Let's put it this way: Right now, I wouldn't be able to prove the crucial theorems :P

Anyways, I have a semi-specific question if you're up for listening

So we had this complex $i\Omega^0\to i\Omega^1\times \Gamma(V_+)\to i\Omega^2_+\times \Gamma(V_-)$, which as fin. dim. cohomology and index/Euler characteristic $-1/4(c_1^2(L_{\cal S})-\sigma(X))$ where $L_{\cal S}$ is supposed to be this characteristic line bundle.

I don't find that very convincing, because to be able to prove the crucial theorems in full detail you need the mass of analysis in Donaldson-Kronheimer.

Sure. You're doing Seiberg-Witten theory, I forgot.

Yeah, right. Not the $SU(2)$ stuff

So if you have the SW configuration space, the space of solutions $Z_\omega$ is identified with $f_\omega^{-1}(0)$, where $f_\omega(A,\Phi)=(F^+_{\hat A}-\sigma(\Phi,\Phi)-\omega,D_A^+\Phi)$ (praying to god that this terminology is standard)

I wouldn't call it the "SU(2) stuff". They're different theories, not the same theory with a different group.

But sure, yes.

@MikeMiller Physicist's mindset... :P

They're literally different theories.

Denoting the cohomology of that complex, for fixed $(A,\Phi)$ (which define the maps) by $H^i_{(A,\Phi)}$, we then saw that if $H^0_{(A,\Phi)}=0=H^2_{(A,\Phi)}$, then a neighborhood of $[A,\Phi]\in\cal M_\omega=Z_\omega/\mathscr G$ is a smooth manifold of dimension equal to the index of the complex.

sure

Now, we briefly discussed the case $H^0\cong \Bbb R$.

Then $H^1$ is of dimension index+1 (actually it's always -index, but okay)

And the lecturer asserted that the neighborhood of $[A,\Phi]$ is now a quotient of a smooth manifold of dimension= dim $\cal M_\omega+1$ by a $U(1)$ action

1 - index, yes; you're taking an alternating sum of the complex

But I didn't quite get that last part

I mean, I see how there's a $U(1)$ coming from the constant maps into $S^1$

But a priori this acts on... just the configuration space. How does it descend to... where do we really want it to descend to?!

@Danu What you do is consider the quotient of $\mathcal Z_\omega$ by a smaller group, the "based gauge group", where you demand that the automorphism is trivial at some point; you prove that this actually acts freely (straightforward).

You have a short exact sequence $\mathcal G_b \to \mathcal G \to U(1)$, where the last term is the value at the point $b$

So when you mod out by $\mathcal G_b$ you're left with some object with a $U(1)$ action

@BalarkaSen Sigh. My laptop lost power last night. So bye-bye hydra game :(

maybe it finally cut it off.

we'd never know anymore

probably not, but who knows

Help me out in this

@MikeMiller Sorry for going AWOL, I biked home.

I'll allow it

So you're viewing $\mathcal G$ as the group of automorphisms of... which principal $U(1)$-bundle? Sorry if this is a stupid question

what do you think $\mathcal G$ is?

ah, you're thinking of it as maps to $U(1)$

Yeah

what is $A$ a connection on? $L_S$?

yeah

...or... Urgh I keep on messing this up

there's $A$ and $\hat A$

one is a connection on $L_S$, the other is a spin-c connection on the spinor bundle

One is on $L_S$

you may as well think about the one on $L_S$

yeah exactly

yeah, sure

@AntonioVargas Here's something fun. I'm at a talk on numerical optics stuff, and one of the posters talks about orthogonal polynomials

you consider $\mathcal G$ as the unitary automorphism group of $L_S$, which turns out to be the same as the set of maps to $U(1)$. then it acts on $(A,\Phi)$ by $g(A,\Phi) = (g^*A, g\Phi)$, where you literally multiply the section by a complex number, and pull back the connection by an automorphism

(their advisor is giving a talk right now, actually. stream is here, if you're curious: ima.umn.edu/IMA-live-stream)

$(g^{-1})^*A$ to make it a right action, but yeah

(orthogonal polynomials haven't shown up yet.)

@Semiclassical ah that is interesting

fine

I plan to ask them if they've done any Riemann-Hilbert stuff if I get the chance :>

@Danu I'm not sure why you want it to be a right action

How do OPS show up for them? I don't know much optics but I assume they use their vector space properties mostly?

but in any case just do it consistently

oh you said they haven't mentioned them in the talk yet, right

@MikeMiller I also don't know... It's how we defined a principal bundle in my course.

