Hi @iwriteonbananas

Balarka!

Whutz happening?

Nothing much. Studying vector bundles.

'atta boy

:P

What about you?

You know more about VB's than me :(

Doing homotopy colimits and model cat's

@iwriteonbananas I doubt this.

But I plan to know more than you :)

If you don't already, you soon will

What's homotopy colimit again? The infinite mapping cylinder construction?

It's a puffed up version of colimits

Right, I think we have the same things in mind.

E.g. to construct the homotopy pushout of $X\leftarrow A\rightarrow Y$, we would replace both maps by cofibrations via the mapping cylinder and then take the normal pushout

(glue the front of X_{i+1} x [0, 1] to the end of X_i x [0, 1] by the i-th structure map)

That's not quite the construction but i think the two may be homotopy equivalent

Hmm

We got some higher dimensional simplices in the general construction

Can you explain your construction so that a mere mortal like me can understand?

Maybe, let's see

So for instance if you have a diagram $A\stackrel{f}{\to} X\stackrel{g}{\to} Y$, you can ask what the homotopy colimit of it is. You said you want to form the double mapping cylinder

I.e. $\operatorname{cyl}(A\hookrightarrow \operatorname{cyl}(f))$

The thing is, that diagram forms a 2-simplex: we also have the composite map $g\circ f$

did you really mean to place the arrows in $A \to X \to Y$ the way you wrote it?

Yes

ok, let me read it then

For the morphisms $f$ and $g$ we get two internals $\Delta^1$ in your double mapping cylinder

ok, got it, go on.

but since the diagram can be arranged in a triangle with the composite $g\circ f$, the general construction also gives a $2$-simples $\Delta^2$

glued to the double mapping cylinder

in some weird way

but the whole thing deformation retracts onto the double mapping cylinder anyway I think

how do you take the information about the 2-simplex formed by the diagonal inside your space?

To the double mapping cylinder, we glue the mapping cylinder of the composite $g\circ f$

and then fill someting out with a 2-simplex

somehow it's hard to visualize

for me at least

huh, I see

A big advantage of homotopy colimits as opposed to colimits is that a natural transformation between two diagrams which is a levelwise weak equivalence yields a weak equivalence of the two colimits.

You've seen this before in disguise

@iwriteonbananas If your diagram is just a sequence of arrows then yeah Balarka's version is a perfectly good realization of the hocolim.

@iwriteonbananas ah, ok, interesting

@MikeMiller Right, it is homotopy equivalent to the one I'm thinking of bc we can just deformation retract the "higher homotopies" to points

Into non-being, which is to say, everything.

Actually I think as long as your diagram is a tree (maybe I should say as long as it's contractible?) you just need to replace each morphism by a cofibration and then take the colimit.

The fancy stuff is for hard diagrams where that's not so easy.

@Bala, how do you feel about Number Theory? (Same question for @Mike if he wants)

@Krij It's pretty cool

and mysterious

@MikeMiller Ah, good point. Just read about that somewhere

@BalarkaSen Mysterious?

Given a diagram $D$ , you can replace it by another diagram $QD$ such that $colim QD = hocolim D$

and in the case of a tree it's just replacing all arrows by cofibrations

hi. let $S(K)=\sum_{i=1}^K f(i,K)$ for some known function $f$. I am interested in finding the order (big oh) of $S(K)$, like $S(K)=O(K^2)$, but I am unable to find a closed form for $S(K)$. are there some other general techniques for that apart from approximating and bounding the sum with inequalities?

too general

@Krijn Lots of nice theorems of which I either have no personal intuition or beautiful geometric pictures (perhaps naive, but exciting to me).

@Krijn Can you ask a more precise question?

Eg, Fermat's two squares theorem. That's pretty surprising and the picture I have is that preimage of points corresponding to 1 mod 4 primes under the map $Spec \Bbb Z[i] \to Spec \Bbb Z$ (dual to the inclusion) breaks up into two points, so the this map branches away from those points.

(on interestingness)

On the more "advanced" level, it surprises me that the etale cohomology theory, the key to Weil conjectures (the statement of which is less interesting to me, admittedly) satisfies the Lefschetz hyperplane theorem and the Lefschetz fixed point theorem.

@BalarkaSen Yes, I was reading up on that but it's too hard for me

(even though I do not know how to define the etale cohomology theory)

I wrote a small paper on it for my end project on Alg Geom

Got an 10/10 for it, but that was ridiculous

More of an 8/10

I'll read it if you can link an online version of it, if there is one

Let's see

@Krijn What are you actually asking me?

@Mike Whether or not you like Number Theory?

