@ACuriousMind I made it out ;)

@dmckee @ChrisWhite Either of you around?

Sorta. What's up?

@dmckee Do you know memory bug debugging in C++?

I'm out of practice and might not know your particular environment, but I've done it in the past.

Do you have a leak, or something more concrete?

Can you make it do it on command or does it pop up unexpectedly?

*** Error in ./ackermann.o': double free or corruption (fasttop): 0x0000000000c`

as soon as it starts

I know where the bug is

I just don't know why it happens

github.com/bemeurer/multi-ackermann/blob/master/ack.cpp

Know how to run with the debugger?

@dmckee I know how to run it, just not what to do with it

First question: does the bug occur with the debugger running?

If no, you probably have an uninitialized variable.

If so, then when you get the crash ask the debugger what line it occurs on.

That's not it I think. The bug only happens when I try to clear memory of variables I no longer need. Let me try this into gdb

All those commented out mpz_clearss?

Yep

Even if I use only the first one (that clears m and n) it breaks

are function arguments constants in C?

no right

No, though changes to them are in local execution context only.

Yeah, so I thought

You don't have a lot of code and aren't calling new/delete (or alloc/free) yourself. SO the problem is likely one of understanding how the library is suppose to work. I've never used GMP so I can't help much.

It doesn't crash on Valgrind .-.

AH

AAAAH

Oh. Uhm. Your m and n are local to main and are loop variables.

Got it :)

you shouldn't call clear in the C++ API

Mmmm. Good. But I'd also think carefully about what you should be doing to those variables; their automatic and shouldn't be delete(free)ed.

@dmckee Yeah, I'm still getting the hang of C++. It's a funny language

Well, it's built on a C foundation, and C is a very raw language. On 1970s hardware there was very nearly a 1-to-1 correspondence between language constructs and features of the chips.

Since then the chips have gained more features, but C still presents almost the same virtual machine.

C is cool, but I just can't get around to using it more, it's too much work to do simple things at times

and string management is dreadful

A lot of pros don't use the stdlib.h and string.h string interfaces for real projects. They bring in safer and more featureful libraries to manage strings for them.

If you do use them you use getline for IO and always use the n variants of the string functions.

What I like about C is that it's so kick in the balls

it takes 0 bullshit home

that makes you better

It will teach you discipline or break your spirit in the attempt.

Lol

Have you guys ever written a toy code to simulate and calculate diffusion coefficients?

In double slit interference, will we see interference pattern, if we send protons from one slit and neutrons from other?

Why not? They both have a de Broglie wavelength.

unicamp.br/~vitiello/f6892s04/nw.pdf

@Mikhail In feynmanlectures.caltech.edu/III_01.html , they had problem in identifying which slit did electron came from. If they pass proton from one slit and neutron from 2nd slit, can't problem be solved. They know which slit did particles come from and also they would see interference pattern?

The experiment was done already with nuetrons, indeed it is not clear if there were some stray protons in the group.

The real problem with using protons it that I expect them to anti-bunch due to the charge.

What is anti bunch?

Was experiment tried actually for electrons?

@ChrisWhite Got it. Thanks for the tip :)

@AnubhavGoel electron interference is the mechanism for electronic (SEM) image formation. Actually people have been able to interfere larger multi-atom structures.

Also the famous bose-einstein condensation interference

@JohnRennie hello

Morning

Or I guess it isn't morning in your part of the world :-)

@JohnRennie Today I learned of a cool way of integrating over a nonorientable Riemannian manifold

I probably don't know enough about the subject to be suitably impressed :-)

When you say integrating over a manifold do you mean an integral like $\int something dx^n$ ?

@JohnRennie Yes.

Why are non-orientable manifolds different from orientable ones in that respect?

@JohnRennie $\mathrm d^nx$ encodes orientation so it's only defined for orientable manifolds.

Aaaaaaaaaaaaah, I get it.

There's now two solutions to the problem that I know of

I remember last year, I got an A- in physics 'cause the teacher hated me

OK so what do you with a non-orientable manifold?

