Mathematics - 2016-10-21 (page 2 of 4)

0celo7

2:00 PM

Haha people wrote "see Munkres" in the analysis homework

Faqruddin

Is Ramanujan here

?

Danu

You can ping people by writing @nameofperson

Hi @MikeMiller!

Ramanujan

@Faqruddin hi

Welcome @Faqruddin on this chat room

@DHMO hi

DHMO

@Ramanujan yes

Ramanujan

@DHMO I have a question can I?

DHMO

2:07 PM

@Ramanujan yes

I already said yes

Ramanujan

How to solve it? @DHMO

@DHMO are u here?

@DHMO what?!?!?!?!?

DHMO

Let me think, will you?

Ramanujan

I will,

DHMO

2:25 PM

Hint: $\displaystyle \tan^{-1} \left( \frac{1}{1+n+n^2} \right) = \tan^{-1} \left( \frac{1}{1+(n)(n+1)} \right) = \tan^{-1} \left( \frac{(n+1)-(n)}{1+(n+1)(n)} \right)$

@Ramanujan here

Ramanujan

You did how my math sir do , great

OK , so it will be 4 th option @DHMO

0celo7

@DHMO lol your hint from the other day

a bunch of people tried to solve that problem using that stuff and got destroyed by the prof

Ramanujan

@0celo7 what?

DHMO

@0celo7 oh, that one

sorry :p

Ramanujan

@DHMO iam totally destroy at this problem

(uploading)

DHMO

2:33 PM

@Ramanujan yes

0celo7

@DHMO The actual proof uses Baire category.

DHMO

Baire category theorem

The Baire category theorem (BCT) is an important tool in general topology and functional analysis. The theorem has two forms, each of which gives sufficient conditions for a topological space to be a Baire space. The theorem was proved by René-Louis Baire in his 1899 doctoral thesis. == Statement of the theorem == A Baire space is a topological space with the following property: for each countable collection of open dense sets , their intersection is dense. (BCT1) Every complete metric space is a Baire space. More generally, every topological space which is homeomorphic to an open subset of a...

you should have posted the question here instead

so I could be flamed by the gods such as @BalarkaSen and @MikeMiller before I misdirect you

0celo7

You didn't misdirect me, I had thought of those "naive" proofs already and concluded the prof was trolling.

And he was.

Ramanujan

Balarka Sen

What's the question.

That was to 0celo, not Ramanujan.

0celo7

2:37 PM

There is no sequence of continuous functions converging to the Dirichlet function on $[0,1]$.

Balarka Sen

ah

Mike Miller

Hi @Danu.

Balarka Sen

also, what norm? you mean pointwise, of course, yes?

0celo7

Right, forgot to say that.

DHMO

@Ramanujan hint: the end term of each bracket (1,3,6,10,...) are the sum from 1 to n

Balarka Sen

2:39 PM

I don't know how to do it off the top of my head, to be honest.

0celo7

@BalarkaSen It's hard, I don't blame you.

Balarka Sen

I believe ya.

0celo7

But a bunch of grad students thought they had it, but they ended up showing they didn't understand how $\epsilon$-$\delta$ proofs work.

DHMO

The Dirichlet Function is Nowhere Continuous - Advanced Calculus Proof

►

@0celo7 how did they do it?

0celo7

I.e. that you FIX $x,\epsilon,N$, whatever.

Balarka Sen

2:40 PM

I don't think this can be done by epsilon delta tweakings. Pointwise norm doesn't have good tools to fiddle.

0celo7

@BalarkaSen Yeah. Turns out that a pointwise limit of cont. functions has a meager set of discontinuities.

Balarka Sen

Nice result.

0celo7

but the Dirichlet function is everywhere discontinuous

Ramanujan

@DHMO I didn't get what it mean by sum from 1 to n

DHMO

@Ramanujan for example, 10 is the sum from 1 to 4

> The Dirichlet function can be written analytically as $$D(x) = \lim_{n \to \infty} \lim_{n \to \infty} \cos^{2n}(m! \pi x)$$
Mind = blown

Ramanujan

2:43 PM

So 10 is last term of 4th bracket , right?

0celo7

@DHMO That's not a counterexample btw.

abenthy

Is there a category whose objects are topological manifolds and those automorphisms are the self-homeomorphisms homotopic to the identity?

DHMO

@Ramanujan yes

@0celo7 what are you referring to?

0celo7

@DHMO It does not contradict what I told Balarka.

But maybe you weren't thinking that.