@BalarkaSen Does this work? drive.google.com/…

yeah

thanks

@BalarkaSen I think you'd like paragraph 3.4 the most

@Krijn Sure.

@Krijn Looks good. I have got it saved, I'll read it later and get back to you.

@Kaumudi The sign depends "mostly" on $a$, because that controls end behavior.

it depends on all of $a,b,c,x$

@Balarka So have you cracked open anything new yet?

that depends on the sign of $D$

well yes that too

either

so what's the sign of $D$ ?

What do you mean by "the sign" of $y(x)$ anyway

at a single point?

in general?

you can adjust $C$ to make $y$ everywhere positive or negative, yes

what do you mean it doesn't depend on $c$

so what is the sign of $D$ ???

what is your textbook saying again

and it also depends on $x$

which are .... ?

it depends

maybe you can

and maybe you can't

first

isn't $y(x)$ dependant on $x$

so

whay would you call the case where the sign of $y(x)$ is independant of $x$

I think you are not realising at all how ABSOLUTELY CRITICAL the sign of $D$ is in order to answer or not answer the question

Then you have a bit of a problem I'm afraid

because if $D > 0$ then the sign of $y(x)$ depends on $x$

for example, with $y(x) = x^2-4$

$y(0) = -4$ which is negative

and $y(3) = 5$ which is positive

so you can't say "well I know the sign of $D = 16$ and I know the sign of $c = -4$ therefore I can deduce the sign of $y(x) = ???$ for all $x$"

because whatever you put in ??? makes your sentence false for some $x$

well if $D$ is positive, knowing the sign of $c$ or $a$ isn't going to help you much because the sign of $y(x)$ ACTUALLY depends on $x$

it will be negative for some $x$ and positive for some other $x$

however

if $D$ is negative, you can do things

wtf am i looking at

in case $D < 0$, then $y(x)$ has no root, which means that the sign of $y(x)$ does not depend on $x$

For example pick $y(x) = -x^2-3$

well $y(0) = -3$ which is negative

$y(1) = -4$ which is also negative

$y(-3) = -12$ which is also negative

in fact in this case, $y(x)$ is negative forall $x$

in this case,

the sign of $y(x)$ is the sign of $y(0)$ forall $x$

and $y(0) = c$

incidentally, the sign of $c$ is also the same as the sign of $a$

which is the sign of $y(x)$ as $x \to \infty$ in general

since $\Delta = b^2-4ac$ and $b^2 > 0$, if $\Delta < 0$ then $ac$ has to be positive, which means that $a$ and $c$ have the same sign

@Danu: Regarding surjectivity of $L$, you're on the right track, but don't use just $\Lambda^{n-k}\alpha$. Throw in some $\star$s as needed in the formula.

Hi @mercio

hi ted

do you think there is a nonconstant bounded real sequence $(a_n)$ such that $\forall n \in \Bbb N, a_n = \sum_{k=0}^\infty 2^{-(k+1)} a_{n+2^k}$ ?

@MikeMiller Learnt the S-W obstruction for immersion of a manifold inside a Euclidean space of a specific dimension. I want to figure out how S-W classes give cobordism invariants (?) before reading it from the book.

g'day @Balarka

Hi @TedShifrin!

@TedShifrin Oh... Okay.

I guess I should've seen that.

@mercio: I have no earthly idea. Is that formula self-consistent? I.e., if you have $a_n$ satisfying that, does it work for $a_{n+2^\ell}$ for any $\ell$?

so @Kaumudi, out of the $16$ sentences of the form "if (a or c) (< or >) 0 and D (< or >) 0, then y(x) (< or >) 0 forall x", which are the true sentences ?

@Danu I guess I don't know how to respond to that other than to say, "uh huh." :D

[still have to work it out]

@TedShifrin I'm not sure what you call self-consistent

@TedShifrin Actually, I'm not sure. How do $*$'s not spoil the bidegree?

Oh, @mercio, I sorta see what's going on. (Well, I said what I meant.) So an easy solution is something like $a_1=a_3=a_5=a_9=\dots=1$, but then I can take $a_2=a_4=a_6=a_{10}=\dots=3$, take the next undefined one and start there with a different constant.

@Danu: You have to work with different spaces of harmonic forms. You start by doing $\star\alpha$ (so we're in a different space ... so what, still harmonic) ...

if the recurrence relation only shifted the index by an even number you could have trivial counter examples like this yes

But you're shifting only by powers of 2 !!

but also by the $0$th power of $2$

which is not even

$a_n = a_{n+1}/2 + a_{n+2}/4 + a_{n+4}/8 + \ldots$

@TedShifrin But I have to find an $\beta$ in a fixed space of harmonic forms to map to my given $\alpha\in \mathcal H^{n-k+p,n-p}$

OK, mercio, you don't need to club me over the head. I see.

