Mathematics - 2021-02-13 (page 2 of 2)

Balarka Sen

19:00

so now i just tell them to run the cellular cosheaf instead

Mike Miller

Oh yeah I don't help people compute that

i just tell them to

Thorgott

@Balarka wanna talk about some differential topology

Balarka Sen

no

im sick of talking to people about differential topology every day

user2103480

@BalarkaSen wanna talk about some long twitter threads on skull measurements that start out scientifically and end up with the usual racist tropes

Mike Miller

You know he does

Thorgott

19:02

lol

user2103480

is that a yes

Alessandro Codenotti

But do you have a moment to talk about our lord and saviour?

geocalc33

Jesus?

user2103480

:57048398 anyways here it is mobile.twitter.com/crimkadid/status/1356181036883992576

enjoy

Balarka Sen

@AlessandroCodenotti what is it

@user2103480 of course i am not reading that

alt right sucks, they're not only bad they're also boring

user2103480

19:05

@BalarkaSen probably smart, this is utter nonsense

Balarka Sen

bad + boring is a bad combination

Mike Miller

its so common :(

Alessandro Codenotti

@BalarkaSen No I don't have an actual topic, I was just curious if that also bores you if we talk about it every day

Thorgott

whats wrong with people who think twitter is the right outlet for longer texts with attempts at substantiality

user2103480

theyre on twitter

Balarka Sen

19:07

@MikeMiller so tired of people still milking the culture of making fun of boring bad people

geocalc33

twitter is powerful engine of destruction

user2103480

thats the prime mistake

Thorgott

also true

Balarka's very tired today

Balarka Sen

nothing to do with today

geocalc33

drink some Brazilian coffee

user2103480

19:08

lmao what

sorry confused that with colombian coffee

Simone

Can you believe this guy?

0

Q: If $-1<\lim _{n\to \infty }\left(\frac{x_{n+1}}{x_n}\right)<1$ then there exists the limit $\lim _{n\to \infty }\left(\sum _{k=0}^n\:\:x_k\right)$

Let be $x_n$ a sequence. Prove that if $$-1<\lim _{n\to \infty }\left(\frac{x_{n+1}}{x_n}\right)<1$$ then there exists the limit of the sequence $$y_n=\sum _{k=1}^nx_k\:$$ and it is finite. I am not one hundred percent that this is completely true. In any event, I managed to show that if $$0<\lim...

limits

Look at his comments.

Thorgott

lol

Ted Shifrin

A @Balarka: I'm a boring bad person. What's your point?

Balarka Sen

Seems doubtful Ted

Mike Miller

He's saying he doesn't want to read a Twitter thread making fun of you, then

Balarka Sen

19:11

lol

Ted Shifrin

@Simone: The guy is an ass, but I suspect the point is to prove the ratio test theorem, not quote it.

BigSocks

@TedShifrin you are not that boring!

Ted Shifrin

Emphasis on that

BigSocks

hahaha

Simone

@TedShifrin I guess, but that's not the question he asked. What an arse though

geocalc33

19:13

probably just didn't have coffee this afternoon

Balarka Sen

lol "your post is irrelevant"

thats a good shutdown

Simone

lmao

Balarka Sen

next time i will ask a question, and when someone responds i will say "your post is irrelevant"

geocalc33

hehe

Ted Shifrin

I would normally be helpful and ask if he knows that absolutely convergent series are convergent, but hell if I'm going to help him.

Simone

19:14

XD

If I use the comparison test to prove the ratio test will he ask me to prove that as well?

is that how you begin writing calculus textbooks?

Ted Shifrin

You're right that you shouldn't do his homework for him, regardless.

Simone

ikr

Ted Shifrin

But I think my comment is the relevant one. He apparently understands the idea of the ratio test for positive series only.

