The h Bar

@0celo7 then how could you say that the fourier transform is an isometry of $L^2$ if you're not sure you can do the integration on any of its functions

0celo7

@yuggib well then I guess I can't do anything

might as well throw out everything because I can't prove basic analysis theorems

yuggib

12:08 AM

no no, I am just saying that you should take time

0celo7

I can't prove that the partial derivative is unique. Might as well throw out all diff geo books

Wait why do partial derivatives even commute?

yuggib

and that I, as the planner of math courses in a US university, would teach some analysis before PDEs

:-P

0celo7

Throw out all books that use calculus

yuggib

you said you know calculus

(or at least you are supposed to)

0celo7

I can't prove any of it

yuggib

12:10 AM

they should have taught you something in calculus 1-3

0celo7

Proofs? No.

Why should they

Engineers don't need them and mathematicians take analysis

yuggib

strange system

0celo7

Unless you go to a school that I can't pay for, that's the way it is (not sure it's different at those schools either)

And I'd probably drop out if I did math like the Europeans do it

yuggib

I don't think so...here they start with the basics usually

0celo7

You want me doing functional analysis!

That's pretty non basic!

yuggib

12:14 AM

we were talking about fourier analysis

0celo7

no we're talking about PDE for a first year student

or a middle schooler in your country

yuggib

yes, that I think it should be done later

0celo7

ok so I'll cancel the class and sit in on surgery theory?

dunno what you want me to do

yuggib

(in the two european countries that I know on an education standpoint, PDEs are done in the first year of master)

0celo7

yeah, here too

yuggib

12:16 AM

@0celo7 I don't want you to do anything

0celo7

but there's an undergrad course for people who are interested

yuggib

ok, but you can understand that there is a lack of first-year level books in PDEs, since all the world do them later on

0celo7

there's plenty

I've listed like 4 or 5 today

How do you think engineers learn PDE?

yuggib

after they have learned a bit of analysis as well?

0celo7

No

They don't need that

yuggib

12:19 AM

that's arguable

0celo7

they built the car you drive in and the plane you fly in without analysis

source: father who designed fighter jets

yuggib

I think they know a bit of analysis anyways

if else they have someone else, or a computer program, to do the dirty work for them

gotta go

it's getting late here on the old continent

and I have a bochner-minlos type theorem to prove (hopefully) tomorrow

(talking about Fourier transforms)

you'll have much more fun in $\sim$2 years when you could do some proper analysis ;-)

alarge

12:46 AM

As if engineers and physicists cared about defining and solving PDEs rigorously... Navier-Stokes says hi. Yeah, we'd be pretty boned going down that route.

Now I do enjoy when maths comes along and gives these enlightening rigorous frameworks. But there's a lot involved in figuring out how to write down new PDEs, doing approximations based on physical intuition and writing down code to give a solution of sorts. It's this latter part that engineers and physicists are usually taught and which is emphasized over the mathy bits.

0celo7

@alarge I want to have experience with both parts

alarge

And I am sure that you will if you keep reading up on stuff at the pace you've been going at thus far.

0celo7

@ACuriousMind Schwarzschild $\cong \mathbb{R}\times (\mathbb{R}^3-\{0\})$, right?

NeuroFuzzy

1:25 AM

@ACuriousMind Just was explaining my homework to my girlfriend. Explained how abstract index notation makes some things that are really simple abstractly hard to write & a bit unclear. Her words: "that sounds awful" :D

Are you proud??? hehe

0celo7

@all Suppose I upvote an answer only to find an error in it. Am I allowed to trivially edit the question so I can remove the upvote?

(Assuming the time has passed during which I can still naturally remove the upvote, of course.)

user54412

"Additionally, all sites were updated ... and now have new profile pages."

0celo7

@ChrisWhite Huh the new profile looks suspiciously like the old one

But cool that we have the new profile!

