The h Bar

John Doe

12:01

@ACuriousMind Yeah it is. I hate this binomial coefficients and identity stuff. It's like high school trig where you just trying stuff until it works.

Phase

Hi, sorry for intruding but I have a really basic, fundamental confusion over something, and put it in a question but still haven't had any responses, and wondered if putting this in chat could help, if not, no worries :^)

skill patrol

@YashasSamaga try this: when you add 2 feet and 2 inches you get 2 feet 2 inches.

2' + 2" = 2'2"

Yashas Samaga

HAHAH! I love that.

skill patrol

:-)

ACuriousMind

@Phase Asking questions in chat is fine, although we prefer you don't just come here to advertise your main site questions.

skill patrol

12:05

Askaway

Phase

Sorry, I can try asking it in new words here:

Yashas Samaga

@skillpatrol it is a contradiction, it is both 4 and 14 so math is wrong

LOL

Phase

I'm just confused because in Lectures they never really went into detail on the Unitary operator, and I feel like I'm misunderstanding it. In my head I can only imagine it as a constant 'rotation', so if it's an orthonormal basis of two basis states, the classical analogue of it in my head would be a constant rate of rotation between two orthogonal basis vectors. And I guess the only real evidence i have of this is that if H is constant, then $\hat U$ is just a function of t

and that if H|psi> is constant then so too would ihbar d/dt |psi>

skill patrol

yesterday, by skill patrol

@YashasSamaga you got it pal. Don't let them drag you down to their level.

ACuriousMind

@Phase 1. It's not "the" unitary operator. The time evolution operator $U$ you get from the Hamiltonian is one particular family of unitary operators, but there are many others.

Yashas Samaga

12:10

It is really surprising that super religious people have become anti-science/math to an extent that they have problems accepting that 2 + 2 = 4.

This is absolutely insane.

ACuriousMind

2. Indeed, a unitary operator is the complex analogue of a rotation in a real vector space

Phase

re: 2) So the 'rate of change of state' [I don't know if that's actually something that makes sense] is constant if H is constant?

ACuriousMind

However, in two complex dimensions you have more than just a simple rotation - the two-dimensional unitary group $\mathrm{SU}(2)$ is almost the same as the three-dimensional group of real rotations $\mathrm{SO}(3)$.

skill patrol

Ingnorance is infinite @YashasSamaga

Yashas Samaga

I wonder how much $ would he pay to the retailer when he buys 2 packs of apples each consisting of 2 apples,

14$ :D

ACuriousMind

12:13

@Phase That...depends on your state. For eigenstates of $H$ that is true, for non-eigenstates the "rate of change" does not really make much sense - the "rate of change" would usually be the time derivative, which is just given by the Hamiltonian applied to the state in that instant (that's what the Schrödinger equation tells us!), and if the state is not an eigenstate that doesn't give you a number but a (non-constant) vector.

Yashas Samaga

"US Senate passes new bill, makes it legal to kill hibernating bear families in den"

I thought Republicans were religious people.

0celouvsky

@YashasSamaga $ goes before the number

Phase

@ACuriousMind thanks, sorry to keep asking basic questions but would it be possible for you to briefly explain the statement about the non-eigenstates?

ACuriousMind

@Phase Well, the operator is $U(t) = \mathrm{e}^{-\mathrm{i}Ht}$, right? (For activating MathJax in chat, look in the upper right corner of the chatroom) So when $\psi$ is an eigenstate, we get $\psi(t) = \mathrm{e}^{-\mathrm{i}E_\psi t} \psi$, i.e. $\psi$ just get's multiplied by a phase where $E_\psi$ is the eigenvalue, i.e. rotates in a very simply way.

If you don't have an eigenstate, but a superposition like $\psi_1+\psi_2$ where the $\psi_i$ are eigenstates, then you get $\exp(-\mathrm{i}E_1t)\psi_1 + \exp(-\mathrm{i}E_2 t)\psi_2$, that is, the different components of your vector rotate with different speeds, and you can't really speak of an overall speed

Phase

OH

Thank you very much

Sorry for the delay was doing the ChatJax stuff. It's like magic

ACuriousMind

12:21

You're welcome :)

Phase

Can I ask a final, different question? I guess this is more of just a newbie question for the site

Is it bad to actually answer homework-style questions on SE?