Balarka Sen

Adam Curtis has a new documentary out

Simone

how do I reach these kids?

user2103480

19:22

@Simone lmao great reference

Simone

XD

geocalc33

has anyone ever played mafia?

it's a social deduction game

Mafia (party game)

Mafia, also known as Werewolf, is a social deduction game, created by Dimitry Davidoff in 1986. The game models a conflict between two groups: an informed minority (the mafiosi or the werewolves), and an uninformed majority (the villagers). At the start of the game, each player is secretly assigned a role affiliated with one of these teams. The game has two alternating phases: first, a night role, during which those with night killing powers may covertly kill other players, and second, a day role, in which surviving players debate the identities of players and vote to eliminate a suspect. The game...

user2103480

@BalarkaSen you into post/prog rock?

geocalc33

here you can read about the game called mafia

Simone

maybe "abduction" not deduction... and yes it's a double entendre

geocalc33

19:25

it's a party game

Simone

ah I thought you meant the videogame

Balarka Sen

@user2103480 yeah

Simone

@BalarkaSen I like Devo. And that's as far as my prog/rock knowledge goes

user2103480

@BalarkaSen Ok I have two songs and I just can't find other stuff like that

VOLA - Inmazes

►

Balarka Sen

oh i dont know devo

Simone

19:28

@BalarkaSen Yes you do, you just don't know that you know

user2103480

The Orwells - Double Feature [Official Audio]

►

Simone

Devo - Whip It (Official Music Video) | Warner Vault

►

user2103480

if you have any recommendations, I'd be thankful

Balarka Sen

@user2103480 listening

@Simone oh alright lol

Simone

XD

Mike Miller

19:31

this is cool

people.maths.ox.ac.uk/lackenby/quasipolynomial-talk.pdf

user2103480

I should add that in both songs the rocky parts I find less cool, but the later parts are dope

Balarka Sen

@user2103480 So this is cool. Check out Haken and TesseracT

user2103480

TesseracT I know, a friend of mine is a total fanboy

Balarka Sen

It reminded me of some parts from Haken's "The Mountain"

user2103480

But being tortured at 9am with 20 minutes live sessions had a negative impact

Balarka Sen

19:34

lol

The rocky stuff is Floydian

but what isn't

user2103480

yeah fair, timeless crap

Oh haken sounds cool, M83 type stuff

Balarka Sen

M83 is good shit

user2103480

The voice is a bit more prominent in Haken songs, it seems. But I see why you thought of that. Listening to the cockroach king rn

Balarka Sen

cockroach king is A+

its heavier than the rest of The Mountain though

just heads up

the albums you should shuffle around to find what you like is Visions, The Mountain, Affinity

user2103480

actually, this heavier style is more in the spirit of what I posted

Balarka Sen

19:39

then you might like Affinity because its on the heavier side

user2103480

but it may be a bit too playful to be melodramatic to

Balarka Sen

Visions has a coherent story so if you like concept albums thats a good choice

user2103480

@BalarkaSen definitely. I'm glad to have caught them on the last album tour

Balarka Sen

@user2103480 wow ok, parts of these are like radiohead and the repetitive postrockish/shoegazing things are like explosions in the sky

check out explosions in the sky debut album

the one where a kid is making a sandcastle on blue beach

user2103480

@BalarkaSen if I could only really get warm with thom yorkes voice

Balarka Sen

19:41

in the cover

user2103480

@BalarkaSen will do

Ted Shifrin

@MichaelAlbanese I gave up fighting with this guy. Any thoughts?

user2103480

Weirdly, anthony gonzalez (M83 guy) cut out the last verse of his song "road blaster" in his live performance. May have overcome his broken heart lmao

Balarka Sen

wait i misremembered. it's not explosions in the sky debut album

altho thats definitely a great album

let me think

frick

user2103480

@BalarkaSen "How strange, innocence?"

Balarka Sen

19:48

yes

thats not it i dont know why i thought that

i will tell you the correct ref give me a minute

user2103480

no hurry

Ted Shifrin

a @Balarka, you can weigh in on the thing I asked Michael about. Just put away your topologist hat.

Balarka Sen

I am afraid I do not know much about holomorphic bundles

Thorgott

so the point of contention is whether the complexification of a real bundle is a holomorphic bundle?

Mike Miller

I think he might be right but I am not sure... I thought I remembered that a holomorphic structure on a bundle over a complex manifold is the same data as an $\overline \partial$ operator on that bundle whose Nijenhuis tensor vanishes. If this guy's is not integrable then N(dbar) ought to be nonzero. May as well just calculate?