ACuriousMind

1:42 AM

@NeuroFuzzy Heh

@0celo7 Yes

ACuriousMind

Hmmm

TeX is behaving weirdly

...nvm

knzhou

2:11 AM

hey, everybody going through grad admissions:

is there any point in taking the math subject GRE?

vzn

2:45 AM

Dark matter and dinosaurs: meet Lisa Randall, America’s superstar scientist / guardian

FenderLesPaul

2:56 AM

@knzhou are you applying for math grad school?

if so you have to take it, you don't have a choice

if not then there's literally no reason to take it

simple as that

there are some schools that don't require the math GRE for math grad school but they are far and few in between

NeuroFuzzy

Does anyone have an idea how to mathematically define $\nabla_a x^b$ where $x^b$ is a coordinate function and $\nabla_a$ is the covariant derivative?

oh

I screwed up because I treated an index as an abstract index

-sigh-

0celo7

3:11 AM

@NeuroFuzzy You know...I've never understood that

Coordinates are not vectors

How the heck is the covariant derivative of them defined?

I know Wald uses it at certain points...

NeuroFuzzy

@0celo7 Oh, you just treat them as scalar functions on your manifold (hehe "just". Keep in mind tthis is coming from a guy who just screwed up while using that, though actually it was my fault and if I had done it correctly I'd have been fine)

on part of your manifold*

Do you ever feel like someone, or a select few people (authors?) are responsible for all the bad GR notation in the world? Which causes almost every student to say "no one really knows what a tensor is"? (I'm quoting a math grad student there)

I think it's someone's fault.

for teaching everyone "a tensor is a list of numbers such that [...]"

0celo7

3:27 AM

@NeuroFuzzy that's actually a valid definition though

Wald's...isn't

basically anything with indices is a tensor for him

3

A: Repeated index in covariant derivative using abstract index notation

As Qmechanic pointed out in the comments, you're mixing Einstein and abstract index notation a bit. To make things absolutely clear, we will use early Latin indices for abstract indices $(abc)$ and Greek indices for component indices $(\mu\nu\rho)$ and will always indicate Einstein summation expl...

See ^

Why did that not make a link

@NeuroFuzzy Basically, Wald abuses the notation that he invented

0537

@0celo7 lol...

we don't have the new profile.

it barely changed.

FenderLesPaul

4:09 AM

is there a link between analyticity of correlators in position space and unitarity of S-matrix?

i.e. is there a position space version of optical theorem that people have developed?

I need to know if my research project has been a total waste of time or not

TanMath

@DanielSank again, I do not know what I should do...I am confused about making a code that just defines matrices, runs a simulation using another module, and plot the results into a bunch of functions with testcases...

FenderLesPaul

Also, anyone here excited about the LIGO "random injection event" rumors about possibly having detected gravitational waves?

I really hope the hype doesn't end up like BICEP

0celo7

4:46 AM

> No laptop computers, cellular phones, MP3 players, or any other personal electronic devices will be welcome in our classroom. If I notice you are using any of the above during class time, twenty-five (25) points will be docked from your next reading quiz.

Holy crap lit prof is hard core

TanMath

5:35 AM

where is DS?

@0celo7 now that is strict!

5:47 AM

11:43 AM

In quantum mechanics, is wavenumber times h bar always equal to momentum?

yuggib

12:12 PM

@Shing momentum is an operator in quantum mechanics, the wavenumber times hbar a number

so you can easily see they are quite different objects

The relation you write is valid only in the following context: the average of the momentum operator for a single free photon is of the form $\hbar k$.

Shing

12:42 PM

@yuggib thanks for the help! :D

0celo7

@Slereah welcome to 2003

0celo7

1:14 PM

PSE is down...is this the big update?

0celo7

1:51 PM

Well, it's not down but I keep getting a 522 error.

0537

4:10 PM

@dmckee Can you please create the tag [Fluid-Mechanics] and have it be synonymous with [Fluid-Dynamics]?

0celo7

@ACuriousMind Can I make a Q&A along the lines of "what the heck is abstract index notation really"

ACuriousMind

@0celo7 I don't know, can you? :P

0celo7

@ACuriousMind I can A it

but will it be closed as off topic, i.e. can I Q it?

0537

ask a specific question relating to it and give a broad answer.

ACuriousMind

Well, I can't say until I see the actual question, but, as such, I don't think that a question like "what exactly do abstract indices mean?" would be closed.

0celo7

4:14 PM

I want to have a canonical answer on abstract index notation that can be linked in the future

@ACuriousMind Something along the lines of: How does abstract index notation relate to traditional tensor notation and coordinate index notation?

@ACuriousMind I'm worried about it being a "pure math" question

@ACuriousMind Does $[n/2]$ mean round up or down?

ACuriousMind

Uh...neither. Rounding down is $\lfloor n/2 \rfloor$, rounding up is $\lceil n/2\rceil$.

