Mathematics - 2016-11-13 (page 2 of 5)

DHMO

05:01

you are welcome

saturatedexpo

@DHMO i think that $(bi)^2=-b^2$ has no additive inverse

Ramanujan

1

Q: How to calculate rectangle 'chord' after rotation?

I am no mathematician so please forgive me for using wrong terms. I am looking to calculate the length of a segment of a rectangle according to its rotation around one of its vertices. For instance: After 45deg rotation I'm interested in knowing the length of the red segment. I came with a pl...

@DHMO looks interesting

DHMO

@saturatedexpo it isn't even closed

saturatedexpo

@DHMO closed under what?

arctic tern

i^2 = -1 is not an element of Zi

DHMO

05:03

multiplication

saturatedexpo

ah

DHMO

@ram not interested

saturatedexpo

i dont even know whats asked

but thats usual

DHMO

lol

saturatedexpo

is it just me or is the question indeed not clear?

DHMO

05:10

the latter

saturatedexpo

@DHMO $Z_p$ is a ring, math.stackexchange.com/questions/786382/… But wether this counts as a closed intervall im not sure..

DHMO

@saturatedexpo what is Z_p?

saturatedexpo

modulo p

where p=prime

DHMO

the definition in the link is very misleading

so Z_5 = {0,1,2,3,4}?

saturatedexpo

i guess so

im not that good in that!

DHMO

05:20

alright

I usually see $\Bbb Z / 5 \Bbb Z$

saturatedexpo

ah

makes sense!

arctic tern

Z_p is standard and common

in many textbooks and papers

DHMO

but the definition in the link is actually not that definition

in the link the ring is a subring of $\Bbb Q$

I believe Z_5 is a field as well

arctic tern

dunno why saturated is linking that one, it's a localization not a quotient ring

saturatedexpo

to be honest

that was just the first google result lol

DHMO

05:23

...

saturatedexpo

im happy that it contains math

as my first search was about pokemon

"z_p ring" was my first search

why is $\mathbb{Z}$ even popular in english?

DHMO

@saturatedexpo e.g.?

saturatedexpo

why not $\mathbb{I}$ for Integer

DHMO

because we like to be universal

imagine what would happen if everyone uses a different symbol

saturatedexpo

yes i just wonder how the english got to Z

zounds?

DHMO

05:32

proofwiki.org/wiki/Definition:Integer#Linguistic_Note

they just borrowed it from the Germans apparently

saturatedexpo

=)

this one looks awesome $\mathcal Z$

DHMO

lol

saturatedexpo

$$\mathcal Z$$

:( wish it where bigger^^

DHMO

$$\Huge\mathcal Z$$

saturatedexpo

ha

new trick learned^^

i think I is not used because of physics

DHMO

05:36

maybe

saturatedexpo

i mean apart from simply borrowing

well physics runs out of variables fast anyway

DHMO

lol

@saturatedexpo flylib.com/books/en/3.190.1.45/1

introduction to abstract algebra

@saturatedexpo Would you be interested in proving $(-1)^2 = 1$?

saturatedexpo

@DHMO yes but not now, and i have to know in what (i.e. group/ring

DHMO

saturatedexpo

@DHMO the first one is nice

DHMO

05:43

@saturatedexpo ring

i thought your message is incomplete

@saturatedexpo because it contains closure xd

saturatedexpo

06:08

@DHMO what you said?

DHMO

@saturatedexpo prove that (-1)^2 = 1 in a ring

Balarka Sen

@AkivaWeinberger That's not a closed surface, is it?

It should be a sphere minus a bunch of punctures.

6 punctures to be precise

saturatedexpo

@DHMO my start would be (-1)*a=1, then find a.

DHMO

@saturatedexpo not sure what you mean and not sure if that would be easier

saturatedexpo

@DHMO oh, no that doesnt work

(ring, not field)

DHMO

06:18

@saturatedexpo how does that work in field?

saturatedexpo

@DHMO does your way include proving that $(-)^2=+$?