Yeah, and I couldn't stare at the poster long---I noticed it just as the talk was about to start.

I find it funny how politicians use "we" haha. In contrast to the "we" in typical math' texts.

@Danu That's fine, but this is a group action, not a principal $\mathcal G$-bundle

They are talking about Laurent polynomials right now, but I dunno if that's where it'll lead :)

We're having $\mathcal G$ act on the configuration space of connections on L_S x sections of spinor bundle

My guess, though, is that their zeros will correspond in some fashion to the eigenvalues of a spectral problem

@Semiclassical actually I'm really interested in laurent polynomials, do they use their zeros?

not yet. But check out the stream if you're curious.

link?

I gave it above

oh I see it above

thx

Is the anti-derivative of the Riemann zeta function 0?

no.

@Semiclassical How do you know?

Because that's constant?

^this

@MikeMiller Right, yeah.

Anyways, we defined it as a right action, no matter.

fine

So the fact that those automorphisms can just be viewed as smooth functions is just because $U(1)$ is commutative right

yes

@Null abstract: "We present a method of burying capitalism"

normally you can view them as a subset of smooth functions which satisfy some conjugation property

doesn't sound quite right, normally they're sections of an automorphism bundle

@AntonioVargas The relevant paper is here: arxiv.org/abs/1307.6838

whose transition functions are given by conjugation, but conjugation in $U(1)$ is trivial

Right, okay

(you're probably thinking of them in terms of functions on the total space of the principal bundle, but that's not quite what you want here)

you've got it exactly right

Now recall that the way this action is defined (I'm going to act on the left, sorry bud) is $g(A,\Phi) = (A - g^{-1}dg, g\Phi)$. Suppose $g(x) = 1$. Then $A - g^{-1}dg = A$ iff $dg = 0$, and that's only possible if $g=1$

So the group of maps to $U(1)$ with $g(x) = 1$ - the based gauge group - acts freely on the configuration space

in general the stabilizer of the whole group is trivial if $\Phi \neq 0$ and $U(1)$ if $\Phi = 0$

Right, I know that

great

@AntonioVargas Oh, hah. I just looked at this prof's website. Take a look at the entries under Math Methods: math.lsu.edu/~shipman/research.html

ah there you go

Yeah.

So the based gauge group is obtained by fixing a point $x$ and then imposing $g(x)=1$ @MikeMiller, and $\mathcal G/\mathcal G_b\cong U(1)$

yup

Wish I could watch more of the talk but I have to get back to work

Could someone take a brief look at my the following homotopy problem? Let $f \colon X \to X$ be a homotopy equivalence and $\alpha$ a path from $x_0$ to $f(x_0)$. Is the self-map of $\pi_1(X,x_0)$ given by $[\gamma] \mapsto [\alpha * (f \circ \gamma) * \overline{\alpha}]$ an automorphism? I very strongly believe so and am trying to prove it, or did I get something wrong here?

@MikeMiller Thanks. So I don't need a $U(1)$ action on the actual cohomology spaces?

Basicly, we know that $\pi_1(f,x_0)$ is an isomorphism $\pi_1(X,x_0) \to \pi_1(X,f(x_0)$ if $f \colon X \to X$ is a homotopy equivalence. The question is, does this still hold if we use $\alpha$ to make the map a selfmap of $\pi_1(X,x_0)$?

@Danu I don't understand the question.

The cohomology space $H^1$ here I think should be identifiable with the tangent space at a point after quotienting by the based gauge group. Thus at a fixed point, it inherits the action of $U(1)$.

Knowing the action on the tangent space tells you the local structure of the moduli space.

@MikeMiller Hmm... quotienting out the image of $L_{(A,\Phi)}$ which maps $\xi\in i\Omega^0$ to $(-d\xi,\xi\Phi)$ is quotienting out the based gauge group?

(I may have messed up that second entry of $L$... I don't remember)

is there an Euler product for the $\eta$ function?

@Danu Yes - recall that we're working at a reducible

So $\Phi = 0$

Give me a second to think

Yeah, so you should be able to prove that the tangent space to the zero set of SW at a reducible (that is, the linearization of the SQ eqns) is preisely the kernel of the second operator, so you want to understand in terms of the equations what the first operator does

Sorry, I'm being a bit slow right now

@Danu Can you remind me what the second operator is again?