@Danu, you'll end up using $\star$ twice and end up in the right space.

@Balarka Calculate these things for surfaces and see what happens.

Or learn connections

@TedShifrin It's clear to me that I'll have to use $*$ 2 times, if at all, but I don't see how it's going to help anything. The only thing to try is of course $*\Lambda*$ up to sign, and that's useless because that's jut $L$. And I'm trying to not take $L$ of $\alpha$, but rather find something that, when I apply $L$, goes to $\alpha$.

I understand that.

I have to make a phone call. I'll be back in a bit.

Everything else in the section on Hodge theory went smoothly, at least...

Hey all, I have a bit of a basic question regarding dimensional analysis and reciprocals. Depending on how the equation is set up, the answer might be a bit different (e.g., 0.5 miles/gallon vs 2 gallons/mile). Is it fair and precise to say these are equivalent (\equiv), or is there a better way of indicating their relationship?

So really there is only one other thing one can try namely $*L^{n-k}*$. At least it's in the right space, and it's up to sign equal to $\Lambda^{n-k}$, hence harmonic... But I'm running in circles.

well it's weird when you say it this way

because someone will come along and see "it depends only on $a$" and he'll think "so it doesn't depend on $c$ ?"

where $c$ can be is $a$'s place just as well

if $D < 0$ then $a$ and $c$ have the same sign and $y(x)$ always has that sign too

I'm not seeing any way to get anything out of $\Lambda, L,*$ that is (i) in the right space (ii) not actually $\Lambda$ or $L$ itself, up to sign @TedShifrin.

Hiya @MikeMiller

hi

@MikeMiller How's your morning?

It's okay. I'm confused.

I'm also confused.

What's confusing you (my confusion is extensively chronicles already ;D)?

I can define like four different kinds of gauge group in this context and I'm not sure which one I should be using.

@MikeMiller Different kinds?

Also welcome back @Ted :)

Yes?

@Danu: Sorry about the delay. First on the phone, then thinking about your quandary. If we know finite-dimensionality, injectivity both ways ($L$ and $\Lambda$) gives you isomorphism. But I think all that's going on here is linear algebra again: Since $\Lambda$ is the adjoint, we're back to $T$ injective $\implies T^*$ surjective.

G'night, @MikeM.

@TedShifrin But I want $T$ surjective

Do I have injectivity of $\Lambda$?

Let me check

Aha. Yes.

Should I be paying attention to this?

No.

@Mike No.

This is totally useless :P

Okay.

But tell me about your gauge groups---perhaps writing it out can be useful to you, too

(unless you don't have time)

I think it all follows from the representation theory in Chapter 1, anyhow, @Danu. Not sure why he doesn't tie back into that and say that $\mathscr H(M)$ becomes an $\mathfrak{sl}(2,\Bbb C)$ module.

@TedShifrin Your curly H is.. the combined spaces of harmonic forms?

Yeah. Pick any one you want.

By any one, you mean what? You need all $(p,q)$ for the action of $L,\Lambda$ not to take you outside of course

You mean w.r.t. $\partial,\bar\partial$ or $d$? I know those coincide

Well, I'll stick to all, then. :)

@Danu It's unlikely that writing it out will help me. I have a connection $A_0$ on a pullback bundle over $Y \times \Bbb R$ that is the pullback of some connections on $Y$, $\eta_{\pm}$, near $\pm \infty$. I want to define the space of $L^2_{k,\delta}$ gauge transformations of this bundle (roughly, those s.t. all their first $k$ derivatives decay faster than $e^{-\delta |t|}$). These will have well-defined limits as gauge trasnformations over $Y$ as I go to $\pm \infty$.

The limits must necessarily be in the isotropy groups of the gauge group on $\eta_{\pm}$. So that leaves me a few flavors: the gauge transformations that go to the identity; those that I place no restrictions on; and those that go to the identity at $+\infty$ but not $-\infty$ (or vice versa).

@MikeMiller OK. That sounds... sorta-within-my-range-of-understanding

@MikeMiller Ah, okay. Boundary conditions!

Isn't letting it go to the identity everywhere the obvious nice choice? :P

Hell if I know.

If you have no clue, then perhaps just go with the most general approach initially, until you run into trouble?

Needing to vent: Absolutely astounding.

@Danu No, precisely one of those will be the one that works. I just need to work out which I want.