0celo7

@ACuriousMind if you saw $\sum_{m=0}^{[n/2]-1}$ what would you think it means

ACuriousMind

4:29 PM

I'd look somewhere else in the document I'm reading for a clue because I couldn't decide on that alone.

knzhou

3

Q: What defines a physical property?

The physical world around us has all sorts of properties, shape, color etc. If you move on to more complex systems, there are even more like some emotional properties etc. Why do we deem only certain of them as physical like mass, length, time etc. and not others?

those are some fighting comments.

0celo7

@ACuriousMind time to hunt a 60s journal article :(

ACuriousMind

@knzhou Ah, just the normal rant of someone who had their question closed. :P

0celo7

@ACuriousMind I'd rant like that too

actually, I'm pretty sure I have

ACuriousMind

Also, that question is the perfect excuse to post

Haddaway - What Is Love

►

again

0celo7

4:32 PM

@ACuriousMind How would one write $R_a{}^bR_b{}_c$ without indices/coordinates

@ACuriousMind can you please post a version that plays in my country

ACuriousMind

@0celo7 How am I supposed to find out whether it plays in your country?

@0celo7 One wouldn't ;)

0celo7

@ACuriousMind but math people would need such a thing

It's a part of an Euler form.

user54412

@knzhou The real question is whose sockpuppet is that?

ACuriousMind

@0celo7 Okay. Then you need to define $\langle T,S\rangle_{ij}$ for applying the natural pairing to the $i$-th slot of $T$ and the $j$-th slot of $S$, and you'd write something like $\langle R^{\sharp_2},R\rangle_{2,1}$. I think.

0celo7

@ACuriousMind Can I steal that as an example for why abstract indices can be nice?

ACuriousMind

4:42 PM

@0celo7 Of course, but note that I haven't ever seen anyone write something like this. It might be that actual geometers have a different way of writing this

0celo7

Also, calling all historians! (@HDE226868 @Danu) Who invented abstract index notation? I've never seen it mentioned before Wald (1984).

Physics Meta

-1

Q: Minimal spherical aberration

How can I calculate the ratio between R1 and R2 (the radii of curvature) at a binconvex lenght to minimize the spherical aberration?It is known the refractive index of the glass which is manufactured lens, n .

discussion

0celo7

@ACuriousMind Jost, who does cohomology classes and all that just uses components for most of the book :D

ACuriousMind

Ah, yes, a fair number of geometers just goes the physicist's way ;)

0celo7

@ACuriousMind In any case, the "natural pairing" is defined using coordinates, no?

You need a basis and then you sum.

ACuriousMind

4:44 PM

@0celo7 No, it's the natural pairing between a vector space and its dual

0celo7

@ACuriousMind How do you do that for two tensors

ofc I understand $\omega_iv^i:=\omega(v)$

as $\omega$ is defined to be a function of $v$

but $R$ is a function of two vectors, not a function of half a tensor

see what I'm saying?

ACuriousMind

For $(v\otimes f)\in V\otimes V^\ast$, you define $C(v\otimes f) = f(v)$. Then you take $R^{\sharp_2}\otimes R \in V\otimes V^\ast \otimes V \otimes V$ and apply the map $\mathrm{id}\otimes C \otimes \mathrm{id}$ to it.

0celo7

o.o

Let me think about that a bit

@ACuriousMind Alright, I haven't done much multilinear algebra. Suppose I have some element of $V\otimes V^*\otimes V\otimes V$ and the map $A\otimes B\otimes C$, how exactly does that work

Is $A$ applied to the part over the first part $V\otimes V^*$?

@ACuriousMind Challenge: write $R^{\sigma\rho}R_{\mu\sigma\nu\rho}$ w/o indices ;)

or $R_\mu{}^{\sigma\rho\lambda}R_{\nu\sigma\rho\lambda}$

ACuriousMind

@0celo7 Okay, I'll write it more carefully: Consider $V\otimes W\otimes X$, where in our case, $V = X$ and $W = V^\ast\otimes V$. I have defined the map $C : W\to\mathbb{R}$. Now it take the identity maps $V\to V$ and $X\to X$. Clearly, $(\mathrm{id}_V,C,\mathrm{id}_X) : V\times W\times X\to V\otimes X, (v,w,x)\mapsto C(w)\cdot (v\otimes x)$ is a multilinear map. By the universal property of the tensor product, this induces a linear map $V\otimes W\otimes X\to V\otimes X$.