Ramanujan

@DHMO let me do it on my own from now,thanks

DHMO

2:44 PM

@0celo7 What does not contradict what you told Balarka?

Balarka Sen

(also, sorry about saying the adjective "norm" after pointwise when I really meant topology)

SteamyRoot

Can't you just do that by determining the baire class of discontinuities?

0celo7

@DHMO your $D(x)$ expression.

@BalarkaSen I knew what you meant.

DHMO

@0celo7 Didn't say it does. I just mean that I am fascinated by the cleverness of whoever came up with that thing

SteamyRoot

Like, Dirichlet's discontinuities are baire class 2, but the discontinuities of the pointwise limit of continuous functions is class 1?

0celo7

2:45 PM

@SteamyRoot Yes, but that's nontrivial.

DHMO

oops, I just spotted a typo in my equation

SteamyRoot

Is there a trivial proof?

0celo7

Yes.

Mike Miller

@abenthy The only obvious one is where there are no maps other than isomorphisms.

SteamyRoot

Oh

Hmmm

abenthy

2:47 PM

@MikeMiller But not every self-homeomorphism is homotopic to the identity, so this doesn't work, does it?

SteamyRoot

Couldn't you like, pick some rational $x_1$ and irrational $x_2$

and then use the intermediate value theorem between them

0celo7

@SteamyRoot If you can write a proof I'll be very happy

But no one could think of one that didn't use the Baire machinery.

DHMO

> This function is continuous at irrational $x$ and discontinuous at rational $x$.

(Not the Dirichlet function)

that's too amazing

0celo7

@BalarkaSen Ok, should I study for QM or read Hatcher?

Balarka Sen

Toss a coin.

0celo7

2:50 PM

Or play Fallout?

Balarka Sen

Play Fallout.

0celo7

Why would you say that?

SteamyRoot

Right... Let $[a,b] \subset [0,1]$ and $(f_n)$ a series of continous functions converging pointwise to Dirichlet. Pick $x_1$ rational and $x_2 > x_1$ irrational in $[a,b]$

There must exist some $N$ such that $f_n(x_1) > 2/3$ and $f_n(x_2) < 1/3$.

DHMO

@0celo7 @BalarkaSen why is the function below continuous at irrational x?
$$\begin{cases}0&\mbox{for x irrational}\\1/b&\mbox{for x = a / b a reduced fraction}\end{cases}$$

SteamyRoot

when $n \geq N$

0celo7

2:54 PM

@SteamyRoot Yes, that's how everyone begins the "proof"

Why is that starred?

Balarka Sen

@0celo7 1) It's obviously the more tempting choice for you 2) you wouldn't blame for making that decision, unlike if I told you to read QM 3) I can work in peace without getting asked a barrage of Hatcher questions from you, and moreover complaints about how completely nonrigorous and totally outrageous Hatcher's book is.

0celo7

@BalarkaSen 1) That's why I wanted you to tell me something else 2) I really need to study QM and that's what I'm actually doing 3) Outrageous?

Balarka Sen

I quit.

0celo7

Lol what?

SteamyRoot

Hence there is some nonempty interval $I \subset [x_1,x_2]$ such that $f_N(I) = [1/3,2/3]$.

(using intermediate value theorem)

And $I \subset [x_1,x_2] \subset [a,b]$. So repeat this with the interval $I$ instead of $[a,b]$

You get a sequence of indices $N$ together with intervals $I_N$, and the sequence of intervals is (non-strictly) decreasing

So if you take the intersection of all those intervals, you get something nonempty.

Mike Miller

3:01 PM

@abenthy You could say that the objects are topological manifolds and the maps are self-homeomorphisms which are homotopic to the identity. That's a perfectly good category!

abenthy

But what is a map "M -> N homotopic to the identity"?

DHMO

@0celo7 could you help me?

Mike Miller

Right, that's the problem. All you have are maps M -> M.

0celo7

@SteamyRoot Interesting.

Mike Miller

That's a category. Just a lame one.

SteamyRoot

3:02 PM

Then take a point $p$ in that intersection, and consider the subsequence of functions $f_N$ of the sequence of functions $f_n$. $\lim f_N(p) \in [1/3,2/3]$.

abenthy

Ah okay, thanks Mike!

SteamyRoot

but this limit must be the Dirichlet function, which doesn't take values there. Contradiction.

0celo7

@SteamyRoot How should I credit you?

I'm writing a note to my prof.

SteamyRoot

"some random dude on Math.stackexchange.com" or so, I don't really care.

