The h Bar

John Rennie

06:00

@SirCumference Good. Being degenerate is one of my specialities :-)

2

Sir Cumference

@JohnRennie I'd agree, but I don't know what this guy is going to pull on us

He's teaching so many things I don't know what to expect

So I'm going to finish the whole textbook tonight

All ~350 remaining pages

user116211

@SirCumference Can you remember everything by a single reading?

Sir Cumference

@MAFIA36790 I'd better

user116211

Very well, then.

Sir Cumference

You know, when I'm tired, I think up strange physics things

Anonymous

06:02

@SirCumference OMG, you better start reading instead of the chat ... LOL....350 pages !

Sir Cumference

Like, why do things even happen? Wouldn't it make more sense if time never moved?

user116211

@SirCumference yes, that is why we need to sleep. Brain feels exhausted and it doesn't work to its optimum level; I see even weird dreams about crazy theorems and endless debate to prove them; all weird.

Sir Cumference

The weirdest part of physics is that things happen

Sigh Im' drifting

@MAFIA36790 Reminds me of a dream I had about this chat

We were in a real bar

0celo7 was the bartender

user116211

@SirCumference yes!! I saw a dream on this very thing; there was 0ce.

user116211

@SirCumference yes!!!

Sir Cumference

06:05

@JohnRennie was there and he was planning to trap us all within the stackexchange network

So me and @HDE226868 and @obe tried to escape to Jupiter

user116211

I dream of the h Bar frequently; this is a common thing now @SirC.

John Rennie

@SirCumference Englishmen are always the villain!

Anonymous

@SirCumference So, John was the villian ? :-P

Anonymous

HAHAHAHHAHHAHAHHAHAHAHAHHAHA

Sir Cumference

@S007 Yeah, and someone else in the chat

Might've been ACM

Anonymous

06:06

Highly reputed villians

Sir Cumference

But we needed someone to make the spaceship, so @Hohmannfan built it with @obe

Anonymous

we would'nt stand a chance against them :-P

user116211

That is going crazy ;)

Sir Cumference

Turns out we ended up suffocating in space because we forgot to bring suits

I tried calling 911 but everytime I called it, I dialed 991

Soon I woke up though

user116211

@SirCumference then Secret attended the call and gave you a mammoth lecture on structures on divisibility by zero. Crazy indeed.

Sir Cumference

06:08

@MAFIA36790 Do tell

DanielSank

@JohnRennie True dat.

Sir Cumference

@JohnRennie I took a course in some kind of drama, they mentioned that kind of thing

It's the reason why scar in the lion king was british

user116211

@SirCumference Mine is always related to the latest topics I read. That dream was related to something in group theory I can't remember.

John Rennie

@SirCumference It's the American inferiority complex at work :-)

2

We English know we're superior so we see no reason to harp on about it :-)

user116211

@JohnRennie 0celot might protest; after all they defeated the English.

user116211

06:11

@DanielSank o/

John Rennie

@MAFIA36790 All part of our long term secret plan.

user116211

ah!

Sir Cumference

Can someone just teach me what social penetration theory is?

Secret

Last night dream, h bar related section:
forgot except at h bar x minecraft, there's a message saying that I "was deeply suspended for deeply disruptive messages throughout the chat". Checking my profile, there is no expiry date for the ban, not even the word NEVER. Meanwhile my friend was sitting next to me as she will be using the computer later.
Later the plot deepens when I realised that after logging out and logging in, I was being restored and I can chat normally. It is then revealed that it is actually my friend's SE account got banned because of her argument with a user called Darks…

user116211

Was Brexit in the plan too? Never mind.

Anonymous

06:11

@JohnRennie Actually seeing your profile picture, I think you might be a great villian in a Marvel movie :-D

John Rennie

@SirCumference That doesn't sound like physics?

user116211

And here we go with @Secret's diary.

Sir Cumference

Oh yeah! And @0celo7 was barred

I woke up thinking how ironic that was

@JohnRennie psychology

user116211

@Secret oh man; that's indeed intricate.

user116211

Anyways, @SirC, all the best for your exam; I need to go back to my book.

Sir Cumference

06:13

@MAFIA36790 Oh god same

Secret

now when you do a reality check on this, you will find there is something interesting. For starters:

in Mathematics, 20 hours ago, by Balarka Sen

I wish there were less division by zero discussions in this chat.