If it seems they've made an effort to get the answer themselves

ACuriousMind

@Phase We generally consider that bad, yes, because it encourages those questions. Complete answers are also usually deleted if they are caught. This holds for questions that are off-topic as homework-like. If you genuinely think the question is on-topic, i.e. shows effort and asks a conceptual question, go ahead and answer it however you like.

Phase

Oh ok, well thanks for all the help! Have a great day : )

skill patrol

Different sites have differing policies

0celouvsky

@ACuriousMind the JEE bubble

ACuriousMind

12:32

@skillpatrol What a profound insight :P

skill patrol

:P

0celouvsky

bookstore.ams.org/surv-216

wat is zis

AMS sent me an email about it

and this bookstore.ams.org/surv-218

skill patrol

in JEE Preparation, yesterday, by Madhuchhanda Mandal

@SirCumference "The engineering test" which is regarded as the toughest entrance examination in the world

0celouvsky

I know what the JEE is

skill patrol

Show some empathy then.

0celouvsky

12:37

@skillpatrol I am grateful that they moved that trash from this room, that's the extent of my empathy right now.

skill patrol

empathy

Noun: empathy (countable and uncountable, plural empathies)

Identification with or understanding of the thoughts, feelings, or emotional state of another person.
She had a lot of empathy for her neighbor; she knew what it was like to lose a parent too.
(parapsychology, science fiction) A paranormal ability to psychically read another person's emotions.

0celouvsky

I don't know what any of those mean.

3. sounds neat.

skill patrol

Yeah, TIL.

Yashas Samaga

lol

user129412

I had a conceptual question about modelling physics on a discretion mesh, but I don't think it warrants an actual question. Say you have some discrete chain of length L. Each site has some onsite energy t, and each site can hop to its neighbour with a hopping energy h. Now, an electron traveling across the chain would be like a semi-free electron.

But what if we now cut the chain into 3 equal parts, in two different ways: either we make the two cuts by setting onsite energies to something like 5*t, creating a potential barrier, or we set the hopping energy across the cut to something like h/5, creating a weak link. How do these scenario's differ, with regions separated by a weak link or by potential barriers?

Yashas Samaga

12:51

"Actually string theory is very close to God speaking this universe into existence. And those "laws" where put in place by Him. No matter how you cut it this universe is highly organized and based on our observance of reality we know information cannot be produced randomly."

:o

0celouvsky

no need to post all of this in the chat

@ACuriousMind why is $P(\alpha)=|(\alpha,\psi)|^2$. Why the $^2$?

ACuriousMind

What kind of question is that? Without the square it would not be a real number!

0celouvsky

@ACuriousMind $P(\alpha)=|(\alpha,\psi)|$ should be real. Guess not...

4 real tho, that's a real number

ACuriousMind

You're gonna have to ask a better question for me to give a better answer :P

Yes, it's a real number- so what?

0celouvsky

@ACuriousMind can you give a better reason for the born rule other than (i) it works (ii) analogy with intensities in EM

ACuriousMind

13:07

@0celouvsky 1. I think "it works" is a most excellent reason. 2. If you view states as functionals on operators by their density matrix, then that's what taking the expectation value of the projector onto $\alpha$ with respect to the state given by $\psi$ yields.

That is, one may see the Born rule as a very specific instance of the rule for expectation values

(you still have to postulate that it's the projector that is related to measurement/probability, but you don't have to postulate the mathematical form of the probability anymore)

0celouvsky

Ah, thanks for reminding me about 2. That's good, assuming I accept density matrices.

Emilio Pisanty

@DanielSank booya

ACuriousMind

You...couldn't make it 10000? :P

Emilio Pisanty

@ACuriousMind I forgot I was that close to the line

I'll make it a round thousand for the 11k pass

0celouvsky

@BernardoMeurer

Hugh Mungus

589 1 12

0

Q: What is the significance of scattering cross section?

I am trying to calculate the scattering cross section for X-ray generation and how it is affected by material thickness, detector angle and how it differs between AlGaAs and GaAs. To calculate the scattering cross section I will be using the following equation: $$σ=〖𝑋−𝑟𝑎𝑦𝑠〗_𝑡𝑜𝑡/(𝑁𝑢𝑚𝑏�...

electrons scattering-cross-section

mmm that formatting

ACuriousMind

13:11

@EmilioPisanty The people-who-are-needlessly-obsessive-over-round-numbers of the world thank you

Emilio Pisanty

anyways, I'm still 3k short of my goal

ACuriousMind

Why is 13k your goal, exactly?