DHMO

@saturatedexpo that would be a stronger result

Balarka Sen

@AkivaWeinberger Take my previous comment back. Actually that's not even a surface. Look at the meeting point of $1$, $2'$ and $B$ - that has a bad neighborhood.

saturatedexpo

@DHMO $(-1)^2=1\iff 0=1-((-1)^2)$. Since only 1 can fit the bracket, (-1)^2=1

wait

that doesnt seem right xd

DHMO

what the hell is that

saturatedexpo

06:32

haha

just thought in my bed about it xd

Ramanujan

07:07

@DHMO is 1^∞ = 1 or it is an indetriminate form?

DHMO

@Ramanujan abuse of notation

Ramanujan

?!?

DHMO

infinity is not a number

Ramanujan

in The h Bar, 13 mins ago, by Mew

but 1^infinity =1

DHMO

physicsists always abuse maths

2

126

A: Why is $1^{\infty}$ considered to be an indeterminate form

Forms are indeterminate because, depending on the specific expressions involved, they can evaluate to different quantities. For example, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers. $$\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n =...

Ramanujan

07:09

I remembered minutephysics did once

DHMO

56 secs ago, by DHMO

physicsists always abuse maths

and minutephysics is all about physics

Ramanujan

One person from numberphile made a video on it

Balarka Sen

@AkivaW But there's an obvious fix by perturbing $1$ and $2'$ and the like to not meet. In which case, I have no earthly idea. Even if you do it for like the square pyramid instead of hexagon, you'll get torus with a puncture. So that might have more genus than I initially thought. Maybe it's a genus 3 - it deformation retracts to the wedge of 6 spheres which is the 1-skeleton of a surface of genus 3.

DHMO

07:23

What would be the consequence if the multiplicative identity of a ring (with unity) equals the additive identity of the ring?

(Note for beginners: the requirement that $0 \ne 1$ is only found in fields)

Balarka Sen

@DHMO The ring would be trivial.

DHMO

@BalarkaSen what does trivial mean?

Balarka Sen

It just has a single element.

DHMO

why?

Balarka Sen

If you multiply something by the additive identity, it gives the additive identity back. In short, it's a zero element.

The multiplicative identity never does that, unless you just have one element.

DHMO

07:29

I don't understand

wait

let a be an element of the ring

then a*0 = 0 because 0 is ring annihilator (i know how to prove this)

also, a*0 = a because 0 is multiplicative identity

since equality is transitive, a = 0

therefore, the ring only has one element @BalarkaSen

thanks

ahorn

07:47

@DHMO Morning. Sorry about hanging up yesterday evening. I'm still working on my problem about how to prove that $x_{n+1}=1+\frac 1 {x_n}$ is convergent. The domain is $\mathbb R$. I don't know if there are some tests for convergence of sequences, or rules to follow, e.g. monotonic and bounded; strong contraction mapping;...

DHMO

@ahorn no idea

0+0 = 0 (a.identity)
a*(0+0) = a*0
a*0 + a*0 = a*0 (distribution)
a*0 = 0 (a.inverse)
a = 0 (m.identity)

Ramanujan

@DHMO if DC's of AB = ( L_1 , M_1,N_1) and DC's of AB = (L_2,M_2,N_2) then DR's of bisector of ∠BAC are?

DHMO

what is DC and DR?

Ramanujan

DC are cosine of angles made by a line wrt x,y,z axes

Dr are direction ratios

DHMO

sorry, no idea

Ramanujan

08:09

Iam my book it says drs of angular bisector are (L_1 ± L_2 , M_1± M_2 , N_1 ± N_2) @DHMO

Ramanujan

08:21

In*

@robjohn hi

DHMO

Is there any special results of the trivial field Z/2Z?

can we prove that 1+1=0 in a field {0,1}?

straightforwardly?

i thought of a handful of proofs by contradiction..

Alessandro

08:42

How did you define Z/2Z? If you defined it as Z modulo the ideal 2Z you can just compute 1+1 and see that it is 0

DHMO

yep, and i realized that it was a bad name on my part

i just meant the field with set {0,1}

Alessandro

All finite fields with $p^n$ elements are isomorphic so any field with 2 elements is essentially Z/2Z

DHMO

i want to use just the axioms

also, above i proved that in a ring if 0=1 then the ring only has one element; is there other defining properties of a trivial ring?