Sir Cumference

Btw, found this funny

Secret

Thus it is interesting in light of Sircumference's dream is I do indeed gave a mammoth seminar on division by zero in the math chat yesterday, initiated by DHMO

user116211

@Secret You zeroed in the maths room, I see.

Secret

and this, is what I believe to be what form the basis of that ban message in last night's dream:

in Mathematics, 16 hours ago, by Balarka Sen

What's more disruptive is the constant use of images with strange scribblings to communicate mathematics, because most of the time they are moderately large and takes up a lot of chat space, putting the messages I want to see far above from my window.

Sir Cumference

06:17

@obe btw

Secret

the message alone reminds of a similar advice from acuriousmind. Thus it must have set up a lot of chain reaction of cross talks in my memory resulting it to become a dream element

Sir Cumference

please tell me how you got that full size pic of my gravatar?

Secret

Perhaps the dream is joining ACM and others on trying to warn me to use less pictures and more text

Sir Cumference

Thanks google...

To be fair, it is a good telescope...

Anonymous

@SirCumference Google likes joking too ;-)

DanielSank

06:36

@MAFIA36790 \o

Anonymous

07:05

@JohnRennie Any idea about this question ? physics.stackexchange.com/questions/293156/…

Anonymous

0

Q: Direction of current in a wire connecting two conducting loops placed concentrically in a time varying magnetic field which is into the plane of paper

Figure shown plane figure made of a conductor located in a magnetic field along the inward normal to the plane of the figure. The magnetic field starts increasing. What will be the direction of current at point R ? I know currents in both the loops will be anticlockwise, but I am not sure about ...

electromagnetism

John Rennie

@S007 I suspect that will fall afoul of our homework policy ...

Anonymous

@JohnRennie Okay I am deleting it then...but could you answer it here in chat ?

Anonymous

Deleted.

John Rennie

That's just Faraday's law isn't it? The induced emf in a closed loop equals the negative of the time rate of change of the magnetic flux through the loop

Anonymous

07:13

I know the direction of current in the loops

Anonymous

I am asking about the point R

John Rennie

The flux through the outer loop is larger so the EMF generated in the outer loop is larger

Anonymous

which connects the two loops by a wire

Secret

For base b, $\{nx_b: n\in [(1_b)_{10},(10_b)_{10}]\}$
$$(b-1)_b=(b-1)_b*0+(0_b+(b-1)_b)$$
$$(b-1)_b*n=(b-1)_b*(n-1)+(n_b+(b-n)_b)$$

where $(b-n)_b=0_b$ for $n=(10_b)_{10}$

I think the reason why it works (at least for integers, not sure how to prove it for arbitrary bases) is because the symbols from the base are elements of $(\mathbb{Z}/b,+)$. Therefore, for each succesor in base b, the unit digit is always the final symbol in that base, which can be decomposed into the sum of first and last digit in that base

Anonymous

@JohnRennie EMF in outer loop is indeed higher, but can we conclude the current flows from outer loop to inner loop due to that ?

Anonymous

07:19

Moreover I thought a current always needs a complete loop to flow....so I am not sure whether current can flow from one loop to another loop

John Rennie

To be honest I can't remember enough electrostatics to be sure what the answer is. I agree with you that there can't be a continuous current flow, but the question might be asking about a transient flow when the field first starts changing.

Secret

@SirCumference This is the reason why the digits of 9n add up to 9. In fact, it applies to all base

reddit.com/r/explainlikeimfive/comments/2s79sd/…

Anonymous

@JohnRennie Anyway, thank you for the effort. But I actually need an answer. Should I undelete the question ? I do not mind a few downvotes :-P

John Rennie

If you undelete the question it will just be closed. It's a blatant homework question.

Anonymous

ahhhhhhhh :-P

Anonymous

07:23

so what should i do ? :-P

Sir Cumference

07:37

@Secret I-I...I thought I discovered something special ;-;

Secret

Don't worry, I have many instances like that, only to find out something have been done before. This is common in research

Sir Cumference

@Secret D:

Imma go cry now

Mew

07:55

hi all

Anonymous

08:08

@JohnRennie Hey! I got the answer " There is also no general potential between the two loops, like there is no general potential between two batteries. Connecting one terminal of the first with one of the second battery brings them to the same potential, and only then, you will have a potential between the unconnected terminals. I.e. without R, there could never be a potential between P and Q"

Anonymous

Correct, no current. There is also no general potential between the two loops, like there is no general potential between two batteries. Connecting one terminal of the first with one of the second battery brings them to the same potential, and only then, you will have a potential between the unconnected terminals. I.e. without R, there could never be a potential between P and Q. — sweber 7 mins ago

Anonymous

0

Q: Direction of current in a wire connecting two conducting loops placed concentrically in a time varying magnetic field which is into the plane of paper

Figure shown plane figure made of a conductor located in a magnetic field along the inward normal to the plane of the figure. The magnetic field starts increasing. What will be the direction of current at point R ? I know currents in both the loops will be anticlockwise, but I am not sure about ...

current electromagnetism

Anonymous

Fortunately EE SE didn't close my question till now :-P

Secret

[Division by zero] Division by zero is hard, but good luck dividing by zero divisors

heather

12:14

hello, anyone around?