Emilio Pisanty

@ACuriousMind well, it's a moving goal

awarding more rep in bounties than Daniel Sank's total rep

ACuriousMind

The goal is "offer 3k more bounties than I already have"? ;)

@EmilioPisanty Oh, lol

0celouvsky

Apparently one cannot dualize a non-densely defined operator.

Dualizabiltiy is equivalent to densely defined

Emilio Pisanty

13:15

0

Q: How physicists can observe events at big scales such as a star birth

I read recently multiple articles about physicists observing birth of a star, or a star swallowed into a Black Hole. However I can't manage to understand how these phenomenas are observable at such scale. Common sense would lead to think that, the bigger the object you observe is, the bigger the...

time stars galaxies

who are these amateur-astronomer medics?

dammit

ACuriousMind

@EmilioPisanty Jerry just edited it :P

Emilio Pisanty

physics.stackexchange.com/posts/322426/revisions

@ACuriousMind beat me to the trigger

ACuriousMind

I was in the course of editing it but I couldn't resist to leave a comment first that physicians would likely observe other types of births, so I was too slow

0celouvsky

lol

Emilio Pisanty

> Astronomy turns out to be full of astronomically large numbers.

zinnnggg!

0celouvsky

13:25

I would really like to write an intro topology/analysis book

Bernardo Meurer

@0celouvsky I want to transcribe my analysis I & II notes

0celouvsky

why?

Bernardo Meurer

Just because

0celouvsky

learn latex

Bernardo Meurer

I know LaTeX

0celouvsky

13:27

then what's the issue?

Bernardo Meurer

Lazyness

0celouvsky

the usual fare for human folly

...who calls the Euclidean norm the $\ell^2$ norm

that's just showing off

Emilio Pisanty

@0celouvsky that's pretty standard fare

0celouvsky

@EmilioPisanty oh please

you don't have to disagree with everything I say you know

Emilio Pisanty

@0celouvsky I mean, depending on context

@0celouvsky I don't

oh the irony

0celouvsky

13:30

what's ironic?

Emilio Pisanty

@0celouvsky that that comment was itself a disagreement

mild irony if you will

0celouvsky

oh i see

AccidentalFourierTransform

no you dont

0celouvsky

@EmilioPisanty the context is: "Let $X$ and $Y$ be $n$-dimensional euclidean spaces normed by the $(l^2)$-norm."

Emilio Pisanty

@0celouvsky that's a bit pompous, yes

0celouvsky

13:33

thank you

Emilio Pisanty

but if you're in $\mathbb R^\mathbb N$, though

0celouvsky

Oh, sure

there's lots of norms on that space, I agree

but on $\Bbb R^n$ you should just say "the usual norm"

lol I bet some Portugese analysis prof (@BernardoMeurer ) only uses the downtown metric on $\Bbb R^n$

Bernardo Meurer

The hell's that?

0celouvsky

$||x||=\sum_{i=1}^n |x_i|$.

It's honestly the best norm for proving things.

Actually $||x||=\max_{i=1,\dotsc, n}|x_i|$ is the best norm

Yashas Samaga

"Since when is "synthetic life" considered life? And did they start from nothing? And doesn't it still entail intelligent design? And didn't they actually do a transplant and label it "life?""

they ran out of arguments

after I sent a link showing artificial life created in the lab

:D victory

AccidentalFourierTransform

13:49

they have a point though

JC died for your sins you know

you're being kinda disrespectful to the son of god

0celouvsky

I need to watch JC superstar again

AccidentalFourierTransform

Yashas Samaga

DYED :O

0celouvsky

do they even have Jesus in India?

AccidentalFourierTransform

JESUS IS EVERYWHERE

Yashas Samaga

13:51

Minority

AccidentalFourierTransform

HE SAW WHAT YOU DID THIS MORNING

Yashas Samaga

@AccidentalFourierTransform you are kidding, right?

0celouvsky

@YashasSamaga He might be kidding but I'm not

Be good or you won't get into heaven

Yashas Samaga

I'm good. I save insects. Feed around 20 cats daily. Sugar for my ants. Water for birds.

0celouvsky

very good

20 cats tho?

skill patrol

13:54

All life is sacred.