Said I asked too many questions above the scope of the class

@JohnRennie It turns out you can define a scalar density known as the Riemannian density

In the orientable case it's just the volume form $\sqrt g dx^1\wedge\cdots\wedge dx^n$

Hello, Anyone here worked on VASP ?

@JohnRennie you basically construct it like this:

you pick a bunch of open sets that cover your manifold, and each one of these open sets is an orientable Riemannian manifold

then you take the volume forms on each of these pieces and glue the whole thing together

@Gowtham I've seen few comments about VASP, but right now is a bad time as most of the site members are asleep. Around 17:00 UTC is the liveliest time.

this isn't really well-defined but it turns out if you integrate it, it is

so you get a density $\mu$ and then you can integrate functions like $f\mapsto\int_M f\mu$.

But there's a really neat way that I learned today

@JohnRennie thanks

The key concept is something called the "cut locus." It's a set of the manifold that basically describes where geodesics fail to minimize length.

The cut locus of a point $p$ is denoted $C(p)$

Google Google

This means that $M-C(p)$ is contractible

By a theorem of topology, any fiber bundle defined on $M-C(p)$ is trivial

In particular, $T(M-C(p))=(M-C(p))\times\Bbb R^n$

By another theorem of topology, this implies that $M-C(p)$ is orientable

$M-C(p)$ is what's left if you snip out the troublesome bits?

@JohnRennie I guess so! I'd have to explicitly calculate it for some nonorientable manifold to be sure!

But it's often said that $C(p)$ carries all topological information of $M$

Oh I forgot

If you give $M$ a Riemannian metric you need a theorem by Nomizu and Ozeki that says there is a complete metric globally conformal to $g$

And we use this conformal metric to calculate $C(p)$

cut locus stuff only works with complete metrics.

So you do the integration on $M-C(p)$ then do some clever trick with the remaining bit?

So, because $C(p)$ has measure zero, $\int_{M-C(p)}\equiv \int_M$

@JohnRennie no, that's the cool part

@0celo7 Aaaaaaaaaaaaahhh (again)

it's like integrating over $\Bbb R^2$ vs. $\Bbb R^2-\{0\}$

there's no difference

That feels somehow like cheating ....

But on ${M-C(p)}$ you can define an orientation

But I guess it works.

So you just restrict your original metric to this set

and find the volume form

and ta-da you used a bunch of high-powered theorems to define the integral in a really cool way

Peter Higgs says Brexit ‘a disaster’ for scientific research / irish times

Leonard Susskind’s Open Letter on “The Lunatic” / aaronson blog

I'd be cautious about anyone's views on the consequences of Brexit. The truth is that no-one knows what's going to happen.

My guess is that things will change less than we think. It's in everyone's interests to find a compromise solution that keeps things much as they are.

@ChrisWhite :-)

Soap is dreadful stuff. It destroys cell membranes. You'll hear people say that soap is very good at killing bacteria. Well yes, that's because it's very good at killing everything including your skin cells. The reason it stings when it gets in your eyes is because it's killing the cells in your eyes!

Cell membranes are basically surfactant bilayers, where the surfactants are phospholipids. Foreign surfactants disrupt this bilayer structure and rupture the cell membrane, which kills the cell.

Soap is a mixture of sodium salts of carboxylic acids, and these happen to be particularly good at disrupting the phospholipid bilayers.

By contrast sodium lauryl ether sulphate doesn't get into the lipid bilayers so it doesn't damage them. Well, that's not strictly true, but it damages them a lot less.

See, colloid science is just awesome! I can't understand why everyone doesn't want to study it :-)

@JohnRennie Because colloid sounds like something nasty

"Yeah he couldn't stop seizing so he got his colloids removed"

@JohnRennie you there?

Hi Kevin

hey

So if I had an idea, can you help me bring it to life?

I have not even computed anything yet , but wanted to discuss it first

Ok, fire away.

Well, I wanted to use maximum entropy-like methods to constrain some possibilities of correlation functions(T_uv stuff) in ads/cft, and have some this should give me some list of plausible gravity duals with special properties.