Mike Miller

I can't remember if I meet with my advisor at 10 or 11.

Balarka Sen

3:05 PM

On a side note, I forget: isn't there a conjecture about all K(G, 1) manifolds having that property?

0celo7

Alzheimers

Mike Miller

@BalarkaSen No.

The property that every homeomorphism is homotopic to the identity? :P

Balarka Sen

Obviously, there's a counterexample for the torus (switch meridian and longitude). Duh

I think that might be for homotopy equivalence.

Mike Miller

Again, no.

You can tell me what the set of self-homotopy equivalences are.

Balarka Sen

Isomorphisms G --> G, of course.

abenthy

3:07 PM

The set of homotopy classes of homotopy equivalences of a K(G,1) is Out(G) btw.

Balarka Sen

Ah, right, I was thinking of basepoint preserving ones

Hmm, I can't remember the right and not obviously stupid statement of the conjecture

Mike Miller

The conjecture you're thinking of interpolates between the homotopy-theoretic information, which is easy, and the topological info, which is hard. If $M, N$ are aspherical topological manifolds, the conjecture is that any map $f: M \to N$ that induces an isomorphism on fundamental groups is homotopic to a homeomorphism, unique up to isotopy.

Balarka Sen

Ah, right.

Mike Miller

-unique, that's unreasonable.

Ramanujan

@DHMO iam getting 1275×638 - 613×1227

abenthy

3:10 PM

It's still not proven, right?

DHMO

@Ramanujan seems good

Mike Miller

@abenthy Not in dimensions greater than three (and when the dimension is that small, I think uniqueness does hold, which is what confused me). There is some progress. It's now basically a K-theory question I think.

Ramanujan

@DHMO will I get answer now?

DHMO

@Ramanujan isn't that the answer?

@Ramanujan it should be 1275x1274/2 - 1225x1224/2 instead, i think

Ramanujan

Answer is in number

DHMO

3:13 PM

since 1275-50=1225

Ramanujan

I did 1275 - 49

By seeing above terms

(brackets)

DHMO

should be 1275 - 50 instead

Ramanujan

If you look 4th (any ) bracket first term in 10-3 not 10-4 @DHMO

DHMO

@Ramanujan yes, but you need to minus an extra 1

Ramanujan

Why?!?!?!?

Mike Miller

3:21 PM

OK, a manifold has a spectrum $\mathbf{L}(X)$ coming from surgery theory. Apparently the Borel conjecture is equivalent to saying that the map (whatever it is, I don't know) $M_+ \wedge \mathbf{L}(\Bbb Z) \to \mathbf{L}(M)$ is a homotopy equivalence when $M$ is aspherical. This is some flavor of K-theory.

Balarka Sen

@MikeMiller: Does the case for dimension 3 use the full machinery of Geometrization, or is it easier? For hyperbolic manifolds Mostow gives this, certainly. I don't know many other examples of aspherical 3-manifolds (take the torus into account and I'm out of examples).

Mike Miller

@BalarkaSen You need geometrization in general. For Haken manifolds, Waldhausen proved it.

Balarka Sen

Oh, I do know

Knot complements :P

Mike Miller

I'm talking closed. If you want to allow boundary, you have to be more careful.

Balarka Sen

@MikeMiller Interesting. Thanks.

Hmm, ok

Mike Miller

3:23 PM

The Borel conjecture doesn't allow boundary, but it's still true for 3-manifolds with boundary. Now it's just any map that induces an iso on $\pi_1$ and sends the "peripheral subgroup" (conjugacy class of image of the pi_1 of the boundary) of the first to that of the second, then it's homotopic to a homeomorphism.

Balarka Sen

I see, nice.

Mike Miller

Otherwise a knot complement would be determined by its $\pi_1$.

Ramanujan

@DHMO I need pure way to solve this problem (uploading)

DHMO

@Ramanujan what is your formula?

so the start term is 1226 and your end term is 1275

what is your next step?

Balarka Sen

Right, good point

Ramanujan

3:26 PM

I got same terms

And now I am adding by using sum of n natural numbers formula

0celo7

@SteamyRoot I wrote "a Belgian grad student showed me this"

Ramanujan

Last term I got from that it is sum from 1 to 50

SteamyRoot

Fair enough :)

Ramanujan

And first term by subtracting 49 from last term@DHMO

DHMO

@Ramanujan yes, you're just repeating what i just said

what is your next step?