Bernardo Meurer

^ Not true

Yashas Samaga

@0celouvsky imgur.com/a/8bpXk

AccidentalFourierTransform

Bernardo Meurer

What does it mean that I can only write good code listening to this:

BRAIN TENTACLES - S/T (Full Album Stream)

►

AccidentalFourierTransform

Richard Hendricks codes to dubstep

0celouvsky

14:00

where do you even find 20 cats

@YashasSamaga so skinny :(

Yashas Samaga

aw are you spying on me?

0celouvsky

what?

Yashas Samaga

you are checking other pictures in my profile? :P

0celouvsky

what?

I clicked on the link

it's a bunch of cats

Bernardo Meurer

@AccidentalFourierTransform Very good series :)

Yashas Samaga

14:02

How do you know I am skinny?

0celouvsky

the cats are skinny

Yashas Samaga

lol

0celouvsky

i.imgur.com/zmgrCET.jpg?1

Yashas Samaga

I have some pics from IISc in my profile too

I thought you were seeing those

skill patrol

I missed that ping @0celouvsky

0celouvsky

14:03

I don't know/care how to get to your profile

@skillpatrol I didn't ping you, my browser was acting on its own

idk what happened

Yashas Samaga

I weigh just 42kg and I am nearly 180cms tall

0celouvsky

eat more

Fawad

@YashasSamaga you don't have math subject in 12th? Why don't you answer on math site?

0celouvsky

you're gonna get blown away in the wind

Bernardo Meurer

@YashasSamaga Impossible

Yashas Samaga

14:07

@Fawad I had Physics, Chemistry, Mathematics, Computer Science, Data Entry, Economics, English + some more

I can't remember

@BernardoMeurer I am not joking. I am serious. My BMI was around 11.4 last month.

Bernardo Meurer

@YashasSamaga That's silly, eat more

Yashas Samaga

Yea, I am severely underweight. I was 37 few months ago.

skill patrol

Skinny people live longer.

Yashas Samaga

:O

Bernardo Meurer

@skillpatrol Not true, slightly overweight people live longer

Yashas Samaga

14:08

I am incredibly skinny. Even girls have more muscles than me.

Bernardo Meurer

^ Sexist

Yashas Samaga

I'd probably lose a fight with a girl :d

@BernardoMeurer Nope. Girls don't have testosterone.

Testosterone causes muscle growth.

skill patrol

@BernardoMeurer prove it.

Fawad

> I'm a nurse who also believes that having a few extra pounds on is better than being too thin. In my experience extremely thin people were just as likely to die as obese people were. I'm aiming for the low overweight range. Healthy but with enough meat on to withstand a catastrophic illness should it come.

sparkpeople.com/…

ACuriousMind

@BalarkaSen Can you fathom how an application of the Seifert-van Kampen theorem to a gluing $X_1\cup_f X_2$ might result in $\pi_1(X_1\cup_f X_2) = \pi_1(X_1)\times \pi_1(X_2)$?

Bernardo Meurer

14:11

@skillpatrol scientificamerican.com/article/…

There was another study by UCSB who agreed with the results IIRC

Yashas Samaga

:(

Balarka Sen

@ACuriousMind Maybe you're killing the commutators while gluing by $f$?

0celouvsky

@YashasSamaga I don't know what point you're trying to make

you're not impressing anyone, not getting sympathy

ACuriousMind

@BalarkaSen Argh, I'm silly. $\mathbb{Z}_p\to \mathbb{Z}$ must be trivial, not the other way around! Sorry to summon you :/

Yashas Samaga

@0celouvsky I was sad becaz slightly obese people are going to live longer and I'm skinny.

0celouvsky

14:15

that's on average, dude

slighly overweight people can still die early

and you might be hit by a car in a month, who knows

I wouldn't worry about death quite yet if I were you

@ACuriousMind so that's not generally true for gluing maps?

ACuriousMind

@0celouvsky Hmm, what?

0celouvsky

@ACuriousMind this is generally not true?

Balarka Sen

No.

0celouvsky

I didn't think so

So what is that question supposed to mean

Balarka Sen

$\pi_1(S^1 \vee S^1)$

ACuriousMind

14:18

@0celouvsky No - if you glue "trivially" you get $\pi_1(X_1)*\pi_1(X_2)$.