I don't know how to quite talk about this but I can draw some diagrams or explain in longer form what I am thinking

That's outside my area of expertise I'm afraid.

Oh darn

Try asking later when more of the regulars are around.

will do

what is your expertise?

susy?

I trained as a colloid scientist, however I do GR for fun. But I only know a little QFT and no string theory at all.

yeah, I know even less, but have some silly ideas on occasion lol

See what happens when physicists get interested in cooking:

I'll have to devise an experimental protocol to explore all the possibilities :-)

@ACuriousMind Here you find the decomposition of two fundamental(and anti-fundamental) of SU(N). Is the expression of the projectors true? I should expect that when I take $N=2$, the projectors are equals $P_1 = P_A$ and $P_S = P_{Adj}$ but it doesn't seems true....

@FrancescoS I'm sorry, I have no idea what exactly the indices on the projectors are supposed to mean.

However, I suppose that they become equal in the $N=2$ case only after you've written down the isomorphism that sends your abstract definition of $\mathbf{N}$ to $\bar{\mathbf{N}}$.

@ACuriousMind Ok, thank you. This was the answer ;)

Shouldn't the tensor product of two fundamentals contain the trivial irrep (the trace), i.e. $\mathbf N\otimes\mathbf N=\mathbf 1\oplus\mathbf A\oplus\mathbf S$?

@Bass You're thinking of $\mathrm{SO}(N)$, this is for $\mathrm{SU}(N)$

Oh right.

Also, nothing there really says that either A or S is irreducible

@ACuriousMind But for $SO(n)$ they are, right? I mean in $\mathbf N\otimes\mathbf N=\mathbf 1\oplus\mathbf A\oplus\mathbf S$.

@Bass I think so, yes.

@ACuriousMind About the relative homology stuff: Isn't the name a bit misleading? The term "relative homology" $H_n(X, Y)$ seems to suggest that it "captures those holes" of $X$ which are "not in $Y$". But $D^n$ has trivial homology, and yet $H_m(D^n, S^{n-1})$ is nontrivial, because $D_n/S^{n-1}\simeq S^n$. Or did I miss something?

@Bass $H_\bullet(X;Y)$ detects not "holes" (cycles that are not boundaries), but chains whose boundary lies in $Y$ and that are not boundaries up to chains in $Y$. That is, you essentially declare all parts of chains that lie in $Y$ to be irrelevant.

The formal statement is that $H_\bullet(X,Y)$ is the ordinary homology of the space you get when you attach a cone to $A$ - since cones are contractible, all the chains living on $A$ become trivial since you can "contract them to the tip of the cone", so all the homologically relevant chains must live outside $A$.

So $H(X,Y)\simeq H(X/Y)$ is always true? You just contract $Y$ to a point and the resulting homology is the relative one.

@Bass That one is only true if $Y$ has a neighbourhood that deformation retracts onto $Y$

@ACuriousMind True, I read that on Wiki, but I could not think of a counter-example. What space has a neighbourhood that doesn't defo-retract to it?

What's always true is $H_\bullet(X,Y)\cong H_\bullet(X\cup_f \mathrm{cone}(Y))$, where $f$ maps $\{1\}\times Y$ to $Y$. More explicitly, $X\cup_f\mathrm{cone}(Y)$ is the space $X\cup ([0,1]\times Y)$ modulo the relations $y\sim (1,y)$ and $(0,y)\sim(0,y')$

Any example of a space $Y\subset X$ that has no neighbourhood that deformation-retracts onto $Y$? In the normal $\mathbb R^n$ topology this doesn't exist, right?

@Bass I think you get only rather pathological examples for that. For example, the Cantor set is totally disconnected, so there are no deformation retracts onto a single point, as deformation retracts preserve connectivity

Hmm, getting confused. Does a neighbourhood $U$ of a closed subspace $Y\subset\mathbb R^n$ defo-retract onto $Y$?