Alessandro

$0$ is the additive identity, so it can't be the inverse of other elements, since $0+a=a\neq0$, but we know $1$ has an inverse and there aren't many choices left

DHMO

ok

@aless any other defining property of a trivial ring?

Alessandro

08:58

I don't know what kind of property you expect, but "having one element" completely characterizes the trivial ring (up to isomorphism)

DHMO

@Alessandro earlier i proved that 0=1 implies degeneracy

do you know any other condition?

Alessandro

Not really, but I'm pretty sure that checking whether it only has one element is the fastest way to decide if it's trivial

DHMO

you still need something to check if it there is only one element

robjohn

09:58

@Ramanujan are you? ;-)

Ramanujan

@robjohn hi,iam here

@robjohn I need help in DC's and drs

2 hours ago, by Ramanujan

@DHMO if DC's of AB = ( L_1 , M_1,N_1) and DC's of AB = (L_2,M_2,N_2) then DR's of bisector of ∠BAC are?

Secret

10:20

Therefore the additive absorber properties of additive identities collapse any such ring where $0=1$ into the trivial ring

Since $a$ in the above cayley table is arbitrary (it could well be n distinct elements), it constructively proved that $0=1$ collapses into a ring of one element, which is trivial up to isomorphism

Alessandro

"proof by paint" :P

Balarka Sen

lol

Alessandro

my topology professor often does "proof by amoebas" before the actual proof to give an intuitive feeling of what's going on

Balarka Sen

I understand what you mean. I draw a lot of pictures too.

Ramanujan

@BalarkaSen do you have some time to help me?

Alessandro

10:31

me too, it's definitely helpful to visualize things

Balarka Sen

@Ramanujan Nope

Ramanujan

@Alessandro have some time to help me?

Alessandro

depends what you need help with

Ramanujan

3 hours ago, by Ramanujan

@DHMO if DC's of AB = ( L_1 , M_1,N_1) and DC's of AB = (L_2,M_2,N_2) then DR's of bisector of ∠BAC are?

Balarka Sen

Oct 20 at 13:02, by Balarka Sen

@Ramanujan Please don't ping random people to answer your questions.

Ramanujan

10:33

I know the concepts of dcs and drs but I have no idea on this

@BalarkaSen @Alessandro was this a random ping?

Alessandro

ugh... geometry... I think I'll pass, I know nothing about it

Ramanujan

@BalarkaSen atleast pinging you,dhmo,allesandro,and some others is not random,iam on this chat for nearly a month

Balarka Sen

You have pinged several people with your question multiple times today. That's frowned upon. If someone wants to answer, they will. But don't spam the chat.

Ramanujan

@Alessandro do you know someone who can help me in geometry?

Alessandro

not specifically, but there's surely someone on the main site if you ask a question there

Ramanujan

10:37

So it will not be put on hold for any reason?

Alessandro

As long as it's well phrased, understandable and show some effort while not being a duplicate I don't see why it should be closed

Ramanujan

@BalarkaSen now if you are not going to help , iam not listen your bla bla if I did something wrong moderators of this site will tell me.(till now Rob John didn't told me any thing)

Balarka Sen

@Ramanujan Since you are part of the community, you're entitled to listen to the people in here, especially if you're going to keeping pinging them demanding to answer your questions. And it's not just me, this message was starred by 13 other chat-users.

2

But now you're being plain rude, so I am going to put you on ignore.

Ramanujan

@BalarkaSen I beg pardon if my pinging looks disturbing, @robjohn please take decision and tell did I pinged random people now?

Qwerp-Derp

$\frac {2}{2}$

Oh, it works here

Ramanujan

10:48

@Qwerp-Derp did you enabled chatjax?

Qwerp-Derp

Yup

Ramanujan

On pc or Android?