DHMO

12:43

@heather hi

@Secret division is not well-defined if zero divisors exist

heather

@DHMO, hello

DHMO

how do you do

Secret

heather

@DHMO, well, and you?

DHMO

well

heather

12:45

quick question: do you have to know multivariable calculus to study differential equations?

Secret

Not if you throw away all the additive inverses >: D

DHMO

@heather depends on the type of differential equations you would like to study

Secret

(although I still yet to check its associativity etc., chekcing them now

heather

normal ones? (is that a thing?)

DHMO

@heather the easy ones do not require multivariable calculus

but i have no idea what you would like to study

any context?

are you planning to study in college or just self-study?

heather

12:46

well, I would like to study in college some day, but currently, I am in middle school, so for now it is just self-study

DHMO

then of course you get to decide your prerequisites

heather

(::shrugs::) I don't know enough to decide prerequisites, which was why I asked =)

DHMO

alright, then you do not need to know multivariable calculus.

heather

okay, thank you

but I assume that's just for ordinary ones?

DHMO

sure

heather

12:48

okay

thanks again

DHMO

@Secret are you using a program to check associativity?

Secret

Nope, I check them by hand. I have a handy way to check them by organising them into an array, so that I am basically end up doing matrix addition

In theory I can automate it in mathematica, but I have not renew my mathematica license yet

DHMO

alright, i'll check associativity and distributivity by program

Secret

(NB, on the writing of this message, there's already a crack developed somewhere. Somehow, one of the 64 distributive law is broken)

Details to be investigated

O wait, that's my careless mistake, 2+2=2 not 2+2=0. I am too used to Z/4...

continue...

DHMO

2+2=2 is indeed correct in your picture?

Secret

12:54

Yeah, I found that when building these crazy algebraic systems, multiplying a+b=c both sides with all the elements to propagate the whole set of equations will minimise chance of introducing contradictions

e.g. begin with 0+0=0, multiply by 3 gives 2+2=2

I suspect it is a common thing for all semirings, the multiplication structure controls the additive structure

DHMO

associativity checked

distributivity checked

my code for your reference:

add_table = [[0,1,2,3],[1,1,1,3],[2,1,2,3],[3,3,3,3]]
mul_table = [[0,0,0,2],[0,1,2,3],[0,2,0,1],[2,3,1,3]]

def add(a,b):
	return add_table[a][b]

def mul(a,b):
	return mul_table[a][b]

print("associativity:")
for a in range(4):
	for b in range(4):
		for c in range(4):
			if add(add(a,b),c) != add(a,add(b,c)):
				print("%d,%d,%d"%(a,b,c))

print("left distributivity:")
for a in range(4):
	for b in range(4):
		for c in range(4):
			if mul(a,add(b,c)) != add(mul(a,b),mul(a,c)):
				print("%d,%d,%d"%(a,b,c))

Secret

yup, distributivity passed also in my manual check. Have not started on associativity yet

DHMO

could you teach me how to check using your method?

Secret

The cayley table is a representation of all possible ways to add or multiply any two elements. If your underlying set is finite, then you can define addition as an elementwise operation on the array of columns of the n elements plus rows of the n elements, The rules of + is then given by the cayley table. Thus

$$\begin{matrix}a & b & c & d \\ a & b & c & d \\ a & b & c & d \\ a & b & c & d\end{matrix}+\begin{matrix}a & a & a & a \\ b & b & b & b \\ c & c & c & c \\ d & d & d & d\end{matrix}$$

Now, associative laws means a+(b+c)=(a+b)+c. This means one of these brackets is going to represent $n^2$ possible outcomes of adding any two elements. E.g. we can pick (a+b) to be the array shown previsouly

Slereah

hey hey

DHMO

13:05

@Slereah hi

Secret

Now, c is left free, thus can be treated as a nxn array of a single element. By running through all possible c s you compute all $n^3$ associative and distributive laws

DHMO

how is that different from checking each case manually?