@TedShifrin wat^wat

I knew all this like two months ago. :(

(Vote to close the question!)

Does it at least have some close votes?!?!

If not, I think someone should post on meta to establish whether or not this should be accepted.

Seems way outta acceptable range to me.

I've fought this battle before and I'm fed up. But thanks for reminding me to vote to close. I rarely do that.

Can I vote to close discussion on our national election, too?

@TedShifrin It's seriously important to be vigilant in that respect.

Voting to close is the most serious form of quality control the site has.

I can't believe the chutzpah of somebody to come post it and say it's an exam question.

@TedShifrin lol. Now you're tempting me...

@TedShifrin Yes, amazing.

There is absolutely no incentive to do so

Anyhow, are we done with that question?

Ehm, I'm sorta okay with just saying: Both $L,\Lambda$ descend to operators on $\mathcal H^{\bullet,\bullet}(X)$ and then invoking that they give rise to an sl(2) rep to say that they're bijective.

OK. But I think it's a pointwise application of the adjoint thingy.

Move on!

Moving on. Hodge decomposition has super nice consequences.

Ayup.

Mhm.

I love these simple proofs like "use Hodge to say $\alpha=\bar\partial\gamma$, then use Hodge again to say $\gamma=\partial \beta+\partial^*\beta'+\beta''$" and then you'll see that $\alpha=\bar\partial\partial\beta$

(I'm talking about the $\partial\bar\partial$-lemma)

Now I'm into the section on Lefschetz theorems

Hmmm, but that shouldn't need such a sledgehammer ... it should be local.

Relationship between Pic(X) and H^{1,1}

@TedShifrin But the local version doesn't require Kahler

The global requires the Hodge decomp

Huybrechts even has a remark on it

Ah, right, because the local is sheafical — you have to shrink open sets.

Glad you're way more on top of it than I am :)

I'm just repeating what I'm told

That's fine :)

Repeat till you live it :)

hey @TedShifrin ...

I starting to like analysis a lot btw

is analysis used in algebraic geometry ?

@TedShifrin ?

You don't need to keep pinging me, Karim.

If you do stuff NOT algebraically, but with differential geometry and analysis, yes.

cool

I like both ways it is very enlightning

I don't understand how some people are able to remember so many theorems and relationships/identities

Using them a lot and understanding rather than memorizing.

I wonder sometimes if I had done number theory in grad school if I would have actually been successful. I have a tough time with algebra. But I don't know if that's just because I haven't spent enough time on it, or if I just have a brain that's wired incorrectly.

My brain has never been naturally algebraic. I found it amusing that my first textbook was algebra.

I struggle with algebra too

Hardly.

Yeah but who the hell remembers stuff like $\nabla\times\left(\mathbf{A}\times\mathbf{B}\right)=\mathbf{A}\left(\nabla\cdot‌​\mathbf{B}\right)-\mathbf{B}\left(\nabla\cdot\mathbf{A}\right)+\left(\mathbf{B}\c‌​dot\nabla\right)\mathbf{A}-\left(\mathbf{A}\cdot\nabla\right)\mathbf{B}$

Not I.

If I used it every day for a month, I'd know it after a week or two.

No need to memorize such things. Just have them at your fingertips if you need to use them.

@Lozansky That looks like a flavor of Leibniz rule, though I have no idea what you mean to do when you just put two vectors next to each other.

Where do I do that?

$A(\nabla \cdot B)$?

Oh, nevermind.

:P

I'm not sure what the difference is between $B(\nabla \cdot A)$ and $(A \cdot \nabla) B$.

The latter is differentiating $\mathbf B$. The former is $(\text{div}\mathbf A)$ times $\mathbf B$.

I hate this notation. But sure.

Suffix notation is much better

Ugh.

Most things are better without reference to a specific coordinate system.

Mm.

But sometimes one has no choice.

Why do I always feel dumb when I study math...

Same reason the rest of us do, probably.

@MikeMiller did you do number theory in grad school ?

I'm in grad school. I applied to do number theory, but ended up pretty quickly becoming a topologist/analyst/differential geometer instead.

I'm taking a number theory class this quarter.

Number theory is cool

I am interested in number theory I am gonna take it next year.

Was it Gauss that said it's the queen of math?

.....

I have to show that each simple group of order less than 100 must have order 60.

a lot of work

(nitpick, but you need an extra adjective there)

What's a quick proof for $\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$?

I'm pretty sure this has been asked several time on the main site

I am skeptic if there is a quick proof. I like the Fourier/complex analytic proofs.