Explicitly, this map sends $v\otimes w\otimes x$ to $C(w)(v\otimes x)$, and if we return to $V\otimes V^\ast\otimes V\otimes V$, it sends $v_1\otimes f\otimes v_3\otimes v_4$ to $f(v_3)(v_1\otimes v_4)$.

0celo7

@ACuriousMind Ah, and the associated linear map is denoted by $\mathrm{id}_V\otimes C\otimes\mathrm{id}_X$?

ACuriousMind

4:56 PM

@0celo7 Yes, that's the map I meant with that

0celo7

@ACuriousMind Ok, I did understand that.

I was really wondering about the order of $V$ and $X$ ;)

ie. were you talking about $\mathrm{id}_V\cdots$ or $\mathrm{id}_X\cdots$

@ACuriousMind Ok, next: Does it make sense to take a functional derivative wrt. an abstract tensor?

i.e. does $$\frac{\delta R}{\delta g_{ab}}=G^{ab}+\text{total divergence}$$ make sense to the most pedantic person

ACuriousMind

@0celo7 I would say yes. The functional derivative of $F[g]$ is simply defined by $\lim_{h\to 0}\frac{1}{h}(F[g+hg'] - F[g])$ where $g'$ is arbitrary (and the result has to be independent of the choice of $g'$ for the derivative to exist).

0celo7

@ACuriousMind Ok

ACuriousMind

Wait

0celo7

@ACuriousMind What are the qualitative differences between smooth and $C^k$ manifolds

ACuriousMind

5:03 PM

That limit there is the definition of $\int \frac{\delta F}{\delta g}(x)g'(x)\mathrm{d}x$.

0celo7

@ACuriousMind Yes.

ACuriousMind

But it should still work

0celo7

@ACuriousMind why did you say "wait"

Does $\int G_{ab}$ have meaning?

ACuriousMind

@0celo7 Because the limit doesn't define $\delta F/\delta g$, but only that integral.

@0celo7 ?

0celo7

@ACuriousMind how would I tell if spacetime is $C^k$ or $C^\infty$

(this is unrelated)

ACuriousMind

5:06 PM

@0celo7 Uh...you just examine the atlas for your manifold and look at up to what $k$ the transition functions are $C^k$?

0celo7

@ACuriousMind I know the textbook definition.

ACuriousMind

But, really, there's no reason to think about $C^k$ manifolds all too much. Every maximal $C^k$-atlas (for $k> 2$ or something) contains a $C^\infty$-atlas, so there really aren't many $C^k$-manifolds which are not $C^\infty$.

0celo7

Oh, we once had that conversation.

I wonder where a proof of that would be...

Don't say "any book on manifolds"

ACuriousMind

Also, your spacetime is smooth, because you can't identify tangent vectors with derivations acting on smooth function if it isn't, i.e. writing $v^\mu \partial_\mu$ for a vector only works on smooth manifolds

0celo7

Huh

@ACuriousMind Do you like the font used in HE?

@ACuriousMind for convenience

ACuriousMind

5:14 PM

@0celo7 With a bit of google-fu, the proof is in here. (It's theorem 1, and proven in the course of the notes, presumably)

0celo7

What on earth did you google

ACuriousMind

@0celo7 A "simple" proof is using the Whitney extension theorem on the $C^k$ transition functions, but for that you need to know the proof of that one, ofc

@0celo7 Googled C^k manifolds which are not smooth, landed here, clicked the link.

0celo7

@ACuriousMind you're too good at Google

ACuriousMind

B)

0celo7

@ACuriousMind HE font! This is the most important question!

ACuriousMind

5:22 PM

I have no intense feeling one way or the other about that font

0celo7

@ACuriousMind Do you have a strong feeling about bolding tensors

ACuriousMind

@0celo7 Yes, don't do it

0celo7

@ACuriousMind :(

Why

I was considering doing it for my Q&A where I'll be switching between three notations

ACuriousMind

@0celo7 What is the bolding supposed to signify? What makes a tensor different from any other map - i.e. why don't you bold other linear maps then, too?

0celo7

Hmm.

0celo7

6:09 PM

@ACuriousMind It's a context thing. You bold linear maps when they're thought of as sections of a tensor bundle.