Secret

There is no difference, but it organise them in a nice way so you can check them without making too many mistakes

at least, no need to write all $n^3$ + signs!

DHMO

I see

Secret

NB I compute things much faster in terms of matrices

DHMO

13:07

then I suspect using codes is better

(self promotion lol)

Secret

because I think much faster vertically than horizontally

DHMO

that is true for me also, I suspect using your approach would be better than manually calculating 128 sums

Secret

And indeed, because of the array structure, it can be automated in code

To a computer, I think it is just doing usual matrix addition with some extra rules

so it should be quite quick

Extra shortcuts can be used e.g. one sided absorbers means they trivially associative and distributive for that corresponding side, so are identities

DHMO

indeed

Secret

13:22

Associativity result=BUSTED
Back to the drawing board!

DHMO

so my code is wrong :o

what is the counter-example?

oh

i didn't check for multiplicative associativity

additive associativity:
multipliciative associativity:
(0*0)*3=2; 0*(0*3)=0
(0*2)*3=2; 0*(2*3)=0
(0*3)*3=1; 0*(3*3)=2
(2*0)*3=2; 2*(0*3)=0
(2*3)*3=3; 2*(3*3)=1
(3*0)*0=0; 3*(0*0)=2
(3*0)*2=0; 3*(0*2)=2
(3*2)*0=0; 3*(2*0)=2
(3*3)*0=2; 3*(3*0)=1
(3*3)*2=1; 3*(3*2)=3
left distributivity:
right distributivity:
check finished

code:

add_table = [[0,1,2,3],[1,1,1,3],[2,1,2,3],[3,3,3,3]]
mul_table = [[0,0,0,2],[0,1,2,3],[0,2,0,1],[2,3,1,3]]

def add(a,b):
	return add_table[a][b]

def mul(a,b):
	return mul_table[a][b]

print("additive associativity:")
for a in range(4):
	for b in range(4):
		for c in range(4):
			if add(add(a,b),c) != add(a,add(b,c)):
				print("%d,%d,%d"%(a,b,c))

print("multipliciative associativity:")
for a in range(4):
	for b in range(4):
		for c in range(4):
			if mul(mul(a,b),c) != mul(a,mul(b,c)):
				print("(%d*%d)*%d=%d; %d*(%d*%d)=%d"%(a,b,c,mul(mul(a,b),c),a,b,c,mul(a,mul(b,c))))

Secret

$$2*0=0$$
Multiply both sides by 3
$$3*2*0=3*0$$
$$1*0=2$$
$$0=2!!!!$$

DHMO

nice use of the fact that $2$ is a fixed point of factorial

Secret

Checking the multiplicative associativity is a MUST for these algebraic structures, because that is where they break most readily

(unless of course, you want them to be nonassociative. However I am not ready for nonassociative algebras)

Nope, I killed the additive inverses, thus zero terms should not vanish to 0 unless it is 0*0=0

however... in the world of division by zeor algebra, there is something else:

Looks like there's an unknown pathway it can trigger indirect collapse

Physics Guy

hey guys.

DHMO

13:29

@PhysicsGuy hi

Secret

Caption: paperwork

Physics Guy

What's about that algebra thing ?

DHMO

@PhysicsGuy we are exploring division-by-zero algebras

Secret

DHMO

0x3 = 2, 2x3 = 1
2+2 = 0x3+0x3 = (0+0)x3 = 0x3 = 2
2x3+2x3=2x3 -> 1x1=1 -> pxp=p (if 1 is unity)

Secret

13:32

I am not sure if what I am going to state is a known result, but I do found semirings (which rings are included as a subset) tend to have the additvie structure controlled by the multiplicatie structure

DHMO

extension of 1+1=1

Secret

finding the rules that controls the multiplicative structure is a very important (I said that, because nobody have done binary divison by zero algebras before) open problem that can give important info on what is possible or impossibel about division by zero algebra (and its generalisation the division by zero divisor algebra)

DHMO

@Secret what are your axioms?