Actually I'm getting confused, you just need that delbar-squared is zero?

Ted Shifrin

19:57

Well, he's starting with a holomorphic vector bundle (over a complex manifold), thinking of it as a real bundle and complexifying. The standard example is the holomorphic tangent bundle of a complex manifold. Complexification then splits as $(1,0)$ plus $(0,1)$.

I think you need compatibility with the base.

Mike Miller

I don't think so. I will try to find a reference that spells out the NN theorem carefully.

Ted Shifrin

So $\bar\partial (fs)$ should be what it should be.

Mike Miller

I am not confident though NN always confused me.

Ted Shifrin

I will look in Chern.

I totally think his argument is wrong.

This is different from just NN ... that's about almost complex structures on a manifold.

Mike Miller

There is an NN variant for holomorphic structures on bundles

Ted Shifrin

20:00

OK, which only makes sense over a base complex manifold.

Mike Miller

Right

Ted Shifrin

Actually, now that you remind me, I answered a tricky question on main relating to that point ... with help from Robert Bryant.

Let me find it.

Balarka Sen

@MikeMiller Some subbundle of the total tangent bundle transverse to the fiber directions should be integrable for this to happen, I assume?

Mike Miller

The statement i found was: suppose E -> M is a [smooth] complex vector bundle over a holomorphic manifold. A pseudo-holomorphic structure on this bundle is an operator delbar: Omega^{p, q}(E) -> Omega^{p, q+1}(E) which is C-linear. Then this bundle has a holomorphic structure (so that projection is holomorphic) and so that the delbar operator of this holomorphic structure coincides with the operator we gave, iff delbar^2 = 0.

Balarka Sen

The delbar operator is not a lift of the delbar operator of the base, yes?

Ted Shifrin

20:03

This was the question on main. Not relevant, really.

Mike Miller

@BalarkaSen Not assumed to be, nor do I know that what that means

Ted Shifrin

That's the question I'm raising, Balarka.

Balarka Sen

It seems like your statement is giving a new complex structure on M apriori

Namely restrict the complex structure on E to M (M embedded as zero section of E)

Ted Shifrin

I want a holomorphic bundle to have holomorphic transition functions. So I need $\bar\partial fs = f\bar\partial s$ when $f$ is actually holomorphic.

Mike Miller

I think the claim is that this follows from $\overline \partial^2 = 0$.

Ted Shifrin

20:07

It won't follow in the case the fellow is using the isomorphism $\bar E\cong E^*$. I am pretty darn sure.

It is very upsetting to me that $E\oplus \bar E$ can be a holomorphic bundle when $E$ is.

Mike Miller

Oh, of course I did not mean to say that a delbar operator is just C-linear. It should also satisfy Leibniz. That is where your compatibiliity comes from.

Ted Shifrin

Right. I'm saying Leibniz from the beginning.

Mike Miller

So does that fail for his formula?

Ted Shifrin

If you're intertwining the bundle isomorphism $\bar E\cong E^*$ using the hermitian metric, it sure will.

BigSocks

"Leibniz". Many are saying this. Some of the best people are interested in Leibniz. It's true.

Mike Miller

20:09

For omega a function we ought to indeed have dbar_E(fsigma) = (dbar_M f) sigma + (-1)^{p+q} f dbar_E sigma.

@TedShifrin Let's just spell it out

Ted Shifrin

No sign.

Mike Miller

Right, thanks

f is a (0,0)-form

Ted Shifrin

nods

Mike Miller

Hold up, I am confused. Can't you take tensor products of holomorphic bundles?

Ted Shifrin

Of course.

Mike Miller

20:11

His bundle is just C tensor E.

Oh, you tensor over C.

This is C o_R E.

Ted Shifrin

No, you tensor $E_\Bbb R$ with $\Bbb C$.

So you lose the holomorphic structure and then recapture it in a subbundle.

I say you cannot simultaneously capture it in both subbundles.

Mike Miller

So a section of this bundle looks like $f \otimes \sigma$ where $f: M \to \Bbb C$ is a real function, and you are allowed to commute real functions across either side. I bet his formula is not well-defined.

Well, sums of those, but don't shoot me about it.

Ted Shifrin

Not shooting you.