ACuriousMind

@0celo7 You just made that up because you want to bold tensors ;)

0celo7

6:33 PM

@ACuriousMind Perhaps

@ACuriousMind I think writing $g_{ab}=\eta_{ab}+h_{ab}$ is also a "sin" because $\eta_{ab}$ (like $\partial_a v_b$) has no "invariant" meaning

(There will be a section called "SINS OF ABSTRACT INDEX NOTATION")

ACuriousMind

@0celo7 Correct (if you think $\eta$ stands for the diagonal -1,1,1,1 metric). One could perhaps rescue this by declaring $\eta$ to be the Minkowski metric.

0celo7

@ACuriousMind Define Minkowski metric?

ACuriousMind

@0celo7 Admittedly tricky for arbitrary manifolds, but perhaps one could simply pick some flat metric?

0celo7

@ACuriousMind Aren't all flat metrics isometric

ACuriousMind

@0celo7 I think so, yes

0celo7

6:44 PM

@ACuriousMind How do you define/find a flat metric on an arbitrary manifold

Solve $\operatorname{Riem}[\eta]=0$?

that only works in a chart

but GR is pretty much done in a chart anyway

ACuriousMind

@0celo7 Ehh...if there isn't one, whatever you're doing with $\eta_{ab}+h_{ab}$ is nonsense anyway, and that has nothing to do with the index notation

0celo7

@ACuriousMind I'm gonna have another epistemological crisis

ACuriousMind

:)

0celo7

I've never seen a flat metric in the sky!

How do I even know they exist

Mike Miller

It is correct that the only simply connected manifolds with zero sectional curvature are $\Bbb R^n$ with the Euclidean metric if that's what was being asked above

ACuriousMind

6:47 PM

@MikeMiller Uh...we're being Lorentzian here

Mike Miller

oops

ACuriousMind

@0celo7 Well, for $\mathbb{R}^{4}$, you know that one exists ;)

0celo7

@ACuriousMind the sky is not $\mathbb{R}^4$!

Physics Meta

0

Q: Car Color-Temperature Duplicate Declined

The question from today How much difference to interior temperature would a white car roof make? says essentially "My car is hot so I'm thinking of painting the roof white. How much would that affect the temperature?" This question from Feb 2014 Temperature behavior over time of black or white c...

discussion

ACuriousMind

@0celo7 True.

0celo7

6:49 PM

@MikeMiller No, that's not what's being asked above.

I'm asking if there's a way to pedantically split the metric into a flat part and a perturbation

Most GR texts just "do" this

I'm not even sure what it means now.

@MikeMiller Ok, how about: can I equip any manifold (i.e. nontrivial fundamental group, etc.) with a flat metric that is globally defined

ACuriousMind

@0celo7 I don't think flat metrics exist for every manifold.

0celo7

@ACuriousMind Let's see if the geometer has any ideas

Mike Miller

Well, I'm not sure if I'm being asked about Lorentzian manifolds here, which I only know trivialities about.

0celo7

If @MikeMiller doesn't know I'll ask an MSE question

@MikeMiller Let's go with Riem. first

user54412

@0celo7 No. Why would you want to?

Mike Miller

6:55 PM

The only flat Riemannian manifolds are products of the flat tori $\Bbb R^n/\Bbb Z^n$ and Euclidean space.

0celo7

@ChrisWhite How is $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ defined?

ACuriousMind

@ChrisWhite Because GR texts usually do a split $g = \eta + h$ when doing e.g. gravitational waves

0celo7

i.e. what is $\eta_{\mu\nu}$ there

user54412

@0celo7 Locally. At a single tangent space. Maybe in a neighborhood too, if you beg enough.

Mike Miller

Well, orientable and complete. You get Klein bottle-ish stuff too otherwise.

0celo7

6:57 PM

@ChrisWhite how is $\eta_{\mu\nu}$ defined on that tangent space?

user54412

@ACuriousMind In those cases didn't we start with $\eta$ and add a small perturbation?

ACuriousMind

@ChrisWhite Yep

I only explained why @0celo7 wants to do that, not that I agree it is a problem ;)

user54412

@0celo7 the components of $\operatorname{diag}(-1,1,1,1)$?

0celo7

@ChrisWhite I can make any metric look like that on a single tangent space

user54412

@0celo7 I didn't give you permission to change coordinates, did I?

0celo7

7:00 PM

what coordinates?