Ramanujan

3:29 PM

1275×638 - 613×1227

But it's too large

DHMO

how do you derive this from the fact that the start term is 1226 and your end term is 1275?

Ramanujan

First I get nth term of 50th bracket to get 1275(from your given hint) and then I came to know that first term will be 1275 - n(of bracket ) -1 from observation

And there is only common difference of 1

So we can apply sum of n natural numbers formula

@DHMO

DHMO

and what is the formula?

Ramanujan

n(n+1)/2

@DHMO one more problem,please I want to get it by pure way ( not by substitute)

DHMO

@Ramanujan yes, n(n+1)/2, but why is it 1275x1276/2 - 1226x1227/2 ?

Ramanujan

3:41 PM

@DHMO sorry, formula is for sum of n natural numbers from 1,but we are having from middle so we are adding from 1 to total and subtracting from 1 to middle to get our required sum

DHMO

yes, that is what i wanted you to say

actually, you only need to subtract from 1 to middle-1

consider summing from 7 to 10

7+8+9+10

it is (1+2+3+4+5+6+7+8+9+10) - (1+2+3+4+5+6)

notice that the end term of the second bracket is just 6

instead of 7

Ramanujan

Oh,igot it now,iwas not concentrating more on that term

DHMO

@Ramanujan q12 or q13?

Ramanujan

Thanks

13

You can do , I will do dinner and come now

DHMO

let that expression be f(n)

f(1) = sqrt(2) right

the problem is not clear on how the ... ends

Ramanujan

3:46 PM

@DHMO continue,

DHMO

the problem is not clear

Ramanujan

What do you mean by problem is not clear?@DHMO

DHMO

is f(1) 2 or sqrt(2)?

Ramanujan

Sar(2)

√2

DHMO

ok

and as we can see, $f(n+1) = \sqrt{2+f(n)}$

Ramanujan

3:48 PM

@DHMO why did you think f(1) =2?

DHMO

@Ramanujan never mind

Ramanujan

Ok

DHMO

sorry, I have no idea how to do it without substitution

Ramanujan

OMG! It's again

DHMO

???

Ramanujan

3:53 PM

@DHMO why there is C_1 as only 1 in area of triangle,remembered?

DHMO

I don't remember

Ramanujan

Area of triangle in detail form

Why c_1 is 1 1 1?

DHMO

oh

because it evaluates to the right answer?

just expand the determinant

to check that the two formulas are identical

Ramanujan

I can't see any way how mathematics can expamd any matrix(before knowing it)

DHMO

what do you mean?

Ramanujan

4:03 PM

I know we get same,but my question was to know how it came in formula and in any mathematicians mind while making that formula

DHMO

ask the mathematician who came up with that formula

Ramanujan

He might not revealed it for us to know the beauty of mathematics

I think so

@DHMO. Will I get answer for that if I post it on MSE?

DHMO

Is $\Bbb R^{\omega}$ countable or uncountable?

@Ramanujan I don't think so.

Ramanujan

@DHMO I will try but, if not I can get upvote too

DHMO

@Ramanujan I don't think you will get upvotes

Ramanujan

4:10 PM

@DHMO why?!?!?

DHMO

it is not really a mathematical question

asking how a formula is derived

Ramanujan

Why?

DHMO

it would be closed as too broad

Ramanujan

So you want to say it can be used as formula?

DHMO

?

Ramanujan

4:13 PM

When we get something like √2+√2+√2+√2+… instead of adding up all , we can write n=4 in formula?

DHMO

yes?

Ramanujan

Haha(lol)

Asking or saying?

DHMO

both

Ramanujan

I think yes

Till we know why cos( π/n) will be

@DHMO one more question?

Can I ask one more question?

DHMO

yes

Ramanujan

4:20 PM

Again without substitute,becoz I got it by substituting n=1

DHMO

4 or 5?

Ramanujan

4

DHMO

are you familiar with geometric series?

Ramanujan

Yes (series which has common ratio between each successive terms)

DHMO

can you express 777....7 n times in terms of such series?

Ramanujan

4:24 PM

Only last term 777777…………ntimes?

yes

r=n^0

???

7 7 7 7 7 has common ratio of 1

DHMO

no...