The linearization of the equations

I guess the key formula I don't remember is what $D^+_{A + \eta}$ is in terms of $D^+_A$ and $\eta$

I'm struggling with the exact same problem right now -_-

Left my notes at the office.

I can't remember two terms

Look in Morgan's book

the bottom right is his term for the Dirac operator, the top right is his term for $\sigma(\psi, psi)$

@MikeMiller Yeah, so it sends $(a,\varphi)$ in the tangent space to $(2d^+a+\sigma(\Phi,\varphi)+\sigma(\varphi,\Phi), D^+_A\Phi+\gamma(a)\cdot \Phi)$

I must've messed somethign up.

@Sophie Well, we have $\eta(s)=(1-2^{1-s})\zeta(s)$ and $\zeta(s)=\prod_p(1-p^{-s})^{-1}$

@Danu Yikes

I think the second entry is right, at least.

Perhaps I'm just fine. Morgan seems to agree.

No extra term

is $\sqrt[0]{5}$ simply 5?

@Sophie So I guess it would be the same product except the $p=2$ factor is different

Well, can I just say that you should try to prove that $H^1$ at a reducible is the same as the tangent space to the zero set mod based gauge group

wait

And you figure out the details? :)

(Identifiable as spaces with $U(1)$-action!)

@Null That's $5^{1/0}$, so

i see @AkivaWeinberger thanks

@MikeMiller I think this is essentially the question I came to you with to start with!

But along the way of this loop you've helped me with some stuff regardless. Thanks!

Do you know Steven Sivek?

He was here to give a talk on Tuesday, it was very nice! Incredibly inspirational.

I might try to start building towards Floer homology in the coming months... Or I might stress otu about my thesis.

@AkivaWeinberger isn't there a way to express it as $\prod_p (1-\omega_p p^{-s})^{-1}$?

Not sure

the $2^{-x}$ dictates the $w_2$ term has to be $-1$ but then the $4^{-x}$ has to be 1

@abenthy That's literally composing $\pi_1(f)$ with the basechange isomorphism $\pi_1(X, f(x_0)) \to \pi_1(X, x_0)$ arising from $\alpha$ right? Each of them are isoms, hence so is the composition.

@Danu But this is just a computation - identify appropriate tangent spaces with the spaces involved in the cohomology groups

As in L should literally just be the differential of th map G -> A

@Sophie $(-1)^n$ is not a completely multiplicative function of $n$, so it cannot be expressed as $\prod_{p^r\| n}\omega_p^r$ for any set $\{\omega_2,\omega_3,\omega_5,\cdots\}$ (where $p^r\| n$ means "$p^r$ is the highest power of the prime $p$ dividing $n$")

in general, $L$-functions can have Euler factors that are more complicated rational functions of $p^{-s}$ than simply $(1-x)^{-1}$.

in this case, one absorbs the $(1-2^{1-s})$ that Akiva mentioned into the $(1-2^{-s})^{-1}$ term

You can put yourself before a typewriter, but that doesn't make you a writer. Similarly you can put yourself before mathematical exercises, but that doesn't make you a mathematician.

@Danu Yes, Sivek is a very good speaker

Understand the 4-manifold stuff really well, including Sobolev spaces, before trying the 3-manifolds. i syggest Morgan

Salamon also has an extremely extensive set of notes

Sivek is very lets say optimistic.

You mean about future work?

I mean about the relationship between the topics hes working on and big open problems.

I don't mean that he oversells his work.

More like he oversells his subfields.

It isn't necessairly a bad thing.

He's just willing to make more non-concrete connections to something like SPC4 (or slice-ribbon) than most.

does $(na_n)_n$ mean every term of the sequence $(a_n)_n$ is multiplied by $n$?

Yup

Kinda

@PVAL Ah yeah that's fair. I have my own conspiracies about those I guess but I wouldn't share.

Hey there, i'm looking to understand better the Glicko rating system. Is there a simplified version of it that is match by match and not period per period ? Like when 1 match = 1 period ?

@PVAL-inactive Hmm. In the talk I saw he stuck to the particular thing he proved a theorem on, which was the existence of a nontrivial homomorphism to $SU(2)$ from the fundamental group of a $3$-manifold (this is the paper).

@MikeMiller Thanks for the recommendations.

@Danu we'be both seen the paper - he was speaking more generally

hi

@ZachHauk ih, woh era uoy?

ih @Null doog, tsuj tog kcab morf loohsc

did uoy nrael gnihtemos gnitsretni ereht?

epon. P:

Hey !

yeH

Simple question

Can I apply the Lagrange Remainder to an interval ?