Qwerp-Derp

I have ChatJax as a tampermonkey script, I was just checking if it worked here because it didn't work on PPCG chat

PC

Alessandro

you don't need all the brackets with simple fractions, $\frac22$ works too

Qwerp-Derp

$\frac23$

doesn't work :(

Alessandro

10:50

I see it rendered properly

Qwerp-Derp

$$\frac {1^{\frac{1}{4}}}{2^{23}}$$

Astyx

So do I

Qwerp-Derp

Wait what

Balarka Sen

@Alessandro What're you working on?

Alessandro

Analytical mechanics @Balarka, more precisely the Lagrangian formulation of classical mechanics (it's an argument I don't particularly enjoy or find interesting, but it's a mandatory course...)

what about you?

Balarka Sen

10:58

good stuff

Secret

Caption: Hmm... Unless I miscalculated something, it seems there can be nontrivial semirings where $0=1=e$. Although they still look pretty trivial since any non identities can only be absorbers

Semiring= throw away additive inverses or cancellation laws

Balarka Sen

@Not anything particular. I should study for my exams, but I am killing time flipping through Fourier theory and/or Morse theory.

Secret

Cannot seemed to find a pathway where it can be shown that $e=a$ so this semiring has at least two elements

(For the green stuff, it is just a convenient way to organise all 8 associative laws for an algebraic structure, which the underlying set has two elements)

Edit: Oops, there's a typo

$$\begin{matrix}a & a \\ a & a\end{matrix}\circ \textrm{anyarray} = \begin{matrix}a & a \\ a & a\end{matrix}$$

where $\circ$ are elementwise $\times$ or $+$ with rules given by the cayley table

Alessandro

I'm afraid I'll have to post on the physics website sooner or later because analytical mechanics is giving me a few headaches

Balarka Sen

yikes

Ramanujan

11:31

@Danu hi

Secret

Update: O wait sorry, I forgot semiring has annihilators stated as an axiom, in that case, triviality trivially results

So whatever that above thing is, it is not a semiring

Danu

12:03

@MikeMiller Do you know how to make sense out of the concept of "degree" for an arbitrary holomorphic vector bundle over a (compact) Riemann surface?

Huybrechts only introduced it for line bundles, but now he suddenly starts using it for general rank.

@Ramanujan Just for clarity: You're not breaking any rules of the site by pinging people with your questions, but it's definitely breaking some implicit social rules, which may cause some people to stop trying to help you.

Ramanujan

12:27

@Danu OK , I will keep in mind,my apologize to all for my behavior

DHMO

Why is Q(sqrt(40)) not a unique factorization domain?

s.harp

because $40=\sqrt{40}^2=40$

$\sqrt{40}$ is irreducible

but here you have written an element in two different ways involving irreducible things, on the one had $1\cdot\sqrt{40}^2$ and on the other hand $40\cdot irr^0$, ie a decomposition which has different irreducible factors (once twice $\sqrt{40}$ and once no irred. at all)

Secret

Update: Mistake again, the above structure is indeed a semiring because the element $a$ act like an annihiilator

s.harp

actually never mind, your thing is a field?

so there are no irreducible elements at all

Balarka Sen

12:43

@s.harp But $40$ is not an irreducible, so that's rather flawed argument. But you already figured out that it is.

Astyx

Are there spaces where there exists a contraction map that does not have a fixed point ?

Could you give an example ?

Danu

Yeah

Alessandro

Q should work

Danu

But they're a bit annoying

One standard example is something like an infinite comb

let me show you

s.harp

why not $x\mapsto 1/2 x$ on $(0,1)$

so basically it would have a fixed point if it were complete, but the where the fixedpoint is there is a hole

Alessandro

12:45

Since it's not complete, a map moving everything closer to $\sqrt{2}$ has no fixed point

Astyx

Sorry, I meant over a close subspace

Danu

The example I have is more interesting in some sense, since the space does not allow any strong deformation retractions to a point, despite being contractible

here it is:

Balarka Sen

@Danu But coming up with a metric on the infinite comb is rather annoying.

Danu

@BalarkaSen Metric?

