balarka Sen

@iwriteonbananas :)

Mike Miller

In the case $E=\Bbb R$, trivial connection $f \mapsto df$, you just recover $d$.

Balarka Sen

double comb space isn't a thing, sorry

:D

Mike Miller

@iwriteonbananas: So if I know your type, you immediately want to try to define cohomology with this, yes?

iwriteonbananas

@MikeMiller Sounds good!

Mike Miller

17:37

I have some good news and some bad news.

iwriteonbananas

Ok...

Balarka Sen

Always start with the good news.

Mike Miller

The good news is that $d_A^2$ is a tensor, meaning that $d_A^2 \omega$ depends only pointwise on $\omega$, not locally (like $d$ depends locally on the input $f$, but not pointwise)

The bad news is that it's usually not zero.

iwriteonbananas

That's sad. Why does it matter that $d^2_A$ is a tensor?

(And what does the $_A$ mean?)

Balarka Sen

@Krijn The rationale behind the example I gave you is that cohomology as groups is not sufficient to tell you that the attaching map of the 4-cell in $\Bbb {CP}^2$, $S^3 \to S^2$, is not nullhomotopic.

Krijn

17:41

@BalarkaSen nullhomotopic?

Balarka Sen

However you can show these two have nonisomorphic ring structure.

@Krijn Homotopic to constant map, i.e.

Mike Miller

@iwriteonbananas: the real question is what does $d_A$ mean; that's just what I renamed my connection to, because we're changing our point of view. What you get is not a connection, it's an actual section of a certain tensor bundle

We call it $F_A = d_A^2 \in \Omega^2(\text{End}(E))$ (that's the endomorphism bundle of your bundle). This is known as the curvature of the connection.

iwriteonbananas

Wow

Krijn

@BalarkaSen I see

Superstar Monica

Dancing maths teacher shows off his moves in the classroom

►

iwriteonbananas

17:43

Gosh, we did define curvature in my lecture, but I haven't gotten to that yet. And it involved taking derivative twice.

Mike Miller

Since $d^2=0$ is an instantiation of equality of mixed partials, the curvature o your connection says to what degree mixed partials are not equal with respect to $d_A$

@iwriteonbananas: I assume you defined curvature of a metric, rather than of a connection, yeah?

iwriteonbananas

@MikeMiller No, we did it for a connection

Mike Miller

oh

well there's this thing called the Riemann curvature tensor

have you heard of it?

it's of the form $R(v,w)x \in T_pM$, where $v,w,x \in T_pM$

iwriteonbananas

Yeah, we defined that thing

Mike Miller

well, it also satisfies $R(w,v)x = -R(v,w)x$... which is the same as $R$ being a 2-form with values in End(E)

x is the thing you plug into the endomorphism R(v,w); R(v,w)x is what you get out

iwriteonbananas

17:45

Oh nice

Mike Miller

and indeed, if you take $E = TM$, and $d_A$ the Levi-Civita connection, $F_A$ is the Riemann curvature tensor!

Balarka Sen

@MikeMiller Speaking of, that conservative fields are irrotational follows from the fact that mixed partials are equal for C^2 functions, assuming the other half of (1). So problem (2) is done.

iwriteonbananas

That's quite awesome

Mike Miller

@BalarkaSen: Correct. Any other progress?

iwriteonbananas

I was gonna start working through everything we did on curvature in my course tomorrow; this will surely help put things in perspective.

Mike Miller

17:47

@iwriteonbananas: A couple more quick things

iwriteonbananas

sure

Mike Miller

1) Connections with $F_A = 0$ are called flat connections. On a Riemannian 4-manifold, I can define a Hodge star $\Omega^2(E) \to \Omega^2(E)$. It's also interesting to study anti-self-dual connections, meaning $*F_A = -F_A$. But that's a topic for another day.

Balarka Sen

@MikeMiller I think I have made some on the other half of (1). I'll think about it more carefully after dinner.