Mike Miller

He wants to say $$\overline \partial_{E'}(f \otimes \sigma) = (\overline \partial f) \otimes \sigma + f \otimes \overline \partial \sigma.$$

First: does this make sense? It seems like you are commuting things past the tensor sign to turn this into a form in Omega^{0,1}(E). But no matter.

Ted Shifrin

Wedges needed, right?

One wedge.

So he's saying: Do $\bar\partial_{\bar E} s = \phi^{-1}\bar\partial_{E^*}(\phi(s))$.

Mike Miller

20:15

I'm getting lost in the symbols.

Ted Shifrin

Where you apply $\phi^{-1}$ to the vector bundle part.

Mike Miller

I think I'm gonna leave this one to the experts. It seems like a good exercise to not get confused though.

Ted Shifrin

LOL ... I wasn't trying to bug you with it. Michael's the right one, but I'm going to bet I'm right. For one thing, someone would have taught me this decades ago if it made sense.

The question is what sort of linearity $\phi$ has over complex things. If $\phi\colon \bar E\to E^*$ comes from the hermitian metric, then ...

$\langle\phi(\bar v),w\rangle = h(w,v)$ is conjugate linear in $v$.

So $\phi(c\bar v) = \bar c\phi(\bar v)$, no?

I think that messes him up.

Mike Miller

The hermitian metric stuff all happened in your comments though. To give an answer convincing to OP you have to write down explicitly what his formula means on the bundle C o_R E, and explain why it is not complex linear.

Probably if you mention the hermitian stuff he will not buy it.

Ted Shifrin

This is a doctoral student in complex geometry and Higgs stuff, apparently. Scary.

Well, I found his attitude condescending, so I'm not going any further with him, but maybe Michael will :P

Thorgott

20:30

scary indeed, why would anyone try to get a phd in complex geometry

Balarka Sen

I would like to understand the story so you can tell me

I do not have any intuition for holomorphic bundles

Maybe a holomorphic bundle (E -> M) is a complex vb E on a complex manifold M with a half-dimensional horizontal foliation on E

Ted Shifrin

I guess Mike nailed it down a bit: We need the bundle isomorphism $\bar E\to E^*$ (the latter having an obvious holomorphic structure) to be $\Bbb C$-linear or else Leibniz won't work. This is equivalent to my wanting holomorphic transition functions in the first place.

No, first and foremost, it's a complex manifold $\pi\colon E\to M$ so that $\pi$ is holomorphic.

You can't do this just topologically.

At any rate, I definitely need $\phi(f\sigma) = f\phi(\sigma)$ when $f$ is holomorphic on the base.

Balarka Sen

Let me make it all pseudo. I have an almost complex base $(M, J_M)$, an almost complex total $(E, J_E)$ with a fiberwise almost-complex structure $J_{E/M}$.

Ted Shifrin

you mean almost?

LOL

I honestly don't think this way.

So you need integrability of $J_E$ and I'm not sure.

This has confused me enough without this.

Balarka Sen

lol i am just screwing around, do not pay heed

Ted Shifrin

20:41

It's just so random that this person thinks I can use any random topological isomorphism to a holomorphic bundle and it should give a holomorphic bundle structure on our crazy thing.

Balarka Sen

What exactly is it that integrates when you go from an almost complex structure to a complex structure?

Ted Shifrin

I understand this in terms of the Nirenberg integrability version of Frobenius (which I referenced in my write-up I linked to earlier).

Balarka Sen

Thanks! Let me read

Ted Shifrin

Here is Nirenberg's.

Balarka Sen

Let me screw around a bit more before digging in. In a local frame $(e^1, \cdots, e^n)$, suppose $Je^i = \sum_j f_{ij} e^j$. $J^2 = -I$ demands $e^i + \sum_{j, k} f_{ij} f_{jk} e^k = 0$. So that means $\sum_{j} f_{ij} f_{jk} = -\delta_{ik}$.

I suppose I could differentiate this system.