For most $Y$ it does. I'm willing to bet there are strange subspaces that don't admit a deformation retract

For example, my example above makes me think that the Cantor set as a subspace of $\mathbb{R}$ also isn't a deformation retract

@ACuriousMind I see.

@ACuriousMind But can an open set defo-retract onto a closed set?

Maybe there are simpler examples...is $\{1/n\mid n\in N\}\cup \{0\} \subset \mathbb{R}$ a deformation retract?

@Bass Sure, $(-0.5,0.5)\times [0,1]\to (-0.5,0.5),(x,y)\mapsto (1-y)x$ deformation retracts the open interval $(-0.5,0.5)$ to the point $\{0\}$.

Oh, of course. Thanks!

What is this gem doing in the "Trash"?

@ACuriousMind How do you translate the idiom "go easy"

Like "go easy on the sauce"

One can go easy on a sauce?

I only know that idiom as going easy on someone, not something

@ACuriousMind It means "don't go overboard"

Then (Sei) langsam mit seems to fit

langsam?

Slow?

wat

To me, someone saying Langsam mit X means that you should be cautious with X

@ACuriousMind Ok, I'm feeling a Döner craving but I want to tell the lady not to drown it like she did last time

Why not just say Wenig Soße, bitte, for that?

Because that's obvious.

@ACuriousMind Apparently it's really hard to show that the holonomy group is a Lie group, but it's easier to show that the restricted holonomy group is a Lie group because it's compact. (a) why is it compact and (b) why does this make it easier?

@ACuriousMind Do the contractible loops form a compact subset of the loop space of the manifold?

(beats me what topology I'm talking about)

@0celo7 (a) no idea (b) no idea

@0celo7 I don't think so - all loops in $\mathbb{R}^n$ are contractible but I don't think the loop space is compact.

@ACuriousMind Do you not need holonomy for your...stringy gauge stuff?

Okay, maybe I do have an idea for (b): If you can show it's compact, then you can probably conclude it's closed, and closed subgroups of Lie groups are Lie groups themselves.

@ACuriousMind Of course that's true given that the holonomy group is a Lie group...but my understanding is that one can show it without that.

And that the straightforward proof that it's a Lie group is easy.

(The full proof is in Kobayashi-Nomizu, I'll check it out.)

@0celo7 In my world, both the holonomy group and the restricted holonomy group are already defined as subgroups of the gauge group

(You didn't ask me a Riemannian geometry question again knowing that I only speak gauge theory, right? :P )

But they probably do everything in terms of principal $G$-bundles, so it would take some work to get it to make sense with the Levi-Civita connection.

@ACuriousMind o.O

DEFINED

You have to prove that, bruh

It's not a hard proof but still

I wonder if the proof is in Helgason.

@0celo7 No, the holonomy at $p$ is $\{g\in G\mid p\sim pg\}$ where the relation $\sim$ is "there is a horizontal curve starting at $p$ and ending at $pg$.

Nothing to prove here :)

@ACuriousMind You have to prove that it's a group ;P

@0celo7 That's obvious :P

The inverses are the paths in reverse direction, and multiplication is concatenation

@ACuriousMind ...I never said it was hard.

@ACuriousMind Hmm, but a generic subgroup of a Lie group is not a Lie group.

So don't you still have to prove that that gives you a closed subgroup?

Yes

Ok, what's the idea there?

No idea, physicists never prove that :P

Off to the wonderful world of Kobayashi-Nomizu!

...which is not in this building.

@ACuriousMind See this is why I need to carry all my books with me at all times

(a) builds mooscles (b) I can look things up when I need them

@ACuriousMind Do you know if there's a relation between holonomy groups on principal bundles and holonomy groups on the associated bundles?

I think the relation is that they're the same? But I don't know that

The proof, if that's true, is probably in Kobayashi-Nomizu.

Black box theorems are fun

Please unhold my question regarding extremely low temperatures measurement. I edited it upon your request.

@ACuriousMind nicekicks.com/air-jordan-12-wool-launches-fall

@LeonKigelman Edited questions automatically enter a reopen review queue. There is no need to post about them in this chat.