Mike Miller

(The cohomologies you get from flat connections are mildly valuable but it's not really worth dwelling on. They're mainly good at telling you about flat connections.)

iwriteonbananas

@MikeMiller Ok, sure

Mike Miller

17:51

2) Given an automorphism $g$ of the bundle $E$, this induces a pushforward of the connection which I'll write as $A$ out of laziness, roughly given by $g A g^{-1} + g^{-1} dg$

this is a pretty darn interesting form for an automorphism to take - not just conjugation, but this weird extra action

(Do not worry about what the formula means too much)

iwriteonbananas

Hmm, ok, I find 2) hard to parse

Mike Miller

but in any case, it's usually pretty common to study connections modulo identifications by these automorphisms. That's because many equations that invoke them - the flatness equation $F_A = 0$, or the ASD equation $*F_A = -F_A$ - are preserved under automorphisms, meaning $F_{g(A)} = 0$ and $*F_{g(A)} = -F_{g(A)}$

@iwriteonbananas: The formula doesn't matter, just that it's weird-looking. It's more to the point that given an isomorphism of bundles (in particular, iso with yourself) you can always push the connection on the first bundle forward, through the iso, to the second

These autos are called gauge transformations, and if there's a gauge transformation taking one connection to another, they're called gauge equivalent

iwriteonbananas

Yeah, right.

Ah, ok.

Mike Miller

Now consider the subset of connections (Which you could, without much harm, topologize) and the subspace of flat connections. Mod out by gauge transformations.

You get a topological space.

What is this space? Well, that's sort of hard to answer. What's easier to answer is: what is the disjoint union of these spaces over all $G$-bundles $E$? (Where $G = SO(n)$ or something, say, to recover oriented real vector bundles of rank $n$)

Fact: $\text{Flat}(M,G)/\text{Gauge}$ - the disjoint union of these over all $G$-bundles over $M$ - is homeomorphic to $\text{Hom}(\pi_1 M, G)/\text{conjugacy}$

we've taken a very differential geometric object and found it's the same as a very representation-theoretic object.

iwriteonbananas

Holy whack, that's hard to fathom

Mike Miller

17:57

This fact led to a LOT of 3- and 4-manifold topology

this fact also causes me much consternation because those spaces kinda blow. usually very singular.

iwriteonbananas

hehe, right.

Mike Miller

there's this thing called the Casson invariant, obtained representation-theoretically... the idea is that $\text{Hom}(\pi_1 M, G)$ is actually obtained as the intersection of two subsets of $\text{Hom}(\pi_1 \Sigma_g, G)$ each isomorphic to $\text{Hom}(F_g, G)$

now one takes $G = SU(2)$ and says "OK, so they don't intersect well, let's perturb them until they do. That is, let's take an intersection number of these subvarieties."

(M here is an oriented 3-manifold. Actually I also demand that $H_1(M) = 0$ but don't worry about that.)

iwriteonbananas

What's $\Sigma_g$ and $F_g$?

Mike Miller

surface of genus g; free group on g letters

iwriteonbananas

Oh, right

Mike Miller

18:02

this comes from a heegaard splitting

iwriteonbananas

Ok

Mike Miller

one of my favorite papers takes this great invariant and says "Well, since this is the same as flat connections mod gauge, which are the critical points of the chern-simons functional, this invariant should really come from gauge theory"

And hence Taubes' "Casson's invariant and gauge theory was born"; and instanton Floer homology was not far off.

speaking of Taubes, I like his differential geometry book a lot

iwriteonbananas

Does it cover the stuff I'm doing right now?

Mike Miller

Yes, but it will end up less focused on Riemannian and more focused on general connections, since those are essential in gauge theory and he is a gauge theorist

It still has quite a lot of geometry in it.

iwriteonbananas

Cool, I'll check it out. I find it useful to draw from multiple sources.

Mike Miller

18:06

me too

when multiple sources exist at least

iwriteonbananas

Luckily I have yet to run into the problem that there don't exist multiple sources

Actually, not true. Lück is the only source for L2 invariants lol

Mike Miller

right

I don't really know your tastes really well. what are you into?

are you an undergrad?

iwriteonbananas

Anyways, I appreciate the lecture. Gives me a fun outlook.

Mike Miller

well, when you want to hear about gauge theory, you're always free to ask

iwriteonbananas

@MikeMiller Yeah, I'm an undergrad. First and foremost, I enjoy algebraic topology. Diffgeo and L2 invariants is fun too.