Balarka Sen

Being a contraction map does not make sense unless you have a metric

Danu

12:46

Here, a single comb is described by

$$X=([0,1]\times \{0\})\cup \bigcup_{r\in \Bbb Q\cap [0,1]}(\{r\}\times [0,1-r])\subset \Bbb R^2$$

Or, in a picture:

DHMO

@s.harp why is 40 irr?

Danu

Astyx

Ok I follow up to there

Balarka Sen

@Astyx I don't know what you mean by closed subspace. Just take $\Bbb R - 0$ and contract by a factor of $1/2$. This has no fixed points.

Danu

So take infinitely many of those, patched together as in my picture. It's contractible but not a defo retract onto any poijnt.

DHMO

12:48

@Secret annihilator is a semiring axiom?

s.harp

@DHMO it isnt, I was wrong, there are no irreducible elements what you have is a field and thus also a unique factoriasation domain

Balarka Sen

On complete metric spaces you always have fixed points.

Alessandro

Fields are trivially UFDs @DHMO

Danu

Right, that's the classic fixed point theorem

Balarka Sen

@Danu I don't think what you're saying is too relevant. A contraction map means a distance-decreasing map, not a deformation retraction to a point.

Secret

12:50

@DHMO Yes, $0x=0$ is inserted into semirings as a convention

DHMO

@s.harp but my source says it isnt

s.harp

@DHMO what source is that

Alessandro

And $\mathbb{Q}[\sqrt{40}]$ is a field since it's isomorphic to $\mathbb{Q}[x]/(x^2-40)$ by the first isomorphism theorem

s.harp

maybe we are using different definitions

DHMO

web.archive.org/web/20140928192643/http://www.math.niu.edu/…

s.harp

12:51

but my usual definition of $\mathbb Q[\sqrt{40}]$ will show that it is a field

Danu

@BalarkaSen Lol, I thought about it as a contraction in the sense of contractible space.

DHMO

@Secret and what are we supposed to show?

Balarka Sen

@DHMO They probably mean Z[sqrt(d)]

DHMO

@BalarkaSen then why isn't it a UFD?

Balarka Sen

shrug

Astyx

12:54

Let me rephrase my question :
Let $E$ be a metric vector space, $A\subset E$ a closed subset.
The fixed point theorem states that if $E$ is complete and if $f:A\rightarrow A$ is a contraction map, $f$ has a unique fixed point.

I'm trying to see why the fact that $E$ is complete is relevant, and am therefore searching for examples of non-complete metric vector spaces where there exists $A\subset E$ a closed subset and $f:A \rightarrow A$ a contraction map such that $f$ has no fixed point.

Secret

@DHMO That all semirings where $0=1$ is trivial

DHMO

@Secret and what are the semiring axioms?

user228700

Hey guys :-) Does anybody know how to find the cube root of a number via long division method?

Secret

@DHMO all ring axioms, throwing away additive inverses and insert the annihilator axiom

DHMO

@BalarkaSen which axioms do Z(sqrt(40)) satisfy? which axioms do a UFD satisfy?

Danu

12:56

@Astyx The examples the others gave all were good

Balarka Sen

@Astyx Take E = R - 0 and A = [-1, 1] - 0, and f be shrinking by a factor of 1/2. That's a map A --> A just fine, but has no fixed points

DHMO

@Secret let a be an element; a*0=a (m.identity); a*0=0 (annihilate); a=0

Astyx

@BalarkaSen R-0 is not a vector space is it ?

Balarka Sen

oh, now you want a vector space.

DHMO

btw a*0=0; a(1-1)=0; a+(-1)a=0; additve inverse exists @secret

Astyx

12:57

Yes I forgot to mention that, sorry

Balarka Sen

Banach's theorem is more general than that: it works on any complete metric space

DHMO

@Secret never mind, i assumed -1 exists

Mike Miller

h

Balarka Sen

but maybe someone can give you an infinite-dimensional example. I can't. finite-dimensional vector spaces are all complete.

@MikeMiller Hi.