Thorgott

20:59

an embedded closed ball in a smooth manifold is always contained in some slightly larger embedded ball, right? extend it by a tubular neighborhood of the boundary sphere

Balarka Sen

smoothly embedded closed balls are, yeah

good luck finding ball neighborhoods of alexander horned ball

(not alexander gored ball, that's the exterior, which is not a ball)

AMDG

Hello! I would like to know what exactly the relationship is of the metallic means to the diagonals of a regular polygon.

Thorgott

hmm yeah, doesn't seem doable for horned ball

glad I'm working in the smooth category

Balarka Sen

it is not because the horned sphere is not flat

Karim Mansour

hi @TedShifrin

Balarka Sen

21:09

If the ac structure is integrable then the eigenspaces of $J$ are integrable subbundles.

Is that sufficient?

Thorgott

it looks to me as if any such neighborhood would have to have sort of self-intersections

Balarka Sen

yes you will get screwed

you will get genus if you try to thicken

it is possible to define the alexander horned sphere as an intersection of nice looking closed subsets (approximating from the outside), which are all handlebodies (1/n-nbhds of the alexander horned sphere), but the number of handles go to infinity

Thorgott

21:32

makes sense, from the inductive pov, you get the horned sphere from the sphere by attaching a handle, cutting out a portion of that handle, attach two handles, cut out portions of these, etc.pp. metrically, these get smaller at each step and taking an epsilon-neighborhood just thickens up what you replaced the cut out portions with after a certain step, making it look like nothing was cut at all. this gets finer as you take smaller neighborhoods, so the smaller the neighborhood, the more handles

robjohn

21:47

@Ted: the rain has completely disappeared here. We were supposed to have some Thursday night, but got less than .01". We were also slated to get some tonight, but that also seems to have evaporated.

Michael Albanese

Hi @TedShifrin. Maybe I'm missing something, but I believe there is a complex linear isomorphism $\Psi : \overline{E} \to E^*$. Fix a hermitian metric $h$ on $E$ (for me, $h$ is linear in the first argument and conjugate linear in the second). For $e \in \overline{E}$, then by definition of the complex-vector bundle structure, $i \cdot e = -ie$ where concatenation on the right is the action of $-i$ on $e$ regarded as an element of $E$.

Then $\Psi(e) = h(\,\cdot\, , e)$ is the desired vector bundle morphism. Note that $h(\,\cdot\, , e) \in E^*$ as the hermitian metric is complex linear in the first argument. The map $\Psi$ is complex linear as $\Psi(i\cdot e) = \Psi(-ie) = h(\,\cdot\, , -ie) = ih(\,\cdot\, , e) = i\Psi(e)$.

Ted Shifrin

22:30

@Michael: That's the map I wrote down with $\phi$ above. I agree that that is an element of $E^*$. Hmm ... So you're agreeing with the OP, then that $E\oplus \bar E$ is a holomorphic vector bundle. Somehow this bothers me a lot.

So you are saying what I said exactly, I think: $\phi(c\bar v) = \bar c\phi(\bar v)$. Something's fishy.

So do I actually get $\bar\partial (f\sigma) = f\bar\partial \sigma$ when $f$ is holomorphic on the base?

Mike Miller

Isn't the complex linear structure on bar E given by i(vbar) = (-iv)bar?

so your formula seems fine?

I dunno too confusing

Ted Shifrin

LOL

I still remember the pain in getting $i$'s sorted out through complex geometry. Even Chern messed it up and I corrected it in the second edition of his beautiful little book.

But I was a lot younger and smarter then.

Karim Mansour

Beautiful

Ted Shifrin

@robjohn We had a tiny bit yesterday.

Mike Miller

I spent a couple hours today trying to understand closely related but slightly different versions of index of some operator which all differ by relatively small amounts (like, some number from 9 to 3) because I kept getting off by small constant errors in various formulae.

Ted Shifrin

22:44

@MichaelAlbanese Is the complex structure on $\bar E$ you're suggesting compatible with the isomorphism $E_{\Bbb R}\otimes\Bbb C \cong E\oplus \bar E$? So, the usual game is to look at eigenspaces of $\text{Id}\pm J$. Now I'm too confused to sort this out.

And, if this is really right, why is that I was never "taught" that $E\otimes\Bbb C$ is again a holomorphic bundle when $E$ is? Were you?