Secret

I usually start with identities, and all the axioms needed for the definitions of some elements (e.g. zero divisors need the axiom ab=0 and also any of the 3 axioms that can define a zero element 0, and a hypothetical inverse will mean I need 1

anything else the table will tell me as I work through it

Currently I worked with finite systems, as they are easier to manage

DHMO

I see

Secret

13:37

finite system means I can pick any equations of the cayley table, assume it is equal to something ,and then I use multiplication to propagate the rules implied form it, and reject rules that lead to a contradiction

Whenever I hit a contradiction, I discard that possibility and then start again

Thus it is kinda trial and error in some way

DHMO

I see

Secret

If I am lucky to stumble upon an important equation, it will pretty much fixed a lot of entries in the cayley table

Howeevr Ihave yet to find an efficient way to predict associative law behaviour

multiplying entries in the + cayley table to get other entries usually mean you implicitly checked the distributive law as you go, thus it is less prone to contradictions

but you cannot do simialr things in general for multiplicative associativity

The only thing you can do is to be alert whenever you get a zero term, because contradiction might be hiding somewhere

Of course, not all zero terms means contradiction. In fact, the consequence of dividing by zero or zero divisors is that you must have zero terms

and (as it seems so far) introduction of an additive absorber

You also frequently lose the induction axiom 1+1=2 due to the fact that 0+0=0 for 0 to be an identity

Ultimately, as Tobias have warned, I am not sure if they will be useful in the end, because there is so much not knwon about them. Given how difficult it is to construct them without them spontaneously collapsing means whatever problems that need them are highly unusual

When I first started all of this, the major motivation is basically morbid curiosity, and I don't have aplications in mind

DHMO

have you constructed any consistent system?

Secret

For division by zero algebra (not division by zero divisors algebras), I have a coupel of examples. They have relatively trivial structuees though

DHMO

like?

Secret

13:49

These are division by zero semirings

DHMO

but they are not commutative

Secret

Commutative is not a concern for me. They are easily covered by the cayley table and many interesting systems are noncommutative

For commutative examples, you have this family of rather useless guy:

where the trivial ring is the first member of them

DHMO

how to generalize it?

Secret

You add an extra layer of absorbers

so that you end up with a semilattice chain

They are not very useful because they really only have an ordered structure and nothing else

ACuriousMind

0

Q: Simple Ricci flow equation

For the $2D$ Kahler manifold, the Ricci flow equation (which is also a one-loop RG equation for the $\sigma$-model on this space) can be written in the form $\frac{\partial^2 \Phi}{\partial u^2}=\frac{\partial \Phi}{\partial u} \frac{\partial \Phi}{\partial \tau}$. Does anybody know how to solve...

renormalization differential-equations

^homework-like or not?

Secret

13:55

ACM: They ask how to solve an equation without stating any conceptual details, thus it is homework like

DHMO

@Secret so both operations are "max" and all required axioms follow

Secret

@DHMO yup, a join semilattices for both operators (which are secretly the same)

DHMO

I see

Kyle Kanos

@ACuriousMind Yes

Secret

ACM: But then, the problem is that I am not sure if Mew's site have resources to handle this high level stuff if it is decided to sent to it due to being homework like.

Kyle Kanos

13:58

Though the comment that I saw there previously was rightfully deleted

DHMO

@Secret do you have more structures with both operators identical?

Secret

not at the moment, and I am not very interested in mutual distibutive laws (they are too restrictive and not many non lattices have them)

DHMO

ok

Secret

Currently my focus is: Group theory, division by zero algebra, division by zero divisor algebra (or proof of its nonexistence)

Why I study group theory have nothing to do with divison by zero, group theory is already interesting all by itself

Emilio Pisanty

@ACuriousMind more of a Too Broad for me

Hey @DHMO, is this your homepage?

dhmo.org

DHMO

14:11

@EmilioPisanty sure :p

Emilio Pisanty

brilliant, that one

DHMO

no i'm just kidding

ACuriousMind

@Secret We don't "send" any questions to that site currently, and whether or not they want a certain question doesn't influence whether it's off-topic here.

@KyleKanos Just doing my job :P

@EmilioPisanty Yeah, I can also see that

Secret

I never said the question is not off topic, but otherwise I agree (in fact, Mew's site's questions are mostly high school to year 1 college level atm)

Emilio Pisanty

@ACuriousMind have you seen that DHMO page?

Bernard Meurer

14:17

@DHMO Your chemistry is nuts

ACuriousMind

@EmilioPisanty Not that specific one, but I've seen the kind of joke before

DHMO

@BernardMeurer lol

Bernard Meurer

How come something called Madelung's constant changes from compound to compound?

It's not a constant then goddamit

Emilio Pisanty

@ACuriousMind love the t-shirts

ACuriousMind

Those "uses" worry me too! We can't allow cult rituals involving DHMO to continue

Emilio Pisanty

14:27

@ACuriousMind yeah. I'm really worried that they completely skipped over its role in Fukushima, too.