I'm doing a seminar presentation on either cohomology operations or homotopy groups of spheres next semester, and I'll probably do my undergrad thesis on a topic related to that.

Mike Miller

18:23

@iwriteonbananas: Maybe you can work to understand the recent paper about $S^{61}$.

So next year is your last?

iwriteonbananas

@MikeMiller Indeed. What paper about $S^{61}$ are you referring to?

Mike Miller

The only one.

iwriteonbananas

lol

Mike Miller

They calculate that $\pi_{61}^{st}/J=0$, hence by classical results that the 61-sphere has a unique smooth stucture.

iwriteonbananas

Oh, this must be it arxiv.org/abs/1601.02184

Pretty funny result

Mike Miller

18:28

Not an isolated theorem

this is the last piece in the poincare conjecture puzzle in odd dimensions.

see here for further discussion

iwriteonbananas

Oh, cool

dREaM

19:02

@DanielFischer, what is the difference between scanf("%d",&A[i]) and scanf("%d",A+i) ??

which is better?

Idle001

@dREaM same

dREaM

the computer does the same thing for both?

Idle001

the pointer is actually at A[0], whether u order the processor to move i steps then addressing the variable, or addressing before moving i steps , same thing

but

between these two loops

for i=0->n A[i]

for i=0->n A=A+1 ;*A

second is faster

Mary Star

Which is a perpendicular vector to the tangent developable of a unit-speed curve γ?

dREaM

so scanf("%d",A+i) is slightly faster=

?

Idle001

19:07

@dREaM i wouldnt claim something beyond my knowledge, i see those are same

dREaM

ok, thanks a lot

Idle001

"slightly" i dunno exactly

u welcom

Balarka Sen

Back.

Daniel Fischer

@dREaM Depends. What is A, a pointer or an array? If it's a pointer, both are the same, if it's an array, they are different.

dREaM

I declared A as an array, int A[100] for example.

Daniel Fischer

19:15

@dREaM Ah, sorry, they are equivalent nevertheless, confused some things.

dREaM

oh, ok, thank you very much.

Mike Miller

@BalarkaSen: So why are conservative vector fields the gradient of a function?

Daniel Fischer

But if the compiler isn't completely brain-dead, both ways give the same machine code, @dREaM, no performance difference.

Idle001

@DanielFischer u were assuming the <array> c++ structure ?

Balarka Sen

@MikeMiller Right, I think I have an idea. If $X : U \to \Bbb R^2$ is the conservative field, then $\int_\gamma X$ is independent of the $\gamma$ we choose as long as we fix the endpoints. Thus, choose $\gamma$ so that $\gamma(0) = 0$ (assuming $U$ contains $0$ WLOG) and $\gamma(1) = x \in U$ for some $x$.

I think $\int_\gamma X : U \to \Bbb R$ is the desired function the gradient of which is $X$.

Daniel Fischer

19:19

@Idle001 No, I do only sane languages, so no C++. But I was confused and thought that for an array you'd add i times the size of the array (confuzzled with &A + i).

Mike Miller

Well, one has to do this on each path-component, but otherwise correct.

How would you prove this?

Idle001

@DanielFischer i would recommend dealing with c++ structures in case of huge arrays, their research integrated methods, value-accessing and sorting all optimised.

Daniel Fischer

@Idle001 Heh, C++ is a pretty powerful and versatile language. Unfortunately, it's mind-bogglingly ugly, and all design decisions were to use the worst possible syntax.

Balarka Sen

Well, each coordinate of $\nabla \int_\gamma X : U \to \Bbb R^n$ is $D_{e_i} \int_\gamma X$. This is $\lim_{t \to 0} 1/t \int_{\gamma_i} X$ where $\gamma_i$ is the image of $\gamma|_{[x, x + te_i]}$, right? $e_i$ is the standard $i$-th basis vector. Once I parametrize that, it seems to boil down to derivative of $\int_0^t X(x + se_i) e_i ds$ with respect to $t$ at $0$ by chain rule. Which, by FTC, is the same as $X(x) e_i$.

Mike Miller

Good answer: pick a local form for the paths (ie, straight lines) that makes the derivarive easy to compute.