Astyx

Ok, thank you for your time ! @Danu @BalarkaSen @Alessandro

DHMO

12:59

@secret btw earlier i did an exercise to prove that (-1)(-1)=1 in a ring with multiplicative identity. maybe u would be interested

Danu

@MikeMiller Hi

Secret

Yes, it is a well known result that additive inverse of the multiplicative identity is involutive in rings

DHMO

I see

Balarka Sen

@DHMO again, that's a googleable question

DHMO

alright

Secret

13:01

2

Q: Additive inverse of multiplicative identity, must it be it's own multiplicative inverse?

Ok another probably very basic algebra question. With real numbers and integers and complex numbers, one is used to $(-1) \cdot (-1) = 1$, i.e. the additive inverse of the multiplicative identity is it's own multiplicative inverse. Does this have to be the case for fields or does it just happen ...

abstract-algebra

DHMO

@secret maybe we can explore what axioms are needed for (-1)(-1)=1 to hold

obviously the multiplicative identity and the additive inverse

but i was able to prove it without commutation

Mike Miller

@Danu what's the degree of a line bundle?

s.harp

I'm wondering if in infinite dimensional representations of a finite group the group elements are diagonalisable.
Although I admit I do not fully know how to think naturally about diagonalisable linear maps without some kind of topology on the vector space

Mike Miller

@Danu if it's c_1, just take c_1

in terms of the splitting principle, c_1 of a direct sum is just the sum of the c_1s, so you're extending the old definition linearly

Danu

13:16

Okay

@MikeMiller I'm not exactly sure how to do this quite yet. The point is that the degree for a line bundle is just an integer, namely if you divisor is given by $\sum n_j p_j$ for points $p_j$, then the degree is $\sum n_j$.

Akiva Weinberger

@BalarkaSen "That's not a closed surface, is it?" It is. It's a surface with boundary. "Actually that's not even a surface. Look at the meeting point of $1$, $2'$ and $B$ - that has a bad neighborhood." No it doesn't. It's just a boundary point, and is thus locally homeomorphic to $\Bbb R_{\ge0}\times\Bbb R$. "In which case, I have no earthly idea." Yes :)

Secret

@DHMO Caption: I think it is the two sided multiplicative identity is the most important reason. Given any square of the form $(a+(-a))^2$, if the cross terms are equal to the additive inverse, and $a$ is idempotent (which is always the case for the (one sided) multiplicative identity), then there will always exists a pair of a and -a to be eliminated, leaving behind one a to force -a to be involutive if a is the identity

Danu

Ah, the point is that on a compact curve $H^2(X;\Bbb Z)\cong \Bbb Z$.

Secret

DHMO

@Secret that requires very many axioms

i'm talking about your caption

Akiva Weinberger

13:22

@BalarkaSen "Even if you do it for like the square pyramid instead of hexagon, you'll get torus with a puncture." Yes. "Maybe it's a genus 3 - it deformation retracts to the wedge of 6 spheres which is the 1-skeleton of a surface of genus 3." You mean six circles? In any case — no, it doesn't.

Secret

@DHMO if a is a two sided identity, then -a must be involutive. So the axiom (other than additive inverse and annihilation) boils down to multiplicative identity

DHMO

@Secret either asso. or comm. have to be true, in my opinion

Secret

asso might be important, commutativity only need to apply for the element a

DHMO

I don't understand your photo

Secret

because a(-a) and (-a)a will equal -a if a is the two sided identity

Ok that photio is too messy, let me redraw

Danu

13:27

Hmm @MikeMiller I'm still a bit confused. I only proved that $c_1(\cal O(\sum n_j p_j))=\sum n_j [p_j]$, i.e. I only have a way of identifying $c_1$ with the degree for some line bundles (the ones that can be expressed as $\cal O(\sum n_j p_j)$.

DHMO

@Secret never mind, i understand your photo now

Mike Miller

@Danu Well, I asked you what degree was.

Danu

11 mins ago, by Danu

@MikeMiller I'm not exactly sure how to do this quite yet. The point is that the degree for a line bundle is just an integer, namely if you divisor is given by $\sum n_j p_j$ for points $p_j$, then the degree is $\sum n_j$.