Yashas Samaga

I asked few people to stop equating anti-athism with anti-science and this is what I got in reply: "Accepting the theory of evolution reflects a deep need to evade the overwhelming evidence of logic that proves the existence of God. Atheism is not based ont he notion of evolution. Evolution is based on the folly of atheism."

ACuriousMind

@0celouvsky I was about to give details on the situation and explain why I thought it couldn't work when I realized the problem I thought there was wasn't there

0celouvsky

@BalarkaSen yes, of course

@ACuriousMind ok

never mind them

Slereah

14:29

Hello

skill patrol

Hi pal

Slereah

Hawking-Ellis totally has the fancy definition of singularities

Although in less details than the Ellis paper

skill patrol

What does Penrose say?

Slereah

I think Penrose is still the old definition of singularities?

Geodesic incompleteness

0celouvsky

14:52

Daily reminder that penrose is insane.

So I need to show that $P^2\#P^2\# P^2\approx K\#P^2$

damn gluings

Balarka Sen

@0celouvsky one of my favorite surface theory results, that

Slereah

The hell is $K$

0celouvsky

Klein bottle

Balarka Sen

Klein bottle

Slereah

Oh

0celouvsky

14:55

@BalarkaSen I know that $K\approx P^2\#P^2$, but it's not working out with the gluings once I concatenate another $P^2$

Slereah

Can you prove that via fundamental polygons?

Balarka Sen

@0celouvsky Oh, wait a second, that's not what I thought my favorite result was. Well, that's obvious.

0celouvsky

What's obvious?

Balarka Sen

K # P^2 = (P^2 # P^2) # P^2

0celouvsky

I know that it's true, but I can't figure out the gluings

Balarka Sen

14:56

You want to show that via polygons?

0celouvsky

I don't want to, that's the assignment.

Slereah

Isn't there an identity $A \approx B \to f(A) \approx f(B)$ for homeomorphisms?

0celouvsky

@Slereah what?

Slereah

Well it's true for equality

So I dunno, might also be true for equivalence relations with some conditions

Balarka Sen

Well, so you'd essentially want to show K = P^2 # P^2 via gluings right?

0celouvsky

14:58

No, that's easy

I can't figure out the moves when there's another thing attached.

Balarka Sen

If you know that then concatenating a P^2 is super-easy

0celouvsky

Super easy for you I guess

I've been here for three hours

ACuriousMind

You don't know that the connected sum is associaive?

Slereah

associative

Balarka Sen

You just have to show K minus a disk is P^2 # P^2 minus a disk, @0celo7

0celouvsky

14:59

what?

ACuriousMind

Well, you have to, otherwise there should be brackets up there, but then I don't get what there is to show

Balarka Sen

Via polygons

0celouvsky

@ACuriousMind I do not know how to get a gluing from $abab^{-1}cc$ to $aabbcc$

I can get one from $abab^{-1}$ to $aabb$

ACuriousMind

@0celouvsky You don't need to, the associativity of the connected sum follows much more generally. Or is the assignment for some reason forbidding you to use that?

Balarka Sen

Cut out the $cc$, get $abab^{-1}$ to $aabb$ and glue back the $cc$

@ACuriousMind Apparently he is assigned to do it by polygons of the surfaces.

Essentially, what you have to do is get $abab^{-1}d$ to $aabbd$ where $d$ is an open edge.

0celouvsky

15:02

I know that

maybe it does work out

I think I needed to reflect this piece and didn't

Balarka Sen

I am not sure exactly where you are having troubling mimicking the moves of $abab^{-1} \to aabb$ for that.

Ah ok

Anyway, my favorite result is that T^2 # P^2 = K # P^2.

0celouvsky

@BalarkaSen that's what I have to prove via gluings

well, that $abca^{-1}b^{-1}c$ is $\#^3 P^2$

but your result is an intermediate step, at least with the scheme I'm using

@BalarkaSen ok it works out now

this is turrble

Balarka Sen

Ahh. Yeah, it's actually not so obvious. Downright surprising when I encountered it actually (twists in K get cancelled after you connected sum with a P^2? Why?!)

@0celouvsky Cool.

0celouvsky

@BalarkaSen yeah these things are strange. $abca^{-1}b^{-1}c^{-1}$ is just $T^2$

Balarka Sen

There's a complex analogue of that result btdubs. $(\Bbb{CP}^1 \times \Bbb{CP}^1) \# \overline{\Bbb{CP}^2} \cong \Bbb{CP}^2 \# 2\overline{\Bbb{CP}^2}$.