I have to know what's the error of a Taylor Polynomial when evaluated in x = [7,9]

suppose I have a function $f(z)=\sum_{n=0}^\infty a_nz^n$ and I'm interested in the rough location of its zeroes,

is there a way I can tell where they are?

I don't think so. You might think of $\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$ and it seems completely unexpected that the zeros are integer multiples of $\pi$

and if it is transcendental chances are it is either $\pi$ or e, so we're halfway there

@ZachHauk consider the Mercator series for the logarithm. All the coefficients are rational but $1$ is a root

@Sophie That's a very vague question. What do you mean by their location? You can for example tell how many zeroes there are inside a preferred loop by argument principle. You can also tell how they grow if you keep making the loop larger.

i suppose its not a polynomial because its an infinite series

Eg, if order of growth of your function is atmost $\ell$, the number of zeroes inside a circle of radius $r$ is at most of order $r^\ell$. That tells you how sparsely they are distributed.

yeah these series aren't elements of $\Bbb Q[x]$

sorry

other than that im not sure what to say

i mean, it depends on the series

@MikeMiller Okay.

does $\frac{1}{n^nsin(n)}$ converge by some theory? what about even bigger factors before sin?

@Null what do you mean by converge?

does $\lim\limits_{n\to\infty}\frac{1}{n^nsin(n)}$ exist?

Yes. Squeeze it.

sheesh, the doubly-graded stuff is maddening

@BalarkaSen why can we say that? sin(n) will eventually be very close to 0. how can we say that any $n$ dependant term will be big enough?

sin(n) will not eventually be very close to 0. it'd fluctuate between -1 and 1 forever

it's just that $n^n$ is too big for that to have any effect. $-n^n \leq n^n \sin(n) \leq n^n$. Reciprocate: $-\frac1{n^n} \geq \frac{1}{n^n \sin(n)} \geq \frac1{n^n}$.

@BalarkaSen What's doubly graded?

Now let $n \to \infty$. Both the two sides go to $0$, so does the middle.

@Danu Spectral sequences!

@BalarkaSen wow, that squeezing made sense to me

Ohh, exciting!

Is it hard? :\

people say so

(i think so, $\leq$ should have been used, but whatever)

I thought you meant you were studying them

I just started reading about them

Aigh't

If there are some insights that are easy enough to explain, I'm all ears!

I'm reading some Riemannian geometry now---Petersen's book. I want to learn something about the relationship(s) between Betti numbers and curvature conditions

I think it's a sort of generalization of cellular homology, at least spectral sequence of a filtration

but I am not sure if it's the right way to think of 'em

@Danu kewl

@BalarkaSen On a more basic note, Petersen blew my mind with his great exposition of the Lie derivative.

To just think about it as the first order term in a suitable Taylor expansion... Really good stuff.

That sounds so generic that it can be applied to any derivative concept---but it's really nicely worked out in the book.

I guess I think of it as an infinitesimal flow which goes along $X$, then $Y$, then backwards along $X$ and then backwards along $Y$

that's precisely what flowing along the Lie derivative does I think

I think so too. But I think this way is much more intuitive

So what Petersen does is the following

You know that for a function, it's the directional derivative.

So in analogy with that ("that" being $f(\phi^t(p))=f(p)+t(L_X f)(p)+o(t)$, where $\phi$ is the flow of $X$)

you want to write something similar for the Lie derivative of a vector field, but the problem is that the vector field evaluated at different points doesn't live in the same space (different tangent spaces)

So what do you do? You push it back with the inverse flow, obtain a path in $T_pM$ (fixed $p$, now!) and do what you did before

i.e. $D_{\phi^t(p)}(\phi^t)^{-1}Y(\phi^t(p))=Y(p)+t(L_XY)(p)+o(t)$

ah ok

the notation is a bit ugly like this: Streamlining it a bit for clarity you have $D\phi^{-t}(Y\circ \phi^t)=Y+tL_XY+o(t)$

This is such a beautiful idea

The formula for $(0,k)$-tensors is then also immediate (now you pullback with $\phi^t$ instead of pushing forward with its inverse)

@Balarka I don't think that's a good way of explaining it - it's about approximating the differential, one bit at a time

And for instance the proof that $L_XY=[X,Y]$ can also be done completely using this way of thinking about it (no need to use the limit definition)

It just turns out that when calculating cellular homology there are no terms "higher" than 0th and 1st order, so you get a complex instead of a spectral sequence

@MikeMiller How is the approximation improved in each step?