Balarka Sen

19:28

Well, actually, I missed a subtlety there. For sufficiently small $t$, I can pick $\gamma|_{[x, x+te_i]}$ to be straightlines.

This is because $U$ is open, and there is a ball around $x$ contained in $U$. A ball is convex so we're done.

Mike Miller

That's fine.

Balarka Sen

@MikeMiller Right.

Mike Miller

So can you find an irrotational, not conservative, field on the pinctured plane?

Idle001

@DanielFischer i know what u mean, and any language which is oo is like this, im still half the way exploing its ambiguous corners

Balarka Sen

Not sure. But that reminds me, I was going to ask whether the normalized version of the field you mentioned, i.e., $X(x, y) = (-y, x)$ (which is well defined on the R^2 - 0), is conservative.

Mike Miller

19:32

What's the normalized version?

Balarka Sen

The vectors of $X$ get larger in magnitude as you go father away from origin. What if you normalize so that each vector has magnitude $1$?

Obviously not continuous on R^2, because of the wonkiness at origin. But well defined on R^2 - 0.

In formulas, I mean $X(\vec{x}) = 1/\|\vec{x}\| (-y, x)$.

Mike Miller

Well, whether that's irrotational or conservative seem like good questions for you to answer.

Balarka Sen

Intuition says it's not irrotational. But not sure about conservative. I'll check it, give me a minute.

Mike Miller

Well, you know conservative implies irrotational, so if you find it's not irrotational...

Balarka Sen

It'd have to be non-conservative, yeah.

Daniel Fischer

19:38

@MikeMiller You know, I first misread, "conservative implies irrational" ;)

Mike Miller

@DanielF: Also fairly easy to prove.

@Balarka so?

Balarka Sen

I'm a bit confused. I think I am messing baby calculus.

It says $X$ is irrotational.

Mike Miller

:)

The infinitesimal "twisiness" of $(-y,x)$ comes from the fact that it's increasing on radial lines. When you stop that, it stops twisting.

Balarka Sen

That's weird.

Mike Miller

So is it conservative?

Balarka Sen

19:55

I dunno. Give me a minute to check.

L33ter

I would like to discuss something in algebraic coding theory

maybe I should post it on main

I have something I don't understand

0

Q: Parity check matrix intuition

I am currently learning algebraic coding theory. There is something I would like to understand, so according to my book the way we generate this parity check matrix is that we pick some matrix A and based on this matrix we generate the parity check matrix H as follows H = (A | I). They state tha...

dREaM

@DanielFischer haha

Balarka Sen

@MikeMiller Sorry, I was away.

Er, wait a second, I think I just computed things with $X(\vec{x}) = 1/\|\vec{x}\|^2 (-y, x)$, not $X(\vec{x}) = 1/\|\vec{x}\| (-y, x)$. The latter is actually not irrotational.

I am confused. Why does vectors increasing in magnitude as it comes closer to origin make the field irrotational, but fixed length makes it non-irrotational, both being twisty?

The geometric picture is not clear.

L33ter

20:16

what are you doing @BalarkaSen?

this looks like vector calc

Balarka Sen

@L33ter Multivariable calculus, probably. Who cares? :D

As long as the exercises are interesting...

Mike Miller

@Balarka: It's the first one. And Eh, I'm going off drawings, not much rigor here.

The physicists have a better understanding of the geometry of irrotational fields. I just use them.

Balarka Sen

lol

Mike Miller

You still have not told me if it's conservative

L33ter

it will be conservative if have the integral along a closed path being zero

Balarka Sen

20:23

@MikeMiller Yeah, it's not conservative.

L33ter

I don't think in this case it will be conservative

Balarka Sen

If I go around the unit circle, it's $2\pi$.

L33ter

you can actually just see it from the formula

Mike Miller

Good.

L33ter

interesting stuff

Balarka Sen

20:24

So I guess this thing represents the first de Rham cohomology of $\Bbb R^2 - 0$ or something?

Mike Miller

Now... Can you prove that if an irrotational vector field integrates to 0 along the unit circle, it's conservative?

L33ter

what is de Rham coholomogy

Balarka Sen

@MikeMiller If a vector field integrates to $0$ along a closed loop it's conservative. I can prove it.