Secret

Caption: I wonder if there is a more geometric way to illustrate these pairing of the elements and their additvie inverses...?

Danu

So if one of the $n_j$'s is negative, I don't know if the degree actually equals $c_1$

Mike Miller

13:31

@Danu It is.

Danu

@MikeMiller How do I prove it?

The proof I had really relies on using the hypersurface

Mike Miller

You can't because you've literally never told me what degree is!

Secret

Actually, the issue is there is no geometric way to illustrate the discrete convolution (a+(-a))(a+(-a))

Danu

I just quoted the definition to you!!!

1 min ago, by Danu

11 mins ago, by Danu

@MikeMiller I'm not exactly sure how to do this quite yet. The point is that the degree for a line bundle is just an integer, namely if you divisor is given by $\sum n_j p_j$ for points $p_j$, then the degree is $\sum n_j$.

Mike Miller

ok, then that's literally exactly the definition of c_1

DHMO

13:32

@Secret there is

use matrices

Danu

@MikeMiller But, there is a problem with that: I don't actually have a way to assign a divisor to any line bundle.

Also, why is that the definition of $c_1$?

Mike Miller

@Danu Then you haven't given me a definition of degree.

Danu

@MikeMiller So I only have one for the case of the bundle being given by a divisor

It's weird how Huybrechts somehow starts using terms he hasn't defined sometimes

Mike Miller

Then you're going to have trouble proving anything, yeah?

Danu

I guess it must be an artifact of the book arising from lecture notes?

Mike Miller

13:35

@Danu c_1 is a homomorphism from line bundles to Z, so it suffices to show that [p_j] mapsto 1

Danu

Hmm

So I just need that $[p_j]\in H^2(X,\Bbb Z)$ corresponds to $1$

Mike Miller

that's Poincare duality

Danu

@MikeMiller Urgh... I'm being a bit slow in seeing this right now.

Mike Miller

a point generates $H_0(X,\Bbb Z)$...

Danu

Can I just say somethingsomething intersection numbers, because $[p_j]$ is PD to a point?

Mike Miller

13:45

sorry, I don't understand the question

Danu

Oh, man

Mike Miller

oh, you mean the line bundle [p_j]

not the homology class of a point

Danu

No, I do. Never mind.

I think you've answered my questions.

I just completely forgot how to think about $H_0$ (?!?!) but now I remember

Balarka Sen

@AkivaWeinberger "It is. It's a surface with boundaries" Those two are not consistent answers. Closed => no boundaries. "No it doesn't" It looks like two sheets of paper glued along a vertex at that point. That vertex is not a manifold point.

Danu

In either case, this suffices again to show things for line bundles given by divisors. But not all of them are... At least not as far as I know @MikeMiller.

Mike Miller

13:48

@Danu It's not reasonable to expect them to be different for anything else.

Danu

@MikeMiller So I guess I just say that $c_1$ is the only natural thing to define the degree as in the more general case.

Mike Miller

mm

Akiva Weinberger

@BalarkaSen Oh, sorry. My bad there. (The first quote.)

@BalarkaSen Second quote — you sure?

Imagine making a cut partway through a sheet of paper, and folding stuff on one side of it up 90deg, and folding stuff on the other side down 90deg.

That's what it looks like there.

Danu

Huybrechts should feel bad for not mentioning how to define degree, though.

I'll put it in my typos list :P

Akiva Weinberger

Remember that 1 and 2' meet B at an edge, not a point…

Mike Miller

13:52

it's a public room, so if you want to say that, sure

Balarka Sen

@AkivaWeinberger Actually the local model I said is not right, but it's still not a manifold point. 1 and 2' meet B at an edge, 1 and 2' meet each other at a point, precisely the point the two edges 1 \cap B and 2' \cap B intersect at, yes? I am saying that point doesn't have a neighborhood homeomorphic to R^2.

And oh, that agrees with your local model.

Akiva Weinberger

Yeah, it's a boundary point

Balarka Sen

Good call. Thanks.

It does have an nbhd diffeom to a disk minus a cut, which is H^2 just fine.

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