DHMO is completely responsible for that disaster.

DHMO

change.org/p/…

You can make a change

Put your name there!

ACuriousMind

0

Q: Why is the $B$-Field an axial Vector?

in trying to understand the Wu-Experiment I wonder why the $B$-Field is an axial vector. I know that $\vec{B} = \vec{\nabla} \times \vec{A}$. Under Parity transformation I'd expect $\vec{A} \rightarrow -\vec{A}$, however I don't know whether $\vec{\nabla} \rightarrow -\vec{\nabla}$.

electromagnetism magnetic-fields parity

Î can't believe this isn't a duplicate, but I can't find an original

Secret

14:44

DHMO: An update to the 4 element stuff: It seems when all the rules are propagated correctly will result in the + structure to be isomorphic to the left or right null semigroup, and the * structure to be isomprhic to $(\mathbb{Z}/4,+)$, i.e. one of the division by zero algebra derived previously. The result is that there is no element called 0 (because the additive identity is destroyed during the propagation of the equations, and the null semigroup means all 4 elements are zero elements (in the form of one sided absorbers)) and hence, there are no zero divisors. Since it happens for any si…

One observation is that during the propagation, the procedure of multiplying each equation by 2 or 3 forms a cycle that repeats

DHMO

@Secret I do not understand what your conjecture means... can you use examples?

Secret

Given $S$ is an associative algebra with additive and multiplicative identity
$\forall \epsilon \in S: \textrm{if }\epsilon^2=0 \textrm{then } \not\exists\epsilon^{-1},\epsilon^{-1}\epsilon=1$

DHMO

but $3^2=0$ and $2\circ3=1$

Secret

The catch here is that: 0 is not zero because after the + cayley table is filled in correctly, you get a left or right zero semigroup, thus all elements are left or right absorbers, and there is no element you can call zero

DHMO

right

@Secret Let $e^2=0$ and $fe=1$.

Secret

14:53

In fact, there's also a curious thing about this structure is it forms a cycle by multiplication by 3 or 2
e.g.
Multiply by 3
$1\rightarrow 3\rightarrow 0 \rightarrow 2 \rightarrow 1$
Multiply by 2
$1\rightarrow 2\rightarrow 0 \rightarrow 1 \rightarrow 1$

This somehow result in all the off diagonal entries of the + cayley table to be filled in such that the resultign table is isomorphic to the null semigroups

The proof of this conjecture need to be more careful, as there exists zero terms. We need to show all choices of zero terms will result in the + table to be isomorphic to the null semigroups to complete the proof

A proof by contradiction will mean we assume 0 is the additive identity, and then show that since the resulting + table is isomorphic to the null semigroup, there is no identity and hence 0 is not an identity, a contradiction

DHMO

@Secret why isn't this a counter-example?

Secret

because it is trivial...? (?_?)

DHMO

your conjecture did not state that it cannot be trivial

Secret

o yeah, bad habit of mine tend to overlook trivial examples...

(cannot edit now, but anyway, update conjecture)

DHMO

15:09

my point is that if there is a counter-example then there must be another

Secret

I am currently redoing the * table by checking the case 2*0=1, it does not lok optimistic as it result in 0*0=/=0 which is not possible because both identities and distributive law are present to collapse that

so the conjecture might be true

DHMO

@Secret

John Rennie

@ACuriousMind it looks OK - post edit. It's not as if it's some trivial question that the OP just couldn't be bothered to answer themselves.

Secret

16:41

Ok, I have the proof of the conjecture

Obliv

@secret which conjecture are you referring to? Has this to do with that division by 0 ring?

Secret

It has to do with the impossibility of dividing by zero divisors in any distributive algebra, associative or not

a very strong result

(note zero is not considered as a zero divisor)

Obliv

wasn't your semiring distributive?

oh

if it's not a zero divisor... then what is?

Secret

I don't know, the existence of the division by zero semirings I have constructed suggested 0 somehow escaped from this conjecture (now a theorem as I will be posting the proof shortly

Obliv

You should make clear what you mean by zero-divisor. I.e what their properties are in the algebra. perhaps the 0 in your semiring wasn't of the same properties as you are conjecturing.