Emilio Pisanty

15:09

@ACuriousMind this one might've benefited from the canned from-review message of 'ask it separately'

Balarka Sen

@0celouvsky Yeah, agreed.

Slereah

more importantly what is this

ACuriousMind

@EmilioPisanty You're right, I've added a comment

@Slereah that's at least a manifold you can write down easily

AccidentalFourierTransform

not nice

ACuriousMind

0celouvsky

15:16

@BalarkaSen I still don't get this overline business for complex manifolds

Balarka Sen

@0celouvsky It's pretty nice actually. $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ and $\Bbb{CP}^2 \# \Bbb{CP}^2$ are not even homotopy equivalent, let alone homeo/diffeomorphic.

(Also: the former is boundary of a 5-manifold, the latter is not)

Yashas Samaga

Whoa! I am fighting against a bunch of people who are defending rape. Religious people defending rape. WTH?

ACuriousMind

The longer I stare at that picture, the less I know if it actually helps understanding what is going on.

John Rennie

@YashasSamaga I really don't want to hear about this

Yashas Samaga

I'm sorry.

This makes me hate the world. I just can't believe what kind of people exist. I won't bother fighting with them again. They are not sane.

pi = 3 and 2 + 2 =5; ignorance taken to the next level.

Balarka Sen

15:23

@YashasSamaga Interesting reply tho

ACuriousMind

Monty Python - Always Look on the Bright Side of Life

►

Yashas Samaga

When they have no answer, they write some rubbish such as 2 + 2 is not 4.

ACuriousMind

Lighten up, people :P

2

skull petrol

^

I warned you.

Balarka Sen

@ACuriousMind In the Scarecrow sense?

ACuriousMind

15:24

@BalarkaSen I don't know which one that is, but I'm gonna guess that's not what I meant ;)

Balarka Sen

Ya need to lighten up.

►

skull petrol

@YashasSamaga remember this is the internet.

Slereah

"Now, you cannot even construct approximate solutions to the initial value problem reliably using numerical methods, since ultrahyperbolic equations do not have finite-speed of propagation, so you cannot "localise" the problem, and small changes around a point x may almost instantaneously affect the solution at a far-away point y."

Fairly bad

skull petrol

Always remember that @YashasSamaga

ACuriousMind

@Slereah Why are you looking at ultrahyperbolic equations?

Slereah

15:27

why not

(They are used for spacetimes of signature $(p,q)$)

AccidentalFourierTransform

why don't you Wick rotate? that fixes everything

ACuriousMind

@Slereah I know. Why do you need more time? Is one time not enough for you?!

Slereah

Is wick rotation well defined in that case?

@ACuriousMind I need all the time in the world!

AccidentalFourierTransform

I guess we have Cauchy for $\mathbb C^n$, right?

ACuriousMind

Wick rotation is not actually a geometric operation but a statement about Euclidean QFT results being amenable to analytic continuation to obtain Minkowskian results.

AccidentalFourierTransform

15:29

$\oint f(z)\mathrm d^nz=0?$

ACuriousMind

I've talked about that and OS reconstruction here before but it escapes my memory with whom

AccidentalFourierTransform

I was there.

ACuriousMind

@AccidentalFourierTransform ...and all you got was not even a lousy T-shirt?

AccidentalFourierTransform

I got a MAGA hat

Slereah

^me

10

Q: Stokes' theorem etc., for non-Hausdorff manifolds

This question is prompted by another one. I want to motivate the definition of a scheme for people who know about manifolds(smooth, or complex analytic). So I define a manifold in the following way. Defn: A smooth $n$-manifold is a pair $(X, \mathcal{O}_X)$, where $X$ is a topological space an...

differential-topology smooth-manifolds

DanielSank

15:54

@EmilioPisanty -_-

0celouvsky

@ACuriousMind that's random

Balarka Sen

@Slereah Eh.

0celouvsky

@ACuriousMind me

Emilio Pisanty

@DanielSank ::grin::

0celouvsky

@BalarkaSen So my strategy for $abca^{-1}b^{-1}c$ is to use a few "obvious" cuts to go to $P^2\#T^2$. Then I go to $P^2\# K$, which is pretty nonobvious using just cutting and pasting, and finally to $\#^3P^2$, which is again easy.

I think this is 7-ish cuts in total

Way better than my original 20!

Transcript for