@0celo7 ???

@ACuriousMind suggested $wag for you

Although the wool might not work well with your current climate

@ACuriousMind here's something for your bad weather nicekicks.com/air-jordan-12-waterproof-neoprene-another-look

So anyway

Thinking about it

A spacetime with compact CTCs but no closed causal geodesics technically avoids TWO chronology protection theorems

The one on energy conditions and the one on semiclassical gravity

Although to be fair, all examples I found still violated the NEC

So odds are good this still applies

Also all the spacetimes I found are like

Using what Krasnikov refers to as square roots of CTCs

Spacetime shortcuts

Which also all violate the NEC

@0celo7 I am (I read part of G&P, too).

@Danu Do you have it handy?

@0celo7 Not physically.

(but yes)

@Danu Can you view page 36?

Specifically the second part of Fig. 1-25

Sure

Instability of non-transversal intersection

@Danu I don't get the second part

Seems like they perturb it and the result is transversal but it wasn't at first

i don't know what they're trying to say

@0celo7 That's the point.

@ChrisWhite What the hell, really?! Does he really think that mathematicians should be doing the job of historians?

I also find it strange that a moderator would go out of his way to call a related SE site tragic/(implicitly) useless

@ChrisWhite Did he just delete those comments?

Did you flag any?

Hmm, strange.

@ChrisWhite I am actually spending quite a lot of time in the chat room, nowadays. Mike Miller, Ted Shifrin and Balarka Sen are pretty nice, and I learn a lot from them.

The main site, however, remains a ghastly place.

:: h bar whistles off-key while staring distractedly into a distance corner of the room ::

Speaking of differences between chat and the main site, this

is actually quite a cute exercise.

I found what appears to be a solution using first semester methods by requiring consistency between the linear and angular accelerations of the CoM, but would like to check it in Lagrangiam mechanics with undetermined multipliers.

@dmckee :)))

Man, now people already leave aggressive comments towards me on their own answers just when they receive a downvote. I need not even leave a critical comment to be the suspect :/

@0celo7 3rd ed. nakahara delayed to february 2017.

crcpress.com/Geometry-Topology-and-Physics-Third-Edition/…

@Danu Ah, so nontransversality is never stable?

I find that hard to believe.

@3075 Ok?

I wasn't planning on reading it

@ACuriousMind Well stop maliciously downvoting people!

@0celo7 I am (mostly) free of malice.

My prof gave me some problems from GP but one of them requires I solve 4 others first -.-

@ACuriousMind Can I, in good faith, take those others for granted?

@0celo7 Uh, how am I supposed to judge that?

@ACuriousMind Since when do rubber ducks speak

::quacks angrily::

Hmm, apparently $(A\times B)\cap (A'\times B')=(A\cap A')\times(B\cap B')$ but this is not true for $\cup$

@ACuriousMind You're as free of malice as my calculator is free of will

and it just asked me for love and care

@yuggib Why does the Bulltin of the AMS have reviews of papers from the 80s

Why not current stuff?

@ACuriousMind I don't see why the above is not true for $\cup$

I proved it for $\cap$ by showing inclusion both ways

And it was pretty trivial

But why does the argument fail for $\cup$

@0celo7 Sets of the form of the r.h.s. $A\times B$ are "rectangles". Intersections of rectangles are smaller rectangles

@ACuriousMind Rectangles if $A,B$ are connected...

But unions of rectangles are not necessarily rectangles

Rectangles should not be allowed to get married

@ACuriousMind Call the LHS $X$ and the RHS $Y$. Let $x\in X$. Then $x=(a,b)$ where $x\in A\times B$ AND $x\in A'\times B'$. Clearly this is only possible if and only if $a\in A\cap A'$ and $b\in B\cap B'$. For suppose $a\in A$ but $a\notin A'$. Then $(a,b)\notin A'\times B'$, and similarly for the other cases.

@ACuriousMind Is that all one needs?

I don't even think you need inclusion both ways

You can just argue an iff directly.