SSes are pretty easy once you understand their origins

@Danu At each stage you allow the part of the differential that decreases filtration by k.

(or rather, "what are the steps"?)

What does "decreases filtration" mean?

Start by learning what a filtration is

I was hoping you'd tell me

Nope.

OK, got it. Seems easy enough (wikipedia)

@MikeMiller right, agreed.

If you want an introduction look at Hutchings' very brief notes

Tim Chow has a note too, but it's rather algebraic

I mean, it's algebra

Eh, true enough. :/

the sore on my face is really becoming a bother. i really hope it's not a cold sore

@BalarkaSen you mean inflaming?

yeah

it's a very tiny blister which popped up a day ago and it's hurting bad.

why would a cold sore be bad?

@Bala Just noticed your message about Eliot

Nice

@Null Because they are literally caused by herpes simplex.

Oh my, what have you been up to

not much. fiddling with spectral sequences

That gives you herpes now?!

Jesus, I need to get checked

lol

yeah, i am sure it's gonna be worse for you

infinity categories are definitely worse than spectral sequences

gonna head to bed for today. will fight with the convergence to page $\infty$ (which seems like the whole point) thing tomorrow.

'night

Have a good night

So I was told that $O(\sum^{log_2(n)}_{i = 0}2^i) = O(n)$. I'm confused. How?

That $O$ is for the big-$O$ notation.

Isn't $\sum_{i=0}^m2^i=2^{m+1}-1$? Then just plug in $m=\log_2(n)$.

@SalehenRahman

$a^{\text{log}_a(b)}=b$?

@Null @AkivaWeinberger yeah, that's right. Just checked with wolfram.

Thanks.

log$_a$(x) means: the exponent which is neccessary to get x mmh

dunno to word this xd

if log$_a(x)=c$, then $a^c=x$?

appearantly

what happens if you take every unprovable statement, choose each to be true or false and take them all as axioms?

i guess you get a lot of contradictions then ;)

so you will dismiss at least some axioms again

but if you take an unprovable statement as an axiom, isn't the new system consistent iff the old one is too?

if you restrict it to conjectures probably. But some conjectures where disproven, so...

(and disproving an axiom means that the system was inconsistent)

@Sophie don't take me super serious as i don't know what i talk about :P

I just proposed taking an uncountably infinite number of statements as axioms. That's pretty silly

@Sophie So you're saying you take your list of axioms, find something unprovable with them, add it to the axioms, find something unprovable in the new list of axioms, add it to the axioms, etcetera?

yes

I'm not really sure why I would want to do that

I think that, if you do it right, you end up with is an infinitely long list of axioms such that (a) nothing is unprovably and (b) no algorithm can tell you whether something is an axiom of not

You can't have (a) without (b), I think, if your system can do arithmetic

I think that's Gödel's incompleteness theorem

"no algorithm can tell you whether something is an axiom of not" is a scary statement

Like, sure, you could also take the language of Peano Arithmetic and make every statement that's true about the naturals an axiom.

But if I give you a random sentence, how do you know if it's an axiom or not?

It's an infinitely long list, and the only rule for it is "if it's true it's an axiom".

if it's not true its negation is an axiom

that looks like a setup for the world's most devious counterexample

procrastination keeps an ace up my sleeve lol

The proof of Gödel's theorem fails in such a case (where there's no algorithm that describes the axioms) because the proof relies on using natural numbers to represent proofs, and there's no way to tell whether a natural number represents an actual proof or not.

axiom: $a$ and $\neg a$ can be true at the same time

(The proof, if I recall correctly, is that you also use natural numbers to represent statements, and then you find a natural number $n$ that represents the statement "For any $x$ that represents a valid proof, the proof represented by $x$ does not prove the statement represented by $n$")

is there such a thing as an empty statement?

how long does that hydra game should take? Hercules has cut 40 thousand heads so far

(Essentially, it's a way to express "This statement is unprovable" in PA.)

@Sophie Dunno, probably lots and lots of digits if I had to guess

it's the new cookie clicker

The version with the dire edges, by the way, is much worse than the original hydra game

I know how to prove that the original hydra game can't last forever (no matter which heads you cut off) but not how to prove the dire edge version

are you listening to anything interesting?

The AC in the room I'm in is making a noise

better than a fly I suppose

hi chat

well i suck at vectors holy heck

so confusing