77777 does not mean 7+7+7+7+7

Ramanujan

4:27 PM

Then there is no common ratio as one term doesn't create any series

DHMO

77777 means 7+70+700+7000+70000

Ramanujan

OMG,how could I miss this (school boy error)

OK so common ratio is 10 @DHMO

DHMO

yes

Ramanujan

OK wait ,let me do it on my own now

DHMO

no....

second line is wrong

$n7$ is right

$n70$ is wrong

Ramanujan

4:33 PM

@DHMO I got till here , also I don't know properly how to apply summation

DHMO

i gave you a hint

you didnt use it

Mike Miller

@BalarkaSen I sent you the Taubes exercises.

Balarka Sen

Thanks a bunch!

Ramanujan

@DHMO oh,i got ,it is (n-1 )70 and so on

DHMO

yes

i said express the n-th term as a geometric series

use the formula on each term

Ramanujan

4:36 PM

DHMO

....

Ramanujan

@DHMO why can't I go in this way?

DHMO

@Ramanujan you can if you know how to

for me, i don't know how to

Ramanujan

OMG!!it's again

DHMO

?!?!

what is this supposed to mean

47 mins ago, by Ramanujan

OMG! It's again

Ramanujan

4:38 PM

A mathematician don't know how to get it

How to solve

DHMO

$7+77+777+\cdots+777...7 = \dfrac{7(10^1-1)}{10-1} + \dfrac{7(10^2-1)}{10-1} + \dfrac{7(10^3-1)}{10-1} + \cdots + \dfrac{7(10^n-1)}{10-1}$

start from here

Ramanujan

But I think nth term will become (n-n)7×10^n =0

DHMO

@Ramanujan yes, if you use the method in your notes

Ramanujan

@DHMO iam not getting how to do it😥 @DHMO

DHMO

$\dfrac{7(10^1-1)}{10-1} + \dfrac{7(10^2-1)}{10-1} + \dfrac{7(10^3-1)}{10-1} + \cdots + \dfrac{7(10^n-1)}{10-1} = \dfrac{7\times10^1-7}{9} + \dfrac{7\times10^2-7}{9} + \dfrac{7\times10^3-7}{9} + \cdots \dfrac{7\times10^n-7}{9}$

Ramanujan

4:43 PM

First term is 7/9?

DHMO

@Ramanujan I think I know how to do it from here now

@Ramanujan who told you that?

Ramanujan

@DHMO wait,dinner is ready (chicken 65) 🍢

DHMO

Let $S=7n+70(n-1)+700(n-2)+\cdots+7\times10^{n-1}(1)+7\times10^{n}(0)$.
$10S=70n+700(n-1)+7000(n-2)+\cdots+7\times10^{n}(1)+7\times10^{n+1}(0)$
Subtracting the first equation from the second:
$9S=-7n+70[n-(n-1)]+700[(n-1)-(n-2)]+7000[(n-2)-(n-3)]+\cdots+7\times10^{n}(1-0)+7\times10^{n+1}(0)$
$9S=-7n+70+700+7000+\cdots+7\times10^{n}$
$9S=-7n+\dfrac{70(10^n-1)}{10-1}$
$S=-\dfrac79n+\dfrac{70(10^n-1)}{81}$
$S=-7\left(9n-\dfrac{10(10^n-1)}{81}\right)$
$S=-\dfrac7{81}(9n-10(10^n-1))$
$S=-\dfrac7{81}(9n-10^{n+1}+10)$

bye, see you later

user379685

Hello could someone help me with the lim as x goes to 2 of (2^x - x^2)/(x-2)

Ramanujan

@DHMO wait

@DHMO I don't know how to properly apply summation

Did I did it in right way

?

Rrjrjtlokrthjji

5:06 PM

hello

SteamyRoot

5:17 PM

@Ramanujan If you're going to put a summation sign, use indices and start/end values

I'm guessing what you want is $\sum_{i=1}^n (i+1) = \sum_{i=1}^n i + \sum_{i=1}^n 1 = \frac{n(n+1)}{2} + n = n\left(\frac{n+1+2n}{2}\right) = \frac{n(3n+1)}{2}$

(note that it's wrong, though. When you place $n$ in front of the brackets, you get $n\left(\frac{n+1+2}{2}\right)$ instead of $n\left(\frac{n+1+2n}{2}\right)$

Danu

@MikeMiller I'm learning about cohomology with bundle-coefficients now. In the standard case (so trivial bundle), there are isomorphisms $*:\cal{H}^{p,q}(X)\to \cal{H}^{n-q,n-p}(X)$ where $\cal H$ are harmonic forms. This, at least in Huybrechts book, follows from the analog for the real case (which is assumed), plus the fact that $*$ maps $p,q$ into $n-q,n-p$. Now, for bundle valued forms, is there also an analogous real statement, making the complex one easy?