Karim Mansour

I don't like to think of tensor product for complexification

I always like to translate it to $E_{\mathbb{R}} \oplus E_{\mathbb{R}}$

geocalc33

Can you analytically continue a function and the answer gives you a function that you can analytically continue again?

Karim Mansour

no by the identity theorem

Thorgott

that question is not well-posed

Ted Shifrin

@Karim: That's not useful at all for complex geometry.

How do you see the complexified cotangent bundle?

Karim Mansour

22:54

I see it as real bundle direct sum with real bundle with a complex structure J on it

Ted Shifrin

That doesn't help me see $T^{1,0}\oplus T^{0,1}$.

Which is essential for complex analysis/geometry.

Karim Mansour

right

geocalc33

What if you analytically continue f(x) to g(x) and then continue g(x) to h(x)?

Astyx

@geocalc33 What does that mean?

geocalc33

$\Phi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns).$

I need help understanding this formula

$\Phi(s)$ is the analytic continuation of something and I don't understand what $\zeta(s)$ means

Astyx

23:03

I suspect it's the zeta function

geocalc33

my question is is it the analytically continued zeta function?

not sure which form it takes

I'm guessing it's in the form $\sum n^{-s}$

Ted Shifrin

@MikeMiller Interesting that integers are "relatively small amounts" :)

geocalc33

$\Phi(s)$ converges for complex $s\ne0$ and Re(s)<1

Mike Miller

@TedShifrin Well, O(1).

Ted Shifrin

LOL

geocalc33

23:17

anybody have insight?

I wonder if the zeta function appears in other analytic continuations

Balarka Sen

Hm

If all prime ideals are principal, the ring is a PID, right?

If not, take the maximal non-principal ideal I. It cannot be prime (otherwise it's principal). It's contained in some prime P (Zorn) and the containment must therefore be proper.

BigSocks

But maximal ideals are prime?

Astyx

maximal non principal doesn't mean maximal

Ted Shifrin

Indeed.

Thorgott

why does that exist

is it clear that the ascending union of non-principal guys is non-principal

Michael Albanese

23:32

@TedShifrin "It depends upon what the meaning of the word 'is' is."

(Regarding whether $E_{\mathbb{C}}$ is holomorphic)

Ted Shifrin

Michael !!

Michael Albanese

I think that by choosing a hermitian metric on $E$, you can view $E_{\mathbb{C}}$ as a holomorphic vector bundle. But it doesn't have a natural holomorphic vector bundle structure without this choice.

Ted Shifrin

I find it disconcerting to think of the complexified tangent bundle as a holomorphic bundle.

Michael Albanese

At least, that's what all of this seems to suggest.

Me too.

BigSocks

@Thorgott now wondering this

Balarka Sen

23:34

@Thorgott Obvious

BigSocks

@Astyx words are cursed, switching to pictures until things have settled down

Ted Shifrin

I'm still worrying, even with your sign choice, if I get the Leibniz rule with holomorphic functions pulling through $\bar\partial$.

Astyx

Balarka's argument isn't finished though? Why does the containment being proper implies that the ideal is principal?

Balarka Sen

Yes, I am trying to finish it. I was thinking, pick a in P \ I. Then look at I + (a)

This is larger than I, so it has to be principal.

Thorgott

for the record, your statement needs fixing

are you assuming domain a priori or are you only asking for PIR as conclusion?

Balarka Sen

23:35

I don't care

It is obvious that ascending union of nonprincipal ideals is nonprincipal. Go learn some basic algebra.

Thorgott

im not thinking

at least not about this

Balarka Sen

Whereof one cannot speak, thereof one must be silent.

BigSocks

-Michael Scott

Ted Shifrin

Well, I guess I need to go relearn complex geometry, too. Or quit.

BigSocks

but yeah if you had the union be principal it means one of the guys had the generator of the principal ideal, which means one of the guys was principal, I am guessing

Balarka Sen

23:39

That's correct.