But I don't see where this fails for $\cup$

@0celo7 Just...try to write the actual argument for $\cup$.

@Danu Mathematicians are like Frenchmen: whatever you say to them they translate into their own language and forthwith it is something completely different. - Goethe.

@ACuriousMind Oh.

So...what is the actual answer

$(A\cup A')\times(B\cup B')=?$

Uh...there is no "actual answer"

You can't rewrite that in a "nicer" form

$(A\times B)\cup(A'\times B)\cup (A\times B')\cup (A'\times B')$?

@ACuriousMind Yes, I think that's correct.

Let's check it with some intervals.

@0celo7 Yes, that's correct.

I'm getting less bad at set theory lol

@0celo7 That's why there are theorems.

@Danu which one specifically?

Consider two lines on top of each other.

There are plenty of perturbations that leave them nontransverse

Homotope one up slightly.

@Danu The stability theorem says that transversality is stable, not that nontransversality is unstable.

@ACuriousMind So, I was trying to prove rigorously that $\Bbb R^n$ is second countable but I needed to handwave. I know that the set of all open balls is a basis (for any metric space). I also know that the balls with rational radii form a subbasis. But it's the rational centers that are giving my trouble. My idea was to show that if $B_r(a)$ is a ball with rational radius and arbitrary origin, I can write it as the union of balls with rational radii and rational centers. Since $B_r(a)$ (cont.)

is open, for each point $x$ we have a ball with rational radius $\epsilon$ contained in this ball

So my idea was to write $$B_r(a)=\bigcup_{x\in B_r(a)\cap\Bbb Q^n}B_\epsilon(x)$$

I'm 99.99% sure this works but I don't know how to rigorously argue equality.

Clearly $\supset$ is true.

It's $\subset$ I'm having trouble with.

@0celo7 What is $\epsilon$?

@ACuriousMind The radius whose existence is guaranteed by the definition of open set

$\epsilon$ depends on $x$

@0celo7 You mean "one of the" not "the" ;)

@ACuriousMind so?

Nothing, it is of no particular importance

$\subset$ follows from density of the rationals in the reals

@ACuriousMind Clearly.

That's what I wrote, but that's too handwavey for what I'm looking for.

@0celo7 usually it has current stuff; sometimes they reproduce also some old "nice" review

@yuggib 2012 is the most recent one in this month's

then 2002

then 1997

It goes downhill from there.

@0celo7 If $x\in B_r(a)$ is already rational, then it is clearly in the r.h.s. If it is irrational, you can find a rational arbitarily close to it - and if you choose any rational within half the distance of $x$ from the boundary, then there is a ball around that rational that lies inside $B_r(a)$ and contains $x$.

@0celo7 I suppose it is a reprint of old nice reviews of the mathscinet

the book reviews are usually new

Where do I find mathscinet reviews, btw?

and the articles as well

(I have access to mathscinet)

@Danu mathscinet and zbmath

@yuggib 2013 and 2014.

in mathscinet you may find some reviews of mine as well ;-P

next I have to do is on chern-simons equation T__T

I know nothing about that

Guess you're learning gauge theory real quick

@ACuriousMind Hmm

@ACuriousMind Thanks, although that's basically what I thought

@ChrisWhite do you know why the proof environment is doing this?

The thing sticking out? It's because it doesn't want to break your math up.

@Danu Why can't it move $f\times g=\cdots$ to the next line

@0celo7 The sentence before is too short.

@Danu What would you do if this were your problem set?

Work on the sentence.

Don't write what $f\times g$ is in components - you don't actually use the components anywhere after that.

You can define that notation somewhere else.

@0celo7 Have you talked to your sister?

@BernardMeurer stop creeping holy shit

@ACuriousMind The solutions are partially for me too.

because I write them in components I can easily see that the derivatives are easy to write down.

@0celo7 Really though

@ACuriousMind can you please tell him to stop being a creeper

@ACuriousMind Is it just an unfortunate coincidence that the tangent functor, which associates a contravariant tensor with each point, is a covariant functor, and that the cotangent functor is a contravariant functor?