I first learned the fourier one when i was in high school, and then I found the one from euler with the factorisation of sin as an infinite product

(i asked my high school teacher about it and he gave me the fourier proof as an exercise)

Cool

I wish we had a high school teacher like that

he was great

The reason I don't like Euler's proof as much is that it takes hard work to justify his proof.

Hey, @Balarka and @mercio :-)

Just sayin', because I had to go through that hard work a few days ago while learning Hadamard factorization.

Oh yeah, Apostol's proof using $\displaystyle \int_{[0, 1]^2} \frac{1}{1-xy} dV$ is great. Completely forgot about that one.

42 answers ....

Hi @Kari

@mercio Did you previously go by a different name?

on this website long ago yes but then I think I forgot my password or something and eventually the accounts got merged

in particular I don't know you by a different name

I don't see anything wrong with that

Is it obvious that $H^{0,2}(X,\Bbb C) = H^2(X,\mathcal O_X)$?

@Danu Dolbeaut theorem

But D theorem is about the cohomology of forms, no?

What is $\Omega^0$?

Herpaderp

I feel like my overall reasoning abilities are going down :P

That's why you do exercises

At least I didn't raise a stupid objection earlier with Ted

Well, I do.

You mean like deadlift?

Too much time pressure, Balarka :(

@Lozansky :P

I have like 2 more weeks for a hundred pages

Plus I don't know for sure yet that my topic will be complex geometry

If it's confirmed I might go through the notes while doing some exercises

Seems like a good idea actually

It does.

Complex geometry is hard.

Something I've wondered: Let $X$ be a smooth simply connected complex surface (projective, if you like) with $b_2$ sufficiently small. (For instance, $b_2^+ = 1$ and $b_2^- = 9$ is too large.) But if $b_2$ is small enough, is $X$ necessarily diffeomorphic to the "standard" complex surface of a fixed $b_2$? That is, if $b_2^+ = 1$ and $b_2^-$ is small - I dunno, smaller than 9? - then is it diffeomorphic to $\Bbb{CP}^2 \# k\overline{\Bbb{CP}^2}$?

I have a feeling $b_2$ means something other than the singular betti number.

It means the same thing.

Um, then what are $b_2^+$ and $b_2^-$?

I do wonder why Freedman's theorem does not help, in any case. Is the corresponding linear algebra statement unhelpful?

Oh, you said diffeomorphic, not homeomorphic.

@Balarka Maximal dimension of a subspace on which the intersection form is positive resp negative definite.

Ah, got it.

Interesting question (I do remember there being an exotic $\Bbb P^2 \# 9\Bbb P^2$).

@Balarka You forgot a bar. The constructions in smaller $b_2$ seem to be symplectic, not complex, so I'm curious if they can't be complex. IIRC if a symplectic manifold has $b_2=1$, it must be isomorphic to one of the standard structures on $\Bbb P^2$.

hey guys, require some help with method

I've found that 12k satisfies this, however I can only achieve this through trial and error, is there a more convinent method I can use?

@MikeMiller Thanks, indeed I did. Googling gives this paper which seems to claim there's an exotic P^2 # 3 \bar P^2

Yes, but not a complex one!

Ah, I see. I am reading your question correctly bit-by-bit by missing the adjectives you have mentioned and saying stupid things, sorry about that.

anybody able to help me

@sylent is that k = 12 or for all multiples of 12?

all multiples of 24**

that are also natural

or maybe its 12.. im not sure, one of them

How did you test for this?

did you really compute $(\frac{1}{2}(1 + i\sqrt{3}))^{12}$?

somehow yes

somehow meaning I am looking back at my work and its just numbers everywhere but yes

@sylent I think it's useful to recognize $(1 + i\sqrt{3})^a$ won't return a real number for any $a$ that is odd.

but honestly I don't know lol

yea

$(1+i\sqrt{3})^k = 2^k$ and then try even numbers.. hmm

and $i^a$ will only return $1$ for $a$ being a multiple of 4

@sylent en.wikipedia.org/wiki/Binomial_theorem maybe this'll help?

our teacher did say matrices can help

@sylent calcul.com/show/calculator/binomial-theorem If you didn't try using a calculator maybe this will help you test different cases.

Hey guys! If I have that $X_i$ is an independent RV (but not iid) in $S_N=X_1+X_2+...+X_N$ where $E[X_k]=m^k$, $m\neq 1$, $S_0=0$ and $N \sim Po(\lambda)$, how can I show that $E[S_N]=\frac{m}{m-1}(e^{\lambda(m-1)}-1)$? I can only figure out how to do this for iid RV..

hey

Hello!