I'm confused

user54412

so am I

0celo7

in what coordinates is $\eta=\operatorname{diag}(-1,1,1,1)$

ACuriousMind

@ChrisWhite: Those indices are abstract indices, there are no coordinates in 0celo7's world right now

user54412

well that would be a problem

0celo7

To say that $g=\operatorname{diag}(-1,1,1,1)+h$ at a point is nonsense

I can eliminate that $h$ by changing coordinates

And there are no holy coordinates that prevent me from doing that

user54412

7:02 PM

is the manifold made of charts of abstract diffeomorphisms, or you know, actual diffeomorphisms?

0celo7

wat

ACuriousMind

^

0celo7

@ChrisWhite What I want to find out is if it makes sense to try to do g-wave stuff in abstract indices or if one necessarily needs coordinates of some type.

user54412

pretty sure nothing makes sense in abstract indices...

2

0celo7

All the books say one needs coordinates to see if $h$ is small compared to $\eta$

But they never intrinsically define $\eta$

user54412

7:05 PM

also, even if you do go and diagonalize g at a point, what about neighborhoods of that point?

ACuriousMind

@0celo7 Well, you don't need coordinates if someone handed you the metric that is being perturbed, but you need either coordinates or a god-given $\eta$.

0celo7

@ACuriousMind if someone hands me $g_{\mu\nu}$=matrix of stuff, I certainly do need coordinates

ACuriousMind

@0celo7 Sure. To compute things you'll always have to choose coordinates. But one can give metrics without choosing coordinates, such as inducing the metric from a larger manifold (e.g. the metric on a sphere in $\mathbb{R}^n$ you can define without choosing coordinates on the sphere)

0celo7

Ok, let's reformulate it

@ACuriousMind I know how pullbacks work

But spacetime is not embedded anywhere

(or immersed or whatever the technical requirement is)

ACuriousMind

@0celo7 Still, there are options to give (some) metrics without choosing coordinates. Isn't Schwarzschild, for instance, the unique rotationally symmetry static mumble mumble metric?

0celo7

7:10 PM

@ACuriousMind Omit the mumble mumble and yeah

But to prove that one needs coordinates ;)

ACuriousMind

Perhaps true.

0celo7

@ACuriousMind Technically you can omit static there

(that's precisely Birkhoff's theorem)

@ChrisWhite Alright, suppose I give you some metric $g$ without coordinates, as ACM wants. Can you split this as $\eta+h$?

user54412

No?

0celo7

@ChrisWhite How about: split the metric on $S^2$ (induced from $\mathbb{R}^3$) as $\delta+h$

@ACuriousMind Is that an equivalent problem?

ACuriousMind

@0celo7 Perhaps. The answer is negative - there is no flat metric on the sphere.

0celo7

7:24 PM

@ACuriousMind Even on an open subset?

Also: why

FenderLesPaul

Question for those who have read the recent Hawking/Strominger paper on soft black hole hair: am I right in saying that for the gravitational case, they don't compute the actual Noether charges in eq.(7.10) of the paper but rather the first variations thereof?

ACuriousMind

@0celo7 On open subsets it works fine

0celo7

@ACuriousMind Since one does GR on open subsets anyway there might not be a conflict then...

ACuriousMind

@0celo7 Integral of the Ricci tensor gives the Euler characteristic of $S^2$, which is non-zero, but for a flat metric, that integral would be zero.

0celo7

@ACuriousMind Oh, right.

ACuriousMind

7:29 PM

@FenderLesPaul "soft black hole hair"...black holes can be fluffy now?

0celo7

@ACuriousMind And why does it work on open sets?

ACuriousMind

@0celo7 Well, not an all open sets, but for your standard ones homeomorphic to open disks it works fine because the flat metric on the disk is just the restriction of the flat metric on $\mathbb{R}^2$

0celo7

@ACuriousMind Hmm...

@ACuriousMind what's a catchy title for the canonical answer on indices

(hopefully canonical, it might turn into a rant)

0celo7

7:52 PM

@ACuriousMind Interesting result I randomly found: If $M$ admits a flat Riem. metric, then $b^k\le {n\choose k}$ where $b^k$ is the Betti number (also a TeX fail)

ACuriousMind

@0celo7 Huh, funny result

FenderLesPaul

@ACuriousMind black holes are really cute little puppies

nature is adorable

0celo7

can I pet a black hole

Transcript for