Mike Miller

Along a closed loop?

Balarka Sen

i.e., if $X : U \to \Bbb R^2$ is a vector field, $\int_\gamma X = 0$ for any $\gamma$ a closed curve ($\gamma(0) = \gamma(1)$), then $X$ is conservative.

dREaM

20:26

Is anyone here participating in the facebook hacker cup?

Mike Miller

Ok, so now you mean over every closed curve. I just said the circle.

Balarka Sen

Oh, I see what you mean.

Erp. I'll have to think.

I presume the vector field is on some open subset containing the unit circle?

Mike Miller

We're still in the punctured plane.

Balarka Sen

Ah, ok.

It's not obvious to me how to do it. I don't have Green, because domain is not simply connected.

Mike Miller

Don't forget I assumed the vector field was already irrotational.

Balarka Sen

20:37

I'll ponder a bit more.

JeSuis

@DanielFischer Hi, about your construction of $f$, how did you get that $f$ is zero on the diagonal ? I got the fact this $f$ symmetric.

Take a basis $B = \{b_1,b_2,\dotsc\}$ and define $f(b_1,b_1,b_2) = 1,\; f(b_1,b_2,b_2) = -1$ (and correspondingly for the permutations to preserve symmetry), and $f(u,v,w) = 0$ for all combinations of basis vectors that involve any other than $b_1$ or $b_2$, and of course $f(b_i,b_i,b_i) = 0$ for $i\in \{1,2\}$. Then extend $f$ to a trilinear map. Check that $f$ is symmetric and $f(x,x,x) = 0$ for all $x$, but $f\not\equiv 0$.

Mary Star

Does someone of you have an idea about my question:

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Q: The parameter curves are asymptotic curves

I am looking at the following exercise: Let $p$ be a hyperbolic point of a surface $S$. Show that there is a patch of $S$ containing $p$ whose parameter curves are asymptotic curves. Show that the second fundamental form of such a patch is of the form $2Mdudv$. $$$$ I have done the foll...

differential-geometry surfaces curvature

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Daniel Fischer

21:05

@JeSuis Oops, it doesn't, sorry :( With $x = \alpha b_1 + \beta b_2 + y$, we are left with $3\alpha\beta(\alpha-\beta)$, which usually is not $0$.

It works over $\mathbb{F}_2$, but not over general fields.

aaadddaaa

Hey can anyone help me prove that e^(1/x) is not uniformly continuous in (0,inf)?

Mike Miller

@Balarka: My phone will die soon. Feel free to ping with questions or answers.

Balarka Sen

Ok. I don't have ideas on how to solve this one.

So probably it'll take some time to get some ideas.

Mike Miller

@Balarka: Green's theorem doesn't just tell you that integrals of irrotational fields are zero in the plane. It tells you that the integrals of irrotational fields only depends on the homotopy class of the loop.

Balarka Sen

Oh, I guess.

Then the rest follows pretty easily. The only loops in R^2 - 0 which do not bound disk are upto homotopy the ones going circles around the origin. Clearly if the integral is 0 for the loop going once, it's 0 for all of them.

Mike Miller

21:21

:)

Balarka Sen

:( Unhappy about not realizing that earlier.

Daniel Fischer

@JeSuis But from that mistake we learn something. In dimension $\leqslant 1$, we already know $f = 0$, so consider dimension $\geqslant 2$. Suppose the characteristic is $\neq 3$. Taking two linearly independent vectors $b_1,b_2$, we must have $f(x,x,x) = 0$ for all $x = \alpha b_1 + \beta b_2$. Using $\alpha = \beta = 1$ we find $f(b_1,b_1,b_2) = - f(b_1,b_2,b_2)$, and with the above we then get $3\alpha\beta(\alpha - \beta)f(b_1,b_1,b_2) = 0$ for all $\alpha,\beta$.

If there are two different nonzero $\alpha,\beta$, we can choose the coefficients so that $3\alpha\beta(\alpha-\beta) \neq 0$, which forces $f(b_1,b_1,b_2) = 0$. Dimension $2$ is thus done. For dimension $\geqslant 3$, this shows that $f(e_i,e_j,e_k) \neq 0$ is only possible when all three basis vectors are different. Then looking an $\alpha e_i + \beta e_j + \gamma e_k$ gives us $6\alpha\beta\gamma f(e_i,e_j,e_k) = 0$ for all $\alpha,\beta,\gamma$.