Michael Albanese

I believe so, because $\Psi$ is $\mathbb{C}$-linear. The proposed Dolbeault operator is given by $\Psi^{-1}\bar{\partial}_{E^*}\Psi$. If $s$ is a section of $\overline{E}$ and $f$ is a complex-valued function on the base, then $$\Psi^{-1}\bar{\partial}_{E^*}\Psi(fs) = \Psi^{-1}\bar{\partial}_{E^*}(f\Psi s) = \Psi^{-1}(f\bar{\partial}_{E^*}(\Psi s) + \Psi(s)\otimes \bar{\partial}f) = f\Psi^{-1}\bar{\partial}_{E^*}\Psi s + s\otimes\bar{\partial} f.$$

Ted Shifrin

Is it really $\Bbb C$-linear?

Balarka Sen

@Astyx Oh, ok, here you go. I is not prime, so there are a, b not in I but ab in I. Consider I + (a) and I + (b).

Both are principal, virtue of being bigger than I. But I = (I + (a))(I + (b)).

So I is also principal.

Ted Shifrin

I guess you're effectively saying that $\bar E \cong E^*$ with the opposite complex structure on $E$.

Michael Albanese

Yes. You can view a hermitian metric as a complex bilinear map $E\otimes\overline{E} \to \varepsilon^1_{\mathbb{C}}$, so it induces complex linear maps (in fact, isomorphisms due to the non-degeneracy) $E \to \overline{E}^*$ and $\overline{E} \to E^*$; the map I called $\Psi$ is the second map.

To me, that's what $\overline{E}$ is. It has the same underlying real bundle, but the complex multiplication is the conjugate one.

Ted Shifrin

23:45

So I guess the point (which I was stuck on in the first place) is that you cannot stay in the holomorphic category to make $E\otimes \Bbb C$ a holomorphic vector bundle.

Michael Albanese

Not canonically.

Ted Shifrin

This is just crazy. Never in my almost 50 years of working with this stuff has it occurred to me or been mentioned that this is a holomorphic vector bundle.

But maybe the physicists do this routinely.

Michael Albanese

Yeah, it seems an unnatural thing to do.

I've never thought about it before.

Ted Shifrin

Of course it shows up in characteristic classes, but there we're using hermitian structures and so who cares. But in the usual complex geometry setting, we want to think of $T^{1,0}$ and $T^{0,1}$, and one is a holomorphic bundle and one sure isn't.

Michael Albanese

Right, holomorphicity is irrelevant for Chern classes, etc.

Ted Shifrin

23:49

The usual place it shows up is Pontryagin classes.

Michael Albanese

I suppose the point is that the holomorphicity of $T^{1,0}$ comes from the complex structure on the manifold, but to identify $T^{0,1}$ with a holomorphic bundle, you need to choose a metric.

Ted Shifrin

And I've used it there.

Michael Albanese

Really? How so?

Ted Shifrin

No, that was to Pontryagin.

Why would I ever want to even think of $d\bar z$ as something holomorphic?

Michael Albanese

I meant how does it show up with Pontryagin classes?

Ted Shifrin

23:50

In fact, the whole confusion in geometry with hermitian metrics is bothersome enough. Most of the time we use it as a bilinear form on the holomorphic tangent bundle, not as a pairing of tangent and conjugate.

Oh, you define $p_j(E) = c_{2j}(E\otimes\Bbb C)$.

Up to sign.

Astyx

@BalarkaSen I'm not totally convinced, we only have $(I+(a))(I+(b)) \subset I$

Michael Albanese

Oh, you were referring to complexification. I thought you meant that holomorphicity was somehow relevant.

Ted Shifrin

No, sorry.

Michael Albanese

No worries.

Ted Shifrin

I wonder if I should go back and look at that post again. He just irritated me, but I guess I irritated him more.

Michael Albanese

23:57

It happens. I've been told that misunderstandings on the internet are quite common.

Ted Shifrin

LOL ... Well, I'm sorta curious how many people think I'm an idiot and how many think I am right :P

Michael Albanese

I don't think anyone on this site thinks you're an idiot. That question has only been viewed 56 times, and about half of those are probably due to me.

Ted Shifrin

So if we change the $J$ for $\bar E$ but not for $E$, of course, then how does this mess with the usual identification of $T^{1,0}$ and $T^{0,1}$ as appropriate eigenspaces?

Astyx

(I just want to point out that those are not completely incompatible :-) )

Balarka Sen

@Astyx @Astyx Oh, good point.

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