Anubhav

@BalarkaSen what exercises?

Ted Shifrin

5:43 PM

@Danu: It's better in the holomorphic category, because $\bar\partial$ is well-defined on holomorphic sections, so you don't even need a connection to make sense of differentiating sections. Check it for yourself.

Danu

Hi Ted!

Ted Shifrin

Oh, I sort of missed the point of your question.

Danu

Your comment doesn't make much sense to me

Ramanujan

math.stackexchange.com/questions/1978908/… can any one convert it into matjax pls.

Ted Shifrin

It's an important comment. In the real category, you cannot differentiate sections of a bundle unless you have a connection on the bundle (or unless the bundle is globally trivial).

Danu

5:46 PM

@TedShifrin I believe that. I think I know that :P

I took a course on "Riemannian geometry" after all

Ted Shifrin

The generalization of the isomorphism you're thinking of is called Serre duality.

Danu

But it was about constructing the Chern-Weil homomorphism (?!) instead

Ted Shifrin

Oh, I forgot about that. Well, I love Chern-Weil.

Danu

I know, Ted:) That's what I'm studying now :D

Anyways

My question was more practical

Ted Shifrin

Anyhow, I answered your question. Wait for Serre duality.

There is no real analogue.

Danu

5:47 PM

Sorry, let me reiterate.

For $*: \mathcal H^{p,q}(X)\to \mathcal H^{n-q,n-p}(X)$, Huybrechts claims it is an isomorphism because (i) on Riemannian manifolds $*:\mathcal H^k(X) \to \mathcal H^{n-k}(X)$ is an iso, plus the fac that $*$ maps $(p,q)$-forms into $(n-q,n-p)$ forms.

Ted Shifrin

OK

Danu

That's sorta not satisfying of course

But never mind that.

I'm fine with it.

Ted Shifrin

You can check it directly from the identities you've done for the various Laplacians in the complex setting.

Danu

Now, I'm onto bundle-valued forms and he claims (without comment) that $\bar *_E$ induces an analogous isomorphism

@TedShifrin Yeah but we're not assuming Kahler!

I'm just wondering if the $\bar *_E$ case is exactly the same story as the $*$ case (probably, since every other aspect of it seems to be...)

Ted Shifrin

Oh, right. OK. No, it's a bit more subtle. The bundle has to get dualized.

Does he not do Serre duality?

You need a plain ol' $(n,n)$-form to integrate, so you need to pair $E$ and $E^*$ ...

So look carefully at what his $\bar\star_E$ is doing.

hi @PVAL

PVAL-inactive

5:52 PM

@ted hiya

Ted Shifrin

Ah, the topology crew has arrived. Just in time for me to leave :P

Mike Miller

@Danu Ask me again at around 1 if you haven't figured stuff out.

I have to meet with Ciprian in five minutes.

Ted Shifrin

LOL @ "around 1" ... we're on so many different time zones. :)

Have a good meeting.

Danu

@TedShifrin Yes, I'm sorry, of course!

Mike Miller

He'll figure it out.

Danu

5:54 PM

He does do Serre duality

Mike Miller

Serre duality is good,

Ted Shifrin

So the meaning of harmonic forms has to be clarified, @Danu. Are we talking $\bar\partial$ Laplacian?

Mike Miller

Arn't we on Kahler

Ted Shifrin

He says we're not.

Danu

@TedShifrin Yes

And we're on Hermitian only

Ted Shifrin

5:56 PM

OK, so it's obviously not possibly going to follow from anything in the real category.

Danu

Well, see my earlier statement

for $*$ it makes a lot of sense

But then again yeah...

The $\bar\partial$ of course doen'st coincide with the $d$ Laplacian

That's actually a bit weird

Ted Shifrin

Uh huh.

Danu

Perhaps he's just being slightly inaccurate

Mike Miller

Time to go. We'll see how it goes.

Ted Shifrin

Even in the Kähler case, you have trouble once the coefficients are in bundles.

Have a good meeting, @MikeM.

Danu

5:57 PM

Right, so @Ted in the discussion of $*$ he does it for the $d$-Laplacian

Ted Shifrin

Oh, so he is actually using hermitian structure and putting on the canonical connection compatible with the holomorphic structure and the hermitian structure. Then you get $\mathscr H^{p,q}(E) \cong \mathscr H^{n-p,n-q}(E^*)$, the iso being conjugate linear.

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$Mathematics$ Mathematics