If the characteristic of the field is also different from $2$, we have $6 \neq 0$, and hence we must have $f(e_i,e_j,e_k) = 0$. So whenever the characteristic of the field is neither $2$ nor $3$, $f$ must be $0$. As seen above, in characteristic $2$ there can be a nonzero such $f$, and also in characteristic $3$.

Balarka Sen

Interesting how $\pi_1$ comes up from this.

But I shouldn't be excited by subsets of $\Bbb R^2$, because of winding numbers. If there is a generalization to subsets of $\Bbb R^n$, then I guess it'd be rather interesting.

Daniel Fischer

@aaadddaaa How does it behave when $x$ tends to $0$?

Mike Miller

@BalarkaSen: Well, the issue is that the notion of "curl" is harder to write down.

Anyway, you just proved that $H^1_{dR}(\Bbb R^2 \setminus \{0\}) \cong \Bbb R$, or more precisely, \cong $\text{Hom}(\Bbb Z,\Bbb R)$.

Balarka Sen

21:29

@MikeMiller I can believe that. I don't even know what curl means in $\Bbb R^2$.

Mike Miller

In general it is reasonably easy to prove that $H^1_{dR}(M) \cong \text{Hom}(\pi_1(M),\Bbb R)$. The proof is not that far off from what you wrote!

Indeed the only hard part is to find a form realizing any given homomorphism.

Balarka Sen

Oh, I guess that makes sense. Covectors are precisely the 1-forms.

Mike Miller

You'll learn all of this soon enough. Here's some geometry you might not learn from Ted's book.

(Which, BTW, get back to work :D )

Let's rewrite everything as $fdx + gdy$. The irrotational equation (now called being 'closed') is $f_y = g_x$. Students often ask me why we don't talk about $f_x$, too.

Well, there's a pretty natural equation coming out of geometry: $f_x = -g_y$. Putting these two eqns together, does this remind you of anything?

Balarka Sen

Cauchy Riemann equations, I think.

Mike Miller

I say that a form is coclosed if it satisfies that equations. And yes.

It's just slightly off. I've swapped some signs. Indeed $f-ig$ is holomorphic iff $fdx+gdy$ is closed and coclosed - which are the so-called "harmonic forms".

aaadddaaa

21:34

@DanielFischer it goes to infinity

Mike Miller

Now, fact (do not try to prove this): In every cohomology class, there is a harmonic representative. This means that I can add an exact form ($df$ for some $f$) to get a harmonic form.

Balarka Sen

Ah, ok.

I wonder what that means in the singular/simplicial context. If there is even an analogous version.

Mike Miller

Wtf does that even mean

Daniel Fischer

@aaadddaaa And can a uniformly continuous function do that?

Mike Miller

What is "that"

Balarka Sen

21:35

@MikeMiller "What's the correct analogue for that fact for singular cohomology?"

aaadddaaa

@DanielFischer guess not,but doesn't really know how to show it as a proof

Daniel Fischer

@aaadddaaa Guessed right, it can't. As for the proof, look at what the definition of uniform continuity is. What is it?

Mike Miller

@BalarkaSen: Of what fact?

aaadddaaa

for every ε>0 exist δ>0 so that for any |x-y|<δ |f(x)-f(y)|<ε

Balarka Sen

@MikeMiller Of the harmonic form fact you mentioned.

Mike Miller

21:39

I don't see why you would think there would be one. I haven't said anything algebraic.

It is a natural question to ask: If $f-ig$ is holomorphic, is there a harmonic function $w$ such that $w_x = f$, $w_y = g$? (You can verify that if this is true, $w$ is harmonic, meaning $w_{xx} + w_{yy} = 0$. Using the above fact about cohomology classes, we know that if we're simply connected, every closed form is exact; in particular, $fdx+gdy$ is. This answers the above in the affirmative.

But if we're not simply connected, picking some harmonic representative of a non-exact but closed cohomology class, we have a form $fdx+gdy$ such that $f-ig$ is holomorphic, but the form is not closed. So there is not always such a $w$.