Secret

16:52

The simplest definition for zero divisors is $ab=0$ for nonzero $a$, $b$

Given, for all $a\in S$, where $S$ is distributive
$$\epsilon^2=0, 0+a=a$$
Suppose there exists $f$ such that
$$f\epsilon = 1$$
Then multiply $0+a=a$ both sides by $\epsilon$
$$\epsilon 0 + \epsilon a = \epsilon a$$
Since $a$ is arbitrary and since $\epsilon a \in S$, therefore $\epsilon a \in S$ is always a right additive absorber for all $a$ and $\epsilon$.
Thus the + structure is isomorphic to the right null semigroup. Hence $0$ is not the unique additive identity and hence no zero elements exists. Hence $\epsilon$ is not a zero divisor if $f$ exists

Obliv

nvm--didn't realize ab = 0 implied 0/a = b

0*1 = 0 and 1*0 = 1 too I think. man that is confusing me

Secret

oops, I screwed up somethign in the proof. LEt me fix that...

Yeah (that's one way to say division by zero divisors)

but it seems there are hints that it is actually impossible, even wth the crazy division by zero algebraic stuff I have been doing

Obliv

hmm but isn't $\epsilon$ its own inverse with $\epsilon^2$? unless you assume $0$ not the identity

Secret

17:07

This is where things get a bit different from the usual algebras we have read in books. When you start doing crazy things like division by zero or division by zero divisors, then you need to consider the possibility of zero terms (i.e. $0x=/=0$ for nonzero x). Sometimes it is possible to define the zero terms in a way to remove the contradiction (which is what happens for the division by zero algebra I have been playing with),
but for this case, it seems to be impossible without making the + structure becoming a one sided null semigroup

If $\epsilon$ is its own inverse, then you will get $1=0$ which will collapse the structure into a boring semillatie chain very quickly

Obliv

if f exists, then $f*(0/\epsilon) = 1$ so that $0/ \epsilon = 1$?

I'm assuming $\epsilon^2 = 0$ implies $\epsilon = 0/\epsilon$

Secret

I am currently on $\epsilon^2=0$. multiplying by $f$ gives $\epsilon = 0f$

based on what you said, then $f=/\epsilon$

Obliv

you mean $f = \epsilon / 0$?

Secret

nope, zero divisors are not dividing by zero. They are generalisation of the concept of cannot divide by zero in usual algebras. e.g. in the ring $\mathbb{Z}/4$, 2 is a zero divisor as $2^2=0$ and hence $/2$ is undefined

It is well known that in rings zero divisors have no inverses or cannot even by inverted because the additive inverses will show that the zero divisor = zero, thus a contradiction

However, it is not known if it holds true in semirings

The experiments of mine and DHMO suggest it might be also true in semirings as well

Obliv

for that example $\mathbb{Z}/4$ don't you have 2/2 defined? So, 2 is not always a zero divisor?

Secret

17:18

at least for nipotent zero divisors

Nope in Z/4, 2*2=4 mod 4 = 0, thus you cannot dividie by 2

else you get 2=0

All $\mathbb{Z}/n$ n not prime contains zero divisors

this is why the finite fields all have prime chaarcteristics

Obliv

if you did allow 2/2 = 1 mod 4, then you have 2 = 2 and 2 = 0/2. but that suggests 0 = 4? i'm confused again lol

DHMO

who is using the word "mod" in algebra?

yes, 0=4 in Z/4Z

Secret

In $\mathbb{Z}/4$ the only elements are 0,1,2,3.

therefore 0=4=8=... and similarly for other integers, forming 4 equivalence classes

ACuriousMind

@Secret They can be inverted in the proper algebraic sense of "inverting" - you can localize at a set containing zero divisors. However, the map to the localization will not be injective.

@DHMO Writing $a = b \mod I$ where $I$ is a subgroup/ideal that you quotiented out of a group/ring is perfectly standard, although you'll also see $[a] = [b]$ and $a + I = b+ I$.

DHMO

@ACuriousMind I see

Obliv

17:36

Nvm I got it.

ACuriousMind

18:05

Arrrrgh. Why on earth does Witten not simply say "normalization factor of the trace" instead of sending me on a wild goose chase what a "dual Coxeter number" is?

Obliv

both of those phrases would send me on a wild goose chase :P @acuriousmind

Kyle Kanos

@ACuriousMind Because obscurity is key to keeping your job

ACuriousMind

@KyleKanos Believe me, there's enough obscure reasoning here that obfuscation through terminology is unnecessary :P

vzn

@KyleKanos lol spking of jobs hows it going? arent you working in (quant) finance?