Daniel Fischer

@aaadddaaa Okay. Now fix an $\varepsilon$, say $\varepsilon = 1$. So there is a $\delta > 0$ such that $\lvert x-y\rvert < \delta \implies \lvert f(x) - f(y)\rvert = 1$. Can you see how that $\delta$ shows that $f$ must remain bounded as $x \to 0$?

Balarka Sen

I guess there isn't any analogue. The motivation is that whenever I hear something in de Rham cohomology which looks vaguely similar, I try listing it down. E.g., analogue for integration of forms is the pairing $H^k(M) \times H_k(M) \to R$ obtained by letting a cochain eat a simplex. Stokes theorem is that this pairing satisfy $\langle \sigma, \partial \varphi\rangle = \langle \partial \sigma, \varphi\rangle$.

Mike Miller

Well, I still haven't said anything special. I could, but I haven't.

Balarka Sen

@MikeMiller Interesting. Please continue.

Mike Miller

Given a closed smooth Riemannian manifold $M$, I can still define harmonic forms. (I need the Riemannian structure.) These are forms that are both closed and coclosed. In this context it is not hard to see that harmonic forms are exact iff they're zero.

In other words, calling the space of harmonic forms (I have not modded out by anything!) $\mathcal H^k(M)$, there is an injection $\mathcal H^k(M) \to H^k_{dR}(M)$ given by passing to cohomology classes.

Balarka Sen

21:46

What does exact mean again?

Mike Miller

It is $d\omega$ for some form $\omega$.

It is the "conservative" of before, just for forms in arbitrary rank,

Balarka Sen

Ah, ok.

@MikeMiller Very cool.

Mike Miller

Now the previous (hard! this is the hard part of everything I'm doing here!) statement, that every cohomology class has a harmonic representative, says that map is an isomorphism.

Balarka Sen

Oh, true.

Mike Miller

This is an algebraic fact, now. I'm saying there's a special set of classes that you pick out that represent the cohomology.

AKA, you can find the homology at the chain level, without passing to quotients.

This has some advantages. When you learn the setup, it immediately implies Poincare duality. It slightly less immediately implies that the cohomology groups are finite-dimensional. It generalizes wildly to situations where you don't just get these facts from algebraic topology.

Balarka Sen

21:50

I am not sure why you're calling this an algebraic fact. You mean it's not strictly restricted to the setting of forms over Riemannian manifolds?

Mike Miller

No... I mean it's a fact about the algebra of differential forms

That I can canonically pick out a special subspace that represents cohomology, instead of having to take kernels and quotient by image etc

Balarka Sen

Ah, ok.

@MikeMiller This is fun.

Mike Miller

That is special. I know of no way to do this for the singular chain complex (other than for $H_n(M)$).

Balarka Sen

Aw. Too bad.

That's what I was hoping for.

Mike Miller

No, not too bad. It tells you that Riemannian geometry can help you with something more topologically flavored.

If there was nothing new or helpful about forms there would be no value to them.

The other thing to note is that $\mathcal H^k(M)$ depends on the Riemannian structure. Obviously they'll be isomorphic, since it's isomorphic to de Rham cohomology which is an invariant of the underlying smooth manifold. But the actual subspace of $\Omega^k(M)$ of harmonic forms will change.

Balarka Sen

21:53

I guess. This is interesting though. It seems to mean that differential forms are a lot more than just a geometric analogue for singular cochains over smooth manifolds.

Mike Miller

Yes, differential forms are geometry

Balarka Sen

@MikeMiller Right. I agree.

Mike Miller

That you can extract topological data from them is exciting. But they are not a bit of topological data.

Actually, I think if you consider $\Omega^*(M)$ as a commutative differential graded algebra, its homotopy type (defined similarly to the homotopy type of a chain complex) determines the rational homotopy type of the underlying compact manifold

Balarka Sen

wow.

Mike Miller

That is not really a fact about forms.

But I think if you want to do that with an arbitrary space you can't just use the singular cochain complex, I think you need to do your best to mimic forms.

This is still topology; there is a lot of geometry hidden in differential forms, especially given that most all of most any kind of geometry can be phrased in terms of forms.

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