ACuriousMind

18:30

Also, these people are writing in TeX. How hard would it be to write "Kähler" instead of "Kahler" or "kahler"?

Secret

Update: It is possible I am just unlucky, cause attempt to prove the conjecture only get a partial result that all the columns in the + cayley table involving $f,\epsilon,0,1$ are indeed all one sided absorbers. Since a 4 element set the only choice of f that is not 0,1 or $\epsilon$ is the remaining 4th element, the + cayley table is then became tiled into the null semigroups. Will see if similar thing happened for larger size...

vzn

@Secret wondering what inspired this prj?

Kyle Kanos

@vzn Going well. I am indeed a quant.

vzn

18:47

@KyleKanos cool good to hear are you working with stocks? options?

Kyle Kanos

Neither: credit derivatives

vzn

@KyleKanos are those sold on NYSE or NASDAQ, some other market?

Kyle Kanos

Though, actually, some of the credit derivatives are options, so I guess I could have said yes to the latter.

Usually it's party-to-party type marketing (e.g., credit default swaps)

vzn

@KyleKanos yeah think heard of those in 2008 o_O ... isnt there still some kind of exchange?

Kyle Kanos

Yes, of course

But I've got errands to run, back later

ACuriousMind

19:01

Does anyone have an explanation of domain walls in QFT/string theory that doesn't assume I know a lot of condensed matter theory? All the stringy papers just casually say "clearly, this object is a domain wall" but I'm not really sure what I'm looking at.

Obliv

@acuriousmind is this what you're looking for?

ACuriousMind

19:21

@Obliv That looked like it would be good until it said something rather nonsensical when describing how branes and domain walls are connected (which is what I am mainly interested in). It says "To describe a domain wall, the D4-branes must start in one vacua [ouch, it's vacuum, but nevermind] at $x_3\to- \infty$ and interpolate to the final vacuum as $x_3\to\infty$."

But $x_3$ is a spatial direction and transverse to the walls, so...how does a brane that does not strech or move in the $x_3$ direction interpolate between vacua at the "infinities" in $x_3$ direction.

Slereah

@ACuriousMind errr isn't it just "Those two regions have a different ground state"

ACuriousMind

@Slereah Well, that's the "useless" explanation :P

Slereah

and the action has some topological invariant that makes it impossible for it to change to a single one

I mean what do you want to know exactly

IIRC the domain wall is just like

ACuriousMind

@Slereah On the one hand, there is this solitonic description of domain walls as field configurations that interpolate between two vacua at infinity (much like instantons, just domain walls interpolate spatially, not temporally) that has a codimension 1 "kink", which is "the wall". On the other hand, there are those M-theorists happily wrapping branes around stuff and claiming they've got a domain wall as soon as the result has codimension 1. I don't get how the two concepts connect.

Slereah

Action has some $Z_n$ thing for the ground state

And then you have some topological charge

then it is a domain wall

Dunno much about strings

so can't help too much

ACuriousMind

19:26

Ahhhhhh

I remember now

I need to reread the "branes as solitons" part of my string theory course.

Obliv

@ACuriousMind what is that 3-dimensional space in that image above the mentioned section? it has $x_4,x_6,x_9$ as axes...

ACuriousMind

@Obliv It's a slice of the full 10D spacetime

Obliv

10 dimensions of space.. yeah okay.. physics gets really weird at some point.

ACuriousMind

Yeah, okay, this is starting to make sense now

Obliv

19:55

Does it make sense to define a scalar function $f(\vec{r}) = x^2 + 3xx_3 - 2x_2 + x_4 - 6$ where $\vec{r} = \langle x(t), x_2(t), ...\rangle$ and evaluate $f(r(2)) = x(2)^2 + 3x(2)x_3(2) + x_4(2) - 6$

DanielSank

20:05

@EmilioPisanty You trying to start some beef up in here?

I challenge you to Smash Brothers.

Obliv

it must have been a pain in the neck to have to type up papers with a type-writer like stephen hawking's thesis and having to insert symbols by hand. Though, I guess it beats having to write it all out by hand.

Kyle Kanos

@DanielSank SSB? SSBM? SSBB? SSB for 3DS/WiiU?

DanielSank

20:41

@KyleKanos melee

Sir Cumference

@DanielSank That game with Roy

Kyle Kanos

@DanielSank On a Wii or Gamecube?

Sir Cumference

20:59

@KyleKanos Gamecube is best

Kyle Kanos

@SirCumference Never had one. I have a Wii.

Though the kids shoved something in the disc slot and now it won't play discs.

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